Generalized Foulkes' Conjecture and Tableaux Construction Citation Vessenes, Rebecca Angel (2004) Generalized Foulkes' Conjecture and Tableaux Construction. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/4C12-SV65. https://resolver.caltech.edu/CaltechETD:etd-05192004-121256 Abstract Foulkes conjectured that for n=ab and a ≤ b, every irreducible module occurring as a constituent in 1 Sb∫Sa Sn occurs with greater or equal multiplicity in 1 Sa∫Sb Sn . We generalize part of this to say those irreducibles also occur in 1 Sd∫Sc Sn , where cd=n and c,d ≥ a. We prove the generalized conjecture for a=2 and a=3, by explicitly constructing the corresponding tableaux. We also prove the multiplicity constraint for certain cases. For these proofs we develop a theory of construction conditions for tableaux giving rise to Sb∫Sa modules and in doing so, completely classify all such tableaux for a=2 and a=3. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: symmetric group; wreath product Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Wales, David B. Thesis Committee: Wales, David B. (chair) Chan, Ada Ramakrishnan, Dinakar Aschbacher, Michael Wilson, Richard M. Defense Date: 29 April 2004 Record Number: CaltechETD:etd-05192004-121256 Persistent URL: https://resolver.caltech.edu/CaltechETD:etd-05192004-121256 DOI: 10.7907/4C12-SV65 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 1870 Collection: CaltechTHESIS Deposited By: Imported from ETD-db Deposited On: 28 May 2004 Last Modified: 22 Jan 2021 01:23 Thesis Files Preview PDF (FullThesis.pdf) - Final Version See Usage Policy. 863kB Preview PDF (TableOfContents.pdf) - Final Version See Usage Policy. 107kB Preview PDF (Chapter1.pdf) - Final Version See Usage Policy. 104kB Preview PDF (Chapter2.pdf) - Final Version See Usage Policy. 134kB Preview PDF (Chapter3.pdf) - Final Version See Usage Policy. 230kB Preview PDF (Chapter4.pdf) - Final Version See Usage Policy. 219kB Preview PDF (Chapter5.pdf) - Final Version See Usage Policy. 145kB Preview PDF (Chapter6.pdf) - Final Version See Usage Policy. 165kB Preview PDF (Chapter7.pdf) - Final Version See Usage Policy. 237kB Preview PDF (Chapter8.pdf) - Final Version See Usage Policy. 265kB Preview PDF (Chapter9.pdf) - Final Version See Usage Policy. 300kB Preview PDF (Chapter10.pdf) - Final Version See Usage Policy. 122kB Preview PDF (Chapter11and12.pdf) - Final Version See Usage Policy. 125kB Preview PDF (AppendixA.pdf) - Final Version See Usage Policy. 126kB Repository Staff Only: item control page

Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004) 194 Appendix A Association between Tableaux Spaces and Irreducibles Let n = ab and H = Sb o Sa . Recall that Wλ,a = {T |T a λ-tableau filled with 1 to a, each b times}. We will explicitly show why the multiplicity of χλ in 1SHn equals the dimension of C{qT |T ∈ Wλ,a }. View H as a subgroup of Sn , where H acts on 1, 2, . . . b|b + 1, . . . , 2b| . . . |(a − 1)b + 1, . . . , ab by Sb on each block and by Sa permuting the blocks. The elements of H are the form (π1 , . . . πa , σ) with πi ∈ Sb , σ ∈ Sa . Now Sa × Sn acts on Wλ,a with Sa acting on the numbers 1 to a and Sn acting on the positions (corresponding to labelling across the rows). Let K = {(σ −1 , (π1 , . . . πa , σ))|πi ∈ Sb , σ ∈ Sa }. So K ≤ Sa × H ≤ Sa × Sn . Let T be the λ-tableau filled across the rows with b 1’s, then b 2’s, etc. Then Sa × Sn acting on T gives Wλ,a and K fixes T . Specifically, StabSa ×Sn (T ) = K. Hence as Sa × Sn Sa ×Sn a ×Sn modules, Wλ,a ≃ 1SStab(T . ) = 1K Proposition A.0.2. Wλ,a ≃ P Sn where ϕSa (µ) is the irreducible µ`a ϕSa (µ)⊗(ϕH (µ)) of Sa indexed by µ and ϕH (µ) is the irreducible of H/(Sb × . . . × Sb ) ≃ Sa indexed by µ. Proof. Since 1 × (Sb × . . . × Sb ) ≤ K, it is in the kernel of 1K . As Sb × . . . × Sb £ H, it 195 Sa ×H is in the kernel of 1SKa ×H . So we can view 1K as an Sa × H/(Sb × . . . × Sb ) ≃ Sa × Sa module. Let D = {(σ −1 , σ)|σ ∈ Sa } be the image of K in Sa × H/(Sb × . . . × Sb ). Sa ×Sa Hence 1SKa ×H ≃ 1SDa ×Sa as Sa × H/(Sb × . . . × Sb ) modules. Thus we can write 1D = P aµ,ν φµ ⊗ φν for µ, ν ` a and some aµ,ν , where φ is the corresponding irreducible of Sa . Sa ×Sa By Frobenius reciprocity aµ,ν = (φµ ⊗ φν , 1D ) = (φµ ⊗ φν |D , 1D )D . Now φµ ⊗ P |Sa | 1 φν |D = φµ φν . So (φµ ⊗ φν |D , 1D )D = |D| σ∈Sa φµ (σ)φν (σ) = |D| · (φµ , φν )Sa . Using   1 if µ = ν row orthogonality and |Sa | = |D|, we have aµ,ν = .  0 otherwise P Sa ×Sa Sa ×H So 1D = we µ`a φµ ⊗ φµ . If we lift back to the original module 1K P have 1SKa ×H = µ φSa (µ) ⊗ φH (µ) where φSa (µ) is the irreducible of Sa indexed by µ and φH (µ) is the irreducible of H/(Sb × · · · × Sb ) ≃ Sa indexed by µ. Since 1SKa ×Sn = (1SKa ×H )Sa ×Sn we get 1SKa ×Sn = ( X φSa (µ) ⊗ φH (µ))Sa ×Sn = µ By this proposition we have Wλ,a ≃ X φSa (µ) ⊗ (φH (µ))Sn . µ P Sn as Sa × Sn modules. µ`a ϕSa (µ) ⊗ (ϕH (µ)) Consider the submodule on which Sa is trivial, that is, µ = (a). This corresponds P P to 1Sa ⊗ (1H )Sn . If 1SHn = ν`n mν χν , this module corresponds to ν`n 1Sa ⊗ mν χν . P P Now eλ = σ∈RT τ ∈CT ²(τ )στ is an idempotent of Sn on λ-tableau T . So the action P P of eλ on 1 ⊗ mν χν is the same as the action of qλ = π∈Sa πeλ on Wλ,a . Then qλ · Wλ,a ≃ mλ (eλ · Sλ ) as Sn modules, as eλ · Sν = 0 for λ 6= ν. Now Sλ is is a cyclic Sn -module generated by eλ (T ). (Correspondingly, the semi-standard tableaux which span Sλ are equivalent under the action of Sn .) Therefore dim(eλ Sλ ) = 1 and dim(qλ Wλ ) = mλ . Hence {qT |T ∈ Wλ,a }, spans a module of dimension mλ , the multiplicity of χλ in 1SSnb oSa . This proof is due to Wales, [22]. 196 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. Doran, IV, On Foulkes’ conjecture, J. Pure Appl. Algebra 130 (1998), no. 1, 85–98. MR 99h:20014 [8] H. O. Foulkes, Concomitants of the quintic and sextic up to degree four in the coefficients of the ground form, J. London Math. Soc. 25 (1950), 205–209. MR 12,236e [9] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.3, 2002, (http://www.gap-system.org). 197 [10] David A. Gay, Characters of the Weyl group of SU (n) on zero weight spaces and centralizers of permutation representations, Rocky Mountain J. Math. 6 (1976), no. 3, 449–455. MR 54 #2886 [11] Larry C. Grove, Groups and characters, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1997, A Wiley-Interscience Publication. MR 98e:20012 [12] Roger Howe, (GLn , GLm )-duality and symmetric plethysm, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, 85–109 (1988). MR 90b:22020 [13] N. F. J. Inglis, R. W. Richardson, and J. Saxl, An explicit model for the complex representations of Sn , Arch. Math. (Basel) 54 (1990), no. 3, 258–259. MR 91d:20017 [14] G. James and A. Kerber, Representation theory of the symmetric group, Encyclopedia of Mathematics and its Applications, vol. 16, Addison-Wesley, Reading, MA, 1981. [15] G. D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, vol. 682, Springer, Berlin, 1978. [16] Serge Lang, Algebra, 3 ed., Addison Wesley, Reading Massachusetts, 1999. [17] I. G. Macdonald, Symmetric functions and Hall polynomials, second ed., Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, New York, 1995, With contributions by A. Zelevinsky, Oxford Science Publications. MR 96h:05207 [18] Bruce E. Sagan, The symmetric group, The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1991, Representations, combinatorial algorithms, and symmetric functions. MR 93f:05102 198 [19] Richard P. Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004) 1 Chapter 1 Introduction and Statement of Main Results Foulkes’ Conjecture is an outstanding problem in the areas of plethysms, rational homotopy theory, multisymmetric functions, and representation theory of symmetric groups. In representation terms, Foulkes’ Conjecture deals with induced permutation characters of wreath products of symmetric groups. The wreath product of symmetric groups Sa and Sb , denoted Sa oSb , is the normalizer of the Young subgroup Sa ×· · ·×Sa (b times) in Sab . Let 1H be the trivial representation of a group H and 1G H the induced representation from H to G. From this we can state Foulkes’ Conjecture: Conjecture 1. (Foulkes’ Conjecture) If a ≤ b then every irreducible character occurwith multiplicity greater than or equal ring as a constituent in 1SSab occurs in 1SSab a oSb b oSa to its multiplicity in 1SSab . b oSa Foulkes made this conjecture in [8] from his work on plethysms. The a = 2 case was contained in Thrall’s 1942 work on symmetrized Kronecker powers, [20], and proved again in [14] by James and Kerber using Gaussian coefficients. Coker also gives a proof for a = 2 using eigenvalues in [5], while Doran used transition matrices in [7]. The a = 3 case was proven by Dent and Siemons using mappings of unordered partitions [6]. In [2], Black and List formulated Foulkes’ Conjecture in terms of matrix incidences and Wu has rephrase it in terms of rational homotopy theory in [23]. Howe, in [12], used a plethystic approach to interpret Foulkes’ Conjecture via canonical morphisms between symmetric power modules. Using this, Brion, [4], showed 2 the conjecture holds for b sufficiently large with respect to a and Briand [3] proved Foulkes’ Conjecture for a = 4. Also using symmetric powers and plethysms, Stanley, [19], places Foulkes’ Conjecture inside a larger body of open positivity conjectures in Algebraic Combinatorics. With a more combinatorial approach, Doran gave additional formulations in [7] using tableaux spaces. Doran also suggested generalizing Foulkes’ Conjecture to: Conjecture 2. (Generalized Foulkes’ Conjecture) Given n = ab, a ≤ b, if c, d are such that cd = n, and c, d ≥ a, then every irreducible character occurring as a constituent in 1SSnb oSa occurs in 1SSnd oSc with multiplicity at least as large. For c = a and d = b this becomes the standard Foulkes’ Conjecture. Note that c, d ≥ a is necessary. This is easily verified by using GAP, [9], which shows that some irreducibles in 1SS12 do not occur in 1SS12 . Conjecture 2 holds for small n, (less than 4 oS3 6 oS2 28), by computer verification also using GAP, [9]. In Chapter 5 we will prove it holds for a = 2 by construction. Namely, we will show: Theorem 1. Given b ≥ 2, let n = 2b. If c, d are such that cd = n, and c, d ≥ 2, then every irreducible occurring in 1SSnb oS2 occurs in 1SSnd oSc with equal or larger multiplicity. We can also prove the following variation on Conjecture 2 for a = 3. The bulk of the proof is discussed in Chapter 6 with supporting details in Chapters 7 and 8. Theorem 2. Let n = 3b = cd, with c, d ≥ 3. Then every irreducible character occurring in 1SSnb oS3 occurs in 1SSnd oSc . This theorem can be strengthened when the irreducibles involved correspond to two row partitions of n. We prove this version in Chapter 9. Theorem 3. Let n = 3b = cd, with c, d ≥ 3 and let λ = [λ1 , λ2 ] be a two row partition of n. Then every irreducible character χλ occurring in 1SSnb oS3 occurs in 1SSnd oSc with multiplicity at least as large. In Chapter 2 we describe the necessary concepts to approach the Foulkes’ Conjecture combinatorially using tableaux. Much of this framework was developed by 3 Doran in [7]. We develop the theory and techniques behind constructing appropriate tableaux in Chapter 3. In Chapter 4 we completely classify and discuss all tableaux that occur in 1SSnb oS3 . The proof of Theorem 1 is given in Chapter 5. We prove Theorem 2 in Chapter 6, though the required tableau constructions are postponed until Chapter 7. In Chapter 8 we show that these tableaux suffice to cover all necessary cases. We prove Theorem 3 in Chapter 9. The corresponding results for the alternating character is given in Chapter 11. Theorems 1 and 3 also can be interpreted in terms of the Gaussian coefficient; this is discussed in Chapter 10. Further implications of all these results are listed in Chapter 12. Some of these results appear in a forthcoming article in the Journal of Algebra, [21], namely, Chapters 1, 2, 3, 5, and portions of Chapters 4 and 6. 197 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. Doran, IV, On Foulkes’ conjecture, J. Pure Appl. Algebra 130 (1998), no. 1, 85–98. MR 99h:20014 [8] H. O. Foulkes, Concomitants of the quintic and sextic up to degree four in the coefficients of the ground form, J. London Math. Soc. 25 (1950), 205–209. MR 12,236e [9] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.3, 2002, (http://www.gap-system.org). 198 [10] David A. Gay, Characters of the Weyl group of SU (n) on zero weight spaces and centralizers of permutation representations, Rocky Mountain J. Math. 6 (1976), no. 3, 449–455. MR 54 #2886 [11] Larry C. Grove, Groups and characters, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1997, A Wiley-Interscience Publication. MR 98e:20012 [12] Roger Howe, (GLn , GLm )-duality and symmetric plethysm, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, 85–109 (1988). MR 90b:22020 [13] N. F. J. Inglis, R. W. Richardson, and J. Saxl, An explicit model for the complex representations of Sn , Arch. Math. (Basel) 54 (1990), no. 3, 258–259. MR 91d:20017 [14] G. James and A. Kerber, Representation theory of the symmetric group, Encyclopedia of Mathematics and its Applications, vol. 16, Addison-Wesley, Reading, MA, 1981. [15] G. D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, vol. 682, Springer, Berlin, 1978. [16] Serge Lang, Algebra, 3 ed., Addison Wesley, Reading Massachusetts, 1999. [17] I. G. Macdonald, Symmetric functions and Hall polynomials, second ed., Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, New York, 1995, With contributions by A. Zelevinsky, Oxford Science Publications. MR 96h:05207 [18] Bruce E. Sagan, The symmetric group, The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1991, Representations, combinatorial algorithms, and symmetric functions. MR 93f:05102 199 [19] Richard P. Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004) 185 Chapter 10 Two Row Partitions and the Gaussian Polynomial Let n = ab with a, b ∈ N and take ` ∈ N such that 1 ≤ ` ≤ b n2 c. Let Pba (`) be the numbers of partitions of n having at most a parts each of size less that or equal to b, that is partitions of n fitting inside a b × a rectangle. Then Pba (`) is the co-efficient ¤ ¤ £ £ , , as in [1]. The Gaussian polynomial, a+b of q ` in the Gaussian polynomial a+b b b q q is also called the Gaussian co-efficient or the generalized q-binomial coefficient. Lemma 10.0.1. Take n = ab and λ = [n − `, `]. Let K = Sb o Sa . The multiplicity of χλ in 1SKn equals Pba (`) − Pba (` − 1). Proof. Let H = Sn−` × S` and H 0 = Sn−`+1 × S`−1 . Then χ[n−`,`] = 1SHn − 1SHn0 by the determinantal formula [14]. So h1SKn , χ[n−`,`] iSn = h1SKn , 1SHn iSn − h1SKn , 1SHn0 iSn . Hence it suffices to show h1SKn , 1SHn iSn = Pba (`). Now h1SKn , 1SHn iSn is the number of orbits of K acting on the cosets of H in Sn [11]. View the numbers 1 to n in blocks of size b, that is |1, 2, . . . , b|b + 1, . . . 2b| · · · |(a − 1)b + 1, . . . ab| The copies of H in Sn correspond to the different ways S` sits in Sn , that is subsets of {1, . . . n} of size `. Given such a subset L (corresponding to a copy of H) it will be broken into a parts by intersection with the blocks above. Let µi be the size of the part of L in the ith block. Since K acts by Sb on each of the blocks, L is equivalent 186 (under K) to a subset L0 where the first µi numbers {i · b + 1, . . . i · b + µi } are chosen from block i (starting with the 0th block). Since K also has the wreath product action by Sa acting on the blocks, L0 is equivalent to the subset L∗ , where the blocks are reordered so the µi ≥ µi+1 . Hence L∗ corresponds to a partition of the number ` into a parts of size at most b and every such partition corresponds to a copy of H in Sn . So every such partition is contained in some orbit of K on Sn /H, and every orbit contains some such partition. Hence it suffices to shows that no two partitions are in the same orbit. Say µ = [µ0 , . . . µa−1 ] and ν = [ν0 , . . . νa−1 ] are partitions of ` where we allow 0 ≤ µi , νi ≤ b. If µ and ν are in the same orbit, then there exists g ∈ K such that g · {i · b + j|0 ≤ i ≤ a − 1, 1 ≤ j ≤ µi } = {i · b + j|0 ≤ i ≤ a − 1, 1 ≤ j ≤ νi }. So g(ib + j) = ki,j b + ci,j . Since g moves complete blocks, we must have ki,j = ki,j 0 for all 1 ≤ j, j 0 ≤ µi . As the action is injective, we must then have ci,j 6= ci,j 0 for j 6= j 0 . Hence looking at the image, we have µi = |{ci,j }| ≤ νki . Take µ ¥ ν, µ 6= ν. There exists i such that µi > νi and µi0 = νi0 for all i0 < i. Then there are i such νi0 with νi0 ≥ µi . But if g · µ = ν then µi = |{ci,j }| ≤ νki implies there are at least i + 1 such νi0 , which is a contradiction. Hence no orbit contains two such partitions, which finishes the proof. The ideas behind this proof are due to J. Saxl stemming from discussions of his paper [13]. Since the multiplicity of irreducibles in induced characters is non-negative [16], this lemma implies the well-known unimodality of the Gaussian coefficients [1]. Now £ ¤ Pba (`) = Pab (`), since a+b = [ a+b a ]q by taking conjugate partitions. Hence this b q lemma shows that Foulkes’ Conjecture always holds for two row partitions, which is discussed in [14]. We can also interpret our results on the generalized Foulkes’ Conjecture in terms of the Gaussian coefficient. From Theorem 1 we have: Theorem 12. If n = 2b = cd, with c, d ≥ 2, then for 1 ≤ ` ≤ b n2 c, Pdc (`) − Pdc (` − 1) ≥ Pb2 (`) − Pb2 (` − 1) 187 Similarly, Theorem 3 gives: Theorem 13. If n = 3b = cd, with c, d ≥ 3, then for 1 ≤ ` ≤ b n2 c, Pdc (`) − Pdc (` − 1) ≥ Pb3 (`) − Pb3 (` − 1) Hence our results give insight into the relationship between the rates of growth of different Gaussian coefficients. 197 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. Doran, IV, On Foulkes’ conjecture, J. Pure Appl. Algebra 130 (1998), no. 1, 85–98. MR 99h:20014 [8] H. O. Foulkes, Concomitants of the quintic and sextic up to degree four in the coefficients of the ground form, J. London Math. Soc. 25 (1950), 205–209. MR 12,236e [9] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.3, 2002, (http://www.gap-system.org). 198 [10] David A. Gay, Characters of the Weyl group of SU (n) on zero weight spaces and centralizers of permutation representations, Rocky Mountain J. Math. 6 (1976), no. 3, 449–455. MR 54 #2886 [11] Larry C. Grove, Groups and characters, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1997, A Wiley-Interscience Publication. MR 98e:20012 [12] Roger Howe, (GLn , GLm )-duality and symmetric plethysm, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, 85–109 (1988). MR 90b:22020 [13] N. F. J. Inglis, R. W. Richardson, and J. Saxl, An explicit model for the complex representations of Sn , Arch. Math. (Basel) 54 (1990), no. 3, 258–259. MR 91d:20017 [14] G. James and A. Kerber, Representation theory of the symmetric group, Encyclopedia of Mathematics and its Applications, vol. 16, Addison-Wesley, Reading, MA, 1981. [15] G. D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, vol. 682, Springer, Berlin, 1978. [16] Serge Lang, Algebra, 3 ed., Addison Wesley, Reading Massachusetts, 1999. [17] I. G. Macdonald, Symmetric functions and Hall polynomials, second ed., Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, New York, 1995, With contributions by A. Zelevinsky, Oxford Science Publications. MR 96h:05207 [18] Bruce E. Sagan, The symmetric group, The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1991, Representations, combinatorial algorithms, and symmetric functions. MR 93f:05102 199 [19] Richard P. Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004) 188 Chapter 11 The Alternating Character Since Foulkes’ Conjecture is based on the trivial character, it is natural to ask whether the ideas hold for the alternating character. The first question is, what we mean by the alternating character in terms of induced modules. Consider, the ‘alternating’ character of the form ((−1)Sab ↓Sa oSb ) ↑Sab , that is the usual alternating character of Sab restricted to the subgroup Sa o Sb , which is induced back up to Sab . A brief computer check of Foulkes’ Conjecture using this character shows it holds for some small values of a and b. In fact, Foulkes’ Conjecture is equivalent to the following conjecture using the alternating character. Conjecture 3 (Foulkes’ Conjecture for Alternating Characters). If a ≤ b then every irreducible character occurring in ((−1)Sab ↓Sb oSa ) ↑Sab occurs in ((−1)Sab ↓Sa oSb ) ↑Sab with multiplicity greater than or equal to its multiplicity in ((−1)Sab ↓Sb oSa ) ↑Sab . Naturally, this conjecture also generalizes to: Conjecture 4 (Generalized Foulkes’ Conjecture for Alternating Characters). Given n = ab, a ≤ b, if c, d are such that cd = n, and c, d ≥ a, then every irreducible character occurring in ((−1)Sn ↓Sb oSa ) ↑Sn occurs in ((−1)Sn ↓Sd oSc ) ↑Sn with multiplicity at least as large. Showing the equivalences of Conjecture 1 or Conjecture 2, (Foulkes’ Conjecture for trivial characters), and Conjectures 3 or 4 (Foulkes’ Conjecture for alternating characters) is straightforward. We will assume Conjecture 1 (or Conjecture 2) holds 189 and prove the alternating character version. The same argument shows the reverse equivalence. Proof. First recall that if S is a subgroup of finite index in G, F an S-module and E a G-module over a field, then there is an isomorphism IndG S (ResS (E) ⊗ F ) ≃ E ⊗ IndG S (F ). (See Chapter XVIII §7 of [16].) Note that here we have used Ind for induction and Res for restriction of modules. Also, let G = Sn , S = Sa o Sb and T = Sb o Sa or Sd o Sc as appropriate. Let E be the G-module corresponding to the character (−1) on G. Since we are working over C, we will use C to denote the trivial module over any group. Sab The characters we’re comparing are χS = IndG and S (ResS (E)) =(−1)Sab ↓Sb oSa ↑ G G χT = IndG T (ResT (E)). Then χS ≃ IndS (ResS (E) ⊗ C) ≃ E ⊗ IndS (C) by the isomor- phism mentioned above. Similarly for χT . Switching notation back to characters, we G get χS ≃ (−1)G 1G S and χT ≃ (−1)G 1T . G Since we’ve assumed Foulkes’ Conjecture on trivial characters, we have 1G S = 1T +ψ G G for some character ψ. Then χS = (−1)G (1G T + ψ) = (−1)T 1T + (−1)G ψ. Hence χS ≤ χT as desired. Since we’ve proven Theorem 1 the argument above shows: Theorem 14. If 2 ≤ b then every irreducible character occurring in ((−1)S2b ↓Sb oS2 ) ↑S2b occurs in ((−1)S2b ↓S2 oSb ) ↑S2b with multiplicity greater than or equal to its multiplicity in ((−1)S2b ↓Sb oS2 ) ↑S2b . Similarly, Theorem 2 gives: Theorem 15. Given n = 3b, 3 ≤ b, if c, d are such that cd = n, and c, d ≥ 3, then every irreducible character occurring in ((−1)Sn ↓Sb oS3 ) ↑Sn occurs in ((−1)Sn ↓Sd oSc ) ↑Sn . While Theorem 3 shows: 190 Theorem 16. Let n = 3b = cd, with c, d ≥ 3 and let λ = [λ1 , λ2 ] be a two row partition of n. Then every irreducible character χλ occurring in ((−1)Sn ↓Sb oS3 ) ↑Sn occurs in ((−1)Sn ↓Sd oSc ) ↑Sn with multiplicity at least as large. Given the success of replacing the trivial character in Foulkes’ Conjecture with this ‘alternating’ character, it is natural to investigate if other definitions of an alternating character yield similar results. One suggestion was to try (−1)SSab − (−1)SSab for an a oAb a oSb in Foulkes’ Conjecture. Alas, a simple induced alternating character in place of 1SSab a oSb computer check via GAP [9] shows Foulkes’ Conjecture for this character fails when a = 3 and b = 4. Other variations on this character, such as (−1)SAaba oSb − (−1)SSab a oSb also fail at those values. 191 Chapter 12 Discussion of General Results Theorems 1, 2, and 3 extend the current research on Foulkes’ Conjecture. Although, the proof used combinatorial techniques on Young tableaux, the results correspondingly apply to areas such as Shur functions, Rational Homotopy Theory, and other means of interpreting Foulkes’ Conjecture. In addition we may interpret the Foulkes’ Conjecture using the alternating character via these theorems, which we discussed in Chapter 11. While the construction of the tableaux themselves are cumbersome, the development of the theory illustrates new approaches to Young tableaux. These concepts could be carried forth in contexts involving tableaux other than its usage here. Although the main theorems are specific to the cases a = 2 and a = 3, some general results arise from this study. The main theoretical techniques of this paper are that of weight-set counting in Theorem 4, the application of Theorem 8, and the use of maximality to show linear independence. The theory and technique of weight-set counting developed in this paper can be implemented in general, as can the concept of maximal form. While we only used tableaux with three or fewer rows, the theoretical foundations of weight-set counting have been laid for tableaux of an arbitrary number of rows. Although the computations are impractical for a random tableau, the counting works smoothly for tableaux with suitable symmetries, particularly those tableaux in maximal form. Moreover, the technique of weight-set counting is not dependent on a filling of 192 content [ba ]. However, for non-uniform contents, one must watch carefully the action of Sa ; the weight-set counting may need to count all rows and the definition of maximality will need adjustment. Similarly, the usage of Theorem 8 in constructing larger tableaux will also work for other contents and row quantities. The use of the Lemma 3.4.9, to show weight-set disjointness by maximality, however, has only been defined for three row partitions. It should be possible to generalize it for other partitions. Tableau maximality is a very useful concept for showing weight-set disjointness and applying Theorem 8. It is vital in proving linear independence of tableaux. Linear independence through tableau maximality should allow more progress on issues such as multiplicity. The methods of proving Theorem 2 could also apply to proving Conjecture 2 with other a’s, not including multiplicities. Unfortunately, the computations are likely to be somewhat cumbersome, especially the establishments of non-zero shapes as done by Theorem 9. However, should those parameters be established through other techniques, the tableaux constructed for Theorem 2 should provide nearly all the needed shapes with three or fewer rows, thus reducing the work substantially. Moreover, the reduction procedures will also apply. Specifically, given any n = ab = cd, if the shapes having multiplicity zero in Sb o Sa are bounded, then to prove the generalized Foulkes’ Conjecture for arbitrary c, d ≥ a, we should only need to prove it for a limited number of c’s. For instance, suppose a shape had multiplicity zero only if λi − λi+1 ≤ fi . (For n = 3b, we had f1 = f2 = 4.) Assume d is even (the odd case, though more cumbersome should follows analogously). Then given a tableau, we can ‘peel off’ a column block of size d with the appropriate row length, for instance, U1 (d) and P1 (d) are the two and three row versions. We can repeat this process so long as λi − λi+1 > d + fi (and there are at least as many elements as there are rows). Hence in the end, we need only construct a tableau with λi − λi+1 ≤ d + fi . This tableau will need at most P 1 i i(d + fi ) elements, hence c will be bounded by this number. This should imply, d given the fi , if the generalized Foulkes’ Conjecture is true for c up to some bound, 193 it is true for all c. (Presumably, if these tableaux exists, we can find versions with maximal/disjoint weight sets as needed.) The existence of the fi seems probable since Theorem 2 implies f1 , f2 ≤ 4 if 3|n. It may be the case, as in Theorem 1 that the parity of λi strongly effects the multiplicity. However, since the ‘peeling off’ does not change the parity, this process should still go through. In addition to this procedure, the investigations of Theorems 1 and 2 yield some and consider the irreducible Sλ correspondgeneral results. Take the character 1SSab a oSb ing to λ = [λ1 , λ2 ]. Then Sλ always has non-zero multiplicity whenever λ1 and λ2 are even. Moreover, this multiplicity is zero whenever λ2 = 1 regardless of the choice of a and b. These theorems also have implications regarding the generalized Gaussian polynomial, as discussed in Chapter 10. Finally, the techniques within the proof of Theorem 3 should extend beyond two row tableaux. Specifically, Theorem 2 can probably be strengthened to include multiplicities for all partitions. Such a result should follow the ideas of Theorem 3, though sufficient linearly independent three row tableaux for c = 4, 5, and 6 must first be established. However, weight-set maximality should be sufficient to demonstrate linear independence. In all, these results provide a strong foundation for those wishing to study the representation theory of wreath products of symmetric groups via tableaux. 197 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. Doran, IV, On Foulkes’ conjecture, J. Pure Appl. Algebra 130 (1998), no. 1, 85–98. MR 99h:20014 [8] H. O. Foulkes, Concomitants of the quintic and sextic up to degree four in the coefficients of the ground form, J. London Math. Soc. 25 (1950), 205–209. MR 12,236e [9] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.3, 2002, (http://www.gap-system.org). 198 [10] David A. Gay, Characters of the Weyl group of SU (n) on zero weight spaces and centralizers of permutation representations, Rocky Mountain J. Math. 6 (1976), no. 3, 449–455. MR 54 #2886 [11] Larry C. Grove, Groups and characters, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1997, A Wiley-Interscience Publication. MR 98e:20012 [12] Roger Howe, (GLn , GLm )-duality and symmetric plethysm, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, 85–109 (1988). MR 90b:22020 [13] N. F. J. Inglis, R. W. Richardson, and J. Saxl, An explicit model for the complex representations of Sn , Arch. Math. (Basel) 54 (1990), no. 3, 258–259. MR 91d:20017 [14] G. James and A. Kerber, Representation theory of the symmetric group, Encyclopedia of Mathematics and its Applications, vol. 16, Addison-Wesley, Reading, MA, 1981. [15] G. D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, vol. 682, Springer, Berlin, 1978. [16] Serge Lang, Algebra, 3 ed., Addison Wesley, Reading Massachusetts, 1999. [17] I. G. Macdonald, Symmetric functions and Hall polynomials, second ed., Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, New York, 1995, With contributions by A. Zelevinsky, Oxford Science Publications. MR 96h:05207 [18] Bruce E. Sagan, The symmetric group, The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1991, Representations, combinatorial algorithms, and symmetric functions. MR 93f:05102 199 [19] Richard P. Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004) 4 Chapter 2 Background 2.1 Tableaux A partition λ = [λ1 , . . . , λ` ] of a number n is an ordered tuple of positive integers P such that λi = n and λi ≥ λi+1 > 0; it is denoted by λ ` n. The length of λ is `. A Ferrers diagram is the set [λ] = {(i, j)|1 ≤ i ≤ `, 1 ≤ j ≤ λi }. We view [λ] as a (left-justified) stack of boxes with row i having λi boxes. A tableau of shape λ is a filling of the Ferrers diagram [λ] with a set of elements, usually the positive integers. It is said to have content α = [α1 , . . . , αk ] if the integer i occurs exactly αi times. A tableau is semi-standard if the entries are weakly increasing across the rows and strictly increasing down the columns. This notation is standard and further discussion can be found in [7] and [18]. Example 2.1.1. Consider the following tableaux: 125 P = 42 3 115 Q= 22 33 112 R= 233 P is a [3, 2, 1]-tableau of content α = [1, 2, 1, 1, 1], while Q has shape λ = [3, 2, 2] and content α = [2, 2, 2, 0, 1]. Similarly, R has shape [3, 3] with content [23 ] = [2, 2, 2]. Both Q and R are semi-standard, but P is not. 5 2.2 Combinatorial Structures There are two different permutation actions on tableaux: an action permuting the entry positions in the tableau and an action permuting the numbers filling the tableau. These actions commute with each other. Let T be a λ-tableau filled with the numbers 1 to a, where λ ` n. The permutation of entry positions corresponds to an action of Sn on T . View the entry positions (i.e., boxes) of T as labelled 1 to n. Then σ ∈ Sn acts on T by permuting the entries in the positions moved by σ. To avoid confusion between entry positions and numbers in T , we will denote all entry positions with the subscript T when necessary. The permutation action of the numbers corresponds to their permutation by π ∈ Sa . Example 2.2.1. Take n = 6, λ = [3, 2, 1], and a = 3. Consider T = 331 1 2 . In 2 1T 2T 3T terms of entry positions, we label T as 4T 5T . For σ = (23)T ∈ S6 , we have 6T 313 221 σT = 1 2 , while the action of π = (23) ∈ S3 gives πT = 1 3 . 2 3 We generally restrict the action of Sn to two subgroups. Let RT be the subgroup of Sn which set-wise fixes the rows of T , namely, the row permutations. Denote this action by σT for σ ∈ RT . Let CT be the subgroup of Sn which set-wise fixes the columns of T , namely, the column permutations. This action is denoted τ T for 0 τ ∈ CT . If λ is the conjugate partition of λ, (i.e., the partition corresponding to column lengths) we have that RT ≈ Sλ1 × · · · × Sλ` and CT ≈ Sλ01 × · · · × Sλ0`0 . Viewing the subgroups under these isomorphisms, we can label the entry positions by labelling each row (resp. column) with 1 to λi (resp. λi0 ). Under these labellings we write σ (resp. τ ) as a direct product of the permutations for each row (resp. column). 112 1 2 3 , then for RT we view T as labelled T T T . 23 1T 2T 1 1 1 Likewise we use the labelling T T T for CT . Applying these actions to T , gives 2T 2T Example 2.2.2. Let T = the following sets: 6 ) ( 121 211 112 121 211 112 23 , 23 , 23 , 32 , 32 , 32 {σT | σ ∈ RT } = with each element occurring twice. ( {τ T | τ ∈ CT } = ) 112 212 132 232 23 , 13 , 21 , 11 ( {πT | π ∈ Sa } = ) 112 221 332 113 223 331 23 , 13 , 21 , 32 , 31 , 12 CT = {()T × ()T × ()T , (12)T × ()T × ()T , ()T × (12)T × ()T , (12)T × (12)T × ()T } Note that the actions of σ and τ do not commute, but π commutes with both σ and τ . Given λ a partition of n = ab, let Wλ,a be the set of all tableaux of shape λ and content [ba ] = [b, . . . , b], where the entries are 1 to a. Let Sλ,a be the set of all semi-standard tableaux in Wλ,a . These sets and the following constructions were developed by Doran in [7]. Note that when b = 1, Sλ give rise to the Specht modules which are discussed extensively in [15] and [18]. The partitions of n index the Specht modules S λ , which in turn correspond to precisely the irreducible modules of Sn . From Wλ,a we can construct the complex vector space W λ,a with the tableaux as a basis. The action of Sa on the tableaux give rise to a permutation representation. Inside W λ,a we construct the following objects. Definition 2.2.3. Let T ∈ Wλ,a . Let ²(τ ) be the sign of τ as a permutation. Inside W λ,a we have: a) eT = P P σ∈RT τ ∈CT ²(τ )στ T 7 b) qT = c) mT = P P π∈Sa P P σ∈RT τ ∈CT ²(τ )πστ T P π∈Sa σ∈RT πσT The tableaux are independent basis for W λ,a , so for T1 , T2 ∈ Wλ,a , we have T1 + T1 = 2T1 but T1 + T2 = T1 + T2 . Example 2.2.4. Let T = ½ eT = 4 × 11 , then 22 11 22 + 22 11 ¾ ½ −2 12 21 21 12 + + + 21 12 21 12 ¾ Here, symmetry gives qT = 2eT . Definition 2.2.5. From eT and mT we can construct the following subspaces of W λ,a : a) S λ,a = C[eT |T ∈ Wλ,a ] b) M λ,a = C[mT |T ∈ Wλ,a ] These spaces are Sa -modules. We have {eT |T ∈ Sλ,a }, a basis for S λ,a , and {mT |T ∈ Sλ,a /Sa }, a basis for M λ,a . The set {qT |T ∈ Wλ,a } generates S λ,a ∩M λ,a , but does not form a basis. Background on these spaces and the proofs of the statements may be found in [7]. In [7], Doran uses Gay’s result from [10]: Lemma 2.2.6 (Gay’s Result). The multiplicity of the irreducible module S λ in 1SSab equals the multiplicity of the trivial representation in S λ,a . b oSa From this, Doran reformulated Foulkes’ Conjecture to: Lemma 2.2.7. The dimension of S λ,a ∩M λ,a equals the multiplicity of the irreducible S λ in 1SSab . b oSa A proof of this lemma in terms of qT ’s is presented in Appendix A. From this, Foulkes’ Conjecture is equivalent to proving dim(S λ,a ∩ M λ,a ) ≤ dim(S λ,b ∩ M λ,b ) for a ≤ b and all λ ` n. Proving Conjecture 2 is equivalent to showing that for all λ ` n, dim(S λ,a ∩ M λ,a ) ≤ dim(S λ,c ∩ M λ,c ) when ab = n = cd with c, d ≥ a. 8 Remark 2.2.8. In terms of tableaux, proving Theorem 1 is equivalent to exhibiting mλ non-zero linearly independent qT , where T has shape λ and content [dc ], with mλ the multiplicity of S λ in 1SS2b . Theorem 2 is provable by exhibiting a non-zero qT b oS2 with T having shape λ and content [dc ] for all λ such that the multiplicity of S λ in 1SS3b is non-zero. For Theorem 3 we want mλ linearly independent tableaux T with b oS3 content [dc ] such that qT is non-zero, where λ = [λ1 , λ2 ] and mλ is the multiplicity of S λ in 1SS3b . b oS3 197 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. Doran, IV, On Foulkes’ conjecture, J. Pure Appl. Algebra 130 (1998), no. 1, 85–98. MR 99h:20014 [8] H. O. Foulkes, Concomitants of the quintic and sextic up to degree four in the coefficients of the ground form, J. London Math. Soc. 25 (1950), 205–209. MR 12,236e [9] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.3, 2002, (http://www.gap-system.org). 198 [10] David A. Gay, Characters of the Weyl group of SU (n) on zero weight spaces and centralizers of permutation representations, Rocky Mountain J. Math. 6 (1976), no. 3, 449–455. MR 54 #2886 [11] Larry C. Grove, Groups and characters, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1997, A Wiley-Interscience Publication. MR 98e:20012 [12] Roger Howe, (GLn , GLm )-duality and symmetric plethysm, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, 85–109 (1988). MR 90b:22020 [13] N. F. J. Inglis, R. W. Richardson, and J. Saxl, An explicit model for the complex representations of Sn , Arch. Math. (Basel) 54 (1990), no. 3, 258–259. MR 91d:20017 [14] G. James and A. Kerber, Representation theory of the symmetric group, Encyclopedia of Mathematics and its Applications, vol. 16, Addison-Wesley, Reading, MA, 1981. [15] G. D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, vol. 682, Springer, Berlin, 1978. [16] Serge Lang, Algebra, 3 ed., Addison Wesley, Reading Massachusetts, 1999. [17] I. G. Macdonald, Symmetric functions and Hall polynomials, second ed., Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, New York, 1995, With contributions by A. Zelevinsky, Oxford Science Publications. MR 96h:05207 [18] Bruce E. Sagan, The symmetric group, The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1991, Representations, combinatorial algorithms, and symmetric functions. MR 93f:05102 199 [19] Richard P. Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004) 9 Chapter 3 Theory of Tableaux Construction Throughout this and subsequent chapters we will use T to represent an arbitrary tableau, σ an element of RT , τ an element of CT , and π an element of Sa for T filled with 1 to a. Remark 3.0.9. a) eτ T = ²(τ )eT b) eπT = πeT c) qπT = πqT = qT d) qT = Σπ∈Sa πeT e) qτ T = ²(τ )qT These are standard computations, which are discussed in [7] and [18]. This remark shows that we can ignore the effects of permuting entries when constructing the tableaux. Also, we may order the columns however we choose at the cost of a sign. 3.1 Filling Tableaux Definition 3.1.1. In T , the weight of a number x in row i, denoted ωi (x), is the number of times x occurs in row i of T . When T is not clear from context, we write ωi (x|T ) in place of ωi (x). We extend this so that ωi (x1 , . . . , xj |T ) = (ωi (x1 |T ), . . . , ωi (xj |T )). 10 Implicitly, we take ωi (T )à = (ωi (1), ! . . . , ωi (a)), which is called the row-weight of row i ω1 (xj ) .. . of T . Similarly, ω(xj ) = is the weight vector of xj of T . Hence ω(T ) is the ω` (xj ) matrix corresponding to ωi (j|T ). Note that row permutations do not effect weight, so ω(σT ) = ω(T ). 122 ³ Example 3.1.2. T = 3 4 4 . We have ω(T ) = 55 1 2 0 0 0 0 0 1 2 0 0 0 0 0 2 ´ . From this we can read that ω2 (3) = 1. Weights for a tableau are only comparable with tableaux of the same shape and content. Recall that Wλ,a is the set of all λ-tableaux with content [ba ]. Let N a be the set of all a-tuples w with non-negative integer entries and N a,` those ` × a matrices. Then we can view the row-weight function as a linear operator ωi : Z[T |T ∈ Wλ,a ] → Z[w|w ∈ N a ], where {w|w ∈ N a } is a Z-basis, or correspondingly, ω : Z[T |T ∈ Wλ,a ] → Z[w|w ∈ N a,` ], with ` = `(λ). This means we treat weights like linearly independent basis in Z[w|w ∈ N a,` ]. Hence ω(T1 + aT2 ) = ω(T1 ) + aω(T2 ). If T1 = 11 2 and T2 = 21 1 then ω1 (T1 ) = (2, 0) and ω1 (T2 ) = (1, 1). So ω1 (T1 + T2 ) = (2, 0) + (1, 1) and ω1 (T1 + T1 ) = (2, 0) + (2, 0) = 2 · (2, 0). For convenience, we take ω(0) = 0. Definition 3.1.3. Let rowsumT (i) be the sum of all the entries in row i of T , P j ωi (j)j. Lemma 3.1.4. eT = 0 iff T has a repeat entry in a column. Proof. ⇒ It suffices to show that if T has no repetitions within a column, then eT 6= 0. By Remark 3.0.9 column permutations only change the sign of eT , so without loss of generality, we may assume the columns of T are strictly increasing. If eT = 0 then there exists σ ∈ RT , τ ∈ CT such that ²(τ ) = −1 and στ T = T , and we say that T cancels in the summation. Since στ T = T , we must have ω(στ T ) = ω(T ) and so ω(τ T ) = ω(T ). This implies that rowsumT = rowsumτ T for all rows. Let r be the first row in which τ moves an entry of T . Let αi1 , . . . , αik be the entries of row r moved by τ . Say τ moves βj to αij . Since r is the first row moved by τ , βj > αij . Then rowsumT (r) = 11 Pk j=1 αij + P i6=ij αi < Pk j=1 βj + P i6=ij αi = rowsumτ T (r). Contradiction. Therefore eT 6= 0. ⇐ If T has a repeated entry in a column, then there exists a transposition τ ∈ CT such that τ T = T . Hence eT = eτ T = ²(τ )eT = −eT and so eT = 0. Knowing that any tableau with repeated numbers in a column makes eT = 0 is very useful for our construction of non-zero tableaux. We summarize this fact as follows: Lemma 3.1.5. Any tableau T filled with the numbers 1 to a having more than a rows will have eT = 0 and qT = 0, as will any T having repetitions within a column. Hence every tableau T filled with the numbers 1, 2, and 3, with qT 6= 0, will have at most three rows and all column entries will be distinct. Due to this, from now on we will assume all tableaux have distinct column entries. Notation: Many of the tableaux we construct will have multiple identical columns. We call a group of such columns, a column block. For both clarity and space we denote a column block by one copy of the column with the number of repetitions listed above. If the number of column copies is omitted, it is assumed to be one. For example, T = has k copies of 1 2 3 111 would be denoted by T = 222 3 k l m n 1112 1 , while T = 2233 2 3 and l copies of 12 , m copies of 13 , and n copies of 23 . We call the columns of T having only one entry the tail of T . When specifying T by column blocks, we may omit the tail, provided the content of T is known. The rest of T is called the body. For instance, if T = 2 11223 and we know that T has content 324 4 12 [3, 3, 3, 3], we can just list the body, T = 2 112 324 4 instead. It is assumed that any entries not specified are contained in the tail. We also use this abbreviated notation when describing elements of CT . We write τ ∈ CT as a direct product of permutations on the column blocks. Since the only permutation possible on the tail is the identity, we omit the permutations corresponding 2 k to the tail. Hence for T listed above, τ is of the form τ1 ×τ2 × τ3 . We write τi if the same permutation τi is to be applied to k columns within a column block. When k is k less than the size of the column block, we understand τi to mean that τi is applied to all k of the columns (determined by context) and the identity permutation is applied to the remaining columns within the block. For instance, on the previous T , there 4114 are two permutations of the form (13)T ×()T × (12)T , which produce 3 3 2 2 and 14 1414 3322 . 41 1 3.2 Showing Tableaux are Non-Zero Definition 3.2.1. A tableau T is said to be non-zero if qT 6= 0. Two tableaux are said to be distinct if qT1 6= ±qT2 , otherwise T1 and T2 are said to be equivalent. Since qT involves many summands, showing qT 6= 0 by direct summation is not practical. Instead, we use a technique called weight-set counting. Weight-set counting involves summing only those tableaux with a given weight; if that sum is non-zero, the entire qT summation must be non zero. Definition 3.2.2. Given a tableau T , let ω(T ) = (x1 , . . . , xa ), where xj is the weight vector ω(j|T ) = xj of the element j in T . A weight assignment of T is a pairing between the set of elements of T with the multiset of weight vectors of ω(T ). We denote the pairing of the weight vector xj with the element kj by ω(kj |T ∗ ) = xj . (The T ∗ represents a possible tableau which has the weight of kj being xj .) Note 13 that if the vectors xj and xj 0 are equal, the weight assignment pairing kj with xj is the same as the weight assignment pairing kj with xj 0 . What matters in a weight assignment is the vector paired with each element, not how we label the vectors. We usually indicate a weight assignment by writing (k1 , . . . , ka ), by which we mean ω(k1 , . . . , ka |T ∗ ) = (x1 , . . . , xa ) = ω(T ). Given a permutation π we can create a weight assignment by assigning the element π(k) to xk , since ω(π(k)|πT ) = ω(k|T ) = xk . Similarly, given such pairing (k1 , . . . , ka ) we can construct a permutation π by taking π = (k1 , . . . , ka ) in one-line notation. 111 ³ ´ 3 0 0 For example, let T = 2 2 3 . Then ω(T ) = 0 2 1 . The weight assignment 3 (3, 1, 2) means ω(3, 1, 2|T ∗ ) = 0 0 1 ³ 3 0 0 0 2 1 0 0 1 ´ for some tableau T ∗ . This weight assignment corresponds to the weight permutation π = (132) in cycle notation (from left to right). Note, however, that such a listing (k1 , . . . , ka ) of a weight assignment is not necessarily unique. For instance, if x1 = x2 then (k1 , k2 , k3 , . . . ka ) and (k2 , k1 , k3 , . . . ka ) represent the same weight assignment (pairing) but give rise to different permutations. The numbers of such permutations corresponding to the same weight assignment depends only on the vector symmetries of ω(T ). This number is denoted s(ω(T )). For example, if T = 1 1 2 2 3 , then ωT = (2, 2, 1). There are three distinct weight assignments of T corresponding to which of the three elements is assigned a weight of 1. Since there are two permutations arising from such an assignment (for instance, ω(213|T ) = (2, 2, 1) as well) we have s(ω(T )) = 2. Definition 3.2.3. A weight assignment (k1 , . . . , ka ) is valid for T if there exists τ ∈ CT such that ω(k1 , . . . , ka |τ T ) = ω(T ), i.e., T ∗ = τ T . If this happens we say τ is valid for (k1 , . . . , ka ), otherwise τ is invalid. Given a valid τ we say τ is positive if ²(τ ) = 1 and negative if ²(τ ) = −1. 1222334 ´ ³ 1 3 2 1 Example 3.2.4. Let T = 3 1 1 . We have ω(T ) = 2 0 1 0 . Recall that for 0 0 0 2 44 1T 1T 1T CT we use the labelling 2T 2T 2T (the tail is omitted since all column permutations 3T 3T 14 on it are trivial). Now (2, 3, 1, 4) is a valid weight assignment since τ = (12)T ×(12)T × 3112334 ³ ´ 2 1 3 1 (12)T ∈ CT has τ T = 1 2 2 with ω(τ T ) = 1 2 0 0 , So ω(2, 3, 1, 4|τ T ) = 0 0 0 2 44 ω(T ). The weight assignment, (2, 3, 1, 4) corresponds to the permutation π = (123) in cycle notation, meaning ω(1, 2, 3, 4|τ T ) = ω(1, 2, 3, 4|πT ) However, (1, 4, 3, 2) is not a valid weight assignment since then we must have ω3 (2|τ T ) = 2 for some τ , but there is no column permutation that will put two 2’s in the third row. We will make frequent use of weight assignments in order to determine when T is non-zero. Definition 3.2.5. Given T , consider the following functions: • P(π(T )) = the number of τ ∈ CT such that ²(τ ) = 1 and ∈ CT such that ²(τ ) = −1 and ω(π −1 (1), . . . , π −1 (a)|τ T ) = ω(T ). • N(π(T )) = the number of τ ω(π −1 (1), . . . , π −1 (a)|τ T ) = ω(T ). • P(T ) = P • N(T ) = π P(π(T )), where π correspond to distinct weight assignments of ω(T ) P π N(π(T )), where π correspond to distinct weight assignments of ω(T ) Theorem 4. (Weight-set Counting) If qT = 0 then P(T ) = N(T ). Proof. Let D be the set permutations corresponding to distinct weight assignments of T . qT = 0 XXX ⇒ π ⇒ ⇒ σ XX π τ π τ XX (3.2.1) ²(τ )ω(σπτ T ) = 0 (3.2.2) τ ²(τ )ω(πτ T ) = 0 (3.2.3) ²(τ )ω(πτ T ) = 0 s.t. ω(1, . . . , a|πτ T ) = ω(T ) (3.2.4) 15 ⇒ XX π ⇒ XX π ²(τ ) = 0 s.t. ω(1, . . . , a|πτ T ) = ω(T ) (3.2.5) ²(τ ) = 0 s.t. ω(π −1 (1), . . . , π −1 (a)|τ T ) = ω(T ) (3.2.6) τ τ ⇒s(ω(T )) X (P(πT ) − N(πT )) = 0 (3.2.7) π∈D ⇒s(ω(T ))(P(T ) − N(T )) (3.2.8) ⇒P(T ) − N(T ) = 0 (3.2.9) If qT = 0, taking the weight of both sides gives Eq. (3.2.2). Since row permutations do not effect weights, we can reduce to Eq. (3.2.3). As distinct weights can not cancel, we can consider only those tableaux with the same weight as ω(T ), hence we must have Eq. (3.2.4). Since we are only summing over tableaux of a fixed weight we may drop the weight from the sum and simply add the sign of τ , for Eq. (3.2.5). By definition of weight assignments, ω(1, . . . , a|πτ T ) = ω(π −1 (1), . . . , π −1 (a)|τ T ) hence we get Eq. (3.2.6). Restricting the sum to distinct weight-sets gives the factor of s(ω(T )), and if we split over the sign of τ 0 we get Eq. (3.2.7), which by definition is the same as Eq. (3.2.8). Factoring out s(ω(T )) yields Eq. (3.2.9). Thus P(T ) = N(T ). We can also write Theorem 4 as: Corollary 5. Let A = {τ |ω(τ T ) = ω(πT ) for some π ∈ Sa }. If P τ ∈A ²(τ ) 6= 0, then qT 6= 0 by weight-set counting on ω(T ). An easy application of this theorem is the following useful lemma. Lemma 3.2.6. If T is a tableau consisting of a single column block (and an arbitrary k tail), for instance, T = 1 , then qT 6= 0 iff k even. 2 Proof. We have w = ω(T ) = ( k0 k0 ). So there are two distinct weight assignments for T , namely, ω(1, 2|T ∗ ) = w and ω(2, 1|T ∗ ) = w. We wish to determine for which τ does T ∗ = τ T satisfy one of these equations. For ω(1, 2|τ T ) = w, we must have k 2’s in the second row, hence none of the columns of T may move. The only τ satisfying 16 k this is τ =()T , which is positive. For ω(2, 1|τ T ) = w, we must have k 1’s in the second k row, thus τ must exchange row 1 and row 2 for every column in T . Thus τ =(12)T and ²(τ ) = (−1)k . Hence when k is even, we have P(T ) = 2 and N(T ) = 0, so qT is non-zero. The same idea applies for T having more than two rows. When k is odd, however, we have P(T ) = N(T ) so the Theorem 4 does not apply. k Instead, let τ =(12)T and let π ∈ Sa be the corresponding entry transposition (in our example π = (12)). Then πτ T = T and ²(τ ) = −1. Thus qπτ t = πqτ T = qτ T = ²(τ )qT = −qT . So qT = 0. It is also true that if qT 6= 0, then it is non-zero by weight-set counting on some weight ω(τ T ). That is we can’t have P(τ T ) = N(τ T ) for all τ ∈ CT and still have qT 6= 0. Theorem 6. If qT 6= 0, then it is non-zero by weight set counting on τ T for some τ ∈ CT . Proof. Recall that qT = P P P π σ τ πσ ²(τ )τ T . Since a tableau may be written in multiple ways (such as T 0 = πστ T = π 0 σ 0 τ 0 T ), we need to be careful of how we denote a tableau. Consider the terms of qT partitioned in to classes {τ T } = {πστ T |π ∈ Sa , σ ∈ RT }, using {τ T } as the class representative. Note that all tableaux in a given class have the same sign and generic weight, i.e., the same weight modulo the action of Sa . Hence if τ T = π 0 σ 0 τ 0 T for some π 0 , σ 0 , τ 0 , with ²(τ τ 0 ) = −1 then πστ T = ππ 0 σσ 0 τ 0 T , so the classes are equal set wise, but of opposite sign. This holds for any element of the class, not just τ T . This means we may view the tableaux which cancel in qT as a matching between equal classes of opposite sign. So if qT 6= 0, there is a set of tableaux T which are not canceled in qT . Now for any τ T ∈ T either the weight of τ T does not cancel, (that is ω(τ T ) 6= πω(π 0 σ 0 τ 0 T ) for all π ∈ Sa and all π 0 σ 0 τ 0 T ∈ T with ²(τ τ 0 ) = −1) or the weight cancels with some π 0 σ 0 τ 0 T having ²(τ τ 0 ) = −1 (that is ω(τ T ) = πω(π 0 σ 0 τ 0 T ) for some π 0 σ 0 τ 0 T ∈ T). If the weight does not cancel for some τ T , the qT 6= 0 by weight set counting on τ T . Assume all the weights do cancel. Then ω(τ T ) = πω(π 0 σ 0 τ 0 T ), so for every row, τ T and ππ 0 σ 0 τ 0 T have the same number of each symbol. Hence there 17 exists σ ∈ RT such that τ T = ππ 0 σσ 0 τ 0 T . But then, since ²(τ τ 0 ) = −1, the classes {τ T } and {τ 0 T } will cancel in qT contradicting τ T, π 0 σ 0 τ 0 T ∈ T. Hence qT 6= 0 by weight set counting on some τ T . We will not need Theorem 6 for our results; it is included for theoretical interest and completeness. Theorem 4 is used quite heavily, however. For instance, it allows use to directly establish the multiplicities of the irreducible characters in 1SS2b . b oS2 Theorem 7. The only irreducible characters occurring in 12b Sb oS2 are those corresponding to all partitions λ = [λ1 , λ2 ] of 2b where λi is even. Moreover, these characters occur with multiplicity 1. Proof. By Remark 2.2.8 we need only consider those shapes with qT non-zero. By Lemma 3.0.9 and Lemma 3.1.5 all distinct non-zero tableaux filled with b 1’s and b k 2’s must be equivalent to T = 1 , (not including tail). Hence there is at most one 2 distinct non-zero tableau for any shape λ. By Lemma 3.2.6, when λ = [2b−k, k] then qT 6= 0 iff k is even. Since hqT i = S λ,2 ∩ M λ,2 we must have dim(S λ,2 ∩ M λ,2 ) = 1 if k is even and zero otherwise. This is a well-known result, appearing in [13] and [17]. The weight-set counting of Theorem 4 is useful for much more complicated tableaux as well. To illustrate the general usage of the theorem, we list here a slightly more involved example. Example 3.2.7. To see directly how weight-set counting works, consider the following example. The tableau Q∗ is listed below using the column block notation, with the conditions on the block size listed to the right. Underneath the tableau we list the weight and shape of the tableau. 18 aabbc 2 2 2 3 1 Q∗ = 1 4 4 1 4 3 3 a+b+c=d 2a + b < d c even a, b > 0 0 0 d ω2,3 = ( a+b 0 0 2a 0 ) λ = [3d − 2a − b, d + a+b, 2a] We want to show that qQ∗ 6= 0, which we do by showing P(Q∗ ) − N(Q∗ ) 6= 0 and applying Theorem 4. Now P(Q∗ ) is the number of τ ∈ CQ∗ with ²(τ ) = 1 such that ω(i1 , i2 , i3 , i4 |τ Q∗ ) = ω(Q∗ ) for some distinct weight assignment (i1 , i2 , i3 , i4 ). (Equivalently, τ is such that {ω(i|τ Q∗ )ki = 1, 2, 3, 4} = {ω(i|Q∗ )ki = 1, 2, 3, 4}.) Similarly, N(Q∗ ) is those with ²(τ ) = −1. The easiest way to count these τ is to use weight assignments. First we determine which weight assignments might be possible using some general properties of the tableau. Then we count how many τ correspond to each weight assignment (i.e., for which τ is the weight assignment valid) and determine ²(τ ). Finally, we add this signed sum to determine P(Q∗ ) − N(Q∗ ). First we want to determine which weight assignments are possible for Q∗ . That is, determine for which 4-tuples x = (i1 , i2 , i3 , i4 ) there might exist τ ∈ CQ∗ such that ∗ 0 0 d a+b 0 0 d ω2,3 (x|τ Q∗ ) = ( a+b 0 0 2a 0 ). Let w = ( 0 0 2a 0 ). Simply looking at Q , there are a few restrictions on what x can be. Notice the body contains d copies of the elements 1 and 4, but fewer than d copies of 2 and 3 since 2a+b < d. Also note that the body of τ Q∗ contains the same elements as the body of Q∗ . This implies that not all elements can have d copies in row two of τ Q∗ for some τ . If ω2,3 (x|τ Q∗ ) = w then either ω2 (1|τ Q∗ ) = d or ω2 (4|τ Q∗ ) = d; namely, only the elements 1 and 4 may have ω2 (i|τ Q∗ ) = d. Hence any valid weight assignment must have i4 = 1 or 4. If ω2 (1|τ Q∗ ) = d, then there is only one other non-zero weight to assign in row two. As b > 0 the remaining columns (the second a block and the first b block) must have the same element in row two, namely, 2 or 19 4. That means we must have either ω2 (2|τ Q∗ ) = a+b or ω2 (4|τ Q∗ ) = a+b, so the weight assignment must have i1 = 2 or 4. Similarly, if ω2 (4|τ Q∗ ) = d, then since b > 0 we must have either ω2 (1|τ Q∗ ) = a+b or ω2 (3|τ Q∗ ) = a+b, that is i1 = 1 or 3. We also consider which elements j may have ω3 (j|τ Q∗ ) = 2a. We find that only with j = 2 or 3 may this occur since these are the only elements for which both a blocks will be the same in row three. (That is if ω2,3 (τ Q∗ ) = w then either ω3 (2|τ Q∗ ) = 2a or ω3 (3|τ Q∗ ) = 2a, so any valid weight assignment has i3 = 2 or 3.) There are six distinct weight assignments x = (i1 , i2 , i3 , i4 ) meeting these conditions: (1, 2, 3, 4), (1, 3, 2, 4), (4, 3, 2, 1), (3, 1, 2, 4), (4, 2, 3, 1), and (2, 4, 3, 1). In the table below, for each weight assignment we list for what type of tableau τ Q∗ it is valid, the form τ used, the number of such τ , and the sign of τ . This is an easy way to summarize the counting of τ and their signs. (We will omit the subscripts Q∗ when writing τ for easy reading. Remember that τ is labeled by the entry positions of Q∗ and not the elements.) 20 ω2,3 (x|τ Q∗ ) τ Q∗ 0 0 d = ( a+b 0 0 2a 0 ) x = (1, 2, 3, 4) x = (1, 3, 2, 4) x = (3, 1, 2, 4) x = (4, 2, 3, 1) x = (4, 3, 2, 1) x = (2, 4, 3, 1) τ aabbc 2 2 2 3 1 1 4 4 1 4 3 3 aabbc 3 3 2 3 1 1 4 4 1 4 2 2 aabbc 1 3 2 1 1 3 4 4 3 4 2 2 aabbc 2 2 2 3 4 1 4 4 1 1 3 3 aabbc 3 3 2 3 4 1 4 4 1 1 2 2 aabbc 2 4 4 3 4 1 2 2 1 1 3 3 a a b b c b b () × () × () × () × () a a c (13) × (13) × () × () × () a a b b b b c c (132) × (13) × () × (12) × () a a () × () × () × () × (12) a a b b c b c (13) × (13) × () × () × (12) a a b () × (12) × (12) × () × (12) # ²(τ ) 1 (−1)0 1 (−1)2a 1 (−1)a+b 1 (−1)c 1 (−1)2a+c 1 (−1)a+b+c To see how we obtain such a table, consider the last row. We want a tableau τ Q∗ such that ω2,3 (2, 4, 3, 1|τ Q∗ ) = w. This means that ω3 (3|τ Q∗ ) = 2a so τ cannot move any entries in row 3. We also have ω2 (4|τ Q∗ ) = 0, so examining Q∗ , we know that τ acts non-trivially on the second column block a, the first column block b, and column a a b b c block c, namely, τ =∗ × (12)T × (12)T × ∗ × (12)T . If τ were to act non-trivially on the first column block a or the second column block b, the number of 1’s in row two of τ Q∗ would decrease. Since we must have ω2 (1|τ Q∗ ) = d, this cannot happen. Hence a a b b c τ =()T × (12)T × (12)T × ()T × (12)T and has been completely determined for us. Then we have ω2,3 (2, 4, 3, 1|τ Q∗ ) = w, so such a τ exists and is unique. For reference, τ and τ Q∗ are listed. Once τ has been determined, computing ²(τ ) = (−1)a+b+c is 21 straightforward. Finally, to compute the weight sum for w, we sum the product of the number of τ with the sign of τ , that is # · ²(τ ). Here the sum is 1 + 1 + (−1)a+b + 1 + 1 + (−1)a+b . This sum is between 2 and 6, depending on the parity of a and b. Since it is non-zero in all cases, Theorem 4 shows qQ∗ 6= 0. 3.3 Joining Tableaux Definition 3.3.1. The join of two tableaux, U and V , denoted U ∨ V is a way of combining tableaux together. If the entries of U and V are not disjoint, renumber V so that they are. For instance, if U contains the numbers 1 to n and V contains the numbers 1 to m we first renumber V with the numbers n + 1 to n + m. Then concatenate the tableaux and sort the columns by length. Note that entries of every column remain fixed, only the order of the columns change. 113 123 567 Example 3.3.2. U = 2 2 4 and V = 2 4 4 We renumber V to get V = 6 8 8 . 34 3 113567 Concatenating gives 2 2 4 6 8 8 . When sorted we get T = U ∨ V = 34 7 7 115367 2 2 6 4 8 8. 347 Note that since applying permutations of Sa to a tableau has no effect on qT and mT , the renumbering of a tableau is irrelevant. Also, any σ ∈ RT that only interchanges columns will commute with all τ ∈ CT . Hence column sorting has no effect of eT , qT , and mT , since there is no sign change for row permutations. This join operation also joins the weight-sets, namely, ω3 (U ∨ V ) = ω3 (U ), ω3 (V ) = (0, 0, 1, 1, 0, 0, 1, 0). Definition 3.3.3. Let T = U ∨ V for tableaux U filled with 1 to m and V filled with m + 1 to a. The weights, ω(U ) and ω(V ) are disjoint (equivalently, ω(T ) splits over U and V ), if every valid weight assignment of ω(T ), can be obtained from concatenating valid weight assignments of ω(U ) and ω(V ). Say ω(1, . . . , m|U ) = (x1 , . . . , xm ) where the xi are weight vectors and ω(m + 1, . . . , a|V ) = (xm+1 , . . . , xa ). So ω(1, . . . , a|T ) = (x1 , . . . , xa ). Consider a valid 22 weight assignment of T assigning to the element j the weight vector xkj . Then ω(1, . . . , a|τ T ) = (xk1 , . . . , xka ) for some τ ∈ CT . This restricts to a valid weight assignment of τ|U U by considering only the elements 1 to m. This restriction is unique because a weight assignment is defined by the vector-element pairing, not the label assigned to the vector. If this restriction corresponds to a weight assignment of ω(U ) (i.e., the weights assigned to elements 1 to m are the same as the weights of ω(U ) as vectors) then the weight assignment of T arose from valid weight assignments of U . Similarly for V . By restricting to a weight assignment of ω(U ) we mean {xki |i = 1 . . . m} = {xi |i = 1 . . . m}, i.e., the weights assigned to U are equivalent to those of ω(U ). If this result is a weight assignment of ω(U ), it is valid for the tableau τ|U U . If this is true for all valid weight assignments of T , then ω(T ) splits and the weights of U and V are disjoint. Example 3.3.4. Consider U = 11 3333 113333 and V = . So T = and 22 4444 224444 ω2 (1, 2, 3, 4|T ) = (0, 2, 0, 4). There are four valid weight assignments possible for T : (Recall that T ∗ represents any possible tableau τ T .) ω2 (1, 2, 3, 4|T ∗ ) = (0, 2, 0, 4) ω2 (1, 2, 3, 4|T ∗ ) = (2, 0, 0, 4) ω2 (1, 2, 3, 4|T ∗ ) = (0, 2, 4, 0) ω2 (1, 2, 3, 4|T ∗ ) = (2, 0, 4, 0) When we restrict these weight assignments to U we get two possible assignments for U , ω2 (1, 2|U ∗ ) = (0, 2) and ω2 (1, 2|U ∗ ) = (2, 0). Since the original weight of U is ω2 (1, 2|U ) = (0, 2), both of these assignments are assignments of (0, 2) and both are valid for U (simply take τU = () and τU = (12)T × (12)T ). A similar argument holds for V . Thus the ω(T ) splits over U and V . However, weight-set disjointness is highly dependent on the filling of T . Consider instead, U = 11 43 1143 and V = . So T = and ω2 (1, 2, 3, 4|T ) = (0, 2, 1, 1). 22 34 2234 23 Then ω2 (1, 2, 3, 4|T ∗ ) = (1, 1, 2, 0) is a valid weight assignment for T by T ∗ = τ T with τ = (12)T × ()T × ()T × (12)T . However, ω2 (1, 2|U ∗ ) = (1, 1) it is not a valid weight assignment of ω2 (U ) = (2, 0), even though there exists τ such that ω2 (τ U ) = (1, 1). Hence the weights are not disjoint. Although this definition of disjointness is a bit involved, in Section 3.4 we will give a sufficient (but not necessary) condition on the tableau which is easier to check. However, we use disjointness here to obtain the full generality of Theorem 8, which is one of the fundamental tools we use to construct non-zero tableaux. Theorem 8. Let U and V be tableaux such that • elements(U ) = {1, . . . m} and elements(V ) = {m + 1, . . . a} (renumber if necessary) • The weights ω(1, . . . , m|U ), ω(m + 1, . . . , a|V ) are such that qU and qV are non-zero by weight-set counting on ω. • The weight assignments corresponding to ω(1, . . . , m|U ) and ω(m + 1, . . . , a|V ) are disjoint. Then for T = U ∨ V , we have qT 6= 0 by weight-set counting on ω(T ). Proof. By weight-set counting on U and V we have P(U ) − N(U ) 6= 0 and P(V ) − N(V ) 6= 0. By Theorem 4, showing P(T ) − N(T ) 6= 0 implies qT 6= 0. Thus it suffices to show P(T ) − N(T ) = (P(U ) − N(U ))(P(V ) − N(V )) or equivalently P(T ) = P(U )P(V ) + N(U )N(V ) and N(T ) = N(U )P(V ) + P(U )N(V ). We will show P(T ) = P(U )P(V ) + N(U )N(V ). The N(T ) claim follows similarly. Consider the weight assignment (k1 , . . . , ka ) of T . ω(1, . . . , a|T ) with ²(τ ) is counted in P(T ). = If ω(k1 , . . . , ka |τ T ) = 1 (i.e., τ is positive for (k1 , . . . , ka )) then it Since the weight splits, we have ω(k1 , . . . , ka |τ T ) = ω(k1 , . . . km |τ|U U )ω(km+1 , . . . ka |τ|V V ) = ω(1, . . . , m|U )ω(m + 1, . . . , a|V ) = ω(1, . . . , a|T ) with ²(τ ) = ²(τ|U )²̇(τ|V ) = 1. Hence either ²(τ|U ) = ²(τ|V ) = 1 or ²(τ|U ) = ²(τ|V ) = −1. If ²(τ|U ) = 1 then since ω(k1 , . . . km |τ|U U ) = ω(U ) is a 24 valid weight assignment (because the weights are disjoint), it is counted in P(U ). Similarly for the P(V ) cases. The ²(τ|U ) = −1 cases are counted in N(U ). Thus P(T ) ≤ P(U )P(V ) + N(U )N(V ). Now any weight assignments ω(k1 , . . . , ka |τ|U U ) = ω(1, . . . , a|U ) in P(U ) and ω(km+1 , . . . , a|τ|V V ) = ω(m + 1, . . . , a|V ) in P(V ) must also correspond to the valid weight assignment ω(k1 , . . . , ka |τ T ) = ω(1, . . . , a|T ) with τ = τ|U × τ|V . Moreover ²(τ ) = ²(τ|U )²̇(τ|V ) = 1. So this weight assignment is in P(T ). Similarly for weight assignments in N(U ) and N(V ). Hence P(T ) ≥ P(U )P(V ) + N(U )N(V ). Thus we have P(T ) = P(U )P(V ) + N(U )N(V ) as desired. Theorem 8 allows us to construct a non-zero tableau from smaller non-zero tableaux. The main difficulty in applying this theorem is showing that the weight-sets are disjoint. To deal with this, we develop an idea of maximality of weights which is sufficient for weight-set disjointness. 3.4 Maximal Weights Since the action of Sa on a tableau T does not change the resulting qT , we generalize the definition of tableau weight to account for this. We put an order on this generic weight, thus defining maximal weights. Definition 3.4.1. Given a tableau T , the generic form of ωi (T ) is wi (T ) = ωi (πT ) = (x1 , . . . , xa ) for any π ∈ Sa such that xj ≥ xj+1 for all j. In essence, wi (T ) is the weights of ωi (T ) listed in decreasing order. This definition works for any row of T . We define the generic form of a weight on the entire tableau (assuming T has at xa most three rows) by, w(T ) = ω2,3 (πT ) = ( xy11 ... ... ya ) for any π ∈ Sa such that yj ≥ yj+1 for all j and if yj = yj+1 then xj ≥ xj+1 . We consider only the weight vectors of the second and third rows and list the vectors so that the row three weights are decreasing. If two vectors have the same weight for row three, we list the vector with the larger weight in row two first. 25 221 For instance, if T = 1 1 2 then ω2,3 (T ) = ( 20 10 03 ). Hence w(T ) = ( 03 20 10 ). 333 Definition 3.4.2. We put an order on generic weights by wi (T1 ) > wi (T2 ) if when wi (T1 ) = (x1 , . . . xa ), wi (T2 ) = (v1 , . . . va ), there exists k ≥ 1 such that xj = vj for j < k and xk > vk . We say w(T1 ) > w(T2 ) if 1. w3 (T1 ) > w3 (T2 ) or 2. w3 (T1 ) = w3 (T2 ) and w2 (T1 ) > w2 (T2 ) or ... ya 3. w3 (T1 ) = w3 (T2 ), w2 (T1 ) = w2 (T2 ), and if we have w(T1 ) = ( xy11 ... xa ) and za w(T2 ) = ( xz11 ... ... xa ) then there exists k ≥ 1 such that yj = zj for j < k and yk > zk . We also apply this ordering to sets of weight vectors, by associating to each set the generic weight vector formed by concatenating the given weights in order. So to ¡¢ ¡¢ the set A = { 02 , 11 } we associate the weight ( 02 11 ). Definition 3.4.3. We define the maximum generic weight of row i of T to be the maximum with respect to > of {wi (τ T )|τ ∈ CT }, where w is the generic weight defined above. Similarly the maximum generic weight of T is the maximum with respect to > of {w(τ T )|τ ∈ CT }. Note that the maximum generic weight of T is based only on the weights of rows two and three. As such we ignore the weight of the first row. 12 . Then the generic weights w2 (τ T ) are (1, 1) and 21 11 22 (2, 0), with (2, 0) (the generic weight of τ T = or ) as maximum. Here 22 11 Example 3.4.4. Let T = we’ve suppressed writing w3 (T ) since T has only two rows. 11444 If T = 2 2 3 3 3 then the maximum generic weight of T (of the second and third 33 rows) is ( 32 20 00 00 ) which is ω2,3 (T ) in generic form, w(T ). Definition 3.4.5. Let T be a tableau having three or fewer rows. Let wm be the maximum generic weight of T . We say wm is the max weight for T if wm occurs in qT . That is qT 6= 0 by weight-set counting on wm . 26 Unlike the maximum weight, the max weight of a tableau may not exist since the 111 weight may not occur in qT . For instance, consider T = 2 2 2 . We know qT = 0 by 333 Lemma 3.2.6, yet T has ( 03 30 00 ) as its maximum weight. The max weight for T is always the maximum weight for row three of T , but it 5233 need not be the maximum weight for row two of T . Consider T = 1 4 4 1 . The 33 maximum generic weight of T is ( 02 20 20 00 ), but the maximum generic weight of row two of T is (4, 0, 0, 0) which is not the generic form of (0, 2, 2, 0). Definition 3.4.6. If wm is the max weight for T , we say T is in maximal form provided ω2,3 (πT ) = wm for some π ∈ Sa . This only requires that some permutation of the weight vectors of ω2,3 (T ) be equal to the max weight of T . While the max weight may not exist for all tableaux, it is easy to show weight-set disjointness for those tableaux which are in maximal form. In order to prove this, we use the following lemmas regarding our ordering. Lemma 3.4.7. Given two weights, W1 and W2 , of the same length, let Ck = ¡ ¢¡ ¢ { xy | xy ∈ Wk }, k = 1, 2 be the multisets of weight vectors in each of these weights. Let A = C1 \ (C1 ∩ C2 ), the weight vectors of W1 not in W2 . Similarly, let B = C2 \ (C1 ∩ C2 ). If W1 ≥ W2 , then A ≥ B. Proof. Without loss of generality, assume Wi is written in maximal form (i.e., is equal to its generic weight). If W1 = W2 then A = B = ∅, so the result holds trivially. So suppose W1 > W2 . Then either W1 differs from W2 at some place in the third row, or the third rows are equal and they differ at some place in the second row (at least vectorwise). ¡ ¢¡ ¢ ¡ ¢¡ ¢ Define Ay = { xyii | xyii ∈ A, yi = y} and B v = { uvii | uvii ∈ B, vi = v}. Defining C1y and C2v similarly, we have Ay = C1y \ (C1y ∩ C2y ) and B v = C2v \ (C1v ∩ C2v ). xm u1 ... um Let W1 = ( xy11 ... ... ym ) and W2 = ( v1 ... vm ). If W1 differs from W2 in the third row, then there exists j such that yj > vj and yi = vi for all i < j. Hence |C1yi | = |C2vi | = |C2yi | for yi > yj , so |Ayi | = |B vi | = |B yi |. Then to show A > B it suffices to show 27 y y |Ayj | > |B yj |. But since yj > vj , we have |C1 j | > |C2 j |. Thus w3 (A) > w3 (B) and the result follows. Note that when the third rows are equal, this argument shows w3 (A) = w3 (B). If W1 and W2 are equal in the third row but the generic weights of their second rows differ, we can apply the same argument as above, where Ax , B x , Cix are the appropriate sets of weight vectors with the second row weight equal to x. This shows w2 (A) > w2 (B). Since we’ve already have w3 (A) = w3 (B), the result follows. If W1 and W2 have the same generic weights in the second and third rows, then by above we know the generic weights of rows two and three of A and B are the same. By definition, A ∩ B = ∅, so the second row of the first vectors in A and B are different. Now W1 and W2 agree in the third row, so the first vectors where they differ in the second row must be the first vector in A and B respectively. Since W1 > W2 we have the large vector occurring in W1 and hence in A. Thus A > B. The following lemma is an obvious property of the ordering, but is included due to the non-standard ordering used. Lemma 3.4.8. Given sets A and B of weight vectors, if A ≥ B and B ≥ A then A = B. xm Proof. View A and B as generic weight vectors WA and WB . Let WA = ( xy11 ... ... ym ) and um WB = ( uv11 ... ... vm ). Since A ≥ B we have (y1 , . . . , ym ) ≥ (v1 , . . . , vm ). So either the rows are equal or there exists j such that yj > vj and yi = vi for all i < j. As B ≥ A we would similarly get vj ≥ yj , which is a contradiction. Hence w3 (A) = w3 (B). A similar argument shows w2 (A) = w2 (B). ¡ ¢ ¡ ¢ If A 6= B, let j be the first place where they differ. Then A ≥ B implies xyjj > uvjj . Since yj = vj , this means xj > uj . But the same argument on B ≥ A implies uj > xj , which is a contradiction. Thus A = B. Now using these lemmas we can show that the max weights of tableaux are necessarily disjoint. 28 Lemma 3.4.9. If U and V are tableaux in maximal form, with U containing b copies of the elements 1 to m and V containing b copies of the elements m + 1 to a (after renumbering as necessary), then ω(U ) and ω(V ) are disjoint. ... xm xm+1 ... xa Proof. Let U and V be in maximal form. Let ω2,3 (U ∨ V ) = ( xy11 ... ym ym+1 ... ya ). To show that U and V are disjoint, we need to show that any valid weight assignment of ´ ³ x ... x x km+1 ... xka k km U ∨ V restricts to a valid weight assignment of U and V . Let yk11 ... ykm ykm+1 ... yka be a valid weight assignment of U ∨ V . That means there exists τ such that ³ x ... x x ´ k km km+1 ... xka ω2,3 (τ [U ∨ V ]) = yk11 ... ykm ykm+1 ... yka . We want to show it restricts to a valid ¡ x ... x m ¢ weight assignment of U , that is ω2,3 (τ|U U ) = ykk11 ... ykkm = ω2,3 (πU ) for some π ∈ Sm . ¡xi ¢ ¡xk ¢ This is equivalent to showing { yi |i = 1 . . . m} = { yki |i = 1 . . . m}. i ¡ ¢ Let C = { xyii |i = 1 . . . m} be set of the weight vectors of ω2,3 (U ) and D = ¡ ¢ { xykki |i = 1 . . . m} the set of weight vectors of ω2,3 (τ|U U ). Define A = C \ (C ∩ D) i and B = D \ (C ∩ D). So A consists of those weight vectors of U assigned to V which are distinct from the weight vectors of V assigned to U under this weight assignment. That is, the vectors in A occur in ω2,3 (U ) but not in ω2,3 (τ|U U ). The set B is the weight vectors are those vectors coming from ω2,3 (τ|U U ) which are not in ω2,3 (U ). To prove disjointness, we need to show that C = D, which is equivalent to showing A = B = ∅. Now U is in maximal form, so ω2,3 (U ) ≥ ω2,3 (τ 0 U ) for all τ 0 . Hence ω2,3 (U ) ≥ ω2,3 (τ|U U ). So by Lemma 3.4.7, we have A ≥ B. But we can also view A as the weight vectors of U in ω2,3 (τ|V V ) which are not in ω2,3 (V ). Similarly B is the set of vectors from ω2,3 (V ) which are not in ω2,3 (τ|V V ). Since V is also in maximal form, ω2,3 (V ) ≥ ω2,3 (τ|V V ). Hence by Lemma 3.4.7, we have B ≥ A. Thus Lemma 3.4.8 shows A = B. However, A ∩ B = ∅ by definition, so A = B = ∅. Thus the weights are disjoint. Example 3.4.10. Suppose U ω2,3 (1, 2, 3, 4|U ) = is a tableau in maximal form such that ( 30 12 00 00 ) and V is a tableau in maximal form with ω2,3 (5, 6, 7, 8, 9, 10, 11|V ) = ( 20 11 11 00 00 00 00 ). Suppose these weights were not disjoint. ¡¢ That means we must be able to assign some weight xy of U to V and some weight 29 ¡x0 ¢ y0 of V to U . First consider the vector ¡1¢ 2 of U (i.e., the vector with the largest weight in row three). Since V is in maximal form, we know that there can be at most one copies of any element in row three of τ V for any τ . Since 2 > 1, this vector cannot be assigned ¡¢ to V . Similarly, once we know that 12 remains a weight of U , the largest row three ¡¢ weight we can assign to U is 0. Hence 11 remains with V . ¡¢ ¡¢ Now consider 30 of U . Having V in maximal form means that when 11 are ¡¢ ¡¢ assigned to V , a vector ∗0 assigned to V must have ∗ ≤ 2. Thus 30 is assigned to U . Therefore the only vectors of U and V that can be assigned to each other are the ¡0¢ vectors. However, since a weight assignment is based only on the vector and not 0 its label, this is the same as a weight assignment arising from U and V . Hence the weights are disjoint. Lemma 3.4.9 shows that if U and V are in maximal form, qU ∨V 6= 0 by Theorem 8. We will apply Theorem 8 repeatedly when constructing tableaux. As such, we want U ∨ V to be in maximal form whenever U and V are. Lemma 3.4.11. If T1 and T2 are maximal tableaux filled with different elements, then T1 ∨ T2 is maximal. Proof. The tableaux have no elements in common so the weight splits over the join. Hence ω(τ [T1 ∨ T2 ]) = ω(τ1 T1 ∨ τ2 T2 ) = ω(τ1 T1 ) ∨ ω(τ2 T2 ). Since each tableau weight was maximal, so too is their join. Hence maximality is preserved under the join operation. Through this join operation, we will construct collections of tableaux. To show these tableaux are linearly independent (over C), we can simply compare max weights. Lemma 3.4.12. Let {Bp } be a set of tableaux in maximal form. If the max weights of these tableaux are distinct, then {qBp } is linearly independent. Proof. Assume {qBp } is not linearly independent. Let Bk be the tableau with the largest weight such that qBk is not linearly independent from the rest of {qBp }. Write 30 qBk = P ap qBp . Then ω(qBk ) = P ap ω(qBp ). Since Bk is in maximal form, ω(Bk ) occurs with non-zero coefficient in ω(qBk ). Hence ω(Bk ) must occur with non-zero coefficient in ω(qBp ) for some p. However, Bk was chosen such that ω(Bk ) ≥ ω(Bp ). By hypothesis these weights are distinct, so the inequality is strict. But the Bp are in maximal form, so ω(Bp ) is the largest weight occurring in qBp . Hence ω(Bk ) does P not occurring in ap ω(qBp ), contradicting the linear dependence. We will use Lemma 3.4.12 heavily in Chapter 9 to prove Theorem 3. To make use of this lemma, we need to have distinctness of max weights. When the tableaux are formed via the join operation, we can sometimes simplify the proof of max weight distinctness via the following lemma: Lemma 3.4.13. Suppose we have the two row tableaux T1 , T2 , T3 , and T4 , were the non-zero max weights are as follows: ω(T1 ) = (A, B), ω(T2 ) = (C, D), ω(T3 ) = (a, b), and ω(T4 ) = (c, d). Assume ω(T1 ) 6= ω(T3 ), ω(T2 ) 6= ω(T4 ), but λ2 (T1 ) = λ2 (T3 ) and λ2 (T2 ) = λ2 (T4 ). If λ2 (T1 ) 6= λ2 (T2 ), then ω(T1 ∨ T2 ) 6= ω(T3 ∨ T4 ). Proof. Since λ2 (T1 ) 6= λ2 (T2 ) and ω(T1 ) 6= ω(T3 ), assume A + B > C + D and A > a. If ω(T1 ∨ T2 ) = ω(T3 ∨ T4 ), either A = c or A = d since a ≥ b by maximality of ω(T3 ). Consider A = c, then B > d and B 6= c because C + D = c + d. Then A > a and A + B = a + b, implies B < b ≤ a. Hence there is no weight equal to B, so this cannot occur. Similarly if A = d then c < B and hence there is no weight equal to B. Thus the weights are distinct. Although Lemma 3.4.13 applies directly to the join of only two tableaux, it may often be applied in a broader context. Namely, if many of the tableaux being joined are the same, the question of distinct weights reduces to looking only at the weights of those tableaux which differ. This approach will be used and discussed in Chapter 9. 197 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. Doran, IV, On Foulkes’ conjecture, J. Pure Appl. Algebra 130 (1998), no. 1, 85–98. MR 99h:20014 [8] H. O. Foulkes, Concomitants of the quintic and sextic up to degree four in the coefficients of the ground form, J. London Math. Soc. 25 (1950), 205–209. MR 12,236e [9] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.3, 2002, (http://www.gap-system.org). 198 [10] David A. 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[16] Serge Lang, Algebra, 3 ed., Addison Wesley, Reading Massachusetts, 1999. [17] I. G. Macdonald, Symmetric functions and Hall polynomials, second ed., Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, New York, 1995, With contributions by A. Zelevinsky, Oxford Science Publications. MR 96h:05207 [18] Bruce E. Sagan, The symmetric group, The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1991, Representations, combinatorial algorithms, and symmetric functions. MR 93f:05102 199 [19] Richard P. Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004) 31 Chapter 4 S The Tableaux of 1S3boS b 3 In this chapter we completely classify and discuss those tableaux occurring in 1SS3b . b oS3 To begin with, we determine exactly which tableaux associated with this space have qT 6= 0. This is done in Section 4.1. In Section 4.2, we use this information and the results from Thrall’s paper, [20], to determine precisely which partitions occur. Finally in Section 4.3, we construct a complete tableau basis for each the tableaux space S λ,3 ∩ M λ,3 associated to the irreducibles of 1SS3b b oS3 4.1 Classification of qT 6= 0, for T filled with 1, 2, 3 Let T be a λ-tableau filled with b copies of the numbers 1, 2, and 3. By Lemma 3.1.5, T has at most three rows. By Remark 3.0.9 entry permutations, column permutations and column exchanges do not change whether qT is non-zero. Hence we may take T to be in the following the form: k l m n o p q T = 1112123 2233 3    k+l+m+o=b       k + l + n + p = b    k+m+n+q=b      l ≥ m ≥ n ≥ 0 32 However, since we know there are exactly b copies of every number in T , we may omit k l m n the tail and simply write T as T = 1112 , retaining the condition l ≥ m ≥ n ≥ 0 2233 3 and assuming the tail. Theorem 9. With T as described above, qT = 0 ⇐⇒    k + l odd       k + n odd l>m=n≥0 l=m≥n≥0    k+m even      k+l even l = m+2, n = m-1 ≥ 0 l = m + 1, m=n ≥ 0 Moreover, when qT 6= 0, it is non-zero by weight-set counting. Proof. ⇐= Using Remark 3.0.9, to show qT = 0 it suffices to exhibit π ∈ S3 , τ ∈ CT , ²(τ ) = −1, such that πτ T = T up to an exchange of columns. k l m n For l > m = n, l + k odd, take τ =(12)T × (12)T × ()T × ()T and π = (12). k l m n k l m n So ²(τ ) = (−1)k+l = −1. Then τ T = 2212 1121 and πτ T = . Since m= n, 1133 2233 3 3 exchanging columns gives T . k l m n For l = m ≥ n ≥ 0, k + n odd use π = (23), τ =(23)T × ()T × ()T × (12)T and interchange columns l and m. k mmm 1 1 1 1 2 Now consider when l = m+1, m = n with k + l even. Then T = . 2 2 2 3 3 3 kmmm 13 23 1 1 1 2 and T1 = . Let T2 = . Take σ 0 ∈ Write T = T ∗ ∨ T1 where T ∗ = 2 1 2 2 3 3 3 k m m m 32 RTi so σ 0 T2 = . Let π 0 = (123) ∈ S3 and τ ∗ =(132) × () × (12) × (12)∈ CT ∗ . 1 33 Then via reordering the columns by σ̃, we have σ̃τ ∗ π 0 σ 0 (T ∗ ∨ T2 ) = T ∗ ∨ T1 . Note that ²(τ ∗ ) = 1 and σ̃, σ 0 , π 0 commute with each other and all τ ∈ CT ∗ . Then qT = X X X σπ ²(τ )τ T σ∈RT π∈S3 τ ∈CT = X X X σπ ²(τ )τ (T ∗ ∨ T1 ) σ∈RT π∈S3 τ ∈CT = X X σ∈RT π∈S3 = X X ²(τ )τ T ∗ ) ∨ ( τ ∈CT ∗ X σπ[( σ∈RT π∈S3 = X σπ[( X X X X τ 0 ∈CT1 ²(τ )τ T ∗ ) ∨ (T1 − T2 )] τ ∈CT ∗ σπ ²(τ )τ (T ∗ ∨ T1 − T ∗ ∨ T2 ) σ∈RT π∈S3 τ ∈CT ∗ = X X X σπ ²(τ )τ T ∗ ∨ T1 − σ∈RT π∈S3 τ ∈CT ∗ = ²(τ 0 )τ 0 T1 )] X X X X X X σ∈RT π∈S3 τ ∈CT ∗ σπ ²(τ )τ σ̃τ ∗ π 0 σ 0 T ∗ ∨ T2 − = σπ ²(τ ) ²(τ ∗ )τ T ∗ ∨ T2 − σ∈RT π∈S3 τ ∈CT ∗ = X X X X X X σπ ²(τ )τ T ∗ ∨ T2 σ∈RT π∈S3 τ ∈CT ∗ σ∈RT π∈S3 τ ∈CT ∗ X X X σπ ²(τ )τ T ∗ ∨ T2 X X X σπ ²(τ )τ T ∗ ∨ T2 σ∈RT π∈S3 τ ∈CT ∗ σπ ²(τ )τ T ∗ ∨ T2 − σ∈RT π∈S3 τ ∈CT ∗ X X X σπ ²(τ )τ T ∗ ∨ T2 σ∈RT π∈S3 τ ∈CT ∗ =0 Note that where appropriate we commute σ̃, σ 0 , and π 0 to combine with σ and π and reparameterize. The ²(τ ∗ ) factor arises from the reparamterization of τ τ ∗ . Now consider when l = m+2, n = m-1 with k + m even. k m-1 m-1 m-1 1 1 1111 1 2 . 2 2 2223 3 3 3 11112333 T1 = . 2223 So T = k m-1 m-1 m-1 1 1 1 2 Write T = T ∗ ∨ T1 where T ∗ = and 2 2 3 3 3 We will use the idea of the previous case to show qT = 0. However, we need to sum over all ²(τ )τ T1 with τ in CT1 . These tableaux have different symmetry relations 34 with each other, so we will list all the ²(τ )τ T and their relations. Note that the tail is omitted for readability. T1 = 1111 2223 T5 = − T9 = 1113 2221 1221 2113 T13 = − 2111 1223 T2 = − 2123 1211 1211 2123 T3 = − 1121 2213 T4 = − T6 = 2211 1123 T7 = 2121 1213 T8 = T10 = 1213 2121 T11 = 1123 2211 T12 = − T14 = − 2213 1121 T15 = − 2221 1113 T16 = 2113 1221 1223 2111 2223 1111 For appropriate σ 0 ∈ RT1 we have the following relations. All permutation listed are from S3 . T1 = −(123)σ 0 T14 T2 = −(123)σ 0 T9 T3 = −(123)σ 0 T6 T4 = −(123)σ 0 T7 T5 = (12)σ 0 T13 T8 = (12)σ 0 T8 T10 = (12)σ 0 T10 T12 = (23)σ 0 T15 T16 = (23)σ 0 T16 T11 = (12)σ 0 T11 We also have some relations on T ∗ , namely that for π ∗ ∈ S3 a transposition, then π ∗ T ∗ = τπ∗ T ∗ for τπ∗ ∈ CT ∗ with ²(τπ∗ ) = (−1)k+m−1 . Also (132)T ∗ = τ(132) T ∗ for τ(132) ∈ CT ∗ with ²(τ(132) ) = (−1)2(m−1) = 1. Then X X X πσ ²(τ )τ (T ∗ ∨ (π ∗ )−1 σ 0 Ti ) σ∈RT π∈S3 τ ∈CT ∗ = X X X πσ ²(τ )τ (π ∗ T ∗ ∨ σ 0 Ti ) σ∈RT π∈S3 τ ∈CT ∗ = X X X πσ ²(τ ) ²(τπ∗ )τ (T ∗ ∨ σ 0 Ti ) σ∈RT π∈S3 τ ∈CT ∗ = X X X σ∈RT π∈S3 τ ∈CT ∗ πσ ²(τ ) ²(τπ∗ )τ (T ∗ ∨ Ti ) 35 Hence if Ti = P P P σ∈RT then π∈S3 P P σ∈RT −(123)σ 0 Tj or Ti τ ∈CT ∗ πσ ²(τ )τ [(T P π∈S3 ∗ = π ∗ σ 0 Tj for π ∗ a transposition, then ∨ Ti ) + (T ∗ ∨ Tj )] = 0. Also, if Ti = π ∗ σ 0 Tj , τ ∈CT ∗ πσ ²(τ )τ (T ∗ ∨ Ti ) = 0. So using the cancellations above, we have qT = X X X πσ ²(τ )τ (T ∗ ∨ T1 ) σ∈RT π∈S3 τ ∈CT = X X πσ[( σ∈RT π∈S3 = X ²(τ )τ T ∗ ) ∨ ( ²(τ 0 )τ T1 )] τ 0 ∈CT1 τ ∈CT ∗ 16 X X X X X πσ ²(τ )τ (T ∗ ∨ Ti ) i=1 σ∈RT π∈S3 τ ∈CT ∗ =0 =⇒ To prove these are the only non-zero cases, we will use weight-set counting of Theorem 4 to show qT 6= 0 in the remaining cases. Given a weight, for every weight assignment π ∈ S3 we will count (with sign) the number of τ ∈ CT such that πω(T ) = ω(τ T ) and show the sum of these numbers is non-zero. In some cases it may be necessary to use ω(τ 0 T ) instead of ω(T ) to show the weight-sum is non-zero. Since applying τ 0 affects only the sign of qT , this will not change our result. Specifically we wish to show if: k + l even l>m=n≥0 k + n even l=m≥n≥0 then qT 6= 0 l>m>n unless l=m+1, m=n, k+l even, or l=m+1, m-1=n, k+m even. Note that if k = l = m = n = 0 then T has only one row and CT = 1. In this case qσT = qT , so there is exactly one distinct T . Since k + n = 0, so the statement holds. Similarly, if l = m = n = 0, applying Lemma 3.2.6 gives k even. Hence we may assume l > 0. 36 We will use the first weight-set to illustrate the technique and notation of weightset counting. This method of argument will be used extensively throughout the rest of the paper. We will list our weight-counting in a table of the following form: ωi = (l, m, n) Tableau # ² τ (x, y, z) T0 j (−1)² l m bound n τ1 × τ2 × τ3 m=n The column headings are: the weight being used, the form of the tableau, the number of τ corresponding to this weight, the sign of the τ , the form of τ , and any bounds required. The subsequent lines correspond to different weight assignments in line i. By (x, y, z), we mean ωi (x, y, z) = (l, m, n). When m=n, there are j distinct l m n τ such that τ =τ1 × τ2 × τ3 . All the τ have sign (−1)² and τ T is of the form T 0 . The following is a standard formula that we will use in computing these weight sums. Its proof is a straightforward inductive application of Pascal’s Identity. For ¡¢ notation purposes, we take ba = 0 for b > a or b < 0. We also use the convention ¡0¢ = 1. 0 Lemma 4.1.1. ¡a+h¢ b ¡¢ P ¡ ¢ − ba = hi=1 a+h−i b−1 Consider the cases where k ≥ 0 and at least l > 0. We will apply weightset counting to rows two and three. For the most part, the table should be self explanatory, though we will discuss the first weight-set table for clarity. k l m n We’ll start with T = 1113 , 2232 3 m ω2,3 = ( 00 k+l+n 0 k) 37 ω2,3 Tableau # ² τ bound 1 (−1)0 ()T × ()T × ()T × ()T 1 (−1)k+n (23)T × ()T × ()T × (12)T ¡m¢ n (−1)k+l (12)T × (12)T × (12)T × (12)T 1 (−1)l+n (123)T × (12)T × ()T × (12)T ¡ l ¢ m−n (−1)l+n (132)T × (12) T × (12)T × ()T 1 (−1)k+m (13)T × ()T × (12)T × ()T k l m n (1, 2, 3) T = 1113 2232 3 k l m l m n k l m n (1, 3, 2) T = 1112 3233 2 k n l=m k l m n (2, 1, 3) T = 2 2 31 2 1 1 13 3 3 k l n n m n k l m n (2, 3, 1) T = 2212 3133 1 k l k l+n−m l=m k l m n (3, 1, 2) T = 3 21 3 3 1 12 1 2 2 m n k l m n (3, 2, 1) T = 3133 2212 1 k l m n To understand how these τ are obtained, first apply the permutations needed to have ω(x) = 0, for the appropriate x. Additionally, apply necessary permutations so that row three of T has the correct weight. Once this is done, there will only be one column block whose permutations have not been specified. Apply the number of permutations needed to get the correct weight. For the first line of this table, we see that T has the desired weight and any column permutations will change this. Thus there is exactly one τ and it is positive. In line two, we have ω3 (2) = k and ω2,3 (1) = 0. Hence we must apply (23)T to column block k. Columns 12 and 13 cannot move since ω2,3 (1) = 0. This gives l 2’s in row two, so we must have l=m and apply (12)T to block n. When l=m, this completely determines τ , and it has sign (−1)k+n . No such τ exists for l > m. 38 Line three counts ω2,3 (2, 1, 3). Since there can be no 2’s in either rows two or k l n three, τ must contain (12)T × (12)T × (12)T . As row three already contains k 3’s, column block k needs no other permutation. Hence column 13 is the only column where τ has not yet been determined. As it stands, we already have k+l 1’s in row n two, hence only n more are required. Thus we need to apply (13)T to 13 . There are ¡m¢ ¡m¢ ways to choose which n columns move within the m block. Hence we get n n distinct τ of the form described. In line four we apply (132)T to block k in order to have ω3 (1) = k and ω2,3 (2) = 0. Additionally, ω2,3 (2) = 0 means we must apply (12)T to block l and (12)T to block n. This gives l 1’s in row two, so we must have l=m and leave block m unchanged. In this case there is one such τ ; it has sign (−1)l+n . Line five is similar to line three. From the constraints ω3 (1) = k and ω2,3 (3) = 0 k m n we have that τ contains (123)T × (12)T × ()T . In order to have the correct weight l+n-m ¡ l ¢ for row two, we need l+n-m more 2’s. There are m-n ways to choose (12) T from block l and all these τ have sign (−1)l+n . k The last line has a similar argument to line one. We need only apply τ =(13)T m × (12)T to get the correct weight and this is the only possible τ . From this table we get the following sums. ¶ ¶ µ l m l+n + (−1)k+m + (−1) 1 + (−1) m−n m−n µ ¶ µ ¶ k+n k+l l l+n l+n l + (−1)k+l + (−1) + (−1) 1 + (−1) + (−1) n n µ k+l l 6= m (4.1.1) l=m (4.1.2) ¡ l ¢ ¡ m ¢ Case I: (l > m > n). Here, (4.1.1) equals zero only if | m−n ± m−n | = 0 or 2. For it to equal 0, we must have m = n or l = m. ¡ l ¢ ¡ m ¢ To have | m−n − m−n | = 2, and (4.1.1) equal to zero, we must have k + m even and k + n odd. Applying Lemma 4.1.1, we get l = m + 1 or l = m + 2. ¡ m ¢ ¡m+1¢ ¡m+1¢ ¡ m ¢ If l = m + 1, then (4.1.1) becomes 1 − m−n + m−n + 1. Now m−n − m−n = 2 ¡ m ¢ only if m−n−1 = 2, that is m = 2 and n = 0. This contradicts k + m even and k + n odd. Hence (4.1.1) is non-zero for l = m + 1. If l = m + 2 and k + m even, (4.1.1) becomes 1 − ¡ m ¢ m−n ¡ ¢ + 1. + (−1)m+n m+2 m−n 39 For this expression to be zero we must have n = m − 1, in which case we’ve already shown qT = 0. Thus (4.1.1) is non-zero unless l = m + 2 and n = m − 1. ¡ l ¢ ¡ m ¢ Finally, m−n + m−n = 2 only when m = n. Case II: (l = m ≥ n, l > 0). We need only show the expression (4.1.2) is non-zero ¡ ¢ ¡ ¢ for k+n even. From this, (4.1.2) becomes 2+(−1)k+l (1+ nl )+(−1)l+n (1+ nl ). The ¡ ¢ parity of k and n is the same, so we reduce to determining when 1+(−1)k+l ((1+ nl ) = ¡ ¢ 0. Since nl > 0, this cannot occur. Case III: (l > m = n). We want a non-zero weight sum for k+l even. Under these conditions, expression (4.1.1) becomes 1 + 1 + (−1)l+n + (−1)k+n . This is non-zero unless k + n is odd. It remains to show qT 6= 0 for l > n = m, k + l even, k + m odd. k l m n For this, consider the following weight-set counting on T = 1112 2233 3 with 0 n+1 ω2,3 = ( k+l+m-1 0 0 k ). Note that this weight-set is not the weight of T . We are counting which permutations τ will correspond to a weight assignment π, where ω2,3 = πω2,3 (τ T ). 40 ω2,3 Tableau # ² τ Bound k l m n (1, 2, 3) T = ¡ m ¢ 2 2 31 2 1 1 13 3 3 m−1 k l m−1 n (−1)k+l+m−1 (12)T × (12)T × (12)T × ()T (−1)m+n+l−1 (132)T × (12)T × (12)T × (12)T m≥1 k l m n (1, 3, 2) T = ¡ l ¢ 3 21 3 3 2 12 1 2 1 l−1 k l−1 m n l≥1 k l m n (2, 1, 3) T = ¡ n ¢ 1 1 1 32 2 2 3 23 3 k (−1)m−1 m−1 l m m≥1 m−1 ()T × ()T × ()T × (12)T m−1≤n k l m n (2, 3, 1) T = 3 21 3 2 2 12 1 3 1 ¡ ¢ l n+1−m (−1)k+1 k n+1−m m n (13)T × (12) T × (12)T × (12)T m≤n+1 n+1−m≤l k l m n (3, 1, 2) T = ¡ n ¢ 1 1 1 32 3 2 3 23 2 (−1)k+n+l+1 l−1 k l m n+1−l (23)T × ()T × ()T × (12) T l≥1 l−1≤n k l m n (3, 2, 1) T = 2 2 31 2 3 1 13 3 1 ¡ ¢ m n+1−l (−1)n+1 k l n+1−l n (123)T × (12)T × (12) T × ()T l≤n+1 n+1−l≤m Since we previously dealt with the l = n + 1 case, the last two lines of the table do not contribute. Hence the table gives the weight sum: (−1)m−1 m + (−1)l−1 l + (−1)m−1 m + (−1)l+1 l l 6= n + 1, m 6= 0 (4.1.3) Consider l > m = n, k + l even, k + n odd. For l 6= n + 1, m 6= 0, (4.1.3) is 2((−1)m−1 m + (−1)l−1 l), which is non-zero as l 6= m. If m = 0, we must have k and l odd. This makes (4.1.3) l + l which is not equal to zero since l 6= 0. For l = n + 1 we’ve already shown qT = 0. Thus we determined all the non-zero tableaux of 1SS3b . b oS3 41 4.2 S The Irreducibles Partitions of 1S3b b oS3 For Theorem 2, we need to know which irreducibles occur (i.e., have non-zero multiplicity) in 1SS3b . We call a shape (or partition) non-zero if the multiplicity of the corb oS3 responding irreducible in 1SS3b is non-zero. By Lemma 2.2.7, the non-zero partitions b oS3 are those partitions where dim(S λ,3 ∩ M λ,3 ) > 0. Since qT generates S λ,a ∩ M λ,a we need only determine from Theorem 9 which partitions have non-zero tableaux. These partitions and their multiplicities were completely determined by Thrall in [20]. We will first derive them from Theorem 9 and then confirm it with Thrall’s result. We will only consider shapes [λ] which are partitions of 3b. Consider T = |{z} | {z } s |{z} r . This labeling will be useful in our later constructions, so we will t k l m n derive the non-zero shapes in terms of it. When T is of the form 1112 , we have 2233 3 t = k, s = l+m+n and the tail r = 3b − 2s − 3t. Since 1, 2, and 3 all occur k = t times in the part of T above t, we will sometimes replace b with b0 = b − k when considering the multiplicity of elements in columns r and s, since the subtableau formed by columns r, s and the tail will have the elements 1, 2, and 3 occurring b0 times each. In partition notation we have λ = [r + s + t, s + t, t]. 4.2.1 Non-Zero Partitions from Theorem 9 Definition 4.2.1. For our purposes, we call a partition (or shape) λ of n required if there is a non-zero λ-tableau T filled with n3 copies of the elements 1, 2, and 3. These are precisely the tableaux determined in Theorem 9. Specifically, a required shape is one for which we must construct an appropriately filled non-zero tableau in order to prove Theorems 1 and 2. These shapes are explicitly determined in Theorem 10. To determine the non-zero [λ] = [r + s + t, s + t, t], we analyze the required partitions that correspond to non-zero tableaux in Theorem 9. We find that: 42 Theorem 10. The only partitions [r + s + t, s + t, t] of n = 3b which do not occur in 1SSnb oS3 are those with s or r = 1 as well as those having s + t odd and s or r ∈ {0, 2, 4}. Equivalently, a partition is non-zero if, for r, s 6= 1, when r or s is in {0, 2, 4}, then s + t is even. Proof. For a given partition λ = [r + s + t, s + t, t], we need only find (l, m, n) with l + n + n = s, k = t, and the conditions of Theorem 9 satisfied to show qT 6= 0. This shows that S λ must occur in 1SS3b . Since all values of t = k can occur when b oS3 l > m > n, (if l 6= m + 2), we consider tableaux of this form. Given t, the elements 1, 2 and 3 will occur b0 = b − t times in the remaining columns l, m, n, and the 0 tail of T . For each t we need to determine which s for 0 ≤ s ≤ b2 yield required non-zero partitions. To do so we will take the following parameterizations of (l, m, n) and determine the corresponding s. (l, m, n) = (i + 2, i + 1, i), we get s = 3i + 3 for = (i + 3, i + 2, i) s = 3i + 5 = (i + 4, i + 3, i) s = 3i + 7 b0 − 3 2 b0 − 5 0≤i≤ 2 b0 − 7 0≤i≤ 2 0≤i≤ For a given parameterization of (l, m, n) by i, we have s = l + m + n. Since there are at most b0 1’s in the r and s sections, we must have l + m] ≤ b0 which gives the upper bound on i. These parameterizations of (l, m, n) are non-zero by Theorem 9 since l > m > n and l 6= m + 2. Moreover, the parameterizations cover all equivalence classes of s (mod 3). Hence this tells us that all partitions with s ≥ 5 or s = 3 are non-zero, leaving aside the upper bound on s for now. When s = 4, the possibilities for (l, m, n) are (4, 0, 0), (2, 2, 0), and (2, 1, 1) which are non-zero only when s + t, (i.e., k) is even. For (3, 1, 0) the tableau is always zero since l = m + 2, n = m − 1. For s = 2 the only possibilities are (2, 0, 0) and (1, 1, 0). The non-zero conditions of Theorem 9 require k = t, and hence s + t, to be even in both cases. Similarly for s = 0, we must have k 43 even for T to be non-zero, as shown in Lemma 3.2.6. For s = 1, the only possibility is (l, m, n) = (1, 0, 0), and such a tableau is always zero. Determining the upper bounds on s corresponds to determining lower bounds on r. We will similarly give parameterizations of (l, m, n) which will cover all equivalence classes of r (mod 3). Using the equation 3b0 − 2s = r we get the corresponding equations and lower bounds. b0 − 3 2 b0 − 5 i≤ 2 b0 − 7 i≤ 2 For s = 3i + 3, i≤ s = 3i + 5, s = 3i + 7, then r ≥ 3, r ≡ 0 (mod 3) (4.2.1) r ≥ 5, r ≡ 2 (mod 3) (4.2.2) r ≥ 7, r ≡ 1 (mod 3) (4.2.3) This is the parameterization table for s rewritten in terms of r. So for r ≥ 5, all partitions are non-zero, provided the conditions on s are met. 0 If r = 0, we must have l = m = n = b2 . By Theorem 9, this shape is non-zero only if k+l is even. Thus only the shapes with s+t = k+l+m+n even are non-zero. For r = 1 we get the constraints l + m = l + n = b0 and m + n = b0 − 1. This means m = n, and l=m+1. These tableaux are always zero. When r = 2, we must have either l = m or m = n. These shapes will be non-zero if k+n or k+l is even respectively. In either case, s + t is even. 0 0 0 , b −2 , b −4 ) . In For r = 4, we must have either m = n, l = m, or (l, m, n) = ( b +2 2 2 2 the first two cases, the non-zero conditions of Theorem 9 force s + t to be even. In the last case, the tableau is always zero. 0 0 If r = 3 or 5 then i = b −3 or i = b −5 is an integer for b0 odd so the parameterization 2 2 listed works. For b0 even, these partitions are not required since 3b0 − 2s = r, would make r even. In addition, Theorem 9 shows the remaining partitions are zero. 44 4.2.2 Partition Multiplicities according to Thrall In [20], Thrall determines the partitions occurring in 1SS3b with multiplicity, which b oS3 he calls f (λ). If λ = [λ1 , λ2 , λ3 ], he gives the following method to compute f (λ): To the minimum of 1 + λ1 − λ2 and 1 + λ2 − λ3 , add whichever of −2, 0, +2 will give a result divisible by 3. If this result is even, divide by 6 to get f (λ). If the result is odd, add or subtract 3 according to λ2 being even or odd and then divide by 6 to get f (λ). Letting λ = [r + s + t, s + t, t] we have min(1 + λ1 − λ2 , 1 + λ2 − λ3 ) = 1 + min(r, s). Then f (λ) = 1+min(r,s)+x+y where x ∈ {−2, 0, 2} such that 1 + min(r, s) + x ≡ 0 6 (mod 3), y ∈ {−3, 0, 3} such that 1 + min(r, s) + x + y ≡ 0 (mod 6), and y ≥ 0 if s + t even and y ≤ 0 if s + t odd. Theorem 11. Writing s = 6k + j, r = 6h + i with 0 ≤ i, j ≤ 5, then by [20], the multiplicity of Sλ in 1SS3b is f (λ), where b oS3 f (λ) =     k        k        k+1        k+1    h        h        h+1        h+1 s ≤ r i = 0, 2, 4, s + t odd s≤r i=1 s ≤ r i = 0, 2, 4, s + t even s ≤ r i = 3, 5 r < s i = 0, 2, 4, s + t odd r<s i=1 r < s i = 0, 2, 4, s + t even r < s i = 3, 5 Hence f (λ) 6= 0 for r, s 6= 1, provided s + t is even when r or s is in {0, 2, 4}. This agrees with our results in Section 4.1. 45 4.3 Construction of Basis Tableaux for c = 3 Recall that the space S λ,c ∩ M λ,c is spanned by {qT } where the T are λ-tableaux filled with the numbers 1 to c. By Lemma 2.2.7, dim(S λ,c ∩ M λ,c ) equals the multiplicity of S λ in 1SSnd oSc . Given a partition λ of n = 3b, we want a set of tableaux {Bp } such that {qBp } is linearly independent and that |{Bp }| is the multiplicity of the irreducible corresponding to λ in 1SSnb oS3 . We call these Bp the basis tableaux for c = 3. These tableaux will be used in Chapter 9 for the proof of Theorem 3. We will build these tableaux from the following components: 223333 111122 M1 = 1 2 3 M2 = 33 M3 = 1 1 22 3323 M4 = 1 1 1 2 2 a-1 a-1 Na = 2 3 2 3 1 1 1 < a ≤ n3 = b In constructing these basis tableaux, we want tableaux filled with only the numbers 1, 2, and 3. We will use Y to denote the joining of tableaux without renumbering them. For example, M2 Y M1 = 223333123 . 111122 Let λ = [r + s + t, s + t, t]. When t is even, write s = 6k + j and r = 6h + i, with 0 ≤ i, j ≤ 5. Let g = min(k, h). Since λ is a partition of n = 3b we have . (So δ = 0 for i = j, δ = 1 3t + 2s + r = 3b. Hence 2j + i ≡ 0 (mod 3). Let δ = i−j 3 for i = j + 3, and δ = −1 for i = j − 3.) When t is odd, we proceed as above, except 0 , and g = min(k 0 , h). let s − 3 = 6k 0 + j 0 , δ = i−j 3 For p = 1, 2, . . . , g, we define the basis tableaux: Bp = 2t M3 Y N6p+j Y (k − p)M2 Y (2h − 2p + δ)M1 (t even) Bp = t−1 M3 Y M4 Y N6p+j 0 Y (k 0 − p)M2 Y (2h − 2p + δ)M1 2 (t odd) 46 Additionally, for j, j 0 6= 1 we have the tableau B0 , which is Bp with p = 0 under certain conditions. B0 = 2t M3 Y Nj Y kM2 Y (2h + δ)M1 , j>1 (t even) B0 = 2t M3 Y kM2 Y (2h + δ)M1 j=0 (t even) B0 = t−1 M3 Y M4 Y Nj 0 Y k 0 M2 Y (2h + δ)M1 , 2 j0 > 1 (t odd) B0 = t−1 M3 ∨ M4 Y k 0 M2 Y (2h + δ)M1 2 j0 = 0 (t odd) Note that if δ = −1, then B0 exists only for h ≥ 1 and Bg exists only for g < h. To demonstrate that the {Bp } is a basis, we need to verify that they: • Have the correct shape, • Are non-zero and maximal, • Are linearly independent, • Span the space. Shape: First consider the shape of these tableaux. We need to show these tableaux have shape λ = [r + s + t, s + t, t]. For t even: λ3 (Bp ) = 2t λ3 (M3 ) + λ3 (N6p+j ) + (k − p)λ3 (M2 ) + (2h − 2p + δ)λ3 (M1 ) = 2t ∗ 2 + (k − p) ∗ 0 + (2h − 2p + δ) ∗ 0 =t λ2 (Bp ) = 2t λ2 (M3 ) + λ2 (N6p+j ) + (k − p)λ2 (M2 ) + (2h − 2p + δ)λ2 (M1 ) = t + 6p + j + (k − p) ∗ 6 + (2h − 2p + δ) ∗ 0 = t + 6k + j =t+s 47 λ1 (Bp ) = 2t λ1 (M3 ) + λ1 (N6p+j ) + (k − p)λ1 (M2 ) + (2h − 2p + δ)λ1 (M1 ) = t + (6p + j) ∗ 2 + (k − p) ∗ 6 + (2h − 2p + δ) ∗ 3 = t + 6k + j + 6h + j + 3δ = t + s + 6h + j + 3δ = t + s + 6h + i =t+s+r When t is odd we have: λ3 (M3 ) + λ3 (M4 ) + λ3 (N6p+j 0 ) + (k 0 − p)λ3 (M2 ) + (2h − 2p + δ)λ3 (M1 ) λ3 (Bp ) = t−1 2 = t−1 ∗ 2 + 1 + (k 0 − p) ∗ 0 + (2h − 2p + δ) ∗ 0 2 =t λ2 (Bp ) = t−1 λ2 (M3 ) + λ2 (M4 ) + λ2 (N6p+j 0 ) + (k 0 − p)λ2 (M2 ) + (2h − 2p + δ)λ2 (M1 ) 2 = t − 1 + 4 + 6p + j + (k − p) ∗ 6 + (2h − 2p + δ) ∗ 0 = t + 3 + 6k 0 + j 0 =t+s λ1 (Bp ) = 2t λ1 (M3 ) + λ1 (M4 ) + λ1 (N6p+j 0 ) + (k 0 − p)λ1 (M2 ) + (2h − 2p + δ)λ1 (M1 ) = t − 1 + 4 + (6p + j 0 ) ∗ 2 + (k − p) ∗ 6 + (2h − 2p + δ) ∗ 3 = t + 3 + 6k 0 + j 0 + 6h + j 0 + 3δ = t + s + 6h + j 0 + 3δ = t + s + 6h + i =t+s+r A similar computation works for the shape of B0 . Hence these tableaux have the correct shape. Moreover, within each component, the same number of 1’s, 2’s and 3’s 48 were used. Hence the Bp have the correct number of 1’s 2’s and 3’s. Maximality: When t is even, a generic basis element (with the tail suppressed) looks like: tabc 32 3 3 Bp = 11 1 2 2 a = 2(k − p) + 6p + j − 1 b = 2(k − p) + 1 c = 2(k − p) ¡ ¢ 2(k−p) 0 Then ω2,3 (Bp ) = t+6p+j+4(k−p) . Since b > c and a > b (for p ≥ 1), this 0 t 0 weight is maximal. Bp is also non-zero, since the only other possible weight assignment ¡ ¢ 0 2(k−p) . This has sign (−1)t+2(k−p) = 1 as t is even. is t+6p+j+4(k−p) 0 0 t When t is even, the tableau B0 is the same as Bp with p = 0 when j > 1. In this case a > b and the above argument holds. There is no B0 for j = 1. When j = 0, we have (suppressing the tail): taaa 32 3 3 B0 = 11 1 2 2 a = 2k 2k 0 Then ω2,3 (B0 ) = ( t+4k 0 t 0 ). This weight is clearly maximal. Although there are many different weight assignments possible for B0 , all weight assignments are positive since each column block is even. Hence B0 is non-zero. When t is odd, a generic basis element (with the tail suppressed) looks like: tabc 32 3 3 Bp = 11 1 2 2 Then ω2,3 (Bp ) = a = 2(k 0 − p) + 6p + j 0 b = 2(k 0 − p) + 2 c = 2(k 0 − p) + 1 ¡ t+6p+j 0 +4(k0 −p)+2 2(k0 −p)+1 0 ¢ 0 t 0 . Since b > c and a ≥ b (for p ≥ 1), this weight is maximal. Bp is also non-zero since the only other possible weight ¡ 0 0 0 −p)+2 0 2(k 0 −p)+1 ¢ assignment is t+6p+j +4(k . This has sign (−1)t+2(k −p)+1 = 1 as t is 0 0 t odd. 49 When t is odd, the tableau B0 is the same as Bp with p = 0 when j 0 > 1. In this case a ≥ b and the above argument holds. There is no B0 for j 0 = 1 (since qB0 is zero when j 0 = 1). When j 0 = 0, we have (suppressing the tail): taaa 32 3 3 B0 = 11 1 2 2 Then ω2,3 (B0 ) = ¡ t+4k0 +2 2k0 +1 0 ¢ 0 t 0 a = 2k 0 + 1 . This weight is clearly maximal. Although there are many different weight assignments possible for B0 , These weight assignments always move exactly two or four column blocks. Since the size of the column blocks is odd, all weight assignments are positive. Hence B0 is non-zero. Linear Independence: To show the tableaux Bp are linearly independent, by Lemma 3.4.12 it suffices to show their max weights are distinct. First consider t even. Say w2,3 (Bp ) = w2,3 (Bp0 ), with p < p0 . Then we must have t + 6p + j + 4(k − p) = 2(k − p0 ), which forces k = p = j = t = 0. Then our partition is just [n], so only one tableau is needed. If w2,3 (Bp ) = w2,3 (B0 ) with p > 0, then we get t + 4k = 2(k − p) which implies t = k = p = 0. Hence the tableaux are linearly independent. When t is odd, the max weights must be distinct, since an argument similar to the one above shows t = 0 which is not possible. Span: Since the tableaux Bp are linearly independent they will span the space S λ,3 ∩ M λ,3 if |{Bp }| = mλ , where mλ = f (λ) as determined by [20]. (We listed f (λ) explicitly in Theorem 11.) First consider the case of t even. Given λ = [r + s + t+, s + t, t] with t even, then mλ depends on the relative sizes of r and s. For s ≤ r, we need k + 1 tableaux for j 6= 1 and k tableaux for j = 1. When s ≤ r, we have g = k. Thus we get k different Bp ’s and when j 6= 1 we have B0 as well. The restriction on the tableaux when δ = −1 occurs only when g = h. However, then h = k and i = j − 3 which contradicts s ≤ r, so this case does not occur here. Hence we have a full set of basis 50 tableaux. If r < s, we have g = h. How many tableaux we have depends on δ. Note that Bh exists only for δ 6= −1, and B0 requires h > 0 for δ = −1. The number of tableaux needed according to Theorem 11 also depends on δ. Since s ≡ j (mod 2) and 3δ = i − j, when δ = 0 we have i ≡ s (mod 2). Then s is even for i = 0, 2, and 4, hence by [20] we need we need h + 1 tableaux for i 6= 1 and h tableaux for i = 1. Since δ 6= −1, there are h distinct Bp . We also have B0 when i 6= 1 since i = j. Hence we have a full set of basis tableaux. If δ = 1, all the tableaux described when δ = 0 occur. Since δ = 1, we have i = j + 3 which implies i = 3, 4 or 5. For i = 4 we need need h tableaux since s is odd, while for i = 3, 5 we need h + 1 tableaux. When i = 4 we have h different Bp (though no B0 since j = 1). When i = 3, 5 then j = 0, 2. Hence B0 exists and we obtain the complete set of basis tableaux. For δ = −1 we have i = j − 3, so i = 0, 1, 2. We need exactly h tableaux since either i = 1 or s is odd. However, we no longer have Bh , so we get only h − 1 tableaux from the Bp . In addition, when h ≥ 1 we have B0 since j = i + 3 ≥ 3 and so j > 1. Hence we have h tableaux for h ≥ 1. If h = 0 then no tableaux are needed since either r = 1 or r = 0 or r = 2 with s odd. Thus the correct number of tableaux is given. Now consider the case when t is odd. When s ≤ r we need k tableaux for j = 0, 1, 2, or 4 and k + 1 tableaux for j = 3 or 5. If j = 3 or 5, then k 0 = k = g so there are k tableaux Bp , in addition to the tableau B0 (since j 0 = j − 3 6= 1). If j < 3 then g = k 0 = k − 1, so there are k − 1 tableaux Bp in addition to the tableau B0 . For j = 4, we have k tableaux Bp , since g = k 0 = k. However since j 0 = 1, B0 does not exists. The restriction on the tableaux when δ = −1 occurs only when g = h. However, then h = k 0 = k and i = j − 3 which contradicts s ≤ r, so this case does not occur here. Hence we have a full set of basis tableaux. When r < s we need h tableaux when i = 1. For i = 0, 2, and 4 we need h tableaux when s is even and h + 1 tableaux when s is odd. For i = 3 or 5 we need h + 1 tableaux. Note that s 6≡ j 0 (mod 2). 51 If δ = 0 then i = j 0 . We have h tableaux Bp , along with B0 for j 0 6= 1. Hence we have h tableaux when i = 1, and h + 1 tableaux otherwise. Since s is odd when i = 0, 2, or 4, this is the correct number. If δ = 1 then i = j 0 + 3, so i = 3, 4 or 5. If i = 3 or 5, then j 0 6= 1 and there are h + 1 tableaux as desired. If i = 4 then j 0 = 1, hence there are only h tableaux Bp . However, only h tableaux are needed here since s is even. If δ = −1 then i = j 0 − 3, so i = 0, 1 or 2. Since s is even when i 6= 1, only h tableaux are required in this case. However, we no longer have Bh , so there are h − 1 tableaux Bp . In addition, when h ≥ 1, we have B0 , since j 0 > 1, so the correct number of tableaux are obtained. If h = 0 then no tableaux are needed since either r = 1 or r = 0 or r = 2 with s + t odd. Thus the correct number of tableaux is given. 197 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. Doran, IV, On Foulkes’ conjecture, J. Pure Appl. Algebra 130 (1998), no. 1, 85–98. MR 99h:20014 [8] H. O. Foulkes, Concomitants of the quintic and sextic up to degree four in the coefficients of the ground form, J. London Math. Soc. 25 (1950), 205–209. MR 12,236e [9] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.3, 2002, (http://www.gap-system.org). 198 [10] David A. Gay, Characters of the Weyl group of SU (n) on zero weight spaces and centralizers of permutation representations, Rocky Mountain J. Math. 6 (1976), no. 3, 449–455. MR 54 #2886 [11] Larry C. Grove, Groups and characters, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1997, A Wiley-Interscience Publication. MR 98e:20012 [12] Roger Howe, (GLn , GLm )-duality and symmetric plethysm, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, 85–109 (1988). MR 90b:22020 [13] N. F. J. Inglis, R. W. Richardson, and J. Saxl, An explicit model for the complex representations of Sn , Arch. Math. (Basel) 54 (1990), no. 3, 258–259. MR 91d:20017 [14] G. James and A. Kerber, Representation theory of the symmetric group, Encyclopedia of Mathematics and its Applications, vol. 16, Addison-Wesley, Reading, MA, 1981. [15] G. D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, vol. 682, Springer, Berlin, 1978. [16] Serge Lang, Algebra, 3 ed., Addison Wesley, Reading Massachusetts, 1999. [17] I. G. Macdonald, Symmetric functions and Hall polynomials, second ed., Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, New York, 1995, With contributions by A. Zelevinsky, Oxford Science Publications. MR 96h:05207 [18] Bruce E. Sagan, The symmetric group, The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1991, Representations, combinatorial algorithms, and symmetric functions. MR 93f:05102 199 [19] Richard P. Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004) 52 Chapter 5 Proof of Theorem 1 Recall, Theorem 1 says that every irreducible occurring in 1SSnb oS2 occurs in 1SSnd oSc with equal or greater multiplicity, where n = 2b = cd and b, c, d ≥ 2. In Section 3.2 we proved Theorem 7, which showed that the irreducibles occurring in 1SSnb oS2 were exactly those corresponding to partitions λ = [n − s, s] for s even and they occur with multiplicity one. (Since n = 2b is even, it suffices to consider only the even values of s.) By Remark 2.2.8, to prove Theorem 1, it suffices to construct a non-zero tableau filled with d copies of c elements for each partition [n − s, s], where 0 ≤ s ≤ n2 , s even and n = cd. To do this we will construct some non-zero generic tableaux that when assembled via Theorem 8 will produce all the shapes and fillings needed. Since we are constructing generic tableaux for many partitions and fillings, we will not use a fixed c. However, we assume that every element listed in the body of the tableau occurs d times, filling out the tail as needed. We apply weight-set counting to prove a tableau is non-zero. The tableaux we need are: Tableau U1 a d-a d-a a U1 = 1 1 2 ∼ 1 2 2 ω2 (U1 ) = (0, a) a even a≤d 53 λ = [2d − a, a] For this first tableau, we listed U1 both with and without the tail. Normally we will suppress the tail when writing these tableaux. U1 is non-zero by Lemma 3.2.6 since a is even. It is maximal since (a, 0) is the largest possible weight-set for this shape. Tableau U2 aabb U2 = 1 3 1 3 2 4 4 2 a+b≤d a, b > 0 ω2 = (0, a+b, 0, a+b) λ = [4d − 2(a+b), 2(a+b)] Examining the filling of U2 and a,b > 0 we find the following constraints on any valid weight assignment: (Recall that U ∗ corresponds to a possible tableau τ U2 .) • If ω2 (1|U ∗ ) = 0 then ω2 (2 and 4|U ∗ ) > 0. • If ω2 (2|U ∗ ) = 0 then ω2 (1 and 3 |U ∗ ) > 0. • We must have ω2 (1 or 2|U ∗ ) > 0 and ω2 (3 or 4|U ∗ ) > 0. Since any valid weight assignment of (0, a + b, 0, a + b) has exactly two zeros, the restrictions above show that (1, 2, 3, 4) and (2, 1, 4, 3) are the only valid weight a a b b assignments. These weights-sets correspond to applying τ =()T × ()T × ()T × ()T a a b b and τ =(12)T × (12)T × (12)T × (12)T respectively. As both of these τ have positive sign, qU2 6= 0. This tableau is maximal since every element x must have ω2 (x) ≤ a+b. 54 Tableau U3 a even abb U3 = 1 1 2 2 3 3 a+b≤d d even b = d2 ω2 = (0, a, d) λ = [2d − a, a + d] To show U3 is non-zero we will use weight-set counting on ω2 = (0, a, d). There are two cases for which we need to determine weight assignments, a + b < d and a + b = d. When a + b < d, only the element 3 may be assigned a row two weight of d. So the distinct weight assignments are (1, 2, 3) and (2, 1, 3), which occur with τ = ()T a b b and τ =(12)T × ()t × ()T respectively. Since a is even, both τ have positive sign. Hence U3 is non-zero. If a + b = d, then a = b = d2 and d ≡ 0 (mod 4). While every permutation corresponds to a distinct weight assignment, every weight assignment can only be obtained by having τ move complete column blocks. Since all of these blocks are even, τ is positive for every weight assignment and hence U3 is non-zero. This tableau is maximal since (d, a, 0) is the largest possible weight. Tableau V (d) V = d 1 ω1 (V ) = (d) λ = [d] 55 This is just a single row with d ones. Since there are no column permutations, this tableau is always non-zero. It is obviously maximal. Having constructed these generic tableaux, we will use the notation Ui (x) to denote the tableau Ui with the parameter a = x or Ui (x, y) for x = a and y = b in Ui . We will use fUi to denote the join of f copies of Ui . Note that these tableaux are all in maximal form. For the proof of Theorem 1, the parity effects the construction process. To simplify notation, we define the ∗ -function. x∗ =   x x even  x − 1 x odd We analyze T by the parameters r = n − 2s and s, where λ = [r + s, s]. For reference, we consider tableau of the following shape, with r and s even. T = | {z } | {z } r s Proof of Theorem 1. To prove Theorem 1 we need to construct a non-zero tableau of shape λ = [n − s, s] for s ≤ n2 , with s even and n = cd, c, d ≥ 2. First we construct ∗ ∗ a general tableau that covers most s. Suppose s ≤ c 2d . We know s is even, so write s = fd∗ + e, where 0 ≤ e < d∗ , e even. Since s, d∗ , and e are even, this is possible by the Euclidean algorithm. Let T = fU1 (d∗ ) ∨ U1 (e). Note that the bound on s guarantees that 2(f + 1) ≤ c when e > 0, and 2f ≤ c when e = 0. This insures that there are at most c distinct elements in T . If there are fewer than c elements in T add all the remaining elements to the tail of T by joining the appropriate number of V (d)’s. Suppressing the tail elements from the U1 ’s and V (d)’s, T looks like: 56 d∗ d∗ · · · d∗ v d ··· d T = 1 3 · · · 2f − 1 2f + 1 2f + 3 · · · c 2 4 · · · 2f 2f + 2 Theorem 8 shows T is non-zero, provided the weight-sets are disjoint. Since the tableaux are in maximal form, the weights must be disjoint by Lemma 3.4.9. This covers the majority of the s. The remaining tableaux will be constructed according to the parity of c and d. ∗ ∗ Case I: (c, d even) In this case c 2d = cd , so T constructed above covers all 2 partitions. Case II: (d even, c odd) By the above construction, we have all tableaux with s up . Thus we only need those even partitions with s = cd−k for 0 ≤ k ≤ d − 2, to (c−1)d 2 2 k ≡ d (mod 4). Take T = c−3 U1 (d) ∨ U3 (a) for 0 ≤ a ≤ d2 with a even. Then 2 s = c−3 d + a + d = cd−d+2a . Thus we have k = d − 2a, which ranges over the correct 2 2 parameters. Since U1 and U3 are in maximal form, Lemma 3.4.9 implies disjointness and Theorem 8 shows T is non-zero. Case III: (c even, d odd) Since r = n − 2s = cd − 2s s we need λ = [r + s, s] for r ≤ cd with r ≡ cd (mod 4). It suffices to construct a non-zero tableau for r < 4d. When r ≥ 4d, let r0 = r − 4dz with r0 < 4d. Then if we construct a λ0 = [s + r0 , s] tableau T 0 filled with d copies of c − 4z elements, we get the needed tableau by T = T 0 ∨ 4zV (d). Hence we will take r < 4d. U2 (d − 1, 1) ∨ U2 (d − 4r − When c ≡ 0 (mod 4) then r ≡ 0 (mod 4). Take T = c−4 4 1, 1). This construction gives the shape c−4 [2d, 2d]+[2d+ 2r , 2d− 2r ] = [ cd + 2r , cd − 2r ] as 4 2 2 desired. The parameters of these tableaux are positive unless r = 4d − 4 since r < 4d, r ≡ 0 (mod 4), and d ≥ 2. If r = 4d − 4 then d − 4r − 1 = 0, so use U1 (2) ∨ 2V (d) instead of of U2 (d − 4r − 1, 1). For c ≡ 2 (mod 4) we will assume r < 2d. When 2d ≤ r < 4d let r0 = r−2d. Then 57 can construct a λ0 = [r0 + s, s] tableau T 0 with c ≡ 0 (mod 4) and use T = T 0 ∨ 2V (d). Take T = c−2 U2 (d−1, 1)∨U1 ( 2d−r ) with V (d)’s as needed. Note that cd ≡ r (mod 4) 4 2 implies that 2d−r is even, while r < 2d insures it is positive. So we get the shape 2 [ cd + 2r , cd − 2r ] as needed. Theorem 8 shows these T ’s are non-zero provided the 2 2 weight-sets are disjoint, which follows from maximality. Note that since cd = n, n even, then c or d is even. Thus we have constructed all cases. Although it is not directly apparent from this construction, c or d even is often a necessary requirement for any non-zero two row tableau with s even to exist. For instance, when c = 3 and d = 7, the shape [11, 10] has s even, but all tableaux are zero by Theorem 9. 197 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. Doran, IV, On Foulkes’ conjecture, J. Pure Appl. Algebra 130 (1998), no. 1, 85–98. MR 99h:20014 [8] H. O. Foulkes, Concomitants of the quintic and sextic up to degree four in the coefficients of the ground form, J. London Math. Soc. 25 (1950), 205–209. MR 12,236e [9] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.3, 2002, (http://www.gap-system.org). 198 [10] David A. Gay, Characters of the Weyl group of SU (n) on zero weight spaces and centralizers of permutation representations, Rocky Mountain J. Math. 6 (1976), no. 3, 449–455. MR 54 #2886 [11] Larry C. Grove, Groups and characters, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1997, A Wiley-Interscience Publication. MR 98e:20012 [12] Roger Howe, (GLn , GLm )-duality and symmetric plethysm, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, 85–109 (1988). MR 90b:22020 [13] N. F. J. Inglis, R. W. Richardson, and J. Saxl, An explicit model for the complex representations of Sn , Arch. Math. (Basel) 54 (1990), no. 3, 258–259. MR 91d:20017 [14] G. James and A. Kerber, Representation theory of the symmetric group, Encyclopedia of Mathematics and its Applications, vol. 16, Addison-Wesley, Reading, MA, 1981. [15] G. D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, vol. 682, Springer, Berlin, 1978. [16] Serge Lang, Algebra, 3 ed., Addison Wesley, Reading Massachusetts, 1999. [17] I. G. Macdonald, Symmetric functions and Hall polynomials, second ed., Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, New York, 1995, With contributions by A. Zelevinsky, Oxford Science Publications. MR 96h:05207 [18] Bruce E. Sagan, The symmetric group, The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1991, Representations, combinatorial algorithms, and symmetric functions. MR 93f:05102 199 [19] Richard P. Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004) 58 Chapter 6 Proof of Theorem 2 From Remark 2.2.8, to prove Theorem 2, it suffices to construct non-zero tableaux filled with d copies of c elements for all required partitions of n = cd, c, d ≥ 3. These partitions were determined in Theorem 10. Our approach is similar to the proof of Theorem 1 in Chapter 5. Using some generic non-zero tableaux (like Ui and V in Chapter 5) with c elements, we join them together by Theorem 8 to form a tableau of the appropriate shape and filling. However, unlike in Chapter 5, a large number of generic tableaux are needed. Since the cataloging of non-zero tableaux is quite tedious, we post-pone the construction until Chapter 7. Namely, our proof here will presuppose the construction of all tableaux of the required shapes for c ≤ 8 . The general idea is to write a tableau T as follows: T = S ∨ T0 ∨ U ∨ V for an appropriate T 0 , where S, U , and V generic constructions based on the parity of d. This reduces the construction of T to a construction of T 0 where the shape parameters, (r, s, and t), of T 0 are small. Thus we only need to construct tableaux for a limited number of cases corresponding to small shapes. The tableaux S, U, and V are based on the following non-zero maximal tableaux: (Here U1 , U2 and V occurred in Chapter 5.) 59 d 1 S0 = P1 (d) = 2 3 d U1 (d) = 1 2 d even, d even, aabbbb 5 6 6 6 5 5 S1 = 4 3 3 4 3 4 1 2 1 1 2 2 aabb U2 = 1 3 1 3 2 4 4 2 V = a = d−x +x 3 b = d−x 3 d≡x (mod 3), x ∈ {0, 1, 2} a=d−1 b=1 d 1 Let λ = [r + s + t, s + t, t]. We can write T = S ∨ T 0 ∨ U ∨ V , for appropriate T 0 provided T 0 is maximail. Then T 0 will be filled with d copies of c0 elements, for some c0 < c, which will eventually allow us to reduce to c ≤ 8. If T 0 is non-zero and maximal then by Lemma 3.4.9 and the Theorem 8 qT 6= 0 as desired. For simplicity, we will base our construction on the parity of d. 6.1 Case: d even To see how to write T as T = S ∨ U ∨ V ∨ T 0 for an appropriate T 0 we first discuss the individual reductions allowing us to write T = S ∨ T 0 , T = U ∨ T 0 , or T = V ∨ T 0 . Then successive applications of these reductions yield our desired decomposition. An analysis of these reductions also computes the resulting bounds on the shape of T 0 . An example application follows the reductions listed below. The reader may wish to refer to Example 6.1.1 while reading these reductions. Reduction 1: Let T be any λ-tableau with λ = (λ1 , λ2 , λ3 ), filled with d copies of c elements. Take f to be the maximum integer such that fd ≤ λ3 and c − 3f ≥ 3. Let S = fP1 (d) be the join of f copies of P1 (d). Then by Theorem 8, we may write T = S ∨ T 0 for T 0 a λ0 = (λ1 − df, λ2 − df, λ3 − df)-tableau filled with d copies of c0 = c − 3f elements, provided the weight-sets of S and T 0 are disjoint. The choice 60 of f means that in T 0 , t0 = λ03 = λ3 − df < d or c0 = c − 3f < 6. Thus we need only consider tableaux with t = λ3 < d or c < 6. The c < 6 condition coresponds to the requirement c − 3f ≥ 3. We need this requirement so that there are at least three elements available with which to fill the remaining tableau, T 0 . Reduction 2: Let T be any λ-tableau with λ = (λ1 , λ2 , λ3 ), filled with d copies of c elements. Take g to be the maximum integer such that gd ≤ λ2 − λ3 = s and c − 2g ≥ 3. Let U = gU1 (d) be the join of g copies of U1 (d). Then by Theorem 8, we may write T = U ∨ T 0 for T 0 a λ0 = (λ1 − dg, λ2 − dg, λ3 )-tableau filled with d copies of c0 = c−2g elements, provided the weight-sets of U and T 0 are disjoint. The choice of g 0 means that in T 0 , s0 = λ02 −λ3 = λ2 −dg−λ0 3 < d or c0 = c−2g < 5. However, we will need the existence of a non-zero T 0 in the specified shape. As was shown in Theorem 9, this is not always the case for some s. Specifically, when s < 5 non-zero tableaux do not exist for certain shapes when c = 3. (Consider λ = [6 + d, 2 + d, 1] = [9, 5, 1] with d = 3 and c = 5. Applying Reduction 2 yields λ0 = [5, 2, 1] with c = 3. All such tableaux are zero by Theorem 10 since s = 1.) To account for this, we modify the construction above to use g − 1 copies of U1 (d) when g > 0 and s0 < 5. In such a case, the modified T 0 now has s0 < d + 5. Thus we need only consider arbitrary tableaux with s < d + 5 or c < 5. Reduction 3: Let T be any λ-tableau with λ = (λ1 , λ2 , λ3 ), filled with d copies of c elements. Take h to be the maximum integer such that hd ≤ λ1 − λ2 = r and c − h ≥ 3. Let V = hV1 (d) be the join of h copies of V1 (d). Then by Theorem 8, we may write T = V ∨ T 0 for T 0 a λ0 = (λ1 − dh, λ2 , λ3 )-tableau filled with d copies of c0 = c − h elements, provided the weight-sets of V and T 0 are disjoint. The choice of h means that in T 0 , r0 = λ01 − λ02 = λ1 − dh − λ2 < d or c0 = c − h < 4. However, we will need the existence of non-zero T 0 in the specified shape. As was shown in Theorem 9, this is not always the case for some r. Specifically, when r < 5 non-zero tableaux do not exist for certain shapes when c = 3. To account for this, we modify the construction above to use h − 1 copies of U1 (d) when h > 0 and r0 < 5. In that 61 case, the modified T 0 now has r0 < d + 5. Thus we need only consider arbitrary tableaux with r < d + 5 or c < 4. Conclusion: When d is even, we can apply these reductions successively. Take an arbitrary λ-tableau T filled with d copies of c elements and assume c ≥ 6. We use T (i) to represent the appropriate T 0 obtained in these reductions. By Reduction 1, T = S ∨ T (1) , where S = fP1 (d) and T (1) has t = λ3 (T (1) ) < d and is filled with c(1) = c − 3f elements. Now, if c(1) ≥ 6 apply Reduction 2 to T (1) . Since c(1) ≥ 6, then by Reduction 2, write T (1) = U ∨ T (2) where U = gU1 (d) and T (2) has t < d (since T (1) does) and s < d + 5. Here T (2) is filled with c(2) = c(1) − 2g elements. Finally if c(2) ≥ 6 apply Reduction 3. This gives T (2) = V ∨T (3) , where V = hV1 (d) and T (3) has t < d, s < d + 5, and r < d + 5. Here T (3) is filled with c(3) = c(2) − h elements. Hence T = S ∨ U ∨ V ∨ T (i) where either T (i) is filled with fewer than 6 elements, or T (i) has t < d, s < d + 5, and r < d + 5. In the second case, T (i) must be filled with 3t + 2s + r = cd elements. This is less than or equal to 3(d − 1) + 2(d + 4) + (d + 4) = 6d + 9 ≤ 8d if d > 4. (If d = 4 we have 6d + 8 ≤ 8d and it’s not possible to have 6d + 9 = 9d when d = 4. For d = 3 additional reductions apply.) Hence we only need those tableaux with c ≤ 8. Moreover, if r or s < 5 in T (i) , then r or s < 5 in T , because the reductions do not reduce r or s to less than 5. Hence T (i) = T 0 has a shape occurring in Theorem 10 since all partitions of n with r and s ≥ 5 are needed. This reduction uses Theorem 8. Our usage only requires verification that the weight-sets are disjoint. However, the tableaux S, U , and V are in maximal form. Hence for appropriately chosen tableaux (i.e., ones in maximal form), an application of Lemma 3.4.9 can easily prove weight-set disjointness. Example 6.1.1. To see how this reduction works, let us consider a specific shape, λ = [9d − 2, 5d, d + 2] where d ≥ 6, d even and c = 15. This shape has t = d + 2, s = 4d − 2, and r = 4d − 2. First we apply Reduction 1, which joins P1 (d) in order 62 to have t < d. [9d − 2, 5d, d + 2] = P1 (d) ∨ [8d − 2, 4d, 2] Then we apply Reduction 2 to the shape [8d − 2, 4d, 2], which has s = 4d − 2 to reduce to s < d + 5 by joining three copies of U1 (d). [8d − 2, 4d, 2] = 3U1 (d) ∨ [5d − 2, d, 2] Applying Reduction 3 to shape [5d − 2, d, 2], which has r = 4d − 2 we normally want to reduce r to be between 5 and d + 5. Here we won’t necessarily reduce r fully, so that the resulting tableau will be familiar. Instead we will reduce to r = 2d − 2 (which may be reduced further depending on d) by joining two copies of V (d). [5d − 2, d, 2] = 2V (d) ∨ [3d − 2, d, 2] Hence, when we combine all these reductions, we get [9d − 2, 5d, d + 2] = P1 (d) ∨ 3U1 (d) ∨ 2V (d) ∨ [3d − 2, d, 2] A non-zero tableau of shape [3d − 2, d, 2] is Q∗ of Example 3.2.7 with a = 1, b = 1, c = d − 2. Therefore, writing T = P1 (d) ∨ Q∗ (1, 1, d − 2) ∨ 3U1 (d) ∨ 2V (d) and omitting the extra tail of Q∗ we have d 1 1 1 1 d-2 d d d d d 1 5 5 5 6 4 8 10 12 14 15 T = 2 4 7 7 4 7 9 11 13 366 As Q∗ is in maximal form, qT 6= 0. 63 6.2 Case: d odd When d is odd, we proceed exactly as in the even case, except the tableaux we use are slightly different. Namely, we use S1 instead of P1 and U2 instead of U1 . These adjustments are necessary for Reductions 1 and 2 since P1 (d) and U1 (d) are zero for d odd. Reduction 3 remains unchanged however. For completeness, we rewrite these reductions in terms of d odd. However, these reductions alone are not enough to reduce to c ≤ 8. So after these reductions, we apply a few more in order to reduce the size of tableaux we need to consider. Reduction 10 : Let T be any λ-tableau with λ = (λ1 , λ2 , λ3 ), filled with d copies of c elements. Take w to be the maximum integer such that w·2d ≤ λ3 and c−6w ≥ 3. Let S = wS1 (d) be the join of w copies of S1 (d). Then by Theorem 8, we may write T = S ∨ T 0 for T 0 a λ0 = (λ1 − 2d · w, λ2 − 2d · w, λ3 − 2d · w)-tableau filled with d copies of c0 = c − 6w elements, provided the weight-sets of S and T 0 are disjoint. The choice of w means that in T 0 , t0 = λ03 = λ3 − 2d · w < 2d or c0 = c − 6w < 9. Thus we need only consider tableaux with t = λ3 < 2d or c < 9. Reduction 20 : Let T be any λ-tableau with λ = (λ1 , λ2 , λ3 ), filled with d copies of c elements. Take v to be the maximum integer such that v · 2d ≤ λ2 − λ3 = s and c − 4v ≥ 3. Let U = vU2 (d) be the join of v copies of U2 (d). Then by Theorem 8, we may write T = U ∨ T 0 for T 0 a λ0 = (λ1 − 2d · v, λ2 − 2d · v, λ3 )-tableau filled with d copies of c0 = c − 4v elements, provided the weight-sets of U and T 0 are disjoint. The choice of v means that in T 0 , s0 = λ02 − λ03 = λ2 − d · v − λ03 < 2d or c0 = c − 4v < 8. As in the even case, to account for the shapes s < 5, we modify this reduction to use v − 1 copies of U1 (d) when v > 0 and s0 < 5. Then the modified T 0 now has s0 < 2d + 5. Thus we need only consider arbitrary tableaux with s < 2d + 5 or c < 8. Summary: The same argument as in the even case works for the d odd cases, though the numbers are adjusted slightly. Take an arbitrary λ-tableau T with filled with d copies of c elements, but this time assume c ≥ 9. Then by applications of 64 Reductions 1’, 2’ and 3, T = S ∨ U ∨ V ∨ T (i) where either T (i) is filled with fewer than 9 elements, or T (i) has t < 2d, s < 2d + 5, and r < d + 5. In the second case, T (i) must be filled with 3t + 2s + r elements, which is less than or equal to 3(2d − 1) + 2(2d + 4) + (d + 4) = 11d + 9 as d ≥ 3. Moreover, if r or s < 5 in T (i) , then r or s < 5 in T . Hence T (i) has a required shape of Theorem 10. However, we wish to have T (i) fillable with c ≤ 8. To do this we have additional reduction techniques. However, these techniques are very sensitive to the parameters in T (i) , so we will categorize them by such. The additional non-zero maximal tableaux we use are d−1 U1 (d − 1) = 1 2 d odd ω2 = (0, d − 1) d−1 1 P1 (d − 1) = 2 3 d odd ω2,3 = d−2 1 113 P4 (d − 2, 1, 1) = 2 232 3 d odd ω2,3 = ¡ 0 d−1 0 ¡0 d 0 0 d−1 1 0 0 d−2 ¢ ¢ Start with a tableau T where t ≤ 2d − 1, s ≤ d + 4, s 6= 1, r ≤ d + 4, r 6= 1, d odd and s + t even if r or s in {0, 2, 4}. (These are the partitions required by Theorem 10 after the previous reductions have been applied.) First consider those tableaux with r ≥ 10, which implies d ≥ 6. Case A: Assume r ≥ 10, s < d + 4, t < d − 1. Then 3t + 2s + r ≤ 3(d − 2) + 2(d + 3) + d + 4 = 6d + 4 ≤ 8d. Hence this case is covered by c ≤ 8. Case B: Assume r ≥ 10, s ≥ d+4, t ≥ d−1. Write T = P1 (d−1)∨U1 (d−1)∨T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r − 5, s − (d + 1), t − (d − 1)). Thus 5 ≤ r0 ≤ d − 1, 5 ≤ s0 ≤ d + 5, and 0 ≤ t0 ≤ d. Then 65 3t + 2s + r ≤ 3d + 2(d + 5) + d − 1 = 6d + 9 ≤ 8d. Note that no exceptional r or s cases occur in T 0 . Hence this case is covered by c ≤ 8. Case C: Assume r ≥ 10, s < d + 4, t ≥ d − 1. Write T = P1 (d − 1) ∨ T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r−3, s, t−(d−1)). Thus 7 ≤ r0 ≤ d + 1, s0 ≤ d + 3, and 0 ≤ t0 ≤ d. Then 3t + 2s + r ≤ 3d + 2(d + 3) + d + 1 = 6d + 7 ≤ 8d. Note that no exceptional r or s cases occur in T 0 . Hence this case is covered by c ≤ 8. Case D: Assume r ≥ 10, s ≥ d + 4, t < d − 1. Write T = U1 (d − 1) ∨ T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r − 2, s − (d − 1), t). Thus 8 ≤ r0 ≤ d + 2, s0 ≤ d + 5, and 0 ≤ t0 ≤ d − 1. Then 3t + 2s + r ≤ 3(d − 1) + 2(d + 5) + d + 2 = 6d + 9 ≤ 8d. Note that no exceptional r or s cases occur in T 0 . Hence this case is covered by c ≤ 8. For r < 10 the arguments depend more on the values of r, but the general idea is the same. Case E: Assume r < 10, s < d + 4, t < d − 1. Then 3t + 2s + r ≤ 3(d − 1) + 2(d + 3) + 9 = 5d + 12 ≤ 8d for d ≥ 4. When d = 3, c = 9 is a possibility. Now, when d = 3, we have r ≤ d + 4 = 7, s ≤ d + 3 = 6, and t ≤ d − 2 = 1. So here 3t + 2s + r ≤ 3 + 2 · 6 + 7 = 22 < 8d. Hence c ≤ 8 tableaux will suffice to cover this case. Note that the s + t parity is preserved in the exceptional r and s cases. Case F: Assume r < 10, s ≥ d + 4, t < d − 1. If r = 9, 8, 7, 5 then write T = U1 (d − 1) ∨ T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r − 2, s − (d − 1), t). Then r0 ∈ {7, 6, 5, 3}, s0 ≤ d + 5, and 0 ≤ t0 ≤ d − 1. Then 3t0 + 2s0 + r0 ≤ 3(d − 2) + 2(d + 5) + 7 = 5d + 11 ≤ 8d for d > 3. When d = 3, then s0 ≤ 8, t0 ≤ 1. We have 3t0 + 2s0 + r0 ≤ 3 + 16 + 7 = 26 < 9d. Hence c ≤ 8 tableaux will suffice to cover these case. If r = 2, 4, or 6, then write T = U1 (d − 1) ∨ T 0 . If (r, s, t) are the parameters of T , 66 then T 0 has parameters (r0 , s0 , t0 ) = (r −2, s−(d−1), t). Thus r0 ∈ {4, 2, 0}, s0 ≤ d+5, and 0 ≤ t0 ≤ d − 1. Then 3t0 + 2s0 + r0 ≤ 3(d − 2) + 2(d + 5) + 4 = 5d + 8 ≤ 8d. Hence c ≤ 8 tableaux will suffice to cover these cases if the tableau exists. Note that U1 (d − 1) preserves the parity of s + t, so if s + t are even (always the case when r = 2 or 4), we will have s0 + t0 even which is necessary for r0 = 0, 2, or 4. Hence this construction works except when r = 6 and s + t odd. Then 3t + 2s + r ≤ 3(d − 2) + 2(2d + 4) + 6 = 7d + 8 ≤ 8d for d ≥ 8 otherwise it is less than 9d unless d = 3. If 3t + 2s + 6 = 9d, then t is odd so write t = d − 2k with k ≥ 1. Then we have s = 3d + 3k − 3. Since s ≤ 2d + 4, we have d + 3k ≤ 7, so d = 3, k = 1 is the only solution. This corresponds to t = 1, s = 9, r = 6. Since s + t even, this case has already been done. For d = 3 we also need to consider c = 10. However, 3 + 20 + 6 < 10d so no such partition will occur. For r = 0 we have s+t even. Then 3t+2s+r ≤ 3(d−2)+2(2d+4) = 7d+2 ≤ 8d, so this case is covered by c ≤ 8 tableaux. For r = 3 we 3t + 2s + r ≤ 3(d − 2) + 2(2d + 4) + 3 = 7d + 5 ≤ 8d for d ≥ 5. When d = 3 we have 3 + 20 + 3 < 9d, so such a partition does not occur. Hence r = 3 is covered by c ≤ 8. Case G: Assume r < 10, s < d + 4, t ≥ d − 1. If r = 9, 8, 6 then write T = P1 (d − 1) ∨ T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r − 3, s, t − (d − 1)). Thus r0 ∈ {6, 5, 3}, s0 ≤ d + 3, and 0 ≤ t0 ≤ d. Then 3t0 + 2s0 + r0 ≤ 3d + 2(d + 3) + 6 = 5d + 12 ≤ 8d for d ≥ 5. When d = 3, then s0 ≤ 6, t0 ≤ 3. Hence we have 3t0 +2s0 +r0 ≤ 9+12+6 = 27 = 9d, so the only solution is t0 = 3, s0 = 6, r0 = 6. This we can further reduce by writing T 0 = V (d) ∨ T 00 where T 00 is a c = 8 tableau of parameters t = 3, s = 6, and r = 3. Hence c ≤ 8 tableaux will suffice. If r = 7, 5 , or 3 and s + t even then write T = P1 (d − 1) ∨ T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r − 3, s, t − (d − 1)) and the parity of s0 + t0 is preserved. Thus r0 ∈ {0, 2, 4}, s0 ≤ d + 3, and 0 ≤ t0 ≤ d. Then 3t0 + 2s0 + r0 ≤ 3d + 2(d + 3) + 4 = 5d + 10 ≤ 8d for d ≥ 5. For d = 3 we have 67 3t0 + 2s0 + r0 ≤ 9 + 12 + 4 = 25 < 9d, so this does not occur. Hence the c ≥ 8 tableaux will suffice. Consider r = 0, 2, 4 with s + t even, or r = 3, 5, 7 with s + t odd. If s ≥ 8 write T = P4 (d − 2, 1, 1) ∨ T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r, s−3, t−(d−2)) and the parity of s0 +t0 is preserved. Thus , 5 ≤ s0 ≤ d, and 0 ≤ t0 ≤ d + 1. Then 3t0 + 2s0 + r0 ≤ 3(d + 1) + 2(d) + 7 = 5d + 10 ≤ 8d for d ≥ 5. For d = 3 we have 3t0 + 2s0 + r0 ≤ 12 + 6 + 7 = 25 < 9d, so this does not occur. Hence the c ≥ 8 tableaux will suffice. For r = 3, 5, or 7, s + t odd, and s ≤ 7, we have 3t + 3s + r ≤ 3(2d − 1) + 2 · 7 + 7 = 6d + 18 ≤ 8d for d ≥ 9. If 3t + 2s + r = 9d, then since r is odd, we have t = 2d − 2k for k ≥ 1. This implies s = d + 3k + d−r . Since s ≤ 7, the only possible solutions are 2 (r, s, t) = (7, 4, 4), (5, 5, 4), (3, 6, 4), (7, 7, 2) when d = 3 and (7, 4, 8) when d = 5. For those with s + t even, the case has already been done. For (r, s, t) = (7, 7, 2) write T = U1 (2) ∨ T 0 where T 0 has parameters (5, 5, 2) and is a c = 7 tableau. Hence we still need (r, s, t) = (5, 5, 4) for d = 3. But this is U1 (2) ∨ T 0 , where T 0 is a c = 7 tableau of parameters (3, 3, 4). We also need to consider those partitions with c = 10. Then we have 3t + 2s + r = . The only 10d which implies t = 2d − 2k − 1 as t is odd. So s = 2d + 3k + 1 − r−1 2 solutions with s ≤ 7 are (r, s, t) = (3, 6, 5), (5, 5, 5), (7, 4, 5), and (7, 7, 3) all with d = 3. The cases with s + t even have been done already. For (r, s, t) = (3, 6, 5), use T = P4 (1, 1, 1) ∨ T 0 where T 0 is a c = 7 tableau with (r0 , s,0 , t0 ) = (3, 3, 4). For (7, 4, 5) we have s = 4 and s + t odd, so this case partition is not needed. When d = 3 we also may have c = 11 or c = 12. Proceeding as above, the only solutions are (r, s, t) = (7, 7, 4) and (7, 7, 5). The second case has s + t even and hence is not needed. The first case can be reduced to T = U1 (2) ∨ T 0 , where T 0 is a c = 9 case with parameters (5, 5, 4). But this is U1 (2) ∨ T 0 , where T 0 is a c = 7 tableau of parameters (3, 3, 4). If r = 0, 2, 4, s + t even, and s ≤ 7, we have 3t + 3s + r ≤ 3(2d − 1) + 2 · 7 + 4 = 6d + 15 ≤ 8d for d ≥ 9. For c = 9, we have t = 2d − 2k − 1 as t is odd. Then s = d + 3k + 1 + d+1−r . Since s ≤ 7 the only solutions are (r, s, t) = (0, 6, 5), (2, 5, 5), (4, 4, 5), 2 68 and (4, 7, 3) with d = 3 and (4, 7, 9). Note that only those with s + t even are needed. For these cases write T = U1 (d − 1) ∨ T 0 , where T 0 is a c0 = 7 tableau with parameters (r − 2, s − (d − 1), t). Since this preserves the parity of s + t and does not cause r0 , s0 = 1, the tableau exists. For d = 3 we may also have c = 10 or 11. Proceeding as above, the only solutions are (r, s, t) = (4, 7, 4) which has s + t odd, hence it is not need, and (4, 7, 5) which is U1 (2) ∨ U1 (2) ∨ T 0 where T 0 is a c = 7 tableau with (r, s, t) = (0, 3, 5). Case H: Assume r < 10, s ≥ d + 4, t ≥ d − 1. If r = 8 or r = 9, 7, 5 and s + t even then write T = P1 (d − 1) ∨ U1 (d − 1) ∨ T 0 where T 0 has parameters (r0 , s0 , t0 ) = (r − 5, s − (d − 1), t − (d − 1)). Thus r0 ≤ 4, s0 ≤ d + 5, and t0 ≤ d. Then 3t0 + 2s0 + r0 ≤ 3d + 2(d + 5) + 4 = 5d + 14 ≤ 8d for d > 3. When d = 3 we can have 3t0 + 2s0 + r0 = 9d only for (r0 , s0 , t0 ) = (2, 8, 3) or (4, 7, 3). But s + t even means only (4, 7, 3) is needed. This is U1 (d − 1) ∨ T 00 where T 00 is a c = 7 tableau with (r, s, t) = (2, 5, 3). Hence we’ve reduced to c ≤ 8 cases. For r = 9, 7, 5 with s+t odd, write T = P4 (d−2, 1, 1)∨U1 (d−1)∨T 0 where T 0 has parameters (r0 , s0 , t0 ) = (r − 2, s − (d − 1) − 3, t − (d − 2)). Hence r0 ≤ 7, 2 ≤ s0 ≤ d + 2, and t0 ≤ d + 1. We will have s0 6∈ {0, 1, 2, 4} provided s 6= d + 4 or d + 6. Then 3t0 +2s0 +r0 ≤ 3(d+1)+2(d+2)+7 = 5d+14 ≤ 8d for d > 3. When s = d+4 or d+6 write T = P4 (d − 2, 1, 1) ∨ T 0 where T 0 has parameters (r0 , s0 , t0 ) = (r, s − 3, t − (d − 2)). Then 3t0 + 2s0 + r ≤ 3(d + 1) + 2(d + 3) + 9 = 5d + 18 ≤ 8d for d ≥ 5. For d = 5, s0 = d + 3 or d + 1 there are no partitions with t0 ≤ d + 1. Now consider d = 3 with s0 ≤ d + 3, t0 ≤ d + 1. Since r ≤ d + 4 we have r = 3, 5, or 7. Given these parameters, solving 3t0 + 2s0 + r0 = 3c for c ≥ 9, the only solutions are (r0 , s0 , t0 ) = (3, 6, 4), (5, 5, 4), and (7, 4, 4). Since s + t odd the only partition we need to construct is (r0 , s0 , t0 ) = (5, 5, 4) with c = 9. But this is U1 (2) ∨ T 0 where T 0 is a c = 7 tableau of parameters (3, 3, 4). For r = 2, 4, 6 with s + t even, write T = P4 (d − 2, 1, 1) ∨ U1 (d − 1) ∨ T 0 where T 0 has parameters (r0 , s0 , t0 ) = (r − 2, s − (d − 1) − 3, t − (d − 2)). Thus r0 ≤ 4, 2 ≤ s0 ≤ d + 2, and t0 ≤ d + 1. Since s + t even, we need only that s0 6= 1. Then 69 3t0 + 2s0 + r0 ≤ 3(d + 1) + 2(d + 2) + 4 = 5d + 11 ≤ 8d for d > 3. For d = 3, d · 4 + 2 · 5 + 4 = 26 < 9d so that case is not needed. For r = 6, s + t odd write T = P1 (d − 1) ∨ T 0 , where T 0 has parameters (r0 , s0 , t0 ) = (3, s, t − (d − 1)). Thus 3t0 + 2s0 + r0 ≤ 3d + 2(2d + 4) + 3 = 7d + 11 ≤ 8d for d ≥ 11. Solving 3t0 +2s0 +r0 = 9d given the parameters, the only solution is (r0 , s0 , t0 ) = (3, 9, 2). This is P4 (1, 1, 1) ∨ T 00 where T 00 is a c = 6 tableau with parameters (3, 6, 1). When c = 10, we also have (3, 9, 3) as a solution, but this does not have s + t odd. For r = 3, write T = P4 (d − 2, 1, 1) ∨ T 0 where T 0 has parameters (r0 , s0 , t0 ) = (3, s − 3, t − (d − 2)). Thus 3t0 + 2s0 + r0 ≤ 3(d + 1) + 2(2d + 1) + 3 = 7d + 8 ≤ 8d for d > 5. (The c = 8 constructions will hold unless s − 3 ∈ {0, 1, 2, 4}, but since s ≥ d + 4 this can only occur with s = d + 1, d = 3. Then our original tableau will have t ≤ 2d − 1, s = d + 1, r = 3 which is satisfiable by a c ≤ 8 tableau.) Solving 3t0 + 2s0 + r0 = 9d given the parameters, the only solutions are (r0 , s0 , t0 ) = (3, 6, 4), (3, 9, 2), and (3, 12, 6) with d = 5. The (3, 6, 4) case is P4 (1, 1, 1) ∨ T 00 where T 00 is a c = 6 tableau with parameters (3, 3, 3). The (3, 9, 2) case is P4 (1, 1, 1) ∨ T 00 where T 00 is a c = 6 tableau with parameters (3, 6, 1). When d = 5, the (3, 12, 6) case is P4 (3, 1, 1) ∨ T 00 where T 00 is a c = 6 tableau with parameters (3, 9, 3). There are no tableaux with c > 9. For r = 0 we have s+t even. Write T = P4 (d−2, 1, 1)∨T 0 where T 0 has parameters (r0 , s0 , t0 ) = (3, s − 3, t − (d − 2)). So 3t0 + 2s0 + r0 ≤ 3(d + 1) + 2(2d + 1) = 7d + 5 ≤ 8d for d > 3. As P4 preserves parity and s + t even, the c = 8 constructions will work provided s0 6= 1. But s ≥ d + 4 forces s0 ≥ d + 1 ≥ 4. When d = 3 we have 3 · 4 + 2 · 7 = 26 < 9d, hence all cases are covered. Conclusion: When d is odd, we need all tableaux with c ≤ 8. The basic reduction requires those tableaux to be disjoint from multiple copies of S1 , U2 (d − 1, 1) and V (d). The further reductions also require the tableau to be disjoint from P1 (d−1), U1 (d−1), P1 (d−1)∨U1 (d−1), P4 (d−2, 1, 1), and U1 (d−1)∨P4 (d−2, 1, 1). For d even we need all tableaux with c ≤ 6, along with those tableaux having t ≤ d − 1, s ≤ d + 4, r ≤ d + 4 when c = 7 or 8. These tableaux need to be 70 disjoint from P1 (d), U1 (d), and V (d). These tableaux will be listed in Chapter 7. In Chapter 8, we verify that all necessary tableaux have been produced. 197 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. Doran, IV, On Foulkes’ conjecture, J. Pure Appl. Algebra 130 (1998), no. 1, 85–98. MR 99h:20014 [8] H. O. Foulkes, Concomitants of the quintic and sextic up to degree four in the coefficients of the ground form, J. London Math. Soc. 25 (1950), 205–209. MR 12,236e [9] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.3, 2002, (http://www.gap-system.org). 198 [10] David A. Gay, Characters of the Weyl group of SU (n) on zero weight spaces and centralizers of permutation representations, Rocky Mountain J. Math. 6 (1976), no. 3, 449–455. MR 54 #2886 [11] Larry C. Grove, Groups and characters, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1997, A Wiley-Interscience Publication. MR 98e:20012 [12] Roger Howe, (GLn , GLm )-duality and symmetric plethysm, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, 85–109 (1988). MR 90b:22020 [13] N. F. J. Inglis, R. W. Richardson, and J. Saxl, An explicit model for the complex representations of Sn , Arch. Math. (Basel) 54 (1990), no. 3, 258–259. MR 91d:20017 [14] G. James and A. Kerber, Representation theory of the symmetric group, Encyclopedia of Mathematics and its Applications, vol. 16, Addison-Wesley, Reading, MA, 1981. [15] G. D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, vol. 682, Springer, Berlin, 1978. [16] Serge Lang, Algebra, 3 ed., Addison Wesley, Reading Massachusetts, 1999. [17] I. G. Macdonald, Symmetric functions and Hall polynomials, second ed., Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, New York, 1995, With contributions by A. Zelevinsky, Oxford Science Publications. MR 96h:05207 [18] Bruce E. Sagan, The symmetric group, The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1991, Representations, combinatorial algorithms, and symmetric functions. MR 93f:05102 199 [19] Richard P. Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004) 71 Chapter 7 Tableaux Construction The proof of Theorem 2 in Chapter 6 requires the construction of non-zero maximal tableaux with c ≤ 8 for the shapes discussed in Theorem 10. In this chapter we construct all the necessary tableaux and show they are both non-zero and maximal. Some basic properties of maximality are listed in Section 7.1 and are used throughout our construction. We construct those tableaux with two rows in Section 7.2. In Sections 7.3, 7.4, 7.5, and 7.6 we construction the necessary tableaux with c = 3, 4, 5, and 6, respectively. Section 7.7 contains additional tableaux need when c = 7 or 8. 7.1 Maximality of Tableaux Maximality, as discussed in Lemma 3.4.9, is an important property of the tableaux we construct to prove Theorem 2. Given certain conditions of a weight, it is easy to verify that a tableau is in maximal form. We discuss maximality here in general, in order to simplify the proof of maximal form for the specific tableaux we construct in the next sections. Recall that a tableau T is in maximal form if qT 6= 0 by weight-set counting on ω(T ) and ω(T ) is the largest weight of all ω(τ T ) for τ ∈ CT . All tableaux that we consider in the next sections are shown to be non-zero by weight-set counting prior to addressing the maximality issue. As such, we only consider the weight condition here. To summarize the basic conditions of Definition 3.4.2, a tableau weight is maximal if it has the largest weight possible for row three, and given this, the largest weight 72 possible for row two. When multiple weights satisfy this, the weight where rows two and three have elements in common (e.g., the element 4 appears in both rows) is considered larger. In general, to determine the maximal form of a tableau, we start with a given filling of T and use column operations to produce tableaux τ T with larger weights. The following procedure provides an overview of how to determine the maximality of a filling. • First maximize the weight of row three. This usually involves having a weight of d for as many elements as possible. (This is discussed more extensively in Lemma 7.1.2.) There may be multiple different fillings having the same maximum weight for row three. All such fillings should be considered for the next step. • Now maximize the weight of row two, given the filling(s) of row three determined previously. To do this, determine the largest possible row two weight of each element, provided those elements assigned to row three are not used. For a given element this is equivalent to the number of copies in the body of T minus the number of copies used in row three. Fill row two with the element having the largest weight, then repeat the procedure with the remaining elements and positions. (This is summarized for weights in Lemma 7.1.4.) There may be multiple such fillings having the largest weight. • If there are multiple fillings after step two that have different generic weights, choose the filling in which an element in row three has the largest weight in row two. (The weight of their common elements is maximized.) There may not be a unique such filling, but all such maximal fillings will have the same generic weight. As such these fillings will differ on by an action of Sa . 73 Example 7.1.1. Consider z+x z z a 3 323 T = Q2 = 4 244 1 112 ω2,3 (T ) = ¡ 0 z 0 d−z+a+x ¢ d a 0 0 where z = d−x , a < z, and some conditions on z, a, and x to insure qT is non-zero. 3 First we want to maximize row three. Since this occurs when the non-zero weights of row three are d and a, we check to see which elements can have weight d in row three. Here, the only option is the element 1. Since any of the remaining elements can have a weight of a, we leave that column unfilled for now. Thus the maximal (third row) form for T is 111∗. To determine the maximal second row form, we consider how many of each element is available given the third row is partially determined. There are zero 1’s available, 2z+a 2’s available, and 2z+a+x 3’s and 4’s available. Hence we can maximally fill the second row with either 3’s or 4’s. Since the tableau is symmetric in 3’s and 4’s, we will use 4’s without loss of generality. Thus we get a maximal (second and third row) form of 4*44 . 111* This form gives rise to two different generic weights depending on whether the ∗’s ¢ ¡ 0 0 d−z+a z ¢ ¡ 0 are the same. These generic weights are d0 az d−z+a 0 0 . The first one, 0 0 and d a corresponding to the same element for both ∗’s, is larger. Hence a maximal (second 4244 . Note that this filling is not unique. We could have 1112 4344 3433 3233 used , , or instead. All would be in maximal form. 1113 1114 1112 and third row) form is Some fillings can be seen as maximal strictly by examining the weight of the tableau. Lemma 7.1.2. If a row has at most one weight not equal to 0 or d, then its row weight is maximal. This follows directly from the ordering on weights and the limit of d copies of an element in a tableau. If we call the weight not equal to 0 or d the non-d weight for the row, we get the following result on tableau maximality. 74 Lemma 7.1.3. A tableau is maximal if rows two and three satisfy Lemma 7.1.2 and either their non-d weights come from the same element or the sum of the non-d weights is greater than d. Proof. Given a tableau satisfying Lemma 7.1.2 for rows two and three, its maximal generic weight must be either ( d0 ab d0 00 ) or ( d0 a0 d0 b0 ) with the appropriate number of 0 and d weights. By our ordering, the first weight is strictly larger than the second weight. Hence when the non-d weights come from the same element the tableau is maximal. If a + b is greater than d, the first weight is not possible and so the second one is maximal. This lemma is directly applicable to the tableau weights. A more generalized form of this, depending on the actual filling, is also useful. The following lemma reflects the technique used in Example 7.1.1. Lemma 7.1.4. A row is maximal if all except one non-zero weight corresponds to the largest weights possible for any elements. This lemma is a generalization of Lemma 7.1.2, where the maximum weight for each element is no longer d. Using this in Lemma 7.1.3 gives: Lemma 7.1.5. A tableau is maximal if row three satisfies Lemma 7.1.2, row two satisfies Lemma 7.1.4, and the non-d weight of row three has the largest possible weight in row two of all such weights satisfying Lemma 7.1.4. Example 7.1.1 represents an appropriate use of Lemma 7.1.5. The weight of row three is (d, a, 0, 0), clearly satisfying Lemma 7.1.2. The weight of row two is (0, z, 0, d − z + a + x). That satisfies Lemma 7.1.4, since we checked that d − z + a + x was the largest weight possible. Finally, ω3 (2) = a is the non-d weight of row three. Since ω2 (2) = z is larger than ω2 (2) = 0, the weight is maximal by Lemma 7.1.5. In general, for the application of Lemma 7.1.5 it is to check that Lemma 7.1.2 and Lemma 7.1.4 apply and then to show that the non-d weight cannot be assigned a larger row two weight without changing the generic weight of row two. In particular 75 the non-d weight conditions of Lemma 7.1.5 are satisfied if all extra (non-third row) copies an element in row three are contained in row two. We will refer to these lemmas for maximality justification of the tableaux constructed in the next section. 7.2 Tableaux for Two Row Partitions In this section we construct all the two row tableaux needed for the proof of Theorem 2. For many of these constructions, the parity of d is relevant. Recall the notation: d∗ =   d if d is even  d − 1 if d is odd Since we are constructing tableaux for many partitions, we will not use a fixed c. However, for every element that is listed in the tableau, we assume it occurs d times, filling out the tail as needed using only those numbers in the body of T . Tableau U1 a d−a d−a a U1 = 1 1 ∼ 1 2 2 2 a even a≤d ω2 (U1 ) = (0, a) λ = [2d − a, a] r = 2d − 2a, s = a, t = 0 We showed U1 non-zero in the proof of Theorem 1. Maximality: This tableau is maximal by Lemma 7.1.4. 76 Tableau U2 aabb U2 = 1 3 1 3 2 4 4 2 a+b≤d a, b > 0 ω2 = (0, a + b, 0, a + b) λ = [4d − 2a − 2b, 2a + 2b] r = 4d − 4a − 4b, s = 2a + 2b, t = 0 We showed U2 non-zero in the proof of Theorem 1. Maximality: This tableau is maximal by Lemma 7.1.4. Tableau U3 a even abb U3 = 1 1 2 2 3 3 a+b≤d 0 < 2b ≤ d b≥a ω2 = (0, a, 2b) λ = [3d − a − 2b, a + 2b] r = 3d − 2a − 4b, s = a + 2b, t = 0 If b > a, any valid weight assignment must have ω2 (3) = 2b. Hence (1, 2, 3) with sign 1 and (2, 1, 3) with sign (−1)a are the only possible weight assignments. Thus U3 is non-zero. If a = b, we may also have weight assignments: (1, 3, 2) with sign (−1)b ; (2, 1, 3) with sign (−1)a+b ; (3, 1, 2) with sign (−1)2b ; and (3, 2, 1) with sign (−1)a+2b . Since a = b and a is even, these are all positive. Hence U3 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.4 since 2b ≥ a + b. 77 Tableau U4 ab U4 = 2 3 1 1 a≥b>0 ω2 = (a + b, 0, 0) λ = [3d − a − b, a + b] r = 3d − 2a − 2b, s = a + b, t = 0 Since only the element 1 can have ω2 = a + b, there is exactly one valid weight assignment, (1, 2, 3). Thus U4 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.4. Tableau U5 a b−2 a−1 b U5 = 1 1 4 4 2 3 3 2 a≥2 b≥3 a+b≤d ω2 = (0, a + b, a + b − 3, 0) λ = [4d − 2a − 2b + 3, 2a + 2b − 3] r = 4d − 4a − 4b + 6, s = 2a + 2b − 3, t = 0 Only the element 2 may have ω2 (2)=a+b. Since a ≥ 2 and b ≥ 3, we must have ω2 (3) = a + b − 3. Hence (1, 2, 3, 4) is the only valid weight assignment, so U5 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.4 since a+b is the largest weight possible. 78 Tableau U6 a=d−2 abc d U6 = 1 3 5 5 5 3 1 3 1 3 5 2 4 424552 b = d+1 2 d ≡ 1 (mod 2) c = d−3 2 d>5 d = d−5 2 ) ω2 = (0, d, 0, d, d−3 2 λ = [3d − d−3 , 2d + d+3 ] 2 2 r = 3, s = 2d + d−3 , t = 0, 2 Only the elements 2 and 4 may have ω2 = d. If ω2 (5) = 0, then ω2 (1 and 3) > 0. Hence (1, 2, 3, 4, 5) is the only valid weight assignment and so U6 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.2. Tableau U7 2 23 2 U7 = 4 3 5 5 3 4 2 3 4 111223 d=5 ω2 = (5, 4, 2, 0, 0) λ = [14, 11] r = 3, s = 11, t = 0, The valid weight assignments are (5, 4, 2, 0, 0, 0), (5, 4, 0, 2, 0), (2, 0, 4, 0, 5), (0, 0, 2, 4, 5), and (0, 0, 4, 2, 5). Since there are an odd number of weight assignments, this tableau is non-zero. Maximality: This tableau is maximal by Lemma 7.1.4. Tableau U8 U8 = 43333122444 11122 d=4 79 ω2 = (3, 2, 0, 0) λ = [11, 5] r = 6, s = 5, t = 0 Then only valid weight assignments are (1, 2, 3, 4), with sign 1, (1, 3, 2, 4), with sign (−1)2 , and (3, 1, 2, 4) with sign (−1)3 . Hence the weight sum is 1 and U8 is non-zero. Maximality: This tableau is not maximal since ω2 = (0, 0, 4, 1) is larger. However, in qT , the weight (0, 0, 4, 1) always cancels. This tableau cannot be put in maximal form, hence we will need to prove directly that it is disjoint from the requisite tableaux. This will be done in Section 8.7. 80 7.3 Tableaux for c = 3 We know by Theorem 9 which tableaux are non-zero for c = 3. However, using Theorem 8 on tableaux requires the tableaux to be non-zero by weight-set counting on ω(T ). We also want the tableaux to be maximal, in order to obtain the disjointness of Lemma 3.4.9. Here we will list the tableaux used, briefly showing they are non-zero and maximal. For all these tableaux, any valid weight assignment corresponds to a unique tableau, so we will not explicitly state how many tableaux correspond to each weight assignment. Tableau P1 a 1 P1 = 2 3 0≤a≤d a even ω2,3 = ( 00 a0 a0 ) λ = [3d − 2a, a, a] r = 3d − 3a, s = 0, t = a P1 is non-zero by the Lemma 3.2.6. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau P2 ab 1 1 P2 = 2 2 3 a+b≤d a, b even 0 ω2,3 = ( 00 a+b 0 a) λ = [3d − 2a − b, a+b, a] r = 3d − 3a − 2b, s = b, t = a 81 Only the elements 1 and 2 may have ω2 = a+b. Hence the valid weight assignments are: (1, 2, 3) with sign 1; (3, 2, 1) with sign (−1)a ; (2, 1, 3) with sign (−1)a+b ; and (3, 1, 2) with sign (−1)b . Since a and b are even, this weight sum is 4 and qP2 6= 0. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau P3 a b+1 b c 2 2 3 2 P3 = 1 1 1 3 3 0≤a≤d 2b + 1 ≤ d − a 0≤c<b a + c even c ω2,3 = ( 00 a+2b+1 0 a) λ = [3d − 2a − 2b − c − 1, a + 2b + c + 1, a] r = 3d − 3a − 4b − 2c − 2, s = 2b + c + 1, t = a Only the element 1 may have ω2 = a + 2b + 1. Hence the valid weight assignments are (1, 2, 3) with sign 1 and (1, 3, 2) with sign (−1)a+c . Since a+c is even, this weight sum is positive and hence P3 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau P4 abbc 2 2 3 2 P4 = 1 1 1 3 3 0≤a≤d 2b ≤ d − a 0≤c≤b a + c even c ω2,3 = ( 00 a+2b 0 a) λ = [3d − 2a − 2b − c, a + 2b + c, a] r = 3d − 3a − 4b − 2c, s = 2b + c, t = a 82 Unless c = b, the only valid weight assignments are (1, 2, 3) with sign 1 and (1, 3, 2) with sign (−1)a+c = 1. If b = c the tableau is symmetric in 1, 2, and 3. Hence we also get the weight assignments: (2, 1, 3) with sign (−1)a+b+b+c = 1; (2, 3, 1) with sign (−1)2b = 1; (3, 2, 1) with sign (−1)a+b = 1; and (3, 1, 2) with sign (−1)2b = 1. Thus the weight sum is always positive and P4 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. 83 7.4 Tableaux for c = 4 Tableau Q1 0≤a<z z+x z z a b c d 3 323 3 2 3 Q1 = 4 244 4 4 2 1 112 b+x≥d c≥d b+c≤z−a z + a + d even c + d even, if d = b + x d + b + x even, if d = c z = d−x 3 d≡x (mod 3) 0 2z+x+a+b+c ω2,3 = ( d0 z+d ) a 0 0 λ = [2d − 2a − b − c − d, d + a + b + c + d, d + a] r = d − 3a − 2b − 2c − 2d, s = b + c + d, t = d + a Since a < z, any valid weight assignment must have ω3 (1) = d. We must have ω2 (4) = 2z + x + a + b + c unless d = c or d = b + x. If d = c, then we may also have ω2 (3) = 2z+x+a+b+c. If d = b+x, then we may also have ω2 (2) = 2z+x+a+b+c. Given these restrictions, we list the valid weight assignments in the table below, along with the sign corresponding to that assignment. Note that there is a unique tableau associated to each of these assignments. Assignment Sign Condition (1, 2, 3, 4) 1 (1, 3, 2, 4) (−1)z+a+d (1, 2, 4, 3) (−1)z+x+a+b+c+d d=c (1, 4, 2, 3) (−1)2z+x+b+d d=c (1, 3, 4, 2) (−1)2z+x+b+c d=b+x (1, 4, 3, 2) (−1)z+a+c d=b+x Given the parity conditions on the parameters, the sign of these terms reduces to 1 in all cases. Hence the weight sum is positive and Q1 is non-zero. Maximality: Because c ≥ d and b + x ≥ d, the weight 2z + x + a + b + c of 4 in row two is the largest possible weight given the Maximality of row three by 84 Lemma 7.1.2. Hence this tableau is maximal by Lemma 7.1.5. Tableau Q2 z+x z z a 3 323 Q2 = 4 244 1 112 ω2,3 = 0<a≤z z + a even z = d−x 3 d ≡ x (mod 3) x ∈ {0, 2, 4} ¡ 0 z 0 d−z+a ¢ d a 0 0 λ = [2d − 2a, d + a, d + a] r = d − 3a, s = 0, t = d + a For a < z any valid weight assignment must ω3 (1) = d. The tableau is symmetric in 3 and 4, as well as 2 if x = 0. This gives the following signed weight table: Assignment Sign Condition (1, 2, 3, 4) 1 (1, 3, 2, 4) (−1)z+a (1, 2, 4, 3) (−1)3z+x+a (1, 4, 2, 3) (−1)2z+x (1, 3, 4, 2) (−1)2z x=0 (1, 4, 3, 2) (−1)z+a x=0 Since x is even, all the terms are positive and Q2 is non-zero. If z = a the tableau is symmetric in 1, 3, and 4, as well as 2 if x = 0. We get all the weight assignments listed above, in addition to those obtained by interchanging rows or allowing ω2,3 (1) = ( zz ). Interchanging rows has sign (−1)d+z = 1 since x is even. The other possibility changes the sign by (−1)z+a = 1. Hence all the terms are positive and Q2 is non-zero. Maximality: This tableau is maximal as shown in Example 7.1.1. 85 Tableau Q3 0<a≤d−b abb 1 4 4 Q3 = 2 3 2 3 2b ≤ d a≥b a even b 0 ω2,3 = ( 00 a+b 0 a 0) λ = [4d − 2a − 2b, a + 2b, a] r = 4d − 3a − 4b, s = 2b, t = a Since a ≥ b only the elements 2 and 3 may have ω2 = a + b. Hence the only valid weight assignments are (1, 2, 3, 4) with sign 1 and (1, 3, 2, 4) with sign (−1)a = 1. Thus all the valid weight assignments are positive and therefore Q3 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau Q4 2a + b + c = d aabccd 3 4 1 3 1 3 Q4 = 2 2 2 2 4 4 1 1 a+c+d≤d a, b, c, d > 0 b + d even a even, if a + c + d = d 0 d 0 c+d ω2,3 = ( 2a 0 0 0 ) λ = [3d − 2a − c − d, d + c + d, 2a] r = 2d − 2a − 2c − 2d, s = b + 2c + d, t = 2a Any weight assignment must have either ω3 (1) = 2a or ω3 (2) = 2a. Unless a + c + d = d, only the elements 1 and 2 may have ω2 = d. This gives the following weight table: 86 Assignment Sign Condition (1, 2, 3, 4) 1 (2, 1, 4, 3) (−1)b+d (2, 3, 4, 1) (−1)a+b+d a+c+d=d (1, 4, 3, 2) (−1)a a+c+d=d By our parity constraints, the weight sum is always positive. Hence Q4 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau Q5 a+b+c+d=d abcdef 4 3 3 4 31 Q5 = 2 2 2 2 44 1 1 a, b, e, f > 0 a+b+f≤d b+c+e<d a+d+e+f<d ω2,3 = ¡ 0 d 0 e+f a+b 0 0 0 If a + b + f = d then: either c, d > 0 or d = 0 and d + c + e even or c = 0 and d + d even ¢ λ = [3d − a − b − e − f, d + e + f, a + b] r = 2d − a − b − 2e − 2f, s = c + d + e + f, t = a + b Unless a + b + f = d, we must have ω2 (2) = d. Then since e, f > 0 we have ω2 (4) = e + f. As a, b > 0, we have ω3 (1) = a + b. Therefore no other valid weight assignments exist. If a + b + f = d, then it is possible to have ω2 (1) = d. However, unless c or d equals zero, there is no element with ω2 = e + f. Hence no such weight assignment can exist. If d = 0 we can have the weight assignment (2, 1, 4, 3) which has sign (−1)d+c+e = 1. If c = 0 then we can have the weight assignment (2, 1, 3, 4) which has sign (−1)d+d = 1. Thus, in either case, the weight sum is positive and Q5 is non-zero. Maximality: Rows two and three are maximal by Lemma 7.1.2. As no other fillings may have this row two weight and common elements between rows two and three, the tableau is maximal. 87 Tableau Q6 a>0 abc 1 4 114 Q6 = 2 2 323 3 a+b≤d−1 b≥c a even, if b = c b, c not both 0 c+1 0 ω2,3 = ( 00 a+b+1 0 a 0) λ = [4d − 2a − b − c − 2, a + b + c + 2, a] r = 4d − 3a − 2b − 2c − 4, s = b + c + 2, t = a Unless b = c, we must have ω2 (2) = a + b + 1. This force ω2,3 (3) = ( c+1 a ) and hence there are no other valid weight assignments. When c=b then the elements 1, 2, or 3 may have ω2 = a+b+1. If ω2 (2) = a+b+1 then only the element 3 may have ω2,3 = ( c+1 a ). If ω2 (1) = a + b + 1, then a valid weight assignment exists only for b = 0. In this case ω2 (3) = c+1. If ω2 (3) = a+b+1 then we may have ω2 (2) = c + 1 or, if c = 0, ω2 (1) = c + 1. Since these conditions are subject to b = c and b and c are never simultaneously zero, the only valid weight assignments are (1, 2, 3, 4) with sign 1 and (1, 3, 2, 4) with sign (−1)a = 1. Hence the weight sum is positive and Q6 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau Q7 a+b=d−2 ab c 2 3 44414 Q7 = 2 23332 1 11 ω2,3 = ¡ 0 d c+2 0 d−1 0 0 0 a+c≤d−2 b+c≤d−3 a, b > 0 ¢ λ = [2d − c − 1, d + c + 2, d − 1] 88 r = d − 2c − 3, s = c + 3, t = d − 1 Any valid weight assignment must have ω3 (1) = d − 1. Therefore ω2 (1) = 0, so we must have ω2 (3) > 0. Unless b = c only the element 2 may have ω2 = d, so the only valid weight assignment is (1, 2, 3, 4) with sign 1. When b = c, we may also have (1, 3, 2, 4) with sign (−1)a and (1, 3, 4, 2) with sign (−1)a+b+2 . Since this sum is odd, we must have Q7 non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. 89 7.5 Tableaux for c = 5 Tableau R1 xzzzab 52455 5 R1 = 33344 3 11112 2 ω2,3 = ¡0 0 d a+b 0<a≤b≤z d + a + z even, if b = z d even, if a = b, x = 0 ¢ 2z+x+b z+a 0 0 0 z = d−x 3 d ≡ x (mod 3) 0 λ = [3d − 2a − 2b, d + a + b, d + a + b] r = 2d − 3a − 3b, s = 0, t = d + a + b First consider the valid weight assignments with a < z. If b < z then we must have ω3 (1) = d; if b = z we may have ω3 (3) = d as well. If we have ω3 (1) = d, then ω3 (2 or 5) = a + b because a, b > 0. Since b ≥ a only the elements 1 and 3 may have ω2 = 2z + x + b unless a = b and x = 0. In that case, the element 4 may also have this weight. If ω2 (3) = 2z + x + b, then ω2 (5 or 4) = z + a. If ω2 (4) = 2z + b, then ω2 (2 or 3) = z + a. These constraints give the following table of weight assignments: Assignment Sign Condition (1, 2, 3, 4, 5) 1 (1, 5, 3, 4, 2) (−1)a+b (1, 2, 3, 5, 4) (−1)z+a (1, 2, 4, 3, 5) (−1)z a = b, x = 0 (1, 5, 4, 3, 2) (−1)z+b+a a = b, x = 0 (1, 5, 4, 2, 3) (−1)2z+a a = b, x = 0 (3, 4, 1, 2, 5) (−1)d+z+a b=z (3, 5, 1, 2, 4) (−1)d b=z (3, 4, 1, 5, 2) (−1)d b=z For b < z and a 6= b + x, we get a weight sum of 1 + (−1)a+b + (−1)z+a which is non-zero. When a = b, x = 0, we get a weight sum of 1+1+(−1)z+a +(−1)z +(−1)z + 90 (−1)a . This is non-zero by the parity constraints which imply z even. Similarly, if a < b = z, then the weight sum is 1 + (−1)a+z + (−1)z+a + (−1)d+z+a + (−1)d + (−1)d , which is also non-zero by the parity constraints. When a = b = z the tableau has many symmetries. Since the weight of rows two and three are the same, interchanging rows yields a new weight assignment with a difference in sign of (−1)d+a+b = 1. Hence we will not count those weight assignments which are inversions of rows two and three. First consider the possible element pairs (x, y) that can have ω3 (x, y) = (d, 2z). These are (1, 2), (1, 5), (3, 4), (3, 5), (5, 3), and (5, 1), along with (2, 1), (2, 4), (4, 3), (4, 2) when x = 0. Two of these pairs make up a weight assignment (not counting row inversion). If x 6= 0 the possible pair assignments are: (1, 2)(3, 4); (1, 2)(3, 5); (1, 2)(5, 3); (1, 5)(3, 4); and (3, 4)(5, 1). Since there is an odd number, this weight sum in non-zero. If x = 0, the condition d even, implies z is even. Hence all the column blocks are even. Since any valid weight assignment moves full column blocks, all weight assignments are positive and thus R1 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau R2 0≤a≤z−c x+z z z a b c c 2 445 4 4 2 R2 = 3 533 5 3 5 1 112 2 ω2,3 = ¡0 0 d b+a 0≤b≤z−c 0<c b≤a ¢ 2z+a+c+x 0 z+b+c 0 0 z = d−x 3 d≡x (mod 3) d even, if b = a + x 0 λ = [3d − 2a − 2b − 2c, d + a + b + 2c, d + a + b] r = 2d − 3a − 3b − 4c, s = 2c, t = d + a + b A valid weight assignment must have ω3 (1) = d since c > 0. Then we must have ω3 (2 or 5) = a + b, or if b = 0, then ω2 (3) = a is also possible. There are not enough 2’s to have ω2 (2) = 2z + a + c + x. If we have ω2 (3) = 2z + a + c + x, then 91 ω2 (5) = z + b + c. If we have ω2 (4) = 2z + a + c + x, then ω2 (2) = z + b + c, and b = a, x = 0. There are not enough 5’s to have ω2 (5) = 2z + a + c + x. Hence there are two possible weight assignments: (1, 2, 3, 4, 5) with sign 1 and (1, 5, 4, 3, 2) with sign (−1)d , which occurs only when b = a, x = 0. By our parity constraint, this sum is positive. If b = 0, we also have the weight assignment (1, 3, 4, 5, 2) when a = b, x = 0. The weight sum is odd and hence the tableau is non-zero. Maximality: The tableau is maximal by Lemma 7.1.5. To see this, note that when row three has the maximum weight d, the largest possible weights for row two are: 0 for the element 1, z + x + a + b + c for the element 2, 2z + x + a + c for the element 3, 2z + b + c for the element 4, z + a + b + c for the element 5. Since z − c ≥ a ≥ b, the weight 2z + x + a + c is the largest. Given this, Lemma 7.1.4 shows that row two is maximal. Moreover rows two and three cannot have elements in common, given the columns remaining after the largest weight elements have been assigned. Hence the tableau is maximal. Tableau R3 z+x z z b a 5 254 5252 R3 = 3 343 4344 1 112 2 ω2,3 = ¡0 0 2z+x+b+1 z+a+2 0 d a+b 0 0 0 0≤a<b≤z−1 z = d−x 3 d≡x (mod 3) ¢ λ = [3d − 2a − 2b − 3, d + a + b + 3, d + a + b] r = 2d − 3a − 3b − 6, s = 3, t = d + a + b, We must have ω3 (1) = d. Since a < b, only the element 3 may have ω2 = 2z + x + b + 1. Then ω2 (4) = z + a + 2. Hence there are no other valid weight assignments. Thus R3 is non-zero. Maximality: Row two is maximal by Lemma 7.1.2 and row three is maximal by Lemma 7.1.4. Inspection shows that it is not possible to have the non-d weights 92 assigned to the same element. Hence the tableau is maximal by Lemma 7.1.5. Tableau R4 d + a + c even, if a = b x z − 1 z+1 z z − a z − a a − 1 b a+1 c 2 2 5 5 4 5 5 2 2 5 R4 = 3 4 4 3 3 3 3 4 3 4 1 1 1 1 2 2 1≤a≤z a + b even, if c = x b≤a z = d−x 3 c≤x d≡x (mod 3) z≥2 ω2,3 = ¡0 0 d 2z+b+c 0 d 2z−2a 0 0 0 ¢ λ = [3d − 4z − b − c + 2a, d + 2z + b + c, d + 2z − 2a] r = 2d − 6z + 2a − 2b − 2c, s = 2a + b + c, t = d + 2z − 2a, By construction, only the element 1 may have ω3 (1) = d. This means any valid weight assignment must have ω3 (2 or 3) = 2z − 2a (unless a = z in which case the weight assignments are equivalent to those using 2 or 3). Furthermore, the elements appearing in row two must either be both 3 and 4 or both 2 and 5. Now we apply weight-set counting. Assignment Sign Condition (1, 2, 3, 4, 5) 1 (1, 2, 4, 3, 5) (−1)z−a b = a, c = x (1, 3, 2, 5, 4) (−1)d+b+c b=a (1, 3, 5, 2, 4) (−1)x+a+b+c c=x By our parity constraints, all these terms are positive. Hence R4 is non-zero. Maximality: Rows two and three are maximal by Lemma 7.1.2. Inspection shows that it is not possible to have the non-d weights assigned to the same element. Hence the tableau is maximal. 93 Tableau R5 y+w−a y−a a a b y y w 4 5 1 4 5 5 4 5 R5 = 3 2 3 2 1 3 2 2 1 1 ω2,3 = ¡ b d d 0 0 d−2a 0 0 0 0 ∗ a even, if a = b y = d2 0≤b≤a<y 2y + w = d ¢ λ = [2d − b + 2a, 2d + b, d − 2a] r = 2a − 2b, s = d + 2a + b, t = d − 2a Since y − a > 0, any weight assignment must have ω3 (1) = d − 2a. If ω2 (3) = d, then ω2 (2) = d and vice versa. Similarly for the elements 4 and 5, however we may only have ω2 (5) = d if a = b. So unless a = b there is only one valid weight assignment. When a = b, we also have (1, 4, 5, 3, 2) which has sign (−1)2d+a . Since a is even in this case, the sum is non-zero. Therefore R5 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau R6 x z z z z − a z − a − 1 a b a+1 c 2255 4 5 5 2 2 5254 R6 = 3443 3 3 3 4 3 4 1111 2 2 ω2,3 = ¡0 0 d 2z+b+c 0 d 2z−2a−1 0 0 0 1≤a≤z−2 z = d−x 3 0≤b<a d≡x c≤x z≥3 (mod 3) ¢ λ = [3d − 4z − b + 2a + 1, d + 2z + b + x, d + 2z − 2a − 1] r = 2d − 6z + 2a − 2b − 2c + 1, s = 2a + b + c + 1, t = d + 2z − 2a − 1 Only the element 1 can have ω3 =d and then any weight assignment must have ω2 (3) = d. This forces ω2 (4) = 2z+a + 1 + x. Hence ω3 (4)=0 which implies ω3 (2)> 0. Thus there are no other valid weight-assignments and R6 is non-zero. Maximality: Rows two and three are maximal by Lemma 7.1.2. Inspection 94 shows that it is non possible to have the non-d weights assigned to the same element. Hence the tableau is maximal. Tableau R7 1≤a<y y − a y − a a+1 a a − 1 y y+1 4 5 1 4 5 4 5 145 R7 = 3 2 3 2 1 3 2 1 1 ω2,3 = ¡ a−1 d d 0 0 d−1−2a 0 0 0 0 y = d−1 2 d odd d≥5 ¢ λ = [2d + a + 2, 2d + a − 1, d − 2a − 1] r = 3, s = d + 3a, t = d − 2a − 1 As y − a > 0, any weight assignment must have ω3 (1) = d − 1 − 2a . Moreover, only the elements 2 and 3 can have ω2 = d. Hence there are no other valid weight assignments, so R7 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau R8 y − a − 1 y − a a+1 a a − 1 y y 4 5 1 4 5 4 5145 R8 = 3 2 3 2 1 3 2 1 1 ω2,3 = ¡ a−1 d d 0 0 d−1−2a 0 0 0 0 1≤a≤y−2 y = d2 d even d≥6 ¢ λ = [2d + a + 2, 2d + a − 1, d − 2a − 1] r = 3, s = d + 3a, t = d − 2a − 1 Since a < y − 1, any weight assignment must have ω3 (1)=d − 1 − 2a. Moreover, 95 the only elements 2 and 3 can both have ω2 = d. Therefore there are no other valid weight assignments, so R8 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau R9 0≤a<z xzzzzabcde 344324 3 4 4 4 R9 = 225552 2 2 3 5 111133 b + c, b + d < z − a z = d−x 3 c+d+e<z−a+x d ≡ x (mod 3) 0≤e≤x x 6= 0 0≤d<x+b ω2,3 = ¡ 0 z+a+b+c+x d 0 d 0 3z+e z+a 0 0 ¢ λ = [3d − 2a − b − c − d − e − 2z, d + z + a + b + c + d + e, d + z + a] r = 2d − 3a − 2b − 2c − 2d − 2e − 3z, s = b + c + d + e, t = d + z + a Any valid weight assignment must have ω3 (1) = d. Only the elements 1 and 5 may have ω2 = 3z + e. If ω2 (5) = 3z + e, then only the element 2 may have ω2 = x + z + a + b + c due to the condition d < b + x. Examining the tableau in light of these constraints shows that (1, 2, 3, 4, 5) is the only valid weight assignment. Thus the tableau is non-zero. Maximality: As the discussion above shows, the tableau weights are as large as possible. Hence by Lemma 7.1.5, this tableau is maximal. 96 Tableau R10 1≤a≤z b ≤ b a2 c x z z z z−a b c d e f x 3443 3 2 4 3 434 R10 = 2255 5 5 3 2 255 1111 2 c, e ≤ b 2z c d ≤ d 2z e + b a2 c f ≤ d a2 e c even ω2,3 = d>c z+b>c+d z−a+b+f>c+e z+f>d+e z = d−x 3 d ≡ x (mod 3) z≥2 ¡ 0 x+z+d+e c 0 3z−a+b+f+x ¢ d z−a 0 0 0 λ = [3d − 2z + 2a − b − c − d − e − f − x, d + z − a + b + c + d + e + f + x, d + z − a] r = d + 3a − 2b − 2c − 2d − 2e − 2f − x, s = b + c + d + e + f + x, t = d + z − a Any valid weight assignment must have ω3 (1) = d. Given the bounding parameters, only the element 5 can have ω2 = 2z + a + b + f + x. When ω2 (5) is maximal, only the element 2 may have ω2 = x + z + d + e. From this we find that the valid weight assignments are (1, 2, 3, 4, 5) with sign 1, and (1, 2, 4, 3, 5) with sign (−1)c = 1. Maximality: The discussion above shows that the row two weights are maximized. Hence by Lemma 7.1.5, this tableau is maximal. Tableau R11 0 < a, b abcdef 3 5 3 5 15315 R11 = 4 2 4 2 21424 1 1 ω2,3 = ¡ f b+d+e+1 a+c+2 0 0 a+b 0 0 0 0 c≤d−a−2 d≤d−b−e−1 0<f≤e ¢ e+f≤d−a−b−1 b+d>f a+c≥e+f 97 λ = [5d − 2a − 2b − c − d − e − f − 3, a + b + c + d + e + f + 3, a + b] r = 5d − 3a − 3b − 2c − 2d − 2e − 2f − 6, s = c + d + e + f + 3, t = a + b Any valid weight assignment must have ω3 (1) = a + b and hence ω2 (1) = f. This implies ω2 (2) > 0. But to have exactly three non-zero weights in row two, we then must have ω2 (4) > 0. Hence no other weight assignments are possible and hence the tableau is non-zero. Maximality: The possible row two weights, given the maximization of row three, are e + f + 1 for the element 1, b + d + e + 1 for the element 2, a + c + 1 for the element 3, a + c + 2 for the element 4, and b + d + f + 1 for the element 5. Given the conditions on these parameters we see that the weights for 2 and 4 are the largest. Hence by Lemma 7.1.5, this tableau is maximal. Tableau R12 y−1+w y y y a w 1 1 5 4 1 5 5 R12 = 3 3 3 2 2 4 2 4 ∗ 0<a≤d−y−1 y = d2 y even, if y = a 2y + w = d ω2,3 = ( 00 d0 d0 a1 00 ) λ = [3d − a − 1, 2d + a, 1] r = d − 2a − 1, s = 2d + a − 1, t = 1 Any valid weight assignment must have two of the elements 1, 3, and 4 having non-zero weight in row two. Hence at least one of them must have ω2 = d. Moreover, we can not have a row two weight of d for both of these elements. If ω2 (1) = d we may not have ω2 (2) = d, so the only option is ω2 (5) = d when a = y. This corresponds to a weight assignment of (3, 1, 5, 4, 2) with sign (−1)2d+a which equals 1 due to the parity constraint. If ω2 (3) = d we may not have ω2 (5) = d, so the only option is ω2 (2) = d. This corresponds to a weight assignment of (1, 2, 3, 4, 5) with 98 sign 1. When ω2 (4) = d, neither the element 2 nor the element 5 may have ω2 = d, so there is no weight assignment with this option. Hence the weight sum is always positive and thus R12 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau R13 abcde 1 3 1 3 1 R13 = 2 4 4 2 2 5 5 ω2,3 = ¡ 0 a+d+e 0 b+c 0 0 0 0 a+d+e≤d d≥c a+b≤d a+e≥b b+d≤d If c = d then a + b + c + d + e even a, b, c, d, e > 0 ¢ 0 If b = a + e then a + b + c + d + e even a+b λ = [5d − 2a − 2b − c − d − e, a + b + c + d + e, a + b] r = 5d − 3a − 3b − 2c − 2d − 2e, , s = c + d + e, t = a + b Since a, b > 0 any valid weight assignment must have ω3 (5) = a + b. We must have ω2 (1 or 2) > 0. If ω2 (2) > 0, then ω2 (4) > 0; if ω2 (1) > 0, then ω2 (3) > 0. This shows the only valid weight assignments are (1, 2, 3, 4, 5) and (2, 1, 4, 3, 5). The second one only occurs when d = c or b = a + e. If that happens, the sign is (−1)a+b+c+d+e = 1. Hence R13 is non-zero. Maximality: The possible maximal weights for row two are: a + c + e for the element 1; a + d + e for the element 2; b + d for the element 3; b + c for the element 4; and 0 for the element 5. By our parameter conditions, a + d + e is the largest. Hence by Lemma 7.1.5 the tableau is maximal. Tableau R14 d−3 2 d−4 1 4 1 5 151 R14 = 2 2 2 3 344 3 d≥5 99 2 0 ω2,3 = ( 00 d0 d−3 1 0 0) λ = [3d, 2d − 1, 1] r = d + 1, s = 2d − 2, t = 1 For d > 5, any valid weight assignment must have ω2 (2) = d. This forces ω2 (3) = d − 3 and ω2 (4) = 2. Hence there are no other valid weight assignments. When d = 5, we may have ω2 (1) = d or ω2 (1) = d−3 as well. Then the possible weight assignments are (1, 2, 3, 4, 5), (3, 2, 1, 5, 4, ), and (3, 1, 2, 5, 4). Since there is an odd number, this sum is non-zero and thus so is R14 . Maximality: This tableau is maximal by Lemma 7.1.5. Tableau R15 z+x z z 2 55222 R15 = 3 43433 1 11 z = d−x 3 d ≡ x (mod 3) z≥2 z+1 0 ω2,3 = ( d0 00 2z+x+2 0 0 0) λ = [3d − 3, d + 3, d] r = 2d − 6, s = 3, t = d Since only the element 1 may have ω3 = d and only the element 3 may have ω2 = 2z + x + 2, there are no other valid weight assignments. Hence the tableau is non-zero. Maximality: Given the row three weight of d, the row two weights are as large as possible. Thus by Lemma 7.1.5 this tableau is maximal. 100 Tableau R16 d−2 a 1 55531 R16 = 2 44242 3 3 ω2,3 = ¡0 d 0 a+2 0 0 0 d−1 0 0 1≤a≤d−4 d≥5 ¢ λ = [3d − a − 1, d + a + 2, d − 1] r = 2d − 2a − 3, s = a + 3, t = d − 1 Any valid weight assignment must have ω3 (3) = d − 1 and ω2 (2) = d. This forces ω2 (4) = a+2. Hence there are no other valid weight assignments and R16 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau R17 x z z z z−1 a z b 22552 5 225254 R17 = 34433 3 434 11114 ω2,3 = 0≤a≤z−2 b≤x z = d−x 3 d ≡ x (mod 3) z≥2 ¡ 0 0 d 2z+a+b 0 ¢ d 0 0 1 0 λ = [3d − 2z − a − 1, d + 2z + a + x, d + 1] r = 2d − 4z − 2a − 2b − 1, s = 2z + a + b − 1, t = d + 1 Only the element 1 can have ω3 =d and then any weight assignment must have ω2 (3) = d. This forces ω2 (4) = 2z+a + 1 + x. Thus there are no other valid weight assignments and R17 is non-zero. Maximality: The tableau is maximal by Lemma 7.1.3. 101 Tableau R18 xzzzz 52442 R18 = 33353 11115 x even ω2,3 = ( d0 00 d0 00 zz ) λ = [3d − 2z, d + z, d + z] r = 2d − 3z, s = 0, t = d + z Only the elements 1 and 3 may have a weight of d. Hence the only valid weight assignments are (1, 2, 3, 4, 5) and (3, 2, 1, 4, 5), both of which have sign 1 since x is even. Hence R18 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau R19 d−2 d−3 1 553 5 1 R19 = 2 422 4 4 3 3 ω2,3 = ¡0 d 0 d−1 0 0 d−1 0 ¢ λ = [2d + 2, 2d − 1, d − 1] r = 3, s = d, t = d − 1 Any valid weight assignment must have ω3 (3) = d. Then ω2 (2) = d. Hence there are no other weight assignments possible. Thus R19 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. 102 7.6 Tableaux for c = 6 Tableau S1 x+z x+z z z a a 5 6 565 6 S1 = 4 3 343 4 1 2 112 2 ω2,3 = ¡0 0≤a≤z z = d−x 3 z + x even, if a = 0 d≡x 0 2z+a+x 2z+a+x 0 0 d z+2a+x 0 0 0 0 (mod 3) ¢ λ = [4d − 2z − 2x − 4a, d + z + 2a + x, d + z + 2a + x] r = 2d − 2x − 6a, s = 0, t = d + 2a+z + x The construction of S1 means that any valid weight assignment with a > 0 must have ωi (j) > 0 if and only if ωi (k) > 0 for the pairs (j, k) = (1, 2), (3, 4), or (5, 6). These constraints show that the only valid weight assignments are those that interchange complete rows in the body of S1 . Since the length of these rows t = 4z+2x+2a is even, all valid weight assignments are positive and hence S1 is non-zero. When a = 0 we must have ω3 (1) = d and ω2 (3, 4) > 0 or ω2 (5, 6) > 0. Hence the valid weight assignments are: (1, 2, 3, 4, 5, 6) with sign 1; (1, 6, 3, 4, 5, 2) with sign (−1)z+x ; (1, 2, 5, 6, 3, 4) with sign (−1)4z+2x ; and (1, 3, 5, 6, 2, 4) with sign (−1)5z+3x . Since z + x is even, these assignments are all positive. Hence the tableau is non-zero. Maximality: By Lemma 7.1.5 this tableau is maximal. Tableau S2 d−1 a 4 222423 S2 = 5 151111 6 3363 ω2,3 = ¡ a+4 0 0 0 0 d 0 0 a+2 0 0 d 0≤a≤d−4 a ≡ d (mod 2) d≥4 ¢ λ = [4d − 2a − 6, d + a + 4, d + a + 2] r = 3d − 3a − 10, s = 2, t = d + a + 2 103 Any valid weight assignment must have ω2 (1) = a + 4 since if ω2 (1) = 0 then we must have ω2 (2, 3) = (a + 4, d) which is not possible. If ω2 (1) = a + 4, then ω2 (5) = d. Thus the only valid weight assignments are (1, 2, 3, 4, 5, 6) with sign 1 and (1, 3, 2, 6, 5, 4) with sign (−1)d+a+2 . Since a ≡ d (mod 2) this sum is positive. Hence S2 is non-zero. Maximality: Rows two and three are maximal by Lemma 7.1.2. Since it is not possible for the non-d element of row three to have a row two weight of a + 2, the tableau is maximal. Tableau S3 d−4 a 2 2 5 11355511 S3 = 2 42442424 6 36633 ω2,3 = ¡0 d 0 a+5 0 0 0 0 a+2 0 0 d 0≤a≤d−5 d≥5 ¢ λ = [4d − 2a − 7, d + a + 5, d + a + 2] r = 3d − 3a − 12, s = 3, t = d + a + 2 Any valid weight assignment must have ω3 (6) = d and ω3 (3) = a + 2 (or ω3 (5) = 2 if a = 0). Then only the element 2 may have ω2 = d (or 4 if a = d − 5). Moreover if ω2 (2) = d, then ω2 (4) = a + 5. Hence the only valid weight assignments are (1, 2, 3, 4, 5, 6) with sign 1 and, when a = 0, (1, 2, 5, 4, 3, 6) with sign (−1)2 . Therefore the weight sum is positive and S3 is non-zero. Maximality: Rows two and three are maximal by Lemma 7.1.2. Since the nond element of row two has weight a + 5 and there are only two copies of the element 3 available for row two, the tableau is maximal. 104 Tableau S4 d−3 a 2 5 153511111 S4 = 2 424422344 6 36633 ω2,3 = ¡0 d 1 a+5 0 0 0 0 a+2 0 0 d 0≤a≤d−5 d≥5 ¢ λ = [4d − 2a − 8, d + a + 6, d + a + 2] r = 3d − 3a − 14, s = 4, t = d + a + 2 Any valid weight assignment must have ω3 (6) = d. Now ω3 (6) = d implies ω3 (3) = a + 2 and ω2 (3) = 1. Unless a = d − 5, we must have ω2 (2) = d and so ω2 (4) = a + 5. When a = d − 5 we may have ω2 (1) = d or ω2 (4) = d. However, if ω2 (1) = d, then there is no element with ω2 = a + 5. If ω2 (4) = d, then ω2 (2) = a + 5. This shows the only weight assignment is (1, 2, 3, 4, 5, 6). Hence S4 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5 and the discussion above. Tableau S5 0 < a, b, c ≤ d − e x + z0 x + z0 z0 z0 z0 z0 a b c 5 6 5 6 5 6 6 5 1 S5 = 4 3 3 4 3 4 2 4 3 1 2 1 1 2 2 a even z0 = e−x 3 c even if a = b e≡x b even if a = c 3≤e<d b, c ≥ a e+b 0 0 ω2,3 = ( 0e ae e+c 0 0 0 0) λ = [6d − 4e − a − b − c, 2e + a + b + c, 2e] r = 6d − 6e − 2a − 2b − 2c, s = a + b + c, t = 2e (mod 3) 105 The construction of S5 forces the following constraints on any valid weight assignment. • ω3 (1) > 0 ⇐⇒ ω3 (2) > 0. • ω3 (3) > 0 ⇐⇒ ω3 (4) > 0. • ω3 (5) > 0 ⇐⇒ ω3 (6) > 0. • If ω2 (j) = a, then ω3 (j) > 0. • If ω2 (1) = a, then ω2 (5, 6) > 0 and a = c. • If ω2 (2) = a, then ω2 (3, 4) > 0. • If ω2 (3) = a, then ω2 (5, 6) > 0 and a = c. • If ω2 (4) = a, then ω2 (1, 2) > 0 and a = b. • If ω2 (5) = a, then ω2 (1, 2) > 0 and a = b. • If ω2 (6) = a, then ω2 (3, 4) > 0. From this we can derive a signed weight table. Assignment Sign Condition (1, 2, 3, 4, 5, 6) 1 (2, 1, 6, 5, 4, 3) (−1)2e+a+b+c a=c (3, 4, 1, 2, 5, 6) (−1)2e+c a=b (4, 3, 6, 5, 1, 2) (−1)4e+a+b a=c (5, 6, 3, 4, 1, 2) (−1)2e+a (6, 5, 1, 2, 3, 4) (−1)4e+b+c a=b Computing the weight sum we obtain 3 + 3(−1)a = 6 when a = b = c as a is always even. For a = b 6= c, the sum is 2 + 2(−1)c . This is non-zero as c is even when a = b. For a = c 6= b, we have 2 + 2(−1)b . This is non-zero as b even when a = c. Finally if a 6= b, c the sum is 1 + (−1)a which is non-zero. Hence S5 is non-zero. 106 Maximality: By Lemma 7.1.4, row three is maximal. Since b, c ≥ a, row two is maximal by Lemma 7.1.4. As the number of row three elements available for row two is bounded by a, this is the largest weight and hence the tableau is maximal. Tableau S6 x + z0 x + z0 z0 z0 z0 z0 − 1 a b c 3 4 4 3 3 4 2 3 4 S6 = 5 6 5 6 5 6 6 5 1 1 2 1 1 2 2 ω2,3 = ¡c 0 e+b e+a−1 0 0 e e−1 0 0 0 0 0 < a, b, c ≤ d − e z0 = e−x 3 a, b ≥ c e≡x b≥a−1 3≤e≤d (mod 3) ¢ λ = [6d − 4e − a − b − c + 2, 2e + a + b + c − 1, 2e − 1] r = 6d − 6e − 2a − 2b − 2c + 3, s = a + b + c, t = 2e − 1 For any valid weight assignment we can have ω3 (j) = e only for j ∈ {1, 3, 5}. Moreover, ω3 (j) = e if and only if ω3 (j + 1) = e − 1. If ω3 (j) = e then ω2 (j) = c. If ω2 (1) = c, then ω2 (5, 6) > 0. If ω2 (3) = c, then ω2 (1, 2) > 0 and b = c. If ω2 (5) = c, then ω2 (1, 2) > 0 and b = c. This means (1, 2, 3, 4, 5, 6) is the only valid weight assignment when b 6= c. If b = c, then we additionally have weight assignments (3, 4, 1, 2, 5, 6) and (5, 6, 1, 2, 3, 4). In either case the weight sum is odd and hence non-zero. Thus S6 is non-zero. Maximality: By Lemma 7.1.4 row three is maximal. Since a, b ≥ c, row two is maximal by Lemma 7.1.4. As the number of row three elements available for row two is bounded by c, this is the largest weight and the tableau is maximal. Tableau S7 abcd e 1 4 3 11135 S7 = 2 2 2 42246 3 3 ω2,3 = ¡0 d 0 0 0 a+b ¢ d+1 0 e 0 0 0 a+b+c=d−2 a, b, d > 0 b+d≤d−2 0≤e≤d a+d<d−2 e even 107 λ = [5d − a − b − d − e − 1, d + d + e + 1, a + b] r = 4d − a − b − 2d − 2e − 2, s = c + d + e + 3, t = a + b Any valid weight assignment must have ω2 (5 or 6) = e and ω3 (3) = a + b. Since a + d < d − 2, we must have ω2 (2) = d. Then, as d > 0 we must have ω2 (4) = d + 1. Hence the only valid weight assignments are (1, 2, 3, 4, 5, 6) with sign 1 and (1, 2, 3, 4, 6, 5) with sign (−1)e . Since is e is even, this weight sum is positive. Hence S7 is non-zero. Maximality: Inspection shows that rows two and three are maximal. Since d > 0 we cannot have any 3’s in a maximal row two. Thus S7 is maximal by Lemma 7.1.5. Tableau S8 d−4 d−2 d−4 2 66 6 3 2 1 23662 S8 = 34 3 4 4 5 55151 11 d≥5 d d 0 ω2,3 = ( 22 00 d−3 0 0 0 0) λ = [3d − 1, 3d − 1, 2] r = 0, s = 3d − 3, t = 2 Examining the tableau shows that we can only have ω2 (j, k) = (d, d) for (j, k) = (2, 3), (2, 6) or (4, 5). Also, we must have ω2,3 (1 or 6) > 0, so ω2 (2, 6) = (d, d) and ω2 (1) = 2. This is possible only when d = 6. Hence the valid weight assignments are: (1, 2, 3, 4, 5, 6) with sign 1; (6, 4, 1, 2, 3, 5) with sign (−1)2d+4 ; (1, 4, 3, 2, 6, 5) with sign (−1)3d−1 when d = 6; and (6, 2, 1, 4, 5, 3) with sign (−1)3 when d = 5. In all cases, the weight sum is positive and hence S8 is non-zero. d d 0 Maximality: This tableau is not maximal since ω2,3 = ( 20 00 d−1 0 0 0 0 ). However, this weight is zero in qT . This tableau cannot be put in maximal form, hence we will need to prove directly that it is disjoint from the requisite tableaux. This will be done in Section 8.7. 108 Tableau S9 a+c+d=d−1 b+f=d−1 abcdefg 1 5 1 5 36 6 36 S9 = 2 4 2 2 54 5 42 3 3 f+g≤d−1 a+b+e≤d−1 b+d+e+g≤d−1 a, b > 0 e, g > 0 or e = 0 and g even ω2,3 = ¡0 d 0 d e+g 0 0 0 a+b 0 0 0 ¢ λ = [4d − a − b − e − g, 2d + e + g, a + b] r = 2d − a − b − 2e − 2g, s = c + d + e + f + g + 2, t = a + b Any valid weight assignment must have ω2 (3) = a + b because a, b > 0. Only the elements 2 and 4 may simultaneously have ω2 = d. Then ω2 (5) = e + g unless e or g is 0. If e = 0 we may also have ω2 (6) = g. Hence the weight assignments are are (1, 2, 3, 4, 5, 6) with sign 1 and, if e = 0, (1, 2, 3, 4, 6, 5) with sign (−1)g = 1. Thus S9 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5 since g > 0. Tableau S10 2 a 155165 S10 = 242243 33 2≤a≤3 d=4 0 0 ω2,3 = ( 00 04 31 a+1 0 0 0) λ = [15 − a, a + 6, 3] r = 9 − 2a, s = a + 3, t = 3 Any valid weight assignment must have ω3 (3) = 3. Other than 3, the only elements 109 that can have ω2 = d is 2, or 4 if a = 3. Thus there are no other valid weight assignments possible and the tableau is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau S11 d−1 d−3 d−1 1 5 6 1 6 5 S11 = 2 2 3 3 4 4 3 d 0 0 ω2,3 = ( 00 d0 d−2 1 0 0 0) λ = [3d + 1, 3d − 2, 1] r = 3, s = 3d − 3, t = 1 Any valid weight assignment must have two of the elements 1, 2, and 3 with ω2 > 0. Since there are not enough 1’s in the body for this to happen, we must have ω2 (2) = d and ω2 (3) = d − 2. This force ω2 (4) = d. Hence there are no other valid weight assignments. Thus the tableau is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau S12 0≤a≤d−3 ab c 33 11135 S12 = 22 42246 4 0≤b≤d−2 a≥b 0≤c≤d c even 0 b+1 0 c ω2,3 = ( 00 a+3 0 0 1 0 0) λ = [6d − a − b − c − 5, a + b + c + 4, 1] r = 6d − 2a − 2b − 2c − 9, s = a + b + c + 3, t = 1 Any valid weight assignment must have ω2 (5 or 6) = c. We also must have 110 ω2 (2) = a + 3 since a ≥ b. Then ω2 (4) = b + 1 unless b = 0, in which case ω2 (3) = 1 is possible. Hence the weight assignments are (1, 2, 3, 4, 5, 6) with sign 1 and (1, 2, 3, 4, 6, 5) with sign (−1)c . When b = 0 we also have (1, 2, 4, 3, 5, 6) with sign (−1)2 , and (1, 2, 4, 3, 6, 5) with sign (−1)c+2 . Since c is even, this sum is positive. Hence S12 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. 7.7 Tableaux for c = 7 and c = 8 For c = 7 and 8, nearly all the required tableaux can be obtained by joining those tableaux already constructed. This is demonstrated in Chapter 8. However, we do need to construct one additional tableau, which is listed below. Tableau W1 xxzzzzzzzza 78678837873 S1 = 45456456456 1211122233 ω2,3 = a≤x z = d−x 3 d ≡ x (mod 3) a = 0, 2 ¡ 0 0 0 d d 2z+a 0 0 ¢ d d 2z 0 0 0 0 0 0 λ = [4d − 4z − a, 2d + 2z + a, 2d + 2z] r = 2d − 6z − 2a, s = a, t = 2d + 2z Only the triples (1, 2, 3) and (4, 5, 6) may have ωi = (d, d, 2z) or larger. Hence the only valid weight assignments are (1, 2, 3, 4, 5, 6, 7, 8) with sign 1, and (4, 5, 6, 1, 2, 3, 7, 8) with sign (−1)2d+2z+a . Since a is even, the weight sum in positive. Hence W1 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. 197 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. Doran, IV, On Foulkes’ conjecture, J. Pure Appl. Algebra 130 (1998), no. 1, 85–98. MR 99h:20014 [8] H. O. Foulkes, Concomitants of the quintic and sextic up to degree four in the coefficients of the ground form, J. London Math. Soc. 25 (1950), 205–209. MR 12,236e [9] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.3, 2002, (http://www.gap-system.org). 198 [10] David A. Gay, Characters of the Weyl group of SU (n) on zero weight spaces and centralizers of permutation representations, Rocky Mountain J. Math. 6 (1976), no. 3, 449–455. MR 54 #2886 [11] Larry C. Grove, Groups and characters, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1997, A Wiley-Interscience Publication. MR 98e:20012 [12] Roger Howe, (GLn , GLm )-duality and symmetric plethysm, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, 85–109 (1988). MR 90b:22020 [13] N. F. J. Inglis, R. W. Richardson, and J. Saxl, An explicit model for the complex representations of Sn , Arch. Math. (Basel) 54 (1990), no. 3, 258–259. MR 91d:20017 [14] G. James and A. Kerber, Representation theory of the symmetric group, Encyclopedia of Mathematics and its Applications, vol. 16, Addison-Wesley, Reading, MA, 1981. [15] G. D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, vol. 682, Springer, Berlin, 1978. [16] Serge Lang, Algebra, 3 ed., Addison Wesley, Reading Massachusetts, 1999. [17] I. G. Macdonald, Symmetric functions and Hall polynomials, second ed., Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, New York, 1995, With contributions by A. Zelevinsky, Oxford Science Publications. MR 96h:05207 [18] Bruce E. Sagan, The symmetric group, The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1991, Representations, combinatorial algorithms, and symmetric functions. MR 93f:05102 199 [19] Richard P. Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004) 111 Chapter 8 Tableau Sufficiency Our proof of Theorem 2 in Chapter 6 presupposed we had constructed all tableaux with c ≤ 8 for the shapes of Theorem 10. In Chapter 7 we constructed many non-zero tableaux. In this chapter we will demonstrate that all the necessary tableaux have been constructed. Specifically we need all shapes in Theorem 10, that is all partitions [r + s + t, s + t, t] of n, with r, s 6= 1, such that if r or s is in {0, 2, 4} then s + t is even. Recall that those required shapes with r or s less than 5 are called exceptional cases. 8.1 Sufficiency when c = 3 The tableaux we will use for c = 3 are the Pi described in Section 7.3. These tableaux are all maximal and non-zero by weight-set counting. We will show that every necessary partition of n = 3d has a corresponding Pi . First consider the exceptional r cases. These are tableaux having s + t even with r ∈ {0, 2, 4}, or r = 3 with no constraints on s and t. Since our shape is a partition of 3d, we have the condition 3t + 2s + r = 3d. Hence for a given r, we need only to , with s 6= 1 are obtained. verify that all the appropriate s in the range 0 ≤ s ≤ 3d−r 2 This condition shows s ≡ r (mod 3). Table 8.1 below lists the tableaux we are using for these cases. The column ‘Parameters’ indicates the restrictions on the tableaux arising from their construction in Section 7.3. The column ‘s, t values’ indicates their values in terms of the tableau 112 parameters, while the ‘s covered’ lists those cases covered by the given tableau. The restrictions given on the s covered reflect the listed conditions on the parameters. r=0 r=2 r=3 r=4 Tableau Parameters s, t values s covered P4 (d − 2b, b, b) 0 ≤ b ≤ d2 s = 3b 0 ≤ s ≤ 3b d2 c d − b even t = d − 2b s + t even P4 (d − 2b, b, b − 1) 1 ≤ b ≤ d2 s = 3b − 1 2 ≤ s ≤ 3b d2 c − 1 d − b odd t = d − 2b s + t even P4 (d − 2b − 1, b, b, b) 0 ≤ b ≤ d−1 2 s = 3b 0 ≤ s ≤ 3b d−1 2 c d − b odd t = d − 2b − 1 s + t even P3 (d − 2b − 1, b, b − 1) 1 ≤ b ≤ d−1 2 s = 3b 3 ≤ s ≤ 3b d−1 2 c d − b even t = d − 2b − 1 s + t odd P4 (d − 2b, b, b − 2) 2 ≤ b ≤ d2 s = 3b − 2 4 ≤ s ≤ 3b d2 c − 2 d − b even t = d − 2b s + t even Table 8.1: Exceptional r cases for c = 3. To see why Table 8.1 reaches the necessary upper bounds on s, we need to consider − 2r , which is obtained the parity of d. For d and r even, the maximum s needed is 3d 2 c = 3d−r−1 . However, s = 3d−r−1 6≡ r in the table. When d is odd, we need s ≤ b 3d−r 2 2 2 (mod 3), thus the largest s we need is s = 3d−r−3 = 3b d2 c − 2r . When r = 3, we 2 need s ≤ 3d−3 , which equals 3b d−1 c when d is odd. But for d even, the largest s ≡ r 2 2 = 3b d−1 c. Hence the s bounds in Table 8.1 are correct. (mod 3) is s = 3d−9 2 2 For the lower bounds, Table 8.1 shows that all the necessary s are covered, except possibly some s < 5. Since s ≡ r (mod 3), all s are covered in the r = 2 case. In the r = 3 case, s = 0 is only necessary when the s + t is even. Similarly, no additional tableaux are needed in the r = 4 case because s = 1 is not a shape of Theorem 10. Now consider the exceptional s cases. These are tableaux having t even with s ∈ {0, 2, 4}, or s = 3 with no constraints on t. We still have the condition 3t+2s+r = 3d, so for a given s we need only verify that all the appropriate t in the range 0 ≤ t ≤ d − d 2s e, (with t even if s 6= 3) are obtained. 3 Table 8.2 belows lists the tableaux we are using for these cases. The columns ‘Parameters’ and ‘Conditions’ indicate the restrictions on the tableaux arising from their construction in Section 7.3. The column ‘t values’ indicates t’s value in terms 113 of the tableau parameters, while the ‘t covered’ lists those cases covered by the given tableau. The restrictions given on the t covered reflect the conditions listed on the parameters. Tableau Parameters Conditions t value s=0 P2 (a, 0) 0≤a≤d a even t=a s=2 P2 (a, 2) 0≤a≤d−2 a even t=a s=3 P3 (a, 1, 0) 0 ≤ a ≤ d − 3 a even t=a P4 (a, 1, 1) 0 ≤ a ≤ d − 2 a odd t=a P2 (a, 4) a even t=a s=4 0≤a≤d−4 t covered 0≤t≤d t even 0≤t≤d−2 t even 0≤t≤d−3 t even 0≤t≤d−2 t odd 0≤t≤d−4 t even Table 8.2: Exceptional s cases for c = 3. Table 8.2 shows that all the necessary t are covered, except possibly when s = 3 or 4. When s = 3, the t = d − 2, t even case does not appear. In this case, r = 0 and s + t is odd, so by Theorem 10 this case is not needed. For s = 4 and t = d − 3, we have r = 1 which is not a required shape. Hence all the exceptional cases have been covered. Finally, consider the general cases remaining. These are tableaux having r, s ≥ 5 and no additional constraints. We still have the condition 3t + 2s + r = 3d, so for a , fixed t we need only verify that all the appropriate s in the range 5 ≤ s ≤ 3d−3t−5 2 are obtained. (This accounts for the bounds both on r and on s.) Table 8.3 belows lists the tableaux we are using for this case. The columns ‘Parameters’ and ‘Conditions’ indicate the restrictions on the tableaux arising from their construction in Section 7.3. The column ‘s values’ indicates its value in terms of the tableau parameters, while the ‘s covered’ lists those cases covered by the given tableau. To see how Table 8.3 covers all the necessary shapes, first consider P3 . As c varies 114 Tableau Parameters Conditions s value s covered P3 (t, b, c) 2 ≤ b ≤ d−t−1 2 t + c even s = 2b + c + 1 5 ≤ s ≤ 3b d−t−1 2 c P4 (t, b, c) 1 ≤ b ≤ d−t 2 t + c even s = 2b + c 2 ≤ s ≤ 3b d−t 2 c 0≤c<b 0≤c≤b s + t odd s + t even Table 8.3: General c = 3 cases. between 0 and b − 1, we get 2b + 1 ≤ s ≤ 3b. When b increases to b + 1, we go from s = 3b to s = 2b + 3 = 2(b + 1) + 1. There is no gap between these provided b ≥ 2. Since b = 2 yields a minimum s = 5, we don’t need any smaller cases. Hence P3 will cover the cases, provided s + t is odd (equivalently t + c is even). Given a case where s = 2b + c + 1 but t + c is odd, use P4 (t, b, c + 1). Then t + c + 1 is even, and c < b implies c + 1 ≤ b, so the conditions of P4 are satisfied. (As similar analysis on P4 c shows all the s in the range do occur.) To see that the upper bound of s ≤ b 3d−3t−5 2 is met, first consider P4 . Since 3b d−t c ≥ b 3d−3t c − 1 = b 3d−3t−2 c, the upper bound 2 2 2 c ≥ 3d−3t−3 when d − t is odd. If d − t is even, we have is obtained. For P3 , 3b d−t−1 2 2 b 3d−3t−5 c = 3d−3t−6 , which is 3b d−t−1 c as desired. Thus all the required shapes are 2 2 2 listed. 8.2 Sufficiency when c = 4 When c = 4, the tableaux we will use are the Qi listed in Section 7.4. We will show that every partition of n = 4d described in Theorem 10 has a corresponding Qi . Note that shapes with r ≥ d + 5 can be obtained by Pi ∨ V (d) for the appropriate Pi filled with c = 3 elements. As such, we will not include these shapes in the following compilation. Throughout, we will use the convention d = 3z + x, where d ≡ x (mod 3). Unless otherwise specified, take x ∈ {0, 1, 2}. We also use the notation d∗ from previous chapters (d∗ = 2b d2 c). First consider the exceptional r cases. These are tableaux having s + t even with r ∈ {0, 2, 4} or r = 3 with no constraints on s and t. Since our shape is a partition of 115 4d we have the condition 3t + 2s + r = 4d. Hence for a given r, we need only verify that all the appropriate t in the range 0 ≤ t ≤ b 4d−r c, (with s + t even for r 6= 3) are 3 obtained. For the exceptional r cases with s + t even, we only need those t with t ≡ r (mod 4). Given a shape (r, s, t) the next shape needed is (r, s − 6, t + 4). Consider Table 8.4. For r = 0 we have all t ≤ 4d when d ≡ 0 (mod 3). When d ≡ 1 (mod 3), 3 is not possible and the table provides all t ≤ d + d−1 − 1. the shape t = d + d−1 3 2 Similarly, when d ≡ 2 (mod 3), the shape t = d + d−2 corresponds to s = 1, while 3 t = d + d−2 − 1 is not possible. This covers the r = 0 cases. Note when d = 5, the 3 only shapes needed are t = d − 1 and t = 0. Moreover, t = d − 1 and t = d − 2 are not possible for d ≤ 4. For r = 2, Table 8.4 provides tableaux for 0 < t ≤ d + z − 2 − x or t = d + z if d ≡ 2 (mod 3). Since t ≡ 2 (mod 4), t < 2 is not needed. When d ≡ 2 (mod 3), t = d + z − 1 and t = d + z − 3 are not congruent to 2 (mod 4). Similarly, t = d + z − x and t = d + z − x − 1 are not congruent to 2 (mod 4) for x = 0 and 1. When r = 3 we no longer have the conditions t ≡ r (mod 4); instead t must be odd. Table 8.5 accounts for all tableaux with 0 < t ≤ d + d3 − 3 for d ≡ 0 (mod 3). − 4 and for t ≤ d + d−2 −5 For d ≡ 1 (mod 3) Table 8.5 accounts for all t ≤ d + d−1 3 3 when d ≡ 2 (mod 3). Tableaux with larger t correspond to shapes having exceptional s cases (s 6= 3). Since t is odd, these shapes are not needed according to Theorem 10. When t ≤ d − 1 and d small, the shapes are either not required by Theorem 10 or are not possible. For r = 4, we need all t ≡ 0 (mod 4), where t ≤ d + z − 4 if x = 0, t ≤ d + z − 1 if x = 1, and t ≤ d + z − 2 if x = 2. Table 8.6 provides all these tableaux. Again, the bounds on d are necessary to produce a valid shape. When d = 5 and t = 4, the shape has s = 2 and can be found in Table 8.7. Hence these tables cover all the exceptional r cases. Now consider the exceptional s cases. These are tableaux having t even with s ∈ {0, 2, 4}, or s = 3 with no constraints on t. When r ≥ d, r 6= d + 1, we may reduced to Pi ∨ V (d) for some Pi filled with c = 3 elements. (If s = 3 we may only r=2 r=0 ∗ Q3 (2, 1) 0 < 2k < d t ≡ 0 (mod 2) t=d−1 t ≡ 0 (mod 4), d ≥ 6 t=0 t=d+z t = d∗ − 2k s = 2d s=0 d=3 t=2 d=3 t=2 t ≡ 2 (mod 4) d≥4 2≤t≤d−2 s=2 t = d∗ − 2k d≥4 W = d − d∗ s = d2 + 3k + 2W − 1 d∗ − 2k ≡ 2 (mod 4) ∗ t ≡ 2 (mod 4) t=d−1 d ≡ 3 (mod 4) 0 < 2k < d∗ t=d−1 t ≡ 2 (mod 4) d≥5 t=d+z−k d ≡ 2 (mod 3) d≤t≤d+z−x−2 t=d+z s = k + k+x 2 −1 t=0 4≤t≤d−2 s = d2 + 3k + 2W ∗ t=d−1 s = d+1 2 x ≤ k ≤ z, 0 < k 3k + x ≡ 2 (mod 4) x=2 W = d − d∗ 2k ≡ d∗ (mod 4) ∗ d ≥ 5, d odd t ≡ 0 (mod 4) d 6= 5 d ≤ t ≤ d + z − 2, x = 2 s = d+3 2 t=d+z−k x ≤ k ≤ z, 0 < k d ≡ 0 (mod 3) t=d+z 3k + x ≡ 0 (mod 4) t=d+z s=0 d ≤ t ≤ d + z − 1, x 6= 2 t covered s, t values s = k + k+x 2 x=0 Parameters Table 8.4: Exceptional r = 0 and r = 2 cases for c = 4. ∗ ∗ , k, k + W, d2 − 1) Q4 ( d −2k 2 d−1 d−3 Q5 ( d−1 2 , 2 , 1, 0, 2 , 1) k+x k+x Q1 (z − k, k−x 2 , 2 , 2 − 1) Q2 (z) U2 (d − 1, 1) Q4 ( d −2k , k, k + W, d2 ) 2 ∗ d−3 d−3 Q7 ( d−1 2 , 2 , 2 ) k+x k+x Q1 (z − k, k−x 2 , 2 , 2 ) Q2 (z) Tableau 116 Q6 (1, d − 2, d − 3) Q5 (2, 1, d − 3, 0, 1, d − 4) d ≡ 1 (mod 3), d ≥ 19 s=8 t=d+z−5 x=4 t = d + z − 2k − 1 x ∈ {0, 2, 4} d 6= 4, 5, 8 t=1 t=1 d≥8 t odd s = 2d − 3 t = 2k + 1 d≥8 5≤t≤d−3 d≥5 s = 2d − 3k − 3 2 ≤ k ≤ d−3 2 t odd d≥5 t=3 t=d−2 d≥5 t=d−2 t=3 s = d+3 2 d odd d≥6 t odd t=d−1 t odd, d 6= 4, 5, 8 d ≤ t ≤ d + z − 7, x = 4 d ≤ t ≤ d + z − 5, x = 2 d ≤ t ≤ d + z − 3, x = 0, d ≡ 1 (mod 3), d ≥ 13 s = 2d − 6 t=d−1 d≥6 d≥5 s = d2 d even x ≤ 2k ≤ z − 1 s = 3k + 2x k + 2x ≡ 1 (mod 2) d ≥ 19 t = d + d−1 3 −6 t=d+z−4 d ≥ 13 t=d+z−4 s=5 x=1 t covered s, t values Parameters Table 8.5: Exceptional r = 3 cases for c = 4. Q5 (k, k + 1, d − 2k − 2, 1, k − 1, d − 2k − 1) d−3 d−3 Q5 ( d−1 2 , 2 , 2, 0, 2 , 1) Q5 ( d2 , d2 , 1, 0, d−4 2 , 1) 2k+x 2k+x Q1 (z − 2k − 1, 2k−x 2 , 2 , 2 ) Q1 (z − 5, 1, 4, 3) Q1 (z − 4, 1, 2, 2) Tableau 117 r=4 U2 (d − 2, 1) t = d + d−1 3 −1 s = k + k+x 2 −2 d 6= 4 3k + x ≡ 0 (mod 4) t=0 s = 2d − 2 t = d∗ − 2k W = d − d∗ d≥6 s = d2 + 3k + 2W − 2 2k ≡ d∗ (mod 4) ∗ t=d−1 d≥7 0 < 2k ≤ d∗ s = d−1 2 d ≡ 1 (mod 4) d≥6 t=d+z−k s=0 x=4 x ≤ k ≤ z, k ≥ 2 s, t values Parameters Table 8.6: Exceptional r = 4 cases for c = 4. ∗ ∗ , k, k + W, d2 − 2) Q4 ( d −2k 2 d−1 d−5 Q5 ( d−1 2 , 2 , 0, 1, 2 , 1) k+x k+x Q1 (z − k, k−x 2 , 2 , 2 − 2) Q2 (z) Tableau t=0 d≥6 t ≡ 0 (mod 4) 4≤t≤d−2 d≥7 t ≡ 0 (mod 4) t=d−1 d≥6 t ≡ 0 (mod 4) d ≤ t ≤ d + z − 4 − x, x 6= 2 d ≤ t ≤ d + z − 2, x = 2 d 6= 4 d ≡ 1 (mod 3) t = d + d−1 3 −1 t covered 118 119 reduce when r ≥ d + 5.) Since we have already listed those table with exceptional r cases, we will take r ≥ 5. Hence for a given s, we need only verify that all the appropriate t in the range d 3d−2s−1 e ≤ t ≤ b 4d−2s−5 c (with t even for s 6= 3) are 3 3 obtained. For the exceptional s cases, consider Table 8.7. When s = 0, all t are covered except t = d + d−1 when x = 4. However, this case is unnecessary as r = 1. If t ≤ d, 3 the shape is reducible to a c = 3 case. For s = 2, all t ≥ d − 1 except t = d + d−x − 1 (with x ∈ {0, 2, 4}) are given in 3 Table 8.7. When that occurs, either the shape is invalid or r < 5. If t ≤ d − 2, the shape is reducible to a c = 3 case. When d ≤ 7 and t ≥ d the shapes are covered by the exceptional r cases or are reducible to c = 3 cases. For s = 3 we want t ≤ d + d−x − 2. However, when t = d + d−x − 2 (x ∈ {0, 2, 4}), 3 3 then r < 5. Hence these cases have already been covered. When t < d − 2 the shape is reducible to a c = 3 case. For d ≤ 6, the cases are covered by c = 3 or exceptional r cases. The bounds on d for t ≥ d are needed to produce a valid shape. − 3 with x ∈ {1, 3, 5}. For s = 4, Table 8.7 provides all tableaux with t ≤ d + d−x 3 Since any t larger than this has r < 5, this covers all shapes not already listed. Note that for d ≤ 12, all necessary shapes have t < d; those shapes with t = d − 1 are not needed for d ≤ 6. Hence all the exceptional s cases are accounted for in Table 8.7. The general cases of r, s ≥ 5 are classified in Table 8.8. When r ≥ d + 5 we can reduce to a c = 3 tableau. Fix t = d + z − k. Since r and s are greater than 5, we need all t with 5 ≤ k ≤ z. For a t of this form, we need all shapes with c. This range is covered in Table 8.8. To see why all such s are 5 ≤ s ≤ k + b x+k−5 2 obtained, note that for any fixed c, we always get 2c ≤ s ≤ 3c−1. Since c ≥ 2, there are no gaps as we increment c. The parameters between the cases are comparable, so writing s = 2c + d + 1 and using the case corresponding to the parity of s + k will yield the appropriate tableau. Since x ≤ 2, we find s ≤ k + b x+k−5 c implies 2 s ≤ k + b k−1 c − 1 for x = 0 or 2, and s ≤ k + b k2 c − 2 for x = 1. Comparing the 2 bounds shows all s are obtained. This takes care of all shapes with t ≥ d. When 0 < t < d, we require all shapes with 5 ≤ s ≤ 2d − t − b t+5 c. The tableaux 2 120 Tableau Parameters t value 0<a≤z s=0 s=2 Q2 (a) z + a even Q1 (a, 1, 1, 0) d<t≤d+z t=d+a t even, d 6= 4 0≤a≤z−2 d≤t≤d+z−2 x ∈ {0, 2, 4} t=d+a d odd Q1 (a, 2, 1, 0) Q1 (a, 1, 1, 1) x ∈ {0, 2, 4} t=d−1 Q5 ( d2 − 1, d2 , 1, 0, 1, 1) t=d+a 0≤a≤z−2 d≤t≤d+z−2 z + a odd t=d+a d odd d≥7 d even d≥6 t=d−1 t=d−1 t=d−2 z + a odd Q3 (a, 2) d odd d≥7 2≤a≤d−2 a even, d ≥ 4 x ∈ {0, 2, 4} t odd, d ≥ 8 t = d − 1, d ≥ 7, d odd t = d − 1, d ≥ 6, d even t=d−2 d≤t≤d+z−3 t=d+a x ∈ {1, 3, 5} Q5 ( d−1 , d−1 , 1, 0, 2, 1) 2 2 x ∈ {0, 2, 4} t even, d ≥ 11 0≤a≤z−3 Q1 (a, 1, 2, 1) t even z + a even Q6 (d − 2, 1, 0) s=4 t=d−1 d≤t≤d+z−3 x ∈ {0, 2, 4} Q5 ( d−1 , d−1 , 0, 1, 1, 1) 2 2 x ∈ {0, 2, 4} t even, d ≥ 8 0≤a≤z−3 s=3 x ∈ {0, 2, 4} x ∈ {0, 2, 4} z + a even Q3 (d − 1, 1) t covered x ∈ {1, 3, 5} t even, d ≥ 12 t=d−1 t=d−1 d odd d≥7 t=a Table 8.7: Exceptional s cases for c = 4. 2≤t≤d−2 t even, d ≥ 4 t=0 U8 U5 (2, k − 2) d=4 5≤k≤d 0≤k ≤d−2 d≥4 b>c≥0 b≤d−t−1 1 ≤ f ≤ d − 2k − 2 1≤e≤k 1 ≤ k ≤ b d−2 c 2 d≥5 1 ≤ f ≤ d − 2k − 2 s=5 s = 2k − 3 s = 2(d − k) s=b+c+2 s = d − 2k + e + f s = 5, d = 4 d ≥ 5, s odd 7 ≤ s ≤ 2d − 3 s even 4 ≤ s ≤ 2d d≥4 5 ≤ s ≤ 2d − 2t − 1 t even d − 2k + 2 ≤ s ≤ 2d − 3k − 2 d≥5 t odd d − 2k + 1 ≤ s ≤ 2d − 3k − 3 s = d − 2k − 1 + e + f 1 ≤ k ≤ b d−3 c 2 1≤e≤k d≥5 d≥5 4 ≤ s ≤ d d2 e + 1 s=e+3 d even 1 ≤ e ≤ d d−2 e−1 2 s=c+3 d even d odd 3 ≤ s ≤ d2 + 1 d odd 1 ≤ c ≤ d2 − 2 d≥7 4 ≤ s ≤ d−1 2 s + k even s ≤ 3b k2 c if x = 0, 2 4 ≤ s ≤ 3b k2 c − 1 s + k odd 5 ≤ s ≤ 3b k−1 c 2 s covered s=e+2 s = 2c + d + 1 s = 2c + d + 1 s value d≥7 1 ≤ e < d−1 −2 2 k 6≡ d (mod 2) d ≤ c + 1 if x = 0, 2 0≤d≤c−2 5≤k≤z 2 ≤ c ≤ b k2 c k ≡ d (mod 2) 0≤d<c 5≤k≤z 2 ≤ c ≤ b k−1 c 2 Parameters Table 8.8: General c = 4 cases. Q6 (t, b, c) 1≤t≤d−3 t = 2k U2 (d − k − 1, 1) Q5 (k, k, d − 2k − 1, 1, e, f) 2≤t≤d−3 t = 2k + 1 Q5 (k, k + 1, d − 2k − 2, 1, e, f) Q5 (b d−2 c, d d−2 e, 1, 1, e, 1) 2 2 Q7 ( d2 − 1, d2 − 1, c) Q5 ( d−1 , d−1 , 0, 1, e, 1) 2 2 Q1 (z − k, c, c, d + 1) Q1 (z − k, c + 1, c, d) 3≤t≤d−3 t=d−2 t=d−1 t=d+z−k d≤t≤d+z−5 Tableau 121 122 listed in Table 8.8 satisfy this. Note that for t ≤ d − 3 the bounds on s for Q5 and Q6 overlap, thus guaranteeing all s are covered. For the small d not listed, the shapes are either not possible, not needed, or result in exceptional cases done previously. When t = 0 we require all d + d−5 ≤ s ≤ 2d − 3. As Table 8.8 shows, this is satisfied. Hence 2 all necessary cases for c = 4 have been covered. 8.3 Sufficiency for c = 5 When c = 5, the tableaux we will use are the Ri listed in Section 7.5. We will show that every partition of n = 5d described in Theorem 10 has a corresponding Ri . Note that shapes with r ≥ d + 5 can be obtained by Qi ∨ V (d) for the appropriate Qi filled with c = 4 elements. As such, we will not include these shapes in the following compilation. Throughout, we will use the convention d = 3z + x, where d ≡ x (mod 3). Unless otherwise specified, take x ∈ {0, 1, 2}. First consider the exceptional r cases. These are tableaux having s + t even with r ∈ {0, 2, 4} or r = 3 with no constraints on s and t. Since our shape is a partition of 5d we have the condition 3t + 2s + r = 5d. Hence for a given r, we need only verify c, (with s + t even if r 6= 3) are that all the appropriate t in the range 0 ≤ t ≤ b 5d−r 3 obtained. For the exceptional r cases with s + t even, we only need those t with t ≡ d + r (mod 4). Hence given a shape (r, s, t) the next shape needed is (r, s − 6, t + 4). , x = 1, 2 Consider Table 8.9. For r = 0, this table provides all the t, except t = 5d−x 3 and t = d + 2z − 1. These are not a possible shapes. When d < 6 we only need those shapes with t ≤ d. For r = 2, Table 8.9 yields all t ≤ d + 2z − 2 with x ∈ {1, 2, 3}. Since t ≡ d + 2 (mod 4). When t = d+2z, for x = 0, 2 we do not get a shape required by Theorem 10; for x = 1 the tableau is listed in the table. Also, t = d + 2z − 1 is not a shape and t = d + 2z − 2 with d odd is not needed (the d even case is listed). If d ≤ 6, then all necessary shapes have t ≤ d − 2. Hence we have accounted for all necessary tableaux with r = 2. 123 Tableau r=0 R1 (z, z) Conditions x=0 d even t value t = 5d 3 z + a even t = 5d 3 d ≡ 0 (mod 3) d even 1≤a≤z R4 (a, a, x) t covered d ≤ t ≤ d + 2z − 2 t = d + 2z − 2a t ≡ d (mod 4) d≥6 d≥6 d∗ r=2 R5 (a, a) 0≤a≤ 2 −1 R12 ( d−1 2 ) d ≡ 1 (mod 4) t=1 U1 (d) ∨ U3 ( d2 , d2 ) d ≡ 0 (mod 4) t=0 R1 (z, z) a even x=1 d even t = d − 2a t = 5d−2 3 2≤t≤d t ≡ d (mod 4) t=1 d ≡ 1 (mod 4) t=0 d ≡ 0 (mod 4) t = 5d−2 3 d ≡ 1 (mod 3) d even t = d + 2d 3 −2 R2 (z − 1, z − 1, 1) x=0 d even t = d + 2d 3 −2 t ≡ d + 2 (mod 4) d even d ≡ 0 (mod 3) x ∈ {1, 2, 3} R4 (a, a, x − 1) 1≤a≤z d≥7 d ≤ t ≤ d + 2z − 2 t = d + 2z − 2a t ≡ d + 2 (mod 4) d≥7 z+a odd ∗ 3≤t≤d−2 R5 (a, a − 1) 1 ≤ a ≤ d2 − 1 t = d − 2a d−1 P4 (1, d−1 2 , 2 ) U1 (d) ∨ U3 ( d2 − 1, d2 ) d ≡ 3 (mod 4) t=1 t = 1, d ≡ 3 (mod 4) d ≡ 2 (mod 4) t=0 t = 0, d ≡ 2 (mod 4) t = 2, d even Table 8.9: Exceptional r = 0 and r = 2 cases for c = 5. For r = 3 consider Table 8.10. In this case, we need t 6≡ d (mod 2). All cases with t ≤ d + 2z − 3 are covered in the table. In addition we need t = d + 2z − 1 for d ≡ 0, 2 (mod 3) with d even, which are listed as well. When d < 9 we have t ≤ d + 1 or t = d + 2z − 1, and so those cases are covered. Finally, when d = 4, t = d + 1 has s = 1, so that shape is no needed. When t = 0 and d = 3 then s is even, so this case 124 is reducible to c = 4 case with r = 0. Hence all necessary tableaux with r = 3 are provided by Table 8.10. For r = 4 we need those tableaux with t ≡ d (mod 4), t ≤ d + 2z − 4. In addition, we need t = d+2z when d ≡ 2 (mod 3) and t = d+2z−2 when d ≡ 1 (mod 3). These are all found in Table 8.10. For d = 5 the needed tableaux are listed individually. For d = 4, t = d + 2z − 2 and t = 0 are the only shapes required, while d = 3 does not need any shapes. Thus all the exceptional r cases are contained in Table 8.9 and Table 8.10. Now consider the exceptional s cases. These are tableaux having t even with s ∈ {0, 2, 4}, or s = 3 with no constraints on t. When r ≥ d, r 6= d + 1, we may reduced to Qi ∨ V (d) for some Qi filled with c = 4 elements. (If s = 3 we may only reduce when r ≥ d + 5.) Since we have already listed the exceptional r cases, we will also take r ≥ 5. Hence for a given s, we need only verify that all the appropriate t in the range d 4d−2s−1 e ≤ t ≤ b 5d−2s−5 c (with t even if s 6= 3) are obtained. 3 3 When s = 0, we need all even t ≤ d + 2z. Table 8.11 provides all tableaux with t ≥ d + 2, t even. For t ≤ d + 1 we can use the reduction to a c = 4 case. When s = 2, we need all tableaux with d + 1 ≤ t ≤ d + 2z − 2 and t even, which Table 8.11 lists. Those with t ≤ d are reducible to a c = 4 case, while the d ≤ 5 cases correspond to exceptional r cases or are similarly reducible. When s = 3 we need those shapes with t ≤ d + 2z − 3 and d + 5 ≥ r ≥ 5. For other r or r = d + 3 we can reduce to a c = 4 case. Table 8.11 provides tableaux for all t ≥ d. When t ≤ d − 1, we can reduce to a c = 4 case when d ≥ 8 or d = 6. The remaining tableaux with r ≥ 5 and d ≤ 7 which are not reducible to a c = 4 case are listed as well. For s = 4 we need all even t ≤ d + 2z − 5, along with t = d + 2z − 4 when d 6≡ 0 (mod 3). Table 8.11 provides all those tableaux with t ≥ d + 1 and d ≥ 9. When t ≤ d we may reduce directly to a c = 4 case when d ≥ 10. Those remaining tableaux with d ≤ 9 have r ≥ d + 2, r = d or r < 5 and hence are either reducible or listed previously. Thus Table 8.11 suffices for the exceptional s cases. The general tableaux with r, s ≥ 5 are classified in Table 8.12 and Table 8.13. 125 Since r, s ≥ 5, we only need those tableaux with 0 ≤ t ≤ d + 2z − 5. For a fixed t, we need all tableaux with 4d−3t−4 ≤ s ≤ 5d−3t−5 because for r ≥ d + 5 we can 2 2 reduce to a c = 4 tableau. First consider t = d + z + a of Table 8.12. We need all s ≤ z − a + x − 2 + b z−a−1 c. Letting the parameters of R9 vary in order over their 2 c, we get all bounds yields all s up to b + z − a + 2x − 3. Since b ranges to b z−a−x+1 2 s ≤ z − a + x − 3 + b z−a+x−1 c. This covers all the necessary s for x > 1. When x = 1 2 c as needed. This tableau requires d ≥ 16, for d < 15 we also get s = z − a − 1 + b z−a 2 we find there are no shapes with t ≥ d + z, 5 ≤ r ≤ d + 4 and s ≥ 5. When d = 15, R6 provides t = d + z, the largest t required. When d ≤ t < d + z, Table 8.12 provides tableau R10 . Fix t = d + z − a, then we need all s ≤ a + z + x + b a+z−5 c. To see why this tableau suffices, begin by 2 taking c = 0. First let b and f vary over their ranges. Taking them maximal, vary d up to its maximum and then take e up to b 2z c − 1. This satisfies all the required inequalities. Then take R10 (a, b a2 c, c, d 2z e + b a2 c, b 2z c − 1, d a2 e) with 0 ≤ c ≤ b a2 c − 1, c even. Similarly, use R10 (a, b a2 c, c, d 2z e + b a2 c − 1, b 2z c − 1, d a2 e) with 0 ≤ c ≤ b a2 c, c even. This gives all 5 ≤ s ≤ a + z + b a2 c + x + b 2z c − 2. Since b a+z−5 c ≤ b a2 c + b 2z c − 2, 2 all the necessary s are obtained. This tableau required d ≥ 6. When d < 6, we find there are no shapes with t ≥ d and r, s within the needed bounds. Table 8.12 also contains t = d − 1, which requires all s ≤ d − 1, as listed. When t ≤ d − 2, consider Table 8.13. For t ≤ d − 2, we need all s ≤ 2d − t + c − 2. Tableau R11 provides this by taking the parameters through their ranges b d−t−1 2 in order, using b = 1 and b = 2 (with e ≥ f). (We need b = 2 in order to obtain s = 5, otherwise b = 1 suffices.) The only snag is when t = 2. Then we cannot have c = 1, hence s = 5 is not obtainable in this case. However, t = 2, s = 5 is needed only when d = 5 and hence is listed separately. Also, d = 3 needs only t = 0, accounting for d ≤ 4. For t = 1, we need all 2d − 2 ≤ s ≤ 2d + d2 − 6. (The s = 2d − 3 case is a r = d + 3 reduction to c = 4.) All these shapes are obtained in Table 8.13. For small d, those tableaux not listed have s < 5. For t = 0 we need all 2d − 2 ≤ s ≤ 2d + b d−1 c − 2. Table 8.13 provides all the 2 126 necessary tableaux. When d is even, we get all 2d ≤ s ≤ 2d + d2 , with s even and 2d − 1 ≤ s ≤ 2d + d2 − 3 with s odd. While s = 2d − 2 is not listed, this shape has r = d + 4 and s + t even, so we may reduce to a c = 4 case. When d is odd, we get all 2d − 2 ≤ s ≤ 2d + d−1 − 2 with s even and 2d − 1 ≤ s ≤ 2d + d−1 − 2 with s 2 2 odd. Hence all the required s are listed. Therefore Tables 8.12 and 8.13 provide the required tableaux for the non-exceptional cases with c = 5. 8.4 Sufficiency for c = 6 When c = 6, the tableaux we will use are the Si listed in Section 7.6. We will show that every partition of n = 6d described in Theorem 10 has a corresponding Si . Note that shapes with r ≥ d + 5 can be obtained by Ri ∨ V (d) for the appropriate Ri filled with c = 5 elements. As such, we will not include these shapes in the following compilation. Throughout, we will use the convention d = 3z + x, where d ≡ x (mod 3). Unless otherwise specified, take x ∈ {0, 1, 2}. First consider the exceptional r cases. These are tableaux having s + t even with r ∈ {0, 2, 4} or r = 3 with no constraints on s and t. Since our shape is a partition of 6d, we have the condition 3t + 2s + r = 6d. Hence for a given r, we need only verify c, (with s + t even for r 6= 3) are that all the appropriate t in the range 0 ≤ t ≤ b 6d−r 3 obtained. For the exceptional r cases with s + t even, we only need those t with t + r ≡ 2d (mod 4). Hence given a shape (r, s, t) the next shape needed is (r, s − 6, t + 4). Moreover, for r even, both s and t must be even. Consider Table 8.14. For r = 0, this table provides all the all the required partitions. When t = 4 we must have d even, while we need d odd for t = 2. For d = 3 and 4, only t = 2d, 2d − 4, and t = 0 are required. Hence all the r = 0 cases are provided. When r = 2, Table 8.14 gives all the necessary tableaux since only those with t ≤ 2d − 2 and t ≡ 2d + 2 (mod 4) are required. For r = 3 consider Table 8.15. In this case we need all odd t ≤ 2d − 3 because t = 2d − 1 has s = 0 and thus is not required by Theorem 10. All such tableaux are 127 listed except d = 3 with t ≥ 3. When t = 3 we reduce to a c = 5 case with s = 3 and r = 0. When t = 5 and d = 3 we have s = 0 and hence the shape is not needed. For r = 4 Table 8.15 provides all necessary tableaux. In this cases we have s and t even with t ≤ 2d − 4, t ≡ 2d (mod 4). When d = 3 and t = 6 we have s = 1, so that shape is not required. Hence all the exceptional r cases are listed. Now consider the exceptional s cases. These are tableaux having t even with s ∈ {0, 2, 4}, or s = 3 with no constraints on t. When r ≥ d, r 6= d + 1, we may reduced to Ri ∨ V (d) for some Ri filled with c = 5 elements. (If s = 3 we may only reduce when r ≥ d + 5.) Since we have already listed the exceptional r cases, we will also take r ≥ 5. Hence for a given s, we need only verify that all the appropriate t in the range d 5d−2s−1 e ≤ t ≤ b 6d−2s−5 c (with t even for s 6= 3) are obtained. 3 3 Consider Table 8.16. When s = 0 this provides all tableaux since those tableaux with t < d∗ have r ≥ d + 2. For s = 2 the table provides all tableaux with t ≥ d + 2. When t < d + 2 we have r ≥ d + 2 provided d ≥ 5. When d = 3 or d = 4 either r < 5 or we can reduce to a c = 5 case as well. Similarly when s = 3, those tableaux with t < d + 2 are reducible to a c = 5 case provided d ≥ 7. When d = 6, t = 7 we may reduce to a c = 5 case with r = 3; for smaller t we have r ≥ d + 5. When d = 5 and t = 5, we may reduce to a c = 5 case with r = 4 as s + t even; for smaller t, r ≥ d + 5. When d = 4 and t = 5 we have r < 5. When d = 3 and t = 2 we may reduce to a c = 5 case with r = 3; for smaller t we have r ≥ d + 5. When s = 4 we need all even t ≤ 2d−3. Table 8.16 lists all tableaux with t ≥ d+2 and d ≥ 5. When t ≤ d + 1, t even, we may reduce to a c = 5 case provided d ≥ 6. When d = 5, t = 6 we have r = 4 and we may reduce using P1 (4) to a c = 3 case. For d = 3 or 4 there are no needed shapes with 5 ≤ r ≤ d + 1. Hence Table 8.16 lists all the exceptional s cases. The general cases of r, s ≥ 5 are classified in Table 8.17 and Table 8.19. Since r, s ≥ 5, we only need those tableaux with 0 ≤ t ≤ 2d − 5. For a fixed t, we need ≤ s ≤ 6d−3t−5 since if r ≥ d + 5 we can reduce to a c = 5 all tableaux with 5d−3t−4 2 2 tableau. When t is odd, write t = 2e − 1 then we need all s ≤ 3d − 3e − 1. Consider 128 S6 of Table 8.17. First we let b, a and then c vary over their parameters This yields all s ≤ 3d − 3e with t ≥ 5. For t = 3, we use S7 . First let d vary to its bound of d − 4. Then take S7 (d − 4, e) and S7 (d − 5, e) with e varying over the even numbers. This yields all the required s. For t = 1 we use S12 . First vary b over its bounds to d − 2. Then use S12 (d − 3, d − 2, c) and S12 (d − 3, d − 3, c) with c even to obtain the needed s. Note the bounds on d in these tableaux are necessary for coherence. Those tableaux with d ≤ 5 and shapes with r, s ≥ 5 which are not reducible to a c = 5 case are listed in Table 8.18. Hence all odd t are covered. When t is even consider Table 8.19. Write t = 2e, so we want all 2d + d2 − 3e − 2 ≤ s ≤ 3d − 3e − 3. Consider S5 . First let b and c vary. Then take S5 (a, d − e, d − e) and S5 (a, d − e, d − e − 1) as a varies over all even numbers up to d − e − 1 and d − e − 2 respectively. This yields all 8 ≤ s ≤ 3d − 3e − 3 as needed when t ≥ 6. For 5 ≤ s ≤ 7 the tableaux are listed individually. For t = 2 or 4, we use S7 which covers all necessary s (as shown above for t = 3). When t = 0 we need 2d+ d2 −2 ≤ s ≤ 3d−3 which is provided in the table. Note the bounds on d in these tableaux are necessary for coherence. Those tableau with d ≤ 5 and shapes with r, s ≥ 5 which are not reducible to a c = 5 case are listed in Table 8.18. Hence all even t are covered by Table 8.19. Therefor all the required tableaux with c = 6 have been listed. 129 Tableau r=3 R1 (z − 1, z) R9 (z − 1, 1, 0, 0, 1) R6 (a, a − 1, x) R17 (z − 2, x) R18 Conditions x=0 d ≥ 6, d odd x=2 1≤a≤z−2 d≥9 t = d + 2z − 1 d ≡ 2 (mod 3) d + 1 ≤ t ≤ d + 2z − 3 d≥9 d = 3 or 5 t=d+1 t = d + 1, d = 3, 5 t=d−1 t=d−1 1 ≤ a ≤ d2 − 2 d even, d ≥ 6 2≤t≤d−3 t = d − 2a − 1 t 6≡ d (mod 2) d odd, d ≥ 5 3≤t≤d−3 t = d − 2a − 1 t 6≡ d (mod 2) d even, d ≥ 6 d even, d 6= 4 t=1 t = 1, d even, d 6= 4 d=4 t=1 t = 1, d = 4 U6 d > 5, d odd t=0 t = 0, d > 5, d odd U7 d=5 t=0 t = 0, d = 5 P4 (1, 1, 1) ∨ U1 (4) r=4 t = d + 2z − 2a − 1 d ≥ 6, d ≡ 3 (mod 4) t = d + 1, d ≥ 6 d odd, d ≥ 5 R12 ( d2 − 2) t = d + 2z − 1 t = 2d + 2z − 1 t=d+1 1 ≤ a ≤ d−1 2 −1 R8 (a) t = d + 2z − 1 t covered d≥6 R19 R7 (a) t value R1 (z, z) R2 (z − 1, z − 1, 1) R4 (a, a − 2, x) x=2 z even x=1 2≤a≤z d≥6 t = d + 2d−4 3 t = d + 2d−4 3 t even d ≡ 2 (mod 3) t = d + 2d−2 3 −2 t = d + 2z − 2a t = d + 2d−2 3 −2 d ≡ 1 (mod 3), d ≤ t ≤ d + 2z − 4 d≥6 ∗ 2 ≤ a ≤ d2 − 1 t = d − 2a 2 ≤ t ≤ d − 4, d ≥ 6 R13 (3, 2, 1, 1, 1) d=5 t=5 t = 5, d = 5 R12 ( d−5 2 ) d≥7 R5 (a, a − 2) d odd t=1 t=1 d≥7 d odd P4 (1, 2, 1) ∨ U1 (4) d=5 t=1 U1 (d) ∨ U3 ( d2 − 2, d2 ) d ≡ 0 (mod 4) t=0 Table 8.10: Exceptional r = 3 and r = 4 cases for c = 5. t = 1, d = 5 t=0 d ≡ 0 (mod 4) s=4 t = d + 2z − 4 t=d+a+b d=4 x = 1, 2 0≤b<a≤z−2 Q6 (2, 1, 0) R2 (a, b, 2) t=3 t=4 Table 8.11: Exceptional s cases for c = 5. d≥9 d≥7 d=4 R13 (1, 2, 1, 1, 1) R2 (z − 2, z − 2, 2) t=2 d=5 Q5 (1, 3, 0, 1, 1, 1) t=6 d=7 d≥6 t=d t=d+a+b 0≤a<b≤z−1 Q5 (3, 3, 0, 1, 1, 1) R3 (a, b) s=3 t=d+a+b d even, if a = b d≥6 R2 (a, b, 1) s=2 t=d+b+a t value 0≤b≤a≤z−1 d + a + b even b≥a 0<a≤b≤z Conditions R15 R1 (a, b) s=0 Tableau d≥9 d + 1 ≤ t ≤ d + 2z − 5 d≥7 d 6≡ 0 (mod 3) t = d + 2z − 4 t = 2, d = 4 t = 3, d = 4 t = 4, d = 5 t = 6, d = 7 t = d, d ≥ 6 d≥6 d + 1 ≤ t ≤ d + 2z − 3 d≥6 t even d + 1 ≤ t ≤ d + 2z − 3 t even d + 2 ≤ t ≤ d + 2z t covered 130 c ≥ 1 if t = 2 d≥4 a+c>e+f b+d>f f≤e 1 ≤ f ≤ b d−a−b−1 c 2 e 1 ≤ e ≤ d d−a−b−1 2 0≤d≤d−e−b−1 0≤c≤d−a−2 b = 1, or 2 1≤a≤d−b d≥5 1≤a≤d−4 d≥6 c even 0 ≤ f ≤ d a2 e 0 ≤ e ≤ b 2z c 1 ≤ d ≤ d 2z e + b a2 c 0 ≤ c ≤ b 2z c 0 ≤ b ≤ b a2 c 1≤a≤z d = 15 x ≥ 1, d ≥ 16 0≤e≤x or d = b + x − 1, e ≤ 1 0≤d≤b+x−2 0≤c≤z−a−b−1 c 1 ≤ b ≤ b z−a−x 2 0≤a≤z−5 Parameters s=c+d+e+f+3 s=a+3 d+e<z+f c+e<z−a+b+f c+d<z+b c<d s=b+c+d+e+f+x t=d+z s=b+c+d+e s value Table 8.12: General c = 5 cases for t > 2. R11 (a, b, c, d, e, f) t=a+b 2≤t≤d−2 R16 (a) R10 (a, b, c, d, e, f) R6 (2, 0, 0) R9 (a, b, c, d, e) t=d−1 d≤t≤d+z−1 t=d+z−a t = d + z, d = 15 d + z ≤ t ≤ d + 2z − 5 t=d+z+a Tableau c s 6= 5 if t = 2, d = 5 d≥4 s ≤ 2d − t + b d−t 2 c+2 5≤s d≥5 4≤s≤d−1 d≥6 3z s ≤ x + b 3a 2 c+b 2 c−1 5≤s t = d + z, d = 15 d ≥ 16 s ≤ 3z−3a−1 if x = 1 2 5 ≤ s ≤ 3z−3a+3x−6 2 s covered 131 t=0 ∗ c even 0 ≤ c ≤ b d−1 2 c−1 a even s=d ∗ + 2b d−1 2 c+c+1 s = 2d + a ∗ s = 2d − 2 s = 2d − 1 Table 8.13: General c = 5 cases for t = 0, 1, and 2. U1 (d ) ∨ P3 (0, b d−1 2 c, c) U1 (d ) ∨ U3 (a, b d2 c) 0 ≤ a ≤ b a2 c d≥5 R14 ∗ d odd P4 (1, d−1 2 , 1) ∨ U1 (d − 1) s = 2d − 1 s = 2d + a − 1 1 ≤ a ≤ b d2 c − 1 d even R12 (a) t=1 s=5 s value d=5 Parameters P4 (1, d2 − 1, 1) ∨ U1 (d) P3 (2, 1, 0) ∨ U1 (2) t=2 Tableau s even s odd s ≤ d∗ + 3b d−1 2 c d + 2b d−1 2 c+1≤s ∗ s ≤ 2d∗ + b d2 c 2d∗ ≤ s d≥5 s = 2d − 2 odd s = 2d − 1 d even s = 2d − 1 s ≤ 2d + b d2 c − 1 2d ≤ s s = 5, d = 5 s covered 132 133 Tableau r=0 Conditions S1 (z) S5 (d − e, d − e, d − e) Q4 (2, 1, d − 5, 3) ∨ U1 (d) P1 (4) ∨ U3 (2, 2) S8 U1 (d) ∨ U1 (d) ∨ U1 (d) 3≤e≤d−2 d ≡ e (mod 2) t value t covered t = 2d t = 2d 6 ≤ t ≤ 2d − 4 t = 2e t ≡ 2d (mod 4) d≥5 d ≥ 6, d even t=4 d=4 t=4 d ≥ 5, d odd t=2 d even t=0 t=4 d ≥ 6, d even t = 4, d = 4 t=2 d ≥ 5, d odd t = 0, d even e 6≡ d (mod 2) r=2 S5 (d − e − 1, d − e, d − e) 3≤e≤d−1 6 ≤ t ≤ 2d − 2 t = 2e t ≡ 2 + 2d (mod 4) d≥4 d−1 d−5 Q4 (2, d−7 2 , 2 , 2 ) ∨ U1 (d − 1) d≥4 t=4 d ≥ 9, d odd t=4 P4 (3, 3, 3) ∨ P4 (1, 2, 1) d=7 t=4 t = 4, d = 7 P4 (3, 1, 1) ∨ P4 (1, 2, 1) d=5 t=4 t = 4, d = 5 Q2 (1) ∨ U1 (2) d=3 t=4 t = 4, d = 3 d−2 d−2 Q4 (1, d−2 2 , 2 , 2 ) ∨ U1 (d) d even t=2 t = 2, d even U2 (d − 1, 1) ∨ U1 (d − 1) d odd t=0 t = 0, d odd Table 8.14: Exceptional r = 0 and r = 2 cases for c = 6. d ≥ 9, d odd 134 Tableau r=3 Conditions 3≤e≤d−1 S6 (d − e, d − e, d − e) d≥4 t value 5 ≤ t ≤ 2d − 3 t = 2e − 1 d≥4 S9 (2, 1, d − 4, 1, d − 4, d − 2, 1) d≥5 t=3 t = 3, d ≥ 5 S10 (3) d=4 t=3 t = 3, d = 4 t=1 t=1 S11 r=4 t covered S4 (d − 6) d≥6 t = 2d − 4 t = 2d − 4 d≥6 3≤e≤d−4 S5 (d − e − 2, d − e, d − e) e ≡ d (mod 2) 6 ≤ t ≤ 2d − 8 t = 2e t ≡ 2d (mod 4) d≥7 Q2 (1) ∨ U1 (4) d≥7 d=5 t=6 d ≥ 6, d even t=4 d=4 t=4 d ≥ 5, d odd t=2 Q3 (2, 1, 1) ∨ U1 (2) d=3 t=2 U1 (d) ∨ U1 (d) ∨ U1 (d − 2) d even t=0 d−4 d−4 Q4 (2, d−4 2 , 2 , 2 ) ∨ U1 (d) P1 (4) ∨ U4 (2, 2) d−1 d−3 Q4 (1, d−3 2 , 2 , 2 ) ∨ U1 (d − 1) t = 6, d = 5 t=4 d ≥ 6, d even t = 4, d = 4 t=2 d ≥ 5, d odd t = 2, d = 3 t=0 d even Table 8.15: Exceptional r = 3 and r = 4 cases for c = 6. s=0 Tableau Conditions t value P1 (d∗ ) ∨ P1 (a) 0≤a≤d t = d∗ + a a even 0≤a≤d−4 s=2 S2 (a) a ≡ d (mod 2) s=4 0≤a≤d−5 S3 (a) d≥5 d∗ ≤ t ≤ 2d∗ t even d + 2 ≤ t ≤ 2d − 2 t=d+a+2 d≥4 s=3 t covered t even d≥4 t=d+a+2 d + 2 ≤ t ≤ 2d − 3 d≥5 P1 (4) ∨ P3 (2, 1, 0) d=5 t=6 t = 6, d = 5 P1 (4) ∨ U4 (2, 1) d=4 t=4 t = 4, d = 4 S4 (a) 0≤a≤d−5 d≥5 t=d+a+2 Table 8.16: Exceptional s cases for c = 6. d + 2 ≤ t ≤ 2d − 3 d≥5 135 Tableau Parameters s value s covered 1 ≤ a, b, c ≤ d − e t = 2e − 1 5 ≤ t ≤ 2d − 5 a, b ≥ c S6 (a, b, c) b≥a−1 t odd s=a+b+c 3≤e≤d−2 5 ≤ s ≤ 3d − 3e d≥5 d≥5 0≤e≤d t=3 S7 (1, 2, d − 5, d, e) e even 1≤d≤d−4 s=d−2+d+e d − 1 ≤ s ≤ 3d − 7 d≥5 d≥5 0≤c≤d t=1 S12 (d − 3, b, c) c even s=d+b+c 0≤b≤d−2 Table 8.17: General c = 6 cases for odd t. d Tableau Shape d=5 P3 (2, 1, 0) ∨ P3 (2, 1, 0) t = 4, s = 6 d=4 S10 (2) t = 3, s = 5 P4 (2, 1, 0) ∨ U4 (2, 1) t = 2, s = 5 U3 (2, 2) ∨ U4 (2, 1) t = 0, s = 9 Table 8.18: General c = 6 cases for d ≤ 5. d ≤ s ≤ 3d − 3 S7 (1, 1, d − 4, d, e) U1 (d∗ ) ∨ U1 (d∗ ) ∨ U1 (a) t=2 t=0 s=7 e≤d−4 s=5 e≤d−3 0≤a≤d s odd, d odd 2d − 3 ≤ s ≤ 3d − 4 s = 2d − 3 + a d odd, a even s odd 2d − 3 ≤ s ≤ 3d − 3 s even 2d∗ ≤ s ≤ 3d∗ d≥5 d ≤ s ≤ 3d − 6 d ≥ 6, d even s = 2d − 3 + a s = 2d∗ + a s=d−1+d+e d≥6 d − 2 ≤ s ≤ 3d − 9 e odd s = 5, d ≥ 5 e even s = 5, d ≥ 5 s = 6, d ≥ 6 e odd s = 7, d ≥ 5 e even s = 7, d ≥ 5 d≥6 8 ≤ s ≤ 3d − 3e − 3 s covered a even 0≤a≤d d ≥ 6, d even a even 0≤a≤d 1≤d≤d−4 e even, d ≥ 5 0≤e≤d 1≤d≤d−5 e even, d ≥ 6 0≤e≤d e odd s=d−3+d+e s=5 e≤d−3 e even s=6 3≤e≤d−3 e odd e even s=7 s=a+b+c s value 0≤e≤d−4 3≤e≤d−3 a even, d ≥ 6 b, c > a 1 ≤ a, b, c ≤ d − e Parameters Table 8.19: General c = 6 cases for even t. d+3 U5 ( d−1 2 , 2 ) ∨ U1 (a) U5 ( d2 , d2 ) ∨ U1 (a) S7 (2, 2, d − 6, d, e) P4 (e + 1, 1, 0) ∨ P4 (e − 1, 1, 0) P3 (e, 1, 0) ∨ P4 (e, 1, 0) S5 (2, 2, 2) P3 (e + 1, 1, 0) ∨ P4 (e − 1, 2, 0) P3 (e, 1, 0) ∨ P4 (e, 2, 0) S5 (a, b, c) t=4 t even 6 ≤ t ≤ 2d − 6 t = 2e Tableau 136 137 8.5 Sufficiency for c > 6, d even In Chapter 6 we algorithmically demonstrate how to reduce an arbitrary tableau to one of those tableaux filled with fewer elements. In the case where d is even, we reduced all tableaux to joins of tableaux with c ≤ 6 to or those tableaux with t < d, s < d + 5, r < d + 5. Previously we showed all tableaux with c ≤ 6 where constructed in Chapter 7. Now we will address the remaining cases with d even. Since any tableau must satisfy cd = 3t + 2s + r, applying the bounds on r, s, and t we find cd ≤ 3(d − 1) + 2(d + 4) + d + 4 = 6d + 9. Hence for d > 9 all such tableaux will have c ≤ 6. For d = 8 and d = 6 it may be possible to have c = 7, while for d = 4, both c = 7 and c = 8 are possible. When d=8 and c=7, only the shape with (r, s, t) = (11, 12, 7) satisfy the constraints. However, we can reduce this shape by V (d) to a c = 6 case with r = 3. When d = 6 and c = 7, only the shapes with (r, s, t) = (10, 10, 4), (9, 9, 5) and (7, 10, 5) satisfy the constraints. For (10, 10, 4) we may reduce by V (d) to a c = 6 case with (r, s, t) = (4, 10, 4) since s + t is even. For (9, 9, 5) we may reduce by V (d) to a c = 6 case with (r, s, t) = (4, 9, 5) since s + t is even. For (7, 10, 5) we may reduce by U1 (4) to a c = 5 case with (r, s, t) = (3, 6, 5). When d = 4 and c = 7, only the shapes with (r, s, t) = (7, 6, 3), (5, 7, 3), (3, 8, 3), (6, 8, 2), and (8, 7, 2) satisfy the constraints. For (7, 6, 3) we may reduce by V (d) to a c = 6 case of (3, 6, 3). For (3, 8, 3) use R19 ∨ U1 (4). For (5, 7, 3) we may reduce by U1 (4) to the c = 5 cases (5, 3, 3). For (6, 8, 2) we may reduce by V (d) to a c = 6 case of (2, 8, 2) since s + t is even. For (8, 7, 2) we may reduce by U1 (4) to the c = 5 case (8, 3, 2). When d = 4 and c = 8, only the shape with (r, s, t) = (7, 8, 3) satisfies the constraints. It can be obtained by S10 (3) ∨ U1 (2). Thus when d is even, all shapes are reducible to tableaux filled with less than or equal to six elements and all the cases with c ≤ 6 were obtained previously. 138 8.6 Sufficiency for c > 6, d odd When d is odd, the reduction techniques of Chapter 6 work to reduce a tableau to c ≤ 8. As we’ve already constructed those tableaux with c ≤ 6, we will focus on those with c = 7 or 8. First, assume t ≥ d − 1. So long as r 6∈ {0, 2, 3, 4, 5, 7} we may use P1 (d − 1) to reduce to a c − 3 case. If t ≥ d − 1 and r ∈ {0, 2, 3, 4, 5, 7} we may use P4 (d − 2, 1, 1) to reduce to a c − 3 case unless s ∈ {0, 2, 3, 4, 5, 7}. For s = 3, 5, 7 we may still reduce by P4 (d − 2, 1, 1) when t is odd. For r = 0, 2, 4, s = 3, 5, 7 and t even, the shape is not needed by Theorem 10. For r = 3, 5, 7 and s = 3, 5, 7 with t even, no shapes are possible for c = 8. When c = 7 we have d ≥ 5 so we may use P3 (d − 3, 1, 0) to reduce to a c = 4 case with s + t even. For s = 0, 2, 4, we only need those shapes with t even. For r = 3, 5, 7 we may reduce by P1 (d − 1) to a c − 3 case. This leaves those cases with s = 0, 2, 4, r = 0, 2, 4 and t even. When c = 7 there are no shapes having s = 0, 2, 4, r = 0, 2, 4 and t even as d is odd. For c = 8, these shapes are obtainable, depending on d (mod 3). We list the appropriate tableaux in Table 8.20. This completes all cases with t ≥ d − 1 If t < d − 1 and s < d + 4, but r = d + 3 or r ≥ d + 5 we may use V (d) to reduce to a c − 1 case. When c = 8, there are no valid shapes with r ≤ d + 4 and t < d − 1, s < d + 4. For c = 7 this is also true provided d ≥ 5. When d = 3 we need (r, s, t) = (6, 6, 1) which is reducible by V (d) to a c = 6 case with r = 3. If t < d − 1 and s ≥ d + 4 we may use U1 (d − 1) to reduce to a c − 2 case, provided r 6∈ {0, 2, 3, 4, 6}. Consider those cases with r ∈ {0, 2, 3, 4, 6} If s ≥ 2d + 5 or s = 2d + 3 we may use U2 (d, d) to reduce to a c − 4 case. For s ≤ 2d − 1 there are no shapes with t ≤ d − 2 and r ≤ 6. If s = 2d, 2d + 2 or 2d + 4, then we may still reduce via U2 (d, d), provided t is even (which always occurs if r = 0, 2, 4). Thus we need only consider those tableaux with s = 2d + 1, or s = 2d, 2d + 2, 2d + 4 with r = 3 or 6. For c = 7 and s ≥ 2d, s 6= 2d + 1 only r = 6, s = 2d, t = d − 2 is possible. This can be obtained by P4 (d − 2, 1, 1) ∨ U5 (2, d − 2) provided d ≥ 5. When d = 3 we may 139 (r, s) d (0, 0) d ≡ 0 (mod 3) Q2 (z) ∨ Q2 (z) (0, 2) d ≡ 2 (mod 3) W1 (2) (0, 4) d ≡ 1 (mod 3) Q1 (z − 1, 0, 1, 1) ∨ Q1 (z − 1, 0, 1, 1) (2, 0) d ≡ 1 (mod 3) W1 (0) (2, 2) d ≡ 0 (mod 3), d > 3 d=3 (2, 4) d ≡ 2 (mod 3), d > 5 Tableau Q2 (z) ∨ Q1 (z − 2, 1, 1, 0) S1 (1) ∨ U1 (2) Q2 (z) ∨ Q1 (z − 2, 0, 2, 2) d=5 S1 (1) ∨ U1 (4) (4, 0) d ≡ 2 (mod 3) Q2 (z) ∨ Q2 (z) (4, 2) d ≡ 1 (mod 3) Q2 (z; x = 4) ∨ Q1 (z − 1, 0, 1, 1; x = 1) (4, 4) d ≡ 0 (mod 3), d > 9 Q2 (z) ∨ Q1 (z − 4, 2, 2, 0) d=9 Q2 (z) ∨ Q5 (4, 4, 1, 0, 2, 1) d=3 Q1 (1) ∨ U1 (2) ∨ U1 (2) Table 8.20: Exceptional r and s cases for c = 8. reduced by V (d) to a c = 6 case with r = 3. For c = 8 and s ≥ 2d, s 6= 2d + 1 only s = 2d + 4, r = 3, d = 5 can occur. In that case, use Q5 (2, 1, 2, 0, 1, 1) ∨ U2 (5, 5). This leaves those tableaux with t < d − 1, s = 2d + 1 and r ∈ {0, 2, 3, 4, 6}. When r = 0, 2, 4, we must have t odd, so in the c = 8 case there are no possible shapes. For c = 7, we get a valid shape only for r = 4, in which case we have t = d − 2. For this use U1 (d − 1) to reduce to a c = 5 case with r = 2, s = d + 1 which will still have s + t even. For r = 3, there are no shapes satisfying t < d − 1 and s = 2d + 1 for either c = 7 or 8; similarly for r = 6. Thus all required shapes may be reduced to those filled with c ≤ 6 elements. 140 8.7 Tableaux Disjointness Our proof of Theorem 2 requires the tableaux we constructed to be disjoint. Since Lemma 3.4.9 showed that maximal tableaux are always disjoint, we need only be concerned with those tableaux which could not be put in maximal form, namely, U8 and S8 . Recall that U8 = 43333122444 11122 has d = 4 and ω2 = (3, 2, 0, 0). When d = 4 the tableaux used in the reduction techniques are P1 (4), U1 (4), and V (d). These are the only tableaux that would be joined with U8 , so it suffices to show U8 is disjoint from these tableaux. Note that the weights of these tableaux consist only of 4’s and 0’s. If U8 were not disjoint from these tableaux, then there is some weight assignment of U8 , not equivalent to ω2 = (3, 2, 0, 0) which uses a weight from at least one of these tableaux. However, since the only additional weights we may use are 4’s and 0’s, there is no way to have a weight assignment of length 5 using 0’s, 2’s, 3’s, and 4’s without being equivalent to (3, 2, 0, 0). Hence the weights are disjoint from U8 . d−4 Recall that S8 = d−2 d−4 2 66 6 3 2 1 23662 34 3 4 4 5 55151 d d 0 with ω2,3 = ( 22 00 d−3 0 0 0 0 ) and 11 d ≥ 5. When d is even, the tableaux used in the reduction techniques are P1 (d), U1 (d), and V (d). When d is odd, the tableaux used in the reduction techniques are S1 (d), U2 (d), whose weights are 0’s and d’s, P1 (d − 1), U1 (d − 1) whose weights are ¡ ¢ 1 0’s and d − 1’s, and P4 (d − 2, 1, 1) with ω2,3 = 00 d0 d−2 . Hence it suffices to show S8 disjoint from these tableaux. Now any weight assigned to S8 must have λ3 = 2. Since d ≥ 5, there are no weights ¡¢ other of the listed tableaux than the weight 22 of S8 for which this is possible. This means if S8 were not disjoint from these tableaux there would be a weight assignment d d 0 of S8 of the form ( 22 0∗ ∗0 ∗0 0∗ 0∗ ) which is not equivalent to ( 22 00 d−3 0 0 0 0 ). The only weights we may use for the ∗’s are the weights of the listed tableaux, namely, 0’s d’s, d − 1’s and d − 3. Since the ∗’s sum to 3d − 3, they would have to be three d − 1’s and two 141 0’s. That is, the weights of U1 (d − 1) or P1 (d − 1). Then the weights of S8 would need to be assigned to other tableaux, so a weight of d would need to be assigned to either U1 (d − 1) or P1 (d − 1), which is not possible as these tableaux are maximal. Hence there is no other weight assignment for S8 and so the tableau is disjoint as required. Thus all the tableaux constructed in the proof of Theorem 2 are disjoint as desired. 197 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. Doran, IV, On Foulkes’ conjecture, J. Pure Appl. Algebra 130 (1998), no. 1, 85–98. MR 99h:20014 [8] H. O. Foulkes, Concomitants of the quintic and sextic up to degree four in the coefficients of the ground form, J. London Math. Soc. 25 (1950), 205–209. MR 12,236e [9] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.3, 2002, (http://www.gap-system.org). 198 [10] David A. Gay, Characters of the Weyl group of SU (n) on zero weight spaces and centralizers of permutation representations, Rocky Mountain J. Math. 6 (1976), no. 3, 449–455. MR 54 #2886 [11] Larry C. Grove, Groups and characters, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1997, A Wiley-Interscience Publication. MR 98e:20012 [12] Roger Howe, (GLn , GLm )-duality and symmetric plethysm, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, 85–109 (1988). MR 90b:22020 [13] N. F. J. Inglis, R. W. Richardson, and J. Saxl, An explicit model for the complex representations of Sn , Arch. Math. (Basel) 54 (1990), no. 3, 258–259. MR 91d:20017 [14] G. James and A. Kerber, Representation theory of the symmetric group, Encyclopedia of Mathematics and its Applications, vol. 16, Addison-Wesley, Reading, MA, 1981. [15] G. D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, vol. 682, Springer, Berlin, 1978. [16] Serge Lang, Algebra, 3 ed., Addison Wesley, Reading Massachusetts, 1999. [17] I. G. Macdonald, Symmetric functions and Hall polynomials, second ed., Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, New York, 1995, With contributions by A. Zelevinsky, Oxford Science Publications. MR 96h:05207 [18] Bruce E. Sagan, The symmetric group, The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1991, Representations, combinatorial algorithms, and symmetric functions. MR 93f:05102 199 [19] Richard P. Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004) 142 Chapter 9 Proof of Theorem 3 Theorem 2 proves the generalized Foulkes’ Conjecture for c = 3 without multiplicities. We can, however, strengthen this result to include multiplicities for those irreducibles corresponding to two row partitions. Namely, Theorem 3. Let n = 3b = cd, with c, d ≥ 3 and let λ = [λ1 , λ2 ] be a two row partition of n. Then every irreducible character χλ occurring in 1SSnb oS3 occurs in 1SSnd oSc with multiplicity at least as large. The proof of this theorem involves constructing the appropriate number of tableaux, primarily out of the basis elements given in Section 4.3. The tableaux constructed are maximal so linear independence is assured from Lemma 3.4.12 provided the weights are distinct. Once we have the basis tableaux for c = 4, 5, and 6, the procedure generalizes to provide the appropriate number of tableaux for any c. Take λ = [r + s, s]. The multiplicity mλ of Sλ in 1SSnb oS3 was determined by Thrall in [20], which we listed in Theorem 11. Since the multiplicity depends on the relative sizes of r and s we will handle these cases separately. Moreover, if s = 6k +j and r = 6h+i, we will often simply construct k + 1 or h + 1 tableaux when possible to avoid detailed case analysis. 9.1 Case: s ≤ r Let λ = [r + s, s] be a partition of n with s ≤ r, where n = 3b = cd. We wish to construct mλ linearly independent tableaux, where mλ is the multiplicity of χλ in 143 1SSnb oS3 as described in Theorem 11. First we will construct these tableaux for c = 4, 5, and 6; then we will use these constructions in proving Theorem 3 for a general c. We will refer to the tableaux constructed in this way as basis tableaux. These constructions will make use of the c = 3 basis tableaux constructed in Section 4.3. 9.1.1 Basis Tableaux for c = 4, s ≤ r Given λ = [r + s, s] a partition of n, we have 2s + r = 4d = 3b. From this equation and s ≤ r, we have s ≤ d + b d3 c. For each λ we will construct mλ linearly independent λ-tableaux filled with the numbers 1 to 4. These will be our c = 4 basis tableaux. When s ≤ r−d, we can use the basis tableaux constructed in Section 4.3. Consider the partition λ0 = [r0 + s, s] where r0 = r − d. Since s ≤ r0 , we have mλ = mλ0 . In Section 4.3 we constructed mλ0 linearly independent Bp , where Bp are the basis tableaux for c = 3. Take Bp ∨ d as the basis tableaux for c = 4. This works for 4 s ≤ r − d, so r ≥ s + d. Hence 4d = 2s + r ≥ 3s + d implies s ≤ d. When d < s ≤ d + b d3 c, write s = d + f , with 1 ≤ f ≤ b d3 c. Consider the tableau a+b≤d abcd T (a, b, c, d) = 4 3 4 3 1 1 2 2 a, b, c, d > 0 a>d b>c or a = d, b = c w(T ) = (a + b, c + d, 0, 0) If a > d and b > c, no other weight assignments are possible for T , hence this tableau is non-zero and maximal. If a = d and b = c, we may also have the tableau obtained by exchanging the rows. However, this has sign (−1)a+b+c+d = 1 and thus the tableau is still non-zero. Let Cp = T (d − 2p, p + 1, p, f − 1). Then λ2 (T ) = a + b + c + d = d − 2p + p + 1 + p + f − 1 = d + f = s. Hence these tableaux have the desired shape. Consider Cp for p = 1, 2, . . . , b d−f c. To insure the Cp are non-zero and maximal we need to check 2 144 that the constraints on T are satisfied. For f 6= 1 all the parameters are greater than zero. Obviously, b = p + 1 > c = p. For a > d, we need d − 2p > f − 1. This is true provided p < d−f2 +1 . Since p ≤ b d−f c this inequality holds. 2 The Cp are linearly independent by Lemma 3.4.12 if their max weights are distinct. We have w(Cp ) = (d − p + 1, f + p − 1, 0, 0). If w(Cp ) = w(Cp0 ) for p > p0 , then we must + d−f −1 = d−f −1. have d−p+1 = f +p0 −1, that is d−f +2 = p+p0 . But p+p0 ≤ d−f 2 2 Hence this cannot occur. Thus the Cp are linearly independent. Since s = d + f , we have mλ ≤ b d+f c + 1, so it suffices to construct b d+f c+1 6 6 tableaux. The Cp provide b d−f c tableaux. To show b d−f c ≥ b d+f c + 1, it suffices to 2 2 6 show d−f − 12 > d+f , or equivalently, d − 32 ≥ 2f . This holds for d ≥ 5 as f ≤ b d3 c. 2 6 When d = 3 or d = 4 then f = b d3 c = 1, which is handled below. Hence for s > d + 1 the Cp provided at least mλ linearly independent tableaux. For s = d + 1, that is f = 1, take Cp = T (d − p − 2, 2, 1, p), for 1 ≤ p < c. Then the conditions on T are satisfied and such p exist for d ≥ 6. We have b d−2 2 w(Cp ) = (d − p, p + 1, 0, 0) so the max weights are distinct. These Cp provide at c − 1 linearly independent tableaux and we need b d+1 c + 1 tableaux. Now least b d−2 2 6 c − 1 ≥ b d+1 c + 1 provided d ≥ 8. When d = 7, then s = 8 and two tableaux b d−2 2 6 are needed. Use T (4, 2, 1, 1) and T (3, 2, 1, 2). When d = 6, then s = 7 and only one tableau is needed. In this case, use Cp described above for p = 1. When d = 5 then s = 6 so two tableaux are needed. Use T (2, 2, 1, 1) and T (2, 1, 1, 2). When d = 4 then s = 5 so one tableaux suffices. However, for d = 4, s = 5 there are no tableaux of maximal form. We will use the tableau U8 constructed in Section 7.2. This tableau is non-zero but not maximal. Here we use maximal form only to show linear independence. Since only one tableau is needed for s = 5, this tableau works. (The d = 4, s = 5 case is actually not needed for the c = 4 basis tableaux since n = 3b = 4d implies 3|d, but we construct the basis tableaux for all d ≥ 3 in order to simplify the construction process in Section 9.1.4. However we will not use U8 in that construction.) When d = 3, then s = 4 and only one tableau is need, hence T (1, 1, 1, 1) suffices. 145 9.1.2 Basis Tableaux for c = 5, s ≤ r Given λ = [r + s, s] a partition of n, we have 2s + r = 5d = 3b. From this equation and s ≤ r, we have s ≤ b 5d c. For each λ we will construct mλ linearly independent 3 λ-tableaux filled with the numbers 1 to 5. These will be our c = 5 basis tableaux. When s ≤ r − d, we can use the c = 4 basis tableaux constructed in Section 9.1.1. Consider the partition λ0 = [r0 +s, s] where r0 = r−d. Since s ≤ r0 , we have mλ = mλ0 . In Section 9.1.1 we constructed mλ0 linearly independent Cp , where Cp are the basis tableaux for c = 4. Take Cp ∨ d as the basis tableaux for c = 5. This works for 5 s ≤ r − d, so r ≥ s + d. Hence 5d = 2s + r ≥ 3s + d implies s ≤ b 4d c. 3 For b 4d c < s ≤ b 5d c consider 3 3 1≤a≤d−2 a b d2 c d d2 e b T1 (a, b) = 4 4 5 4 3 4 5 222 1 1 33 1 ≤ b ≤ b d2 c − 3 a>b s≥d+6 d≥8 w(T1 ) = (d, a + 2, b + 1, 0, 0) T1 is defined for s ≥ d+6, d ≥ 8. (When d < 8 then s < d+6 since s ≤ b 5d c.) Since 3 a > b, there are no other weight assignments possible and the tableau T1 is maximal. c−2. For Ep to be non-zero and maximal Let Ep = T1 (s−d−3−p, p) for 1 ≤ p ≤ b s−d 2 we need the conditions on T1 to be satisfied. We have a > b since p ≤ b s−d c − 2, 2 while p ≥ 1 implies a = s − d − 3 − p < d − 2. Since d ≥ 8, then b ≤ b d2 c − 3. Note a > b implies these max weights are distinct. Hence this construction provides b s−d c − 2 distinct, linearly independent tableaux when s ≥ d + 6, d ≥ 8. 2 We will need an additional tableau, so take: a b T2 (a, b) = 5 3 5 4 1122 w(T2 ) = (a + 1, b + 1, 0, 0) 1 ≤ a, b ≤ d − 1 s≥4 146 Clearly T2 is non-zero and maximal. Given s, take E0q = T2 (d − q, s − 2 − d + q) for q = 1, 2, . . . , d −b s−2 c. To insure E0 q is non-zero and maximal we need the conditions 2 of T2 are satisfied. Since b 4d c < s and d ≥ 3, then s ≥ 4. We have 1 ≤ a ≤ d − 1 3 because q < d since s ≤ b 5d c. Similarly the bounds on s show 1 ≤ b ≤ d − 1. The E0 q 3 c. This provides d − b 2s c + 1 are linearly independent because a < b for q < d − b s−2 2 tableaux. We need mλ = b 6s c + 1 linearly independent tableaux. By Lemma 3.4.12 the tableaux Ep and E0 q are linearly independent since they have different max weights. First consider s ≥ d + 6 which implies d ≥ 8. Then we have both Ep and E0q , for a total of b s−d c − 2 + d − b 2s c + 1 tableaux. This is greater than or equal to b 6s c + 1, 2 since s ≤ b 5d c and d ≥ 8. 3 For s < d + 6 we only have E0q , which provides d − b 2s c + 1 tableaux. This is greater than or equal to b 6s c + 1 provided d ≥ 5, since s ≤ d + 5. When d = 4, then s ≤ b 5d c = 6 so the two E0 q suffice. When d = 3 then s ≤ 5 and hence one tableau, 3 E01 is sufficient. Hence we have constructed at least mλ tableaux as desired. 9.1.3 Basis Tableaux for c = 6, s ≤ r Given λ = [r + s, s] a partition of n, we have 2s + r = 6d = 3b. From this equation and s ≤ r, we have s ≤ 2d. For each λ we will construct mλ linearly independent λ-tableaux filled with the numbers 1 to 6. These will be our c = 6 basis tableaux. When s ≤ r − d, we can use the c = 5 basis tableaux constructed in Section 9.1.2. Consider the partition λ0 = [r0 +s, s] where r0 = r−d. Since s ≤ r0 , we have mλ = mλ0 . In Section 9.1.2 we constructed mλ0 linearly independent Ep , where Ep are the basis tableaux for c = 5. Take Ep ∨ d as the basis tableaux for c = 6. This works for 6 s ≤ r − d, so r ≥ s + d. Hence 6d = 2s + r ≥ 3s + d implies s ≤ d + b 2d c. 3 For d + b 2d c < s ≤ 2d we want to construct mλ ≤ b 6s c + 1 linearly independent 3 tableaux. We will do this primarily by joining two c = 3 basis tableaux. In addition, we will use the following tableaux: 147 d−2 f G = 5 3 46 1 1 22 1 1≤f <d ω(G1 ) = (d − 1, f + 1, 0, 0, 0, 0) d−2 f −2 2 G = 4 5 4 66 1 1 2 23 2 4≤f <d d≥5 ω(G2 ) = (d − 1, f − 1, 2, 0, 0, 0) 2 d−4 d−3 2 G = 6 4 5 646 1 1 2 233 3 d>6 ω(G3 ) = (d − 2, d − 2, 3, 0, 0, 0) d−2 2 d−5 2 4 6 5 56 1 1 2 2 33 G4 = 6 d>7 ω(G4 ) = (d − 1, d − 3, 3, 0, 0, 0) These tableaux are all maximal. Except for G2 , no other weight assignments are possible, hence these tableaux are non-zero. For G2 , the weight assignment (d − 1, f − 1, 0, 0, 0, 2) is also valid. Since this has sign (−1)2 , G2 is also non-zero. Consider b 5d c < s < 2d. (The s = 2d case will be handled separately.) Write 3 s = f + d, then b 2d c < f < d. This means f > 2 always, and f < 6 only for d ≥ 7. 3 Moreover f ≥ 4 for d ≥ 5 so the conditions on G2 are satisfied. We will use the c = 3 basis tableaux for much of our construction. Let Dp0 represent the basis elements for c = 3 of shape [2d, d] as described in Section 4.3. Let Fp represent the basis elements for c = 3 of shape [3d−f, f ] as described in Section 4.3. We will use the Dp0 , Fp , and the Gi to construct the mλ linearly independent tableaux 148 for c = 6. This construction will depend on d and f mod 6. Write d = 6k 0 + i0 and f = 6k + i, with 0 ≤ i0 , i ≤ 5. For i0 6= 1 there exist m[2d,d] = k 0 + 1 linearly independent Dp0 , with p0 = 0, 1, . . . , k 0 , and k 0 tableaux for i0 = 1. Similarly, there exists m[3d−f,f ] = k + 1 linearly independent Fp with p = 0, 1, . . . , k for i 6= 1 and k when i = 1. Now s = d + f = 6(k 0 + k) + i0 + i = 6k + i, so mλ ≤ k + 1. Since k ≤ k 0 + k + 1 , it suffices to construct k 0 + k + 2 linearly independent tableaux. We will first consider d > 6, f ≥ 6. Since k 0 , k > 0 consider: G p = D k 0 ∨ Fp p = 0, 1, . . . , k Gp0 = Dp0 ∨ Fk p0 = 0, 1, . . . , k 0 − 1 G 0 = D 0 ∨ F0 There are k + 1 + k 0 + 1 = k + k 0 + 2 tableaux listed here. By Lemma 3.4.12 if their max weights are distinct, these tableaux are linearly independent. Since these tableaux are in maximal form we can simply compare their weights. We have: ω(Gp ) = (d, 0, 0, 2p + i + 4k, 2(k − p), 0) p = 0, 1, . . . , k ω(Gp0 ) = (2p0 + i0 + 4k 0 , 2(k 0 − p0 ), 0, f, 0, 0) p0 = 0, 1, . . . , k 0 − 1 ω(G0 ) = (4k 0 + i0 , 2k 0 , 0, 4k + i, 2k, 0) Now d > f and p0 ≤ k 0 − 1, so we have ω(Gp ) 6= ω(Gp0 ) since ω(Gp0 ) does not contain a weight of d. Both of these weights are distinct from ω(G0 ), since they each contain at least three 0’s while ω(G0 ) contains only two 0’s. The weights within each of these collections of tableaux are distinct because each collection {Dp0 } and {Fp } are linearly independent by Section 4.3. When all these tableaux G exist we have a set of basis tableaux for c = 6. However, depending on the conditions on d and f , we may only have k 0 or k basis tableaux to work with. In those situtations we will need to use the appropriate Gi to complete our set of tableaux. Recall that we are taking d > 6 and that f ≥ 6. When d 6≡ 1 (mod 6) and f 6≡ 1 (mod 6), all the tableaux G exist. Hence we have 149 the k 0 + k + 2 linearly independent tableaux required. When d 6≡ 1 (mod 6), f ≡ 1 (mod 6), there exist only k linearly independent Fp with p = 1, 2, . . . , k. Hence we have all the tableaux listed above except for those with p = 0 and G0 . In place of those basis elements, use G1 and G2 . This provided k + k 0 +2 tableaux. They are linearly independent by Lemma 3.4.12 provided the weights of G1 and G2 are distinct from the weights of Gp . We have ω(G1 ) = (d − 1, f + 1, 0, 0, 0, 0) and ω(G2 ) = (d − 1, f − 1, 2, 0, 0, 0). Clearly these weights are distinct from eachother. Now ω(G1 ) 6= ω(Gp0 ) since the number of 0’s differ. If ω(G1 ) = ω(Gp ) we must have f + 1 = d and p = k. Similarly ω(Gp0 ) will be distinct from ω(G2 ) unless d − 1 = f , p0 = k 0 − 1. We have ω(G2 ) distinct from ω(Gp ) since d does not occur in its weight. Hence for f 6= d − 1 these k + k 0 + 2 tableaux are linearly independent. When f = d − 1, use G3 and G4 in place of G1 and G2 . Since ω(G3 ) contains neither d nor f it is clearly distinct from the weights of Gp . Similarly, ω(G4 ) is distinct from ω(Gp ) since it does not contain d. While ω(G4 ) does contain f , Gp0 cannot contain the weight 3 since d > 7. Thus we have sufficient linearly independent tableaux. Now consider d ≡ 1 (mod 6) and f 6≡ 1 (mod 6). We have the Gp and Gp0 listed earlier, for p0 6= 0, along with G1 and G2 . The discussion in the f ≡ 1 (mod 6) case above shows these are linearly independent provided f 6= d − 1. Similarly when f = d − 1 we can replace G1 and G2 with G3 and G4 . If d = 7 then G4 does not exist. However, then s = 13 on only two tableaux, D1 ∨ F1 and G3 , are needed. When d ≡ 1 (mod 6) and f ≡ 1 (mod 6), we can write d = 6k 0 + 1, f = 6k + 1. Then s = 6(k 0 + k) + 2 so k 0 + k + 1 tableaux suffice. Use the Gp and Gp0 listed earlier, for p, p0 6= 0, along with G1 and G2 . This provides the requisite number of tableaux. They are linearly independent by the previous discussion since f 6= d − 1, as d ≡ f (mod 6). Now consider 3 ≤ d ≤ 6. Then f > b 2d c implies f > 2. Since s = d + f , we must 3 have s ≤ 11. Hence two linearly independent tableaux will suffice, namely, D0 ∨ F0 and G1 . As in previous discussions, these weights are distinct provided f 6= d − 1. When f = d − 1, s = 2d − 1. We need two tableaux only when d = 5 or d = 6. Thus 150 D0 ∨ F0 suffices for d = 3 and 4. When d = 5 or 6 use D0 ∨ F0 and T, where 24221 T= 56456 11322 d = 6, ω = (6, 3, 2, 0, 0, 0) 422 T= 5654 1123 d=5 ω = (5, 2, 2, 0, 0, 0) These tableaux are clearly maximal and non-zero. Since f = d − 1 is not a weight of T, we have that T and D0 ∨ F0 are linearly independent. When f < 6 we have d ≤ 7 because f > b 2d c. Since all d ≤ 6 cases were done 3 above, only d = 7 remains. In this case we have only f = 5. Thus s = 12 and three tableaux are required. We can use D1 ∨ F0 , G1 , and G2 . These tableaux are linearly independent by previous discussions. Now consider s = 2d. Write d = 6k 0 + i, so s = 6(2k 0 ) + 2i0 . Hence 2k 0 + 2 linearly independent tableaux will suffice. Let A1 = d−2 d−2 2 4 5 5 46 1 1 2 23 ω(A1 ) = (d − 1, d − 1, 2, 0, 0, 0) This tableau is maximal. Although there are many valid weight assignments, all such assignments have positive sign, hence A1 is non-zero. When d 6≡ 1 (mod 6) we have k 0 + 1 linearly independent tableaux Dp0 . Hence we can use: Dp ∨ Dp p = 0, 1, . . . , k 0 Dk0 ∨ Dp0 p0 = 0, 1, . . . , k 0 − 1 A1 151 with weights: ω(Dp ∨ Dp ) = (2p + i0 + 4k 0 , 2(k 0 − p), 0, 2p + i0 + 4k 0 , 2(k 0 − p), 0) ω(Dk0 ∨ Dp0 ) = (d, 0, 0, 2p + i0 + 4k 0 , 2(k 0 − p), 0) ω(A1 ) = (d − 1, d − 1, 2, 0, 0, 0) This provides 2k 0 + 2 tableaux, provided k 0 > 0. Their weights are clearly distinct so they are linearly independent by Lemma 3.4.12. When d ≡ 1 (mod 6), we have s = 6(2k 0 ) + 2 so only 2k 0 + 1 tableaux are needed. All the tableaux listed above work except for p = 0 and p0 = 0, providing 2k 0 linearly independent tableaux. In addition use: 2 d−3 d−4 2 2 A2 = 5 4 5 664 1 1 2 233 d≥6 ω(A2 ) = (d − 1, d − 2, 3, 0, 0, 0) This tableau is maximal and non-zero. Since d ≡ 1 (mod 6) we have d > 6. Thus A2 provides the additional tableau and its weight is distinct from the other tableaux, insuring linear independence. When k 0 = 0 we have d < 6. Then s < 10 so two tableaux, A1 and D0 ∨ D0 , suffice. 9.1.4 Basis Tableaux for c > 6, s ≤ r Let λ = [r + s, s] be a partition of n, with s ≤ r, where 2s + r = cd = n. We want to construct mλ linearly independent basis tableaux for an arbitrary c. In Sections 4.3, 9.1.1, 9.1.2, and 9.1.3 we constructed basis tableaux for c ≤ 6, which we will make use of in this construction. In addition we will use the following tableaux: 152 d−2 d−2 2 A1 = 4 5 5 46 1 1 2 23 ω(A1 ) = (d − 1, d − 1, 2, 0, 0, 0) 2 d−3 d−4 2 2 A2 = 5 4 5 664 1 1 2 233 d≥6 ω(A2 ) = (d − 1, d − 2, 3, 0, 0, 0) 2 d−4 d−4 2 2 2 A3 = 5 5 6 644 1 2 1 332 d≥6 ω(A3 ) = (d − 2, d − 2, 4, 0, 0, 0) 2 d−4 d−5 2 3 2 4 5 664 1 1 2 233 A4 = 5 d≥5 ω(A4 ) = (d − 2, d − 3, 5, 0, 0, 0) 422 A5 = 4 5 6 123 d=4 ω(A5 ) = (4, 2, 2, 0, 0, 0) These tableaux are all maximal. Although some of these tableaux have additional valid weight assignments, all such assignments have positive sign. Hence these tableaux are non-zero. 153 To construct λ = [r + s, s]-tableaux with s ≤ r, write s = md + f , with 0 ≤ f < d and r = md + pd + g with 0 ≤ g ≤ d. Then 3md + pd + 2f + g = cd, so 2f + g = xd for some x. This means c = 3m + p + x. If p + x ≥ 3, a λ-tableau may be written D ∨ B ∨ (c − 3m − 3)V (d), where D is a tableau of shape [2md, md] filled with 3m elements, B is a [3d − f, f ] tableau filled with 3 elements, and V (d) is the one row tableau. We will first consider this case and handle the p + x < 3 case later. We have s = md + f , so writing d = 6k 0 + i0 , f = 6k + i, with 0 ≤ i0 , i ≤ 5 0 gives s = 6(mk 0 + k) + mi0 + i. Since b mi6+i c ≤ m, it suffices to construct b 6s c + 1 ≤ mk 0 + k + m + 1 linearly independent tableaux. If m < 2 we may simply use the basis tableaux constructed for c = 6 along with V (d)0 s, so assume m ≥ 2. Let Dp be the c = 3 basis tableaux of shape [2d, d] described in Section 4.3. There are m[2d,d] = k 0 + 1 such tableaux when i0 6= 1 and k 0 for i0 = 1. Let Bq be the c = 3 basis tableaux of shape [3d − f, f ] constructed in Section 4.3. This tableau has s ≤ r since f < d. There are m[3d−f,f ] = k 0 + 1 such tableaux for i 6= 1 and k 0 tableaux when i = 1. Take f > 1 and d ≥ 6. The d < 6 and f ≤ 1 cases will be handled separately. Consider the following tableaux forms (with the appropriate number of V (d)’s as necessary): ` = 1, 2, . . . , m I. `Dp ∨ (m − `)Dk0 ∨ Bq p = 0, 1, . . . , k 0 − 1 q = 0, 1, . . . , k II. mDk0 ∨ Bq q = 0, 1, . . . , k ` = 1, 2, . . . , b m2 c III. `Aj ∨ (m − 2`)Dk0 ∨ Bq q = 0, 1, . . . , k j = 1, 2, 3, 4 d ≥ 6 if j 6= 1 154 Note that those tableaux with p = 0 or q = 0 exist only when i0 6= 0 or i 6= 0 respectively. We will not make use of these tableaux unless necessary. However, even when k = 0, at least one Bq exists, so if we regard k as the number of Bq ’s, we may assume k ≥ 1. When d 6≡ 1 (mod 6) we need mk 0 + k + m + 1 tableaux. The list above (taking q 6= 0) provides at least mk 0 k + k + 4b m2 ck tableaux, which suffices since 4b m2 c ≥ m + 1. When d ≡ 1 (mod 6), we need only mk 0 + b m6 c + 2 tableaux. If k 0 ≥ 2 the list above provides at least m(k 0 − 1)k + k + 4b m2 c tableaux. This is sufficient for m 6= 3. When m = 3 we need 3k 0 + k + 2 tableaux when i = 3, 4, or 5 and 3k 0 + k + 1 tableaux for i = 0, 1, and 2. When i ≤ 2 the tableaux listed suffice. For i ≥ 3 we need an additional tableau so use the tableau of the Form II with q = 0. When k 0 = 1, the tableaux of the Form I don’t exist. Hence we have only k + 4b m2 c tableaux when q 6= 0 and we need m + k + b m6 c + 2. For m 6= 3 this is sufficient. However, one additional tableau is needed for m = 3, when i = 3, 4, or 5. In this case we may use q = 0 in Form II for the remaining tableau. To show linear independence of these tableaux it suffices, by Lemma 3.4.12, to show that the max weights are distinct. For max weights we have: I. (4k 0 + 2p + i0 , 2(k − p), 0)` ∨ (d, 0, 0)(m−`) ∨ (4k + 2q + i, 2(k − q), 0) II. (d, 0, 0)m ∨ (4k + 2q + i, 2(k − q), 0) III − 1. (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ (4k + 2q + i, 2(k − q), 0) III − 2. (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ (4k + 2q + i, 2(k − q), 0) III − 3. (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ (4k + 2q + i, 2(k − q), 0) III − 4. (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ (4k + 2q + i, 2(k − q), 0) Consider those tableaux of Form I. If ω(I(`, p, q)) = ω(I(`0 , p0 , q 0 )), then counting the number of d’s shows ` = `0 , while counting the number of 4k 0 + p + i0 ’s indicates ` = 1. If p = p0 then q = q 0 , which is not possible for distinct tableaux. Hence by Lemma 3.4.13, the weights are distinct because f < d. Those of Form II are distinct 155 due to the distinct weights of Bq . Similarly those of the Form III − j are distinct by the number of d’s and distinctness of ω(Bq ). If ω(III − i(`, q)) = ω(III − j(`0 , q 0 )), then by counting the number of d’s we have ` = `0 . Then by counting the number of 0’s, d − 1’s, and d − 2’s, we find the weights must be distinct for d > 6. (If d = 6 the Forms III − 2 and III − 4 have the same weights.) To see the different forms have distinct weights, first count the number of d’s. Obviously ω(II) 6= ω(I) or ω(III). If ω(I(`)) = ω(III(`0 )) then ` = 2`0 . However, counting the number of 0’s shows the weights are distinct. When d = 6 then s ≤ 6m + 5, so m + 1 tableaux suffice. Since there are two D’s and one B, Forms I and II provide the requisite number of linearly independent tableaux. Now consider the case where d < 6. Here s = mi0 + i so m + 1 tableaux suffices. In this case the Aj , j = 2, 3 do not exist. However for d = 5 we have A4 and for d = 4 we have A5 . Moreover, we have exactly one D and one B. Hence the appropriate Forms II, III − 1, and III − 4 or III − 5 provied 2b m2 c + 1 tableaux. This suffices for even m. When m odd we need one additional tableau. For m ≥ 5 use A1 ∨ Ai ∨ (m − 4)D0 ∨ B0 where i = 4 or 5 as appropriate. If m = 3, then s ≤ 19 so the three tableaux listed will suffice except when d = 5 and f = 3. In this case also use the non-zero maximal tableau: 4 4 44 T= 596978 112234 ω(T) = (5, 5, 4, 4, 0, 0, , 0, 0, 0) When d = 3, we have s = 3m + 2 since 1 < f < d. Hence b 3m+2 c + 1 tableaux 6 are needed. We have mD0 ∨ B0 and `A1 ∨ (m − 2`)D0 ∨ B0 , which provide b m2 c + 1 tableaux. Since m ≥ 2 we have b m2 c ≥ b 3m+2 c, so these tableaux suffice. 6 For linear independence, we need only check distinctness of max weights by Lemma 3.4.12. First consider d = 4 or 5. Since ω(A5 ) = ω(A3 ) our discus- 156 sion on linear independence for d ≥ 6 still holds. The only additional tableau is A1 ∨ Ai ∨ (m − 4)D0 ∨ B0 which clearly has a distinct weight. Similarly, by counting the number of 5’s and 4’s, the weight of T is also distinct. Hence these tableaux are linearly independent. When d = 3 the tableaux listed are a subset of the tableaux for d ≥ 6 and hence are linearly independent by the prior discussion. This covers all cases with p + x ≥ 3 provided f > 1. The f = 0 and f = 1 cases will be handled after the p + x < 3 case. Now assume that p + x < 3. Recall that if s = md + f , with 0 ≤ f < d and r = md + pd + g, then 3md + pd + 2f + g = cd and 2f + g = xd for some x. This means c = 3m + p + x, so if p + x < 3 then λ-tableau with s ≤ r may be written as D ∨ F ∨ (c − 3m − 3)V (d). Here D is a tableau of shape [2(m − 1)d, (m − 1)d] filled with 3(m − 1) elements and F is a [(3 + x − 1)d − f, f + d] tableau filled with 3 + x elements. When f > 0, we have x > 0, which means F is a c = 4 or c = 5 tableau. We will first consider the case where m ≥ 3, d ≥ 6 and f > 1. If d = 6k 0 + i0 and f = 6k + i, with 0 ≤ i0 , i ≤ 5, then s = md + f , so we still need b 6s c + 1 ≤ mk 0 + k + m + 1 tableaux. The number of tableaux D is the same as before. Let Fq be the c = 4 or 5 basis tableaux of shape [cd − f − d, f + f ]. There are at least k 0 + k such tableaux. Moreover since d ≥ 6 and f > 1 there is always at least 2 such tableaux. Consider the tableaux of the following forms: ` = 1, 2, . . . , m − 1 I. `Dp ∨ (m − ` − 1)Dk0 ∨ Fq p = 0, 1, . . . , k 0 − 1 q = 1, 2, . . . , k 0 + k II. (m − 1)Dk0 ∨ Fq q = 1, 2, . . . , k 0 + k 157 ` = 1, 2, . . . , b m−1 c 2 III. `Aj ∨ (m − 2` − 1)Dk0 ∨ Fq q = 1, 2, . . . , k 0 + k j = 1, 2, 3, 4 d ≥ 6 if j 6= 1 As before, the tableaux with p = 0 exist only for d 6≡ 1 (mod 6). When d 6≡ 1 (mod 6), the construction above provides (m − 1)k 0 (k 0 + k) + k + k 0 + 4b m−1 c(k 0 + k). 2 Now k 0 + k denotes the number of F’s, which we can assume is at least 2. Then since 4b m−1 c ≥ m + 1 for m 6= 4, this construction provides sufficient tableaux. If m = 4 2 then 4b m−1 c(k 0 + k) ≥ 8, so this construction is sufficient. 2 When d ≡ 1 (mod 6), we need only mk 0 + k + b m6 c + 2 tableaux. For k 0 ≥ 2, we have (m − 1)(k 0 − 1)(k 0 + k) + k + k 0 + 4b m−1 c(k 0 + k) tableaux by the construction 2 above, which is sufficient. If k 0 = 1, the tableaux of Form I do not exist, so we have only k 0 + k + 4b m−1 c(k 0 + k) tableaux. However, since we know that there are always 2 at least two Fq ’s, we have k 0 + k ≥ 2. Thus the tableaux listed are sufficient. When d < 6, the tableaux Aj with j = 2 and 3 do not exist. However we do have A4 for d = 5 and A5 for d = 4. Also, since d < 6, we may no longer assume that there are at least two Fq ’s (unless d = 5 and f = 3 or 4). However there is always at least one. In addition, there is only one D. Under these constraints we have tableaux of the Forms II, III − 1, and III − 4 or III − 5 when d = 5 or 4 respectively. For m ≥ 5 also use A1 ∨ Aj ∨ (m − 5)D0 ∨ F0 . This provides 2b m−1 c + 2 tableaux for m ≥ 5. For 2 d = 4 we need at most b 4m+3 c + 1, hence this this suffices. For d = 5 and f = 2 we 2 need at most b 5m+2 c + 1 which we have. Our construction suffices except for m = 3 6 and 4. When m = 3, then s ≤ 17 and three tableaux, II, III − 1, and III − 4 or III − 5 suffice. For m = 4 we need one additional non-zero maximal tableau, so use d−1 1 d−1 1 d−1 1 d−1 1 2 T= 6 7 8 7 9 12 10 12 11 1 1 2 2 3 3 4 4 5 ω(T) = (d, d, d, d, 2, 0, 0, 0) 158 When d = 5, f ≥ 3 there are two Fq ’s hence the Forms II, III − 1 and III − 4 give 4b m−1 c + 1 tableaux. Since at most b 5m+4 c + 1 tableaux are needed, this suffices. 2 6 When d = 3 we need b 3m+2 c + 1 tableaux. From Forms II and III − 1, we have 6 b m−1 c + 1 tableaux. This is sufficient except for m odd. When m even we need one 2 additional tableau, so use T listed above. Now to consider linear independence. By Lemma 3.4.12 it suffices to show that the max weights are distinct. First consider when the F are c = 4 basis tableaux. The max weights are: I. (4k 0 + 2p + i0 , 2(k 0 − p), 0)` ∨ (d, 0, 0, )m−`−1 ∨ (d − q + 1, f + q − 1, 0, 0) II. (d, 0, 0)m−1 ∨ (d − q + 1, f + q − 1, 0, 0) III − 1. (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f + q − 1, 0, 0) III − 2. (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f + q − 1, 0, 0) III − 3. (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f + q − 1, 0, 0) III − 4. (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f + q − 1, 0, 0) Consider those tableaux of Form I. If ω(I(`, p, q)) = ω(I(`0 , p0 , q 0 ), then counting the number of 0’s shows ` = `0 . By counting the number of different weights we find ` = 1. Now if p = p0 then we must have q = q 0 , which is not possible for distinct tableaux. Hence by Lemma 3.4.13, these weights are distinct. The weights of Form II are distinct by the construction of the F’s in Section 9.1.1. The weights of Form III − j are distinct by counting the number of 0’s and by the distinctness of the F’s. If ω(III − j) = ω(III 0 − i) counting the number of 0’s and different numbers shows ` = `0 = 1. So if the weights are not distinct, then ω(Aj ∨ Fq ) = ω(Ai ∨ Fq0 ), which implies Ai and Aj must have one non-zero element in common. Then by Lemma 3.4.13, these weights are distinct if f > 2. When f = 2 the tableaux have the same length and the lemma does not apply. If f = 2 then ω(III − 3(` = 1, q = 4)) = ω(III − 4(` = 1, q = 3)), though all other weights are 159 distinct. We have ω(I) 6= ω(II), by counting the number of 0’s. Forms II and III are also distinct by counting the number of 0’s. If ω(I) = ω(III 0 ), counting the number of 0’s shows that ` = `0 . Then counting the number of d’s shows that ` = 1, q 0 = 1, and q 6= 1. If the weights are equal, then (4k 0 + 2p + i0 , 2(k 0 − p), d − q + 1, f + q − 1) = (ω(Aj ), f ). Since q ≤ b d−f c, these weights are distinct unless f = 2. When f = 2 2 then ω(I(` = 1, p = k 0 − 1, q = j) = ω(III − j(`0 = 1, q 0 = 1)), j 6= 1. Hence for f > 2 all the tableaux listed are linearly independent. When f = 2 some of the tableaux in our list have the same max weights, and hence may be linearly dependent. If we eliminate these tableaux with duplicate max weights from our list we have (m−1)k 0 (k 0 +k)+k 0 +k+4b m−1 c(k 0 +k)−4 linearly independent 2 0 0 tableaux when d 6≡ 1 (mod 6). We need at most b m(6k 6+i )+2 c+1 ≤ mk 0 +m tableaux, which we have since we may still take k 0 + k ≥ 2. If d ≡ 1 (mod 6) then we have (m − 1)(k 0 − 1)(k 0 + k) + k 0 + k + 4b m−1 c(k 0 + k) − 4 linearly independent tableaux 2 for k 0 > 1. This is sufficient since only mk 0 + b m+2 c + 1 tableaux are needed. When 6 k 0 = 1 those tableaux of Form I don’t exist, so we have k 0 + k + 4b m−1 c(k 0 + k) − 1 2 linearly independent tableaux, which is sufficient. Thus the d ≥ 6, f > 1, m ≥ 3 case is finished for c = 4. When d < 6 the same tableaux are used with the substitution of A5 . However, A5 has the same weight as A3 so the above argument applies. (Note that by the conditions c ≤ 1, so the max weight duplication does not occur.) In addition we of Fq , q ≤ b d−f 2 also use the tableaux A1 ∨ Aj ∨ (m − 5)D0 ∨ F0 , for j = 4 or 5. However, counting the number of 0’s and d’s shows that this tableau is distinct from our previous collection; otherwise ω(A1 ∨ Aj ∨ F0 ) = ω(Ai ∨ Ai ∨ Fq ) which is impossible. In addition, when m = 4, d = 5 or d = 3, m even, we also have the tableau T. Counting the number of 0’s and d’s shows this tableau has a distinct max weight as well. Hence when F is a c = 4 tableau, we have linear independence for f > 1 and m ≥ 3. 160 Now consider the case where F is a c = 5 basis tableau. The c = 5 basis tableaux were constructed in Section 9.1.2. These tableaux have two different types of max weights corresponding to the T1 and T2 tableaux constructions used. Those of form T1 have weights (d, f − q − 1, q + 1, 0, 0) where q = 1, 2, . . . , b f2 c − 2 (when f ≥ 6). Those of the form T2 have weights (d − q + 1, f − 1 + q, 0, 0, 0) with q = 1, 2, d − b d+f2 −2 c. We will refer to the basis tableaux using the T1 tableaux as q-forms and those using T2 as q-forms. Then the max weights for the general tableaux have the following forms: I(q). (4k 0 + 2p + i0 , 2(k 0 − p), 0)` ∨ (d, 0, 0, )m−`−1 ∨ (d, f − q − 1, q + 1, 0, 0) I(q). (4k 0 + 2p + i0 , 2(k 0 − p), 0)` ∨ (d, 0, 0, )m−`−1 ∨ (d − q + 1, f − 1 + q, 0, 0, 0) II(q). (d, 0, 0)m−1 ∨ (d, f − q − 1, q + 1, 0, 0) II(q). (d, 0, 0)m−1 ∨ (d − q + 1, f − 1 + q, 0, 0, 0) III − 1(q). (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d, f − q − 1, q + 1, 0, 0) III − 1(q). (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f − 1 + q, 0, 0, 0) III − 2(q). (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d, f − q − 1, q + 1, 0, 0) III − 2(q). (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f − 1 + q, 0, 0, 0) III − 3(q). (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d, f − q − 1, q + 1, 0, 0) III − 3(q). (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f − 1 + q, 0, 0, 0) III − 4(q). (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d, f − q − 1, q + 1, 0, 0) III − 4(q). (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f − 1 + q, 0, 0, 0) These tableaux are linearly independent by Lemma 3.4.12 provided their max weights are distinct. To show these weights are distinct we will often count the number of 0’s and d’s in each weight. For convenience we will list these values below: Consider those tableaux of Form I(q). If ω(I(`, p, q)) = I(`0 , p0 , q 0 ), then Table 9.1.4 shows that ` = `0 . shows ` = 1. Then counting the number of different numbers If p = p0 then we must have q = q 0 and vice versa, but this 161 Form I(q). # 0’s 2m − ` I(q). 2m − `+ II(q). 2m II(q). 2m + 1 III − j(q). 2m − ` III − j(q). 2m − ` + 1 # d’s m−` m − ` − 1 q 6= 1 m−` q=1 m m − 1 q 6= 1 m q=1 m − 2` m − 2` − 1 q 6= 1 m − 2` q=1 Table 9.1: Weights of 0 or d. can’t happen since the tableaux are different. Then since f < d, the argument of Lemma 3.4.13 shows the weights are distinct. The same reason holds for Form I(q). Table 9.1.4 also shows ω(I(q)) 6= ω(I(q)). The tableaux of Form II are distinct since the max weights of the F are distinct for c = 5 by Section 9.1.2. Moreover, ω(II(q)) 6= ω(II(q)) by Table 9.1.4. The distinctness of max weights for Form III − j(q) or III − j(q) follows from Table 9.1.4 and distinctness of c = 5 basis tableaux max weights. Also from Table 9.1.4 we have ω(III − j(q)) 6= ω(III − j(q)). Now suppose ω(III − j(q, `)) = ω(III − i(q 0 , `0 )). Table 9.1.4 shows that ` = `0 and counting the number of different numbers shows ` = 1. Since i 6= j, the weights of Ai and Aj must have exactly one non-zero weight in common. Thus the argument of Lemma 3.4.13 applies and hence the weights are distinct. Similarly ω(III − j(q, `)) 6= ω(III − i(q 0 , `0 )) for f > 2 by Lemma 3.4.13. (When f = 2, ` = 1 the conditions of Lemma 3.4.13 are not met.) The values of Table 9.1.4 are sufficient to show ω(III − j(q, `)) 6= ω(III − i(q 0 , `0 )). This shows the max weights within each tableau form are distinct for f > 2. Showing that ω(I) 6= ω(II) 6= ω(III) follows directly from Table 9.1.4. When f = 2 our discussion on linear independence holds except for a few tableaux of the Form III(q). Specifically, ω(III − 2(` = 1, q = 3)) = ω(III − 3(` = 1, q = 2)), ω(III − 2(` = 1, q = 4)) = ω(III − 4(` = 1, q = 2)), and ω(III − 3(` = 1, q = 4)) = ω(III −4(` = 1, q = 3)). Thus we have three fewer linearly independent tableaux than 162 we originally calculated. If we eliminate these tableaux with duplicate max weights from our list we have (m−1)k 0 (k 0 +k)+k 0 +k+4b m−1 c(k 0 +k)−3 linearly independent 2 0 0 tableaux when d 6≡ 1 (mod 6). We need at most b m(6k 6+i )+2 c+1 ≤ mk 0 +m tableaux, which we have since we may still take k 0 + k ≥ 2. If d ≡ 1 (mod 6) then we have c(k 0 + k) − 3 linearly independent tableaux (m − 1)(k 0 − 1)(k 0 + k) + k 0 + k + 4b m−1 2 for k 0 > 1. This is sufficient since only mk 0 + b m+2 c + 1 tableaux are needed. When 6 c−3 k 0 = 1 those tableaux of the Form I don’t exist, but we still have 2 + 8b m−1 2 linearly independent tableaux, which is sufficient. Hence we have enough tableaux when f = 2. Thus the d ≥ 6, f > 1, m ≥ 3 case is finished for c = 5. When d < 6 only the q-form tableaux exist for c = 5. These tableaux have the same max weights as the c = 4 tableaux F (with the exception of an extra zero.) Thus by the same argument as in that case, these tableaux are linearly independent. Hence when F is a c = 5 tableau, we have linear independence for f > 1 and m ≥ 3. These constructions assumed m ≥ 3. If m = 1, The c = 4 or c = 5 basis tableaux (joined with sufficient V (d)’s) suffice. However, when m = 2, we need tableaux with c = 7 or 8 elements. This case must be dealt with separately. (We will still assume f > 1.) Tableaux of Form I listed previously (Dp ∨ Fq ) still work for this case. These tableaux are linearly independent by our previous discussion. As before, let d = 6k 0 +i0 , f = 6k+i, with 0 ≤ i0 , i ≤ 5. Since s = 2d+f , we need at most 2k 0 +k+3 tableaux. The tableaux Fq have λ2 = d+f so there are at least k 0 +k such tableaux. So for d 6≡ 1 (mod 6), we have (k 0 +1)(k 0 +k) tableaux. This provides sufficient tableaux unless k 0 = 1, k = 1, or k = 0, k 0 < 2. When k 0 = k = 1, then d + f ≥ 14 so at least three Fq ’s exists. Hence the Form I construction is sufficient. In the remaining cases, computing exactly how many Fq exist and precisely how many tableaux are needed shows the Form I tableaux are sufficient except for: d = 11, f = 2, d = 3, f = 2, d = 4, f = 3, and d = 5, f = 2, 3, 4. (For instance, when d = 6 and f = 5 there are two F’s and two D’s, so Form I provides 4 tableaux. Since s = 17, only three tableaux are needed.) 163 When d = 11 and f = 2, there are two D’s and F’s for a total of 4 Form I tableaux. Since s = 24, five tableaux are required. In addition to the Form I tableaux, use 10 10 4 T= 4 5 6 1 2 3 ω(T) = (10, 10, 4, 0, 0, 0) which is non-zero, maximal, and linearly independent. For d = 3, f = 2, we have c = 8. Since s = 8 two tableaux are needed. In addition to D0 ∨ F0 , use the non-zero maximal tableau: 2222 T= 5678 1234 ω(T) = (2, 2, 2, 2, 0, 0, 0, 0) When d = 4 and f = 3, then we must have c = 9 as s ≤ r. Since s = 11, we need two tableaux. Use D0 ∨ F0 and 2 2 2 2 T= 5969788 1122334 ω(T ) = (3, 3, 3, 2, 0, 0, 0, 0) Counting the number of 4’s shows these tableaux are linearly independent. When d = 5 we have c = 8 and need three tableaux (except for f = 3 when two tableaux suffice). In addition to D0 ∨ F0 use: 4 4 4 f −2 T= 567 8 123 4 ω(T) = (4, 4, 4, f − 2, 0, 0, 0, 0) 4 f 22 T = 566768 112234 0 f 6= 3 164 ω(T 0 ) = (5, f + 1, 2, 2) These tableaux are non-zero. Linear independence follows by counting the number of 5’s and 2’s. Thus we have sufficient tableaux for d 6≡ 1 (mod 6). When d ≡ 1 (mod 6) we need 2k 0 + k + 2 tableaux. The Form I construction Dp ∨ Fq discussed earlier provides at least k 0 (k 0 + k + 1) linearly independent tableaux for f 6≡ 0 (mod 6). (When f 6≡ 0 (mod 6) there are k 0 + k + 1 tableaux Fq .) This is sufficient for k 0 ≥ 2. When f ≡ 0 (mod 6) Form I provides k 0 (k 0 + k) tableaux, but only 2k 0 + k + 2 tableaux are needed. This is sufficient for k 0 ≥ 2 since f 6= 0. Since d ≥ 3, only d = 7 remains. We need at most 4 tableaux, as s ≤ 20. Consider the tableaux Bp ∨ Cq where the Bp are c = 3 basis tableaux with λ2 = d − 1 and the Cq are c = 4 or c = 5 basis tableaux with λ2 = d + f + 1. There are two tableaux each for Bp and Cq so this construction is sufficient. The max weights for Bp are (6, 0, 0) and (4, 2, 0). The max weights for Cq are (7, f + 1, 0, 0) and (6, f + 2, 0, 0). Hence the weights of our construction are distinct unless f = 5. However when f = 5 only three tableaux are needed and the Form I construction D ∨ Fq provides three in this case. When f = 1 we can proceed as in the p + x < 3 case. We have s = md + 1 and r = md + pd + g. Then 3md + pd + 2f + g = cd, so 2 + g = xd for some x. Hence x = 1. This means that for both the p + x ≥ 3 and p + x < 3 cases we can use the c = 4 basis tableaux, proceeding as in the p + x < 3 case when f > 1. As before, write d = 6k 0 + i0 with 0 ≤ i0 ≤ 5. We have at least k 0 linearly independent λ = [3d − 1, d + 1] tableaux Fq , which we use for the following forms: ` = 1, 2, . . . , m − 1 I. `Dp ∨ (m − ` − 1)Dk0 ∨ Fq p = 0, 1, . . . , k 0 − 1 q = 1, 2, . . . , k 0 II. (m − 1)Dk0 ∨ Fq q = 1, 2, . . . , k 0 165 ` = 1, 2, . . . , b m−1 c 2 III. q = 1, 2, . . . , k 0 `Aj ∨ (m − 2` − 1)Dk0 ∨ Fq j = 1, 2, 3, 4 d ≥ 6 if j 6= 1 These tableaux have the following max weights: I. (4k 0 + 2p + i0 , 2(k 0 − p), 0)` ∨ (d, 0, 0, )m−`−1 ∨ (d − q, q + 1, 0, 0) II. (d, 0, 0)m−1 ∨ (d − q, q + 1, 0, 0) III − 1. (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q, q + 1, 0, 0) III − 2. (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q, q + 1, 0, 0) III − 3. (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q, q + 1, 0, 0) III − 4. (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q, q + 1, 0, 0) Counting the number of zeros and d’s in these weights gives: Form #0’s #d’s I. 2m − ` m−`−1 II. 2m m−1 2m − ` m − 2` − 1 III − j. Then by arguments similar to previous cases and Lemma 3.4.13, these tableaux have distinct max weights except for III − 1(` = 1, q = 2) and III − 2(` = 1, q = 1) whose weights are the same. Thus by Lemma 3.4.12, these tableaux (omitting III − 1(` = 1, q = 2)) are linearly independent. When d ≥ 6 and m ≥ 3, we need mk 0 + b 5m+1 c + 1 tableaux. When k 0 ≥ 2, 6 the forms listed above provided (m − 1)k 0 k 0 + k 0 + 4b m−1 ck 0 − 1, which is sufficient. 2 When k 0 = 1 and d 6≡ 1 (mod 6), we still have these tableaux. If d = 7, then there is 166 exactly one D so Form I does not exist. However, there are two distinct Fq ’s. Thus Forms II and III provide 8b m−1 c tableaux. Since only m + b m+1 c + 1 tableaux are 2 6 needed, this suffices. If m = 2 then only Form I and II tableaux exist. Since s = 2d + 1, then 2k 0 + 2 tableaux suffice. Forms I and II provided at least (k 0 − 1)k 0 + k 0 , tableaux. When k 0 ≥ 2 this is enough. For d > 7, four tableaux are needed. Since there are at least two D’s and two F’s, this construction suffices. When d = 6, there are two D’s, but only one F. However, s = 13 so only two tableaux are needed, which we have. If d = 7, three tableaux are needed. Since there is only one D and two F’s, an additional tableau is required. Use: 66 2 H= 4567 ω(H) = (6, 6, 3, 0, 0, 0, 0) 1233 This tableau is clearly non-zero and maximal and its max weight does not contain d, so it is linearly independent from our previously constructed tableaux. When d = 5, there are two distinct Fq ’s, which have weights of the form listed above. However, there is only one D. Thus Forms II, III − 1, and III − 4 provide c tableaux. Since only b 5m+1 c + 1 tableaux are needed, this suffices at least 1 + 4b m−1 2 6 for m ≥ 3. When m = 2 then s = 11, so the two tableaux of Form II are sufficient. When d = 3 there is only one F and one D. Moreover, only A1 exists. However, all c + 1 tableaux and we need the tableaux are linearly independent. Thus we have b m−1 2 b 3m+1 c + 1 tableaux. This is sufficient for m odd. When m even then s = 3m + 1 ≡ 1 6 (mod 6) so only b 3m+1 c tableaux are required, which we have when m ≥ 3. If m = 2 6 then s = 7, which means the single tableau of Form II is sufficient. When d = 4 we need to proceed differently since the c = 4 basis tableaux F is not in maximal form. First consider m ≥ 4. We will replace F in the previous discussion 167 with the following c = 7 tableaux, Hq . 42 2 H0 = 4 5 6 7 1223 2 2 ω(H0 ) = (4, 3, 2, 0, 0, 0, 0) 2 H1 = 4 7 5 4 6 5 ω(H0 ) = (3, 3, 3, 0, 0, 0, 0) 112233 The the tableaux we will use are: II. (m − 2)D0 ∨ Hq q = 0, 1 c ` = 1, 2, . . . , b m−2 2 III. `Aj ∨ (m − 2` − 2)D0 ∨ Hq q = 0, 1 j = 1, 5 Counting the number of 0’s and 4’s in the max weights of these tableaux show they are linearly independent by Lemma 3.4.12. This can be seen from Table 9.1.4 below: Form II. III − j. # 0’s 2m 2m 2m − ` 2m − ` # 4’s m−1 m−2 m − 2` − 1 m − 2` − 2 q 0 1 0 1 Table 9.2: Weights of 0 and 4. Hence for m ≥ 4 this construction provides 4b m−2 c + 2 linearly independent 2 tableaux. Since only b 4m+1 c + 1 tableaux are required, this suffices. 6 When m = 3, then s = 13, so only two tableaux are needed. The two tableaux of Form II suffices. Hence we have constructed sufficient linearly independent tableaux when f = 1 and m ≥ 3. 168 For m = 2, two tableaux are needed, in which case use H0 and H1 . Hence all f = 1 cases are accounted for, since m = 1 may be handled by c ≤ 6 basis tableaux and V (d)’s. When f = 0 the construction is similar to earlier cases, particularly the f = 1 case. However, instead of the F tableaux, we use only the c = 3 basis tableaux D of shape [2d, d]. Again, we may take m ≥ 2, since m = 1 may be handled by c ≤ 6 basis tableaux and V (d)’s. Writing d = 6k 0 + i0 , 0 ≤ i0 ≤ 5, our general tableaux are: I. `Dp ∨ (m − `)Dk0 II. mDk0 III. `Aj ∨ (m − 2`)Dk0 p = 0 . . . k 0 − 1, ` = 1 . . . m ` = 1 . . . b m2 c, j = 1 . . . 4, d ≥ 6 These tableaux have max weights: I. (4k 0 + 2p + i0 , 2(k 0 − p), 0)` ∨ (d, 0, 0, )m−` II. (d, 0, 0)m III − 1. (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)m−2` III − 2. (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)m−2` III − 3. (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)m−2` III − 4. (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)m−2` Counting the number of zeros and d’s in these weights gives: Form #0’s #d’s I. 2m − ` m−` II. 2m m 2m − ` m − 2` III − j. 169 Then by arguments similar to previous cases and Lemma 3.4.13, these tableaux have distinct max weights and are linearly independent by Lemma 3.4.12. Consider d ≥ 6. If d 6≡ 1 (mod 6), we need mk 0 + m tableaux. The above construction provides mk 0 + 1 + 4b m2 c tableaux, which is sufficient. When d ≡ 1 (mod 6) we need mk 0 + b m6 c + 1 tableaux. In this case, D0 does not exists, so p > 0. For k 0 6= 1, we have m(k 0 − 1) + 1 + 4b m2 c which is sufficient. For k 0 = 1, we have 4b m2 c + 1 tableaux since Form I is no longer valid. However, this suffices. When d < 6 not all of the Aj exist and there is exactly one D. For d = 4 and d = 5 we have Forms II, III − 1, and III − 5 or III − 4 respectively. This provides 2b m2 c + 1 tableaux. Since at most b 5m c + 1 tableaux are needed, this is sufficient. 6 When d = 3 we have Forms II and III − 1, which yield b m2 c + 1 tableaux. Since b 3m c + 1 tableaux are required, we have enough. This construction work for m ≥ 2, 6 hence all f = 0 cases are accounted for. 9.2 Case: r < s This proof of this case follows similarly to the s ≤ r case. Let λ = [r + s, s] be a partition of n with r < s, where n = 3b = cd. We wish to construct mλ linearly independent tableaux, where mλ is the multiplicity of χλ in 1SSnb oS3 as described in Theorem 11. First we will construct these tableaux for c = 4, 5, and 6; then we will use these constructions in proving Theorem 3 for a general c. We will refer to the tableaux constructed in this way as basis tableaux. These constructions will make use of the c = 3 basis tableaux constructed in Section 4.3 as well. 9.2.1 Basis Tableaux for c = 4, r < s Given λ = [r + s, s], a partition of n, we have 2s + r = 4d = 3b. From this equation and r < s, we have d + b d3 c < s ≤ 2d. For each λ we will construct mλ linearly independent λ-tableaux filled with the numbers 1 to 4. These will be our c = 4 basis tableaux. 170 When d + b d3 c ≤ s ≤ 2d, write s = d + f , with b d3 c ≤ f ≤ d. Consider the tableau T from Section 9.1.1. This tableau is non-zero and maximal. a+b≤d a, b, c, d > 0 abcd T (a, b, c, d) = 4 3 4 3 1 1 2 2 a>d b>c or a = d, b = c w(T ) = (a + b, c + d, 0, 0) Let Cp = T (d d2 e − p, b d2 c − p, b f2 c + p, d f2 e + p) for 0 ≤ p ≤ b d−f c. The constraints 4 on T are satisfied for all p provided f ≥ 2 when p = 0. Hence Cp is non-zero for these c−1. parameters. Let C0p = T (d d2 e−p+1, b d2 c−p, b f2 c+p, d f2 e+p−1), with 0 ≤ p ≤ b d−f 4 Here the constraints on T are satisfied providedand d − f ≥ 4 and f ≥ 4 when p = 0. Hence C0p is non-zero for these parameters. We have ω(Cp ) = (d − 2p, f + 2p, 0, 0) and ω(C0p ) = (d − 2p + 1, f + 2p − 1, 0, 0). Thus the weights are distinct and Lemma 3.4.12 the tableaux are linearly independent. c+1 ≤ By Theorem 11, mλ = b 6r c+1. Hence it suffices to construct b 6r c+1 ≤ b 2d−2f 6 d−f +1 linearly independent tableaux since r = 4d−2s and s = d+f . If 4 ≤ f ≤ d−4 3 then Cp and C0p together provide 2b d−f c + 1 linearly independent tableaux. Since 4 c ≥ b d−f c, this is sufficient. When f > d − 4 we have r ∈ {0, 2, 4, 6} with s 2b d−f 4 3 even for r = 0, 4 and odd for r = 2, 6. Therefor, at most one tableau is needed when f > d − 4, which is C0 . Now consider f < 4. Since f ≥ b d3 c, when d ≥ 3, only f = 2 and f = 3 remain. For d = 3, then f = 1 and we have r = s which was done in Section 9.1.1. If f = 2, then d ≤ 6. However, by Theorem 11, mλ = 0 for d = 3. When d = 4 or 5, mλ = 1 and hence C0 suffices. For d = 6, the tableaux C0 and C1 suffice. If f = 3, then d ≤ 9. For d = 9 we have mλ = 3, for d = 7, we have mλ = 2, while mλ ≤ 1 for the remaining d ≤ 8. Now C0 suffices for those cases with d 6= 7 or 9. When d = 7 we need an additional tableau, however, C1 exists. For d = 9, then 171 r = s and so this case was done in Section 9.1.1. Hence we have sufficient c = 4 basis tableaux for all partitions with r < s. 9.2.2 Basis Tableaux for c = 5, r < s Given λ = [r +s, s] a partition of n, we have 2s+r = 5d = 3b. From this equation and r < s, we have 0 ≤ r ≤ b 5d c. For each λ we will construct mλ linearly independent 3 λ-tableaux filled with the numbers 1 to 5. These will be our c = 5 basis tableaux. First consider 0 ≤ r ≤ d. If d is even, then s−d ≡ s (mod 2). Moreover, s−d ≥ r. Hence if λ0 = [r + s − d, s − d], then there are mλ0 linearly independent c = 3 basis tableaux and mλ0 = mλ since λ1 = λ2 = λ01 − λ02 . Therefore we can use U1 (d) ∨ T , were T are the c = 3 basis tableaux of shape λ0 , as the c = 5 basis tableaux. If d is odd, then s ≡ d − 1 (mod 2) and s − d + 1 ≥ r. Let T be the c = 3 basis tableaux of shape λ0 = [r + s − d − 1, s − d + 1]. Consider the tableaux U1 (d − 1) ∨ T . There are mλ0 such tableaux. Now mλ = mλ0 when r 6≡ 0, 3 (mod 6) since then λ1 − λ2 − 2 = λ01 − λ02 . Thus for r 6≡ 0, 3 (mod 6) we have constructed sufficient tableaux. However, 2s + r = 5d and d odd implies r is odd, hence only r ≡ 3 (mod 6) remains. In that case, mλ0 = mλ − 1, so only one additional tableau is needed. For d, r > 9 use U1 (d − 3) ∨ B0 where B0 is the c = 3 basis tableau of shape 00 λ = [r + s − d − 6, s − d + 3]. To show that U1 (d − 3) ∨ B0 is linearly independent from the U1 (d − 1) ∨ T it suffices, by Lemma 3.4.12, to show that their max weights are distinct. Since d − 1 is a weight of U1 (d − 1) ∨ T for all T , we need only show that 00 d − 1 is not a weight of U1 (d − 3) ∨ B0 . If we write λ = [r0 + s0 , s0 ] and s0 = 6k + j, 0 ≤ j ≤ 5, then ω(U1 (d − 3) ∨ B0 ) = (d − 3, 4k + j, 2k, 0, 0). If d − 1 is in this weight then that means d − 1 = 4k + j. Since 2s0 + r0 = 3d, we then must have j = 0 and r0 = 3. As r0 = r − 6, this implies r = 9. Thus for r > 9, d > 3 these tableaux are linearly independent. If d = 3, then only r = 3 is needed because r ≤ d. Only one tableau is required, which is U2 (2, 1). When r = 9 we need two tableaux basis tableaux. The U1 (d−3)∨B0 172 we constructed above has weight (d − 1, d − 3, d−1 ) In addition, use 2 1 a d−a−1 b+1 d−b−2 a b T = 43 5 3 4 5 4 11 1 2 2 3 3 a = b d−7 c, b = d d−7 e, d ≥ 11, d odd 4 4 which has weight (d, d − 1, d−7 , 0, 0). For d < 11, only d = 9 is needed since 9 = r ≤ d 2 and d is odd. In that case, s = 2d so use U1 (d − 3) ∨ B0 and d−1 d−1 T = 5 3 5 4 1 1 2 2 Thus we have sufficient tableaux for r ≤ d. Now consider d < r ≤ b 5d c. To construct the c = 5 basis tableaux for these r we 3 will use two different types of tableaux. These are the same tableaux that were used in Section 9.1.2 and hence are non-zero and maximal. 1≤a≤d−2 a b d2 c d d2 e b T1 (a, b) = 4 4 5 4 3 4 5 222 1 1 33 1 ≤ b ≤ b d2 c − 3 a>b s≥d+6 d≥8 w(T1 ) = (d, a + 2, b + 1, 0, 0) a b T2 (a, b) = 5 3 5 4 1122 1 ≤ a, b ≤ d − 1 s≥4 w(T2 ) = (a + 1, b + 1, 0, 0) Since d < r ≤ b 5d c, then 2d > s ≥ d 5d e. Consider Ep = T1 (s − d − 3 − p, p) 3 3 c − 2. For s < 2d − 1, the parameters on T1 are satisfied provided for 1 ≤ p ≤ b s−d 2 d ≥ 8. (When s = 2d − 1, we take p ≤ b s−d c − 3.) This provides b s−d c−2 2 2 173 linearly independent tableaux. We will also use E0q = T2 (d − q, s − 2 − d + q) for 1 ≤ q ≤ d − b s−2 c. The parameters on T2 are satisfied provided s ≥ d + 2 (which 2 holds for d > 3). Together, Ep and E0q provide b s−d c−2+d−b s−2 c linearly independent 2 2 tableaux when d ≥ 8. We need b 6r c + 1 = b 5d−2s c + 1 tableaux. When s ≥ d + 6, we 6 c + 1 ≤ b s−d c − 2 + d − b s−2 c so these tableaux are enough. Since s ≥ d 5d e have b 5d−2s 6 2 2 3 and d ≥ 8, we have s ≥ d + 6. Thus for d ≥ 8, s < 2d − 1 these tableaux suffice When s = 2d − 1, d ≥ 8 we have b s−d c − 3 + d − b s−2 c = b d−1 c − 2. This is greater 2 2 2 than or equal to b 6r c + 1 = b d+2 c + 1 provided d > 8. When d = 8, s = 2d − 1 we 6 have r = 10and only one tableau is needed, which E0q provides. When d = 8 we also have r = 12. In that case two tableaux are needed, which E0q provides. Now consider d < 8 with s ≤ 2d − 1. For d ≤ 5, or d = 6 and r = 8 only one tableau is needed. Here E0q provides this tableau except if d = 3. When d = 3, then r = s which was done in Section 9.1.2. When d = 7 or d = 6 and r = 10 two tableaux are needed. In these cases, E0q suffices. Hence for d < 8 all tableaux are provided. 9.2.3 Basis Tableaux for c = 6, r < s Given λ = [r + s, s] a partition of n, we have 2s + r = 6d = 3b. From this equation and r < s, we have 0 ≤ r < 2d. For each λ we will construct mλ linearly independent λ-tableaux filled with the numbers 1 to 6. These will be our c = 6 basis tableaux. First consider 0 ≤ r ≤ d. If d is even, then s − d ≡ s (mod 2). Moreover, s − d ≥ r. Hence if λ0 = [r + s − d, s − d], then there are at least mλ0 linearly independent c = 4 basis tableaux and mλ0 = mλ since λ1 = λ2 = λ01 − λ02 . Therefore we can use U1 (d) ∨ Cp , were Cp are the c = 4 basis tableaux of shape λ0 , as the c = 6 basis tableaux. If d is odd, then s ≡ d − 1 (mod 2) and s − d + 1 ≥ r. Let Cp be the c = 4 basis tableaux of shape λ0 = [r + s − d − 1, s − d + 1]. Consider the tableaux U1 (d − 1) ∨ Cp . There are mλ0 such tableaux. Now mλ = mλ0 when r 6≡ 0, 3 (mod 6) since then λ1 − λ2 − 2 = λ01 − λ02 . Thus for r 6≡ 0, 3 (mod 6) we have constructed sufficient tableaux. However, 2s + r = 6d implies r is even, hence only r ≡ 0 (mod 6) remains. 174 In that case, mλ0 = mλ − 1, so only one additional tableau is needed. Since 0 ≤ r ≤ d, we have b 5d c ≤ s ≤ 3d. For r 6= 0 we can write s = 2d + f with b d2 c ≤ f ≤ d − 3. 2 Then T = d−2 d−1 f −1 4 63 5 3 6 4 1 11 2 2 3 3 ω(T ) = (d, d, f, 0, 0, 0) provides the additional tableau needed. Since T does not have a weight of d − 1 it must be linearly independent from U1 (d − 1) ∨ Cp by Lemma 3.4.12. When r = 0, then s = 3d. However, since d is odd, then so is s and hence no tableaux are required. Now consider d < r < 2d for arbitrary d. We have 2d < s ≤ b 5d c. Write s = 2d+f 2 for 1 ≤ f ≤ b d2 c. Consider d − p − 1 p + 1 d − 2p p p f Hp = 4 6 5 456 1 1 2 233 0 < p ≤ b d−f c 2 ω(Hp ) = (d, d − p, p + f ) c, Hp are linearly Now Hp is non-zero and maximal. Moreover, for 0 < p ≤ b d−f 2 independent by Lemma 3.4.12. Hence we have b d−f c tableaux. We need b 6r c + 1 = 2 b 2d−2f c + 1. Hence it suffices to show that b d−f c > b d−f c. This holds except for 6 2 3 d = 6, f = 3; d = 5, f = 2; and d = 4, f = 1. For d = 4 and 6, only one tableau is needed, so H1 suffices. When d = 5, f = 2, two tableaux are required. Use H1 and U1 (d − 1) ∨ U1 (d − 1) ∨ U1 (d − 1). Thus all the necessary c = 6 basis tableaux have been constructed. 9.2.4 Basis Tableaux for c > 6, r < s Let λ = [r + s, s] be a partition of n, with r < s, where 2s + r = cd = n. We want to construct mλ ≤ b 6r c + 1 linearly independent basis tableaux for an arbitrary c. First write s = s0 + 2dq such that s0 − 2d < r ≤ s0 . If c − 4q ≥ 3 then a λ-tableau [r + s, s] may be written as T = qU2 (d − 1, 1) ∨ T 0 , where T 0 is a λ0 = [r + s0 , s0 ] tableau filled with c − 4q elements. Since r ≥ s0 in λ0 , we have mλ = mλ0 . Hence it suffices to 175 construct mλ tableaux T 0 . If r = s0 , the tableaux T 0 were constructed in Section 9.1.4. Hence we only need to consider partitions λ = [r + s, s] with s − 2d < r < s and the case where c − 4q < 3. If c − 4q < 3 we may write T = (q − 1)U2 (d − 1, 1) ∨ T ∗ where T ∗ is a λ0 = [r + s0 + d, s0 + d] tableau filled with at most 6 elements. Since at least mλ = mλ0 tableaux T ∗ were constructed for c ≤ 6 in previous sections, no additional construction is needed for this case. To construct λ = [r + s, s] tableaux with s − 2d < r < s, write r = md + f , with 0 ≤ f < d and s = md + pd + g with 0 ≤ g < d. Then 3md + 2pd + f + 2g = cd, so f +2g = xd for some x. This means c = 3m+2p+x. If 2p+x ≥ 3, a λ-tableau may be written D ∨ F, where D is a tableau of shape [2md, md] filled with 3m elements and F is a [pd + g + f, pd + g] tableau filled with 2p + x elements. Given the constraints on r, we have 2p + x ≤ 6. Hence we can use the basis tableaux constructed in Sections 4.3, 9.2.1, 9.2.2, 9.2.3 as the F tableaux. We will first consider this case where 2p + x ≥ 3 and handle the 2p + x < 3 case later. We have r = md + f , so writing d = 6k 0 + i0 , f = 6k + i, with 0 ≤ i0 , i ≤ 5 0 gives r = 6(mk 0 + k) + mi0 + i. Since b mi6+i c ≤ m, it suffices to construct b 6s c + 1 ≤ mk 0 + k + m + 1 linearly independent tableaux. If m ≤ 1 we have c ≤ 9. For c ≤ 6 we have constructed the necessary basis tableaux in previous section. c = 7, 8 and 9 will be handled later. Hence assume m ≥ 2 Let Dp be the c = 3 basis tableaux of shape [2d, d] described in Section 4.3. There are m[2d,d] = k 0 + 1 such tableaux when i0 6= 1 and k 0 for i0 = 1. Let Fq be the c = 2p + x basis tableaux of shape [pd + g + f, pd + g] constructed in Sections 4.3, 9.2.1, 9.2.2, 9.2.3. At least one such tableaux will always exist, provided f 6= 0, 1, 2, 4. There are at least m[pd+g+f,pd+g] ≤ k 0 such tableaux, given the constraints on f . Take f 6= 0, 1, 2, 4 and d ≥ 6. The d < 6 and f cases will be handled separately. Consider the following tableaux forms (where the Ai were defined in Section 9.1.4). By Lemma 3.4.12, these tableaux are linearly independent provided their max weights are distinct. The max weights of these tableaux are: 176 I. `Dp ∨ (m − `)Dk0 ∨ Fq ` = 1, 2, . . . , m p = 0, 1, . . . , k 0 − 1 q = 1, 2, . . . , k II. mDk0 ∨ Fq q = 1, 2, . . . , k `Aj ∨ (m − 2`)Dk0 ∨ Fq ` = 1, 2, . . . , b m2 c q = 1, 2, . . . , k j = 1, 2, 3, 4 d ≥ 6 if j 6= 1 III. Table 9.3: Tableaux Forms. I. (4k 0 + 2p + i0 , 2(k 0 − p), 0)` ∨ (d, 0, 0)(m−`) ∨ ω(Fq ) II. (d, 0, 0)m ∨ ω(Fq ) III − 1. (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ ω(Fq ) III − 2. (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ ω(Fq ) III − 3. (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ ω(Fq ) III − 4. (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ ω(Fq ) The weights of Fq depend on which basis tableaux we are using for F. We categorize these weights by the number of elements in F. 177 c ω(Fq ) Range Conditions 3 (4h + 2q + j, 2(h − q), 0) q = 1, . . . k 0 s = 6h + j 4 (d − q, d−f 2 + q, 0, 0) q = 0, . . . , 2b d+f 8 c 5 (d, 4h + 2q + j, 2(h − q), 0, 0) q = 1, . . . , k 0 s − d = 6h + j, r ≤ d, d even (d − 1, 4h + 2q + j, 2(h − q), 0, 0) q = 1, . . . , k 0 s − d + 1 = 6h + j, d odd (d − 3, 4h + j, 2h, 0, 0) r ≤ d.r > 9if r ≡ 3 (mod 6) (d − 1, d − 3, d−1 2 , 0, 0) r = 9, d ≥ 11, d odd (d, d − 1, d−7 2 , 0, 0) (d, d, 0, 0, 0) r = 9, d = 9 (d − 1, d − 3, d−1 2 , 0, 0) 6 (d, d − f2 − q − 1, q + 1, 0, 0) q = 1, . . . , b 2d−f 4 c−2 (d − q 0 + 1, d − f2 − 1 + q 0 , 0, 0, 0) q 0 = 1, . . . , b f4 c + 1 (d, d − q, d−f 2 + q, 0, 0, 0) q = 0, . . . , 2b d+f 8 c r ≤ d, d even (d − 1, d − q, d−f 2 + q + 1, 0, 0, 0) q = 0, . . . , 2b d+f 8 c r ≤ d, d odd (d, d, d−f 2 , 0, 0, 0) r>d r ≡ 0 (mod 6) (d − 1, d − q, d−f 2 + q + 1, 0, 0, 0) q = 0, . . . , 2b d+f 8 c r ≤ d, d odd (d, d − q, q + d−f 2 , 0, 0, 0) q = 1, . . . , b d+f 4 c r>d The weights of Forms I, II, III − j are distinct for each Fq listed provided d > 7, except for the following cases. When F is a c = 5 tableau, we have ω(I(` = 1, p = k 0 − 1, F1 )) = ω(III − 4(` = 1, F2 )) when d = 11, f = 9. When F is c = 4 or c = 6, we have some duplicate tableau weights if f = d − 4 or f = d − 2. These weights are ω(I(` = 1, p = k 0 − 1, q = j − 1)) = ω(III − j(` = 1, q = 0)) for j = 2, 3, and 4. Also, when f = d − 4, we have ω(III − 2(` = 1, q = 2)) = ω(III − 3(` = 1, q = 1)), ω(III − 2(` = 1, q = 3)) = ω(III − 4(` = 1, q = 1)), and ω(III − 3(` = 1, q = 3)) = ω(III − 4(` = 1, q = 2)). When f = d − 2 we have ω(III − 1(` = 1, q = 2)) = ω(III − 2(` = 1, q = 1)) as well. In these cases we have (at most 6) fewer linearly independent tableaux available than listed. These will be called the constrained cases. The linear independence of the remaining tableaux can be seen by counting the number of d’s, 0’s in each tableau, determining the number of distinct elements in each tableau, and 178 applying Lemma 3.4.13 where appropriate. When d = 6, the Forms III − 2 and III − 4 have the same weight. When d = 7 the Forms III − 3 and III − 4 have the same weight. Hence for d = 6 and 7, Form III − 4 will not be used. We wish to have b 6r c + 1 linearly independent tableaux. Recall that r = md + f , d = 6k 0 + i0 and f = 6k + i, so it suffices to construct mk 0 + k + m + 1 linearly independent tableaux. We have f 6= 0, 1, 2, 4, d ≥ 6 and m ≥ 2. Note that there is always at least one F since f 6= 0, 1, 2, 4, so we will take k ≥ 1. When d > 6, d 6≡ 1 (mod 6), all the tableaux listed in Table 9.2.4 exist and are linearly 0 independent. This provides mk 0 k + k + 4b m 2 ck, which is at least mk + k + m + 1. Hence sufficient linearly independent tableaux exist. In the constrained cases, using the full set of F listed (as opposed to only the first k 0 ) will provide sufficient tableaux. When d = 6, the tableaux of Forms III − 2 and III − 4 are not linearly independent. Hence we have mk+k+3b m 2 ck linearly independent tableaux. Then r = 6m+f , so m+k+1 tableaux are sufficient. Since we have k = 1, we have listed sufficient tableaux. When d ≡ 1 (mod 6) the tableau D0 does not exist. Hence Table 9.2.4 provides m(k 0 − 0 0 1)k+k+4b m 2 ck linearly independent tableaux, provided k > 1. In this case r = 6k m+m+f , so mk 0 + k + b m 6 c + 2 tableaux suffices. Thus we have enough tableaux unless m = 3, k = 1. In that case, specifically checking the number of tableaux needed and the number of Fq that exist, shows this construction is sufficient. When k 0 = 1, there are no tableaux of Form I and we do not use Form III − 4. Hence we have 1 + 3b m 2 c tableaux, which is sufficient except in the following cases, where an additional tableau is needed If m = 7, f = 5, c = 26 use 6U1 (6) ∨ U1 (4), if m = 3, f = 3, c = 12 use 5U1 (6), if m = 3, f = 5, c = 14 use 6U1 (6), if m = 3, f = 6, c = 13 use 4U1 (6) ∨ U4 (3, 2), and if m = 3, f = 6, c = 15 use 6U1 (6). In the constrained cases, using the full set of F listed (as opposed to only the first k 0 ) will provide sufficient tableaux. Hence for d ≥ 6, f 6= 0, 1, 2, 4, 2p + x ≥ 3, we have constructed the requisite number of linearly independent tableaux. We will consider the f = 0, 1, 2, 4 case after doing the 2p + x < 3 case. When 2p + x < 3, the procedure described above require Fq to have fewer than three elements. In that case we use m − 1 in place of m in our construction and take Fq to be basis tableaux filled with 2p + x + 3 elements with shape [pd + 2d + g + f, pd + d + g]. There 179 are at least k 0 + k such tableaux, though k may equal 0. Take d ≥ 6, 2p + x < 3, m ≥ 3 Hence from the tableaux of Table 9.2.4 we get Table 9.2.4. I. `Dp ∨ (m − ` − 1)Dk0 ∨ Fq ` = 1, 2, . . . , m − 1 p = 0, 1, . . . , k 0 − 1 q = 1, 2, . . . , k 0 + k II. (m − 1)Dk0 ∨ Fq q = 1, 2, . . . , k 0 + k `Aj ∨ (m − 2` − 1)Dk0 ∨ Fq ` = 1, 2, . . . , b m−1 c 2 q = 1, 2, . . . , k 0 + k j = 1, 2, 3, 4 d ≥ 6 if j 6= 1 III. Table 9.4: Tableaux Forms As before, these tableaux are linearly independent, except in the constrained cases where the same weight equalities occur. We do not use Form III − 4 when d = 6 or 7. For the tableaux of Table 9.2.4, we require m ≥ 3 for tableaux of Form III to exist. Since we still have r = md + f , we want mk 0 + k + m + 1 linearly independent tableaux, when d 6≡ 1 (mod 6). When d > 6, d 6≡ 1 (mod 6), this construction provides (m − 1)k 0 (k + k 0 ) + k + 0 0 k 0 + 4b m−1 2 c(k + k ) linearly independent tableaux. This is larger than mk + k + m + 1, except when m = 4, k 0 = 1, k = 0. However, computing precisely how many Fq exist in this case and the minimum number of tableaux required, shows these tableaux suffice. In the constrained cases, using the full set of F listed (as opposed to only the first k 0 + k) will provide sufficient tableau. When d = 6, Forms III −2 and III −4 are the same so we have only (m−1)+1+3b m−1 2 c linearly independent tableaux. Since r = 6m + f , f < d, we need m + 1 tableaux, which we have. In the f = d − 4 and f = d − 2 cases using the full set of F listed will provide sufficient tableau. If d ≡ 1 (mod 6), the tableau D0 does not exist. Hence Table 9.2.4 provides (m−1)(k 0 − 0 0 1)(k + k 0 ) + k + k 0 + 4b m−1 2 c(k + k ) linearly independent tableaux when k > 1. In this case r = 6m + m + f , so mk 0 + k + b m 6 c + 2 tableaux suffice, which we have. In the constrained cases, using the full set of F listed will provide sufficient tableau. When k 0 = 1, there are 180 no tableaux of Form I or III − 4. Hence we have k + 1 + 3b m−1 2 c(k + 1). This is at least as large as mk 0 + k + b m 6 c + 2 unless m ≤ 8. However, computing precisely how many Fq exist in this case and the minimum number of tableaux required, shows these tableaux suffice in most cases. When m = 6, f = 3, c = 19 we need an additional tableau; use 6U1 (6) ∨ U1 (2). When m = 4, f = 3 we also need an additional tableau; use U4 (3, 2) ∨ 7U1 (6) ∨ U1 (4). In the constrained cases, using the full set of F listed will provide sufficient tableau. Thus for d ≥ 6, 2p + x < 3, m > 2, all necessary tableaux have been constructed. When 2p + x < 3 and m = 2, we find that c = 7 or c = 8. In these cases, consider the tableaux of Forms I and II in Table 9.2.4. We have r = 2d + f , so 2k 0 + k + 3 tableaux are sufficient. For d 6≡ 1 (mod 6), we have at least (k 0 + 1)(k 0 + k) linearly independent tableaux of Forms I and II. When this isn’t enough, checking precisely how many tableaux are needed and how many tableaux of Form I exist, shows that this construction is sufficient except for the following cases. When c = 8, d = 8, f = 2 three tableaux are needed; use 2U1 (8) ∨ U4 (6, 1), U1 (8) ∨ U3 (2, 3) ∨ U4 (6, 1), and U1 (8) ∨ U3 (2, 4) ∨ U4 (4, 1). When c = 8, d = 11, f = 2 five tableaux are needed. They are 4U1 (8), U2 (10, 1)∨U1 (10), U2 (10, 1)∨∨U1 (8)∨U1 (2), U2 (10, 1)∨U1 (6)∨ U1 (4), and U2 (10, 1) ∨ U2 (2, 3). When c = 7, if d = 8 and f = 2, three tableaux are needed, but Form I provides only two. In addition to those tableaux, use P3 (0, 3, 2) ∨ U1 (8) ∨ U1 (2). If c = 7, d = 9 and f = 1, three tableaux are needed, but Form I provides only two. In addition to those tableaux, use U1 (8) ∨ U1 (8) ∨ U1 (6). If d = 10 and f = 0, three tableaux are needed, but Form I provides only two. In addition to those tableaux, use P3 (0, 4, 2) ∨ U1 (10) ∨ U1 (4). When d ≡ 1 (mod 6), it is sufficient to construct 2k 0 + k + 2 tableaux. We have at least (k 0 )(k 0 + k) linearly independent tableaux of Forms I and II. When this is less than 2k 0 + k + 2, checking precisely how many tableaux are needed and how many tableaux of Forms I and II exist, shows that this construction is sufficient except for the following cases. When c = 8, d = 7, f = 4 we need three tableaux; use U2 (6, 1) ∨ U4 (4, 1), 2U1 (6) ∨ U4 (6, 1), and U1 (6) ∨ U3 (2, 3) ∨ U4 (4, 1). When d = 7, f = 2 three tableaux are needed; use U2 (6, 1) ∨ U1 (6), U2 (6, 1) ∨ U1 (4) ∨ U1 (2), and 3U1 (6) ∨ U1 (2). When d = 7, and f = 0, two tableaux are needed, but Form II provides only one. In addition to that tableau, use P3 (0, 3, 2) ∨ 2U1 (6). 181 For c = 7 if d = 13 and f = 1, five tableaux are needed, but Forms I and II provide only four. In addition to those tableaux, use U1 (12) ∨ U1 (12) ∨ U1 (8). If d = 7 and f = 1, three tableaux are needed, but Form I provides only two. In addition to those tableaux, use P3 (0, 3, 2) ∨ U1 (6) ∨ U1 (2). This completes the cases of c = 7 and c = 8 for d ≥ 6. Now consider c = 7 and c = 8 when d < 6. When c = 7, d = 5 we need two tableaux of shapes [23, 12] and [22, 13], and one tableau of shape [21, 14]. Use 3U1 (4) and U2 (4, 1) ∨ U1 (2), for shape [23, 12]; U2 (4, 1) ∨ U4 (2, 1) and U4 (3, 2) ∨ 2U1 (4) for shape [22, 13]; and U2 (4, 1) ∨ U1 (4) for shape [21, 14]. When c = 7an d = 4 we need two tableaux of shape [18, 10], 2U1 (4) ∨ U1 (2) and U2 (2, 1) ∨ U1 (4), and one tableau of shape [17, 11], U4 (2, 1) ∨ 2U1 (4). When c = 7 and d = 3, we need only one tableau of shape [13, 8], U2 (2, 1) ∨ U1 (2). When c = 8 and d = 5, three tableaux are needed for shape [26, 14], one tableau for shape [25, 15], and two tableaux for [24, 16]. Use 3U1 (4) ∨ U1 (2), U2 (4, 1) ∨ U1 (4), and U2 (2, 1) ∨ 2U1 (4) for shape [26, 14]. For shape [25, 15] use U2 (4, 1) ∨ U4 (3, 2), while for shape [24, 16] use 4U1 (4) and U2 (4, 1) ∨ U1 (4) ∨ U1 (2). When c = 8 and d = 4 we need one tableau for shapes [21, 11] and [19, 13] and two tableaux for shape [20, 12]. In the first case, use U4 (2, 1) ∨ U2 (3, 1) and U1 (4) ∨ U4 (2, 1) ∨ P4 (0, 2, 2), respectively. For [20, 12] use 3U1 (4) and 2U2 (2, 1). When c = 8 and d = 3 we need only one tableau of shape [15, 9], which is U2 (2, 1) ∨ U4 (2, 1). This completes the m = 2, 2p + x < 3 case. Now consider d ≥ 6, 2p + x ≥ 3, f = 0, 1, 2, 4. In this case we cannot guarantee that the tableaux Fq of shape [2dp + g + f, 2dp + g] exists. However, since d ≥ 6, tableaux F of shape [pd + d + g + f, pd + d + g] will exist. Moreover, since s < r + 2d, this shape is fillable with c ≤ 6 elements. Hence we can simply use the tableaux constructed in the 2p + x < 3 case with this F. Since we did not apply any restrictions of f in that case, those computations hold. Now take d < 6. We have r = dm + f , f < d, s − 2d < r < s. Since d is small, only a few s are possible for each r. We will consider each case according to the value of d. If d = 3, we have r = 3m + f , s = 3m + 3 + f , and f < 3. First consider f = 0 or 2. If m = 2 we have the shapes [15, 9] and [19, 11]. In both cases, only one tableau is needed. Use 182 U2 (2, 1) ∨ U4 (2, 1) and U2 (2, 1) ∨ U4 (2, 1) ∨ U1 (2). Now take m ≥ 3. Let F be the tableau of shape [9 + 2f, 6 + f ]. Since f = 0 or 2, one such F always exists. Then consider the tableaux of Form II and III − 1 in Table 9.2.4. They have weight (2, 2, 2, 0, 0, 0)` ∨ (3, 0, 0)m−2`−1 ∨ ω(F) for ` = 0, 1, . . . , b m−1 2 c and hence are linearly independent by Lemma 3.4.12. This 3m+2 construction provides b m−1 2 c + 1 tableaux. Since we need at most b 6 c + 1 tableaux for m odd and b 3m+2 6 c for m even, this suffices. When f = 1, let F = U2 (2, 1)∨U1 (2)∨U1 (2). If m = 2 or 3, one tableau will suffice. In those cases use F and F ∨ U4 (2, 1). When m ≥ 4 we m will use the tableaux `A1 ∨ (m − 2` − 2)U4 (2, 1) ∨ F for ` = 0, 1, . . . , b m−2 2 c. There are b 2 c such tableaux and they are linearly independent. Since only b m 2 c are needed, this suffices. When d = 4 we have r = 4m + f , s = 4m + h + f , and f < 3. Since s − 8 < r < s, the only possibilities are f = 0, h = 2, 4, 6 and f = 2, h = 1, 3, 5, 7. Let F be a tableau of shape [8 + 2f + h, 4 + f + h]. Since F needs at most 8 elements it has already been constructed. If m = 2, we need one tableau when f = 2, so F suffices. When f = 0, two tableaux are needed. Use U1 (4) ∨ U1 (4) ∨ U1 (2) and U2 (2, 1) ∨ U1 (4); U1 (4) ∨ U1 (4) ∨ U1 (4) and U2 (2, 1) ∨ U1 (4) ∨ U1 (2); or U1 (4) ∨ U1 (4) ∨ U1 (4) ∨ U1 (2) and U2 (2, 1) ∨ U1 (4) ∨ U1 (4); depending on h. For m ≥ 3, use the tableaux of Forms II and III − 1 of Table 9.2.4 along with those of form III − 1, with A5 (given in Section 9.1.4) in place of A1 . This provides 1 + 2b m−1 2 c linearly independent tableaux, which is sufficient. When d = 5 we have r = 5m + f , s = 5m + h + f , with f < 5, 0 < h < 10, and h ≡ f (mod 5). Let F be a tableau of shape [10 + h + 2f, 5 + h + f ]. When m ≥ 3, consider the tableaux of Forms II, III − 1, III − 4 of Table 9.2.4. This provides 1 + 2b m−1 2 c linearly independent tableaux. Computing precisely how many tableaux are needed for each f and h we find that this is sufficient except in the following cases. When f = 1, m = 4 we need one additional tableau. For h = 1 use 4U1 (4) ∨ U1 (2) and for h = 6 use 3U4 (3, 2) ∨ 3 ∨ U1 (4). When f = 2 and h = 2 we need an additional tableau for m = 8, 6, and 4. Use 11U1 (4), 8U1 (4) ∨ U1 (2), and 6U1 (4) respectively. When f = 2 and h = 7, this construction suffices. When f = 3 and h = 3, one additional tableau is needed for m = 6 and m = 4; use 9U1 (4) and 6U1 (4) ∨ U1 (2), respectively. When f = 3 and h = 8 we can take F to be a tableau of shape [14, 11] and use the tableaux of Forms II, III − 1, and III − 4 of Table 9.2.4. This provides 1 + 2b m 2 c tableaux, which suffices except for m = 3. In that case three tableaux are needed; use U2 (4, 1) ∨ 3U1 (2), U2 (4, 1) ∨ 2U4 (2, 1), and U2 (4, 1) ∨ U1 (4) ∨ U1 (2). When f = 4 and h = 9 this construction suffices except when m = 4. In that case we need an additional 183 tableau, so use 3U2 (4, 1) ∨ U4 (3, 2). However, when h = 9 the tableau F has 9 elements. Since we’ve only constructed the basis tableaux for c ≤ 8, use F = U2 (4, 1) ∨ 2U1 (4). When f = 4 and h = 4 we can take F to be a tableau of shape [12, 8] and use the tableaux of Forms II, III − 1, and III − 4 of Table 9.2.4. This provides 1 + 2b m 2 c tableaux, which suffices. Now consider m = 2 for d = 5. The tableaux will have c ≤ 9 elements unless f = 2, h = 7, f = 3, h = 8, or f = 4. The c ≤ 9 will be constructed later. If f = 2, h = 7 we need two tableaux, U2 (4, 1) ∨ U4 (3, 2) ∨ U1 (4) and U2 (4, 1) ∨ U4 (2, 1) ∨ P4 (0, 2, 2). If f = 3, h = 8, two tableaux are required. Let F be a basis tableaux of shape [14, 11]; use A1 ∨ F and A4 ∨ F. If f = 4, h = 9, two tableaux are needed, 2U2 (4, 1) ∨ U4 (2, 1) and U2 (4, 1) ∨ U4 (2, 1) ∨ P4 (0, 2, 2) ∨ U1 (4). If f = 4, h = 4 we need three tableaux. Let F be the basis tableau of shape [12, 8]. Then A4 ∨ F, A1 ∨ F and 2U4 (3, 2) ∨ F provide the requisite tableaux. Thus all necessary tableaux for d = 5 have been constructed. Now consider when m < 2 for arbitrary d. If r < d then since s − 2d < r < s we have c ≤ 6, which has been done. If r = d + f then we must have c = 7, 8 or 9. First consider when c = 7 with r = d + f . We get s = 3d − f2 . For d even use U1 (d) ∨ F where F are the c = 5 tableaux with r = d + f . When d is odd use U1 (d − 1) ∨ F where F are the c = 5 tableaux with r = d + f − 2. This is sufficient unless d ≡ 1 (mod 6), f ≡ 2 (mod 6), d ≡ 3 (mod 6), f ≡ 0 (mod 6), or d ≡ 5 (mod 6), f ≡ 4 (mod 6), in which case we need one additional tableau. For that, use U2 (d − 1, 1) ∨ B where B is a c = 3 basis tableau with s ≤ r. This construction holds for d > 3. When d = 3, only the shape [12, 9] is needed. One tableau, U2 (2, 1) ∨ U4 (2, 1), suffices. Thus the c = 7 case is complete. Now take c = 8. Since s − 2d < r, we have f > b d3 c and s = 3d + d−f 2 . For d even we can d−f use U1 (d) ∨ F where F are the c = 6 basis tableaux of shape [3d − d+f 2 , 2d + 2 ]. When d is d−f odd, use U1 (d − 1) ∨ F where F is the c = 6 tableaux of shape [3d − d+f 2 − 1, 2d + 2 + 1]. This construction suffices unless d + f ≡ 0 (mod 6). (Since d3 < f , d ≡ f (mod 2) and d ≥ 6, such an F always exists.) If d + f ≡ 0 (mod 6) use the tableau of Forms I and d−f II of Table 9.2.4, with F the c = 5 tableaux of shape [2d + d−f 2 + f, 2d + 2 ]. This suffices unless d = 13, f = 11, d ≡ 1 (mod 6), f = 5, or d ≡ 5 (mod 6), f = 7. In the first case, five tableaux are needed; use 2U2 (9, 1), U2 (9, 1) ∨ U1 (12) ∨ U1 (8), 3U1 (12) ∨ U1 (4), 2U1 (12)∨U1 (10)∨U1 (6), U2 (12, 1)∨U1 (12)∨U1 (2). In the f = 5 case use the tableaux 184 Bp ∨ F where Bp are the c = 3 basis tableaux of shape [2d + 1, d − 1] and F is the c = 5 d−f tableau with r = 3. This suffices for d−f 2 even. When 2 odd, an additional tableau, U2 (d − 1, 1) ∨ U1 (d − 1) ∨ U1 ( d−f 2 + 1), is needed. In the f = 7 case use the tableaux Bp ∨ F where Bp are the c = 3 basis tableaux of shape [2d − 1, d + 1] and F are the c = 5 tableaux with r = 9. Now consider when c = 9 with r = d + f . We get s = 4d − f2 . Since s − 2d < r < s we d have 2d 3 < f < 2 and hence this case does not occur. This completes the r < s case. Hence we have proven Theorem 3. 197 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. Doran, IV, On Foulkes’ conjecture, J. Pure Appl. Algebra 130 (1998), no. 1, 85–98. MR 99h:20014 [8] H. O. Foulkes, Concomitants of the quintic and sextic up to degree four in the coefficients of the ground form, J. London Math. Soc. 25 (1950), 205–209. MR 12,236e [9] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.3, 2002, (http://www.gap-system.org). [10] David A. Gay, Characters of the Weyl group of SU (n) on zero weight spaces and centralizers of permutation representations, Rocky Mountain J. Math. 6 (1976), no. 3, 449–455. MR 54 #2886 198 [11] Larry C. Grove, Groups and characters, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1997, A Wiley-Interscience Publication. MR 98e:20012 [12] Roger Howe, (GLn , GLm )-duality and symmetric plethysm, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, 85–109 (1988). MR 90b:22020 [13] N. F. J. Inglis, R. W. Richardson, and J. Saxl, An explicit model for the complex representations of Sn , Arch. Math. (Basel) 54 (1990), no. 3, 258–259. MR 91d:20017 [14] G. James and A. Kerber, Representation theory of the symmetric group, Encyclopedia of Mathematics and its Applications, vol. 16, Addison-Wesley, Reading, MA, 1981. [15] G. D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, vol. 682, Springer, Berlin, 1978. [16] Serge Lang, Algebra, 3 ed., Addison Wesley, Reading Massachusetts, 1999. [17] I. G. Macdonald, Symmetric functions and Hall polynomials, second ed., Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, New York, 1995, With contributions by A. Zelevinsky, Oxford Science Publications. MR 96h:05207 [18] Bruce E. Sagan, The symmetric group, The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1991, Representations, combinatorial algorithms, and symmetric functions. MR 93f:05102 [19] Richard P. Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. 199 [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 28, 2004) ii c 2004 ° Rebecca Vessenes All Rights Reserved iii To God through whom all things, even this, are possible iv Acknowledgements I wish to thank my advisor, David Wales, for his guidance and support. I also want to thank Cherie Galvez for her help and advice, and the math department staff in general for fostering a friendly environment. In addition, I appreciate the assistance of Rowan Killip in dealing with the intricacies of LATEX and creating specific macros for my use. I enjoyed my time here at Caltech, especially the many long and varied discussions during tea; my thanks to all those who participated. Finally, I am grateful for the love and support of my husband, Ted. I also want to recognize my undergraduate alma mater, the University of Chicago, for providing me with a strong background in mathematics. I particularly wish to thank Diane Herrmann for her support and advice during those years. In addition, I wish to recognize the University of Minnesota Talented Youth Mathematics Program (UMTYMP) for giving me a strong start in mathematics during my adolescent years. v Abstract Foulkes conjectured that for n = ab and a ≤ b, every irreducible module occurring as a constituent in 1SSnb oSa occurs with greater or equal multiplicity in 1SSna oSb . We generalize part of this to say those irreducibles also occur in 1SSnd oSc , where cd = n and c, d ≥ a. We prove the generalized conjecture for a = 2 and a = 3, by explicitly constructing the corresponding tableaux. We also prove the multiplicity constraint for certain cases. For these proofs we develop a theory of construction conditions for tableaux giving rise to Sb o Sa modules and in doing so, completely classify all such tableaux for a = 2 and a = 3. vi Contents Acknowledgements iv Abstract v 1 Introduction and Statement of Main Results 1 Theorem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Theorem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Theorem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 Background 4 2.1 Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.2 Combinatorial Structures . . . . . . . . . . . . . . . . . . . . . . . . . 5 3 Theory of Tableaux Construction 9 3.1 Filling Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3.2 Showing Tableaux are Non-Zero . . . . . . . . . . . . . . . . . . . . . 12 Example: Weight-set Counting . . . . . . . . . . . . . . . . . . . . . 17 3.3 Joining Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.4 Maximal Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 4 The Tableaux of 1SS3b b oS3 31 4.1 Classification of qT 6= 0, for T filled with 1, 2, 3 . . . . . . . . . . . . 31 4.2 The Irreducibles Partitions of 1SS3b . . . . . . . . . . . . . . . . . . . b oS3 41 4.2.1 Non-Zero Partitions from Theorem 9 . . . . . . . . . . . . . . 41 4.2.2 Partition Multiplicities according to Thrall . . . . . . . . . . . 44 vii 4.3 Construction of Basis Tableaux for c = 3 . . . . . . . . . . . . . . . . 45 5 Proof of Theorem 1 52 6 Proof of Theorem 2 58 6.1 6.2 Case: d even . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Example: Tableau Reduction . . . . . . . . . . . . . . . . . . . . . . 61 Case: d odd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 7 Tableaux Construction 7.1 71 Maximality of Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . 71 Example: Maximizing a Tableau . . . . . . . . . . . . . . . . . . . . . 72 7.2 Tableaux for Two Row Partitions . . . . . . . . . . . . . . . . . . . . 75 7.3 Tableaux for c = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 7.4 Tableaux for c = 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 7.5 Tableaux for c = 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 7.6 Tableaux for c = 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 7.7 Tableaux for c = 7 and c = 8 . . . . . . . . . . . . . . . . . . . . . . . 110 8 Tableau Sufficiency 111 8.1 Sufficiency when c = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . 111 8.2 Sufficiency when c = 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 114 8.3 Sufficiency for c = 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 8.4 Sufficiency for c = 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 8.5 Sufficiency for c > 6, d even . . . . . . . . . . . . . . . . . . . . . . . 137 8.6 Sufficiency for c > 6, d odd . . . . . . . . . . . . . . . . . . . . . . . . 138 8.7 Tableaux Disjointness . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 9 Proof of Theorem 3 9.1 142 Case: s ≤ r . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 9.1.1 Basis Tableaux for c = 4, s ≤ r . . . . . . . . . . . . . . . . . 143 9.1.2 Basis Tableaux for c = 5, s ≤ r . . . . . . . . . . . . . . . . . 145 viii 9.2 9.1.3 Basis Tableaux for c = 6, s ≤ r . . . . . . . . . . . . . . . . . 146 9.1.4 Basis Tableaux for c > 6, s ≤ r . . . . . . . . . . . . . . . . . 152 Case: r < s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 9.2.1 Basis Tableaux for c = 4, r < s . . . . . . . . . . . . . . . . . 170 9.2.2 Basis Tableaux for c = 5, r < s . . . . . . . . . . . . . . . . . 172 9.2.3 Basis Tableaux for c = 6, r < s . . . . . . . . . . . . . . . . . 174 9.2.4 Basis Tableaux for c > 6, r < s . . . . . . . . . . . . . . . . . 175 10 Two Row Partitions and the Gaussian Polynomial 186 11 The Alternating Character 189 12 Discussion of General Results 192 A Association between Tableaux Spaces and Irreducibles 195 Bibliography 197 ix List of Tables 8.1 Exceptional r cases for c = 3. . . . . . . . . . . . . . . . . . . . . . . . 112 8.2 Exceptional s cases for c = 3. . . . . . . . . . . . . . . . . . . . . . . . 113 8.3 General c = 3 cases. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 8.4 Exceptional r = 0 and r = 2 cases for c = 4. . . . . . . . . . . . . . . . 116 8.5 Exceptional r = 3 cases for c = 4. . . . . . . . . . . . . . . . . . . . . . 117 8.6 Exceptional r = 4 cases for c = 4. . . . . . . . . . . . . . . . . . . . . . 118 8.7 Exceptional s cases for c = 4. . . . . . . . . . . . . . . . . . . . . . . . 120 8.8 General c = 4 cases. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 8.9 Exceptional r = 0 and r = 2 cases for c = 5. . . . . . . . . . . . . . . . 123 8.10 Exceptional r = 3 and r = 4 cases for c = 5. . . . . . . . . . . . . . . . 129 8.11 Exceptional s cases for c = 5. . . . . . . . . . . . . . . . . . . . . . . . 130 8.12 General c = 5 cases for t > 2. . . . . . . . . . . . . . . . . . . . . . . . 131 8.13 General c = 5 cases for t = 0, 1, and 2. . . . . . . . . . . . . . . . . . . 132 8.14 Exceptional r = 0 and r = 2 cases for c = 6. . . . . . . . . . . . . . . . 133 8.15 Exceptional r = 3 and r = 4 cases for c = 6. . . . . . . . . . . . . . . . 134 8.16 Exceptional s cases for c = 6. . . . . . . . . . . . . . . . . . . . . . . . 134 8.17 General c = 6 cases for odd t. . . . . . . . . . . . . . . . . . . . . . . . 135 8.18 General c = 6 cases for d ≤ 5. . . . . . . . . . . . . . . . . . . . . . . . 135 8.19 General c = 6 cases for even t. . . . . . . . . . . . . . . . . . . . . . . . 136 8.20 Exceptional r and s cases for c = 8. . . . . . . . . . . . . . . . . . . . . 139 9.1 Weights of 0 or d. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 9.2 Weights of 0 and 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 9.3 Tableaux Forms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 x 9.4 Tableaux Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 1 Chapter 1 Introduction and Statement of Main Results Foulkes’ Conjecture is an outstanding problem in the areas of plethysms, rational homotopy theory, multisymmetric functions, and representation theory of symmetric groups. In representation terms, Foulkes’ Conjecture deals with induced permutation characters of wreath products of symmetric groups. The wreath product of symmetric groups Sa and Sb , denoted Sa oSb , is the normalizer of the Young subgroup Sa ×· · ·×Sa (b times) in Sab . Let 1H be the trivial representation of a group H and 1G H the induced representation from H to G. From this we can state Foulkes’ Conjecture: Conjecture 1. (Foulkes’ Conjecture) If a ≤ b then every irreducible character occurwith multiplicity greater than or equal ring as a constituent in 1SSab occurs in 1SSab a oSb b oSa to its multiplicity in 1SSab . b oSa Foulkes made this conjecture in [8] from his work on plethysms. The a = 2 case was contained in Thrall’s 1942 work on symmetrized Kronecker powers, [20], and proved again in [14] by James and Kerber using Gaussian coefficients. Coker also gives a proof for a = 2 using eigenvalues in [5], while Doran used transition matrices in [7]. The a = 3 case was proven by Dent and Siemons using mappings of unordered partitions [6]. In [2], Black and List formulated Foulkes’ Conjecture in terms of matrix incidences and Wu has rephrase it in terms of rational homotopy theory in [23]. Howe, in [12], used a plethystic approach to interpret Foulkes’ Conjecture via canonical morphisms between symmetric power modules. Using this, Brion, [4], showed 2 the conjecture holds for b sufficiently large with respect to a and Briand [3] proved Foulkes’ Conjecture for a = 4. Also using symmetric powers and plethysms, Stanley, [19], places Foulkes’ Conjecture inside a larger body of open positivity conjectures in Algebraic Combinatorics. With a more combinatorial approach, Doran gave additional formulations in [7] using tableaux spaces. Doran also suggested generalizing Foulkes’ Conjecture to: Conjecture 2. (Generalized Foulkes’ Conjecture) Given n = ab, a ≤ b, if c, d are such that cd = n, and c, d ≥ a, then every irreducible character occurring as a constituent in 1SSnb oSa occurs in 1SSnd oSc with multiplicity at least as large. For c = a and d = b this becomes the standard Foulkes’ Conjecture. Note that c, d ≥ a is necessary. This is easily verified by using GAP, [9], which shows that some irreducibles in 1SS12 do not occur in 1SS12 . Conjecture 2 holds for small n, (less than 4 oS3 6 oS2 28), by computer verification also using GAP, [9]. In Chapter 5 we will prove it holds for a = 2 by construction. Namely, we will show: Theorem 1. Given b ≥ 2, let n = 2b. If c, d are such that cd = n, and c, d ≥ 2, then every irreducible occurring in 1SSnb oS2 occurs in 1SSnd oSc with equal or larger multiplicity. We can also prove the following variation on Conjecture 2 for a = 3. The bulk of the proof is discussed in Chapter 6 with supporting details in Chapters 7 and 8. Theorem 2. Let n = 3b = cd, with c, d ≥ 3. Then every irreducible character occurring in 1SSnb oS3 occurs in 1SSnd oSc . This theorem can be strengthened when the irreducibles involved correspond to two row partitions of n. We prove this version in Chapter 9. Theorem 3. Let n = 3b = cd, with c, d ≥ 3 and let λ = [λ1 , λ2 ] be a two row partition of n. Then every irreducible character χλ occurring in 1SSnb oS3 occurs in 1SSnd oSc with multiplicity at least as large. In Chapter 2 we describe the necessary concepts to approach the Foulkes’ Conjecture combinatorially using tableaux. Much of this framework was developed by 3 Doran in [7]. We develop the theory and techniques behind constructing appropriate tableaux in Chapter 3. In Chapter 4 we completely classify and discuss all tableaux that occur in 1SSnb oS3 . The proof of Theorem 1 is given in Chapter 5. We prove Theorem 2 in Chapter 6, though the required tableau constructions are postponed until Chapter 7. In Chapter 8 we show that these tableaux suffice to cover all necessary cases. We prove Theorem 3 in Chapter 9. The corresponding results for the alternating character is given in Chapter 11. Theorems 1 and 3 also can be interpreted in terms of the Gaussian coefficient; this is discussed in Chapter 10. Further implications of all these results are listed in Chapter 12. Some of these results appear in a forthcoming article in the Journal of Algebra, [21], namely, Chapters 1, 2, 3, 5, and portions of Chapters 4 and 6. 4 Chapter 2 Background 2.1 Tableaux A partition λ = [λ1 , . . . , λ` ] of a number n is an ordered tuple of positive integers P such that λi = n and λi ≥ λi+1 > 0; it is denoted by λ ` n. The length of λ is `. A Ferrers diagram is the set [λ] = {(i, j)|1 ≤ i ≤ `, 1 ≤ j ≤ λi }. We view [λ] as a (left-justified) stack of boxes with row i having λi boxes. A tableau of shape λ is a filling of the Ferrers diagram [λ] with a set of elements, usually the positive integers. It is said to have content α = [α1 , . . . , αk ] if the integer i occurs exactly αi times. A tableau is semi-standard if the entries are weakly increasing across the rows and strictly increasing down the columns. This notation is standard and further discussion can be found in [7] and [18]. Example 2.1.1. Consider the following tableaux: 125 P = 42 3 115 Q= 22 33 112 R= 233 P is a [3, 2, 1]-tableau of content α = [1, 2, 1, 1, 1], while Q has shape λ = [3, 2, 2] and content α = [2, 2, 2, 0, 1]. Similarly, R has shape [3, 3] with content [23 ] = [2, 2, 2]. Both Q and R are semi-standard, but P is not. 5 2.2 Combinatorial Structures There are two different permutation actions on tableaux: an action permuting the entry positions in the tableau and an action permuting the numbers filling the tableau. These actions commute with each other. Let T be a λ-tableau filled with the numbers 1 to a, where λ ` n. The permutation of entry positions corresponds to an action of Sn on T . View the entry positions (i.e., boxes) of T as labelled 1 to n. Then σ ∈ Sn acts on T by permuting the entries in the positions moved by σ. To avoid confusion between entry positions and numbers in T , we will denote all entry positions with the subscript T when necessary. The permutation action of the numbers corresponds to their permutation by π ∈ Sa . Example 2.2.1. Take n = 6, λ = [3, 2, 1], and a = 3. Consider T = 331 1 2 . In 2 1T 2T 3T terms of entry positions, we label T as 4T 5T . For σ = (23)T ∈ S6 , we have 6T 313 221 σT = 1 2 , while the action of π = (23) ∈ S3 gives πT = 1 3 . 2 3 We generally restrict the action of Sn to two subgroups. Let RT be the subgroup of Sn which set-wise fixes the rows of T , namely, the row permutations. Denote this action by σT for σ ∈ RT . Let CT be the subgroup of Sn which set-wise fixes the columns of T , namely, the column permutations. This action is denoted τ T for 0 τ ∈ CT . If λ is the conjugate partition of λ, (i.e., the partition corresponding to column lengths) we have that RT ≈ Sλ1 × · · · × Sλ` and CT ≈ Sλ01 × · · · × Sλ0`0 . Viewing the subgroups under these isomorphisms, we can label the entry positions by labelling each row (resp. column) with 1 to λi (resp. λi0 ). Under these labellings we write σ (resp. τ ) as a direct product of the permutations for each row (resp. column). 112 1 2 3 , then for RT we view T as labelled T T T . 23 1T 2T 1 1 1 Likewise we use the labelling T T T for CT . Applying these actions to T , gives 2T 2T Example 2.2.2. Let T = the following sets: 6 ) ( 121 211 112 121 211 112 23 , 23 , 23 , 32 , 32 , 32 {σT | σ ∈ RT } = with each element occurring twice. ( {τ T | τ ∈ CT } = ) 112 212 132 232 23 , 13 , 21 , 11 ( {πT | π ∈ Sa } = ) 112 221 332 113 223 331 23 , 13 , 21 , 32 , 31 , 12 CT = {()T × ()T × ()T , (12)T × ()T × ()T , ()T × (12)T × ()T , (12)T × (12)T × ()T } Note that the actions of σ and τ do not commute, but π commutes with both σ and τ . Given λ a partition of n = ab, let Wλ,a be the set of all tableaux of shape λ and content [ba ] = [b, . . . , b], where the entries are 1 to a. Let Sλ,a be the set of all semi-standard tableaux in Wλ,a . These sets and the following constructions were developed by Doran in [7]. Note that when b = 1, Sλ give rise to the Specht modules which are discussed extensively in [15] and [18]. The partitions of n index the Specht modules S λ , which in turn correspond to precisely the irreducible modules of Sn . From Wλ,a we can construct the complex vector space W λ,a with the tableaux as a basis. The action of Sa on the tableaux give rise to a permutation representation. Inside W λ,a we construct the following objects. Definition 2.2.3. Let T ∈ Wλ,a . Let ²(τ ) be the sign of τ as a permutation. Inside W λ,a we have: a) eT = P P σ∈RT τ ∈CT ²(τ )στ T 7 b) qT = c) mT = P P π∈Sa P P σ∈RT τ ∈CT ²(τ )πστ T P π∈Sa σ∈RT πσT The tableaux are independent basis for W λ,a , so for T1 , T2 ∈ Wλ,a , we have T1 + T1 = 2T1 but T1 + T2 = T1 + T2 . Example 2.2.4. Let T = ½ eT = 4 × 11 , then 22 11 22 + 22 11 ¾ ½ −2 12 21 21 12 + + + 21 12 21 12 ¾ Here, symmetry gives qT = 2eT . Definition 2.2.5. From eT and mT we can construct the following subspaces of W λ,a : a) S λ,a = C[eT |T ∈ Wλ,a ] b) M λ,a = C[mT |T ∈ Wλ,a ] These spaces are Sa -modules. We have {eT |T ∈ Sλ,a }, a basis for S λ,a , and {mT |T ∈ Sλ,a /Sa }, a basis for M λ,a . The set {qT |T ∈ Wλ,a } generates S λ,a ∩M λ,a , but does not form a basis. Background on these spaces and the proofs of the statements may be found in [7]. In [7], Doran uses Gay’s result from [10]: Lemma 2.2.6 (Gay’s Result). The multiplicity of the irreducible module S λ in 1SSab equals the multiplicity of the trivial representation in S λ,a . b oSa From this, Doran reformulated Foulkes’ Conjecture to: Lemma 2.2.7. The dimension of S λ,a ∩M λ,a equals the multiplicity of the irreducible S λ in 1SSab . b oSa A proof of this lemma in terms of qT ’s is presented in Appendix A. From this, Foulkes’ Conjecture is equivalent to proving dim(S λ,a ∩ M λ,a ) ≤ dim(S λ,b ∩ M λ,b ) for a ≤ b and all λ ` n. Proving Conjecture 2 is equivalent to showing that for all λ ` n, dim(S λ,a ∩ M λ,a ) ≤ dim(S λ,c ∩ M λ,c ) when ab = n = cd with c, d ≥ a. 8 Remark 2.2.8. In terms of tableaux, proving Theorem 1 is equivalent to exhibiting mλ non-zero linearly independent qT , where T has shape λ and content [dc ], with mλ the multiplicity of S λ in 1SS2b . Theorem 2 is provable by exhibiting a non-zero qT b oS2 with T having shape λ and content [dc ] for all λ such that the multiplicity of S λ in 1SS3b is non-zero. For Theorem 3 we want mλ linearly independent tableaux T with b oS3 content [dc ] such that qT is non-zero, where λ = [λ1 , λ2 ] and mλ is the multiplicity of S λ in 1SS3b . b oS3 9 Chapter 3 Theory of Tableaux Construction Throughout this and subsequent chapters we will use T to represent an arbitrary tableau, σ an element of RT , τ an element of CT , and π an element of Sa for T filled with 1 to a. Remark 3.0.9. a) eτ T = ²(τ )eT b) eπT = πeT c) qπT = πqT = qT d) qT = Σπ∈Sa πeT e) qτ T = ²(τ )qT These are standard computations, which are discussed in [7] and [18]. This remark shows that we can ignore the effects of permuting entries when constructing the tableaux. Also, we may order the columns however we choose at the cost of a sign. 3.1 Filling Tableaux Definition 3.1.1. In T , the weight of a number x in row i, denoted ωi (x), is the number of times x occurs in row i of T . When T is not clear from context, we write ωi (x|T ) in place of ωi (x). We extend this so that ωi (x1 , . . . , xj |T ) = (ωi (x1 |T ), . . . , ωi (xj |T )). 10 Implicitly, we take ωi (T )à = (ωi (1), ! . . . , ωi (a)), which is called the row-weight of row i ω1 (xj ) .. . of T . Similarly, ω(xj ) = is the weight vector of xj of T . Hence ω(T ) is the ω` (xj ) matrix corresponding to ωi (j|T ). Note that row permutations do not effect weight, so ω(σT ) = ω(T ). 122 ³ Example 3.1.2. T = 3 4 4 . We have ω(T ) = 55 1 2 0 0 0 0 0 1 2 0 0 0 0 0 2 ´ . From this we can read that ω2 (3) = 1. Weights for a tableau are only comparable with tableaux of the same shape and content. Recall that Wλ,a is the set of all λ-tableaux with content [ba ]. Let N a be the set of all a-tuples w with non-negative integer entries and N a,` those ` × a matrices. Then we can view the row-weight function as a linear operator ωi : Z[T |T ∈ Wλ,a ] → Z[w|w ∈ N a ], where {w|w ∈ N a } is a Z-basis, or correspondingly, ω : Z[T |T ∈ Wλ,a ] → Z[w|w ∈ N a,` ], with ` = `(λ). This means we treat weights like linearly independent basis in Z[w|w ∈ N a,` ]. Hence ω(T1 + aT2 ) = ω(T1 ) + aω(T2 ). If T1 = 11 2 and T2 = 21 1 then ω1 (T1 ) = (2, 0) and ω1 (T2 ) = (1, 1). So ω1 (T1 + T2 ) = (2, 0) + (1, 1) and ω1 (T1 + T1 ) = (2, 0) + (2, 0) = 2 · (2, 0). For convenience, we take ω(0) = 0. Definition 3.1.3. Let rowsumT (i) be the sum of all the entries in row i of T , P j ωi (j)j. Lemma 3.1.4. eT = 0 iff T has a repeat entry in a column. Proof. ⇒ It suffices to show that if T has no repetitions within a column, then eT 6= 0. By Remark 3.0.9 column permutations only change the sign of eT , so without loss of generality, we may assume the columns of T are strictly increasing. If eT = 0 then there exists σ ∈ RT , τ ∈ CT such that ²(τ ) = −1 and στ T = T , and we say that T cancels in the summation. Since στ T = T , we must have ω(στ T ) = ω(T ) and so ω(τ T ) = ω(T ). This implies that rowsumT = rowsumτ T for all rows. Let r be the first row in which τ moves an entry of T . Let αi1 , . . . , αik be the entries of row r moved by τ . Say τ moves βj to αij . Since r is the first row moved by τ , βj > αij . Then rowsumT (r) = 11 Pk j=1 αij + P i6=ij αi < Pk j=1 βj + P i6=ij αi = rowsumτ T (r). Contradiction. Therefore eT 6= 0. ⇐ If T has a repeated entry in a column, then there exists a transposition τ ∈ CT such that τ T = T . Hence eT = eτ T = ²(τ )eT = −eT and so eT = 0. Knowing that any tableau with repeated numbers in a column makes eT = 0 is very useful for our construction of non-zero tableaux. We summarize this fact as follows: Lemma 3.1.5. Any tableau T filled with the numbers 1 to a having more than a rows will have eT = 0 and qT = 0, as will any T having repetitions within a column. Hence every tableau T filled with the numbers 1, 2, and 3, with qT 6= 0, will have at most three rows and all column entries will be distinct. Due to this, from now on we will assume all tableaux have distinct column entries. Notation: Many of the tableaux we construct will have multiple identical columns. We call a group of such columns, a column block. For both clarity and space we denote a column block by one copy of the column with the number of repetitions listed above. If the number of column copies is omitted, it is assumed to be one. For example, T = has k copies of 1 2 3 111 would be denoted by T = 222 3 k l m n 1112 1 , while T = 2233 2 3 and l copies of 12 , m copies of 13 , and n copies of 23 . We call the columns of T having only one entry the tail of T . When specifying T by column blocks, we may omit the tail, provided the content of T is known. The rest of T is called the body. For instance, if T = 2 11223 and we know that T has content 324 4 12 [3, 3, 3, 3], we can just list the body, T = 2 112 324 4 instead. It is assumed that any entries not specified are contained in the tail. We also use this abbreviated notation when describing elements of CT . We write τ ∈ CT as a direct product of permutations on the column blocks. Since the only permutation possible on the tail is the identity, we omit the permutations corresponding 2 k to the tail. Hence for T listed above, τ is of the form τ1 ×τ2 × τ3 . We write τi if the same permutation τi is to be applied to k columns within a column block. When k is k less than the size of the column block, we understand τi to mean that τi is applied to all k of the columns (determined by context) and the identity permutation is applied to the remaining columns within the block. For instance, on the previous T , there 4114 are two permutations of the form (13)T ×()T × (12)T , which produce 3 3 2 2 and 14 1414 3322 . 41 1 3.2 Showing Tableaux are Non-Zero Definition 3.2.1. A tableau T is said to be non-zero if qT 6= 0. Two tableaux are said to be distinct if qT1 6= ±qT2 , otherwise T1 and T2 are said to be equivalent. Since qT involves many summands, showing qT 6= 0 by direct summation is not practical. Instead, we use a technique called weight-set counting. Weight-set counting involves summing only those tableaux with a given weight; if that sum is non-zero, the entire qT summation must be non zero. Definition 3.2.2. Given a tableau T , let ω(T ) = (x1 , . . . , xa ), where xj is the weight vector ω(j|T ) = xj of the element j in T . A weight assignment of T is a pairing between the set of elements of T with the multiset of weight vectors of ω(T ). We denote the pairing of the weight vector xj with the element kj by ω(kj |T ∗ ) = xj . (The T ∗ represents a possible tableau which has the weight of kj being xj .) Note 13 that if the vectors xj and xj 0 are equal, the weight assignment pairing kj with xj is the same as the weight assignment pairing kj with xj 0 . What matters in a weight assignment is the vector paired with each element, not how we label the vectors. We usually indicate a weight assignment by writing (k1 , . . . , ka ), by which we mean ω(k1 , . . . , ka |T ∗ ) = (x1 , . . . , xa ) = ω(T ). Given a permutation π we can create a weight assignment by assigning the element π(k) to xk , since ω(π(k)|πT ) = ω(k|T ) = xk . Similarly, given such pairing (k1 , . . . , ka ) we can construct a permutation π by taking π = (k1 , . . . , ka ) in one-line notation. 111 ³ ´ 3 0 0 For example, let T = 2 2 3 . Then ω(T ) = 0 2 1 . The weight assignment 3 (3, 1, 2) means ω(3, 1, 2|T ∗ ) = 0 0 1 ³ 3 0 0 0 2 1 0 0 1 ´ for some tableau T ∗ . This weight assignment corresponds to the weight permutation π = (132) in cycle notation (from left to right). Note, however, that such a listing (k1 , . . . , ka ) of a weight assignment is not necessarily unique. For instance, if x1 = x2 then (k1 , k2 , k3 , . . . ka ) and (k2 , k1 , k3 , . . . ka ) represent the same weight assignment (pairing) but give rise to different permutations. The numbers of such permutations corresponding to the same weight assignment depends only on the vector symmetries of ω(T ). This number is denoted s(ω(T )). For example, if T = 1 1 2 2 3 , then ωT = (2, 2, 1). There are three distinct weight assignments of T corresponding to which of the three elements is assigned a weight of 1. Since there are two permutations arising from such an assignment (for instance, ω(213|T ) = (2, 2, 1) as well) we have s(ω(T )) = 2. Definition 3.2.3. A weight assignment (k1 , . . . , ka ) is valid for T if there exists τ ∈ CT such that ω(k1 , . . . , ka |τ T ) = ω(T ), i.e., T ∗ = τ T . If this happens we say τ is valid for (k1 , . . . , ka ), otherwise τ is invalid. Given a valid τ we say τ is positive if ²(τ ) = 1 and negative if ²(τ ) = −1. 1222334 ´ ³ 1 3 2 1 Example 3.2.4. Let T = 3 1 1 . We have ω(T ) = 2 0 1 0 . Recall that for 0 0 0 2 44 1T 1T 1T CT we use the labelling 2T 2T 2T (the tail is omitted since all column permutations 3T 3T 14 on it are trivial). Now (2, 3, 1, 4) is a valid weight assignment since τ = (12)T ×(12)T × 3112334 ³ ´ 2 1 3 1 (12)T ∈ CT has τ T = 1 2 2 with ω(τ T ) = 1 2 0 0 , So ω(2, 3, 1, 4|τ T ) = 0 0 0 2 44 ω(T ). The weight assignment, (2, 3, 1, 4) corresponds to the permutation π = (123) in cycle notation, meaning ω(1, 2, 3, 4|τ T ) = ω(1, 2, 3, 4|πT ) However, (1, 4, 3, 2) is not a valid weight assignment since then we must have ω3 (2|τ T ) = 2 for some τ , but there is no column permutation that will put two 2’s in the third row. We will make frequent use of weight assignments in order to determine when T is non-zero. Definition 3.2.5. Given T , consider the following functions: • P(π(T )) = the number of τ ∈ CT such that ²(τ ) = 1 and ∈ CT such that ²(τ ) = −1 and ω(π −1 (1), . . . , π −1 (a)|τ T ) = ω(T ). • N(π(T )) = the number of τ ω(π −1 (1), . . . , π −1 (a)|τ T ) = ω(T ). • P(T ) = P • N(T ) = π P(π(T )), where π correspond to distinct weight assignments of ω(T ) P π N(π(T )), where π correspond to distinct weight assignments of ω(T ) Theorem 4. (Weight-set Counting) If qT = 0 then P(T ) = N(T ). Proof. Let D be the set permutations corresponding to distinct weight assignments of T . qT = 0 XXX ⇒ π ⇒ ⇒ σ XX π τ π τ XX (3.2.1) ²(τ )ω(σπτ T ) = 0 (3.2.2) τ ²(τ )ω(πτ T ) = 0 (3.2.3) ²(τ )ω(πτ T ) = 0 s.t. ω(1, . . . , a|πτ T ) = ω(T ) (3.2.4) 15 ⇒ XX π ⇒ XX π ²(τ ) = 0 s.t. ω(1, . . . , a|πτ T ) = ω(T ) (3.2.5) ²(τ ) = 0 s.t. ω(π −1 (1), . . . , π −1 (a)|τ T ) = ω(T ) (3.2.6) τ τ ⇒s(ω(T )) X (P(πT ) − N(πT )) = 0 (3.2.7) π∈D ⇒s(ω(T ))(P(T ) − N(T )) (3.2.8) ⇒P(T ) − N(T ) = 0 (3.2.9) If qT = 0, taking the weight of both sides gives Eq. (3.2.2). Since row permutations do not effect weights, we can reduce to Eq. (3.2.3). As distinct weights can not cancel, we can consider only those tableaux with the same weight as ω(T ), hence we must have Eq. (3.2.4). Since we are only summing over tableaux of a fixed weight we may drop the weight from the sum and simply add the sign of τ , for Eq. (3.2.5). By definition of weight assignments, ω(1, . . . , a|πτ T ) = ω(π −1 (1), . . . , π −1 (a)|τ T ) hence we get Eq. (3.2.6). Restricting the sum to distinct weight-sets gives the factor of s(ω(T )), and if we split over the sign of τ 0 we get Eq. (3.2.7), which by definition is the same as Eq. (3.2.8). Factoring out s(ω(T )) yields Eq. (3.2.9). Thus P(T ) = N(T ). We can also write Theorem 4 as: Corollary 5. Let A = {τ |ω(τ T ) = ω(πT ) for some π ∈ Sa }. If P τ ∈A ²(τ ) 6= 0, then qT 6= 0 by weight-set counting on ω(T ). An easy application of this theorem is the following useful lemma. Lemma 3.2.6. If T is a tableau consisting of a single column block (and an arbitrary k tail), for instance, T = 1 , then qT 6= 0 iff k even. 2 Proof. We have w = ω(T ) = ( k0 k0 ). So there are two distinct weight assignments for T , namely, ω(1, 2|T ∗ ) = w and ω(2, 1|T ∗ ) = w. We wish to determine for which τ does T ∗ = τ T satisfy one of these equations. For ω(1, 2|τ T ) = w, we must have k 2’s in the second row, hence none of the columns of T may move. The only τ satisfying 16 k this is τ =()T , which is positive. For ω(2, 1|τ T ) = w, we must have k 1’s in the second k row, thus τ must exchange row 1 and row 2 for every column in T . Thus τ =(12)T and ²(τ ) = (−1)k . Hence when k is even, we have P(T ) = 2 and N(T ) = 0, so qT is non-zero. The same idea applies for T having more than two rows. When k is odd, however, we have P(T ) = N(T ) so the Theorem 4 does not apply. k Instead, let τ =(12)T and let π ∈ Sa be the corresponding entry transposition (in our example π = (12)). Then πτ T = T and ²(τ ) = −1. Thus qπτ t = πqτ T = qτ T = ²(τ )qT = −qT . So qT = 0. It is also true that if qT 6= 0, then it is non-zero by weight-set counting on some weight ω(τ T ). That is we can’t have P(τ T ) = N(τ T ) for all τ ∈ CT and still have qT 6= 0. Theorem 6. If qT 6= 0, then it is non-zero by weight set counting on τ T for some τ ∈ CT . Proof. Recall that qT = P P P π σ τ πσ ²(τ )τ T . Since a tableau may be written in multiple ways (such as T 0 = πστ T = π 0 σ 0 τ 0 T ), we need to be careful of how we denote a tableau. Consider the terms of qT partitioned in to classes {τ T } = {πστ T |π ∈ Sa , σ ∈ RT }, using {τ T } as the class representative. Note that all tableaux in a given class have the same sign and generic weight, i.e., the same weight modulo the action of Sa . Hence if τ T = π 0 σ 0 τ 0 T for some π 0 , σ 0 , τ 0 , with ²(τ τ 0 ) = −1 then πστ T = ππ 0 σσ 0 τ 0 T , so the classes are equal set wise, but of opposite sign. This holds for any element of the class, not just τ T . This means we may view the tableaux which cancel in qT as a matching between equal classes of opposite sign. So if qT 6= 0, there is a set of tableaux T which are not canceled in qT . Now for any τ T ∈ T either the weight of τ T does not cancel, (that is ω(τ T ) 6= πω(π 0 σ 0 τ 0 T ) for all π ∈ Sa and all π 0 σ 0 τ 0 T ∈ T with ²(τ τ 0 ) = −1) or the weight cancels with some π 0 σ 0 τ 0 T having ²(τ τ 0 ) = −1 (that is ω(τ T ) = πω(π 0 σ 0 τ 0 T ) for some π 0 σ 0 τ 0 T ∈ T). If the weight does not cancel for some τ T , the qT 6= 0 by weight set counting on τ T . Assume all the weights do cancel. Then ω(τ T ) = πω(π 0 σ 0 τ 0 T ), so for every row, τ T and ππ 0 σ 0 τ 0 T have the same number of each symbol. Hence there 17 exists σ ∈ RT such that τ T = ππ 0 σσ 0 τ 0 T . But then, since ²(τ τ 0 ) = −1, the classes {τ T } and {τ 0 T } will cancel in qT contradicting τ T, π 0 σ 0 τ 0 T ∈ T. Hence qT 6= 0 by weight set counting on some τ T . We will not need Theorem 6 for our results; it is included for theoretical interest and completeness. Theorem 4 is used quite heavily, however. For instance, it allows use to directly establish the multiplicities of the irreducible characters in 1SS2b . b oS2 Theorem 7. The only irreducible characters occurring in 12b Sb oS2 are those corresponding to all partitions λ = [λ1 , λ2 ] of 2b where λi is even. Moreover, these characters occur with multiplicity 1. Proof. By Remark 2.2.8 we need only consider those shapes with qT non-zero. By Lemma 3.0.9 and Lemma 3.1.5 all distinct non-zero tableaux filled with b 1’s and b k 2’s must be equivalent to T = 1 , (not including tail). Hence there is at most one 2 distinct non-zero tableau for any shape λ. By Lemma 3.2.6, when λ = [2b−k, k] then qT 6= 0 iff k is even. Since hqT i = S λ,2 ∩ M λ,2 we must have dim(S λ,2 ∩ M λ,2 ) = 1 if k is even and zero otherwise. This is a well-known result, appearing in [13] and [17]. The weight-set counting of Theorem 4 is useful for much more complicated tableaux as well. To illustrate the general usage of the theorem, we list here a slightly more involved example. Example 3.2.7. To see directly how weight-set counting works, consider the following example. The tableau Q∗ is listed below using the column block notation, with the conditions on the block size listed to the right. Underneath the tableau we list the weight and shape of the tableau. 18 aabbc 2 2 2 3 1 Q∗ = 1 4 4 1 4 3 3 a+b+c=d 2a + b < d c even a, b > 0 0 0 d ω2,3 = ( a+b 0 0 2a 0 ) λ = [3d − 2a − b, d + a+b, 2a] We want to show that qQ∗ 6= 0, which we do by showing P(Q∗ ) − N(Q∗ ) 6= 0 and applying Theorem 4. Now P(Q∗ ) is the number of τ ∈ CQ∗ with ²(τ ) = 1 such that ω(i1 , i2 , i3 , i4 |τ Q∗ ) = ω(Q∗ ) for some distinct weight assignment (i1 , i2 , i3 , i4 ). (Equivalently, τ is such that {ω(i|τ Q∗ )ki = 1, 2, 3, 4} = {ω(i|Q∗ )ki = 1, 2, 3, 4}.) Similarly, N(Q∗ ) is those with ²(τ ) = −1. The easiest way to count these τ is to use weight assignments. First we determine which weight assignments might be possible using some general properties of the tableau. Then we count how many τ correspond to each weight assignment (i.e., for which τ is the weight assignment valid) and determine ²(τ ). Finally, we add this signed sum to determine P(Q∗ ) − N(Q∗ ). First we want to determine which weight assignments are possible for Q∗ . That is, determine for which 4-tuples x = (i1 , i2 , i3 , i4 ) there might exist τ ∈ CQ∗ such that ∗ 0 0 d a+b 0 0 d ω2,3 (x|τ Q∗ ) = ( a+b 0 0 2a 0 ). Let w = ( 0 0 2a 0 ). Simply looking at Q , there are a few restrictions on what x can be. Notice the body contains d copies of the elements 1 and 4, but fewer than d copies of 2 and 3 since 2a+b < d. Also note that the body of τ Q∗ contains the same elements as the body of Q∗ . This implies that not all elements can have d copies in row two of τ Q∗ for some τ . If ω2,3 (x|τ Q∗ ) = w then either ω2 (1|τ Q∗ ) = d or ω2 (4|τ Q∗ ) = d; namely, only the elements 1 and 4 may have ω2 (i|τ Q∗ ) = d. Hence any valid weight assignment must have i4 = 1 or 4. If ω2 (1|τ Q∗ ) = d, then there is only one other non-zero weight to assign in row two. As b > 0 the remaining columns (the second a block and the first b block) must have the same element in row two, namely, 2 or 19 4. That means we must have either ω2 (2|τ Q∗ ) = a+b or ω2 (4|τ Q∗ ) = a+b, so the weight assignment must have i1 = 2 or 4. Similarly, if ω2 (4|τ Q∗ ) = d, then since b > 0 we must have either ω2 (1|τ Q∗ ) = a+b or ω2 (3|τ Q∗ ) = a+b, that is i1 = 1 or 3. We also consider which elements j may have ω3 (j|τ Q∗ ) = 2a. We find that only with j = 2 or 3 may this occur since these are the only elements for which both a blocks will be the same in row three. (That is if ω2,3 (τ Q∗ ) = w then either ω3 (2|τ Q∗ ) = 2a or ω3 (3|τ Q∗ ) = 2a, so any valid weight assignment has i3 = 2 or 3.) There are six distinct weight assignments x = (i1 , i2 , i3 , i4 ) meeting these conditions: (1, 2, 3, 4), (1, 3, 2, 4), (4, 3, 2, 1), (3, 1, 2, 4), (4, 2, 3, 1), and (2, 4, 3, 1). In the table below, for each weight assignment we list for what type of tableau τ Q∗ it is valid, the form τ used, the number of such τ , and the sign of τ . This is an easy way to summarize the counting of τ and their signs. (We will omit the subscripts Q∗ when writing τ for easy reading. Remember that τ is labeled by the entry positions of Q∗ and not the elements.) 20 ω2,3 (x|τ Q∗ ) τ Q∗ 0 0 d = ( a+b 0 0 2a 0 ) x = (1, 2, 3, 4) x = (1, 3, 2, 4) x = (3, 1, 2, 4) x = (4, 2, 3, 1) x = (4, 3, 2, 1) x = (2, 4, 3, 1) τ aabbc 2 2 2 3 1 1 4 4 1 4 3 3 aabbc 3 3 2 3 1 1 4 4 1 4 2 2 aabbc 1 3 2 1 1 3 4 4 3 4 2 2 aabbc 2 2 2 3 4 1 4 4 1 1 3 3 aabbc 3 3 2 3 4 1 4 4 1 1 2 2 aabbc 2 4 4 3 4 1 2 2 1 1 3 3 a a b b c b b () × () × () × () × () a a c (13) × (13) × () × () × () a a b b b b c c (132) × (13) × () × (12) × () a a () × () × () × () × (12) a a b b c b c (13) × (13) × () × () × (12) a a b () × (12) × (12) × () × (12) # ²(τ ) 1 (−1)0 1 (−1)2a 1 (−1)a+b 1 (−1)c 1 (−1)2a+c 1 (−1)a+b+c To see how we obtain such a table, consider the last row. We want a tableau τ Q∗ such that ω2,3 (2, 4, 3, 1|τ Q∗ ) = w. This means that ω3 (3|τ Q∗ ) = 2a so τ cannot move any entries in row 3. We also have ω2 (4|τ Q∗ ) = 0, so examining Q∗ , we know that τ acts non-trivially on the second column block a, the first column block b, and column a a b b c block c, namely, τ =∗ × (12)T × (12)T × ∗ × (12)T . If τ were to act non-trivially on the first column block a or the second column block b, the number of 1’s in row two of τ Q∗ would decrease. Since we must have ω2 (1|τ Q∗ ) = d, this cannot happen. Hence a a b b c τ =()T × (12)T × (12)T × ()T × (12)T and has been completely determined for us. Then we have ω2,3 (2, 4, 3, 1|τ Q∗ ) = w, so such a τ exists and is unique. For reference, τ and τ Q∗ are listed. Once τ has been determined, computing ²(τ ) = (−1)a+b+c is 21 straightforward. Finally, to compute the weight sum for w, we sum the product of the number of τ with the sign of τ , that is # · ²(τ ). Here the sum is 1 + 1 + (−1)a+b + 1 + 1 + (−1)a+b . This sum is between 2 and 6, depending on the parity of a and b. Since it is non-zero in all cases, Theorem 4 shows qQ∗ 6= 0. 3.3 Joining Tableaux Definition 3.3.1. The join of two tableaux, U and V , denoted U ∨ V is a way of combining tableaux together. If the entries of U and V are not disjoint, renumber V so that they are. For instance, if U contains the numbers 1 to n and V contains the numbers 1 to m we first renumber V with the numbers n + 1 to n + m. Then concatenate the tableaux and sort the columns by length. Note that entries of every column remain fixed, only the order of the columns change. 113 123 567 Example 3.3.2. U = 2 2 4 and V = 2 4 4 We renumber V to get V = 6 8 8 . 34 3 113567 Concatenating gives 2 2 4 6 8 8 . When sorted we get T = U ∨ V = 34 7 7 115367 2 2 6 4 8 8. 347 Note that since applying permutations of Sa to a tableau has no effect on qT and mT , the renumbering of a tableau is irrelevant. Also, any σ ∈ RT that only interchanges columns will commute with all τ ∈ CT . Hence column sorting has no effect of eT , qT , and mT , since there is no sign change for row permutations. This join operation also joins the weight-sets, namely, ω3 (U ∨ V ) = ω3 (U ), ω3 (V ) = (0, 0, 1, 1, 0, 0, 1, 0). Definition 3.3.3. Let T = U ∨ V for tableaux U filled with 1 to m and V filled with m + 1 to a. The weights, ω(U ) and ω(V ) are disjoint (equivalently, ω(T ) splits over U and V ), if every valid weight assignment of ω(T ), can be obtained from concatenating valid weight assignments of ω(U ) and ω(V ). Say ω(1, . . . , m|U ) = (x1 , . . . , xm ) where the xi are weight vectors and ω(m + 1, . . . , a|V ) = (xm+1 , . . . , xa ). So ω(1, . . . , a|T ) = (x1 , . . . , xa ). Consider a valid 22 weight assignment of T assigning to the element j the weight vector xkj . Then ω(1, . . . , a|τ T ) = (xk1 , . . . , xka ) for some τ ∈ CT . This restricts to a valid weight assignment of τ|U U by considering only the elements 1 to m. This restriction is unique because a weight assignment is defined by the vector-element pairing, not the label assigned to the vector. If this restriction corresponds to a weight assignment of ω(U ) (i.e., the weights assigned to elements 1 to m are the same as the weights of ω(U ) as vectors) then the weight assignment of T arose from valid weight assignments of U . Similarly for V . By restricting to a weight assignment of ω(U ) we mean {xki |i = 1 . . . m} = {xi |i = 1 . . . m}, i.e., the weights assigned to U are equivalent to those of ω(U ). If this result is a weight assignment of ω(U ), it is valid for the tableau τ|U U . If this is true for all valid weight assignments of T , then ω(T ) splits and the weights of U and V are disjoint. Example 3.3.4. Consider U = 11 3333 113333 and V = . So T = and 22 4444 224444 ω2 (1, 2, 3, 4|T ) = (0, 2, 0, 4). There are four valid weight assignments possible for T : (Recall that T ∗ represents any possible tableau τ T .) ω2 (1, 2, 3, 4|T ∗ ) = (0, 2, 0, 4) ω2 (1, 2, 3, 4|T ∗ ) = (2, 0, 0, 4) ω2 (1, 2, 3, 4|T ∗ ) = (0, 2, 4, 0) ω2 (1, 2, 3, 4|T ∗ ) = (2, 0, 4, 0) When we restrict these weight assignments to U we get two possible assignments for U , ω2 (1, 2|U ∗ ) = (0, 2) and ω2 (1, 2|U ∗ ) = (2, 0). Since the original weight of U is ω2 (1, 2|U ) = (0, 2), both of these assignments are assignments of (0, 2) and both are valid for U (simply take τU = () and τU = (12)T × (12)T ). A similar argument holds for V . Thus the ω(T ) splits over U and V . However, weight-set disjointness is highly dependent on the filling of T . Consider instead, U = 11 43 1143 and V = . So T = and ω2 (1, 2, 3, 4|T ) = (0, 2, 1, 1). 22 34 2234 23 Then ω2 (1, 2, 3, 4|T ∗ ) = (1, 1, 2, 0) is a valid weight assignment for T by T ∗ = τ T with τ = (12)T × ()T × ()T × (12)T . However, ω2 (1, 2|U ∗ ) = (1, 1) it is not a valid weight assignment of ω2 (U ) = (2, 0), even though there exists τ such that ω2 (τ U ) = (1, 1). Hence the weights are not disjoint. Although this definition of disjointness is a bit involved, in Section 3.4 we will give a sufficient (but not necessary) condition on the tableau which is easier to check. However, we use disjointness here to obtain the full generality of Theorem 8, which is one of the fundamental tools we use to construct non-zero tableaux. Theorem 8. Let U and V be tableaux such that • elements(U ) = {1, . . . m} and elements(V ) = {m + 1, . . . a} (renumber if necessary) • The weights ω(1, . . . , m|U ), ω(m + 1, . . . , a|V ) are such that qU and qV are non-zero by weight-set counting on ω. • The weight assignments corresponding to ω(1, . . . , m|U ) and ω(m + 1, . . . , a|V ) are disjoint. Then for T = U ∨ V , we have qT 6= 0 by weight-set counting on ω(T ). Proof. By weight-set counting on U and V we have P(U ) − N(U ) 6= 0 and P(V ) − N(V ) 6= 0. By Theorem 4, showing P(T ) − N(T ) 6= 0 implies qT 6= 0. Thus it suffices to show P(T ) − N(T ) = (P(U ) − N(U ))(P(V ) − N(V )) or equivalently P(T ) = P(U )P(V ) + N(U )N(V ) and N(T ) = N(U )P(V ) + P(U )N(V ). We will show P(T ) = P(U )P(V ) + N(U )N(V ). The N(T ) claim follows similarly. Consider the weight assignment (k1 , . . . , ka ) of T . ω(1, . . . , a|T ) with ²(τ ) is counted in P(T ). = If ω(k1 , . . . , ka |τ T ) = 1 (i.e., τ is positive for (k1 , . . . , ka )) then it Since the weight splits, we have ω(k1 , . . . , ka |τ T ) = ω(k1 , . . . km |τ|U U )ω(km+1 , . . . ka |τ|V V ) = ω(1, . . . , m|U )ω(m + 1, . . . , a|V ) = ω(1, . . . , a|T ) with ²(τ ) = ²(τ|U )²̇(τ|V ) = 1. Hence either ²(τ|U ) = ²(τ|V ) = 1 or ²(τ|U ) = ²(τ|V ) = −1. If ²(τ|U ) = 1 then since ω(k1 , . . . km |τ|U U ) = ω(U ) is a 24 valid weight assignment (because the weights are disjoint), it is counted in P(U ). Similarly for the P(V ) cases. The ²(τ|U ) = −1 cases are counted in N(U ). Thus P(T ) ≤ P(U )P(V ) + N(U )N(V ). Now any weight assignments ω(k1 , . . . , ka |τ|U U ) = ω(1, . . . , a|U ) in P(U ) and ω(km+1 , . . . , a|τ|V V ) = ω(m + 1, . . . , a|V ) in P(V ) must also correspond to the valid weight assignment ω(k1 , . . . , ka |τ T ) = ω(1, . . . , a|T ) with τ = τ|U × τ|V . Moreover ²(τ ) = ²(τ|U )²̇(τ|V ) = 1. So this weight assignment is in P(T ). Similarly for weight assignments in N(U ) and N(V ). Hence P(T ) ≥ P(U )P(V ) + N(U )N(V ). Thus we have P(T ) = P(U )P(V ) + N(U )N(V ) as desired. Theorem 8 allows us to construct a non-zero tableau from smaller non-zero tableaux. The main difficulty in applying this theorem is showing that the weight-sets are disjoint. To deal with this, we develop an idea of maximality of weights which is sufficient for weight-set disjointness. 3.4 Maximal Weights Since the action of Sa on a tableau T does not change the resulting qT , we generalize the definition of tableau weight to account for this. We put an order on this generic weight, thus defining maximal weights. Definition 3.4.1. Given a tableau T , the generic form of ωi (T ) is wi (T ) = ωi (πT ) = (x1 , . . . , xa ) for any π ∈ Sa such that xj ≥ xj+1 for all j. In essence, wi (T ) is the weights of ωi (T ) listed in decreasing order. This definition works for any row of T . We define the generic form of a weight on the entire tableau (assuming T has at xa most three rows) by, w(T ) = ω2,3 (πT ) = ( xy11 ... ... ya ) for any π ∈ Sa such that yj ≥ yj+1 for all j and if yj = yj+1 then xj ≥ xj+1 . We consider only the weight vectors of the second and third rows and list the vectors so that the row three weights are decreasing. If two vectors have the same weight for row three, we list the vector with the larger weight in row two first. 25 221 For instance, if T = 1 1 2 then ω2,3 (T ) = ( 20 10 03 ). Hence w(T ) = ( 03 20 10 ). 333 Definition 3.4.2. We put an order on generic weights by wi (T1 ) > wi (T2 ) if when wi (T1 ) = (x1 , . . . xa ), wi (T2 ) = (v1 , . . . va ), there exists k ≥ 1 such that xj = vj for j < k and xk > vk . We say w(T1 ) > w(T2 ) if 1. w3 (T1 ) > w3 (T2 ) or 2. w3 (T1 ) = w3 (T2 ) and w2 (T1 ) > w2 (T2 ) or ... ya 3. w3 (T1 ) = w3 (T2 ), w2 (T1 ) = w2 (T2 ), and if we have w(T1 ) = ( xy11 ... xa ) and za w(T2 ) = ( xz11 ... ... xa ) then there exists k ≥ 1 such that yj = zj for j < k and yk > zk . We also apply this ordering to sets of weight vectors, by associating to each set the generic weight vector formed by concatenating the given weights in order. So to ¡¢ ¡¢ the set A = { 02 , 11 } we associate the weight ( 02 11 ). Definition 3.4.3. We define the maximum generic weight of row i of T to be the maximum with respect to > of {wi (τ T )|τ ∈ CT }, where w is the generic weight defined above. Similarly the maximum generic weight of T is the maximum with respect to > of {w(τ T )|τ ∈ CT }. Note that the maximum generic weight of T is based only on the weights of rows two and three. As such we ignore the weight of the first row. 12 . Then the generic weights w2 (τ T ) are (1, 1) and 21 11 22 (2, 0), with (2, 0) (the generic weight of τ T = or ) as maximum. Here 22 11 Example 3.4.4. Let T = we’ve suppressed writing w3 (T ) since T has only two rows. 11444 If T = 2 2 3 3 3 then the maximum generic weight of T (of the second and third 33 rows) is ( 32 20 00 00 ) which is ω2,3 (T ) in generic form, w(T ). Definition 3.4.5. Let T be a tableau having three or fewer rows. Let wm be the maximum generic weight of T . We say wm is the max weight for T if wm occurs in qT . That is qT 6= 0 by weight-set counting on wm . 26 Unlike the maximum weight, the max weight of a tableau may not exist since the 111 weight may not occur in qT . For instance, consider T = 2 2 2 . We know qT = 0 by 333 Lemma 3.2.6, yet T has ( 03 30 00 ) as its maximum weight. The max weight for T is always the maximum weight for row three of T , but it 5233 need not be the maximum weight for row two of T . Consider T = 1 4 4 1 . The 33 maximum generic weight of T is ( 02 20 20 00 ), but the maximum generic weight of row two of T is (4, 0, 0, 0) which is not the generic form of (0, 2, 2, 0). Definition 3.4.6. If wm is the max weight for T , we say T is in maximal form provided ω2,3 (πT ) = wm for some π ∈ Sa . This only requires that some permutation of the weight vectors of ω2,3 (T ) be equal to the max weight of T . While the max weight may not exist for all tableaux, it is easy to show weight-set disjointness for those tableaux which are in maximal form. In order to prove this, we use the following lemmas regarding our ordering. Lemma 3.4.7. Given two weights, W1 and W2 , of the same length, let Ck = ¡ ¢¡ ¢ { xy | xy ∈ Wk }, k = 1, 2 be the multisets of weight vectors in each of these weights. Let A = C1 \ (C1 ∩ C2 ), the weight vectors of W1 not in W2 . Similarly, let B = C2 \ (C1 ∩ C2 ). If W1 ≥ W2 , then A ≥ B. Proof. Without loss of generality, assume Wi is written in maximal form (i.e., is equal to its generic weight). If W1 = W2 then A = B = ∅, so the result holds trivially. So suppose W1 > W2 . Then either W1 differs from W2 at some place in the third row, or the third rows are equal and they differ at some place in the second row (at least vectorwise). ¡ ¢¡ ¢ ¡ ¢¡ ¢ Define Ay = { xyii | xyii ∈ A, yi = y} and B v = { uvii | uvii ∈ B, vi = v}. Defining C1y and C2v similarly, we have Ay = C1y \ (C1y ∩ C2y ) and B v = C2v \ (C1v ∩ C2v ). xm u1 ... um Let W1 = ( xy11 ... ... ym ) and W2 = ( v1 ... vm ). If W1 differs from W2 in the third row, then there exists j such that yj > vj and yi = vi for all i < j. Hence |C1yi | = |C2vi | = |C2yi | for yi > yj , so |Ayi | = |B vi | = |B yi |. Then to show A > B it suffices to show 27 y y |Ayj | > |B yj |. But since yj > vj , we have |C1 j | > |C2 j |. Thus w3 (A) > w3 (B) and the result follows. Note that when the third rows are equal, this argument shows w3 (A) = w3 (B). If W1 and W2 are equal in the third row but the generic weights of their second rows differ, we can apply the same argument as above, where Ax , B x , Cix are the appropriate sets of weight vectors with the second row weight equal to x. This shows w2 (A) > w2 (B). Since we’ve already have w3 (A) = w3 (B), the result follows. If W1 and W2 have the same generic weights in the second and third rows, then by above we know the generic weights of rows two and three of A and B are the same. By definition, A ∩ B = ∅, so the second row of the first vectors in A and B are different. Now W1 and W2 agree in the third row, so the first vectors where they differ in the second row must be the first vector in A and B respectively. Since W1 > W2 we have the large vector occurring in W1 and hence in A. Thus A > B. The following lemma is an obvious property of the ordering, but is included due to the non-standard ordering used. Lemma 3.4.8. Given sets A and B of weight vectors, if A ≥ B and B ≥ A then A = B. xm Proof. View A and B as generic weight vectors WA and WB . Let WA = ( xy11 ... ... ym ) and um WB = ( uv11 ... ... vm ). Since A ≥ B we have (y1 , . . . , ym ) ≥ (v1 , . . . , vm ). So either the rows are equal or there exists j such that yj > vj and yi = vi for all i < j. As B ≥ A we would similarly get vj ≥ yj , which is a contradiction. Hence w3 (A) = w3 (B). A similar argument shows w2 (A) = w2 (B). ¡ ¢ ¡ ¢ If A 6= B, let j be the first place where they differ. Then A ≥ B implies xyjj > uvjj . Since yj = vj , this means xj > uj . But the same argument on B ≥ A implies uj > xj , which is a contradiction. Thus A = B. Now using these lemmas we can show that the max weights of tableaux are necessarily disjoint. 28 Lemma 3.4.9. If U and V are tableaux in maximal form, with U containing b copies of the elements 1 to m and V containing b copies of the elements m + 1 to a (after renumbering as necessary), then ω(U ) and ω(V ) are disjoint. ... xm xm+1 ... xa Proof. Let U and V be in maximal form. Let ω2,3 (U ∨ V ) = ( xy11 ... ym ym+1 ... ya ). To show that U and V are disjoint, we need to show that any valid weight assignment of ´ ³ x ... x x km+1 ... xka k km U ∨ V restricts to a valid weight assignment of U and V . Let yk11 ... ykm ykm+1 ... yka be a valid weight assignment of U ∨ V . That means there exists τ such that ³ x ... x x ´ k km km+1 ... xka ω2,3 (τ [U ∨ V ]) = yk11 ... ykm ykm+1 ... yka . We want to show it restricts to a valid ¡ x ... x m ¢ weight assignment of U , that is ω2,3 (τ|U U ) = ykk11 ... ykkm = ω2,3 (πU ) for some π ∈ Sm . ¡xi ¢ ¡xk ¢ This is equivalent to showing { yi |i = 1 . . . m} = { yki |i = 1 . . . m}. i ¡ ¢ Let C = { xyii |i = 1 . . . m} be set of the weight vectors of ω2,3 (U ) and D = ¡ ¢ { xykki |i = 1 . . . m} the set of weight vectors of ω2,3 (τ|U U ). Define A = C \ (C ∩ D) i and B = D \ (C ∩ D). So A consists of those weight vectors of U assigned to V which are distinct from the weight vectors of V assigned to U under this weight assignment. That is, the vectors in A occur in ω2,3 (U ) but not in ω2,3 (τ|U U ). The set B is the weight vectors are those vectors coming from ω2,3 (τ|U U ) which are not in ω2,3 (U ). To prove disjointness, we need to show that C = D, which is equivalent to showing A = B = ∅. Now U is in maximal form, so ω2,3 (U ) ≥ ω2,3 (τ 0 U ) for all τ 0 . Hence ω2,3 (U ) ≥ ω2,3 (τ|U U ). So by Lemma 3.4.7, we have A ≥ B. But we can also view A as the weight vectors of U in ω2,3 (τ|V V ) which are not in ω2,3 (V ). Similarly B is the set of vectors from ω2,3 (V ) which are not in ω2,3 (τ|V V ). Since V is also in maximal form, ω2,3 (V ) ≥ ω2,3 (τ|V V ). Hence by Lemma 3.4.7, we have B ≥ A. Thus Lemma 3.4.8 shows A = B. However, A ∩ B = ∅ by definition, so A = B = ∅. Thus the weights are disjoint. Example 3.4.10. Suppose U ω2,3 (1, 2, 3, 4|U ) = is a tableau in maximal form such that ( 30 12 00 00 ) and V is a tableau in maximal form with ω2,3 (5, 6, 7, 8, 9, 10, 11|V ) = ( 20 11 11 00 00 00 00 ). Suppose these weights were not disjoint. ¡¢ That means we must be able to assign some weight xy of U to V and some weight 29 ¡x0 ¢ y0 of V to U . First consider the vector ¡1¢ 2 of U (i.e., the vector with the largest weight in row three). Since V is in maximal form, we know that there can be at most one copies of any element in row three of τ V for any τ . Since 2 > 1, this vector cannot be assigned ¡¢ to V . Similarly, once we know that 12 remains a weight of U , the largest row three ¡¢ weight we can assign to U is 0. Hence 11 remains with V . ¡¢ ¡¢ Now consider 30 of U . Having V in maximal form means that when 11 are ¡¢ ¡¢ assigned to V , a vector ∗0 assigned to V must have ∗ ≤ 2. Thus 30 is assigned to U . Therefore the only vectors of U and V that can be assigned to each other are the ¡0¢ vectors. However, since a weight assignment is based only on the vector and not 0 its label, this is the same as a weight assignment arising from U and V . Hence the weights are disjoint. Lemma 3.4.9 shows that if U and V are in maximal form, qU ∨V 6= 0 by Theorem 8. We will apply Theorem 8 repeatedly when constructing tableaux. As such, we want U ∨ V to be in maximal form whenever U and V are. Lemma 3.4.11. If T1 and T2 are maximal tableaux filled with different elements, then T1 ∨ T2 is maximal. Proof. The tableaux have no elements in common so the weight splits over the join. Hence ω(τ [T1 ∨ T2 ]) = ω(τ1 T1 ∨ τ2 T2 ) = ω(τ1 T1 ) ∨ ω(τ2 T2 ). Since each tableau weight was maximal, so too is their join. Hence maximality is preserved under the join operation. Through this join operation, we will construct collections of tableaux. To show these tableaux are linearly independent (over C), we can simply compare max weights. Lemma 3.4.12. Let {Bp } be a set of tableaux in maximal form. If the max weights of these tableaux are distinct, then {qBp } is linearly independent. Proof. Assume {qBp } is not linearly independent. Let Bk be the tableau with the largest weight such that qBk is not linearly independent from the rest of {qBp }. Write 30 qBk = P ap qBp . Then ω(qBk ) = P ap ω(qBp ). Since Bk is in maximal form, ω(Bk ) occurs with non-zero coefficient in ω(qBk ). Hence ω(Bk ) must occur with non-zero coefficient in ω(qBp ) for some p. However, Bk was chosen such that ω(Bk ) ≥ ω(Bp ). By hypothesis these weights are distinct, so the inequality is strict. But the Bp are in maximal form, so ω(Bp ) is the largest weight occurring in qBp . Hence ω(Bk ) does P not occurring in ap ω(qBp ), contradicting the linear dependence. We will use Lemma 3.4.12 heavily in Chapter 9 to prove Theorem 3. To make use of this lemma, we need to have distinctness of max weights. When the tableaux are formed via the join operation, we can sometimes simplify the proof of max weight distinctness via the following lemma: Lemma 3.4.13. Suppose we have the two row tableaux T1 , T2 , T3 , and T4 , were the non-zero max weights are as follows: ω(T1 ) = (A, B), ω(T2 ) = (C, D), ω(T3 ) = (a, b), and ω(T4 ) = (c, d). Assume ω(T1 ) 6= ω(T3 ), ω(T2 ) 6= ω(T4 ), but λ2 (T1 ) = λ2 (T3 ) and λ2 (T2 ) = λ2 (T4 ). If λ2 (T1 ) 6= λ2 (T2 ), then ω(T1 ∨ T2 ) 6= ω(T3 ∨ T4 ). Proof. Since λ2 (T1 ) 6= λ2 (T2 ) and ω(T1 ) 6= ω(T3 ), assume A + B > C + D and A > a. If ω(T1 ∨ T2 ) = ω(T3 ∨ T4 ), either A = c or A = d since a ≥ b by maximality of ω(T3 ). Consider A = c, then B > d and B 6= c because C + D = c + d. Then A > a and A + B = a + b, implies B < b ≤ a. Hence there is no weight equal to B, so this cannot occur. Similarly if A = d then c < B and hence there is no weight equal to B. Thus the weights are distinct. Although Lemma 3.4.13 applies directly to the join of only two tableaux, it may often be applied in a broader context. Namely, if many of the tableaux being joined are the same, the question of distinct weights reduces to looking only at the weights of those tableaux which differ. This approach will be used and discussed in Chapter 9. 31 Chapter 4 S The Tableaux of 1S3boS b 3 In this chapter we completely classify and discuss those tableaux occurring in 1SS3b . b oS3 To begin with, we determine exactly which tableaux associated with this space have qT 6= 0. This is done in Section 4.1. In Section 4.2, we use this information and the results from Thrall’s paper, [20], to determine precisely which partitions occur. Finally in Section 4.3, we construct a complete tableau basis for each the tableaux space S λ,3 ∩ M λ,3 associated to the irreducibles of 1SS3b b oS3 4.1 Classification of qT 6= 0, for T filled with 1, 2, 3 Let T be a λ-tableau filled with b copies of the numbers 1, 2, and 3. By Lemma 3.1.5, T has at most three rows. By Remark 3.0.9 entry permutations, column permutations and column exchanges do not change whether qT is non-zero. Hence we may take T to be in the following the form: k l m n o p q T = 1112123 2233 3    k+l+m+o=b       k + l + n + p = b    k+m+n+q=b      l ≥ m ≥ n ≥ 0 32 However, since we know there are exactly b copies of every number in T , we may omit k l m n the tail and simply write T as T = 1112 , retaining the condition l ≥ m ≥ n ≥ 0 2233 3 and assuming the tail. Theorem 9. With T as described above, qT = 0 ⇐⇒    k + l odd       k + n odd l>m=n≥0 l=m≥n≥0    k+m even      k+l even l = m+2, n = m-1 ≥ 0 l = m + 1, m=n ≥ 0 Moreover, when qT 6= 0, it is non-zero by weight-set counting. Proof. ⇐= Using Remark 3.0.9, to show qT = 0 it suffices to exhibit π ∈ S3 , τ ∈ CT , ²(τ ) = −1, such that πτ T = T up to an exchange of columns. k l m n For l > m = n, l + k odd, take τ =(12)T × (12)T × ()T × ()T and π = (12). k l m n k l m n So ²(τ ) = (−1)k+l = −1. Then τ T = 2212 1121 and πτ T = . Since m= n, 1133 2233 3 3 exchanging columns gives T . k l m n For l = m ≥ n ≥ 0, k + n odd use π = (23), τ =(23)T × ()T × ()T × (12)T and interchange columns l and m. k mmm 1 1 1 1 2 Now consider when l = m+1, m = n with k + l even. Then T = . 2 2 2 3 3 3 kmmm 13 23 1 1 1 2 and T1 = . Let T2 = . Take σ 0 ∈ Write T = T ∗ ∨ T1 where T ∗ = 2 1 2 2 3 3 3 k m m m 32 RTi so σ 0 T2 = . Let π 0 = (123) ∈ S3 and τ ∗ =(132) × () × (12) × (12)∈ CT ∗ . 1 33 Then via reordering the columns by σ̃, we have σ̃τ ∗ π 0 σ 0 (T ∗ ∨ T2 ) = T ∗ ∨ T1 . Note that ²(τ ∗ ) = 1 and σ̃, σ 0 , π 0 commute with each other and all τ ∈ CT ∗ . Then qT = X X X σπ ²(τ )τ T σ∈RT π∈S3 τ ∈CT = X X X σπ ²(τ )τ (T ∗ ∨ T1 ) σ∈RT π∈S3 τ ∈CT = X X σ∈RT π∈S3 = X X ²(τ )τ T ∗ ) ∨ ( τ ∈CT ∗ X σπ[( σ∈RT π∈S3 = X σπ[( X X X X τ 0 ∈CT1 ²(τ )τ T ∗ ) ∨ (T1 − T2 )] τ ∈CT ∗ σπ ²(τ )τ (T ∗ ∨ T1 − T ∗ ∨ T2 ) σ∈RT π∈S3 τ ∈CT ∗ = X X X σπ ²(τ )τ T ∗ ∨ T1 − σ∈RT π∈S3 τ ∈CT ∗ = ²(τ 0 )τ 0 T1 )] X X X X X X σ∈RT π∈S3 τ ∈CT ∗ σπ ²(τ )τ σ̃τ ∗ π 0 σ 0 T ∗ ∨ T2 − = σπ ²(τ ) ²(τ ∗ )τ T ∗ ∨ T2 − σ∈RT π∈S3 τ ∈CT ∗ = X X X X X X σπ ²(τ )τ T ∗ ∨ T2 σ∈RT π∈S3 τ ∈CT ∗ σ∈RT π∈S3 τ ∈CT ∗ X X X σπ ²(τ )τ T ∗ ∨ T2 X X X σπ ²(τ )τ T ∗ ∨ T2 σ∈RT π∈S3 τ ∈CT ∗ σπ ²(τ )τ T ∗ ∨ T2 − σ∈RT π∈S3 τ ∈CT ∗ X X X σπ ²(τ )τ T ∗ ∨ T2 σ∈RT π∈S3 τ ∈CT ∗ =0 Note that where appropriate we commute σ̃, σ 0 , and π 0 to combine with σ and π and reparameterize. The ²(τ ∗ ) factor arises from the reparamterization of τ τ ∗ . Now consider when l = m+2, n = m-1 with k + m even. k m-1 m-1 m-1 1 1 1111 1 2 . 2 2 2223 3 3 3 11112333 T1 = . 2223 So T = k m-1 m-1 m-1 1 1 1 2 Write T = T ∗ ∨ T1 where T ∗ = and 2 2 3 3 3 We will use the idea of the previous case to show qT = 0. However, we need to sum over all ²(τ )τ T1 with τ in CT1 . These tableaux have different symmetry relations 34 with each other, so we will list all the ²(τ )τ T and their relations. Note that the tail is omitted for readability. T1 = 1111 2223 T5 = − T9 = 1113 2221 1221 2113 T13 = − 2111 1223 T2 = − 2123 1211 1211 2123 T3 = − 1121 2213 T4 = − T6 = 2211 1123 T7 = 2121 1213 T8 = T10 = 1213 2121 T11 = 1123 2211 T12 = − T14 = − 2213 1121 T15 = − 2221 1113 T16 = 2113 1221 1223 2111 2223 1111 For appropriate σ 0 ∈ RT1 we have the following relations. All permutation listed are from S3 . T1 = −(123)σ 0 T14 T2 = −(123)σ 0 T9 T3 = −(123)σ 0 T6 T4 = −(123)σ 0 T7 T5 = (12)σ 0 T13 T8 = (12)σ 0 T8 T10 = (12)σ 0 T10 T12 = (23)σ 0 T15 T16 = (23)σ 0 T16 T11 = (12)σ 0 T11 We also have some relations on T ∗ , namely that for π ∗ ∈ S3 a transposition, then π ∗ T ∗ = τπ∗ T ∗ for τπ∗ ∈ CT ∗ with ²(τπ∗ ) = (−1)k+m−1 . Also (132)T ∗ = τ(132) T ∗ for τ(132) ∈ CT ∗ with ²(τ(132) ) = (−1)2(m−1) = 1. Then X X X πσ ²(τ )τ (T ∗ ∨ (π ∗ )−1 σ 0 Ti ) σ∈RT π∈S3 τ ∈CT ∗ = X X X πσ ²(τ )τ (π ∗ T ∗ ∨ σ 0 Ti ) σ∈RT π∈S3 τ ∈CT ∗ = X X X πσ ²(τ ) ²(τπ∗ )τ (T ∗ ∨ σ 0 Ti ) σ∈RT π∈S3 τ ∈CT ∗ = X X X σ∈RT π∈S3 τ ∈CT ∗ πσ ²(τ ) ²(τπ∗ )τ (T ∗ ∨ Ti ) 35 Hence if Ti = P P P σ∈RT then π∈S3 P P σ∈RT −(123)σ 0 Tj or Ti τ ∈CT ∗ πσ ²(τ )τ [(T P π∈S3 ∗ = π ∗ σ 0 Tj for π ∗ a transposition, then ∨ Ti ) + (T ∗ ∨ Tj )] = 0. Also, if Ti = π ∗ σ 0 Tj , τ ∈CT ∗ πσ ²(τ )τ (T ∗ ∨ Ti ) = 0. So using the cancellations above, we have qT = X X X πσ ²(τ )τ (T ∗ ∨ T1 ) σ∈RT π∈S3 τ ∈CT = X X πσ[( σ∈RT π∈S3 = X ²(τ )τ T ∗ ) ∨ ( ²(τ 0 )τ T1 )] τ 0 ∈CT1 τ ∈CT ∗ 16 X X X X X πσ ²(τ )τ (T ∗ ∨ Ti ) i=1 σ∈RT π∈S3 τ ∈CT ∗ =0 =⇒ To prove these are the only non-zero cases, we will use weight-set counting of Theorem 4 to show qT 6= 0 in the remaining cases. Given a weight, for every weight assignment π ∈ S3 we will count (with sign) the number of τ ∈ CT such that πω(T ) = ω(τ T ) and show the sum of these numbers is non-zero. In some cases it may be necessary to use ω(τ 0 T ) instead of ω(T ) to show the weight-sum is non-zero. Since applying τ 0 affects only the sign of qT , this will not change our result. Specifically we wish to show if: k + l even l>m=n≥0 k + n even l=m≥n≥0 then qT 6= 0 l>m>n unless l=m+1, m=n, k+l even, or l=m+1, m-1=n, k+m even. Note that if k = l = m = n = 0 then T has only one row and CT = 1. In this case qσT = qT , so there is exactly one distinct T . Since k + n = 0, so the statement holds. Similarly, if l = m = n = 0, applying Lemma 3.2.6 gives k even. Hence we may assume l > 0. 36 We will use the first weight-set to illustrate the technique and notation of weightset counting. This method of argument will be used extensively throughout the rest of the paper. We will list our weight-counting in a table of the following form: ωi = (l, m, n) Tableau # ² τ (x, y, z) T0 j (−1)² l m bound n τ1 × τ2 × τ3 m=n The column headings are: the weight being used, the form of the tableau, the number of τ corresponding to this weight, the sign of the τ , the form of τ , and any bounds required. The subsequent lines correspond to different weight assignments in line i. By (x, y, z), we mean ωi (x, y, z) = (l, m, n). When m=n, there are j distinct l m n τ such that τ =τ1 × τ2 × τ3 . All the τ have sign (−1)² and τ T is of the form T 0 . The following is a standard formula that we will use in computing these weight sums. Its proof is a straightforward inductive application of Pascal’s Identity. For ¡¢ notation purposes, we take ba = 0 for b > a or b < 0. We also use the convention ¡0¢ = 1. 0 Lemma 4.1.1. ¡a+h¢ b ¡¢ P ¡ ¢ − ba = hi=1 a+h−i b−1 Consider the cases where k ≥ 0 and at least l > 0. We will apply weightset counting to rows two and three. For the most part, the table should be self explanatory, though we will discuss the first weight-set table for clarity. k l m n We’ll start with T = 1113 , 2232 3 m ω2,3 = ( 00 k+l+n 0 k) 37 ω2,3 Tableau # ² τ bound 1 (−1)0 ()T × ()T × ()T × ()T 1 (−1)k+n (23)T × ()T × ()T × (12)T ¡m¢ n (−1)k+l (12)T × (12)T × (12)T × (12)T 1 (−1)l+n (123)T × (12)T × ()T × (12)T ¡ l ¢ m−n (−1)l+n (132)T × (12) T × (12)T × ()T 1 (−1)k+m (13)T × ()T × (12)T × ()T k l m n (1, 2, 3) T = 1113 2232 3 k l m l m n k l m n (1, 3, 2) T = 1112 3233 2 k n l=m k l m n (2, 1, 3) T = 2 2 31 2 1 1 13 3 3 k l n n m n k l m n (2, 3, 1) T = 2212 3133 1 k l k l+n−m l=m k l m n (3, 1, 2) T = 3 21 3 3 1 12 1 2 2 m n k l m n (3, 2, 1) T = 3133 2212 1 k l m n To understand how these τ are obtained, first apply the permutations needed to have ω(x) = 0, for the appropriate x. Additionally, apply necessary permutations so that row three of T has the correct weight. Once this is done, there will only be one column block whose permutations have not been specified. Apply the number of permutations needed to get the correct weight. For the first line of this table, we see that T has the desired weight and any column permutations will change this. Thus there is exactly one τ and it is positive. In line two, we have ω3 (2) = k and ω2,3 (1) = 0. Hence we must apply (23)T to column block k. Columns 12 and 13 cannot move since ω2,3 (1) = 0. This gives l 2’s in row two, so we must have l=m and apply (12)T to block n. When l=m, this completely determines τ , and it has sign (−1)k+n . No such τ exists for l > m. 38 Line three counts ω2,3 (2, 1, 3). Since there can be no 2’s in either rows two or k l n three, τ must contain (12)T × (12)T × (12)T . As row three already contains k 3’s, column block k needs no other permutation. Hence column 13 is the only column where τ has not yet been determined. As it stands, we already have k+l 1’s in row n two, hence only n more are required. Thus we need to apply (13)T to 13 . There are ¡m¢ ¡m¢ ways to choose which n columns move within the m block. Hence we get n n distinct τ of the form described. In line four we apply (132)T to block k in order to have ω3 (1) = k and ω2,3 (2) = 0. Additionally, ω2,3 (2) = 0 means we must apply (12)T to block l and (12)T to block n. This gives l 1’s in row two, so we must have l=m and leave block m unchanged. In this case there is one such τ ; it has sign (−1)l+n . Line five is similar to line three. From the constraints ω3 (1) = k and ω2,3 (3) = 0 k m n we have that τ contains (123)T × (12)T × ()T . In order to have the correct weight l+n-m ¡ l ¢ for row two, we need l+n-m more 2’s. There are m-n ways to choose (12) T from block l and all these τ have sign (−1)l+n . k The last line has a similar argument to line one. We need only apply τ =(13)T m × (12)T to get the correct weight and this is the only possible τ . From this table we get the following sums. ¶ ¶ µ l m l+n + (−1)k+m + (−1) 1 + (−1) m−n m−n µ ¶ µ ¶ k+n k+l l l+n l+n l + (−1)k+l + (−1) + (−1) 1 + (−1) + (−1) n n µ k+l l 6= m (4.1.1) l=m (4.1.2) ¡ l ¢ ¡ m ¢ Case I: (l > m > n). Here, (4.1.1) equals zero only if | m−n ± m−n | = 0 or 2. For it to equal 0, we must have m = n or l = m. ¡ l ¢ ¡ m ¢ To have | m−n − m−n | = 2, and (4.1.1) equal to zero, we must have k + m even and k + n odd. Applying Lemma 4.1.1, we get l = m + 1 or l = m + 2. ¡ m ¢ ¡m+1¢ ¡m+1¢ ¡ m ¢ If l = m + 1, then (4.1.1) becomes 1 − m−n + m−n + 1. Now m−n − m−n = 2 ¡ m ¢ only if m−n−1 = 2, that is m = 2 and n = 0. This contradicts k + m even and k + n odd. Hence (4.1.1) is non-zero for l = m + 1. If l = m + 2 and k + m even, (4.1.1) becomes 1 − ¡ m ¢ m−n ¡ ¢ + 1. + (−1)m+n m+2 m−n 39 For this expression to be zero we must have n = m − 1, in which case we’ve already shown qT = 0. Thus (4.1.1) is non-zero unless l = m + 2 and n = m − 1. ¡ l ¢ ¡ m ¢ Finally, m−n + m−n = 2 only when m = n. Case II: (l = m ≥ n, l > 0). We need only show the expression (4.1.2) is non-zero ¡ ¢ ¡ ¢ for k+n even. From this, (4.1.2) becomes 2+(−1)k+l (1+ nl )+(−1)l+n (1+ nl ). The ¡ ¢ parity of k and n is the same, so we reduce to determining when 1+(−1)k+l ((1+ nl ) = ¡ ¢ 0. Since nl > 0, this cannot occur. Case III: (l > m = n). We want a non-zero weight sum for k+l even. Under these conditions, expression (4.1.1) becomes 1 + 1 + (−1)l+n + (−1)k+n . This is non-zero unless k + n is odd. It remains to show qT 6= 0 for l > n = m, k + l even, k + m odd. k l m n For this, consider the following weight-set counting on T = 1112 2233 3 with 0 n+1 ω2,3 = ( k+l+m-1 0 0 k ). Note that this weight-set is not the weight of T . We are counting which permutations τ will correspond to a weight assignment π, where ω2,3 = πω2,3 (τ T ). 40 ω2,3 Tableau # ² τ Bound k l m n (1, 2, 3) T = ¡ m ¢ 2 2 31 2 1 1 13 3 3 m−1 k l m−1 n (−1)k+l+m−1 (12)T × (12)T × (12)T × ()T (−1)m+n+l−1 (132)T × (12)T × (12)T × (12)T m≥1 k l m n (1, 3, 2) T = ¡ l ¢ 3 21 3 3 2 12 1 2 1 l−1 k l−1 m n l≥1 k l m n (2, 1, 3) T = ¡ n ¢ 1 1 1 32 2 2 3 23 3 k (−1)m−1 m−1 l m m≥1 m−1 ()T × ()T × ()T × (12)T m−1≤n k l m n (2, 3, 1) T = 3 21 3 2 2 12 1 3 1 ¡ ¢ l n+1−m (−1)k+1 k n+1−m m n (13)T × (12) T × (12)T × (12)T m≤n+1 n+1−m≤l k l m n (3, 1, 2) T = ¡ n ¢ 1 1 1 32 3 2 3 23 2 (−1)k+n+l+1 l−1 k l m n+1−l (23)T × ()T × ()T × (12) T l≥1 l−1≤n k l m n (3, 2, 1) T = 2 2 31 2 3 1 13 3 1 ¡ ¢ m n+1−l (−1)n+1 k l n+1−l n (123)T × (12)T × (12) T × ()T l≤n+1 n+1−l≤m Since we previously dealt with the l = n + 1 case, the last two lines of the table do not contribute. Hence the table gives the weight sum: (−1)m−1 m + (−1)l−1 l + (−1)m−1 m + (−1)l+1 l l 6= n + 1, m 6= 0 (4.1.3) Consider l > m = n, k + l even, k + n odd. For l 6= n + 1, m 6= 0, (4.1.3) is 2((−1)m−1 m + (−1)l−1 l), which is non-zero as l 6= m. If m = 0, we must have k and l odd. This makes (4.1.3) l + l which is not equal to zero since l 6= 0. For l = n + 1 we’ve already shown qT = 0. Thus we determined all the non-zero tableaux of 1SS3b . b oS3 41 4.2 S The Irreducibles Partitions of 1S3b b oS3 For Theorem 2, we need to know which irreducibles occur (i.e., have non-zero multiplicity) in 1SS3b . We call a shape (or partition) non-zero if the multiplicity of the corb oS3 responding irreducible in 1SS3b is non-zero. By Lemma 2.2.7, the non-zero partitions b oS3 are those partitions where dim(S λ,3 ∩ M λ,3 ) > 0. Since qT generates S λ,a ∩ M λ,a we need only determine from Theorem 9 which partitions have non-zero tableaux. These partitions and their multiplicities were completely determined by Thrall in [20]. We will first derive them from Theorem 9 and then confirm it with Thrall’s result. We will only consider shapes [λ] which are partitions of 3b. Consider T = |{z} | {z } s |{z} r . This labeling will be useful in our later constructions, so we will t k l m n derive the non-zero shapes in terms of it. When T is of the form 1112 , we have 2233 3 t = k, s = l+m+n and the tail r = 3b − 2s − 3t. Since 1, 2, and 3 all occur k = t times in the part of T above t, we will sometimes replace b with b0 = b − k when considering the multiplicity of elements in columns r and s, since the subtableau formed by columns r, s and the tail will have the elements 1, 2, and 3 occurring b0 times each. In partition notation we have λ = [r + s + t, s + t, t]. 4.2.1 Non-Zero Partitions from Theorem 9 Definition 4.2.1. For our purposes, we call a partition (or shape) λ of n required if there is a non-zero λ-tableau T filled with n3 copies of the elements 1, 2, and 3. These are precisely the tableaux determined in Theorem 9. Specifically, a required shape is one for which we must construct an appropriately filled non-zero tableau in order to prove Theorems 1 and 2. These shapes are explicitly determined in Theorem 10. To determine the non-zero [λ] = [r + s + t, s + t, t], we analyze the required partitions that correspond to non-zero tableaux in Theorem 9. We find that: 42 Theorem 10. The only partitions [r + s + t, s + t, t] of n = 3b which do not occur in 1SSnb oS3 are those with s or r = 1 as well as those having s + t odd and s or r ∈ {0, 2, 4}. Equivalently, a partition is non-zero if, for r, s 6= 1, when r or s is in {0, 2, 4}, then s + t is even. Proof. For a given partition λ = [r + s + t, s + t, t], we need only find (l, m, n) with l + n + n = s, k = t, and the conditions of Theorem 9 satisfied to show qT 6= 0. This shows that S λ must occur in 1SS3b . Since all values of t = k can occur when b oS3 l > m > n, (if l 6= m + 2), we consider tableaux of this form. Given t, the elements 1, 2 and 3 will occur b0 = b − t times in the remaining columns l, m, n, and the 0 tail of T . For each t we need to determine which s for 0 ≤ s ≤ b2 yield required non-zero partitions. To do so we will take the following parameterizations of (l, m, n) and determine the corresponding s. (l, m, n) = (i + 2, i + 1, i), we get s = 3i + 3 for = (i + 3, i + 2, i) s = 3i + 5 = (i + 4, i + 3, i) s = 3i + 7 b0 − 3 2 b0 − 5 0≤i≤ 2 b0 − 7 0≤i≤ 2 0≤i≤ For a given parameterization of (l, m, n) by i, we have s = l + m + n. Since there are at most b0 1’s in the r and s sections, we must have l + m] ≤ b0 which gives the upper bound on i. These parameterizations of (l, m, n) are non-zero by Theorem 9 since l > m > n and l 6= m + 2. Moreover, the parameterizations cover all equivalence classes of s (mod 3). Hence this tells us that all partitions with s ≥ 5 or s = 3 are non-zero, leaving aside the upper bound on s for now. When s = 4, the possibilities for (l, m, n) are (4, 0, 0), (2, 2, 0), and (2, 1, 1) which are non-zero only when s + t, (i.e., k) is even. For (3, 1, 0) the tableau is always zero since l = m + 2, n = m − 1. For s = 2 the only possibilities are (2, 0, 0) and (1, 1, 0). The non-zero conditions of Theorem 9 require k = t, and hence s + t, to be even in both cases. Similarly for s = 0, we must have k 43 even for T to be non-zero, as shown in Lemma 3.2.6. For s = 1, the only possibility is (l, m, n) = (1, 0, 0), and such a tableau is always zero. Determining the upper bounds on s corresponds to determining lower bounds on r. We will similarly give parameterizations of (l, m, n) which will cover all equivalence classes of r (mod 3). Using the equation 3b0 − 2s = r we get the corresponding equations and lower bounds. b0 − 3 2 b0 − 5 i≤ 2 b0 − 7 i≤ 2 For s = 3i + 3, i≤ s = 3i + 5, s = 3i + 7, then r ≥ 3, r ≡ 0 (mod 3) (4.2.1) r ≥ 5, r ≡ 2 (mod 3) (4.2.2) r ≥ 7, r ≡ 1 (mod 3) (4.2.3) This is the parameterization table for s rewritten in terms of r. So for r ≥ 5, all partitions are non-zero, provided the conditions on s are met. 0 If r = 0, we must have l = m = n = b2 . By Theorem 9, this shape is non-zero only if k+l is even. Thus only the shapes with s+t = k+l+m+n even are non-zero. For r = 1 we get the constraints l + m = l + n = b0 and m + n = b0 − 1. This means m = n, and l=m+1. These tableaux are always zero. When r = 2, we must have either l = m or m = n. These shapes will be non-zero if k+n or k+l is even respectively. In either case, s + t is even. 0 0 0 , b −2 , b −4 ) . In For r = 4, we must have either m = n, l = m, or (l, m, n) = ( b +2 2 2 2 the first two cases, the non-zero conditions of Theorem 9 force s + t to be even. In the last case, the tableau is always zero. 0 0 If r = 3 or 5 then i = b −3 or i = b −5 is an integer for b0 odd so the parameterization 2 2 listed works. For b0 even, these partitions are not required since 3b0 − 2s = r, would make r even. In addition, Theorem 9 shows the remaining partitions are zero. 44 4.2.2 Partition Multiplicities according to Thrall In [20], Thrall determines the partitions occurring in 1SS3b with multiplicity, which b oS3 he calls f (λ). If λ = [λ1 , λ2 , λ3 ], he gives the following method to compute f (λ): To the minimum of 1 + λ1 − λ2 and 1 + λ2 − λ3 , add whichever of −2, 0, +2 will give a result divisible by 3. If this result is even, divide by 6 to get f (λ). If the result is odd, add or subtract 3 according to λ2 being even or odd and then divide by 6 to get f (λ). Letting λ = [r + s + t, s + t, t] we have min(1 + λ1 − λ2 , 1 + λ2 − λ3 ) = 1 + min(r, s). Then f (λ) = 1+min(r,s)+x+y where x ∈ {−2, 0, 2} such that 1 + min(r, s) + x ≡ 0 6 (mod 3), y ∈ {−3, 0, 3} such that 1 + min(r, s) + x + y ≡ 0 (mod 6), and y ≥ 0 if s + t even and y ≤ 0 if s + t odd. Theorem 11. Writing s = 6k + j, r = 6h + i with 0 ≤ i, j ≤ 5, then by [20], the multiplicity of Sλ in 1SS3b is f (λ), where b oS3 f (λ) =     k        k        k+1        k+1    h        h        h+1        h+1 s ≤ r i = 0, 2, 4, s + t odd s≤r i=1 s ≤ r i = 0, 2, 4, s + t even s ≤ r i = 3, 5 r < s i = 0, 2, 4, s + t odd r<s i=1 r < s i = 0, 2, 4, s + t even r < s i = 3, 5 Hence f (λ) 6= 0 for r, s 6= 1, provided s + t is even when r or s is in {0, 2, 4}. This agrees with our results in Section 4.1. 45 4.3 Construction of Basis Tableaux for c = 3 Recall that the space S λ,c ∩ M λ,c is spanned by {qT } where the T are λ-tableaux filled with the numbers 1 to c. By Lemma 2.2.7, dim(S λ,c ∩ M λ,c ) equals the multiplicity of S λ in 1SSnd oSc . Given a partition λ of n = 3b, we want a set of tableaux {Bp } such that {qBp } is linearly independent and that |{Bp }| is the multiplicity of the irreducible corresponding to λ in 1SSnb oS3 . We call these Bp the basis tableaux for c = 3. These tableaux will be used in Chapter 9 for the proof of Theorem 3. We will build these tableaux from the following components: 223333 111122 M1 = 1 2 3 M2 = 33 M3 = 1 1 22 3323 M4 = 1 1 1 2 2 a-1 a-1 Na = 2 3 2 3 1 1 1 < a ≤ n3 = b In constructing these basis tableaux, we want tableaux filled with only the numbers 1, 2, and 3. We will use Y to denote the joining of tableaux without renumbering them. For example, M2 Y M1 = 223333123 . 111122 Let λ = [r + s + t, s + t, t]. When t is even, write s = 6k + j and r = 6h + i, with 0 ≤ i, j ≤ 5. Let g = min(k, h). Since λ is a partition of n = 3b we have . (So δ = 0 for i = j, δ = 1 3t + 2s + r = 3b. Hence 2j + i ≡ 0 (mod 3). Let δ = i−j 3 for i = j + 3, and δ = −1 for i = j − 3.) When t is odd, we proceed as above, except 0 , and g = min(k 0 , h). let s − 3 = 6k 0 + j 0 , δ = i−j 3 For p = 1, 2, . . . , g, we define the basis tableaux: Bp = 2t M3 Y N6p+j Y (k − p)M2 Y (2h − 2p + δ)M1 (t even) Bp = t−1 M3 Y M4 Y N6p+j 0 Y (k 0 − p)M2 Y (2h − 2p + δ)M1 2 (t odd) 46 Additionally, for j, j 0 6= 1 we have the tableau B0 , which is Bp with p = 0 under certain conditions. B0 = 2t M3 Y Nj Y kM2 Y (2h + δ)M1 , j>1 (t even) B0 = 2t M3 Y kM2 Y (2h + δ)M1 j=0 (t even) B0 = t−1 M3 Y M4 Y Nj 0 Y k 0 M2 Y (2h + δ)M1 , 2 j0 > 1 (t odd) B0 = t−1 M3 ∨ M4 Y k 0 M2 Y (2h + δ)M1 2 j0 = 0 (t odd) Note that if δ = −1, then B0 exists only for h ≥ 1 and Bg exists only for g < h. To demonstrate that the {Bp } is a basis, we need to verify that they: • Have the correct shape, • Are non-zero and maximal, • Are linearly independent, • Span the space. Shape: First consider the shape of these tableaux. We need to show these tableaux have shape λ = [r + s + t, s + t, t]. For t even: λ3 (Bp ) = 2t λ3 (M3 ) + λ3 (N6p+j ) + (k − p)λ3 (M2 ) + (2h − 2p + δ)λ3 (M1 ) = 2t ∗ 2 + (k − p) ∗ 0 + (2h − 2p + δ) ∗ 0 =t λ2 (Bp ) = 2t λ2 (M3 ) + λ2 (N6p+j ) + (k − p)λ2 (M2 ) + (2h − 2p + δ)λ2 (M1 ) = t + 6p + j + (k − p) ∗ 6 + (2h − 2p + δ) ∗ 0 = t + 6k + j =t+s 47 λ1 (Bp ) = 2t λ1 (M3 ) + λ1 (N6p+j ) + (k − p)λ1 (M2 ) + (2h − 2p + δ)λ1 (M1 ) = t + (6p + j) ∗ 2 + (k − p) ∗ 6 + (2h − 2p + δ) ∗ 3 = t + 6k + j + 6h + j + 3δ = t + s + 6h + j + 3δ = t + s + 6h + i =t+s+r When t is odd we have: λ3 (M3 ) + λ3 (M4 ) + λ3 (N6p+j 0 ) + (k 0 − p)λ3 (M2 ) + (2h − 2p + δ)λ3 (M1 ) λ3 (Bp ) = t−1 2 = t−1 ∗ 2 + 1 + (k 0 − p) ∗ 0 + (2h − 2p + δ) ∗ 0 2 =t λ2 (Bp ) = t−1 λ2 (M3 ) + λ2 (M4 ) + λ2 (N6p+j 0 ) + (k 0 − p)λ2 (M2 ) + (2h − 2p + δ)λ2 (M1 ) 2 = t − 1 + 4 + 6p + j + (k − p) ∗ 6 + (2h − 2p + δ) ∗ 0 = t + 3 + 6k 0 + j 0 =t+s λ1 (Bp ) = 2t λ1 (M3 ) + λ1 (M4 ) + λ1 (N6p+j 0 ) + (k 0 − p)λ1 (M2 ) + (2h − 2p + δ)λ1 (M1 ) = t − 1 + 4 + (6p + j 0 ) ∗ 2 + (k − p) ∗ 6 + (2h − 2p + δ) ∗ 3 = t + 3 + 6k 0 + j 0 + 6h + j 0 + 3δ = t + s + 6h + j 0 + 3δ = t + s + 6h + i =t+s+r A similar computation works for the shape of B0 . Hence these tableaux have the correct shape. Moreover, within each component, the same number of 1’s, 2’s and 3’s 48 were used. Hence the Bp have the correct number of 1’s 2’s and 3’s. Maximality: When t is even, a generic basis element (with the tail suppressed) looks like: tabc 32 3 3 Bp = 11 1 2 2 a = 2(k − p) + 6p + j − 1 b = 2(k − p) + 1 c = 2(k − p) ¡ ¢ 2(k−p) 0 Then ω2,3 (Bp ) = t+6p+j+4(k−p) . Since b > c and a > b (for p ≥ 1), this 0 t 0 weight is maximal. Bp is also non-zero, since the only other possible weight assignment ¡ ¢ 0 2(k−p) . This has sign (−1)t+2(k−p) = 1 as t is even. is t+6p+j+4(k−p) 0 0 t When t is even, the tableau B0 is the same as Bp with p = 0 when j > 1. In this case a > b and the above argument holds. There is no B0 for j = 1. When j = 0, we have (suppressing the tail): taaa 32 3 3 B0 = 11 1 2 2 a = 2k 2k 0 Then ω2,3 (B0 ) = ( t+4k 0 t 0 ). This weight is clearly maximal. Although there are many different weight assignments possible for B0 , all weight assignments are positive since each column block is even. Hence B0 is non-zero. When t is odd, a generic basis element (with the tail suppressed) looks like: tabc 32 3 3 Bp = 11 1 2 2 Then ω2,3 (Bp ) = a = 2(k 0 − p) + 6p + j 0 b = 2(k 0 − p) + 2 c = 2(k 0 − p) + 1 ¡ t+6p+j 0 +4(k0 −p)+2 2(k0 −p)+1 0 ¢ 0 t 0 . Since b > c and a ≥ b (for p ≥ 1), this weight is maximal. Bp is also non-zero since the only other possible weight ¡ 0 0 0 −p)+2 0 2(k 0 −p)+1 ¢ assignment is t+6p+j +4(k . This has sign (−1)t+2(k −p)+1 = 1 as t is 0 0 t odd. 49 When t is odd, the tableau B0 is the same as Bp with p = 0 when j 0 > 1. In this case a ≥ b and the above argument holds. There is no B0 for j 0 = 1 (since qB0 is zero when j 0 = 1). When j 0 = 0, we have (suppressing the tail): taaa 32 3 3 B0 = 11 1 2 2 Then ω2,3 (B0 ) = ¡ t+4k0 +2 2k0 +1 0 ¢ 0 t 0 a = 2k 0 + 1 . This weight is clearly maximal. Although there are many different weight assignments possible for B0 , These weight assignments always move exactly two or four column blocks. Since the size of the column blocks is odd, all weight assignments are positive. Hence B0 is non-zero. Linear Independence: To show the tableaux Bp are linearly independent, by Lemma 3.4.12 it suffices to show their max weights are distinct. First consider t even. Say w2,3 (Bp ) = w2,3 (Bp0 ), with p < p0 . Then we must have t + 6p + j + 4(k − p) = 2(k − p0 ), which forces k = p = j = t = 0. Then our partition is just [n], so only one tableau is needed. If w2,3 (Bp ) = w2,3 (B0 ) with p > 0, then we get t + 4k = 2(k − p) which implies t = k = p = 0. Hence the tableaux are linearly independent. When t is odd, the max weights must be distinct, since an argument similar to the one above shows t = 0 which is not possible. Span: Since the tableaux Bp are linearly independent they will span the space S λ,3 ∩ M λ,3 if |{Bp }| = mλ , where mλ = f (λ) as determined by [20]. (We listed f (λ) explicitly in Theorem 11.) First consider the case of t even. Given λ = [r + s + t+, s + t, t] with t even, then mλ depends on the relative sizes of r and s. For s ≤ r, we need k + 1 tableaux for j 6= 1 and k tableaux for j = 1. When s ≤ r, we have g = k. Thus we get k different Bp ’s and when j 6= 1 we have B0 as well. The restriction on the tableaux when δ = −1 occurs only when g = h. However, then h = k and i = j − 3 which contradicts s ≤ r, so this case does not occur here. Hence we have a full set of basis 50 tableaux. If r < s, we have g = h. How many tableaux we have depends on δ. Note that Bh exists only for δ 6= −1, and B0 requires h > 0 for δ = −1. The number of tableaux needed according to Theorem 11 also depends on δ. Since s ≡ j (mod 2) and 3δ = i − j, when δ = 0 we have i ≡ s (mod 2). Then s is even for i = 0, 2, and 4, hence by [20] we need we need h + 1 tableaux for i 6= 1 and h tableaux for i = 1. Since δ 6= −1, there are h distinct Bp . We also have B0 when i 6= 1 since i = j. Hence we have a full set of basis tableaux. If δ = 1, all the tableaux described when δ = 0 occur. Since δ = 1, we have i = j + 3 which implies i = 3, 4 or 5. For i = 4 we need need h tableaux since s is odd, while for i = 3, 5 we need h + 1 tableaux. When i = 4 we have h different Bp (though no B0 since j = 1). When i = 3, 5 then j = 0, 2. Hence B0 exists and we obtain the complete set of basis tableaux. For δ = −1 we have i = j − 3, so i = 0, 1, 2. We need exactly h tableaux since either i = 1 or s is odd. However, we no longer have Bh , so we get only h − 1 tableaux from the Bp . In addition, when h ≥ 1 we have B0 since j = i + 3 ≥ 3 and so j > 1. Hence we have h tableaux for h ≥ 1. If h = 0 then no tableaux are needed since either r = 1 or r = 0 or r = 2 with s odd. Thus the correct number of tableaux is given. Now consider the case when t is odd. When s ≤ r we need k tableaux for j = 0, 1, 2, or 4 and k + 1 tableaux for j = 3 or 5. If j = 3 or 5, then k 0 = k = g so there are k tableaux Bp , in addition to the tableau B0 (since j 0 = j − 3 6= 1). If j < 3 then g = k 0 = k − 1, so there are k − 1 tableaux Bp in addition to the tableau B0 . For j = 4, we have k tableaux Bp , since g = k 0 = k. However since j 0 = 1, B0 does not exists. The restriction on the tableaux when δ = −1 occurs only when g = h. However, then h = k 0 = k and i = j − 3 which contradicts s ≤ r, so this case does not occur here. Hence we have a full set of basis tableaux. When r < s we need h tableaux when i = 1. For i = 0, 2, and 4 we need h tableaux when s is even and h + 1 tableaux when s is odd. For i = 3 or 5 we need h + 1 tableaux. Note that s 6≡ j 0 (mod 2). 51 If δ = 0 then i = j 0 . We have h tableaux Bp , along with B0 for j 0 6= 1. Hence we have h tableaux when i = 1, and h + 1 tableaux otherwise. Since s is odd when i = 0, 2, or 4, this is the correct number. If δ = 1 then i = j 0 + 3, so i = 3, 4 or 5. If i = 3 or 5, then j 0 6= 1 and there are h + 1 tableaux as desired. If i = 4 then j 0 = 1, hence there are only h tableaux Bp . However, only h tableaux are needed here since s is even. If δ = −1 then i = j 0 − 3, so i = 0, 1 or 2. Since s is even when i 6= 1, only h tableaux are required in this case. However, we no longer have Bh , so there are h − 1 tableaux Bp . In addition, when h ≥ 1, we have B0 , since j 0 > 1, so the correct number of tableaux are obtained. If h = 0 then no tableaux are needed since either r = 1 or r = 0 or r = 2 with s + t odd. Thus the correct number of tableaux is given. 52 Chapter 5 Proof of Theorem 1 Recall, Theorem 1 says that every irreducible occurring in 1SSnb oS2 occurs in 1SSnd oSc with equal or greater multiplicity, where n = 2b = cd and b, c, d ≥ 2. In Section 3.2 we proved Theorem 7, which showed that the irreducibles occurring in 1SSnb oS2 were exactly those corresponding to partitions λ = [n − s, s] for s even and they occur with multiplicity one. (Since n = 2b is even, it suffices to consider only the even values of s.) By Remark 2.2.8, to prove Theorem 1, it suffices to construct a non-zero tableau filled with d copies of c elements for each partition [n − s, s], where 0 ≤ s ≤ n2 , s even and n = cd. To do this we will construct some non-zero generic tableaux that when assembled via Theorem 8 will produce all the shapes and fillings needed. Since we are constructing generic tableaux for many partitions and fillings, we will not use a fixed c. However, we assume that every element listed in the body of the tableau occurs d times, filling out the tail as needed. We apply weight-set counting to prove a tableau is non-zero. The tableaux we need are: Tableau U1 a d-a d-a a U1 = 1 1 2 ∼ 1 2 2 ω2 (U1 ) = (0, a) a even a≤d 53 λ = [2d − a, a] For this first tableau, we listed U1 both with and without the tail. Normally we will suppress the tail when writing these tableaux. U1 is non-zero by Lemma 3.2.6 since a is even. It is maximal since (a, 0) is the largest possible weight-set for this shape. Tableau U2 aabb U2 = 1 3 1 3 2 4 4 2 a+b≤d a, b > 0 ω2 = (0, a+b, 0, a+b) λ = [4d − 2(a+b), 2(a+b)] Examining the filling of U2 and a,b > 0 we find the following constraints on any valid weight assignment: (Recall that U ∗ corresponds to a possible tableau τ U2 .) • If ω2 (1|U ∗ ) = 0 then ω2 (2 and 4|U ∗ ) > 0. • If ω2 (2|U ∗ ) = 0 then ω2 (1 and 3 |U ∗ ) > 0. • We must have ω2 (1 or 2|U ∗ ) > 0 and ω2 (3 or 4|U ∗ ) > 0. Since any valid weight assignment of (0, a + b, 0, a + b) has exactly two zeros, the restrictions above show that (1, 2, 3, 4) and (2, 1, 4, 3) are the only valid weight a a b b assignments. These weights-sets correspond to applying τ =()T × ()T × ()T × ()T a a b b and τ =(12)T × (12)T × (12)T × (12)T respectively. As both of these τ have positive sign, qU2 6= 0. This tableau is maximal since every element x must have ω2 (x) ≤ a+b. 54 Tableau U3 a even abb U3 = 1 1 2 2 3 3 a+b≤d d even b = d2 ω2 = (0, a, d) λ = [2d − a, a + d] To show U3 is non-zero we will use weight-set counting on ω2 = (0, a, d). There are two cases for which we need to determine weight assignments, a + b < d and a + b = d. When a + b < d, only the element 3 may be assigned a row two weight of d. So the distinct weight assignments are (1, 2, 3) and (2, 1, 3), which occur with τ = ()T a b b and τ =(12)T × ()t × ()T respectively. Since a is even, both τ have positive sign. Hence U3 is non-zero. If a + b = d, then a = b = d2 and d ≡ 0 (mod 4). While every permutation corresponds to a distinct weight assignment, every weight assignment can only be obtained by having τ move complete column blocks. Since all of these blocks are even, τ is positive for every weight assignment and hence U3 is non-zero. This tableau is maximal since (d, a, 0) is the largest possible weight. Tableau V (d) V = d 1 ω1 (V ) = (d) λ = [d] 55 This is just a single row with d ones. Since there are no column permutations, this tableau is always non-zero. It is obviously maximal. Having constructed these generic tableaux, we will use the notation Ui (x) to denote the tableau Ui with the parameter a = x or Ui (x, y) for x = a and y = b in Ui . We will use fUi to denote the join of f copies of Ui . Note that these tableaux are all in maximal form. For the proof of Theorem 1, the parity effects the construction process. To simplify notation, we define the ∗ -function. x∗ =   x x even  x − 1 x odd We analyze T by the parameters r = n − 2s and s, where λ = [r + s, s]. For reference, we consider tableau of the following shape, with r and s even. T = | {z } | {z } r s Proof of Theorem 1. To prove Theorem 1 we need to construct a non-zero tableau of shape λ = [n − s, s] for s ≤ n2 , with s even and n = cd, c, d ≥ 2. First we construct ∗ ∗ a general tableau that covers most s. Suppose s ≤ c 2d . We know s is even, so write s = fd∗ + e, where 0 ≤ e < d∗ , e even. Since s, d∗ , and e are even, this is possible by the Euclidean algorithm. Let T = fU1 (d∗ ) ∨ U1 (e). Note that the bound on s guarantees that 2(f + 1) ≤ c when e > 0, and 2f ≤ c when e = 0. This insures that there are at most c distinct elements in T . If there are fewer than c elements in T add all the remaining elements to the tail of T by joining the appropriate number of V (d)’s. Suppressing the tail elements from the U1 ’s and V (d)’s, T looks like: 56 d∗ d∗ · · · d∗ v d ··· d T = 1 3 · · · 2f − 1 2f + 1 2f + 3 · · · c 2 4 · · · 2f 2f + 2 Theorem 8 shows T is non-zero, provided the weight-sets are disjoint. Since the tableaux are in maximal form, the weights must be disjoint by Lemma 3.4.9. This covers the majority of the s. The remaining tableaux will be constructed according to the parity of c and d. ∗ ∗ Case I: (c, d even) In this case c 2d = cd , so T constructed above covers all 2 partitions. Case II: (d even, c odd) By the above construction, we have all tableaux with s up . Thus we only need those even partitions with s = cd−k for 0 ≤ k ≤ d − 2, to (c−1)d 2 2 k ≡ d (mod 4). Take T = c−3 U1 (d) ∨ U3 (a) for 0 ≤ a ≤ d2 with a even. Then 2 s = c−3 d + a + d = cd−d+2a . Thus we have k = d − 2a, which ranges over the correct 2 2 parameters. Since U1 and U3 are in maximal form, Lemma 3.4.9 implies disjointness and Theorem 8 shows T is non-zero. Case III: (c even, d odd) Since r = n − 2s = cd − 2s s we need λ = [r + s, s] for r ≤ cd with r ≡ cd (mod 4). It suffices to construct a non-zero tableau for r < 4d. When r ≥ 4d, let r0 = r − 4dz with r0 < 4d. Then if we construct a λ0 = [s + r0 , s] tableau T 0 filled with d copies of c − 4z elements, we get the needed tableau by T = T 0 ∨ 4zV (d). Hence we will take r < 4d. U2 (d − 1, 1) ∨ U2 (d − 4r − When c ≡ 0 (mod 4) then r ≡ 0 (mod 4). Take T = c−4 4 1, 1). This construction gives the shape c−4 [2d, 2d]+[2d+ 2r , 2d− 2r ] = [ cd + 2r , cd − 2r ] as 4 2 2 desired. The parameters of these tableaux are positive unless r = 4d − 4 since r < 4d, r ≡ 0 (mod 4), and d ≥ 2. If r = 4d − 4 then d − 4r − 1 = 0, so use U1 (2) ∨ 2V (d) instead of of U2 (d − 4r − 1, 1). For c ≡ 2 (mod 4) we will assume r < 2d. When 2d ≤ r < 4d let r0 = r−2d. Then 57 can construct a λ0 = [r0 + s, s] tableau T 0 with c ≡ 0 (mod 4) and use T = T 0 ∨ 2V (d). Take T = c−2 U2 (d−1, 1)∨U1 ( 2d−r ) with V (d)’s as needed. Note that cd ≡ r (mod 4) 4 2 implies that 2d−r is even, while r < 2d insures it is positive. So we get the shape 2 [ cd + 2r , cd − 2r ] as needed. Theorem 8 shows these T ’s are non-zero provided the 2 2 weight-sets are disjoint, which follows from maximality. Note that since cd = n, n even, then c or d is even. Thus we have constructed all cases. Although it is not directly apparent from this construction, c or d even is often a necessary requirement for any non-zero two row tableau with s even to exist. For instance, when c = 3 and d = 7, the shape [11, 10] has s even, but all tableaux are zero by Theorem 9. 58 Chapter 6 Proof of Theorem 2 From Remark 2.2.8, to prove Theorem 2, it suffices to construct non-zero tableaux filled with d copies of c elements for all required partitions of n = cd, c, d ≥ 3. These partitions were determined in Theorem 10. Our approach is similar to the proof of Theorem 1 in Chapter 5. Using some generic non-zero tableaux (like Ui and V in Chapter 5) with c elements, we join them together by Theorem 8 to form a tableau of the appropriate shape and filling. However, unlike in Chapter 5, a large number of generic tableaux are needed. Since the cataloging of non-zero tableaux is quite tedious, we post-pone the construction until Chapter 7. Namely, our proof here will presuppose the construction of all tableaux of the required shapes for c ≤ 8 . The general idea is to write a tableau T as follows: T = S ∨ T0 ∨ U ∨ V for an appropriate T 0 , where S, U , and V generic constructions based on the parity of d. This reduces the construction of T to a construction of T 0 where the shape parameters, (r, s, and t), of T 0 are small. Thus we only need to construct tableaux for a limited number of cases corresponding to small shapes. The tableaux S, U, and V are based on the following non-zero maximal tableaux: (Here U1 , U2 and V occurred in Chapter 5.) 59 d 1 S0 = P1 (d) = 2 3 d U1 (d) = 1 2 d even, d even, aabbbb 5 6 6 6 5 5 S1 = 4 3 3 4 3 4 1 2 1 1 2 2 aabb U2 = 1 3 1 3 2 4 4 2 V = a = d−x +x 3 b = d−x 3 d≡x (mod 3), x ∈ {0, 1, 2} a=d−1 b=1 d 1 Let λ = [r + s + t, s + t, t]. We can write T = S ∨ T 0 ∨ U ∨ V , for appropriate T 0 provided T 0 is maximail. Then T 0 will be filled with d copies of c0 elements, for some c0 < c, which will eventually allow us to reduce to c ≤ 8. If T 0 is non-zero and maximal then by Lemma 3.4.9 and the Theorem 8 qT 6= 0 as desired. For simplicity, we will base our construction on the parity of d. 6.1 Case: d even To see how to write T as T = S ∨ U ∨ V ∨ T 0 for an appropriate T 0 we first discuss the individual reductions allowing us to write T = S ∨ T 0 , T = U ∨ T 0 , or T = V ∨ T 0 . Then successive applications of these reductions yield our desired decomposition. An analysis of these reductions also computes the resulting bounds on the shape of T 0 . An example application follows the reductions listed below. The reader may wish to refer to Example 6.1.1 while reading these reductions. Reduction 1: Let T be any λ-tableau with λ = (λ1 , λ2 , λ3 ), filled with d copies of c elements. Take f to be the maximum integer such that fd ≤ λ3 and c − 3f ≥ 3. Let S = fP1 (d) be the join of f copies of P1 (d). Then by Theorem 8, we may write T = S ∨ T 0 for T 0 a λ0 = (λ1 − df, λ2 − df, λ3 − df)-tableau filled with d copies of c0 = c − 3f elements, provided the weight-sets of S and T 0 are disjoint. The choice 60 of f means that in T 0 , t0 = λ03 = λ3 − df < d or c0 = c − 3f < 6. Thus we need only consider tableaux with t = λ3 < d or c < 6. The c < 6 condition coresponds to the requirement c − 3f ≥ 3. We need this requirement so that there are at least three elements available with which to fill the remaining tableau, T 0 . Reduction 2: Let T be any λ-tableau with λ = (λ1 , λ2 , λ3 ), filled with d copies of c elements. Take g to be the maximum integer such that gd ≤ λ2 − λ3 = s and c − 2g ≥ 3. Let U = gU1 (d) be the join of g copies of U1 (d). Then by Theorem 8, we may write T = U ∨ T 0 for T 0 a λ0 = (λ1 − dg, λ2 − dg, λ3 )-tableau filled with d copies of c0 = c−2g elements, provided the weight-sets of U and T 0 are disjoint. The choice of g 0 means that in T 0 , s0 = λ02 −λ3 = λ2 −dg−λ0 3 < d or c0 = c−2g < 5. However, we will need the existence of a non-zero T 0 in the specified shape. As was shown in Theorem 9, this is not always the case for some s. Specifically, when s < 5 non-zero tableaux do not exist for certain shapes when c = 3. (Consider λ = [6 + d, 2 + d, 1] = [9, 5, 1] with d = 3 and c = 5. Applying Reduction 2 yields λ0 = [5, 2, 1] with c = 3. All such tableaux are zero by Theorem 10 since s = 1.) To account for this, we modify the construction above to use g − 1 copies of U1 (d) when g > 0 and s0 < 5. In such a case, the modified T 0 now has s0 < d + 5. Thus we need only consider arbitrary tableaux with s < d + 5 or c < 5. Reduction 3: Let T be any λ-tableau with λ = (λ1 , λ2 , λ3 ), filled with d copies of c elements. Take h to be the maximum integer such that hd ≤ λ1 − λ2 = r and c − h ≥ 3. Let V = hV1 (d) be the join of h copies of V1 (d). Then by Theorem 8, we may write T = V ∨ T 0 for T 0 a λ0 = (λ1 − dh, λ2 , λ3 )-tableau filled with d copies of c0 = c − h elements, provided the weight-sets of V and T 0 are disjoint. The choice of h means that in T 0 , r0 = λ01 − λ02 = λ1 − dh − λ2 < d or c0 = c − h < 4. However, we will need the existence of non-zero T 0 in the specified shape. As was shown in Theorem 9, this is not always the case for some r. Specifically, when r < 5 non-zero tableaux do not exist for certain shapes when c = 3. To account for this, we modify the construction above to use h − 1 copies of U1 (d) when h > 0 and r0 < 5. In that 61 case, the modified T 0 now has r0 < d + 5. Thus we need only consider arbitrary tableaux with r < d + 5 or c < 4. Conclusion: When d is even, we can apply these reductions successively. Take an arbitrary λ-tableau T filled with d copies of c elements and assume c ≥ 6. We use T (i) to represent the appropriate T 0 obtained in these reductions. By Reduction 1, T = S ∨ T (1) , where S = fP1 (d) and T (1) has t = λ3 (T (1) ) < d and is filled with c(1) = c − 3f elements. Now, if c(1) ≥ 6 apply Reduction 2 to T (1) . Since c(1) ≥ 6, then by Reduction 2, write T (1) = U ∨ T (2) where U = gU1 (d) and T (2) has t < d (since T (1) does) and s < d + 5. Here T (2) is filled with c(2) = c(1) − 2g elements. Finally if c(2) ≥ 6 apply Reduction 3. This gives T (2) = V ∨T (3) , where V = hV1 (d) and T (3) has t < d, s < d + 5, and r < d + 5. Here T (3) is filled with c(3) = c(2) − h elements. Hence T = S ∨ U ∨ V ∨ T (i) where either T (i) is filled with fewer than 6 elements, or T (i) has t < d, s < d + 5, and r < d + 5. In the second case, T (i) must be filled with 3t + 2s + r = cd elements. This is less than or equal to 3(d − 1) + 2(d + 4) + (d + 4) = 6d + 9 ≤ 8d if d > 4. (If d = 4 we have 6d + 8 ≤ 8d and it’s not possible to have 6d + 9 = 9d when d = 4. For d = 3 additional reductions apply.) Hence we only need those tableaux with c ≤ 8. Moreover, if r or s < 5 in T (i) , then r or s < 5 in T , because the reductions do not reduce r or s to less than 5. Hence T (i) = T 0 has a shape occurring in Theorem 10 since all partitions of n with r and s ≥ 5 are needed. This reduction uses Theorem 8. Our usage only requires verification that the weight-sets are disjoint. However, the tableaux S, U , and V are in maximal form. Hence for appropriately chosen tableaux (i.e., ones in maximal form), an application of Lemma 3.4.9 can easily prove weight-set disjointness. Example 6.1.1. To see how this reduction works, let us consider a specific shape, λ = [9d − 2, 5d, d + 2] where d ≥ 6, d even and c = 15. This shape has t = d + 2, s = 4d − 2, and r = 4d − 2. First we apply Reduction 1, which joins P1 (d) in order 62 to have t < d. [9d − 2, 5d, d + 2] = P1 (d) ∨ [8d − 2, 4d, 2] Then we apply Reduction 2 to the shape [8d − 2, 4d, 2], which has s = 4d − 2 to reduce to s < d + 5 by joining three copies of U1 (d). [8d − 2, 4d, 2] = 3U1 (d) ∨ [5d − 2, d, 2] Applying Reduction 3 to shape [5d − 2, d, 2], which has r = 4d − 2 we normally want to reduce r to be between 5 and d + 5. Here we won’t necessarily reduce r fully, so that the resulting tableau will be familiar. Instead we will reduce to r = 2d − 2 (which may be reduced further depending on d) by joining two copies of V (d). [5d − 2, d, 2] = 2V (d) ∨ [3d − 2, d, 2] Hence, when we combine all these reductions, we get [9d − 2, 5d, d + 2] = P1 (d) ∨ 3U1 (d) ∨ 2V (d) ∨ [3d − 2, d, 2] A non-zero tableau of shape [3d − 2, d, 2] is Q∗ of Example 3.2.7 with a = 1, b = 1, c = d − 2. Therefore, writing T = P1 (d) ∨ Q∗ (1, 1, d − 2) ∨ 3U1 (d) ∨ 2V (d) and omitting the extra tail of Q∗ we have d 1 1 1 1 d-2 d d d d d 1 5 5 5 6 4 8 10 12 14 15 T = 2 4 7 7 4 7 9 11 13 366 As Q∗ is in maximal form, qT 6= 0. 63 6.2 Case: d odd When d is odd, we proceed exactly as in the even case, except the tableaux we use are slightly different. Namely, we use S1 instead of P1 and U2 instead of U1 . These adjustments are necessary for Reductions 1 and 2 since P1 (d) and U1 (d) are zero for d odd. Reduction 3 remains unchanged however. For completeness, we rewrite these reductions in terms of d odd. However, these reductions alone are not enough to reduce to c ≤ 8. So after these reductions, we apply a few more in order to reduce the size of tableaux we need to consider. Reduction 10 : Let T be any λ-tableau with λ = (λ1 , λ2 , λ3 ), filled with d copies of c elements. Take w to be the maximum integer such that w·2d ≤ λ3 and c−6w ≥ 3. Let S = wS1 (d) be the join of w copies of S1 (d). Then by Theorem 8, we may write T = S ∨ T 0 for T 0 a λ0 = (λ1 − 2d · w, λ2 − 2d · w, λ3 − 2d · w)-tableau filled with d copies of c0 = c − 6w elements, provided the weight-sets of S and T 0 are disjoint. The choice of w means that in T 0 , t0 = λ03 = λ3 − 2d · w < 2d or c0 = c − 6w < 9. Thus we need only consider tableaux with t = λ3 < 2d or c < 9. Reduction 20 : Let T be any λ-tableau with λ = (λ1 , λ2 , λ3 ), filled with d copies of c elements. Take v to be the maximum integer such that v · 2d ≤ λ2 − λ3 = s and c − 4v ≥ 3. Let U = vU2 (d) be the join of v copies of U2 (d). Then by Theorem 8, we may write T = U ∨ T 0 for T 0 a λ0 = (λ1 − 2d · v, λ2 − 2d · v, λ3 )-tableau filled with d copies of c0 = c − 4v elements, provided the weight-sets of U and T 0 are disjoint. The choice of v means that in T 0 , s0 = λ02 − λ03 = λ2 − d · v − λ03 < 2d or c0 = c − 4v < 8. As in the even case, to account for the shapes s < 5, we modify this reduction to use v − 1 copies of U1 (d) when v > 0 and s0 < 5. Then the modified T 0 now has s0 < 2d + 5. Thus we need only consider arbitrary tableaux with s < 2d + 5 or c < 8. Summary: The same argument as in the even case works for the d odd cases, though the numbers are adjusted slightly. Take an arbitrary λ-tableau T with filled with d copies of c elements, but this time assume c ≥ 9. Then by applications of 64 Reductions 1’, 2’ and 3, T = S ∨ U ∨ V ∨ T (i) where either T (i) is filled with fewer than 9 elements, or T (i) has t < 2d, s < 2d + 5, and r < d + 5. In the second case, T (i) must be filled with 3t + 2s + r elements, which is less than or equal to 3(2d − 1) + 2(2d + 4) + (d + 4) = 11d + 9 as d ≥ 3. Moreover, if r or s < 5 in T (i) , then r or s < 5 in T . Hence T (i) has a required shape of Theorem 10. However, we wish to have T (i) fillable with c ≤ 8. To do this we have additional reduction techniques. However, these techniques are very sensitive to the parameters in T (i) , so we will categorize them by such. The additional non-zero maximal tableaux we use are d−1 U1 (d − 1) = 1 2 d odd ω2 = (0, d − 1) d−1 1 P1 (d − 1) = 2 3 d odd ω2,3 = d−2 1 113 P4 (d − 2, 1, 1) = 2 232 3 d odd ω2,3 = ¡ 0 d−1 0 ¡0 d 0 0 d−1 1 0 0 d−2 ¢ ¢ Start with a tableau T where t ≤ 2d − 1, s ≤ d + 4, s 6= 1, r ≤ d + 4, r 6= 1, d odd and s + t even if r or s in {0, 2, 4}. (These are the partitions required by Theorem 10 after the previous reductions have been applied.) First consider those tableaux with r ≥ 10, which implies d ≥ 6. Case A: Assume r ≥ 10, s < d + 4, t < d − 1. Then 3t + 2s + r ≤ 3(d − 2) + 2(d + 3) + d + 4 = 6d + 4 ≤ 8d. Hence this case is covered by c ≤ 8. Case B: Assume r ≥ 10, s ≥ d+4, t ≥ d−1. Write T = P1 (d−1)∨U1 (d−1)∨T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r − 5, s − (d + 1), t − (d − 1)). Thus 5 ≤ r0 ≤ d − 1, 5 ≤ s0 ≤ d + 5, and 0 ≤ t0 ≤ d. Then 65 3t + 2s + r ≤ 3d + 2(d + 5) + d − 1 = 6d + 9 ≤ 8d. Note that no exceptional r or s cases occur in T 0 . Hence this case is covered by c ≤ 8. Case C: Assume r ≥ 10, s < d + 4, t ≥ d − 1. Write T = P1 (d − 1) ∨ T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r−3, s, t−(d−1)). Thus 7 ≤ r0 ≤ d + 1, s0 ≤ d + 3, and 0 ≤ t0 ≤ d. Then 3t + 2s + r ≤ 3d + 2(d + 3) + d + 1 = 6d + 7 ≤ 8d. Note that no exceptional r or s cases occur in T 0 . Hence this case is covered by c ≤ 8. Case D: Assume r ≥ 10, s ≥ d + 4, t < d − 1. Write T = U1 (d − 1) ∨ T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r − 2, s − (d − 1), t). Thus 8 ≤ r0 ≤ d + 2, s0 ≤ d + 5, and 0 ≤ t0 ≤ d − 1. Then 3t + 2s + r ≤ 3(d − 1) + 2(d + 5) + d + 2 = 6d + 9 ≤ 8d. Note that no exceptional r or s cases occur in T 0 . Hence this case is covered by c ≤ 8. For r < 10 the arguments depend more on the values of r, but the general idea is the same. Case E: Assume r < 10, s < d + 4, t < d − 1. Then 3t + 2s + r ≤ 3(d − 1) + 2(d + 3) + 9 = 5d + 12 ≤ 8d for d ≥ 4. When d = 3, c = 9 is a possibility. Now, when d = 3, we have r ≤ d + 4 = 7, s ≤ d + 3 = 6, and t ≤ d − 2 = 1. So here 3t + 2s + r ≤ 3 + 2 · 6 + 7 = 22 < 8d. Hence c ≤ 8 tableaux will suffice to cover this case. Note that the s + t parity is preserved in the exceptional r and s cases. Case F: Assume r < 10, s ≥ d + 4, t < d − 1. If r = 9, 8, 7, 5 then write T = U1 (d − 1) ∨ T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r − 2, s − (d − 1), t). Then r0 ∈ {7, 6, 5, 3}, s0 ≤ d + 5, and 0 ≤ t0 ≤ d − 1. Then 3t0 + 2s0 + r0 ≤ 3(d − 2) + 2(d + 5) + 7 = 5d + 11 ≤ 8d for d > 3. When d = 3, then s0 ≤ 8, t0 ≤ 1. We have 3t0 + 2s0 + r0 ≤ 3 + 16 + 7 = 26 < 9d. Hence c ≤ 8 tableaux will suffice to cover these case. If r = 2, 4, or 6, then write T = U1 (d − 1) ∨ T 0 . If (r, s, t) are the parameters of T , 66 then T 0 has parameters (r0 , s0 , t0 ) = (r −2, s−(d−1), t). Thus r0 ∈ {4, 2, 0}, s0 ≤ d+5, and 0 ≤ t0 ≤ d − 1. Then 3t0 + 2s0 + r0 ≤ 3(d − 2) + 2(d + 5) + 4 = 5d + 8 ≤ 8d. Hence c ≤ 8 tableaux will suffice to cover these cases if the tableau exists. Note that U1 (d − 1) preserves the parity of s + t, so if s + t are even (always the case when r = 2 or 4), we will have s0 + t0 even which is necessary for r0 = 0, 2, or 4. Hence this construction works except when r = 6 and s + t odd. Then 3t + 2s + r ≤ 3(d − 2) + 2(2d + 4) + 6 = 7d + 8 ≤ 8d for d ≥ 8 otherwise it is less than 9d unless d = 3. If 3t + 2s + 6 = 9d, then t is odd so write t = d − 2k with k ≥ 1. Then we have s = 3d + 3k − 3. Since s ≤ 2d + 4, we have d + 3k ≤ 7, so d = 3, k = 1 is the only solution. This corresponds to t = 1, s = 9, r = 6. Since s + t even, this case has already been done. For d = 3 we also need to consider c = 10. However, 3 + 20 + 6 < 10d so no such partition will occur. For r = 0 we have s+t even. Then 3t+2s+r ≤ 3(d−2)+2(2d+4) = 7d+2 ≤ 8d, so this case is covered by c ≤ 8 tableaux. For r = 3 we 3t + 2s + r ≤ 3(d − 2) + 2(2d + 4) + 3 = 7d + 5 ≤ 8d for d ≥ 5. When d = 3 we have 3 + 20 + 3 < 9d, so such a partition does not occur. Hence r = 3 is covered by c ≤ 8. Case G: Assume r < 10, s < d + 4, t ≥ d − 1. If r = 9, 8, 6 then write T = P1 (d − 1) ∨ T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r − 3, s, t − (d − 1)). Thus r0 ∈ {6, 5, 3}, s0 ≤ d + 3, and 0 ≤ t0 ≤ d. Then 3t0 + 2s0 + r0 ≤ 3d + 2(d + 3) + 6 = 5d + 12 ≤ 8d for d ≥ 5. When d = 3, then s0 ≤ 6, t0 ≤ 3. Hence we have 3t0 +2s0 +r0 ≤ 9+12+6 = 27 = 9d, so the only solution is t0 = 3, s0 = 6, r0 = 6. This we can further reduce by writing T 0 = V (d) ∨ T 00 where T 00 is a c = 8 tableau of parameters t = 3, s = 6, and r = 3. Hence c ≤ 8 tableaux will suffice. If r = 7, 5 , or 3 and s + t even then write T = P1 (d − 1) ∨ T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r − 3, s, t − (d − 1)) and the parity of s0 + t0 is preserved. Thus r0 ∈ {0, 2, 4}, s0 ≤ d + 3, and 0 ≤ t0 ≤ d. Then 3t0 + 2s0 + r0 ≤ 3d + 2(d + 3) + 4 = 5d + 10 ≤ 8d for d ≥ 5. For d = 3 we have 67 3t0 + 2s0 + r0 ≤ 9 + 12 + 4 = 25 < 9d, so this does not occur. Hence the c ≥ 8 tableaux will suffice. Consider r = 0, 2, 4 with s + t even, or r = 3, 5, 7 with s + t odd. If s ≥ 8 write T = P4 (d − 2, 1, 1) ∨ T 0 . If (r, s, t) are the parameters of T , then T 0 has parameters (r0 , s0 , t0 ) = (r, s−3, t−(d−2)) and the parity of s0 +t0 is preserved. Thus , 5 ≤ s0 ≤ d, and 0 ≤ t0 ≤ d + 1. Then 3t0 + 2s0 + r0 ≤ 3(d + 1) + 2(d) + 7 = 5d + 10 ≤ 8d for d ≥ 5. For d = 3 we have 3t0 + 2s0 + r0 ≤ 12 + 6 + 7 = 25 < 9d, so this does not occur. Hence the c ≥ 8 tableaux will suffice. For r = 3, 5, or 7, s + t odd, and s ≤ 7, we have 3t + 3s + r ≤ 3(2d − 1) + 2 · 7 + 7 = 6d + 18 ≤ 8d for d ≥ 9. If 3t + 2s + r = 9d, then since r is odd, we have t = 2d − 2k for k ≥ 1. This implies s = d + 3k + d−r . Since s ≤ 7, the only possible solutions are 2 (r, s, t) = (7, 4, 4), (5, 5, 4), (3, 6, 4), (7, 7, 2) when d = 3 and (7, 4, 8) when d = 5. For those with s + t even, the case has already been done. For (r, s, t) = (7, 7, 2) write T = U1 (2) ∨ T 0 where T 0 has parameters (5, 5, 2) and is a c = 7 tableau. Hence we still need (r, s, t) = (5, 5, 4) for d = 3. But this is U1 (2) ∨ T 0 , where T 0 is a c = 7 tableau of parameters (3, 3, 4). We also need to consider those partitions with c = 10. Then we have 3t + 2s + r = . The only 10d which implies t = 2d − 2k − 1 as t is odd. So s = 2d + 3k + 1 − r−1 2 solutions with s ≤ 7 are (r, s, t) = (3, 6, 5), (5, 5, 5), (7, 4, 5), and (7, 7, 3) all with d = 3. The cases with s + t even have been done already. For (r, s, t) = (3, 6, 5), use T = P4 (1, 1, 1) ∨ T 0 where T 0 is a c = 7 tableau with (r0 , s,0 , t0 ) = (3, 3, 4). For (7, 4, 5) we have s = 4 and s + t odd, so this case partition is not needed. When d = 3 we also may have c = 11 or c = 12. Proceeding as above, the only solutions are (r, s, t) = (7, 7, 4) and (7, 7, 5). The second case has s + t even and hence is not needed. The first case can be reduced to T = U1 (2) ∨ T 0 , where T 0 is a c = 9 case with parameters (5, 5, 4). But this is U1 (2) ∨ T 0 , where T 0 is a c = 7 tableau of parameters (3, 3, 4). If r = 0, 2, 4, s + t even, and s ≤ 7, we have 3t + 3s + r ≤ 3(2d − 1) + 2 · 7 + 4 = 6d + 15 ≤ 8d for d ≥ 9. For c = 9, we have t = 2d − 2k − 1 as t is odd. Then s = d + 3k + 1 + d+1−r . Since s ≤ 7 the only solutions are (r, s, t) = (0, 6, 5), (2, 5, 5), (4, 4, 5), 2 68 and (4, 7, 3) with d = 3 and (4, 7, 9). Note that only those with s + t even are needed. For these cases write T = U1 (d − 1) ∨ T 0 , where T 0 is a c0 = 7 tableau with parameters (r − 2, s − (d − 1), t). Since this preserves the parity of s + t and does not cause r0 , s0 = 1, the tableau exists. For d = 3 we may also have c = 10 or 11. Proceeding as above, the only solutions are (r, s, t) = (4, 7, 4) which has s + t odd, hence it is not need, and (4, 7, 5) which is U1 (2) ∨ U1 (2) ∨ T 0 where T 0 is a c = 7 tableau with (r, s, t) = (0, 3, 5). Case H: Assume r < 10, s ≥ d + 4, t ≥ d − 1. If r = 8 or r = 9, 7, 5 and s + t even then write T = P1 (d − 1) ∨ U1 (d − 1) ∨ T 0 where T 0 has parameters (r0 , s0 , t0 ) = (r − 5, s − (d − 1), t − (d − 1)). Thus r0 ≤ 4, s0 ≤ d + 5, and t0 ≤ d. Then 3t0 + 2s0 + r0 ≤ 3d + 2(d + 5) + 4 = 5d + 14 ≤ 8d for d > 3. When d = 3 we can have 3t0 + 2s0 + r0 = 9d only for (r0 , s0 , t0 ) = (2, 8, 3) or (4, 7, 3). But s + t even means only (4, 7, 3) is needed. This is U1 (d − 1) ∨ T 00 where T 00 is a c = 7 tableau with (r, s, t) = (2, 5, 3). Hence we’ve reduced to c ≤ 8 cases. For r = 9, 7, 5 with s+t odd, write T = P4 (d−2, 1, 1)∨U1 (d−1)∨T 0 where T 0 has parameters (r0 , s0 , t0 ) = (r − 2, s − (d − 1) − 3, t − (d − 2)). Hence r0 ≤ 7, 2 ≤ s0 ≤ d + 2, and t0 ≤ d + 1. We will have s0 6∈ {0, 1, 2, 4} provided s 6= d + 4 or d + 6. Then 3t0 +2s0 +r0 ≤ 3(d+1)+2(d+2)+7 = 5d+14 ≤ 8d for d > 3. When s = d+4 or d+6 write T = P4 (d − 2, 1, 1) ∨ T 0 where T 0 has parameters (r0 , s0 , t0 ) = (r, s − 3, t − (d − 2)). Then 3t0 + 2s0 + r ≤ 3(d + 1) + 2(d + 3) + 9 = 5d + 18 ≤ 8d for d ≥ 5. For d = 5, s0 = d + 3 or d + 1 there are no partitions with t0 ≤ d + 1. Now consider d = 3 with s0 ≤ d + 3, t0 ≤ d + 1. Since r ≤ d + 4 we have r = 3, 5, or 7. Given these parameters, solving 3t0 + 2s0 + r0 = 3c for c ≥ 9, the only solutions are (r0 , s0 , t0 ) = (3, 6, 4), (5, 5, 4), and (7, 4, 4). Since s + t odd the only partition we need to construct is (r0 , s0 , t0 ) = (5, 5, 4) with c = 9. But this is U1 (2) ∨ T 0 where T 0 is a c = 7 tableau of parameters (3, 3, 4). For r = 2, 4, 6 with s + t even, write T = P4 (d − 2, 1, 1) ∨ U1 (d − 1) ∨ T 0 where T 0 has parameters (r0 , s0 , t0 ) = (r − 2, s − (d − 1) − 3, t − (d − 2)). Thus r0 ≤ 4, 2 ≤ s0 ≤ d + 2, and t0 ≤ d + 1. Since s + t even, we need only that s0 6= 1. Then 69 3t0 + 2s0 + r0 ≤ 3(d + 1) + 2(d + 2) + 4 = 5d + 11 ≤ 8d for d > 3. For d = 3, d · 4 + 2 · 5 + 4 = 26 < 9d so that case is not needed. For r = 6, s + t odd write T = P1 (d − 1) ∨ T 0 , where T 0 has parameters (r0 , s0 , t0 ) = (3, s, t − (d − 1)). Thus 3t0 + 2s0 + r0 ≤ 3d + 2(2d + 4) + 3 = 7d + 11 ≤ 8d for d ≥ 11. Solving 3t0 +2s0 +r0 = 9d given the parameters, the only solution is (r0 , s0 , t0 ) = (3, 9, 2). This is P4 (1, 1, 1) ∨ T 00 where T 00 is a c = 6 tableau with parameters (3, 6, 1). When c = 10, we also have (3, 9, 3) as a solution, but this does not have s + t odd. For r = 3, write T = P4 (d − 2, 1, 1) ∨ T 0 where T 0 has parameters (r0 , s0 , t0 ) = (3, s − 3, t − (d − 2)). Thus 3t0 + 2s0 + r0 ≤ 3(d + 1) + 2(2d + 1) + 3 = 7d + 8 ≤ 8d for d > 5. (The c = 8 constructions will hold unless s − 3 ∈ {0, 1, 2, 4}, but since s ≥ d + 4 this can only occur with s = d + 1, d = 3. Then our original tableau will have t ≤ 2d − 1, s = d + 1, r = 3 which is satisfiable by a c ≤ 8 tableau.) Solving 3t0 + 2s0 + r0 = 9d given the parameters, the only solutions are (r0 , s0 , t0 ) = (3, 6, 4), (3, 9, 2), and (3, 12, 6) with d = 5. The (3, 6, 4) case is P4 (1, 1, 1) ∨ T 00 where T 00 is a c = 6 tableau with parameters (3, 3, 3). The (3, 9, 2) case is P4 (1, 1, 1) ∨ T 00 where T 00 is a c = 6 tableau with parameters (3, 6, 1). When d = 5, the (3, 12, 6) case is P4 (3, 1, 1) ∨ T 00 where T 00 is a c = 6 tableau with parameters (3, 9, 3). There are no tableaux with c > 9. For r = 0 we have s+t even. Write T = P4 (d−2, 1, 1)∨T 0 where T 0 has parameters (r0 , s0 , t0 ) = (3, s − 3, t − (d − 2)). So 3t0 + 2s0 + r0 ≤ 3(d + 1) + 2(2d + 1) = 7d + 5 ≤ 8d for d > 3. As P4 preserves parity and s + t even, the c = 8 constructions will work provided s0 6= 1. But s ≥ d + 4 forces s0 ≥ d + 1 ≥ 4. When d = 3 we have 3 · 4 + 2 · 7 = 26 < 9d, hence all cases are covered. Conclusion: When d is odd, we need all tableaux with c ≤ 8. The basic reduction requires those tableaux to be disjoint from multiple copies of S1 , U2 (d − 1, 1) and V (d). The further reductions also require the tableau to be disjoint from P1 (d−1), U1 (d−1), P1 (d−1)∨U1 (d−1), P4 (d−2, 1, 1), and U1 (d−1)∨P4 (d−2, 1, 1). For d even we need all tableaux with c ≤ 6, along with those tableaux having t ≤ d − 1, s ≤ d + 4, r ≤ d + 4 when c = 7 or 8. These tableaux need to be 70 disjoint from P1 (d), U1 (d), and V (d). These tableaux will be listed in Chapter 7. In Chapter 8, we verify that all necessary tableaux have been produced. 71 Chapter 7 Tableaux Construction The proof of Theorem 2 in Chapter 6 requires the construction of non-zero maximal tableaux with c ≤ 8 for the shapes discussed in Theorem 10. In this chapter we construct all the necessary tableaux and show they are both non-zero and maximal. Some basic properties of maximality are listed in Section 7.1 and are used throughout our construction. We construct those tableaux with two rows in Section 7.2. In Sections 7.3, 7.4, 7.5, and 7.6 we construction the necessary tableaux with c = 3, 4, 5, and 6, respectively. Section 7.7 contains additional tableaux need when c = 7 or 8. 7.1 Maximality of Tableaux Maximality, as discussed in Lemma 3.4.9, is an important property of the tableaux we construct to prove Theorem 2. Given certain conditions of a weight, it is easy to verify that a tableau is in maximal form. We discuss maximality here in general, in order to simplify the proof of maximal form for the specific tableaux we construct in the next sections. Recall that a tableau T is in maximal form if qT 6= 0 by weight-set counting on ω(T ) and ω(T ) is the largest weight of all ω(τ T ) for τ ∈ CT . All tableaux that we consider in the next sections are shown to be non-zero by weight-set counting prior to addressing the maximality issue. As such, we only consider the weight condition here. To summarize the basic conditions of Definition 3.4.2, a tableau weight is maximal if it has the largest weight possible for row three, and given this, the largest weight 72 possible for row two. When multiple weights satisfy this, the weight where rows two and three have elements in common (e.g., the element 4 appears in both rows) is considered larger. In general, to determine the maximal form of a tableau, we start with a given filling of T and use column operations to produce tableaux τ T with larger weights. The following procedure provides an overview of how to determine the maximality of a filling. • First maximize the weight of row three. This usually involves having a weight of d for as many elements as possible. (This is discussed more extensively in Lemma 7.1.2.) There may be multiple different fillings having the same maximum weight for row three. All such fillings should be considered for the next step. • Now maximize the weight of row two, given the filling(s) of row three determined previously. To do this, determine the largest possible row two weight of each element, provided those elements assigned to row three are not used. For a given element this is equivalent to the number of copies in the body of T minus the number of copies used in row three. Fill row two with the element having the largest weight, then repeat the procedure with the remaining elements and positions. (This is summarized for weights in Lemma 7.1.4.) There may be multiple such fillings having the largest weight. • If there are multiple fillings after step two that have different generic weights, choose the filling in which an element in row three has the largest weight in row two. (The weight of their common elements is maximized.) There may not be a unique such filling, but all such maximal fillings will have the same generic weight. As such these fillings will differ on by an action of Sa . 73 Example 7.1.1. Consider z+x z z a 3 323 T = Q2 = 4 244 1 112 ω2,3 (T ) = ¡ 0 z 0 d−z+a+x ¢ d a 0 0 where z = d−x , a < z, and some conditions on z, a, and x to insure qT is non-zero. 3 First we want to maximize row three. Since this occurs when the non-zero weights of row three are d and a, we check to see which elements can have weight d in row three. Here, the only option is the element 1. Since any of the remaining elements can have a weight of a, we leave that column unfilled for now. Thus the maximal (third row) form for T is 111∗. To determine the maximal second row form, we consider how many of each element is available given the third row is partially determined. There are zero 1’s available, 2z+a 2’s available, and 2z+a+x 3’s and 4’s available. Hence we can maximally fill the second row with either 3’s or 4’s. Since the tableau is symmetric in 3’s and 4’s, we will use 4’s without loss of generality. Thus we get a maximal (second and third row) form of 4*44 . 111* This form gives rise to two different generic weights depending on whether the ∗’s ¢ ¡ 0 0 d−z+a z ¢ ¡ 0 are the same. These generic weights are d0 az d−z+a 0 0 . The first one, 0 0 and d a corresponding to the same element for both ∗’s, is larger. Hence a maximal (second 4244 . Note that this filling is not unique. We could have 1112 4344 3433 3233 used , , or instead. All would be in maximal form. 1113 1114 1112 and third row) form is Some fillings can be seen as maximal strictly by examining the weight of the tableau. Lemma 7.1.2. If a row has at most one weight not equal to 0 or d, then its row weight is maximal. This follows directly from the ordering on weights and the limit of d copies of an element in a tableau. If we call the weight not equal to 0 or d the non-d weight for the row, we get the following result on tableau maximality. 74 Lemma 7.1.3. A tableau is maximal if rows two and three satisfy Lemma 7.1.2 and either their non-d weights come from the same element or the sum of the non-d weights is greater than d. Proof. Given a tableau satisfying Lemma 7.1.2 for rows two and three, its maximal generic weight must be either ( d0 ab d0 00 ) or ( d0 a0 d0 b0 ) with the appropriate number of 0 and d weights. By our ordering, the first weight is strictly larger than the second weight. Hence when the non-d weights come from the same element the tableau is maximal. If a + b is greater than d, the first weight is not possible and so the second one is maximal. This lemma is directly applicable to the tableau weights. A more generalized form of this, depending on the actual filling, is also useful. The following lemma reflects the technique used in Example 7.1.1. Lemma 7.1.4. A row is maximal if all except one non-zero weight corresponds to the largest weights possible for any elements. This lemma is a generalization of Lemma 7.1.2, where the maximum weight for each element is no longer d. Using this in Lemma 7.1.3 gives: Lemma 7.1.5. A tableau is maximal if row three satisfies Lemma 7.1.2, row two satisfies Lemma 7.1.4, and the non-d weight of row three has the largest possible weight in row two of all such weights satisfying Lemma 7.1.4. Example 7.1.1 represents an appropriate use of Lemma 7.1.5. The weight of row three is (d, a, 0, 0), clearly satisfying Lemma 7.1.2. The weight of row two is (0, z, 0, d − z + a + x). That satisfies Lemma 7.1.4, since we checked that d − z + a + x was the largest weight possible. Finally, ω3 (2) = a is the non-d weight of row three. Since ω2 (2) = z is larger than ω2 (2) = 0, the weight is maximal by Lemma 7.1.5. In general, for the application of Lemma 7.1.5 it is to check that Lemma 7.1.2 and Lemma 7.1.4 apply and then to show that the non-d weight cannot be assigned a larger row two weight without changing the generic weight of row two. In particular 75 the non-d weight conditions of Lemma 7.1.5 are satisfied if all extra (non-third row) copies an element in row three are contained in row two. We will refer to these lemmas for maximality justification of the tableaux constructed in the next section. 7.2 Tableaux for Two Row Partitions In this section we construct all the two row tableaux needed for the proof of Theorem 2. For many of these constructions, the parity of d is relevant. Recall the notation: d∗ =   d if d is even  d − 1 if d is odd Since we are constructing tableaux for many partitions, we will not use a fixed c. However, for every element that is listed in the tableau, we assume it occurs d times, filling out the tail as needed using only those numbers in the body of T . Tableau U1 a d−a d−a a U1 = 1 1 ∼ 1 2 2 2 a even a≤d ω2 (U1 ) = (0, a) λ = [2d − a, a] r = 2d − 2a, s = a, t = 0 We showed U1 non-zero in the proof of Theorem 1. Maximality: This tableau is maximal by Lemma 7.1.4. 76 Tableau U2 aabb U2 = 1 3 1 3 2 4 4 2 a+b≤d a, b > 0 ω2 = (0, a + b, 0, a + b) λ = [4d − 2a − 2b, 2a + 2b] r = 4d − 4a − 4b, s = 2a + 2b, t = 0 We showed U2 non-zero in the proof of Theorem 1. Maximality: This tableau is maximal by Lemma 7.1.4. Tableau U3 a even abb U3 = 1 1 2 2 3 3 a+b≤d 0 < 2b ≤ d b≥a ω2 = (0, a, 2b) λ = [3d − a − 2b, a + 2b] r = 3d − 2a − 4b, s = a + 2b, t = 0 If b > a, any valid weight assignment must have ω2 (3) = 2b. Hence (1, 2, 3) with sign 1 and (2, 1, 3) with sign (−1)a are the only possible weight assignments. Thus U3 is non-zero. If a = b, we may also have weight assignments: (1, 3, 2) with sign (−1)b ; (2, 1, 3) with sign (−1)a+b ; (3, 1, 2) with sign (−1)2b ; and (3, 2, 1) with sign (−1)a+2b . Since a = b and a is even, these are all positive. Hence U3 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.4 since 2b ≥ a + b. 77 Tableau U4 ab U4 = 2 3 1 1 a≥b>0 ω2 = (a + b, 0, 0) λ = [3d − a − b, a + b] r = 3d − 2a − 2b, s = a + b, t = 0 Since only the element 1 can have ω2 = a + b, there is exactly one valid weight assignment, (1, 2, 3). Thus U4 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.4. Tableau U5 a b−2 a−1 b U5 = 1 1 4 4 2 3 3 2 a≥2 b≥3 a+b≤d ω2 = (0, a + b, a + b − 3, 0) λ = [4d − 2a − 2b + 3, 2a + 2b − 3] r = 4d − 4a − 4b + 6, s = 2a + 2b − 3, t = 0 Only the element 2 may have ω2 (2)=a+b. Since a ≥ 2 and b ≥ 3, we must have ω2 (3) = a + b − 3. Hence (1, 2, 3, 4) is the only valid weight assignment, so U5 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.4 since a+b is the largest weight possible. 78 Tableau U6 a=d−2 abc d U6 = 1 3 5 5 5 3 1 3 1 3 5 2 4 424552 b = d+1 2 d ≡ 1 (mod 2) c = d−3 2 d>5 d = d−5 2 ) ω2 = (0, d, 0, d, d−3 2 λ = [3d − d−3 , 2d + d+3 ] 2 2 r = 3, s = 2d + d−3 , t = 0, 2 Only the elements 2 and 4 may have ω2 = d. If ω2 (5) = 0, then ω2 (1 and 3) > 0. Hence (1, 2, 3, 4, 5) is the only valid weight assignment and so U6 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.2. Tableau U7 2 23 2 U7 = 4 3 5 5 3 4 2 3 4 111223 d=5 ω2 = (5, 4, 2, 0, 0) λ = [14, 11] r = 3, s = 11, t = 0, The valid weight assignments are (5, 4, 2, 0, 0, 0), (5, 4, 0, 2, 0), (2, 0, 4, 0, 5), (0, 0, 2, 4, 5), and (0, 0, 4, 2, 5). Since there are an odd number of weight assignments, this tableau is non-zero. Maximality: This tableau is maximal by Lemma 7.1.4. Tableau U8 U8 = 43333122444 11122 d=4 79 ω2 = (3, 2, 0, 0) λ = [11, 5] r = 6, s = 5, t = 0 Then only valid weight assignments are (1, 2, 3, 4), with sign 1, (1, 3, 2, 4), with sign (−1)2 , and (3, 1, 2, 4) with sign (−1)3 . Hence the weight sum is 1 and U8 is non-zero. Maximality: This tableau is not maximal since ω2 = (0, 0, 4, 1) is larger. However, in qT , the weight (0, 0, 4, 1) always cancels. This tableau cannot be put in maximal form, hence we will need to prove directly that it is disjoint from the requisite tableaux. This will be done in Section 8.7. 80 7.3 Tableaux for c = 3 We know by Theorem 9 which tableaux are non-zero for c = 3. However, using Theorem 8 on tableaux requires the tableaux to be non-zero by weight-set counting on ω(T ). We also want the tableaux to be maximal, in order to obtain the disjointness of Lemma 3.4.9. Here we will list the tableaux used, briefly showing they are non-zero and maximal. For all these tableaux, any valid weight assignment corresponds to a unique tableau, so we will not explicitly state how many tableaux correspond to each weight assignment. Tableau P1 a 1 P1 = 2 3 0≤a≤d a even ω2,3 = ( 00 a0 a0 ) λ = [3d − 2a, a, a] r = 3d − 3a, s = 0, t = a P1 is non-zero by the Lemma 3.2.6. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau P2 ab 1 1 P2 = 2 2 3 a+b≤d a, b even 0 ω2,3 = ( 00 a+b 0 a) λ = [3d − 2a − b, a+b, a] r = 3d − 3a − 2b, s = b, t = a 81 Only the elements 1 and 2 may have ω2 = a+b. Hence the valid weight assignments are: (1, 2, 3) with sign 1; (3, 2, 1) with sign (−1)a ; (2, 1, 3) with sign (−1)a+b ; and (3, 1, 2) with sign (−1)b . Since a and b are even, this weight sum is 4 and qP2 6= 0. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau P3 a b+1 b c 2 2 3 2 P3 = 1 1 1 3 3 0≤a≤d 2b + 1 ≤ d − a 0≤c<b a + c even c ω2,3 = ( 00 a+2b+1 0 a) λ = [3d − 2a − 2b − c − 1, a + 2b + c + 1, a] r = 3d − 3a − 4b − 2c − 2, s = 2b + c + 1, t = a Only the element 1 may have ω2 = a + 2b + 1. Hence the valid weight assignments are (1, 2, 3) with sign 1 and (1, 3, 2) with sign (−1)a+c . Since a+c is even, this weight sum is positive and hence P3 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau P4 abbc 2 2 3 2 P4 = 1 1 1 3 3 0≤a≤d 2b ≤ d − a 0≤c≤b a + c even c ω2,3 = ( 00 a+2b 0 a) λ = [3d − 2a − 2b − c, a + 2b + c, a] r = 3d − 3a − 4b − 2c, s = 2b + c, t = a 82 Unless c = b, the only valid weight assignments are (1, 2, 3) with sign 1 and (1, 3, 2) with sign (−1)a+c = 1. If b = c the tableau is symmetric in 1, 2, and 3. Hence we also get the weight assignments: (2, 1, 3) with sign (−1)a+b+b+c = 1; (2, 3, 1) with sign (−1)2b = 1; (3, 2, 1) with sign (−1)a+b = 1; and (3, 1, 2) with sign (−1)2b = 1. Thus the weight sum is always positive and P4 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. 83 7.4 Tableaux for c = 4 Tableau Q1 0≤a<z z+x z z a b c d 3 323 3 2 3 Q1 = 4 244 4 4 2 1 112 b+x≥d c≥d b+c≤z−a z + a + d even c + d even, if d = b + x d + b + x even, if d = c z = d−x 3 d≡x (mod 3) 0 2z+x+a+b+c ω2,3 = ( d0 z+d ) a 0 0 λ = [2d − 2a − b − c − d, d + a + b + c + d, d + a] r = d − 3a − 2b − 2c − 2d, s = b + c + d, t = d + a Since a < z, any valid weight assignment must have ω3 (1) = d. We must have ω2 (4) = 2z + x + a + b + c unless d = c or d = b + x. If d = c, then we may also have ω2 (3) = 2z+x+a+b+c. If d = b+x, then we may also have ω2 (2) = 2z+x+a+b+c. Given these restrictions, we list the valid weight assignments in the table below, along with the sign corresponding to that assignment. Note that there is a unique tableau associated to each of these assignments. Assignment Sign Condition (1, 2, 3, 4) 1 (1, 3, 2, 4) (−1)z+a+d (1, 2, 4, 3) (−1)z+x+a+b+c+d d=c (1, 4, 2, 3) (−1)2z+x+b+d d=c (1, 3, 4, 2) (−1)2z+x+b+c d=b+x (1, 4, 3, 2) (−1)z+a+c d=b+x Given the parity conditions on the parameters, the sign of these terms reduces to 1 in all cases. Hence the weight sum is positive and Q1 is non-zero. Maximality: Because c ≥ d and b + x ≥ d, the weight 2z + x + a + b + c of 4 in row two is the largest possible weight given the Maximality of row three by 84 Lemma 7.1.2. Hence this tableau is maximal by Lemma 7.1.5. Tableau Q2 z+x z z a 3 323 Q2 = 4 244 1 112 ω2,3 = 0<a≤z z + a even z = d−x 3 d ≡ x (mod 3) x ∈ {0, 2, 4} ¡ 0 z 0 d−z+a ¢ d a 0 0 λ = [2d − 2a, d + a, d + a] r = d − 3a, s = 0, t = d + a For a < z any valid weight assignment must ω3 (1) = d. The tableau is symmetric in 3 and 4, as well as 2 if x = 0. This gives the following signed weight table: Assignment Sign Condition (1, 2, 3, 4) 1 (1, 3, 2, 4) (−1)z+a (1, 2, 4, 3) (−1)3z+x+a (1, 4, 2, 3) (−1)2z+x (1, 3, 4, 2) (−1)2z x=0 (1, 4, 3, 2) (−1)z+a x=0 Since x is even, all the terms are positive and Q2 is non-zero. If z = a the tableau is symmetric in 1, 3, and 4, as well as 2 if x = 0. We get all the weight assignments listed above, in addition to those obtained by interchanging rows or allowing ω2,3 (1) = ( zz ). Interchanging rows has sign (−1)d+z = 1 since x is even. The other possibility changes the sign by (−1)z+a = 1. Hence all the terms are positive and Q2 is non-zero. Maximality: This tableau is maximal as shown in Example 7.1.1. 85 Tableau Q3 0<a≤d−b abb 1 4 4 Q3 = 2 3 2 3 2b ≤ d a≥b a even b 0 ω2,3 = ( 00 a+b 0 a 0) λ = [4d − 2a − 2b, a + 2b, a] r = 4d − 3a − 4b, s = 2b, t = a Since a ≥ b only the elements 2 and 3 may have ω2 = a + b. Hence the only valid weight assignments are (1, 2, 3, 4) with sign 1 and (1, 3, 2, 4) with sign (−1)a = 1. Thus all the valid weight assignments are positive and therefore Q3 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau Q4 2a + b + c = d aabccd 3 4 1 3 1 3 Q4 = 2 2 2 2 4 4 1 1 a+c+d≤d a, b, c, d > 0 b + d even a even, if a + c + d = d 0 d 0 c+d ω2,3 = ( 2a 0 0 0 ) λ = [3d − 2a − c − d, d + c + d, 2a] r = 2d − 2a − 2c − 2d, s = b + 2c + d, t = 2a Any weight assignment must have either ω3 (1) = 2a or ω3 (2) = 2a. Unless a + c + d = d, only the elements 1 and 2 may have ω2 = d. This gives the following weight table: 86 Assignment Sign Condition (1, 2, 3, 4) 1 (2, 1, 4, 3) (−1)b+d (2, 3, 4, 1) (−1)a+b+d a+c+d=d (1, 4, 3, 2) (−1)a a+c+d=d By our parity constraints, the weight sum is always positive. Hence Q4 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau Q5 a+b+c+d=d abcdef 4 3 3 4 31 Q5 = 2 2 2 2 44 1 1 a, b, e, f > 0 a+b+f≤d b+c+e<d a+d+e+f<d ω2,3 = ¡ 0 d 0 e+f a+b 0 0 0 If a + b + f = d then: either c, d > 0 or d = 0 and d + c + e even or c = 0 and d + d even ¢ λ = [3d − a − b − e − f, d + e + f, a + b] r = 2d − a − b − 2e − 2f, s = c + d + e + f, t = a + b Unless a + b + f = d, we must have ω2 (2) = d. Then since e, f > 0 we have ω2 (4) = e + f. As a, b > 0, we have ω3 (1) = a + b. Therefore no other valid weight assignments exist. If a + b + f = d, then it is possible to have ω2 (1) = d. However, unless c or d equals zero, there is no element with ω2 = e + f. Hence no such weight assignment can exist. If d = 0 we can have the weight assignment (2, 1, 4, 3) which has sign (−1)d+c+e = 1. If c = 0 then we can have the weight assignment (2, 1, 3, 4) which has sign (−1)d+d = 1. Thus, in either case, the weight sum is positive and Q5 is non-zero. Maximality: Rows two and three are maximal by Lemma 7.1.2. As no other fillings may have this row two weight and common elements between rows two and three, the tableau is maximal. 87 Tableau Q6 a>0 abc 1 4 114 Q6 = 2 2 323 3 a+b≤d−1 b≥c a even, if b = c b, c not both 0 c+1 0 ω2,3 = ( 00 a+b+1 0 a 0) λ = [4d − 2a − b − c − 2, a + b + c + 2, a] r = 4d − 3a − 2b − 2c − 4, s = b + c + 2, t = a Unless b = c, we must have ω2 (2) = a + b + 1. This force ω2,3 (3) = ( c+1 a ) and hence there are no other valid weight assignments. When c=b then the elements 1, 2, or 3 may have ω2 = a+b+1. If ω2 (2) = a+b+1 then only the element 3 may have ω2,3 = ( c+1 a ). If ω2 (1) = a + b + 1, then a valid weight assignment exists only for b = 0. In this case ω2 (3) = c+1. If ω2 (3) = a+b+1 then we may have ω2 (2) = c + 1 or, if c = 0, ω2 (1) = c + 1. Since these conditions are subject to b = c and b and c are never simultaneously zero, the only valid weight assignments are (1, 2, 3, 4) with sign 1 and (1, 3, 2, 4) with sign (−1)a = 1. Hence the weight sum is positive and Q6 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau Q7 a+b=d−2 ab c 2 3 44414 Q7 = 2 23332 1 11 ω2,3 = ¡ 0 d c+2 0 d−1 0 0 0 a+c≤d−2 b+c≤d−3 a, b > 0 ¢ λ = [2d − c − 1, d + c + 2, d − 1] 88 r = d − 2c − 3, s = c + 3, t = d − 1 Any valid weight assignment must have ω3 (1) = d − 1. Therefore ω2 (1) = 0, so we must have ω2 (3) > 0. Unless b = c only the element 2 may have ω2 = d, so the only valid weight assignment is (1, 2, 3, 4) with sign 1. When b = c, we may also have (1, 3, 2, 4) with sign (−1)a and (1, 3, 4, 2) with sign (−1)a+b+2 . Since this sum is odd, we must have Q7 non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. 89 7.5 Tableaux for c = 5 Tableau R1 xzzzab 52455 5 R1 = 33344 3 11112 2 ω2,3 = ¡0 0 d a+b 0<a≤b≤z d + a + z even, if b = z d even, if a = b, x = 0 ¢ 2z+x+b z+a 0 0 0 z = d−x 3 d ≡ x (mod 3) 0 λ = [3d − 2a − 2b, d + a + b, d + a + b] r = 2d − 3a − 3b, s = 0, t = d + a + b First consider the valid weight assignments with a < z. If b < z then we must have ω3 (1) = d; if b = z we may have ω3 (3) = d as well. If we have ω3 (1) = d, then ω3 (2 or 5) = a + b because a, b > 0. Since b ≥ a only the elements 1 and 3 may have ω2 = 2z + x + b unless a = b and x = 0. In that case, the element 4 may also have this weight. If ω2 (3) = 2z + x + b, then ω2 (5 or 4) = z + a. If ω2 (4) = 2z + b, then ω2 (2 or 3) = z + a. These constraints give the following table of weight assignments: Assignment Sign Condition (1, 2, 3, 4, 5) 1 (1, 5, 3, 4, 2) (−1)a+b (1, 2, 3, 5, 4) (−1)z+a (1, 2, 4, 3, 5) (−1)z a = b, x = 0 (1, 5, 4, 3, 2) (−1)z+b+a a = b, x = 0 (1, 5, 4, 2, 3) (−1)2z+a a = b, x = 0 (3, 4, 1, 2, 5) (−1)d+z+a b=z (3, 5, 1, 2, 4) (−1)d b=z (3, 4, 1, 5, 2) (−1)d b=z For b < z and a 6= b + x, we get a weight sum of 1 + (−1)a+b + (−1)z+a which is non-zero. When a = b, x = 0, we get a weight sum of 1+1+(−1)z+a +(−1)z +(−1)z + 90 (−1)a . This is non-zero by the parity constraints which imply z even. Similarly, if a < b = z, then the weight sum is 1 + (−1)a+z + (−1)z+a + (−1)d+z+a + (−1)d + (−1)d , which is also non-zero by the parity constraints. When a = b = z the tableau has many symmetries. Since the weight of rows two and three are the same, interchanging rows yields a new weight assignment with a difference in sign of (−1)d+a+b = 1. Hence we will not count those weight assignments which are inversions of rows two and three. First consider the possible element pairs (x, y) that can have ω3 (x, y) = (d, 2z). These are (1, 2), (1, 5), (3, 4), (3, 5), (5, 3), and (5, 1), along with (2, 1), (2, 4), (4, 3), (4, 2) when x = 0. Two of these pairs make up a weight assignment (not counting row inversion). If x 6= 0 the possible pair assignments are: (1, 2)(3, 4); (1, 2)(3, 5); (1, 2)(5, 3); (1, 5)(3, 4); and (3, 4)(5, 1). Since there is an odd number, this weight sum in non-zero. If x = 0, the condition d even, implies z is even. Hence all the column blocks are even. Since any valid weight assignment moves full column blocks, all weight assignments are positive and thus R1 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau R2 0≤a≤z−c x+z z z a b c c 2 445 4 4 2 R2 = 3 533 5 3 5 1 112 2 ω2,3 = ¡0 0 d b+a 0≤b≤z−c 0<c b≤a ¢ 2z+a+c+x 0 z+b+c 0 0 z = d−x 3 d≡x (mod 3) d even, if b = a + x 0 λ = [3d − 2a − 2b − 2c, d + a + b + 2c, d + a + b] r = 2d − 3a − 3b − 4c, s = 2c, t = d + a + b A valid weight assignment must have ω3 (1) = d since c > 0. Then we must have ω3 (2 or 5) = a + b, or if b = 0, then ω2 (3) = a is also possible. There are not enough 2’s to have ω2 (2) = 2z + a + c + x. If we have ω2 (3) = 2z + a + c + x, then 91 ω2 (5) = z + b + c. If we have ω2 (4) = 2z + a + c + x, then ω2 (2) = z + b + c, and b = a, x = 0. There are not enough 5’s to have ω2 (5) = 2z + a + c + x. Hence there are two possible weight assignments: (1, 2, 3, 4, 5) with sign 1 and (1, 5, 4, 3, 2) with sign (−1)d , which occurs only when b = a, x = 0. By our parity constraint, this sum is positive. If b = 0, we also have the weight assignment (1, 3, 4, 5, 2) when a = b, x = 0. The weight sum is odd and hence the tableau is non-zero. Maximality: The tableau is maximal by Lemma 7.1.5. To see this, note that when row three has the maximum weight d, the largest possible weights for row two are: 0 for the element 1, z + x + a + b + c for the element 2, 2z + x + a + c for the element 3, 2z + b + c for the element 4, z + a + b + c for the element 5. Since z − c ≥ a ≥ b, the weight 2z + x + a + c is the largest. Given this, Lemma 7.1.4 shows that row two is maximal. Moreover rows two and three cannot have elements in common, given the columns remaining after the largest weight elements have been assigned. Hence the tableau is maximal. Tableau R3 z+x z z b a 5 254 5252 R3 = 3 343 4344 1 112 2 ω2,3 = ¡0 0 2z+x+b+1 z+a+2 0 d a+b 0 0 0 0≤a<b≤z−1 z = d−x 3 d≡x (mod 3) ¢ λ = [3d − 2a − 2b − 3, d + a + b + 3, d + a + b] r = 2d − 3a − 3b − 6, s = 3, t = d + a + b, We must have ω3 (1) = d. Since a < b, only the element 3 may have ω2 = 2z + x + b + 1. Then ω2 (4) = z + a + 2. Hence there are no other valid weight assignments. Thus R3 is non-zero. Maximality: Row two is maximal by Lemma 7.1.2 and row three is maximal by Lemma 7.1.4. Inspection shows that it is not possible to have the non-d weights 92 assigned to the same element. Hence the tableau is maximal by Lemma 7.1.5. Tableau R4 d + a + c even, if a = b x z − 1 z+1 z z − a z − a a − 1 b a+1 c 2 2 5 5 4 5 5 2 2 5 R4 = 3 4 4 3 3 3 3 4 3 4 1 1 1 1 2 2 1≤a≤z a + b even, if c = x b≤a z = d−x 3 c≤x d≡x (mod 3) z≥2 ω2,3 = ¡0 0 d 2z+b+c 0 d 2z−2a 0 0 0 ¢ λ = [3d − 4z − b − c + 2a, d + 2z + b + c, d + 2z − 2a] r = 2d − 6z + 2a − 2b − 2c, s = 2a + b + c, t = d + 2z − 2a, By construction, only the element 1 may have ω3 (1) = d. This means any valid weight assignment must have ω3 (2 or 3) = 2z − 2a (unless a = z in which case the weight assignments are equivalent to those using 2 or 3). Furthermore, the elements appearing in row two must either be both 3 and 4 or both 2 and 5. Now we apply weight-set counting. Assignment Sign Condition (1, 2, 3, 4, 5) 1 (1, 2, 4, 3, 5) (−1)z−a b = a, c = x (1, 3, 2, 5, 4) (−1)d+b+c b=a (1, 3, 5, 2, 4) (−1)x+a+b+c c=x By our parity constraints, all these terms are positive. Hence R4 is non-zero. Maximality: Rows two and three are maximal by Lemma 7.1.2. Inspection shows that it is not possible to have the non-d weights assigned to the same element. Hence the tableau is maximal. 93 Tableau R5 y+w−a y−a a a b y y w 4 5 1 4 5 5 4 5 R5 = 3 2 3 2 1 3 2 2 1 1 ω2,3 = ¡ b d d 0 0 d−2a 0 0 0 0 ∗ a even, if a = b y = d2 0≤b≤a<y 2y + w = d ¢ λ = [2d − b + 2a, 2d + b, d − 2a] r = 2a − 2b, s = d + 2a + b, t = d − 2a Since y − a > 0, any weight assignment must have ω3 (1) = d − 2a. If ω2 (3) = d, then ω2 (2) = d and vice versa. Similarly for the elements 4 and 5, however we may only have ω2 (5) = d if a = b. So unless a = b there is only one valid weight assignment. When a = b, we also have (1, 4, 5, 3, 2) which has sign (−1)2d+a . Since a is even in this case, the sum is non-zero. Therefore R5 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau R6 x z z z z − a z − a − 1 a b a+1 c 2255 4 5 5 2 2 5254 R6 = 3443 3 3 3 4 3 4 1111 2 2 ω2,3 = ¡0 0 d 2z+b+c 0 d 2z−2a−1 0 0 0 1≤a≤z−2 z = d−x 3 0≤b<a d≡x c≤x z≥3 (mod 3) ¢ λ = [3d − 4z − b + 2a + 1, d + 2z + b + x, d + 2z − 2a − 1] r = 2d − 6z + 2a − 2b − 2c + 1, s = 2a + b + c + 1, t = d + 2z − 2a − 1 Only the element 1 can have ω3 =d and then any weight assignment must have ω2 (3) = d. This forces ω2 (4) = 2z+a + 1 + x. Hence ω3 (4)=0 which implies ω3 (2)> 0. Thus there are no other valid weight-assignments and R6 is non-zero. Maximality: Rows two and three are maximal by Lemma 7.1.2. Inspection 94 shows that it is non possible to have the non-d weights assigned to the same element. Hence the tableau is maximal. Tableau R7 1≤a<y y − a y − a a+1 a a − 1 y y+1 4 5 1 4 5 4 5 145 R7 = 3 2 3 2 1 3 2 1 1 ω2,3 = ¡ a−1 d d 0 0 d−1−2a 0 0 0 0 y = d−1 2 d odd d≥5 ¢ λ = [2d + a + 2, 2d + a − 1, d − 2a − 1] r = 3, s = d + 3a, t = d − 2a − 1 As y − a > 0, any weight assignment must have ω3 (1) = d − 1 − 2a . Moreover, only the elements 2 and 3 can have ω2 = d. Hence there are no other valid weight assignments, so R7 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau R8 y − a − 1 y − a a+1 a a − 1 y y 4 5 1 4 5 4 5145 R8 = 3 2 3 2 1 3 2 1 1 ω2,3 = ¡ a−1 d d 0 0 d−1−2a 0 0 0 0 1≤a≤y−2 y = d2 d even d≥6 ¢ λ = [2d + a + 2, 2d + a − 1, d − 2a − 1] r = 3, s = d + 3a, t = d − 2a − 1 Since a < y − 1, any weight assignment must have ω3 (1)=d − 1 − 2a. Moreover, 95 the only elements 2 and 3 can both have ω2 = d. Therefore there are no other valid weight assignments, so R8 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau R9 0≤a<z xzzzzabcde 344324 3 4 4 4 R9 = 225552 2 2 3 5 111133 b + c, b + d < z − a z = d−x 3 c+d+e<z−a+x d ≡ x (mod 3) 0≤e≤x x 6= 0 0≤d<x+b ω2,3 = ¡ 0 z+a+b+c+x d 0 d 0 3z+e z+a 0 0 ¢ λ = [3d − 2a − b − c − d − e − 2z, d + z + a + b + c + d + e, d + z + a] r = 2d − 3a − 2b − 2c − 2d − 2e − 3z, s = b + c + d + e, t = d + z + a Any valid weight assignment must have ω3 (1) = d. Only the elements 1 and 5 may have ω2 = 3z + e. If ω2 (5) = 3z + e, then only the element 2 may have ω2 = x + z + a + b + c due to the condition d < b + x. Examining the tableau in light of these constraints shows that (1, 2, 3, 4, 5) is the only valid weight assignment. Thus the tableau is non-zero. Maximality: As the discussion above shows, the tableau weights are as large as possible. Hence by Lemma 7.1.5, this tableau is maximal. 96 Tableau R10 1≤a≤z b ≤ b a2 c x z z z z−a b c d e f x 3443 3 2 4 3 434 R10 = 2255 5 5 3 2 255 1111 2 c, e ≤ b 2z c d ≤ d 2z e + b a2 c f ≤ d a2 e c even ω2,3 = d>c z+b>c+d z−a+b+f>c+e z+f>d+e z = d−x 3 d ≡ x (mod 3) z≥2 ¡ 0 x+z+d+e c 0 3z−a+b+f+x ¢ d z−a 0 0 0 λ = [3d − 2z + 2a − b − c − d − e − f − x, d + z − a + b + c + d + e + f + x, d + z − a] r = d + 3a − 2b − 2c − 2d − 2e − 2f − x, s = b + c + d + e + f + x, t = d + z − a Any valid weight assignment must have ω3 (1) = d. Given the bounding parameters, only the element 5 can have ω2 = 2z + a + b + f + x. When ω2 (5) is maximal, only the element 2 may have ω2 = x + z + d + e. From this we find that the valid weight assignments are (1, 2, 3, 4, 5) with sign 1, and (1, 2, 4, 3, 5) with sign (−1)c = 1. Maximality: The discussion above shows that the row two weights are maximized. Hence by Lemma 7.1.5, this tableau is maximal. Tableau R11 0 < a, b abcdef 3 5 3 5 15315 R11 = 4 2 4 2 21424 1 1 ω2,3 = ¡ f b+d+e+1 a+c+2 0 0 a+b 0 0 0 0 c≤d−a−2 d≤d−b−e−1 0<f≤e ¢ e+f≤d−a−b−1 b+d>f a+c≥e+f 97 λ = [5d − 2a − 2b − c − d − e − f − 3, a + b + c + d + e + f + 3, a + b] r = 5d − 3a − 3b − 2c − 2d − 2e − 2f − 6, s = c + d + e + f + 3, t = a + b Any valid weight assignment must have ω3 (1) = a + b and hence ω2 (1) = f. This implies ω2 (2) > 0. But to have exactly three non-zero weights in row two, we then must have ω2 (4) > 0. Hence no other weight assignments are possible and hence the tableau is non-zero. Maximality: The possible row two weights, given the maximization of row three, are e + f + 1 for the element 1, b + d + e + 1 for the element 2, a + c + 1 for the element 3, a + c + 2 for the element 4, and b + d + f + 1 for the element 5. Given the conditions on these parameters we see that the weights for 2 and 4 are the largest. Hence by Lemma 7.1.5, this tableau is maximal. Tableau R12 y−1+w y y y a w 1 1 5 4 1 5 5 R12 = 3 3 3 2 2 4 2 4 ∗ 0<a≤d−y−1 y = d2 y even, if y = a 2y + w = d ω2,3 = ( 00 d0 d0 a1 00 ) λ = [3d − a − 1, 2d + a, 1] r = d − 2a − 1, s = 2d + a − 1, t = 1 Any valid weight assignment must have two of the elements 1, 3, and 4 having non-zero weight in row two. Hence at least one of them must have ω2 = d. Moreover, we can not have a row two weight of d for both of these elements. If ω2 (1) = d we may not have ω2 (2) = d, so the only option is ω2 (5) = d when a = y. This corresponds to a weight assignment of (3, 1, 5, 4, 2) with sign (−1)2d+a which equals 1 due to the parity constraint. If ω2 (3) = d we may not have ω2 (5) = d, so the only option is ω2 (2) = d. This corresponds to a weight assignment of (1, 2, 3, 4, 5) with 98 sign 1. When ω2 (4) = d, neither the element 2 nor the element 5 may have ω2 = d, so there is no weight assignment with this option. Hence the weight sum is always positive and thus R12 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau R13 abcde 1 3 1 3 1 R13 = 2 4 4 2 2 5 5 ω2,3 = ¡ 0 a+d+e 0 b+c 0 0 0 0 a+d+e≤d d≥c a+b≤d a+e≥b b+d≤d If c = d then a + b + c + d + e even a, b, c, d, e > 0 ¢ 0 If b = a + e then a + b + c + d + e even a+b λ = [5d − 2a − 2b − c − d − e, a + b + c + d + e, a + b] r = 5d − 3a − 3b − 2c − 2d − 2e, , s = c + d + e, t = a + b Since a, b > 0 any valid weight assignment must have ω3 (5) = a + b. We must have ω2 (1 or 2) > 0. If ω2 (2) > 0, then ω2 (4) > 0; if ω2 (1) > 0, then ω2 (3) > 0. This shows the only valid weight assignments are (1, 2, 3, 4, 5) and (2, 1, 4, 3, 5). The second one only occurs when d = c or b = a + e. If that happens, the sign is (−1)a+b+c+d+e = 1. Hence R13 is non-zero. Maximality: The possible maximal weights for row two are: a + c + e for the element 1; a + d + e for the element 2; b + d for the element 3; b + c for the element 4; and 0 for the element 5. By our parameter conditions, a + d + e is the largest. Hence by Lemma 7.1.5 the tableau is maximal. Tableau R14 d−3 2 d−4 1 4 1 5 151 R14 = 2 2 2 3 344 3 d≥5 99 2 0 ω2,3 = ( 00 d0 d−3 1 0 0) λ = [3d, 2d − 1, 1] r = d + 1, s = 2d − 2, t = 1 For d > 5, any valid weight assignment must have ω2 (2) = d. This forces ω2 (3) = d − 3 and ω2 (4) = 2. Hence there are no other valid weight assignments. When d = 5, we may have ω2 (1) = d or ω2 (1) = d−3 as well. Then the possible weight assignments are (1, 2, 3, 4, 5), (3, 2, 1, 5, 4, ), and (3, 1, 2, 5, 4). Since there is an odd number, this sum is non-zero and thus so is R14 . Maximality: This tableau is maximal by Lemma 7.1.5. Tableau R15 z+x z z 2 55222 R15 = 3 43433 1 11 z = d−x 3 d ≡ x (mod 3) z≥2 z+1 0 ω2,3 = ( d0 00 2z+x+2 0 0 0) λ = [3d − 3, d + 3, d] r = 2d − 6, s = 3, t = d Since only the element 1 may have ω3 = d and only the element 3 may have ω2 = 2z + x + 2, there are no other valid weight assignments. Hence the tableau is non-zero. Maximality: Given the row three weight of d, the row two weights are as large as possible. Thus by Lemma 7.1.5 this tableau is maximal. 100 Tableau R16 d−2 a 1 55531 R16 = 2 44242 3 3 ω2,3 = ¡0 d 0 a+2 0 0 0 d−1 0 0 1≤a≤d−4 d≥5 ¢ λ = [3d − a − 1, d + a + 2, d − 1] r = 2d − 2a − 3, s = a + 3, t = d − 1 Any valid weight assignment must have ω3 (3) = d − 1 and ω2 (2) = d. This forces ω2 (4) = a+2. Hence there are no other valid weight assignments and R16 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau R17 x z z z z−1 a z b 22552 5 225254 R17 = 34433 3 434 11114 ω2,3 = 0≤a≤z−2 b≤x z = d−x 3 d ≡ x (mod 3) z≥2 ¡ 0 0 d 2z+a+b 0 ¢ d 0 0 1 0 λ = [3d − 2z − a − 1, d + 2z + a + x, d + 1] r = 2d − 4z − 2a − 2b − 1, s = 2z + a + b − 1, t = d + 1 Only the element 1 can have ω3 =d and then any weight assignment must have ω2 (3) = d. This forces ω2 (4) = 2z+a + 1 + x. Thus there are no other valid weight assignments and R17 is non-zero. Maximality: The tableau is maximal by Lemma 7.1.3. 101 Tableau R18 xzzzz 52442 R18 = 33353 11115 x even ω2,3 = ( d0 00 d0 00 zz ) λ = [3d − 2z, d + z, d + z] r = 2d − 3z, s = 0, t = d + z Only the elements 1 and 3 may have a weight of d. Hence the only valid weight assignments are (1, 2, 3, 4, 5) and (3, 2, 1, 4, 5), both of which have sign 1 since x is even. Hence R18 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau R19 d−2 d−3 1 553 5 1 R19 = 2 422 4 4 3 3 ω2,3 = ¡0 d 0 d−1 0 0 d−1 0 ¢ λ = [2d + 2, 2d − 1, d − 1] r = 3, s = d, t = d − 1 Any valid weight assignment must have ω3 (3) = d. Then ω2 (2) = d. Hence there are no other weight assignments possible. Thus R19 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. 102 7.6 Tableaux for c = 6 Tableau S1 x+z x+z z z a a 5 6 565 6 S1 = 4 3 343 4 1 2 112 2 ω2,3 = ¡0 0≤a≤z z = d−x 3 z + x even, if a = 0 d≡x 0 2z+a+x 2z+a+x 0 0 d z+2a+x 0 0 0 0 (mod 3) ¢ λ = [4d − 2z − 2x − 4a, d + z + 2a + x, d + z + 2a + x] r = 2d − 2x − 6a, s = 0, t = d + 2a+z + x The construction of S1 means that any valid weight assignment with a > 0 must have ωi (j) > 0 if and only if ωi (k) > 0 for the pairs (j, k) = (1, 2), (3, 4), or (5, 6). These constraints show that the only valid weight assignments are those that interchange complete rows in the body of S1 . Since the length of these rows t = 4z+2x+2a is even, all valid weight assignments are positive and hence S1 is non-zero. When a = 0 we must have ω3 (1) = d and ω2 (3, 4) > 0 or ω2 (5, 6) > 0. Hence the valid weight assignments are: (1, 2, 3, 4, 5, 6) with sign 1; (1, 6, 3, 4, 5, 2) with sign (−1)z+x ; (1, 2, 5, 6, 3, 4) with sign (−1)4z+2x ; and (1, 3, 5, 6, 2, 4) with sign (−1)5z+3x . Since z + x is even, these assignments are all positive. Hence the tableau is non-zero. Maximality: By Lemma 7.1.5 this tableau is maximal. Tableau S2 d−1 a 4 222423 S2 = 5 151111 6 3363 ω2,3 = ¡ a+4 0 0 0 0 d 0 0 a+2 0 0 d 0≤a≤d−4 a ≡ d (mod 2) d≥4 ¢ λ = [4d − 2a − 6, d + a + 4, d + a + 2] r = 3d − 3a − 10, s = 2, t = d + a + 2 103 Any valid weight assignment must have ω2 (1) = a + 4 since if ω2 (1) = 0 then we must have ω2 (2, 3) = (a + 4, d) which is not possible. If ω2 (1) = a + 4, then ω2 (5) = d. Thus the only valid weight assignments are (1, 2, 3, 4, 5, 6) with sign 1 and (1, 3, 2, 6, 5, 4) with sign (−1)d+a+2 . Since a ≡ d (mod 2) this sum is positive. Hence S2 is non-zero. Maximality: Rows two and three are maximal by Lemma 7.1.2. Since it is not possible for the non-d element of row three to have a row two weight of a + 2, the tableau is maximal. Tableau S3 d−4 a 2 2 5 11355511 S3 = 2 42442424 6 36633 ω2,3 = ¡0 d 0 a+5 0 0 0 0 a+2 0 0 d 0≤a≤d−5 d≥5 ¢ λ = [4d − 2a − 7, d + a + 5, d + a + 2] r = 3d − 3a − 12, s = 3, t = d + a + 2 Any valid weight assignment must have ω3 (6) = d and ω3 (3) = a + 2 (or ω3 (5) = 2 if a = 0). Then only the element 2 may have ω2 = d (or 4 if a = d − 5). Moreover if ω2 (2) = d, then ω2 (4) = a + 5. Hence the only valid weight assignments are (1, 2, 3, 4, 5, 6) with sign 1 and, when a = 0, (1, 2, 5, 4, 3, 6) with sign (−1)2 . Therefore the weight sum is positive and S3 is non-zero. Maximality: Rows two and three are maximal by Lemma 7.1.2. Since the nond element of row two has weight a + 5 and there are only two copies of the element 3 available for row two, the tableau is maximal. 104 Tableau S4 d−3 a 2 5 153511111 S4 = 2 424422344 6 36633 ω2,3 = ¡0 d 1 a+5 0 0 0 0 a+2 0 0 d 0≤a≤d−5 d≥5 ¢ λ = [4d − 2a − 8, d + a + 6, d + a + 2] r = 3d − 3a − 14, s = 4, t = d + a + 2 Any valid weight assignment must have ω3 (6) = d. Now ω3 (6) = d implies ω3 (3) = a + 2 and ω2 (3) = 1. Unless a = d − 5, we must have ω2 (2) = d and so ω2 (4) = a + 5. When a = d − 5 we may have ω2 (1) = d or ω2 (4) = d. However, if ω2 (1) = d, then there is no element with ω2 = a + 5. If ω2 (4) = d, then ω2 (2) = a + 5. This shows the only weight assignment is (1, 2, 3, 4, 5, 6). Hence S4 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5 and the discussion above. Tableau S5 0 < a, b, c ≤ d − e x + z0 x + z0 z0 z0 z0 z0 a b c 5 6 5 6 5 6 6 5 1 S5 = 4 3 3 4 3 4 2 4 3 1 2 1 1 2 2 a even z0 = e−x 3 c even if a = b e≡x b even if a = c 3≤e<d b, c ≥ a e+b 0 0 ω2,3 = ( 0e ae e+c 0 0 0 0) λ = [6d − 4e − a − b − c, 2e + a + b + c, 2e] r = 6d − 6e − 2a − 2b − 2c, s = a + b + c, t = 2e (mod 3) 105 The construction of S5 forces the following constraints on any valid weight assignment. • ω3 (1) > 0 ⇐⇒ ω3 (2) > 0. • ω3 (3) > 0 ⇐⇒ ω3 (4) > 0. • ω3 (5) > 0 ⇐⇒ ω3 (6) > 0. • If ω2 (j) = a, then ω3 (j) > 0. • If ω2 (1) = a, then ω2 (5, 6) > 0 and a = c. • If ω2 (2) = a, then ω2 (3, 4) > 0. • If ω2 (3) = a, then ω2 (5, 6) > 0 and a = c. • If ω2 (4) = a, then ω2 (1, 2) > 0 and a = b. • If ω2 (5) = a, then ω2 (1, 2) > 0 and a = b. • If ω2 (6) = a, then ω2 (3, 4) > 0. From this we can derive a signed weight table. Assignment Sign Condition (1, 2, 3, 4, 5, 6) 1 (2, 1, 6, 5, 4, 3) (−1)2e+a+b+c a=c (3, 4, 1, 2, 5, 6) (−1)2e+c a=b (4, 3, 6, 5, 1, 2) (−1)4e+a+b a=c (5, 6, 3, 4, 1, 2) (−1)2e+a (6, 5, 1, 2, 3, 4) (−1)4e+b+c a=b Computing the weight sum we obtain 3 + 3(−1)a = 6 when a = b = c as a is always even. For a = b 6= c, the sum is 2 + 2(−1)c . This is non-zero as c is even when a = b. For a = c 6= b, we have 2 + 2(−1)b . This is non-zero as b even when a = c. Finally if a 6= b, c the sum is 1 + (−1)a which is non-zero. Hence S5 is non-zero. 106 Maximality: By Lemma 7.1.4, row three is maximal. Since b, c ≥ a, row two is maximal by Lemma 7.1.4. As the number of row three elements available for row two is bounded by a, this is the largest weight and hence the tableau is maximal. Tableau S6 x + z0 x + z0 z0 z0 z0 z0 − 1 a b c 3 4 4 3 3 4 2 3 4 S6 = 5 6 5 6 5 6 6 5 1 1 2 1 1 2 2 ω2,3 = ¡c 0 e+b e+a−1 0 0 e e−1 0 0 0 0 0 < a, b, c ≤ d − e z0 = e−x 3 a, b ≥ c e≡x b≥a−1 3≤e≤d (mod 3) ¢ λ = [6d − 4e − a − b − c + 2, 2e + a + b + c − 1, 2e − 1] r = 6d − 6e − 2a − 2b − 2c + 3, s = a + b + c, t = 2e − 1 For any valid weight assignment we can have ω3 (j) = e only for j ∈ {1, 3, 5}. Moreover, ω3 (j) = e if and only if ω3 (j + 1) = e − 1. If ω3 (j) = e then ω2 (j) = c. If ω2 (1) = c, then ω2 (5, 6) > 0. If ω2 (3) = c, then ω2 (1, 2) > 0 and b = c. If ω2 (5) = c, then ω2 (1, 2) > 0 and b = c. This means (1, 2, 3, 4, 5, 6) is the only valid weight assignment when b 6= c. If b = c, then we additionally have weight assignments (3, 4, 1, 2, 5, 6) and (5, 6, 1, 2, 3, 4). In either case the weight sum is odd and hence non-zero. Thus S6 is non-zero. Maximality: By Lemma 7.1.4 row three is maximal. Since a, b ≥ c, row two is maximal by Lemma 7.1.4. As the number of row three elements available for row two is bounded by c, this is the largest weight and the tableau is maximal. Tableau S7 abcd e 1 4 3 11135 S7 = 2 2 2 42246 3 3 ω2,3 = ¡0 d 0 0 0 a+b ¢ d+1 0 e 0 0 0 a+b+c=d−2 a, b, d > 0 b+d≤d−2 0≤e≤d a+d<d−2 e even 107 λ = [5d − a − b − d − e − 1, d + d + e + 1, a + b] r = 4d − a − b − 2d − 2e − 2, s = c + d + e + 3, t = a + b Any valid weight assignment must have ω2 (5 or 6) = e and ω3 (3) = a + b. Since a + d < d − 2, we must have ω2 (2) = d. Then, as d > 0 we must have ω2 (4) = d + 1. Hence the only valid weight assignments are (1, 2, 3, 4, 5, 6) with sign 1 and (1, 2, 3, 4, 6, 5) with sign (−1)e . Since is e is even, this weight sum is positive. Hence S7 is non-zero. Maximality: Inspection shows that rows two and three are maximal. Since d > 0 we cannot have any 3’s in a maximal row two. Thus S7 is maximal by Lemma 7.1.5. Tableau S8 d−4 d−2 d−4 2 66 6 3 2 1 23662 S8 = 34 3 4 4 5 55151 11 d≥5 d d 0 ω2,3 = ( 22 00 d−3 0 0 0 0) λ = [3d − 1, 3d − 1, 2] r = 0, s = 3d − 3, t = 2 Examining the tableau shows that we can only have ω2 (j, k) = (d, d) for (j, k) = (2, 3), (2, 6) or (4, 5). Also, we must have ω2,3 (1 or 6) > 0, so ω2 (2, 6) = (d, d) and ω2 (1) = 2. This is possible only when d = 6. Hence the valid weight assignments are: (1, 2, 3, 4, 5, 6) with sign 1; (6, 4, 1, 2, 3, 5) with sign (−1)2d+4 ; (1, 4, 3, 2, 6, 5) with sign (−1)3d−1 when d = 6; and (6, 2, 1, 4, 5, 3) with sign (−1)3 when d = 5. In all cases, the weight sum is positive and hence S8 is non-zero. d d 0 Maximality: This tableau is not maximal since ω2,3 = ( 20 00 d−1 0 0 0 0 ). However, this weight is zero in qT . This tableau cannot be put in maximal form, hence we will need to prove directly that it is disjoint from the requisite tableaux. This will be done in Section 8.7. 108 Tableau S9 a+c+d=d−1 b+f=d−1 abcdefg 1 5 1 5 36 6 36 S9 = 2 4 2 2 54 5 42 3 3 f+g≤d−1 a+b+e≤d−1 b+d+e+g≤d−1 a, b > 0 e, g > 0 or e = 0 and g even ω2,3 = ¡0 d 0 d e+g 0 0 0 a+b 0 0 0 ¢ λ = [4d − a − b − e − g, 2d + e + g, a + b] r = 2d − a − b − 2e − 2g, s = c + d + e + f + g + 2, t = a + b Any valid weight assignment must have ω2 (3) = a + b because a, b > 0. Only the elements 2 and 4 may simultaneously have ω2 = d. Then ω2 (5) = e + g unless e or g is 0. If e = 0 we may also have ω2 (6) = g. Hence the weight assignments are are (1, 2, 3, 4, 5, 6) with sign 1 and, if e = 0, (1, 2, 3, 4, 6, 5) with sign (−1)g = 1. Thus S9 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5 since g > 0. Tableau S10 2 a 155165 S10 = 242243 33 2≤a≤3 d=4 0 0 ω2,3 = ( 00 04 31 a+1 0 0 0) λ = [15 − a, a + 6, 3] r = 9 − 2a, s = a + 3, t = 3 Any valid weight assignment must have ω3 (3) = 3. Other than 3, the only elements 109 that can have ω2 = d is 2, or 4 if a = 3. Thus there are no other valid weight assignments possible and the tableau is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. Tableau S11 d−1 d−3 d−1 1 5 6 1 6 5 S11 = 2 2 3 3 4 4 3 d 0 0 ω2,3 = ( 00 d0 d−2 1 0 0 0) λ = [3d + 1, 3d − 2, 1] r = 3, s = 3d − 3, t = 1 Any valid weight assignment must have two of the elements 1, 2, and 3 with ω2 > 0. Since there are not enough 1’s in the body for this to happen, we must have ω2 (2) = d and ω2 (3) = d − 2. This force ω2 (4) = d. Hence there are no other valid weight assignments. Thus the tableau is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. Tableau S12 0≤a≤d−3 ab c 33 11135 S12 = 22 42246 4 0≤b≤d−2 a≥b 0≤c≤d c even 0 b+1 0 c ω2,3 = ( 00 a+3 0 0 1 0 0) λ = [6d − a − b − c − 5, a + b + c + 4, 1] r = 6d − 2a − 2b − 2c − 9, s = a + b + c + 3, t = 1 Any valid weight assignment must have ω2 (5 or 6) = c. We also must have 110 ω2 (2) = a + 3 since a ≥ b. Then ω2 (4) = b + 1 unless b = 0, in which case ω2 (3) = 1 is possible. Hence the weight assignments are (1, 2, 3, 4, 5, 6) with sign 1 and (1, 2, 3, 4, 6, 5) with sign (−1)c . When b = 0 we also have (1, 2, 4, 3, 5, 6) with sign (−1)2 , and (1, 2, 4, 3, 6, 5) with sign (−1)c+2 . Since c is even, this sum is positive. Hence S12 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.5. 7.7 Tableaux for c = 7 and c = 8 For c = 7 and 8, nearly all the required tableaux can be obtained by joining those tableaux already constructed. This is demonstrated in Chapter 8. However, we do need to construct one additional tableau, which is listed below. Tableau W1 xxzzzzzzzza 78678837873 S1 = 45456456456 1211122233 ω2,3 = a≤x z = d−x 3 d ≡ x (mod 3) a = 0, 2 ¡ 0 0 0 d d 2z+a 0 0 ¢ d d 2z 0 0 0 0 0 0 λ = [4d − 4z − a, 2d + 2z + a, 2d + 2z] r = 2d − 6z − 2a, s = a, t = 2d + 2z Only the triples (1, 2, 3) and (4, 5, 6) may have ωi = (d, d, 2z) or larger. Hence the only valid weight assignments are (1, 2, 3, 4, 5, 6, 7, 8) with sign 1, and (4, 5, 6, 1, 2, 3, 7, 8) with sign (−1)2d+2z+a . Since a is even, the weight sum in positive. Hence W1 is non-zero. Maximality: This tableau is maximal by Lemma 7.1.3. 111 Chapter 8 Tableau Sufficiency Our proof of Theorem 2 in Chapter 6 presupposed we had constructed all tableaux with c ≤ 8 for the shapes of Theorem 10. In Chapter 7 we constructed many non-zero tableaux. In this chapter we will demonstrate that all the necessary tableaux have been constructed. Specifically we need all shapes in Theorem 10, that is all partitions [r + s + t, s + t, t] of n, with r, s 6= 1, such that if r or s is in {0, 2, 4} then s + t is even. Recall that those required shapes with r or s less than 5 are called exceptional cases. 8.1 Sufficiency when c = 3 The tableaux we will use for c = 3 are the Pi described in Section 7.3. These tableaux are all maximal and non-zero by weight-set counting. We will show that every necessary partition of n = 3d has a corresponding Pi . First consider the exceptional r cases. These are tableaux having s + t even with r ∈ {0, 2, 4}, or r = 3 with no constraints on s and t. Since our shape is a partition of 3d, we have the condition 3t + 2s + r = 3d. Hence for a given r, we need only to , with s 6= 1 are obtained. verify that all the appropriate s in the range 0 ≤ s ≤ 3d−r 2 This condition shows s ≡ r (mod 3). Table 8.1 below lists the tableaux we are using for these cases. The column ‘Parameters’ indicates the restrictions on the tableaux arising from their construction in Section 7.3. The column ‘s, t values’ indicates their values in terms of the tableau 112 parameters, while the ‘s covered’ lists those cases covered by the given tableau. The restrictions given on the s covered reflect the listed conditions on the parameters. r=0 r=2 r=3 r=4 Tableau Parameters s, t values s covered P4 (d − 2b, b, b) 0 ≤ b ≤ d2 s = 3b 0 ≤ s ≤ 3b d2 c d − b even t = d − 2b s + t even P4 (d − 2b, b, b − 1) 1 ≤ b ≤ d2 s = 3b − 1 2 ≤ s ≤ 3b d2 c − 1 d − b odd t = d − 2b s + t even P4 (d − 2b − 1, b, b, b) 0 ≤ b ≤ d−1 2 s = 3b 0 ≤ s ≤ 3b d−1 2 c d − b odd t = d − 2b − 1 s + t even P3 (d − 2b − 1, b, b − 1) 1 ≤ b ≤ d−1 2 s = 3b 3 ≤ s ≤ 3b d−1 2 c d − b even t = d − 2b − 1 s + t odd P4 (d − 2b, b, b − 2) 2 ≤ b ≤ d2 s = 3b − 2 4 ≤ s ≤ 3b d2 c − 2 d − b even t = d − 2b s + t even Table 8.1: Exceptional r cases for c = 3. To see why Table 8.1 reaches the necessary upper bounds on s, we need to consider − 2r , which is obtained the parity of d. For d and r even, the maximum s needed is 3d 2 c = 3d−r−1 . However, s = 3d−r−1 6≡ r in the table. When d is odd, we need s ≤ b 3d−r 2 2 2 (mod 3), thus the largest s we need is s = 3d−r−3 = 3b d2 c − 2r . When r = 3, we 2 need s ≤ 3d−3 , which equals 3b d−1 c when d is odd. But for d even, the largest s ≡ r 2 2 = 3b d−1 c. Hence the s bounds in Table 8.1 are correct. (mod 3) is s = 3d−9 2 2 For the lower bounds, Table 8.1 shows that all the necessary s are covered, except possibly some s < 5. Since s ≡ r (mod 3), all s are covered in the r = 2 case. In the r = 3 case, s = 0 is only necessary when the s + t is even. Similarly, no additional tableaux are needed in the r = 4 case because s = 1 is not a shape of Theorem 10. Now consider the exceptional s cases. These are tableaux having t even with s ∈ {0, 2, 4}, or s = 3 with no constraints on t. We still have the condition 3t+2s+r = 3d, so for a given s we need only verify that all the appropriate t in the range 0 ≤ t ≤ d − d 2s e, (with t even if s 6= 3) are obtained. 3 Table 8.2 belows lists the tableaux we are using for these cases. The columns ‘Parameters’ and ‘Conditions’ indicate the restrictions on the tableaux arising from their construction in Section 7.3. The column ‘t values’ indicates t’s value in terms 113 of the tableau parameters, while the ‘t covered’ lists those cases covered by the given tableau. The restrictions given on the t covered reflect the conditions listed on the parameters. Tableau Parameters Conditions t value s=0 P2 (a, 0) 0≤a≤d a even t=a s=2 P2 (a, 2) 0≤a≤d−2 a even t=a s=3 P3 (a, 1, 0) 0 ≤ a ≤ d − 3 a even t=a P4 (a, 1, 1) 0 ≤ a ≤ d − 2 a odd t=a P2 (a, 4) a even t=a s=4 0≤a≤d−4 t covered 0≤t≤d t even 0≤t≤d−2 t even 0≤t≤d−3 t even 0≤t≤d−2 t odd 0≤t≤d−4 t even Table 8.2: Exceptional s cases for c = 3. Table 8.2 shows that all the necessary t are covered, except possibly when s = 3 or 4. When s = 3, the t = d − 2, t even case does not appear. In this case, r = 0 and s + t is odd, so by Theorem 10 this case is not needed. For s = 4 and t = d − 3, we have r = 1 which is not a required shape. Hence all the exceptional cases have been covered. Finally, consider the general cases remaining. These are tableaux having r, s ≥ 5 and no additional constraints. We still have the condition 3t + 2s + r = 3d, so for a , fixed t we need only verify that all the appropriate s in the range 5 ≤ s ≤ 3d−3t−5 2 are obtained. (This accounts for the bounds both on r and on s.) Table 8.3 belows lists the tableaux we are using for this case. The columns ‘Parameters’ and ‘Conditions’ indicate the restrictions on the tableaux arising from their construction in Section 7.3. The column ‘s values’ indicates its value in terms of the tableau parameters, while the ‘s covered’ lists those cases covered by the given tableau. To see how Table 8.3 covers all the necessary shapes, first consider P3 . As c varies 114 Tableau Parameters Conditions s value s covered P3 (t, b, c) 2 ≤ b ≤ d−t−1 2 t + c even s = 2b + c + 1 5 ≤ s ≤ 3b d−t−1 2 c P4 (t, b, c) 1 ≤ b ≤ d−t 2 t + c even s = 2b + c 2 ≤ s ≤ 3b d−t 2 c 0≤c<b 0≤c≤b s + t odd s + t even Table 8.3: General c = 3 cases. between 0 and b − 1, we get 2b + 1 ≤ s ≤ 3b. When b increases to b + 1, we go from s = 3b to s = 2b + 3 = 2(b + 1) + 1. There is no gap between these provided b ≥ 2. Since b = 2 yields a minimum s = 5, we don’t need any smaller cases. Hence P3 will cover the cases, provided s + t is odd (equivalently t + c is even). Given a case where s = 2b + c + 1 but t + c is odd, use P4 (t, b, c + 1). Then t + c + 1 is even, and c < b implies c + 1 ≤ b, so the conditions of P4 are satisfied. (As similar analysis on P4 c shows all the s in the range do occur.) To see that the upper bound of s ≤ b 3d−3t−5 2 is met, first consider P4 . Since 3b d−t c ≥ b 3d−3t c − 1 = b 3d−3t−2 c, the upper bound 2 2 2 c ≥ 3d−3t−3 when d − t is odd. If d − t is even, we have is obtained. For P3 , 3b d−t−1 2 2 b 3d−3t−5 c = 3d−3t−6 , which is 3b d−t−1 c as desired. Thus all the required shapes are 2 2 2 listed. 8.2 Sufficiency when c = 4 When c = 4, the tableaux we will use are the Qi listed in Section 7.4. We will show that every partition of n = 4d described in Theorem 10 has a corresponding Qi . Note that shapes with r ≥ d + 5 can be obtained by Pi ∨ V (d) for the appropriate Pi filled with c = 3 elements. As such, we will not include these shapes in the following compilation. Throughout, we will use the convention d = 3z + x, where d ≡ x (mod 3). Unless otherwise specified, take x ∈ {0, 1, 2}. We also use the notation d∗ from previous chapters (d∗ = 2b d2 c). First consider the exceptional r cases. These are tableaux having s + t even with r ∈ {0, 2, 4} or r = 3 with no constraints on s and t. Since our shape is a partition of 115 4d we have the condition 3t + 2s + r = 4d. Hence for a given r, we need only verify that all the appropriate t in the range 0 ≤ t ≤ b 4d−r c, (with s + t even for r 6= 3) are 3 obtained. For the exceptional r cases with s + t even, we only need those t with t ≡ r (mod 4). Given a shape (r, s, t) the next shape needed is (r, s − 6, t + 4). Consider Table 8.4. For r = 0 we have all t ≤ 4d when d ≡ 0 (mod 3). When d ≡ 1 (mod 3), 3 is not possible and the table provides all t ≤ d + d−1 − 1. the shape t = d + d−1 3 2 Similarly, when d ≡ 2 (mod 3), the shape t = d + d−2 corresponds to s = 1, while 3 t = d + d−2 − 1 is not possible. This covers the r = 0 cases. Note when d = 5, the 3 only shapes needed are t = d − 1 and t = 0. Moreover, t = d − 1 and t = d − 2 are not possible for d ≤ 4. For r = 2, Table 8.4 provides tableaux for 0 < t ≤ d + z − 2 − x or t = d + z if d ≡ 2 (mod 3). Since t ≡ 2 (mod 4), t < 2 is not needed. When d ≡ 2 (mod 3), t = d + z − 1 and t = d + z − 3 are not congruent to 2 (mod 4). Similarly, t = d + z − x and t = d + z − x − 1 are not congruent to 2 (mod 4) for x = 0 and 1. When r = 3 we no longer have the conditions t ≡ r (mod 4); instead t must be odd. Table 8.5 accounts for all tableaux with 0 < t ≤ d + d3 − 3 for d ≡ 0 (mod 3). − 4 and for t ≤ d + d−2 −5 For d ≡ 1 (mod 3) Table 8.5 accounts for all t ≤ d + d−1 3 3 when d ≡ 2 (mod 3). Tableaux with larger t correspond to shapes having exceptional s cases (s 6= 3). Since t is odd, these shapes are not needed according to Theorem 10. When t ≤ d − 1 and d small, the shapes are either not required by Theorem 10 or are not possible. For r = 4, we need all t ≡ 0 (mod 4), where t ≤ d + z − 4 if x = 0, t ≤ d + z − 1 if x = 1, and t ≤ d + z − 2 if x = 2. Table 8.6 provides all these tableaux. Again, the bounds on d are necessary to produce a valid shape. When d = 5 and t = 4, the shape has s = 2 and can be found in Table 8.7. Hence these tables cover all the exceptional r cases. Now consider the exceptional s cases. These are tableaux having t even with s ∈ {0, 2, 4}, or s = 3 with no constraints on t. When r ≥ d, r 6= d + 1, we may reduced to Pi ∨ V (d) for some Pi filled with c = 3 elements. (If s = 3 we may only r=2 r=0 ∗ Q3 (2, 1) 0 < 2k < d t ≡ 0 (mod 2) t=d−1 t ≡ 0 (mod 4), d ≥ 6 t=0 t=d+z t = d∗ − 2k s = 2d s=0 d=3 t=2 d=3 t=2 t ≡ 2 (mod 4) d≥4 2≤t≤d−2 s=2 t = d∗ − 2k d≥4 W = d − d∗ s = d2 + 3k + 2W − 1 d∗ − 2k ≡ 2 (mod 4) ∗ t ≡ 2 (mod 4) t=d−1 d ≡ 3 (mod 4) 0 < 2k < d∗ t=d−1 t ≡ 2 (mod 4) d≥5 t=d+z−k d ≡ 2 (mod 3) d≤t≤d+z−x−2 t=d+z s = k + k+x 2 −1 t=0 4≤t≤d−2 s = d2 + 3k + 2W ∗ t=d−1 s = d+1 2 x ≤ k ≤ z, 0 < k 3k + x ≡ 2 (mod 4) x=2 W = d − d∗ 2k ≡ d∗ (mod 4) ∗ d ≥ 5, d odd t ≡ 0 (mod 4) d 6= 5 d ≤ t ≤ d + z − 2, x = 2 s = d+3 2 t=d+z−k x ≤ k ≤ z, 0 < k d ≡ 0 (mod 3) t=d+z 3k + x ≡ 0 (mod 4) t=d+z s=0 d ≤ t ≤ d + z − 1, x 6= 2 t covered s, t values s = k + k+x 2 x=0 Parameters Table 8.4: Exceptional r = 0 and r = 2 cases for c = 4. ∗ ∗ , k, k + W, d2 − 1) Q4 ( d −2k 2 d−1 d−3 Q5 ( d−1 2 , 2 , 1, 0, 2 , 1) k+x k+x Q1 (z − k, k−x 2 , 2 , 2 − 1) Q2 (z) U2 (d − 1, 1) Q4 ( d −2k , k, k + W, d2 ) 2 ∗ d−3 d−3 Q7 ( d−1 2 , 2 , 2 ) k+x k+x Q1 (z − k, k−x 2 , 2 , 2 ) Q2 (z) Tableau 116 Q6 (1, d − 2, d − 3) Q5 (2, 1, d − 3, 0, 1, d − 4) d ≡ 1 (mod 3), d ≥ 19 s=8 t=d+z−5 x=4 t = d + z − 2k − 1 x ∈ {0, 2, 4} d 6= 4, 5, 8 t=1 t=1 d≥8 t odd s = 2d − 3 t = 2k + 1 d≥8 5≤t≤d−3 d≥5 s = 2d − 3k − 3 2 ≤ k ≤ d−3 2 t odd d≥5 t=3 t=d−2 d≥5 t=d−2 t=3 s = d+3 2 d odd d≥6 t odd t=d−1 t odd, d 6= 4, 5, 8 d ≤ t ≤ d + z − 7, x = 4 d ≤ t ≤ d + z − 5, x = 2 d ≤ t ≤ d + z − 3, x = 0, d ≡ 1 (mod 3), d ≥ 13 s = 2d − 6 t=d−1 d≥6 d≥5 s = d2 d even x ≤ 2k ≤ z − 1 s = 3k + 2x k + 2x ≡ 1 (mod 2) d ≥ 19 t = d + d−1 3 −6 t=d+z−4 d ≥ 13 t=d+z−4 s=5 x=1 t covered s, t values Parameters Table 8.5: Exceptional r = 3 cases for c = 4. Q5 (k, k + 1, d − 2k − 2, 1, k − 1, d − 2k − 1) d−3 d−3 Q5 ( d−1 2 , 2 , 2, 0, 2 , 1) Q5 ( d2 , d2 , 1, 0, d−4 2 , 1) 2k+x 2k+x Q1 (z − 2k − 1, 2k−x 2 , 2 , 2 ) Q1 (z − 5, 1, 4, 3) Q1 (z − 4, 1, 2, 2) Tableau 117 r=4 U2 (d − 2, 1) t = d + d−1 3 −1 s = k + k+x 2 −2 d 6= 4 3k + x ≡ 0 (mod 4) t=0 s = 2d − 2 t = d∗ − 2k W = d − d∗ d≥6 s = d2 + 3k + 2W − 2 2k ≡ d∗ (mod 4) ∗ t=d−1 d≥7 0 < 2k ≤ d∗ s = d−1 2 d ≡ 1 (mod 4) d≥6 t=d+z−k s=0 x=4 x ≤ k ≤ z, k ≥ 2 s, t values Parameters Table 8.6: Exceptional r = 4 cases for c = 4. ∗ ∗ , k, k + W, d2 − 2) Q4 ( d −2k 2 d−1 d−5 Q5 ( d−1 2 , 2 , 0, 1, 2 , 1) k+x k+x Q1 (z − k, k−x 2 , 2 , 2 − 2) Q2 (z) Tableau t=0 d≥6 t ≡ 0 (mod 4) 4≤t≤d−2 d≥7 t ≡ 0 (mod 4) t=d−1 d≥6 t ≡ 0 (mod 4) d ≤ t ≤ d + z − 4 − x, x 6= 2 d ≤ t ≤ d + z − 2, x = 2 d 6= 4 d ≡ 1 (mod 3) t = d + d−1 3 −1 t covered 118 119 reduce when r ≥ d + 5.) Since we have already listed those table with exceptional r cases, we will take r ≥ 5. Hence for a given s, we need only verify that all the appropriate t in the range d 3d−2s−1 e ≤ t ≤ b 4d−2s−5 c (with t even for s 6= 3) are 3 3 obtained. For the exceptional s cases, consider Table 8.7. When s = 0, all t are covered except t = d + d−1 when x = 4. However, this case is unnecessary as r = 1. If t ≤ d, 3 the shape is reducible to a c = 3 case. For s = 2, all t ≥ d − 1 except t = d + d−x − 1 (with x ∈ {0, 2, 4}) are given in 3 Table 8.7. When that occurs, either the shape is invalid or r < 5. If t ≤ d − 2, the shape is reducible to a c = 3 case. When d ≤ 7 and t ≥ d the shapes are covered by the exceptional r cases or are reducible to c = 3 cases. For s = 3 we want t ≤ d + d−x − 2. However, when t = d + d−x − 2 (x ∈ {0, 2, 4}), 3 3 then r < 5. Hence these cases have already been covered. When t < d − 2 the shape is reducible to a c = 3 case. For d ≤ 6, the cases are covered by c = 3 or exceptional r cases. The bounds on d for t ≥ d are needed to produce a valid shape. − 3 with x ∈ {1, 3, 5}. For s = 4, Table 8.7 provides all tableaux with t ≤ d + d−x 3 Since any t larger than this has r < 5, this covers all shapes not already listed. Note that for d ≤ 12, all necessary shapes have t < d; those shapes with t = d − 1 are not needed for d ≤ 6. Hence all the exceptional s cases are accounted for in Table 8.7. The general cases of r, s ≥ 5 are classified in Table 8.8. When r ≥ d + 5 we can reduce to a c = 3 tableau. Fix t = d + z − k. Since r and s are greater than 5, we need all t with 5 ≤ k ≤ z. For a t of this form, we need all shapes with c. This range is covered in Table 8.8. To see why all such s are 5 ≤ s ≤ k + b x+k−5 2 obtained, note that for any fixed c, we always get 2c ≤ s ≤ 3c−1. Since c ≥ 2, there are no gaps as we increment c. The parameters between the cases are comparable, so writing s = 2c + d + 1 and using the case corresponding to the parity of s + k will yield the appropriate tableau. Since x ≤ 2, we find s ≤ k + b x+k−5 c implies 2 s ≤ k + b k−1 c − 1 for x = 0 or 2, and s ≤ k + b k2 c − 2 for x = 1. Comparing the 2 bounds shows all s are obtained. This takes care of all shapes with t ≥ d. When 0 < t < d, we require all shapes with 5 ≤ s ≤ 2d − t − b t+5 c. The tableaux 2 120 Tableau Parameters t value 0<a≤z s=0 s=2 Q2 (a) z + a even Q1 (a, 1, 1, 0) d<t≤d+z t=d+a t even, d 6= 4 0≤a≤z−2 d≤t≤d+z−2 x ∈ {0, 2, 4} t=d+a d odd Q1 (a, 2, 1, 0) Q1 (a, 1, 1, 1) x ∈ {0, 2, 4} t=d−1 Q5 ( d2 − 1, d2 , 1, 0, 1, 1) t=d+a 0≤a≤z−2 d≤t≤d+z−2 z + a odd t=d+a d odd d≥7 d even d≥6 t=d−1 t=d−1 t=d−2 z + a odd Q3 (a, 2) d odd d≥7 2≤a≤d−2 a even, d ≥ 4 x ∈ {0, 2, 4} t odd, d ≥ 8 t = d − 1, d ≥ 7, d odd t = d − 1, d ≥ 6, d even t=d−2 d≤t≤d+z−3 t=d+a x ∈ {1, 3, 5} Q5 ( d−1 , d−1 , 1, 0, 2, 1) 2 2 x ∈ {0, 2, 4} t even, d ≥ 11 0≤a≤z−3 Q1 (a, 1, 2, 1) t even z + a even Q6 (d − 2, 1, 0) s=4 t=d−1 d≤t≤d+z−3 x ∈ {0, 2, 4} Q5 ( d−1 , d−1 , 0, 1, 1, 1) 2 2 x ∈ {0, 2, 4} t even, d ≥ 8 0≤a≤z−3 s=3 x ∈ {0, 2, 4} x ∈ {0, 2, 4} z + a even Q3 (d − 1, 1) t covered x ∈ {1, 3, 5} t even, d ≥ 12 t=d−1 t=d−1 d odd d≥7 t=a Table 8.7: Exceptional s cases for c = 4. 2≤t≤d−2 t even, d ≥ 4 t=0 U8 U5 (2, k − 2) d=4 5≤k≤d 0≤k ≤d−2 d≥4 b>c≥0 b≤d−t−1 1 ≤ f ≤ d − 2k − 2 1≤e≤k 1 ≤ k ≤ b d−2 c 2 d≥5 1 ≤ f ≤ d − 2k − 2 s=5 s = 2k − 3 s = 2(d − k) s=b+c+2 s = d − 2k + e + f s = 5, d = 4 d ≥ 5, s odd 7 ≤ s ≤ 2d − 3 s even 4 ≤ s ≤ 2d d≥4 5 ≤ s ≤ 2d − 2t − 1 t even d − 2k + 2 ≤ s ≤ 2d − 3k − 2 d≥5 t odd d − 2k + 1 ≤ s ≤ 2d − 3k − 3 s = d − 2k − 1 + e + f 1 ≤ k ≤ b d−3 c 2 1≤e≤k d≥5 d≥5 4 ≤ s ≤ d d2 e + 1 s=e+3 d even 1 ≤ e ≤ d d−2 e−1 2 s=c+3 d even d odd 3 ≤ s ≤ d2 + 1 d odd 1 ≤ c ≤ d2 − 2 d≥7 4 ≤ s ≤ d−1 2 s + k even s ≤ 3b k2 c if x = 0, 2 4 ≤ s ≤ 3b k2 c − 1 s + k odd 5 ≤ s ≤ 3b k−1 c 2 s covered s=e+2 s = 2c + d + 1 s = 2c + d + 1 s value d≥7 1 ≤ e < d−1 −2 2 k 6≡ d (mod 2) d ≤ c + 1 if x = 0, 2 0≤d≤c−2 5≤k≤z 2 ≤ c ≤ b k2 c k ≡ d (mod 2) 0≤d<c 5≤k≤z 2 ≤ c ≤ b k−1 c 2 Parameters Table 8.8: General c = 4 cases. Q6 (t, b, c) 1≤t≤d−3 t = 2k U2 (d − k − 1, 1) Q5 (k, k, d − 2k − 1, 1, e, f) 2≤t≤d−3 t = 2k + 1 Q5 (k, k + 1, d − 2k − 2, 1, e, f) Q5 (b d−2 c, d d−2 e, 1, 1, e, 1) 2 2 Q7 ( d2 − 1, d2 − 1, c) Q5 ( d−1 , d−1 , 0, 1, e, 1) 2 2 Q1 (z − k, c, c, d + 1) Q1 (z − k, c + 1, c, d) 3≤t≤d−3 t=d−2 t=d−1 t=d+z−k d≤t≤d+z−5 Tableau 121 122 listed in Table 8.8 satisfy this. Note that for t ≤ d − 3 the bounds on s for Q5 and Q6 overlap, thus guaranteeing all s are covered. For the small d not listed, the shapes are either not possible, not needed, or result in exceptional cases done previously. When t = 0 we require all d + d−5 ≤ s ≤ 2d − 3. As Table 8.8 shows, this is satisfied. Hence 2 all necessary cases for c = 4 have been covered. 8.3 Sufficiency for c = 5 When c = 5, the tableaux we will use are the Ri listed in Section 7.5. We will show that every partition of n = 5d described in Theorem 10 has a corresponding Ri . Note that shapes with r ≥ d + 5 can be obtained by Qi ∨ V (d) for the appropriate Qi filled with c = 4 elements. As such, we will not include these shapes in the following compilation. Throughout, we will use the convention d = 3z + x, where d ≡ x (mod 3). Unless otherwise specified, take x ∈ {0, 1, 2}. First consider the exceptional r cases. These are tableaux having s + t even with r ∈ {0, 2, 4} or r = 3 with no constraints on s and t. Since our shape is a partition of 5d we have the condition 3t + 2s + r = 5d. Hence for a given r, we need only verify c, (with s + t even if r 6= 3) are that all the appropriate t in the range 0 ≤ t ≤ b 5d−r 3 obtained. For the exceptional r cases with s + t even, we only need those t with t ≡ d + r (mod 4). Hence given a shape (r, s, t) the next shape needed is (r, s − 6, t + 4). , x = 1, 2 Consider Table 8.9. For r = 0, this table provides all the t, except t = 5d−x 3 and t = d + 2z − 1. These are not a possible shapes. When d < 6 we only need those shapes with t ≤ d. For r = 2, Table 8.9 yields all t ≤ d + 2z − 2 with x ∈ {1, 2, 3}. Since t ≡ d + 2 (mod 4). When t = d+2z, for x = 0, 2 we do not get a shape required by Theorem 10; for x = 1 the tableau is listed in the table. Also, t = d + 2z − 1 is not a shape and t = d + 2z − 2 with d odd is not needed (the d even case is listed). If d ≤ 6, then all necessary shapes have t ≤ d − 2. Hence we have accounted for all necessary tableaux with r = 2. 123 Tableau r=0 R1 (z, z) Conditions x=0 d even t value t = 5d 3 z + a even t = 5d 3 d ≡ 0 (mod 3) d even 1≤a≤z R4 (a, a, x) t covered d ≤ t ≤ d + 2z − 2 t = d + 2z − 2a t ≡ d (mod 4) d≥6 d≥6 d∗ r=2 R5 (a, a) 0≤a≤ 2 −1 R12 ( d−1 2 ) d ≡ 1 (mod 4) t=1 U1 (d) ∨ U3 ( d2 , d2 ) d ≡ 0 (mod 4) t=0 R1 (z, z) a even x=1 d even t = d − 2a t = 5d−2 3 2≤t≤d t ≡ d (mod 4) t=1 d ≡ 1 (mod 4) t=0 d ≡ 0 (mod 4) t = 5d−2 3 d ≡ 1 (mod 3) d even t = d + 2d 3 −2 R2 (z − 1, z − 1, 1) x=0 d even t = d + 2d 3 −2 t ≡ d + 2 (mod 4) d even d ≡ 0 (mod 3) x ∈ {1, 2, 3} R4 (a, a, x − 1) 1≤a≤z d≥7 d ≤ t ≤ d + 2z − 2 t = d + 2z − 2a t ≡ d + 2 (mod 4) d≥7 z+a odd ∗ 3≤t≤d−2 R5 (a, a − 1) 1 ≤ a ≤ d2 − 1 t = d − 2a d−1 P4 (1, d−1 2 , 2 ) U1 (d) ∨ U3 ( d2 − 1, d2 ) d ≡ 3 (mod 4) t=1 t = 1, d ≡ 3 (mod 4) d ≡ 2 (mod 4) t=0 t = 0, d ≡ 2 (mod 4) t = 2, d even Table 8.9: Exceptional r = 0 and r = 2 cases for c = 5. For r = 3 consider Table 8.10. In this case, we need t 6≡ d (mod 2). All cases with t ≤ d + 2z − 3 are covered in the table. In addition we need t = d + 2z − 1 for d ≡ 0, 2 (mod 3) with d even, which are listed as well. When d < 9 we have t ≤ d + 1 or t = d + 2z − 1, and so those cases are covered. Finally, when d = 4, t = d + 1 has s = 1, so that shape is no needed. When t = 0 and d = 3 then s is even, so this case 124 is reducible to c = 4 case with r = 0. Hence all necessary tableaux with r = 3 are provided by Table 8.10. For r = 4 we need those tableaux with t ≡ d (mod 4), t ≤ d + 2z − 4. In addition, we need t = d+2z when d ≡ 2 (mod 3) and t = d+2z−2 when d ≡ 1 (mod 3). These are all found in Table 8.10. For d = 5 the needed tableaux are listed individually. For d = 4, t = d + 2z − 2 and t = 0 are the only shapes required, while d = 3 does not need any shapes. Thus all the exceptional r cases are contained in Table 8.9 and Table 8.10. Now consider the exceptional s cases. These are tableaux having t even with s ∈ {0, 2, 4}, or s = 3 with no constraints on t. When r ≥ d, r 6= d + 1, we may reduced to Qi ∨ V (d) for some Qi filled with c = 4 elements. (If s = 3 we may only reduce when r ≥ d + 5.) Since we have already listed the exceptional r cases, we will also take r ≥ 5. Hence for a given s, we need only verify that all the appropriate t in the range d 4d−2s−1 e ≤ t ≤ b 5d−2s−5 c (with t even if s 6= 3) are obtained. 3 3 When s = 0, we need all even t ≤ d + 2z. Table 8.11 provides all tableaux with t ≥ d + 2, t even. For t ≤ d + 1 we can use the reduction to a c = 4 case. When s = 2, we need all tableaux with d + 1 ≤ t ≤ d + 2z − 2 and t even, which Table 8.11 lists. Those with t ≤ d are reducible to a c = 4 case, while the d ≤ 5 cases correspond to exceptional r cases or are similarly reducible. When s = 3 we need those shapes with t ≤ d + 2z − 3 and d + 5 ≥ r ≥ 5. For other r or r = d + 3 we can reduce to a c = 4 case. Table 8.11 provides tableaux for all t ≥ d. When t ≤ d − 1, we can reduce to a c = 4 case when d ≥ 8 or d = 6. The remaining tableaux with r ≥ 5 and d ≤ 7 which are not reducible to a c = 4 case are listed as well. For s = 4 we need all even t ≤ d + 2z − 5, along with t = d + 2z − 4 when d 6≡ 0 (mod 3). Table 8.11 provides all those tableaux with t ≥ d + 1 and d ≥ 9. When t ≤ d we may reduce directly to a c = 4 case when d ≥ 10. Those remaining tableaux with d ≤ 9 have r ≥ d + 2, r = d or r < 5 and hence are either reducible or listed previously. Thus Table 8.11 suffices for the exceptional s cases. The general tableaux with r, s ≥ 5 are classified in Table 8.12 and Table 8.13. 125 Since r, s ≥ 5, we only need those tableaux with 0 ≤ t ≤ d + 2z − 5. For a fixed t, we need all tableaux with 4d−3t−4 ≤ s ≤ 5d−3t−5 because for r ≥ d + 5 we can 2 2 reduce to a c = 4 tableau. First consider t = d + z + a of Table 8.12. We need all s ≤ z − a + x − 2 + b z−a−1 c. Letting the parameters of R9 vary in order over their 2 c, we get all bounds yields all s up to b + z − a + 2x − 3. Since b ranges to b z−a−x+1 2 s ≤ z − a + x − 3 + b z−a+x−1 c. This covers all the necessary s for x > 1. When x = 1 2 c as needed. This tableau requires d ≥ 16, for d < 15 we also get s = z − a − 1 + b z−a 2 we find there are no shapes with t ≥ d + z, 5 ≤ r ≤ d + 4 and s ≥ 5. When d = 15, R6 provides t = d + z, the largest t required. When d ≤ t < d + z, Table 8.12 provides tableau R10 . Fix t = d + z − a, then we need all s ≤ a + z + x + b a+z−5 c. To see why this tableau suffices, begin by 2 taking c = 0. First let b and f vary over their ranges. Taking them maximal, vary d up to its maximum and then take e up to b 2z c − 1. This satisfies all the required inequalities. Then take R10 (a, b a2 c, c, d 2z e + b a2 c, b 2z c − 1, d a2 e) with 0 ≤ c ≤ b a2 c − 1, c even. Similarly, use R10 (a, b a2 c, c, d 2z e + b a2 c − 1, b 2z c − 1, d a2 e) with 0 ≤ c ≤ b a2 c, c even. This gives all 5 ≤ s ≤ a + z + b a2 c + x + b 2z c − 2. Since b a+z−5 c ≤ b a2 c + b 2z c − 2, 2 all the necessary s are obtained. This tableau required d ≥ 6. When d < 6, we find there are no shapes with t ≥ d and r, s within the needed bounds. Table 8.12 also contains t = d − 1, which requires all s ≤ d − 1, as listed. When t ≤ d − 2, consider Table 8.13. For t ≤ d − 2, we need all s ≤ 2d − t + c − 2. Tableau R11 provides this by taking the parameters through their ranges b d−t−1 2 in order, using b = 1 and b = 2 (with e ≥ f). (We need b = 2 in order to obtain s = 5, otherwise b = 1 suffices.) The only snag is when t = 2. Then we cannot have c = 1, hence s = 5 is not obtainable in this case. However, t = 2, s = 5 is needed only when d = 5 and hence is listed separately. Also, d = 3 needs only t = 0, accounting for d ≤ 4. For t = 1, we need all 2d − 2 ≤ s ≤ 2d + d2 − 6. (The s = 2d − 3 case is a r = d + 3 reduction to c = 4.) All these shapes are obtained in Table 8.13. For small d, those tableaux not listed have s < 5. For t = 0 we need all 2d − 2 ≤ s ≤ 2d + b d−1 c − 2. Table 8.13 provides all the 2 126 necessary tableaux. When d is even, we get all 2d ≤ s ≤ 2d + d2 , with s even and 2d − 1 ≤ s ≤ 2d + d2 − 3 with s odd. While s = 2d − 2 is not listed, this shape has r = d + 4 and s + t even, so we may reduce to a c = 4 case. When d is odd, we get all 2d − 2 ≤ s ≤ 2d + d−1 − 2 with s even and 2d − 1 ≤ s ≤ 2d + d−1 − 2 with s 2 2 odd. Hence all the required s are listed. Therefore Tables 8.12 and 8.13 provide the required tableaux for the non-exceptional cases with c = 5. 8.4 Sufficiency for c = 6 When c = 6, the tableaux we will use are the Si listed in Section 7.6. We will show that every partition of n = 6d described in Theorem 10 has a corresponding Si . Note that shapes with r ≥ d + 5 can be obtained by Ri ∨ V (d) for the appropriate Ri filled with c = 5 elements. As such, we will not include these shapes in the following compilation. Throughout, we will use the convention d = 3z + x, where d ≡ x (mod 3). Unless otherwise specified, take x ∈ {0, 1, 2}. First consider the exceptional r cases. These are tableaux having s + t even with r ∈ {0, 2, 4} or r = 3 with no constraints on s and t. Since our shape is a partition of 6d, we have the condition 3t + 2s + r = 6d. Hence for a given r, we need only verify c, (with s + t even for r 6= 3) are that all the appropriate t in the range 0 ≤ t ≤ b 6d−r 3 obtained. For the exceptional r cases with s + t even, we only need those t with t + r ≡ 2d (mod 4). Hence given a shape (r, s, t) the next shape needed is (r, s − 6, t + 4). Moreover, for r even, both s and t must be even. Consider Table 8.14. For r = 0, this table provides all the all the required partitions. When t = 4 we must have d even, while we need d odd for t = 2. For d = 3 and 4, only t = 2d, 2d − 4, and t = 0 are required. Hence all the r = 0 cases are provided. When r = 2, Table 8.14 gives all the necessary tableaux since only those with t ≤ 2d − 2 and t ≡ 2d + 2 (mod 4) are required. For r = 3 consider Table 8.15. In this case we need all odd t ≤ 2d − 3 because t = 2d − 1 has s = 0 and thus is not required by Theorem 10. All such tableaux are 127 listed except d = 3 with t ≥ 3. When t = 3 we reduce to a c = 5 case with s = 3 and r = 0. When t = 5 and d = 3 we have s = 0 and hence the shape is not needed. For r = 4 Table 8.15 provides all necessary tableaux. In this cases we have s and t even with t ≤ 2d − 4, t ≡ 2d (mod 4). When d = 3 and t = 6 we have s = 1, so that shape is not required. Hence all the exceptional r cases are listed. Now consider the exceptional s cases. These are tableaux having t even with s ∈ {0, 2, 4}, or s = 3 with no constraints on t. When r ≥ d, r 6= d + 1, we may reduced to Ri ∨ V (d) for some Ri filled with c = 5 elements. (If s = 3 we may only reduce when r ≥ d + 5.) Since we have already listed the exceptional r cases, we will also take r ≥ 5. Hence for a given s, we need only verify that all the appropriate t in the range d 5d−2s−1 e ≤ t ≤ b 6d−2s−5 c (with t even for s 6= 3) are obtained. 3 3 Consider Table 8.16. When s = 0 this provides all tableaux since those tableaux with t < d∗ have r ≥ d + 2. For s = 2 the table provides all tableaux with t ≥ d + 2. When t < d + 2 we have r ≥ d + 2 provided d ≥ 5. When d = 3 or d = 4 either r < 5 or we can reduce to a c = 5 case as well. Similarly when s = 3, those tableaux with t < d + 2 are reducible to a c = 5 case provided d ≥ 7. When d = 6, t = 7 we may reduce to a c = 5 case with r = 3; for smaller t we have r ≥ d + 5. When d = 5 and t = 5, we may reduce to a c = 5 case with r = 4 as s + t even; for smaller t, r ≥ d + 5. When d = 4 and t = 5 we have r < 5. When d = 3 and t = 2 we may reduce to a c = 5 case with r = 3; for smaller t we have r ≥ d + 5. When s = 4 we need all even t ≤ 2d−3. Table 8.16 lists all tableaux with t ≥ d+2 and d ≥ 5. When t ≤ d + 1, t even, we may reduce to a c = 5 case provided d ≥ 6. When d = 5, t = 6 we have r = 4 and we may reduce using P1 (4) to a c = 3 case. For d = 3 or 4 there are no needed shapes with 5 ≤ r ≤ d + 1. Hence Table 8.16 lists all the exceptional s cases. The general cases of r, s ≥ 5 are classified in Table 8.17 and Table 8.19. Since r, s ≥ 5, we only need those tableaux with 0 ≤ t ≤ 2d − 5. For a fixed t, we need ≤ s ≤ 6d−3t−5 since if r ≥ d + 5 we can reduce to a c = 5 all tableaux with 5d−3t−4 2 2 tableau. When t is odd, write t = 2e − 1 then we need all s ≤ 3d − 3e − 1. Consider 128 S6 of Table 8.17. First we let b, a and then c vary over their parameters This yields all s ≤ 3d − 3e with t ≥ 5. For t = 3, we use S7 . First let d vary to its bound of d − 4. Then take S7 (d − 4, e) and S7 (d − 5, e) with e varying over the even numbers. This yields all the required s. For t = 1 we use S12 . First vary b over its bounds to d − 2. Then use S12 (d − 3, d − 2, c) and S12 (d − 3, d − 3, c) with c even to obtain the needed s. Note the bounds on d in these tableaux are necessary for coherence. Those tableaux with d ≤ 5 and shapes with r, s ≥ 5 which are not reducible to a c = 5 case are listed in Table 8.18. Hence all odd t are covered. When t is even consider Table 8.19. Write t = 2e, so we want all 2d + d2 − 3e − 2 ≤ s ≤ 3d − 3e − 3. Consider S5 . First let b and c vary. Then take S5 (a, d − e, d − e) and S5 (a, d − e, d − e − 1) as a varies over all even numbers up to d − e − 1 and d − e − 2 respectively. This yields all 8 ≤ s ≤ 3d − 3e − 3 as needed when t ≥ 6. For 5 ≤ s ≤ 7 the tableaux are listed individually. For t = 2 or 4, we use S7 which covers all necessary s (as shown above for t = 3). When t = 0 we need 2d+ d2 −2 ≤ s ≤ 3d−3 which is provided in the table. Note the bounds on d in these tableaux are necessary for coherence. Those tableau with d ≤ 5 and shapes with r, s ≥ 5 which are not reducible to a c = 5 case are listed in Table 8.18. Hence all even t are covered by Table 8.19. Therefor all the required tableaux with c = 6 have been listed. 129 Tableau r=3 R1 (z − 1, z) R9 (z − 1, 1, 0, 0, 1) R6 (a, a − 1, x) R17 (z − 2, x) R18 Conditions x=0 d ≥ 6, d odd x=2 1≤a≤z−2 d≥9 t = d + 2z − 1 d ≡ 2 (mod 3) d + 1 ≤ t ≤ d + 2z − 3 d≥9 d = 3 or 5 t=d+1 t = d + 1, d = 3, 5 t=d−1 t=d−1 1 ≤ a ≤ d2 − 2 d even, d ≥ 6 2≤t≤d−3 t = d − 2a − 1 t 6≡ d (mod 2) d odd, d ≥ 5 3≤t≤d−3 t = d − 2a − 1 t 6≡ d (mod 2) d even, d ≥ 6 d even, d 6= 4 t=1 t = 1, d even, d 6= 4 d=4 t=1 t = 1, d = 4 U6 d > 5, d odd t=0 t = 0, d > 5, d odd U7 d=5 t=0 t = 0, d = 5 P4 (1, 1, 1) ∨ U1 (4) r=4 t = d + 2z − 2a − 1 d ≥ 6, d ≡ 3 (mod 4) t = d + 1, d ≥ 6 d odd, d ≥ 5 R12 ( d2 − 2) t = d + 2z − 1 t = 2d + 2z − 1 t=d+1 1 ≤ a ≤ d−1 2 −1 R8 (a) t = d + 2z − 1 t covered d≥6 R19 R7 (a) t value R1 (z, z) R2 (z − 1, z − 1, 1) R4 (a, a − 2, x) x=2 z even x=1 2≤a≤z d≥6 t = d + 2d−4 3 t = d + 2d−4 3 t even d ≡ 2 (mod 3) t = d + 2d−2 3 −2 t = d + 2z − 2a t = d + 2d−2 3 −2 d ≡ 1 (mod 3), d ≤ t ≤ d + 2z − 4 d≥6 ∗ 2 ≤ a ≤ d2 − 1 t = d − 2a 2 ≤ t ≤ d − 4, d ≥ 6 R13 (3, 2, 1, 1, 1) d=5 t=5 t = 5, d = 5 R12 ( d−5 2 ) d≥7 R5 (a, a − 2) d odd t=1 t=1 d≥7 d odd P4 (1, 2, 1) ∨ U1 (4) d=5 t=1 U1 (d) ∨ U3 ( d2 − 2, d2 ) d ≡ 0 (mod 4) t=0 Table 8.10: Exceptional r = 3 and r = 4 cases for c = 5. t = 1, d = 5 t=0 d ≡ 0 (mod 4) s=4 t = d + 2z − 4 t=d+a+b d=4 x = 1, 2 0≤b<a≤z−2 Q6 (2, 1, 0) R2 (a, b, 2) t=3 t=4 Table 8.11: Exceptional s cases for c = 5. d≥9 d≥7 d=4 R13 (1, 2, 1, 1, 1) R2 (z − 2, z − 2, 2) t=2 d=5 Q5 (1, 3, 0, 1, 1, 1) t=6 d=7 d≥6 t=d t=d+a+b 0≤a<b≤z−1 Q5 (3, 3, 0, 1, 1, 1) R3 (a, b) s=3 t=d+a+b d even, if a = b d≥6 R2 (a, b, 1) s=2 t=d+b+a t value 0≤b≤a≤z−1 d + a + b even b≥a 0<a≤b≤z Conditions R15 R1 (a, b) s=0 Tableau d≥9 d + 1 ≤ t ≤ d + 2z − 5 d≥7 d 6≡ 0 (mod 3) t = d + 2z − 4 t = 2, d = 4 t = 3, d = 4 t = 4, d = 5 t = 6, d = 7 t = d, d ≥ 6 d≥6 d + 1 ≤ t ≤ d + 2z − 3 d≥6 t even d + 1 ≤ t ≤ d + 2z − 3 t even d + 2 ≤ t ≤ d + 2z t covered 130 c ≥ 1 if t = 2 d≥4 a+c>e+f b+d>f f≤e 1 ≤ f ≤ b d−a−b−1 c 2 e 1 ≤ e ≤ d d−a−b−1 2 0≤d≤d−e−b−1 0≤c≤d−a−2 b = 1, or 2 1≤a≤d−b d≥5 1≤a≤d−4 d≥6 c even 0 ≤ f ≤ d a2 e 0 ≤ e ≤ b 2z c 1 ≤ d ≤ d 2z e + b a2 c 0 ≤ c ≤ b 2z c 0 ≤ b ≤ b a2 c 1≤a≤z d = 15 x ≥ 1, d ≥ 16 0≤e≤x or d = b + x − 1, e ≤ 1 0≤d≤b+x−2 0≤c≤z−a−b−1 c 1 ≤ b ≤ b z−a−x 2 0≤a≤z−5 Parameters s=c+d+e+f+3 s=a+3 d+e<z+f c+e<z−a+b+f c+d<z+b c<d s=b+c+d+e+f+x t=d+z s=b+c+d+e s value Table 8.12: General c = 5 cases for t > 2. R11 (a, b, c, d, e, f) t=a+b 2≤t≤d−2 R16 (a) R10 (a, b, c, d, e, f) R6 (2, 0, 0) R9 (a, b, c, d, e) t=d−1 d≤t≤d+z−1 t=d+z−a t = d + z, d = 15 d + z ≤ t ≤ d + 2z − 5 t=d+z+a Tableau c s 6= 5 if t = 2, d = 5 d≥4 s ≤ 2d − t + b d−t 2 c+2 5≤s d≥5 4≤s≤d−1 d≥6 3z s ≤ x + b 3a 2 c+b 2 c−1 5≤s t = d + z, d = 15 d ≥ 16 s ≤ 3z−3a−1 if x = 1 2 5 ≤ s ≤ 3z−3a+3x−6 2 s covered 131 t=0 ∗ c even 0 ≤ c ≤ b d−1 2 c−1 a even s=d ∗ + 2b d−1 2 c+c+1 s = 2d + a ∗ s = 2d − 2 s = 2d − 1 Table 8.13: General c = 5 cases for t = 0, 1, and 2. U1 (d ) ∨ P3 (0, b d−1 2 c, c) U1 (d ) ∨ U3 (a, b d2 c) 0 ≤ a ≤ b a2 c d≥5 R14 ∗ d odd P4 (1, d−1 2 , 1) ∨ U1 (d − 1) s = 2d − 1 s = 2d + a − 1 1 ≤ a ≤ b d2 c − 1 d even R12 (a) t=1 s=5 s value d=5 Parameters P4 (1, d2 − 1, 1) ∨ U1 (d) P3 (2, 1, 0) ∨ U1 (2) t=2 Tableau s even s odd s ≤ d∗ + 3b d−1 2 c d + 2b d−1 2 c+1≤s ∗ s ≤ 2d∗ + b d2 c 2d∗ ≤ s d≥5 s = 2d − 2 odd s = 2d − 1 d even s = 2d − 1 s ≤ 2d + b d2 c − 1 2d ≤ s s = 5, d = 5 s covered 132 133 Tableau r=0 Conditions S1 (z) S5 (d − e, d − e, d − e) Q4 (2, 1, d − 5, 3) ∨ U1 (d) P1 (4) ∨ U3 (2, 2) S8 U1 (d) ∨ U1 (d) ∨ U1 (d) 3≤e≤d−2 d ≡ e (mod 2) t value t covered t = 2d t = 2d 6 ≤ t ≤ 2d − 4 t = 2e t ≡ 2d (mod 4) d≥5 d ≥ 6, d even t=4 d=4 t=4 d ≥ 5, d odd t=2 d even t=0 t=4 d ≥ 6, d even t = 4, d = 4 t=2 d ≥ 5, d odd t = 0, d even e 6≡ d (mod 2) r=2 S5 (d − e − 1, d − e, d − e) 3≤e≤d−1 6 ≤ t ≤ 2d − 2 t = 2e t ≡ 2 + 2d (mod 4) d≥4 d−1 d−5 Q4 (2, d−7 2 , 2 , 2 ) ∨ U1 (d − 1) d≥4 t=4 d ≥ 9, d odd t=4 P4 (3, 3, 3) ∨ P4 (1, 2, 1) d=7 t=4 t = 4, d = 7 P4 (3, 1, 1) ∨ P4 (1, 2, 1) d=5 t=4 t = 4, d = 5 Q2 (1) ∨ U1 (2) d=3 t=4 t = 4, d = 3 d−2 d−2 Q4 (1, d−2 2 , 2 , 2 ) ∨ U1 (d) d even t=2 t = 2, d even U2 (d − 1, 1) ∨ U1 (d − 1) d odd t=0 t = 0, d odd Table 8.14: Exceptional r = 0 and r = 2 cases for c = 6. d ≥ 9, d odd 134 Tableau r=3 Conditions 3≤e≤d−1 S6 (d − e, d − e, d − e) d≥4 t value 5 ≤ t ≤ 2d − 3 t = 2e − 1 d≥4 S9 (2, 1, d − 4, 1, d − 4, d − 2, 1) d≥5 t=3 t = 3, d ≥ 5 S10 (3) d=4 t=3 t = 3, d = 4 t=1 t=1 S11 r=4 t covered S4 (d − 6) d≥6 t = 2d − 4 t = 2d − 4 d≥6 3≤e≤d−4 S5 (d − e − 2, d − e, d − e) e ≡ d (mod 2) 6 ≤ t ≤ 2d − 8 t = 2e t ≡ 2d (mod 4) d≥7 Q2 (1) ∨ U1 (4) d≥7 d=5 t=6 d ≥ 6, d even t=4 d=4 t=4 d ≥ 5, d odd t=2 Q3 (2, 1, 1) ∨ U1 (2) d=3 t=2 U1 (d) ∨ U1 (d) ∨ U1 (d − 2) d even t=0 d−4 d−4 Q4 (2, d−4 2 , 2 , 2 ) ∨ U1 (d) P1 (4) ∨ U4 (2, 2) d−1 d−3 Q4 (1, d−3 2 , 2 , 2 ) ∨ U1 (d − 1) t = 6, d = 5 t=4 d ≥ 6, d even t = 4, d = 4 t=2 d ≥ 5, d odd t = 2, d = 3 t=0 d even Table 8.15: Exceptional r = 3 and r = 4 cases for c = 6. s=0 Tableau Conditions t value P1 (d∗ ) ∨ P1 (a) 0≤a≤d t = d∗ + a a even 0≤a≤d−4 s=2 S2 (a) a ≡ d (mod 2) s=4 0≤a≤d−5 S3 (a) d≥5 d∗ ≤ t ≤ 2d∗ t even d + 2 ≤ t ≤ 2d − 2 t=d+a+2 d≥4 s=3 t covered t even d≥4 t=d+a+2 d + 2 ≤ t ≤ 2d − 3 d≥5 P1 (4) ∨ P3 (2, 1, 0) d=5 t=6 t = 6, d = 5 P1 (4) ∨ U4 (2, 1) d=4 t=4 t = 4, d = 4 S4 (a) 0≤a≤d−5 d≥5 t=d+a+2 Table 8.16: Exceptional s cases for c = 6. d + 2 ≤ t ≤ 2d − 3 d≥5 135 Tableau Parameters s value s covered 1 ≤ a, b, c ≤ d − e t = 2e − 1 5 ≤ t ≤ 2d − 5 a, b ≥ c S6 (a, b, c) b≥a−1 t odd s=a+b+c 3≤e≤d−2 5 ≤ s ≤ 3d − 3e d≥5 d≥5 0≤e≤d t=3 S7 (1, 2, d − 5, d, e) e even 1≤d≤d−4 s=d−2+d+e d − 1 ≤ s ≤ 3d − 7 d≥5 d≥5 0≤c≤d t=1 S12 (d − 3, b, c) c even s=d+b+c 0≤b≤d−2 Table 8.17: General c = 6 cases for odd t. d Tableau Shape d=5 P3 (2, 1, 0) ∨ P3 (2, 1, 0) t = 4, s = 6 d=4 S10 (2) t = 3, s = 5 P4 (2, 1, 0) ∨ U4 (2, 1) t = 2, s = 5 U3 (2, 2) ∨ U4 (2, 1) t = 0, s = 9 Table 8.18: General c = 6 cases for d ≤ 5. d ≤ s ≤ 3d − 3 S7 (1, 1, d − 4, d, e) U1 (d∗ ) ∨ U1 (d∗ ) ∨ U1 (a) t=2 t=0 s=7 e≤d−4 s=5 e≤d−3 0≤a≤d s odd, d odd 2d − 3 ≤ s ≤ 3d − 4 s = 2d − 3 + a d odd, a even s odd 2d − 3 ≤ s ≤ 3d − 3 s even 2d∗ ≤ s ≤ 3d∗ d≥5 d ≤ s ≤ 3d − 6 d ≥ 6, d even s = 2d − 3 + a s = 2d∗ + a s=d−1+d+e d≥6 d − 2 ≤ s ≤ 3d − 9 e odd s = 5, d ≥ 5 e even s = 5, d ≥ 5 s = 6, d ≥ 6 e odd s = 7, d ≥ 5 e even s = 7, d ≥ 5 d≥6 8 ≤ s ≤ 3d − 3e − 3 s covered a even 0≤a≤d d ≥ 6, d even a even 0≤a≤d 1≤d≤d−4 e even, d ≥ 5 0≤e≤d 1≤d≤d−5 e even, d ≥ 6 0≤e≤d e odd s=d−3+d+e s=5 e≤d−3 e even s=6 3≤e≤d−3 e odd e even s=7 s=a+b+c s value 0≤e≤d−4 3≤e≤d−3 a even, d ≥ 6 b, c > a 1 ≤ a, b, c ≤ d − e Parameters Table 8.19: General c = 6 cases for even t. d+3 U5 ( d−1 2 , 2 ) ∨ U1 (a) U5 ( d2 , d2 ) ∨ U1 (a) S7 (2, 2, d − 6, d, e) P4 (e + 1, 1, 0) ∨ P4 (e − 1, 1, 0) P3 (e, 1, 0) ∨ P4 (e, 1, 0) S5 (2, 2, 2) P3 (e + 1, 1, 0) ∨ P4 (e − 1, 2, 0) P3 (e, 1, 0) ∨ P4 (e, 2, 0) S5 (a, b, c) t=4 t even 6 ≤ t ≤ 2d − 6 t = 2e Tableau 136 137 8.5 Sufficiency for c > 6, d even In Chapter 6 we algorithmically demonstrate how to reduce an arbitrary tableau to one of those tableaux filled with fewer elements. In the case where d is even, we reduced all tableaux to joins of tableaux with c ≤ 6 to or those tableaux with t < d, s < d + 5, r < d + 5. Previously we showed all tableaux with c ≤ 6 where constructed in Chapter 7. Now we will address the remaining cases with d even. Since any tableau must satisfy cd = 3t + 2s + r, applying the bounds on r, s, and t we find cd ≤ 3(d − 1) + 2(d + 4) + d + 4 = 6d + 9. Hence for d > 9 all such tableaux will have c ≤ 6. For d = 8 and d = 6 it may be possible to have c = 7, while for d = 4, both c = 7 and c = 8 are possible. When d=8 and c=7, only the shape with (r, s, t) = (11, 12, 7) satisfy the constraints. However, we can reduce this shape by V (d) to a c = 6 case with r = 3. When d = 6 and c = 7, only the shapes with (r, s, t) = (10, 10, 4), (9, 9, 5) and (7, 10, 5) satisfy the constraints. For (10, 10, 4) we may reduce by V (d) to a c = 6 case with (r, s, t) = (4, 10, 4) since s + t is even. For (9, 9, 5) we may reduce by V (d) to a c = 6 case with (r, s, t) = (4, 9, 5) since s + t is even. For (7, 10, 5) we may reduce by U1 (4) to a c = 5 case with (r, s, t) = (3, 6, 5). When d = 4 and c = 7, only the shapes with (r, s, t) = (7, 6, 3), (5, 7, 3), (3, 8, 3), (6, 8, 2), and (8, 7, 2) satisfy the constraints. For (7, 6, 3) we may reduce by V (d) to a c = 6 case of (3, 6, 3). For (3, 8, 3) use R19 ∨ U1 (4). For (5, 7, 3) we may reduce by U1 (4) to the c = 5 cases (5, 3, 3). For (6, 8, 2) we may reduce by V (d) to a c = 6 case of (2, 8, 2) since s + t is even. For (8, 7, 2) we may reduce by U1 (4) to the c = 5 case (8, 3, 2). When d = 4 and c = 8, only the shape with (r, s, t) = (7, 8, 3) satisfies the constraints. It can be obtained by S10 (3) ∨ U1 (2). Thus when d is even, all shapes are reducible to tableaux filled with less than or equal to six elements and all the cases with c ≤ 6 were obtained previously. 138 8.6 Sufficiency for c > 6, d odd When d is odd, the reduction techniques of Chapter 6 work to reduce a tableau to c ≤ 8. As we’ve already constructed those tableaux with c ≤ 6, we will focus on those with c = 7 or 8. First, assume t ≥ d − 1. So long as r 6∈ {0, 2, 3, 4, 5, 7} we may use P1 (d − 1) to reduce to a c − 3 case. If t ≥ d − 1 and r ∈ {0, 2, 3, 4, 5, 7} we may use P4 (d − 2, 1, 1) to reduce to a c − 3 case unless s ∈ {0, 2, 3, 4, 5, 7}. For s = 3, 5, 7 we may still reduce by P4 (d − 2, 1, 1) when t is odd. For r = 0, 2, 4, s = 3, 5, 7 and t even, the shape is not needed by Theorem 10. For r = 3, 5, 7 and s = 3, 5, 7 with t even, no shapes are possible for c = 8. When c = 7 we have d ≥ 5 so we may use P3 (d − 3, 1, 0) to reduce to a c = 4 case with s + t even. For s = 0, 2, 4, we only need those shapes with t even. For r = 3, 5, 7 we may reduce by P1 (d − 1) to a c − 3 case. This leaves those cases with s = 0, 2, 4, r = 0, 2, 4 and t even. When c = 7 there are no shapes having s = 0, 2, 4, r = 0, 2, 4 and t even as d is odd. For c = 8, these shapes are obtainable, depending on d (mod 3). We list the appropriate tableaux in Table 8.20. This completes all cases with t ≥ d − 1 If t < d − 1 and s < d + 4, but r = d + 3 or r ≥ d + 5 we may use V (d) to reduce to a c − 1 case. When c = 8, there are no valid shapes with r ≤ d + 4 and t < d − 1, s < d + 4. For c = 7 this is also true provided d ≥ 5. When d = 3 we need (r, s, t) = (6, 6, 1) which is reducible by V (d) to a c = 6 case with r = 3. If t < d − 1 and s ≥ d + 4 we may use U1 (d − 1) to reduce to a c − 2 case, provided r 6∈ {0, 2, 3, 4, 6}. Consider those cases with r ∈ {0, 2, 3, 4, 6} If s ≥ 2d + 5 or s = 2d + 3 we may use U2 (d, d) to reduce to a c − 4 case. For s ≤ 2d − 1 there are no shapes with t ≤ d − 2 and r ≤ 6. If s = 2d, 2d + 2 or 2d + 4, then we may still reduce via U2 (d, d), provided t is even (which always occurs if r = 0, 2, 4). Thus we need only consider those tableaux with s = 2d + 1, or s = 2d, 2d + 2, 2d + 4 with r = 3 or 6. For c = 7 and s ≥ 2d, s 6= 2d + 1 only r = 6, s = 2d, t = d − 2 is possible. This can be obtained by P4 (d − 2, 1, 1) ∨ U5 (2, d − 2) provided d ≥ 5. When d = 3 we may 139 (r, s) d (0, 0) d ≡ 0 (mod 3) Q2 (z) ∨ Q2 (z) (0, 2) d ≡ 2 (mod 3) W1 (2) (0, 4) d ≡ 1 (mod 3) Q1 (z − 1, 0, 1, 1) ∨ Q1 (z − 1, 0, 1, 1) (2, 0) d ≡ 1 (mod 3) W1 (0) (2, 2) d ≡ 0 (mod 3), d > 3 d=3 (2, 4) d ≡ 2 (mod 3), d > 5 Tableau Q2 (z) ∨ Q1 (z − 2, 1, 1, 0) S1 (1) ∨ U1 (2) Q2 (z) ∨ Q1 (z − 2, 0, 2, 2) d=5 S1 (1) ∨ U1 (4) (4, 0) d ≡ 2 (mod 3) Q2 (z) ∨ Q2 (z) (4, 2) d ≡ 1 (mod 3) Q2 (z; x = 4) ∨ Q1 (z − 1, 0, 1, 1; x = 1) (4, 4) d ≡ 0 (mod 3), d > 9 Q2 (z) ∨ Q1 (z − 4, 2, 2, 0) d=9 Q2 (z) ∨ Q5 (4, 4, 1, 0, 2, 1) d=3 Q1 (1) ∨ U1 (2) ∨ U1 (2) Table 8.20: Exceptional r and s cases for c = 8. reduced by V (d) to a c = 6 case with r = 3. For c = 8 and s ≥ 2d, s 6= 2d + 1 only s = 2d + 4, r = 3, d = 5 can occur. In that case, use Q5 (2, 1, 2, 0, 1, 1) ∨ U2 (5, 5). This leaves those tableaux with t < d − 1, s = 2d + 1 and r ∈ {0, 2, 3, 4, 6}. When r = 0, 2, 4, we must have t odd, so in the c = 8 case there are no possible shapes. For c = 7, we get a valid shape only for r = 4, in which case we have t = d − 2. For this use U1 (d − 1) to reduce to a c = 5 case with r = 2, s = d + 1 which will still have s + t even. For r = 3, there are no shapes satisfying t < d − 1 and s = 2d + 1 for either c = 7 or 8; similarly for r = 6. Thus all required shapes may be reduced to those filled with c ≤ 6 elements. 140 8.7 Tableaux Disjointness Our proof of Theorem 2 requires the tableaux we constructed to be disjoint. Since Lemma 3.4.9 showed that maximal tableaux are always disjoint, we need only be concerned with those tableaux which could not be put in maximal form, namely, U8 and S8 . Recall that U8 = 43333122444 11122 has d = 4 and ω2 = (3, 2, 0, 0). When d = 4 the tableaux used in the reduction techniques are P1 (4), U1 (4), and V (d). These are the only tableaux that would be joined with U8 , so it suffices to show U8 is disjoint from these tableaux. Note that the weights of these tableaux consist only of 4’s and 0’s. If U8 were not disjoint from these tableaux, then there is some weight assignment of U8 , not equivalent to ω2 = (3, 2, 0, 0) which uses a weight from at least one of these tableaux. However, since the only additional weights we may use are 4’s and 0’s, there is no way to have a weight assignment of length 5 using 0’s, 2’s, 3’s, and 4’s without being equivalent to (3, 2, 0, 0). Hence the weights are disjoint from U8 . d−4 Recall that S8 = d−2 d−4 2 66 6 3 2 1 23662 34 3 4 4 5 55151 d d 0 with ω2,3 = ( 22 00 d−3 0 0 0 0 ) and 11 d ≥ 5. When d is even, the tableaux used in the reduction techniques are P1 (d), U1 (d), and V (d). When d is odd, the tableaux used in the reduction techniques are S1 (d), U2 (d), whose weights are 0’s and d’s, P1 (d − 1), U1 (d − 1) whose weights are ¡ ¢ 1 0’s and d − 1’s, and P4 (d − 2, 1, 1) with ω2,3 = 00 d0 d−2 . Hence it suffices to show S8 disjoint from these tableaux. Now any weight assigned to S8 must have λ3 = 2. Since d ≥ 5, there are no weights ¡¢ other of the listed tableaux than the weight 22 of S8 for which this is possible. This means if S8 were not disjoint from these tableaux there would be a weight assignment d d 0 of S8 of the form ( 22 0∗ ∗0 ∗0 0∗ 0∗ ) which is not equivalent to ( 22 00 d−3 0 0 0 0 ). The only weights we may use for the ∗’s are the weights of the listed tableaux, namely, 0’s d’s, d − 1’s and d − 3. Since the ∗’s sum to 3d − 3, they would have to be three d − 1’s and two 141 0’s. That is, the weights of U1 (d − 1) or P1 (d − 1). Then the weights of S8 would need to be assigned to other tableaux, so a weight of d would need to be assigned to either U1 (d − 1) or P1 (d − 1), which is not possible as these tableaux are maximal. Hence there is no other weight assignment for S8 and so the tableau is disjoint as required. Thus all the tableaux constructed in the proof of Theorem 2 are disjoint as desired. 142 Chapter 9 Proof of Theorem 3 Theorem 2 proves the generalized Foulkes’ Conjecture for c = 3 without multiplicities. We can, however, strengthen this result to include multiplicities for those irreducibles corresponding to two row partitions. Namely, Theorem 3. Let n = 3b = cd, with c, d ≥ 3 and let λ = [λ1 , λ2 ] be a two row partition of n. Then every irreducible character χλ occurring in 1SSnb oS3 occurs in 1SSnd oSc with multiplicity at least as large. The proof of this theorem involves constructing the appropriate number of tableaux, primarily out of the basis elements given in Section 4.3. The tableaux constructed are maximal so linear independence is assured from Lemma 3.4.12 provided the weights are distinct. Once we have the basis tableaux for c = 4, 5, and 6, the procedure generalizes to provide the appropriate number of tableaux for any c. Take λ = [r + s, s]. The multiplicity mλ of Sλ in 1SSnb oS3 was determined by Thrall in [20], which we listed in Theorem 11. Since the multiplicity depends on the relative sizes of r and s we will handle these cases separately. Moreover, if s = 6k +j and r = 6h+i, we will often simply construct k + 1 or h + 1 tableaux when possible to avoid detailed case analysis. 9.1 Case: s ≤ r Let λ = [r + s, s] be a partition of n with s ≤ r, where n = 3b = cd. We wish to construct mλ linearly independent tableaux, where mλ is the multiplicity of χλ in 143 1SSnb oS3 as described in Theorem 11. First we will construct these tableaux for c = 4, 5, and 6; then we will use these constructions in proving Theorem 3 for a general c. We will refer to the tableaux constructed in this way as basis tableaux. These constructions will make use of the c = 3 basis tableaux constructed in Section 4.3. 9.1.1 Basis Tableaux for c = 4, s ≤ r Given λ = [r + s, s] a partition of n, we have 2s + r = 4d = 3b. From this equation and s ≤ r, we have s ≤ d + b d3 c. For each λ we will construct mλ linearly independent λ-tableaux filled with the numbers 1 to 4. These will be our c = 4 basis tableaux. When s ≤ r−d, we can use the basis tableaux constructed in Section 4.3. Consider the partition λ0 = [r0 + s, s] where r0 = r − d. Since s ≤ r0 , we have mλ = mλ0 . In Section 4.3 we constructed mλ0 linearly independent Bp , where Bp are the basis tableaux for c = 3. Take Bp ∨ d 4 as the basis tableaux for c = 4. This works for s ≤ r − d, so r ≥ s + d. Hence 4d = 2s + r ≥ 3s + d implies s ≤ d. When d < s ≤ d + b d3 c, write s = d + f , with 1 ≤ f ≤ b d3 c. Consider the tableau a+b≤d abcd T (a, b, c, d) = 4 3 4 3 1 1 2 2 a, b, c, d > 0 a>d b>c or a = d, b = c w(T ) = (a + b, c + d, 0, 0) If a > d and b > c, no other weight assignments are possible for T , hence this tableau is non-zero and maximal. If a = d and b = c, we may also have the tableau obtained by exchanging the rows. However, this has sign (−1)a+b+c+d = 1 and thus the tableau is still non-zero. Let Cp = T (d − 2p, p + 1, p, f − 1). Then λ2 (T ) = a + b + c + d = d − 2p + p + 144 1 + p + f − 1 = d + f = s. Hence these tableaux have the desired shape. Consider Cp for p = 1, 2, . . . , b d−f c. To insure the Cp are non-zero and maximal we need to check 2 that the constraints on T are satisfied. For f 6= 1 all the parameters are greater than zero. Obviously, b = p + 1 > c = p. For a > d, we need d − 2p > f − 1. This is true c this inequality holds. provided p < d−f2 +1 . Since p ≤ b d−f 2 The Cp are linearly independent by Lemma 3.4.12 if their max weights are distinct. We have w(Cp ) = (d − p + 1, f + p − 1, 0, 0). If w(Cp ) = w(Cp0 ) for p > p0 , then we must have d−p+1 = f +p0 −1, that is d−f +2 = p+p0 . But p+p0 ≤ d−f + d−f −1 = d−f −1. 2 2 Hence this cannot occur. Thus the Cp are linearly independent. Since s = d + f , we have mλ ≤ b d+f c + 1, so it suffices to construct b d+f c+1 6 6 tableaux. The Cp provide b d−f c tableaux. To show b d−f c ≥ b d+f c + 1, it suffices to 2 2 6 show d−f − 12 > d+f , or equivalently, d − 32 ≥ 2f . This holds for d ≥ 5 as f ≤ b d3 c. 2 6 When d = 3 or d = 4 then f = b d3 c = 1, which is handled below. Hence for s > d + 1 the Cp provided at least mλ linearly independent tableaux. For s = d + 1, that is f = 1, take Cp = T (d − p − 2, 2, 1, p), for 1 ≤ p < c. Then the conditions on T are satisfied and such p exist for d ≥ 6. We have b d−2 2 w(Cp ) = (d − p, p + 1, 0, 0) so the max weights are distinct. These Cp provide at c − 1 linearly independent tableaux and we need b d+1 c + 1 tableaux. Now least b d−2 2 6 b d−2 c − 1 ≥ b d+1 c + 1 provided d ≥ 8. When d = 7, then s = 8 and two tableaux 2 6 are needed. Use T (4, 2, 1, 1) and T (3, 2, 1, 2). When d = 6, then s = 7 and only one tableau is needed. In this case, use Cp described above for p = 1. When d = 5 then s = 6 so two tableaux are needed. Use T (2, 2, 1, 1) and T (2, 1, 1, 2). When d = 4 then s = 5 so one tableaux suffices. However, for d = 4, s = 5 there are no tableaux of maximal form. We will use the tableau U8 constructed in Section 7.2. This tableau is non-zero but not maximal. Here we use maximal form only to show linear independence. Since only one tableau is needed for s = 5, this tableau works. (The d = 4, s = 5 case is actually not needed for the c = 4 basis tableaux since n = 3b = 4d implies 3|d, but we construct the basis tableaux for all d ≥ 3 in order to simplify the construction process in Section 9.1.4. However we will not use U8 in that construction.) When d = 3, then s = 4 and only one tableau is need, hence 145 T (1, 1, 1, 1) suffices. 9.1.2 Basis Tableaux for c = 5, s ≤ r Given λ = [r + s, s] a partition of n, we have 2s + r = 5d = 3b. From this equation and s ≤ r, we have s ≤ b 5d c. For each λ we will construct mλ linearly independent 3 λ-tableaux filled with the numbers 1 to 5. These will be our c = 5 basis tableaux. When s ≤ r − d, we can use the c = 4 basis tableaux constructed in Section 9.1.1. Consider the partition λ0 = [r0 +s, s] where r0 = r−d. Since s ≤ r0 , we have mλ = mλ0 . In Section 9.1.1 we constructed mλ0 linearly independent Cp , where Cp are the basis tableaux for c = 4. Take Cp ∨ d 5 as the basis tableaux for c = 5. This works for s ≤ r − d, so r ≥ s + d. Hence 5d = 2s + r ≥ 3s + d implies s ≤ b 4d c. 3 For b 4d c < s ≤ b 5d c consider 3 3 1≤a≤d−2 a b d2 c d d2 e b T1 (a, b) = 4 4 5 4 3 45 222 1 1 33 1 ≤ b ≤ b d2 c − 3 a>b s≥d+6 d≥8 w(T1 ) = (d, a + 2, b + 1, 0, 0) T1 is defined for s ≥ d+6, d ≥ 8. (When d < 8 then s < d+6 since s ≤ b 5d c.) Since 3 a > b, there are no other weight assignments possible and the tableau T1 is maximal. Let Ep = T1 (s−d−3−p, p) for 1 ≤ p ≤ b s−d c−2. For Ep to be non-zero and maximal 2 we need the conditions on T1 to be satisfied. We have a > b since p ≤ b s−d c − 2, 2 while p ≥ 1 implies a = s − d − 3 − p < d − 2. Since d ≥ 8, then b ≤ b d2 c − 3. Note a > b implies these max weights are distinct. Hence this construction provides b s−d c − 2 distinct, linearly independent tableaux when s ≥ d + 6, d ≥ 8. 2 146 We will need an additional tableau, so take: a b T2 (a, b) = 5 3 5 4 1122 1 ≤ a, b ≤ d − 1 s≥4 w(T2 ) = (a + 1, b + 1, 0, 0) Clearly T2 is non-zero and maximal. Given s, take E0q = T2 (d − q, s − 2 − d + q) for c. To insure E0 q is non-zero and maximal we need the conditions q = 1, 2, . . . , d −b s−2 2 of T2 are satisfied. Since b 4d c < s and d ≥ 3, then s ≥ 4. We have 1 ≤ a ≤ d − 1 3 c. Similarly the bounds on s show 1 ≤ b ≤ d − 1. The E0 q because q < d since s ≤ b 5d 3 are linearly independent because a < b for q < d − b s−2 c. This provides d − b 2s c + 1 2 tableaux. We need mλ = b 6s c + 1 linearly independent tableaux. By Lemma 3.4.12 the tableaux Ep and E0 q are linearly independent since they have different max weights. First consider s ≥ d + 6 which implies d ≥ 8. Then we have both Ep and E0q , for a c − 2 + d − b 2s c + 1 tableaux. This is greater than or equal to b 6s c + 1, total of b s−d 2 c and d ≥ 8. since s ≤ b 5d 3 For s < d + 6 we only have E0q , which provides d − b 2s c + 1 tableaux. This is greater than or equal to b 6s c + 1 provided d ≥ 5, since s ≤ d + 5. When d = 4, then c = 6 so the two E0 q suffice. When d = 3 then s ≤ 5 and hence one tableau, s ≤ b 5d 3 E01 is sufficient. Hence we have constructed at least mλ tableaux as desired. 9.1.3 Basis Tableaux for c = 6, s ≤ r Given λ = [r + s, s] a partition of n, we have 2s + r = 6d = 3b. From this equation and s ≤ r, we have s ≤ 2d. For each λ we will construct mλ linearly independent λ-tableaux filled with the numbers 1 to 6. These will be our c = 6 basis tableaux. When s ≤ r − d, we can use the c = 5 basis tableaux constructed in Section 9.1.2. Consider the partition λ0 = [r0 +s, s] where r0 = r−d. Since s ≤ r0 , we have mλ = mλ0 . In Section 9.1.2 we constructed mλ0 linearly independent Ep , where Ep are the basis 147 d tableaux for c = 5. Take Ep ∨ 6 as the basis tableaux for c = 6. This works for s ≤ r − d, so r ≥ s + d. Hence 6d = 2s + r ≥ 3s + d implies s ≤ d + b 2d c. 3 For d + b 2d c < s ≤ 2d we want to construct mλ ≤ b 6s c + 1 linearly independent 3 tableaux. We will do this primarily by joining two c = 3 basis tableaux. In addition, we will use the following tableaux: d−2 f 1 G = 5 3 46 1 1 22 1≤f <d ω(G1 ) = (d − 1, f + 1, 0, 0, 0, 0) d−2 f −2 2 2 G = 4 5 4 66 1 1 2 23 4≤f <d d≥5 ω(G2 ) = (d − 1, f − 1, 2, 0, 0, 0) 2 d−4 d−3 3 2 G = 6 4 5 646 1 1 2 233 d>6 ω(G3 ) = (d − 2, d − 2, 3, 0, 0, 0) d−2 2 d−5 4 2 G = 6 4 6 5 56 1 1 2 2 33 d>7 ω(G4 ) = (d − 1, d − 3, 3, 0, 0, 0) These tableaux are all maximal. Except for G2 , no other weight assignments are possible, hence these tableaux are non-zero. For G2 , the weight assignment (d − 1, f − 1, 0, 0, 0, 2) is also valid. Since this has sign (−1)2 , G2 is also non-zero. 148 Consider b 5d c < s < 2d. (The s = 2d case will be handled separately.) Write 3 s = f + d, then b 2d c < f < d. This means f > 2 always, and f < 6 only for d ≥ 7. 3 Moreover f ≥ 4 for d ≥ 5 so the conditions on G2 are satisfied. We will use the c = 3 basis tableaux for much of our construction. Let Dp0 represent the basis elements for c = 3 of shape [2d, d] as described in Section 4.3. Let Fp represent the basis elements for c = 3 of shape [3d−f, f ] as described in Section 4.3. We will use the Dp0 , Fp , and the Gi to construct the mλ linearly independent tableaux for c = 6. This construction will depend on d and f mod 6. Write d = 6k 0 + i0 and f = 6k + i, with 0 ≤ i0 , i ≤ 5. For i0 6= 1 there exist m[2d,d] = k 0 + 1 linearly independent Dp0 , with p0 = 0, 1, . . . , k 0 , and k 0 tableaux for i0 = 1. Similarly, there exists m[3d−f,f ] = k + 1 linearly independent Fp with p = 0, 1, . . . , k for i 6= 1 and k when i = 1. Now s = d + f = 6(k 0 + k) + i0 + i = 6k + i, so mλ ≤ k + 1. Since k ≤ k 0 + k + 1 , it suffices to construct k 0 + k + 2 linearly independent tableaux. We will first consider d > 6, f ≥ 6. Since k 0 , k > 0 consider: G p = D k 0 ∨ Fp p = 0, 1, . . . , k Gp0 = Dp0 ∨ Fk p0 = 0, 1, . . . , k 0 − 1 G 0 = D 0 ∨ F0 There are k + 1 + k 0 + 1 = k + k 0 + 2 tableaux listed here. By Lemma 3.4.12 if their max weights are distinct, these tableaux are linearly independent. Since these tableaux are in maximal form we can simply compare their weights. We have: ω(Gp ) = (d, 0, 0, 2p + i + 4k, 2(k − p), 0) p = 0, 1, . . . , k ω(Gp0 ) = (2p0 + i0 + 4k 0 , 2(k 0 − p0 ), 0, f, 0, 0) p0 = 0, 1, . . . , k 0 − 1 ω(G0 ) = (4k 0 + i0 , 2k 0 , 0, 4k + i, 2k, 0) Now d > f and p0 ≤ k 0 − 1, so we have ω(Gp ) 6= ω(Gp0 ) since ω(Gp0 ) does not contain a weight of d. Both of these weights are distinct from ω(G0 ), since they each contain at least three 0’s while ω(G0 ) contains only two 0’s. The weights within each 149 of these collections of tableaux are distinct because each collection {Dp0 } and {Fp } are linearly independent by Section 4.3. When all these tableaux G exist we have a set of basis tableaux for c = 6. However, depending on the conditions on d and f , we may only have k 0 or k basis tableaux to work with. In those situtations we will need to use the appropriate Gi to complete our set of tableaux. Recall that we are taking d > 6 and that f ≥ 6. When d 6≡ 1 (mod 6) and f 6≡ 1 (mod 6), all the tableaux G exist. Hence we have the k 0 + k + 2 linearly independent tableaux required. When d 6≡ 1 (mod 6), f ≡ 1 (mod 6), there exist only k linearly independent Fp with p = 1, 2, . . . , k. Hence we have all the tableaux listed above except for those with p = 0 and G0 . In place of those basis elements, use G1 and G2 . This provided k + k 0 +2 tableaux. They are linearly independent by Lemma 3.4.12 provided the weights of G1 and G2 are distinct from the weights of Gp . We have ω(G1 ) = (d − 1, f + 1, 0, 0, 0, 0) and ω(G2 ) = (d − 1, f − 1, 2, 0, 0, 0). Clearly these weights are distinct from eachother. Now ω(G1 ) 6= ω(Gp0 ) since the number of 0’s differ. If ω(G1 ) = ω(Gp ) we must have f + 1 = d and p = k. Similarly ω(Gp0 ) will be distinct from ω(G2 ) unless d − 1 = f , p0 = k 0 − 1. We have ω(G2 ) distinct from ω(Gp ) since d does not occur in its weight. Hence for f 6= d − 1 these k + k 0 + 2 tableaux are linearly independent. When f = d − 1, use G3 and G4 in place of G1 and G2 . Since ω(G3 ) contains neither d nor f it is clearly distinct from the weights of Gp . Similarly, ω(G4 ) is distinct from ω(Gp ) since it does not contain d. While ω(G4 ) does contain f , Gp0 cannot contain the weight 3 since d > 7. Thus we have sufficient linearly independent tableaux. Now consider d ≡ 1 (mod 6) and f 6≡ 1 (mod 6). We have the Gp and Gp0 listed earlier, for p0 6= 0, along with G1 and G2 . The discussion in the f ≡ 1 (mod 6) case above shows these are linearly independent provided f 6= d − 1. Similarly when f = d − 1 we can replace G1 and G2 with G3 and G4 . If d = 7 then G4 does not exist. However, then s = 13 on only two tableaux, D1 ∨ F1 and G3 , are needed. When d ≡ 1 (mod 6) and f ≡ 1 (mod 6), we can write d = 6k 0 + 1, f = 6k + 1. Then s = 6(k 0 + k) + 2 so k 0 + k + 1 tableaux suffice. Use the Gp and Gp0 listed earlier, for p, p0 6= 0, along with G1 and G2 . This provides the requisite number of tableaux. 150 They are linearly independent by the previous discussion since f 6= d − 1, as d ≡ f (mod 6). c implies f > 2. Since s = d + f , we must Now consider 3 ≤ d ≤ 6. Then f > b 2d 3 have s ≤ 11. Hence two linearly independent tableaux will suffice, namely, D0 ∨ F0 and G1 . As in previous discussions, these weights are distinct provided f 6= d − 1. When f = d − 1, s = 2d − 1. We need two tableaux only when d = 5 or d = 6. Thus D0 ∨ F0 suffices for d = 3 and 4. When d = 5 or 6 use D0 ∨ F0 and T, where 24221 T= 56456 d = 6, ω = (6, 3, 2, 0, 0, 0) d=5 ω = (5, 2, 2, 0, 0, 0) 11322 422 T= 5654 1123 These tableaux are clearly maximal and non-zero. Since f = d − 1 is not a weight of T, we have that T and D0 ∨ F0 are linearly independent. When f < 6 we have d ≤ 7 because f > b 2d c. Since all d ≤ 6 cases were done 3 above, only d = 7 remains. In this case we have only f = 5. Thus s = 12 and three tableaux are required. We can use D1 ∨ F0 , G1 , and G2 . These tableaux are linearly independent by previous discussions. Now consider s = 2d. Write d = 6k 0 + i, so s = 6(2k 0 ) + 2i0 . Hence 2k 0 + 2 linearly independent tableaux will suffice. Let d−2 A1 = d−2 2 4 5 5 46 1 1 2 23 ω(A1 ) = (d − 1, d − 1, 2, 0, 0, 0) This tableau is maximal. Although there are many valid weight assignments, all such 151 assignments have positive sign, hence A1 is non-zero. When d 6≡ 1 (mod 6) we have k 0 + 1 linearly independent tableaux Dp0 . Hence we can use: Dp ∨ Dp p = 0, 1, . . . , k 0 Dk0 ∨ Dp0 p0 = 0, 1, . . . , k 0 − 1 A1 with weights: ω(Dp ∨ Dp ) = (2p + i0 + 4k 0 , 2(k 0 − p), 0, 2p + i0 + 4k 0 , 2(k 0 − p), 0) ω(Dk0 ∨ Dp0 ) = (d, 0, 0, 2p + i0 + 4k 0 , 2(k 0 − p), 0) ω(A1 ) = (d − 1, d − 1, 2, 0, 0, 0) This provides 2k 0 + 2 tableaux, provided k 0 > 0. Their weights are clearly distinct so they are linearly independent by Lemma 3.4.12. When d ≡ 1 (mod 6), we have s = 6(2k 0 ) + 2 so only 2k 0 + 1 tableaux are needed. All the tableaux listed above work except for p = 0 and p0 = 0, providing 2k 0 linearly independent tableaux. In addition use: 2 d−3 d−4 2 2 A2 = 5 4 5 664 1 1 2 233 d≥6 ω(A2 ) = (d − 1, d − 2, 3, 0, 0, 0) This tableau is maximal and non-zero. Since d ≡ 1 (mod 6) we have d > 6. Thus A2 provides the additional tableau and its weight is distinct from the other tableaux, insuring linear independence. When k 0 = 0 we have d < 6. Then s < 10 so two tableaux, A1 and D0 ∨ D0 , suffice. 152 9.1.4 Basis Tableaux for c > 6, s ≤ r Let λ = [r + s, s] be a partition of n, with s ≤ r, where 2s + r = cd = n. We want to construct mλ linearly independent basis tableaux for an arbitrary c. In Sections 4.3, 9.1.1, 9.1.2, and 9.1.3 we constructed basis tableaux for c ≤ 6, which we will make use of in this construction. In addition we will use the following tableaux: d−2 A1 = d−2 2 4 5 5 46 1 1 2 23 ω(A1 ) = (d − 1, d − 1, 2, 0, 0, 0) 2 d−3 d−4 2 2 A2 = 5 4 5 664 1 1 2 233 d≥6 ω(A2 ) = (d − 1, d − 2, 3, 0, 0, 0) 2 d−4 d−4 2 2 2 A3 = 5 5 6 644 1 2 1 332 d≥6 ω(A3 ) = (d − 2, d − 2, 4, 0, 0, 0) 2 d−4 d−5 2 3 2 A4 = 5 4 5 664 1 1 2 233 d≥5 ω(A4 ) = (d − 2, d − 3, 5, 0, 0, 0) 422 A5 = 4 5 6 123 d=4 153 ω(A5 ) = (4, 2, 2, 0, 0, 0) These tableaux are all maximal. Although some of these tableaux have additional valid weight assignments, all such assignments have positive sign. Hence these tableaux are non-zero. To construct λ = [r + s, s]-tableaux with s ≤ r, write s = md + f , with 0 ≤ f < d and r = md + pd + g with 0 ≤ g ≤ d. Then 3md + pd + 2f + g = cd, so 2f + g = xd for some x. This means c = 3m + p + x. If p + x ≥ 3, a λ-tableau may be written D ∨ B ∨ (c − 3m − 3)V (d), where D is a tableau of shape [2md, md] filled with 3m elements, B is a [3d − f, f ] tableau filled with 3 elements, and V (d) is the one row tableau. We will first consider this case and handle the p + x < 3 case later. We have s = md + f , so writing d = 6k 0 + i0 , f = 6k + i, with 0 ≤ i0 , i ≤ 5 0 gives s = 6(mk 0 + k) + mi0 + i. Since b mi6+i c ≤ m, it suffices to construct b 6s c + 1 ≤ mk 0 + k + m + 1 linearly independent tableaux. If m < 2 we may simply use the basis tableaux constructed for c = 6 along with V (d)0 s, so assume m ≥ 2. Let Dp be the c = 3 basis tableaux of shape [2d, d] described in Section 4.3. There are m[2d,d] = k 0 + 1 such tableaux when i0 6= 1 and k 0 for i0 = 1. Let Bq be the c = 3 basis tableaux of shape [3d − f, f ] constructed in Section 4.3. This tableau has s ≤ r since f < d. There are m[3d−f,f ] = k 0 + 1 such tableaux for i 6= 1 and k 0 tableaux when i = 1. Take f > 1 and d ≥ 6. The d < 6 and f ≤ 1 cases will be handled separately. Consider the following tableaux forms (with the appropriate number of V (d)’s as necessary): ` = 1, 2, . . . , m I. `Dp ∨ (m − `)Dk0 ∨ Bq p = 0, 1, . . . , k 0 − 1 q = 0, 1, . . . , k 154 II. mDk0 ∨ Bq q = 0, 1, . . . , k ` = 1, 2, . . . , b m2 c III. `Aj ∨ (m − 2`)Dk0 ∨ Bq q = 0, 1, . . . , k j = 1, 2, 3, 4 d ≥ 6 if j 6= 1 Note that those tableaux with p = 0 or q = 0 exist only when i0 6= 0 or i 6= 0 respectively. We will not make use of these tableaux unless necessary. However, even when k = 0, at least one Bq exists, so if we regard k as the number of Bq ’s, we may assume k ≥ 1. When d 6≡ 1 (mod 6) we need mk 0 + k + m + 1 tableaux. The list above (taking q 6= 0) provides at least mk 0 k + k + 4b m2 ck tableaux, which suffices since 4b m2 c ≥ m + 1. When d ≡ 1 (mod 6), we need only mk 0 + b m6 c + 2 tableaux. If k 0 ≥ 2 the list above provides at least m(k 0 − 1)k + k + 4b m2 c tableaux. This is sufficient for m 6= 3. When m = 3 we need 3k 0 + k + 2 tableaux when i = 3, 4, or 5 and 3k 0 + k + 1 tableaux for i = 0, 1, and 2. When i ≤ 2 the tableaux listed suffice. For i ≥ 3 we need an additional tableau so use the tableau of the Form II with q = 0. When k 0 = 1, the tableaux of the Form I don’t exist. Hence we have only k + 4b m2 c tableaux when q 6= 0 and we need m + k + b m6 c + 2. For m 6= 3 this is sufficient. However, one additional tableau is needed for m = 3, when i = 3, 4, or 5. In this case we may use q = 0 in Form II for the remaining tableau. To show linear independence of these tableaux it suffices, by Lemma 3.4.12, to show that the max weights are distinct. For max weights we have: I. (4k 0 + 2p + i0 , 2(k − p), 0)` ∨ (d, 0, 0)(m−`) ∨ (4k + 2q + i, 2(k − q), 0) II. (d, 0, 0)m ∨ (4k + 2q + i, 2(k − q), 0) 155 III − 1. (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ (4k + 2q + i, 2(k − q), 0) III − 2. (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ (4k + 2q + i, 2(k − q), 0) III − 3. (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ (4k + 2q + i, 2(k − q), 0) III − 4. (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ (4k + 2q + i, 2(k − q), 0) Consider those tableaux of Form I. If ω(I(`, p, q)) = ω(I(`0 , p0 , q 0 )), then counting the number of d’s shows ` = `0 , while counting the number of 4k 0 + p + i0 ’s indicates ` = 1. If p = p0 then q = q 0 , which is not possible for distinct tableaux. Hence by Lemma 3.4.13, the weights are distinct because f < d. Those of Form II are distinct due to the distinct weights of Bq . Similarly those of the Form III − j are distinct by the number of d’s and distinctness of ω(Bq ). If ω(III − i(`, q)) = ω(III − j(`0 , q 0 )), then by counting the number of d’s we have ` = `0 . Then by counting the number of 0’s, d − 1’s, and d − 2’s, we find the weights must be distinct for d > 6. (If d = 6 the Forms III − 2 and III − 4 have the same weights.) To see the different forms have distinct weights, first count the number of d’s. Obviously ω(II) 6= ω(I) or ω(III). If ω(I(`)) = ω(III(`0 )) then ` = 2`0 . However, counting the number of 0’s shows the weights are distinct. When d = 6 then s ≤ 6m + 5, so m + 1 tableaux suffice. Since there are two D’s and one B, Forms I and II provide the requisite number of linearly independent tableaux. Now consider the case where d < 6. Here s = mi0 + i so m + 1 tableaux suffices. In this case the Aj , j = 2, 3 do not exist. However for d = 5 we have A4 and for d = 4 we have A5 . Moreover, we have exactly one D and one B. Hence the appropriate Forms II, III − 1, and III − 4 or III − 5 provied 2b m2 c + 1 tableaux. This suffices for even m. When m odd we need one additional tableau. For m ≥ 5 use A1 ∨ Ai ∨ (m − 4)D0 ∨ B0 where i = 4 or 5 as appropriate. If m = 3, then s ≤ 19 so the three tableaux listed will suffice except when d = 5 and f = 3. In this case 156 also use the non-zero maximal tableau: 4 4 44 T= 596978 112234 ω(T) = (5, 5, 4, 4, 0, 0, , 0, 0, 0) When d = 3, we have s = 3m + 2 since 1 < f < d. Hence b 3m+2 c + 1 tableaux 6 are needed. We have mD0 ∨ B0 and `A1 ∨ (m − 2`)D0 ∨ B0 , which provide b m2 c + 1 tableaux. Since m ≥ 2 we have b m2 c ≥ b 3m+2 c, so these tableaux suffice. 6 For linear independence, we need only check distinctness of max weights by Lemma 3.4.12. First consider d = 4 or 5. Since ω(A5 ) = ω(A3 ) our discus- sion on linear independence for d ≥ 6 still holds. The only additional tableau is A1 ∨ Ai ∨ (m − 4)D0 ∨ B0 which clearly has a distinct weight. Similarly, by counting the number of 5’s and 4’s, the weight of T is also distinct. Hence these tableaux are linearly independent. When d = 3 the tableaux listed are a subset of the tableaux for d ≥ 6 and hence are linearly independent by the prior discussion. This covers all cases with p + x ≥ 3 provided f > 1. The f = 0 and f = 1 cases will be handled after the p + x < 3 case. Now assume that p + x < 3. Recall that if s = md + f , with 0 ≤ f < d and r = md + pd + g, then 3md + pd + 2f + g = cd and 2f + g = xd for some x. This means c = 3m + p + x, so if p + x < 3 then λ-tableau with s ≤ r may be written as D ∨ F ∨ (c − 3m − 3)V (d). Here D is a tableau of shape [2(m − 1)d, (m − 1)d] filled with 3(m − 1) elements and F is a [(3 + x − 1)d − f, f + d] tableau filled with 3 + x elements. When f > 0, we have x > 0, which means F is a c = 4 or c = 5 tableau. We will first consider the case where m ≥ 3, d ≥ 6 and f > 1. If d = 6k 0 + i0 and f = 6k + i, with 0 ≤ i0 , i ≤ 5, then s = md + f , so we still need b 6s c + 1 ≤ mk 0 + k + m + 1 tableaux. The number of tableaux D is the same as before. Let Fq be the c = 4 or 5 basis tableaux of shape [cd − f − d, f + f ]. There are at least k 0 + k such tableaux. Moreover since d ≥ 6 and f > 1 there is always at 157 least 2 such tableaux. Consider the tableaux of the following forms: ` = 1, 2, . . . , m − 1 I. `Dp ∨ (m − ` − 1)Dk0 ∨ Fq p = 0, 1, . . . , k 0 − 1 q = 1, 2, . . . , k 0 + k II. (m − 1)Dk0 ∨ Fq q = 1, 2, . . . , k 0 + k c ` = 1, 2, . . . , b m−1 2 III. `Aj ∨ (m − 2` − 1)D ∨ Fq q = 1, 2, . . . , k 0 + k k0 j = 1, 2, 3, 4 d ≥ 6 if j 6= 1 As before, the tableaux with p = 0 exist only for d 6≡ 1 (mod 6). When d 6≡ 1 (mod 6), the construction above provides (m − 1)k 0 (k 0 + k) + k + k 0 + 4b m−1 c(k 0 + k). 2 Now k 0 + k denotes the number of F’s, which we can assume is at least 2. Then since c ≥ m + 1 for m 6= 4, this construction provides sufficient tableaux. If m = 4 4b m−1 2 then 4b m−1 c(k 0 + k) ≥ 8, so this construction is sufficient. 2 When d ≡ 1 (mod 6), we need only mk 0 + k + b m6 c + 2 tableaux. For k 0 ≥ 2, we have (m − 1)(k 0 − 1)(k 0 + k) + k + k 0 + 4b m−1 c(k 0 + k) tableaux by the construction 2 above, which is sufficient. If k 0 = 1, the tableaux of Form I do not exist, so we have only k 0 + k + 4b m−1 c(k 0 + k) tableaux. However, since we know that there are always 2 at least two Fq ’s, we have k 0 + k ≥ 2. Thus the tableaux listed are sufficient. When d < 6, the tableaux Aj with j = 2 and 3 do not exist. However we do have A4 for d = 5 and A5 for d = 4. Also, since d < 6, we may no longer assume that there are at least two Fq ’s (unless d = 5 and f = 3 or 4). However there is always at least one. In addition, there is only one D. Under these constraints we have tableaux of 158 the Forms II, III − 1, and III − 4 or III − 5 when d = 5 or 4 respectively. For m ≥ 5 also use A1 ∨ Aj ∨ (m − 5)D0 ∨ F0 . This provides 2b m−1 c + 2 tableaux for m ≥ 5. For 2 d = 4 we need at most b 4m+3 c + 1, hence this this suffices. For d = 5 and f = 2 we 2 need at most b 5m+2 c + 1 which we have. Our construction suffices except for m = 3 6 and 4. When m = 3, then s ≤ 17 and three tableaux, II, III − 1, and III − 4 or III − 5 suffice. For m = 4 we need one additional non-zero maximal tableau, so use d−1 1 d−1 1 d−1 1 d−1 1 2 T= 6 7 8 7 9 12 10 12 11 1 1 2 2 3 3 4 4 5 ω(T) = (d, d, d, d, 2, 0, 0, 0) When d = 5, f ≥ 3 there are two Fq ’s hence the Forms II, III − 1 and III − 4 c + 1 tableaux. Since at most b 5m+4 c + 1 tableaux are needed, this suffices. give 4b m−1 2 6 c + 1 tableaux. From Forms II and III − 1, we have When d = 3 we need b 3m+2 6 b m−1 c + 1 tableaux. This is sufficient except for m odd. When m even we need one 2 additional tableau, so use T listed above. Now to consider linear independence. By Lemma 3.4.12 it suffices to show that the max weights are distinct. First consider when the F are c = 4 basis tableaux. The max weights are: I. (4k 0 + 2p + i0 , 2(k 0 − p), 0)` ∨ (d, 0, 0, )m−`−1 ∨ (d − q + 1, f + q − 1, 0, 0) II. (d, 0, 0)m−1 ∨ (d − q + 1, f + q − 1, 0, 0) III − 1. (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f + q − 1, 0, 0) III − 2. (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f + q − 1, 0, 0) III − 3. (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f + q − 1, 0, 0) III − 4. (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f + q − 1, 0, 0) 159 Consider those tableaux of Form I. If ω(I(`, p, q)) = ω(I(`0 , p0 , q 0 ), then counting the number of 0’s shows ` = `0 . By counting the number of different weights we find ` = 1. Now if p = p0 then we must have q = q 0 , which is not possible for distinct tableaux. Hence by Lemma 3.4.13, these weights are distinct. The weights of Form II are distinct by the construction of the F’s in Section 9.1.1. The weights of Form III − j are distinct by counting the number of 0’s and by the distinctness of the F’s. If ω(III − j) = ω(III 0 − i) counting the number of 0’s and different numbers shows ` = `0 = 1. So if the weights are not distinct, then ω(Aj ∨ Fq ) = ω(Ai ∨ Fq0 ), which implies Ai and Aj must have one non-zero element in common. Then by Lemma 3.4.13, these weights are distinct if f > 2. When f = 2 the tableaux have the same length and the lemma does not apply. If f = 2 then ω(III − 3(` = 1, q = 4)) = ω(III − 4(` = 1, q = 3)), though all other weights are distinct. We have ω(I) 6= ω(II), by counting the number of 0’s. Forms II and III are also distinct by counting the number of 0’s. If ω(I) = ω(III 0 ), counting the number of 0’s shows that ` = `0 . Then counting the number of d’s shows that ` = 1, q 0 = 1, and q 6= 1. If the weights are equal, then (4k 0 + 2p + i0 , 2(k 0 − p), d − q + 1, f + q − 1) = c, these weights are distinct unless f = 2. When f = 2 (ω(Aj ), f ). Since q ≤ b d−f 2 then ω(I(` = 1, p = k 0 − 1, q = j) = ω(III − j(`0 = 1, q 0 = 1)), j 6= 1. Hence for f > 2 all the tableaux listed are linearly independent. When f = 2 some of the tableaux in our list have the same max weights, and hence may be linearly dependent. If we eliminate these tableaux with duplicate max weights c(k 0 +k)−4 linearly independent from our list we have (m−1)k 0 (k 0 +k)+k 0 +k+4b m−1 2 0 0 tableaux when d 6≡ 1 (mod 6). We need at most b m(6k 6+i )+2 c+1 ≤ mk 0 +m tableaux, which we have since we may still take k 0 + k ≥ 2. If d ≡ 1 (mod 6) then we have (m − 1)(k 0 − 1)(k 0 + k) + k 0 + k + 4b m−1 c(k 0 + k) − 4 linearly independent tableaux 2 c + 1 tableaux are needed. When for k 0 > 1. This is sufficient since only mk 0 + b m+2 6 k 0 = 1 those tableaux of Form I don’t exist, so we have k 0 + k + 4b m−1 c(k 0 + k) − 1 2 linearly independent tableaux, which is sufficient. Thus the d ≥ 6, f > 1, m ≥ 3 case is finished for c = 4. 160 When d < 6 the same tableaux are used with the substitution of A5 . However, A5 has the same weight as A3 so the above argument applies. (Note that by the conditions of Fq , q ≤ b d−f c ≤ 1, so the max weight duplication does not occur.) In addition we 2 also use the tableaux A1 ∨ Aj ∨ (m − 5)D0 ∨ F0 , for j = 4 or 5. However, counting the number of 0’s and d’s shows that this tableau is distinct from our previous collection; otherwise ω(A1 ∨ Aj ∨ F0 ) = ω(Ai ∨ Ai ∨ Fq ) which is impossible. In addition, when m = 4, d = 5 or d = 3, m even, we also have the tableau T. Counting the number of 0’s and d’s shows this tableau has a distinct max weight as well. Hence when F is a c = 4 tableau, we have linear independence for f > 1 and m ≥ 3. Now consider the case where F is a c = 5 basis tableau. The c = 5 basis tableaux were constructed in Section 9.1.2. These tableaux have two different types of max weights corresponding to the T1 and T2 tableaux constructions used. Those of form T1 have weights (d, f − q − 1, q + 1, 0, 0) where q = 1, 2, . . . , b f2 c − 2 (when f ≥ 6). Those of the form T2 have weights (d − q + 1, f − 1 + q, 0, 0, 0) with q = 1, 2, d − b d+f2 −2 c. We will refer to the basis tableaux using the T1 tableaux as q-forms and those using T2 as q-forms. Then the max weights for the general tableaux have the following forms: I(q). (4k 0 + 2p + i0 , 2(k 0 − p), 0)` ∨ (d, 0, 0, )m−`−1 ∨ (d, f − q − 1, q + 1, 0, 0) I(q). (4k 0 + 2p + i0 , 2(k 0 − p), 0)` ∨ (d, 0, 0, )m−`−1 ∨ (d − q + 1, f − 1 + q, 0, 0, 0) II(q). (d, 0, 0)m−1 ∨ (d, f − q − 1, q + 1, 0, 0) II(q). (d, 0, 0)m−1 ∨ (d − q + 1, f − 1 + q, 0, 0, 0) III − 1(q). (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d, f − q − 1, q + 1, 0, 0) III − 1(q). (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f − 1 + q, 0, 0, 0) III − 2(q). (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d, f − q − 1, q + 1, 0, 0) III − 2(q). (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f − 1 + q, 0, 0, 0) III − 3(q). (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d, f − q − 1, q + 1, 0, 0) III − 3(q). (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f − 1 + q, 0, 0, 0) 161 III − 4(q). (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d, f − q − 1, q + 1, 0, 0) III − 4(q). (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q + 1, f − 1 + q, 0, 0, 0) These tableaux are linearly independent by Lemma 3.4.12 provided their max weights are distinct. To show these weights are distinct we will often count the number of 0’s and d’s in each weight. For convenience we will list these values below: Form # 0’s # d’s I(q). 2m − ` m−` I(q). 2m − `+ II(q). 2m II(q). 2m + 1 III − j(q). 2m − ` III − j(q). 2m − ` + 1 m − ` − 1 q 6= 1 m−` q=1 m m − 1 q 6= 1 m q=1 m − 2` m − 2` − 1 q 6= 1 m − 2` q=1 Table 9.1: Weights of 0 or d. Consider those tableaux of Form I(q). If ω(I(`, p, q)) = I(`0 , p0 , q 0 ), then Table 9.1.4 shows that ` = `0 . shows ` = 1. Then counting the number of different numbers If p = p0 then we must have q = q 0 and vice versa, but this can’t happen since the tableaux are different. Then since f < d, the argument of Lemma 3.4.13 shows the weights are distinct. The same reason holds for Form I(q). Table 9.1.4 also shows ω(I(q)) 6= ω(I(q)). The tableaux of Form II are distinct since the max weights of the F are distinct for c = 5 by Section 9.1.2. Moreover, ω(II(q)) 6= ω(II(q)) by Table 9.1.4. The distinctness of max weights for Form III − j(q) or III − j(q) follows from Table 9.1.4 and distinctness of c = 5 basis tableaux max weights. Also from Table 9.1.4 we have ω(III − j(q)) 6= ω(III − j(q)). Now suppose ω(III − j(q, `)) = ω(III − i(q 0 , `0 )). Table 9.1.4 shows that ` = `0 162 and counting the number of different numbers shows ` = 1. Since i 6= j, the weights of Ai and Aj must have exactly one non-zero weight in common. Thus the argument of Lemma 3.4.13 applies and hence the weights are distinct. Similarly ω(III − j(q, `)) 6= ω(III − i(q 0 , `0 )) for f > 2 by Lemma 3.4.13. (When f = 2, ` = 1 the conditions of Lemma 3.4.13 are not met.) The values of Table 9.1.4 are sufficient to show ω(III − j(q, `)) 6= ω(III − i(q 0 , `0 )). This shows the max weights within each tableau form are distinct for f > 2. Showing that ω(I) 6= ω(II) 6= ω(III) follows directly from Table 9.1.4. When f = 2 our discussion on linear independence holds except for a few tableaux of the Form III(q). Specifically, ω(III − 2(` = 1, q = 3)) = ω(III − 3(` = 1, q = 2)), ω(III − 2(` = 1, q = 4)) = ω(III − 4(` = 1, q = 2)), and ω(III − 3(` = 1, q = 4)) = ω(III −4(` = 1, q = 3)). Thus we have three fewer linearly independent tableaux than we originally calculated. If we eliminate these tableaux with duplicate max weights from our list we have (m−1)k 0 (k 0 +k)+k 0 +k+4b m−1 c(k 0 +k)−3 linearly independent 2 0 0 tableaux when d 6≡ 1 (mod 6). We need at most b m(6k 6+i )+2 c+1 ≤ mk 0 +m tableaux, which we have since we may still take k 0 + k ≥ 2. If d ≡ 1 (mod 6) then we have c(k 0 + k) − 3 linearly independent tableaux (m − 1)(k 0 − 1)(k 0 + k) + k 0 + k + 4b m−1 2 c + 1 tableaux are needed. When for k 0 > 1. This is sufficient since only mk 0 + b m+2 6 k 0 = 1 those tableaux of the Form I don’t exist, but we still have 2 + 8b m−1 c−3 2 linearly independent tableaux, which is sufficient. Hence we have enough tableaux when f = 2. Thus the d ≥ 6, f > 1, m ≥ 3 case is finished for c = 5. When d < 6 only the q-form tableaux exist for c = 5. These tableaux have the same max weights as the c = 4 tableaux F (with the exception of an extra zero.) Thus by the same argument as in that case, these tableaux are linearly independent. Hence when F is a c = 5 tableau, we have linear independence for f > 1 and m ≥ 3. These constructions assumed m ≥ 3. If m = 1, The c = 4 or c = 5 basis tableaux (joined with sufficient V (d)’s) suffice. However, when m = 2, we need tableaux with c = 7 or 8 elements. This case must be dealt with separately. (We will still assume 163 f > 1.) Tableaux of Form I listed previously (Dp ∨ Fq ) still work for this case. These tableaux are linearly independent by our previous discussion. As before, let d = 6k 0 +i0 , f = 6k+i, with 0 ≤ i0 , i ≤ 5. Since s = 2d+f , we need at most 2k 0 +k+3 tableaux. The tableaux Fq have λ2 = d+f so there are at least k 0 +k such tableaux. So for d 6≡ 1 (mod 6), we have (k 0 +1)(k 0 +k) tableaux. This provides sufficient tableaux unless k 0 = 1, k = 1, or k = 0, k 0 < 2. When k 0 = k = 1, then d + f ≥ 14 so at least three Fq ’s exists. Hence the Form I construction is sufficient. In the remaining cases, computing exactly how many Fq exist and precisely how many tableaux are needed shows the Form I tableaux are sufficient except for: d = 11, f = 2, d = 3, f = 2, d = 4, f = 3, and d = 5, f = 2, 3, 4. (For instance, when d = 6 and f = 5 there are two F’s and two D’s, so Form I provides 4 tableaux. Since s = 17, only three tableaux are needed.) When d = 11 and f = 2, there are two D’s and F’s for a total of 4 Form I tableaux. Since s = 24, five tableaux are required. In addition to the Form I tableaux, use 10 10 4 T= 4 5 6 ω(T) = (10, 10, 4, 0, 0, 0) 1 2 3 which is non-zero, maximal, and linearly independent. For d = 3, f = 2, we have c = 8. Since s = 8 two tableaux are needed. In addition to D0 ∨ F0 , use the non-zero maximal tableau: 2222 T= 5678 1234 ω(T) = (2, 2, 2, 2, 0, 0, 0, 0) When d = 4 and f = 3, then we must have c = 9 as s ≤ r. Since s = 11, we need 164 two tableaux. Use D0 ∨ F0 and 2 2 2 2 T= 5969788 ω(T ) = (3, 3, 3, 2, 0, 0, 0, 0) 1122334 Counting the number of 4’s shows these tableaux are linearly independent. When d = 5 we have c = 8 and need three tableaux (except for f = 3 when two tableaux suffice). In addition to D0 ∨ F0 use: 4 4 4 f −2 T= 567 8 123 4 f 6= 3 ω(T) = (4, 4, 4, f − 2, 0, 0, 0, 0) 4 f 22 0 T = 566768 112234 ω(T 0 ) = (5, f + 1, 2, 2) These tableaux are non-zero. Linear independence follows by counting the number of 5’s and 2’s. Thus we have sufficient tableaux for d 6≡ 1 (mod 6). When d ≡ 1 (mod 6) we need 2k 0 + k + 2 tableaux. The Form I construction Dp ∨ Fq discussed earlier provides at least k 0 (k 0 + k + 1) linearly independent tableaux for f 6≡ 0 (mod 6). (When f 6≡ 0 (mod 6) there are k 0 + k + 1 tableaux Fq .) This is sufficient for k 0 ≥ 2. When f ≡ 0 (mod 6) Form I provides k 0 (k 0 + k) tableaux, but only 2k 0 + k + 2 tableaux are needed. This is sufficient for k 0 ≥ 2 since f 6= 0. Since d ≥ 3, only d = 7 remains. We need at most 4 tableaux, as s ≤ 20. Consider the tableaux Bp ∨ Cq where the Bp are c = 3 basis tableaux with λ2 = d − 1 and the Cq are c = 4 or c = 5 basis tableaux with λ2 = d + f + 1. There are two tableaux each for Bp and Cq so this construction is sufficient. The max weights for Bp are (6, 0, 0) and (4, 2, 0). The max weights for Cq are (7, f + 1, 0, 0) and (6, f + 2, 0, 0). Hence 165 the weights of our construction are distinct unless f = 5. However when f = 5 only three tableaux are needed and the Form I construction D ∨ Fq provides three in this case. When f = 1 we can proceed as in the p + x < 3 case. We have s = md + 1 and r = md + pd + g. Then 3md + pd + 2f + g = cd, so 2 + g = xd for some x. Hence x = 1. This means that for both the p + x ≥ 3 and p + x < 3 cases we can use the c = 4 basis tableaux, proceeding as in the p + x < 3 case when f > 1. As before, write d = 6k 0 + i0 with 0 ≤ i0 ≤ 5. We have at least k 0 linearly independent λ = [3d − 1, d + 1] tableaux Fq , which we use for the following forms: ` = 1, 2, . . . , m − 1 I. `Dp ∨ (m − ` − 1)Dk0 ∨ Fq p = 0, 1, . . . , k 0 − 1 q = 1, 2, . . . , k 0 II. (m − 1)Dk0 ∨ Fq q = 1, 2, . . . , k 0 c ` = 1, 2, . . . , b m−1 2 III. `Aj ∨ (m − 2` − 1)Dk0 ∨ Fq q = 1, 2, . . . , k 0 j = 1, 2, 3, 4 d ≥ 6 if j 6= 1 These tableaux have the following max weights: I. (4k 0 + 2p + i0 , 2(k 0 − p), 0)` ∨ (d, 0, 0, )m−`−1 ∨ (d − q, q + 1, 0, 0) II. (d, 0, 0)m−1 ∨ (d − q, q + 1, 0, 0) III − 1. (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q, q + 1, 0, 0) III − 2. (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q, q + 1, 0, 0) 166 III − 3. (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q, q + 1, 0, 0) III − 4. (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)m−2`−1 ∨ (d − q, q + 1, 0, 0) Counting the number of zeros and d’s in these weights gives: Form #0’s #d’s I. 2m − ` m−`−1 II. 2m m−1 2m − ` m − 2` − 1 III − j. Then by arguments similar to previous cases and Lemma 3.4.13, these tableaux have distinct max weights except for III − 1(` = 1, q = 2) and III − 2(` = 1, q = 1) whose weights are the same. Thus by Lemma 3.4.12, these tableaux (omitting III − 1(` = 1, q = 2)) are linearly independent. c + 1 tableaux. When k 0 ≥ 2, When d ≥ 6 and m ≥ 3, we need mk 0 + b 5m+1 6 the forms listed above provided (m − 1)k 0 k 0 + k 0 + 4b m−1 ck 0 − 1, which is sufficient. 2 When k 0 = 1 and d 6≡ 1 (mod 6), we still have these tableaux. If d = 7, then there is exactly one D so Form I does not exist. However, there are two distinct Fq ’s. Thus c tableaux. Since only m + b m+1 c + 1 tableaux are Forms II and III provide 8b m−1 2 6 needed, this suffices. If m = 2 then only Form I and II tableaux exist. Since s = 2d + 1, then 2k 0 + 2 tableaux suffice. Forms I and II provided at least (k 0 − 1)k 0 + k 0 , tableaux. When k 0 ≥ 2 this is enough. For d > 7, four tableaux are needed. Since there are at least two D’s and two F’s, this construction suffices. When d = 6, there are two D’s, but only one F. However, s = 13 so only two tableaux are needed, which we have. If d = 7, three tableaux are needed. Since there is only one D and two F’s, an additional 167 tableau is required. Use: 66 2 H= 4567 ω(H) = (6, 6, 3, 0, 0, 0, 0) 1233 This tableau is clearly non-zero and maximal and its max weight does not contain d, so it is linearly independent from our previously constructed tableaux. When d = 5, there are two distinct Fq ’s, which have weights of the form listed above. However, there is only one D. Thus Forms II, III − 1, and III − 4 provide at least 1 + 4b m−1 c tableaux. Since only b 5m+1 c + 1 tableaux are needed, this suffices 2 6 for m ≥ 3. When m = 2 then s = 11, so the two tableaux of Form II are sufficient. When d = 3 there is only one F and one D. Moreover, only A1 exists. However, all c + 1 tableaux and we need the tableaux are linearly independent. Thus we have b m−1 2 b 3m+1 c + 1 tableaux. This is sufficient for m odd. When m even then s = 3m + 1 ≡ 1 6 (mod 6) so only b 3m+1 c tableaux are required, which we have when m ≥ 3. If m = 2 6 then s = 7, which means the single tableau of Form II is sufficient. When d = 4 we need to proceed differently since the c = 4 basis tableaux F is not in maximal form. First consider m ≥ 4. We will replace F in the previous discussion with the following c = 7 tableaux, Hq . 42 2 H0 = 4 5 6 7 ω(H0 ) = (4, 3, 2, 0, 0, 0, 0) 1223 2 2 2 H1 = 4 7 5 4 6 5 ω(H0 ) = (3, 3, 3, 0, 0, 0, 0) 112233 The the tableaux we will use are: II. (m − 2)D0 ∨ Hq q = 0, 1 168 c ` = 1, 2, . . . , b m−2 2 III. `Aj ∨ (m − 2` − 2)D0 ∨ Hq q = 0, 1 j = 1, 5 Counting the number of 0’s and 4’s in the max weights of these tableaux show they are linearly independent by Lemma 3.4.12. This can be seen from Table 9.1.4 below: Form # 0’s # 4’s q II. 2m m−1 0 2m m−2 1 2m − ` m − 2` − 1 0 2m − ` m − 2` − 2 1 III − j. Table 9.2: Weights of 0 and 4. c + 2 linearly independent Hence for m ≥ 4 this construction provides 4b m−2 2 tableaux. Since only b 4m+1 c + 1 tableaux are required, this suffices. 6 When m = 3, then s = 13, so only two tableaux are needed. The two tableaux of Form II suffices. Hence we have constructed sufficient linearly independent tableaux when f = 1 and m ≥ 3. For m = 2, two tableaux are needed, in which case use H0 and H1 . Hence all f = 1 cases are accounted for, since m = 1 may be handled by c ≤ 6 basis tableaux and V (d)’s. When f = 0 the construction is similar to earlier cases, particularly the f = 1 case. However, instead of the F tableaux, we use only the c = 3 basis tableaux D of shape [2d, d]. Again, we may take m ≥ 2, since m = 1 may be handled by c ≤ 6 basis tableaux and V (d)’s. Writing d = 6k 0 + i0 , 0 ≤ i0 ≤ 5, our general tableaux are: 169 I. `Dp ∨ (m − `)Dk0 II. mDk0 III. `Aj ∨ (m − 2`)Dk0 p = 0 . . . k 0 − 1, ` = 1 . . . m ` = 1 . . . b m2 c, j = 1 . . . 4, d ≥ 6 These tableaux have max weights: I. (4k 0 + 2p + i0 , 2(k 0 − p), 0)` ∨ (d, 0, 0, )m−` II. (d, 0, 0)m III − 1. (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)m−2` III − 2. (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)m−2` III − 3. (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)m−2` III − 4. (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)m−2` Counting the number of zeros and d’s in these weights gives: Form #0’s #d’s I. 2m − ` m−` II. 2m m 2m − ` m − 2` III − j. Then by arguments similar to previous cases and Lemma 3.4.13, these tableaux have distinct max weights and are linearly independent by Lemma 3.4.12. Consider d ≥ 6. If d 6≡ 1 (mod 6), we need mk 0 + m tableaux. The above construction provides mk 0 + 1 + 4b m2 c tableaux, which is sufficient. When d ≡ 1 (mod 6) we need mk 0 + b m6 c + 1 tableaux. In this case, D0 does not exists, so p > 0. For k 0 6= 1, we have m(k 0 − 1) + 1 + 4b m2 c which is sufficient. For k 0 = 1, we have 170 4b m2 c + 1 tableaux since Form I is no longer valid. However, this suffices. When d < 6 not all of the Aj exist and there is exactly one D. For d = 4 and d = 5 we have Forms II, III − 1, and III − 5 or III − 4 respectively. This provides 2b m2 c + 1 tableaux. Since at most b 5m c + 1 tableaux are needed, this is sufficient. 6 When d = 3 we have Forms II and III − 1, which yield b m2 c + 1 tableaux. Since b 3m c + 1 tableaux are required, we have enough. This construction work for m ≥ 2, 6 hence all f = 0 cases are accounted for. 9.2 Case: r < s This proof of this case follows similarly to the s ≤ r case. Let λ = [r + s, s] be a partition of n with r < s, where n = 3b = cd. We wish to construct mλ linearly independent tableaux, where mλ is the multiplicity of χλ in 1SSnb oS3 as described in Theorem 11. First we will construct these tableaux for c = 4, 5, and 6; then we will use these constructions in proving Theorem 3 for a general c. We will refer to the tableaux constructed in this way as basis tableaux. These constructions will make use of the c = 3 basis tableaux constructed in Section 4.3 as well. 9.2.1 Basis Tableaux for c = 4, r < s Given λ = [r + s, s], a partition of n, we have 2s + r = 4d = 3b. From this equation and r < s, we have d + b d3 c < s ≤ 2d. For each λ we will construct mλ linearly independent λ-tableaux filled with the numbers 1 to 4. These will be our c = 4 basis tableaux. When d + b d3 c ≤ s ≤ 2d, write s = d + f , with b d3 c ≤ f ≤ d. Consider the tableau T from Section 9.1.1. This tableau is non-zero and maximal. 171 a+b≤d a, b, c, d > 0 abcd T (a, b, c, d) = 4 3 4 3 a>d 1 1 2 2 b>c or a = d, b = c w(T ) = (a + b, c + d, 0, 0) Let Cp = T (d d2 e − p, b d2 c − p, b f2 c + p, d f2 e + p) for 0 ≤ p ≤ b d−f c. The constraints 4 on T are satisfied for all p provided f ≥ 2 when p = 0. Hence Cp is non-zero for these parameters. Let C0p = T (d d2 e−p+1, b d2 c−p, b f2 c+p, d f2 e+p−1), with 0 ≤ p ≤ b d−f c−1. 4 Here the constraints on T are satisfied providedand d − f ≥ 4 and f ≥ 4 when p = 0. Hence C0p is non-zero for these parameters. We have ω(Cp ) = (d − 2p, f + 2p, 0, 0) and ω(C0p ) = (d − 2p + 1, f + 2p − 1, 0, 0). Thus the weights are distinct and Lemma 3.4.12 the tableaux are linearly independent. c+1 ≤ By Theorem 11, mλ = b 6r c+1. Hence it suffices to construct b 6r c+1 ≤ b 2d−2f 6 d−f +1 linearly independent tableaux since r = 4d−2s and s = d+f . If 4 ≤ f ≤ d−4 3 then Cp and C0p together provide 2b d−f c + 1 linearly independent tableaux. Since 4 c ≥ b d−f c, this is sufficient. When f > d − 4 we have r ∈ {0, 2, 4, 6} with s 2b d−f 4 3 even for r = 0, 4 and odd for r = 2, 6. Therefor, at most one tableau is needed when f > d − 4, which is C0 . Now consider f < 4. Since f ≥ b d3 c, when d ≥ 3, only f = 2 and f = 3 remain. For d = 3, then f = 1 and we have r = s which was done in Section 9.1.1. If f = 2, then d ≤ 6. However, by Theorem 11, mλ = 0 for d = 3. When d = 4 or 5, mλ = 1 and hence C0 suffices. For d = 6, the tableaux C0 and C1 suffice. If f = 3, then d ≤ 9. For d = 9 we have mλ = 3, for d = 7, we have mλ = 2, while mλ ≤ 1 for the remaining d ≤ 8. Now C0 suffices for those cases with d 6= 7 or 9. When d = 7 we need an additional tableau, however, C1 exists. For d = 9, then r = s and so this case was done in Section 9.1.1. Hence we have sufficient c = 4 basis 172 tableaux for all partitions with r < s. 9.2.2 Basis Tableaux for c = 5, r < s Given λ = [r +s, s] a partition of n, we have 2s+r = 5d = 3b. From this equation and r < s, we have 0 ≤ r ≤ b 5d c. For each λ we will construct mλ linearly independent 3 λ-tableaux filled with the numbers 1 to 5. These will be our c = 5 basis tableaux. First consider 0 ≤ r ≤ d. If d is even, then s−d ≡ s (mod 2). Moreover, s−d ≥ r. Hence if λ0 = [r + s − d, s − d], then there are mλ0 linearly independent c = 3 basis tableaux and mλ0 = mλ since λ1 = λ2 = λ01 − λ02 . Therefore we can use U1 (d) ∨ T , were T are the c = 3 basis tableaux of shape λ0 , as the c = 5 basis tableaux. If d is odd, then s ≡ d − 1 (mod 2) and s − d + 1 ≥ r. Let T be the c = 3 basis tableaux of shape λ0 = [r + s − d − 1, s − d + 1]. Consider the tableaux U1 (d − 1) ∨ T . There are mλ0 such tableaux. Now mλ = mλ0 when r 6≡ 0, 3 (mod 6) since then λ1 − λ2 − 2 = λ01 − λ02 . Thus for r 6≡ 0, 3 (mod 6) we have constructed sufficient tableaux. However, 2s + r = 5d and d odd implies r is odd, hence only r ≡ 3 (mod 6) remains. In that case, mλ0 = mλ − 1, so only one additional tableau is needed. For d, r > 9 use U1 (d − 3) ∨ B0 where B0 is the c = 3 basis tableau of shape 00 λ = [r + s − d − 6, s − d + 3]. To show that U1 (d − 3) ∨ B0 is linearly independent from the U1 (d − 1) ∨ T it suffices, by Lemma 3.4.12, to show that their max weights are distinct. Since d − 1 is a weight of U1 (d − 1) ∨ T for all T , we need only show that 00 d − 1 is not a weight of U1 (d − 3) ∨ B0 . If we write λ = [r0 + s0 , s0 ] and s0 = 6k + j, 0 ≤ j ≤ 5, then ω(U1 (d − 3) ∨ B0 ) = (d − 3, 4k + j, 2k, 0, 0). If d − 1 is in this weight then that means d − 1 = 4k + j. Since 2s0 + r0 = 3d, we then must have j = 0 and r0 = 3. As r0 = r − 6, this implies r = 9. Thus for r > 9, d > 3 these tableaux are linearly independent. If d = 3, then only r = 3 is needed because r ≤ d. Only one tableau is required, which is U2 (2, 1). When r = 9 we need two tableaux basis tableaux. The U1 (d−3)∨B0 173 we constructed above has weight (d − 1, d − 3, d−1 ) In addition, use 2 1 a d−a−1 b+1 d−b−2 a b T = 43 5 3 4 5 4 11 1 2 2 3 3 c, b = d d−7 e, d ≥ 11, d odd a = b d−7 4 4 which has weight (d, d − 1, d−7 , 0, 0). For d < 11, only d = 9 is needed since 9 = r ≤ d 2 and d is odd. In that case, s = 2d so use U1 (d − 3) ∨ B0 and d−1 d−1 T = 5 3 5 4 1 1 2 2 Thus we have sufficient tableaux for r ≤ d. Now consider d < r ≤ b 5d c. To construct the c = 5 basis tableaux for these r we 3 will use two different types of tableaux. These are the same tableaux that were used in Section 9.1.2 and hence are non-zero and maximal. 1≤a≤d−2 a b d2 c d d2 e b T1 (a, b) = 4 4 5 4 3 45 222 1 1 33 1 ≤ b ≤ b d2 c − 3 a>b s≥d+6 d≥8 w(T1 ) = (d, a + 2, b + 1, 0, 0) a b T2 (a, b) = 5 3 5 4 1122 w(T2 ) = (a + 1, b + 1, 0, 0) 1 ≤ a, b ≤ d − 1 s≥4 174 Since d < r ≤ b 5d c, then 2d > s ≥ d 5d e. Consider Ep = T1 (s − d − 3 − p, p) 3 3 for 1 ≤ p ≤ b s−d c − 2. For s < 2d − 1, the parameters on T1 are satisfied provided 2 d ≥ 8. (When s = 2d − 1, we take p ≤ b s−d c − 3.) This provides b s−d c−2 2 2 linearly independent tableaux. We will also use E0q = T2 (d − q, s − 2 − d + q) for c. The parameters on T2 are satisfied provided s ≥ d + 2 (which 1 ≤ q ≤ d − b s−2 2 holds for d > 3). Together, Ep and E0q provide b s−d c−2+d−b s−2 c linearly independent 2 2 c + 1 tableaux. When s ≥ d + 6, we tableaux when d ≥ 8. We need b 6r c + 1 = b 5d−2s 6 have b 5d−2s c + 1 ≤ b s−d c − 2 + d − b s−2 c so these tableaux are enough. Since s ≥ d 5d e 6 2 2 3 and d ≥ 8, we have s ≥ d + 6. Thus for d ≥ 8, s < 2d − 1 these tableaux suffice When s = 2d − 1, d ≥ 8 we have b s−d c − 3 + d − b s−2 c = b d−1 c − 2. This is greater 2 2 2 than or equal to b 6r c + 1 = b d+2 c + 1 provided d > 8. When d = 8, s = 2d − 1 we 6 have r = 10and only one tableau is needed, which E0q provides. When d = 8 we also have r = 12. In that case two tableaux are needed, which E0q provides. Now consider d < 8 with s ≤ 2d − 1. For d ≤ 5, or d = 6 and r = 8 only one tableau is needed. Here E0q provides this tableau except if d = 3. When d = 3, then r = s which was done in Section 9.1.2. When d = 7 or d = 6 and r = 10 two tableaux are needed. In these cases, E0q suffices. Hence for d < 8 all tableaux are provided. 9.2.3 Basis Tableaux for c = 6, r < s Given λ = [r + s, s] a partition of n, we have 2s + r = 6d = 3b. From this equation and r < s, we have 0 ≤ r < 2d. For each λ we will construct mλ linearly independent λ-tableaux filled with the numbers 1 to 6. These will be our c = 6 basis tableaux. First consider 0 ≤ r ≤ d. If d is even, then s − d ≡ s (mod 2). Moreover, s − d ≥ r. Hence if λ0 = [r + s − d, s − d], then there are at least mλ0 linearly independent c = 4 basis tableaux and mλ0 = mλ since λ1 = λ2 = λ01 − λ02 . Therefore we can use U1 (d) ∨ Cp , were Cp are the c = 4 basis tableaux of shape λ0 , as the c = 6 basis tableaux. If d is odd, then s ≡ d − 1 (mod 2) and s − d + 1 ≥ r. Let Cp be the c = 4 basis tableaux of shape λ0 = [r + s − d − 1, s − d + 1]. Consider the tableaux U1 (d − 1) ∨ Cp . 175 There are mλ0 such tableaux. Now mλ = mλ0 when r 6≡ 0, 3 (mod 6) since then λ1 − λ2 − 2 = λ01 − λ02 . Thus for r 6≡ 0, 3 (mod 6) we have constructed sufficient tableaux. However, 2s + r = 6d implies r is even, hence only r ≡ 0 (mod 6) remains. In that case, mλ0 = mλ − 1, so only one additional tableau is needed. Since 0 ≤ r ≤ d, c ≤ s ≤ 3d. For r 6= 0 we can write s = 2d + f with b d2 c ≤ f ≤ d − 3. we have b 5d 2 Then d−2 T = d−1 f −1 4 63 5 3 6 4 1 11 2 2 3 3 ω(T ) = (d, d, f, 0, 0, 0) provides the additional tableau needed. Since T does not have a weight of d − 1 it must be linearly independent from U1 (d − 1) ∨ Cp by Lemma 3.4.12. When r = 0, then s = 3d. However, since d is odd, then so is s and hence no tableaux are required. Now consider d < r < 2d for arbitrary d. We have 2d < s ≤ b 5d c. Write s = 2d+f 2 for 1 ≤ f ≤ b d2 c. Consider d − p − 1 p + 1 d − 2p p p f Hp = 4 6 5 456 1 1 2 233 0 < p ≤ b d−f c 2 ω(Hp ) = (d, d − p, p + f ) Now Hp is non-zero and maximal. Moreover, for 0 < p ≤ b d−f c, Hp are linearly 2 independent by Lemma 3.4.12. Hence we have b d−f c tableaux. We need b 6r c + 1 = 2 c + 1. Hence it suffices to show that b d−f c > b d−f c. This holds except for b 2d−2f 6 2 3 d = 6, f = 3; d = 5, f = 2; and d = 4, f = 1. For d = 4 and 6, only one tableau is needed, so H1 suffices. When d = 5, f = 2, two tableaux are required. Use H1 and U1 (d − 1) ∨ U1 (d − 1) ∨ U1 (d − 1). Thus all the necessary c = 6 basis tableaux have been constructed. 9.2.4 Basis Tableaux for c > 6, r < s Let λ = [r + s, s] be a partition of n, with r < s, where 2s + r = cd = n. We want to construct mλ ≤ b 6r c + 1 linearly independent basis tableaux for an arbitrary c. 176 First write s = s0 + 2dq such that s0 − 2d < r ≤ s0 . If c − 4q ≥ 3 then a λ-tableau [r + s, s] may be written as T = qU2 (d − 1, 1) ∨ T 0 , where T 0 is a λ0 = [r + s0 , s0 ] tableau filled with c − 4q elements. Since r ≥ s0 in λ0 , we have mλ = mλ0 . Hence it suffices to construct mλ tableaux T 0 . If r = s0 , the tableaux T 0 were constructed in Section 9.1.4. Hence we only need to consider partitions λ = [r + s, s] with s − 2d < r < s and the case where c − 4q < 3. If c − 4q < 3 we may write T = (q − 1)U2 (d − 1, 1) ∨ T ∗ where T ∗ is a λ0 = [r + s0 + d, s0 + d] tableau filled with at most 6 elements. Since at least mλ = mλ0 tableaux T ∗ were constructed for c ≤ 6 in previous sections, no additional construction is needed for this case. To construct λ = [r + s, s] tableaux with s − 2d < r < s, write r = md + f , with 0 ≤ f < d and s = md + pd + g with 0 ≤ g < d. Then 3md + 2pd + f + 2g = cd, so f +2g = xd for some x. This means c = 3m+2p+x. If 2p+x ≥ 3, a λ-tableau may be written D ∨ F, where D is a tableau of shape [2md, md] filled with 3m elements and F is a [pd + g + f, pd + g] tableau filled with 2p + x elements. Given the constraints on r, we have 2p + x ≤ 6. Hence we can use the basis tableaux constructed in Sections 4.3, 9.2.1, 9.2.2, 9.2.3 as the F tableaux. We will first consider this case where 2p + x ≥ 3 and handle the 2p + x < 3 case later. We have r = md + f , so writing d = 6k 0 + i0 , f = 6k + i, with 0 ≤ i0 , i ≤ 5 0 gives r = 6(mk 0 + k) + mi0 + i. Since b mi6+i c ≤ m, it suffices to construct b 6s c + 1 ≤ mk 0 + k + m + 1 linearly independent tableaux. If m ≤ 1 we have c ≤ 9. For c ≤ 6 we have constructed the necessary basis tableaux in previous section. c = 7, 8 and 9 will be handled later. Hence assume m ≥ 2 Let Dp be the c = 3 basis tableaux of shape [2d, d] described in Section 4.3. There are m[2d,d] = k 0 + 1 such tableaux when i0 6= 1 and k 0 for i0 = 1. Let Fq be the c = 2p + x basis tableaux of shape [pd + g + f, pd + g] constructed in Sections 4.3, 9.2.1, 9.2.2, 9.2.3. At least one such tableaux will always exist, provided f 6= 0, 1, 2, 4. There are at least m[pd+g+f,pd+g] ≤ k 0 such tableaux, given the constraints on f . Take f 6= 0, 1, 2, 4 and d ≥ 6. The d < 6 and f cases will be handled separately. Consider the following tableaux forms (where the Ai were defined in Section 9.1.4). 177 ` = 1, 2, . . . , m I. `Dp ∨ (m − `)Dk0 ∨ Fq p = 0, 1, . . . , k 0 − 1 q = 1, 2, . . . , k II. mDk0 ∨ Fq q = 1, 2, . . . , k ` = 1, 2, . . . , b m2 c III. `Aj ∨ (m − 2`)Dk0 ∨ Fq q = 1, 2, . . . , k j = 1, 2, 3, 4 d ≥ 6 if j 6= 1 Table 9.3: Tableaux Forms. By Lemma 3.4.12, these tableaux are linearly independent provided their max weights are distinct. The max weights of these tableaux are: I. (4k 0 + 2p + i0 , 2(k 0 − p), 0)` ∨ (d, 0, 0)(m−`) ∨ ω(Fq ) II. (d, 0, 0)m ∨ ω(Fq ) III − 1. (d − 1, d − 1, 2, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ ω(Fq ) III − 2. (d − 1, d − 2, 3, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ ω(Fq ) III − 3. (d − 2, d − 2, 4, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ ω(Fq ) III − 4. (d − 2, d − 3, 5, 0, 0, 0)` ∨ (d, 0, 0)(m−2`) ∨ ω(Fq ) The weights of Fq depend on which basis tableaux we are using for F. We categorize these weights by the number of elements in F. 178 c ω(Fq ) Range Conditions 3 (4h + 2q + j, 2(h − q), 0) q = 1, . . . k 0 s = 6h + j 4 (d − q, d−f 2 + q, 0, 0) q = 0, . . . , 2b d+f 8 c 5 (d, 4h + 2q + j, 2(h − q), 0, 0) q = 1, . . . , k 0 s − d = 6h + j, r ≤ d, d even (d − 1, 4h + 2q + j, 2(h − q), 0, 0) q = 1, . . . , k 0 s − d + 1 = 6h + j, d odd (d − 3, 4h + j, 2h, 0, 0) r ≤ d.r > 9if r ≡ 3 (mod 6) (d − 1, d − 3, d−1 2 , 0, 0) r = 9, d ≥ 11, d odd (d, d − 1, d−7 2 , 0, 0) (d, d, 0, 0, 0) r = 9, d = 9 (d − 1, d − 3, d−1 2 , 0, 0) 6 (d, d − f2 − q − 1, q + 1, 0, 0) q = 1, . . . , b 2d−f 4 c−2 (d − q 0 + 1, d − f2 − 1 + q 0 , 0, 0, 0) q 0 = 1, . . . , b f4 c + 1 (d, d − q, d−f 2 + q, 0, 0, 0) q = 0, . . . , 2b d+f 8 c r ≤ d, d even (d − 1, d − q, d−f 2 + q + 1, 0, 0, 0) q = 0, . . . , 2b d+f 8 c r ≤ d, d odd (d, d, d−f 2 , 0, 0, 0) r>d r ≡ 0 (mod 6) (d − 1, d − q, d−f 2 + q + 1, 0, 0, 0) q = 0, . . . , 2b d+f 8 c r ≤ d, d odd (d, d − q, q + d−f 2 , 0, 0, 0) q = 1, . . . , b d+f 4 c r>d The weights of Forms I, II, III − j are distinct for each Fq listed provided d > 7, except for the following cases. When F is a c = 5 tableau, we have ω(I(` = 1, p = k 0 − 1, F1 )) = ω(III − 4(` = 1, F2 )) when d = 11, f = 9. When F is c = 4 or c = 6, we have some duplicate tableau weights if f = d − 4 or f = d − 2. These weights are ω(I(` = 1, p = k 0 − 1, q = j − 1)) = ω(III − j(` = 1, q = 0)) for j = 2, 3, and 4. Also, when f = d − 4, we have ω(III − 2(` = 1, q = 2)) = ω(III − 3(` = 1, q = 1)), ω(III − 2(` = 1, q = 3)) = ω(III − 4(` = 1, q = 1)), and ω(III − 3(` = 1, q = 3)) = ω(III − 4(` = 1, q = 2)). When f = d − 2 we have ω(III − 1(` = 1, q = 2)) = ω(III − 2(` = 1, q = 1)) as well. In these cases we have (at most 6) fewer linearly independent tableaux available than listed. 179 These will be called the constrained cases. The linear independence of the remaining tableaux can be seen by counting the number of d’s, 0’s in each tableau, determining the number of distinct elements in each tableau, and applying Lemma 3.4.13 where appropriate. When d = 6, the Forms III − 2 and III − 4 have the same weight. When d = 7 the Forms III − 3 and III − 4 have the same weight. Hence for d = 6 and 7, Form III − 4 will not be used. We wish to have b 6r c + 1 linearly independent tableaux. Recall that r = md + f , d = 6k 0 + i0 and f = 6k + i, so it suffices to construct mk 0 + k + m + 1 linearly independent tableaux. We have f 6= 0, 1, 2, 4, d ≥ 6 and m ≥ 2. Note that there is always at least one F since f 6= 0, 1, 2, 4, so we will take k ≥ 1. When d > 6, d 6≡ 1 (mod 6), all the tableaux listed in Table 9.2.4 exist and are linearly 0 independent. This provides mk 0 k + k + 4b m 2 ck, which is at least mk + k + m + 1. Hence sufficient linearly independent tableaux exist. In the constrained cases, using the full set of F listed (as opposed to only the first k 0 ) will provide sufficient tableaux. When d = 6, the tableaux of Forms III − 2 and III − 4 are not linearly independent. Hence we have mk+k+3b m 2 ck linearly independent tableaux. Then r = 6m+f , so m+k+1 tableaux are sufficient. Since we have k = 1, we have listed sufficient tableaux. When d ≡ 1 (mod 6) the tableau D0 does not exist. Hence Table 9.2.4 provides m(k 0 − 0 0 1)k+k+4b m 2 ck linearly independent tableaux, provided k > 1. In this case r = 6k m+m+f , so mk 0 + k + b m 6 c + 2 tableaux suffices. Thus we have enough tableaux unless m = 3, k = 1. In that case, specifically checking the number of tableaux needed and the number of Fq that exist, shows this construction is sufficient. When k 0 = 1, there are no tableaux of Form I and we do not use Form III − 4. Hence we have 1 + 3b m 2 c tableaux, which is sufficient except in the following cases, where an additional tableau is needed If m = 7, f = 5, c = 26 use 6U1 (6) ∨ U1 (4), if m = 3, f = 3, c = 12 use 5U1 (6), if m = 3, f = 5, c = 14 use 6U1 (6), if m = 3, f = 6, c = 13 use 4U1 (6) ∨ U4 (3, 2), and if m = 3, f = 6, c = 15 use 6U1 (6). In the constrained cases, using the full set of F listed (as opposed to only the first k 0 ) will provide sufficient tableaux. Hence for d ≥ 6, f 6= 0, 1, 2, 4, 2p + x ≥ 3, we have constructed the requisite number of linearly independent tableaux. We will consider the f = 0, 1, 2, 4 case after doing the 2p + x < 3 case. 180 When 2p + x < 3, the procedure described above require Fq to have fewer than three elements. In that case we use m − 1 in place of m in our construction and take Fq to be basis tableaux filled with 2p + x + 3 elements with shape [pd + 2d + g + f, pd + d + g]. There are at least k 0 + k such tableaux, though k may equal 0. Take d ≥ 6, 2p + x < 3, m ≥ 3 Hence from the tableaux of Table 9.2.4 we get Table 9.2.4. ` = 1, 2, . . . , m − 1 I. `Dp ∨ (m − ` − 1)Dk0 ∨ Fq p = 0, 1, . . . , k 0 − 1 q = 1, 2, . . . , k 0 + k II. (m − 1)Dk0 ∨ Fq q = 1, 2, . . . , k 0 + k c ` = 1, 2, . . . , b m−1 2 III. `Aj ∨ (m − 2` − 1)Dk0 ∨ Fq q = 1, 2, . . . , k 0 + k j = 1, 2, 3, 4 d ≥ 6 if j 6= 1 Table 9.4: Tableaux Forms As before, these tableaux are linearly independent, except in the constrained cases where the same weight equalities occur. We do not use Form III − 4 when d = 6 or 7. For the tableaux of Table 9.2.4, we require m ≥ 3 for tableaux of Form III to exist. Since we still have r = md + f , we want mk 0 + k + m + 1 linearly independent tableaux, when d 6≡ 1 (mod 6). When d > 6, d 6≡ 1 (mod 6), this construction provides (m − 1)k 0 (k + k 0 ) + k + 0 0 k 0 + 4b m−1 2 c(k + k ) linearly independent tableaux. This is larger than mk + k + m + 1, except when m = 4, k 0 = 1, k = 0. However, computing precisely how many Fq exist in this case and the minimum number of tableaux required, shows these tableaux suffice. In the constrained cases, using the full set of F listed (as opposed to only the first k 0 + k) will provide sufficient tableau. When d = 6, Forms III −2 and III −4 are the same so we have only (m−1)+1+3b m−1 2 c linearly independent tableaux. Since r = 6m + f , f < d, we need m + 1 tableaux, which we have. In the f = d − 4 and f = d − 2 cases using the full set of F listed will provide sufficient tableau. 181 If d ≡ 1 (mod 6), the tableau D0 does not exist. Hence Table 9.2.4 provides (m−1)(k 0 − 0 0 1)(k + k 0 ) + k + k 0 + 4b m−1 2 c(k + k ) linearly independent tableaux when k > 1. In this case r = 6m + m + f , so mk 0 + k + b m 6 c + 2 tableaux suffice, which we have. In the constrained cases, using the full set of F listed will provide sufficient tableau. When k 0 = 1, there are no tableaux of Form I or III − 4. Hence we have k + 1 + 3b m−1 2 c(k + 1). This is at least as large as mk 0 + k + b m 6 c + 2 unless m ≤ 8. However, computing precisely how many Fq exist in this case and the minimum number of tableaux required, shows these tableaux suffice in most cases. When m = 6, f = 3, c = 19 we need an additional tableau; use 6U1 (6) ∨ U1 (2). When m = 4, f = 3 we also need an additional tableau; use U4 (3, 2) ∨ 7U1 (6) ∨ U1 (4). In the constrained cases, using the full set of F listed will provide sufficient tableau. Thus for d ≥ 6, 2p + x < 3, m > 2, all necessary tableaux have been constructed. When 2p + x < 3 and m = 2, we find that c = 7 or c = 8. In these cases, consider the tableaux of Forms I and II in Table 9.2.4. We have r = 2d + f , so 2k 0 + k + 3 tableaux are sufficient. For d 6≡ 1 (mod 6), we have at least (k 0 + 1)(k 0 + k) linearly independent tableaux of Forms I and II. When this isn’t enough, checking precisely how many tableaux are needed and how many tableaux of Form I exist, shows that this construction is sufficient except for the following cases. When c = 8, d = 8, f = 2 three tableaux are needed; use 2U1 (8) ∨ U4 (6, 1), U1 (8) ∨ U3 (2, 3) ∨ U4 (6, 1), and U1 (8) ∨ U3 (2, 4) ∨ U4 (4, 1). When c = 8, d = 11, f = 2 five tableaux are needed. They are 4U1 (8), U2 (10, 1)∨U1 (10), U2 (10, 1)∨∨U1 (8)∨U1 (2), U2 (10, 1)∨U1 (6)∨ U1 (4), and U2 (10, 1) ∨ U2 (2, 3). When c = 7, if d = 8 and f = 2, three tableaux are needed, but Form I provides only two. In addition to those tableaux, use P3 (0, 3, 2) ∨ U1 (8) ∨ U1 (2). If c = 7, d = 9 and f = 1, three tableaux are needed, but Form I provides only two. In addition to those tableaux, use U1 (8) ∨ U1 (8) ∨ U1 (6). If d = 10 and f = 0, three tableaux are needed, but Form I provides only two. In addition to those tableaux, use P3 (0, 4, 2) ∨ U1 (10) ∨ U1 (4). When d ≡ 1 (mod 6), it is sufficient to construct 2k 0 + k + 2 tableaux. We have at least (k 0 )(k 0 + k) linearly independent tableaux of Forms I and II. When this is less than 2k 0 + k + 2, checking precisely how many tableaux are needed and how many tableaux of Forms I and II exist, shows that this construction is sufficient except for the following cases. When c = 8, d = 7, f = 4 we need three tableaux; use U2 (6, 1) ∨ U4 (4, 1), 2U1 (6) ∨ 182 U4 (6, 1), and U1 (6) ∨ U3 (2, 3) ∨ U4 (4, 1). When d = 7, f = 2 three tableaux are needed; use U2 (6, 1) ∨ U1 (6), U2 (6, 1) ∨ U1 (4) ∨ U1 (2), and 3U1 (6) ∨ U1 (2). When d = 7, and f = 0, two tableaux are needed, but Form II provides only one. In addition to that tableau, use P3 (0, 3, 2) ∨ 2U1 (6). For c = 7 if d = 13 and f = 1, five tableaux are needed, but Forms I and II provide only four. In addition to those tableaux, use U1 (12) ∨ U1 (12) ∨ U1 (8). If d = 7 and f = 1, three tableaux are needed, but Form I provides only two. In addition to those tableaux, use P3 (0, 3, 2) ∨ U1 (6) ∨ U1 (2). This completes the cases of c = 7 and c = 8 for d ≥ 6. Now consider c = 7 and c = 8 when d < 6. When c = 7, d = 5 we need two tableaux of shapes [23, 12] and [22, 13], and one tableau of shape [21, 14]. Use 3U1 (4) and U2 (4, 1) ∨ U1 (2), for shape [23, 12]; U2 (4, 1) ∨ U4 (2, 1) and U4 (3, 2) ∨ 2U1 (4) for shape [22, 13]; and U2 (4, 1) ∨ U1 (4) for shape [21, 14]. When c = 7an d = 4 we need two tableaux of shape [18, 10], 2U1 (4) ∨ U1 (2) and U2 (2, 1) ∨ U1 (4), and one tableau of shape [17, 11], U4 (2, 1) ∨ 2U1 (4). When c = 7 and d = 3, we need only one tableau of shape [13, 8], U2 (2, 1) ∨ U1 (2). When c = 8 and d = 5, three tableaux are needed for shape [26, 14], one tableau for shape [25, 15], and two tableaux for [24, 16]. Use 3U1 (4) ∨ U1 (2), U2 (4, 1) ∨ U1 (4), and U2 (2, 1) ∨ 2U1 (4) for shape [26, 14]. For shape [25, 15] use U2 (4, 1) ∨ U4 (3, 2), while for shape [24, 16] use 4U1 (4) and U2 (4, 1) ∨ U1 (4) ∨ U1 (2). When c = 8 and d = 4 we need one tableau for shapes [21, 11] and [19, 13] and two tableaux for shape [20, 12]. In the first case, use U4 (2, 1) ∨ U2 (3, 1) and U1 (4) ∨ U4 (2, 1) ∨ P4 (0, 2, 2), respectively. For [20, 12] use 3U1 (4) and 2U2 (2, 1). When c = 8 and d = 3 we need only one tableau of shape [15, 9], which is U2 (2, 1) ∨ U4 (2, 1). This completes the m = 2, 2p + x < 3 case. Now consider d ≥ 6, 2p + x ≥ 3, f = 0, 1, 2, 4. In this case we cannot guarantee that the tableaux Fq of shape [2dp + g + f, 2dp + g] exists. However, since d ≥ 6, tableaux F of shape [pd + d + g + f, pd + d + g] will exist. Moreover, since s < r + 2d, this shape is fillable with c ≤ 6 elements. Hence we can simply use the tableaux constructed in the 2p + x < 3 case with this F. Since we did not apply any restrictions of f in that case, those computations hold. 183 Now take d < 6. We have r = dm + f , f < d, s − 2d < r < s. Since d is small, only a few s are possible for each r. We will consider each case according to the value of d. If d = 3, we have r = 3m + f , s = 3m + 3 + f , and f < 3. First consider f = 0 or 2. If m = 2 we have the shapes [15, 9] and [19, 11]. In both cases, only one tableau is needed. Use U2 (2, 1) ∨ U4 (2, 1) and U2 (2, 1) ∨ U4 (2, 1) ∨ U1 (2). Now take m ≥ 3. Let F be the tableau of shape [9 + 2f, 6 + f ]. Since f = 0 or 2, one such F always exists. Then consider the tableaux of Form II and III − 1 in Table 9.2.4. They have weight (2, 2, 2, 0, 0, 0)` ∨ (3, 0, 0)m−2`−1 ∨ ω(F) for ` = 0, 1, . . . , b m−1 2 c and hence are linearly independent by Lemma 3.4.12. This 3m+2 construction provides b m−1 2 c + 1 tableaux. Since we need at most b 6 c + 1 tableaux for m odd and b 3m+2 6 c for m even, this suffices. When f = 1, let F = U2 (2, 1)∨U1 (2)∨U1 (2). If m = 2 or 3, one tableau will suffice. In those cases use F and F ∨ U4 (2, 1). When m ≥ 4 we m will use the tableaux `A1 ∨ (m − 2` − 2)U4 (2, 1) ∨ F for ` = 0, 1, . . . , b m−2 2 c. There are b 2 c such tableaux and they are linearly independent. Since only b m 2 c are needed, this suffices. When d = 4 we have r = 4m + f , s = 4m + h + f , and f < 3. Since s − 8 < r < s, the only possibilities are f = 0, h = 2, 4, 6 and f = 2, h = 1, 3, 5, 7. Let F be a tableau of shape [8 + 2f + h, 4 + f + h]. Since F needs at most 8 elements it has already been constructed. If m = 2, we need one tableau when f = 2, so F suffices. When f = 0, two tableaux are needed. Use U1 (4) ∨ U1 (4) ∨ U1 (2) and U2 (2, 1) ∨ U1 (4); U1 (4) ∨ U1 (4) ∨ U1 (4) and U2 (2, 1) ∨ U1 (4) ∨ U1 (2); or U1 (4) ∨ U1 (4) ∨ U1 (4) ∨ U1 (2) and U2 (2, 1) ∨ U1 (4) ∨ U1 (4); depending on h. For m ≥ 3, use the tableaux of Forms II and III − 1 of Table 9.2.4 along with those of form III − 1, with A5 (given in Section 9.1.4) in place of A1 . This provides 1 + 2b m−1 2 c linearly independent tableaux, which is sufficient. When d = 5 we have r = 5m + f , s = 5m + h + f , with f < 5, 0 < h < 10, and h ≡ f (mod 5). Let F be a tableau of shape [10 + h + 2f, 5 + h + f ]. When m ≥ 3, consider the tableaux of Forms II, III − 1, III − 4 of Table 9.2.4. This provides 1 + 2b m−1 2 c linearly independent tableaux. Computing precisely how many tableaux are needed for each f and h we find that this is sufficient except in the following cases. When f = 1, m = 4 we need one additional tableau. For h = 1 use 4U1 (4) ∨ U1 (2) and for h = 6 use 3U4 (3, 2) ∨ 3 ∨ U1 (4). When f = 2 and h = 2 we need an additional tableau for m = 8, 6, and 4. Use 11U1 (4), 8U1 (4) ∨ U1 (2), and 6U1 (4) respectively. When f = 2 and h = 7, this construction suffices. When f = 3 and h = 3, one additional tableau is needed for m = 6 and m = 4; use 9U1 (4) and 6U1 (4) ∨ U1 (2), respectively. When f = 3 and h = 8 we can take F to be a tableau of 184 shape [14, 11] and use the tableaux of Forms II, III − 1, and III − 4 of Table 9.2.4. This provides 1 + 2b m 2 c tableaux, which suffices except for m = 3. In that case three tableaux are needed; use U2 (4, 1) ∨ 3U1 (2), U2 (4, 1) ∨ 2U4 (2, 1), and U2 (4, 1) ∨ U1 (4) ∨ U1 (2). When f = 4 and h = 9 this construction suffices except when m = 4. In that case we need an additional tableau, so use 3U2 (4, 1) ∨ U4 (3, 2). However, when h = 9 the tableau F has 9 elements. Since we’ve only constructed the basis tableaux for c ≤ 8, use F = U2 (4, 1) ∨ 2U1 (4). When f = 4 and h = 4 we can take F to be a tableau of shape [12, 8] and use the tableaux of Forms II, III − 1, and III − 4 of Table 9.2.4. This provides 1 + 2b m 2 c tableaux, which suffices. Now consider m = 2 for d = 5. The tableaux will have c ≤ 9 elements unless f = 2, h = 7, f = 3, h = 8, or f = 4. The c ≤ 9 will be constructed later. If f = 2, h = 7 we need two tableaux, U2 (4, 1) ∨ U4 (3, 2) ∨ U1 (4) and U2 (4, 1) ∨ U4 (2, 1) ∨ P4 (0, 2, 2). If f = 3, h = 8, two tableaux are required. Let F be a basis tableaux of shape [14, 11]; use A1 ∨ F and A4 ∨ F. If f = 4, h = 9, two tableaux are needed, 2U2 (4, 1) ∨ U4 (2, 1) and U2 (4, 1) ∨ U4 (2, 1) ∨ P4 (0, 2, 2) ∨ U1 (4). If f = 4, h = 4 we need three tableaux. Let F be the basis tableau of shape [12, 8]. Then A4 ∨ F, A1 ∨ F and 2U4 (3, 2) ∨ F provide the requisite tableaux. Thus all necessary tableaux for d = 5 have been constructed. Now consider when m < 2 for arbitrary d. If r < d then since s − 2d < r < s we have c ≤ 6, which has been done. If r = d + f then we must have c = 7, 8 or 9. First consider when c = 7 with r = d + f . We get s = 3d − f2 . For d even use U1 (d) ∨ F where F are the c = 5 tableaux with r = d + f . When d is odd use U1 (d − 1) ∨ F where F are the c = 5 tableaux with r = d + f − 2. This is sufficient unless d ≡ 1 (mod 6), f ≡ 2 (mod 6), d ≡ 3 (mod 6), f ≡ 0 (mod 6), or d ≡ 5 (mod 6), f ≡ 4 (mod 6), in which case we need one additional tableau. For that, use U2 (d − 1, 1) ∨ B where B is a c = 3 basis tableau with s ≤ r. This construction holds for d > 3. When d = 3, only the shape [12, 9] is needed. One tableau, U2 (2, 1) ∨ U4 (2, 1), suffices. Thus the c = 7 case is complete. Now take c = 8. Since s − 2d < r, we have f > b d3 c and s = 3d + d−f 2 . For d even we can d−f use U1 (d) ∨ F where F are the c = 6 basis tableaux of shape [3d − d+f 2 , 2d + 2 ]. When d is d−f odd, use U1 (d − 1) ∨ F where F is the c = 6 tableaux of shape [3d − d+f 2 − 1, 2d + 2 + 1]. This construction suffices unless d + f ≡ 0 (mod 6). (Since d3 < f , d ≡ f (mod 2) and d ≥ 6, such an F always exists.) If d + f ≡ 0 (mod 6) use the tableau of Forms I and 185 d−f II of Table 9.2.4, with F the c = 5 tableaux of shape [2d + d−f 2 + f, 2d + 2 ]. This suffices unless d = 13, f = 11, d ≡ 1 (mod 6), f = 5, or d ≡ 5 (mod 6), f = 7. In the first case, five tableaux are needed; use 2U2 (9, 1), U2 (9, 1) ∨ U1 (12) ∨ U1 (8), 3U1 (12) ∨ U1 (4), 2U1 (12)∨U1 (10)∨U1 (6), U2 (12, 1)∨U1 (12)∨U1 (2). In the f = 5 case use the tableaux Bp ∨ F where Bp are the c = 3 basis tableaux of shape [2d + 1, d − 1] and F is the c = 5 d−f tableau with r = 3. This suffices for d−f 2 even. When 2 odd, an additional tableau, U2 (d − 1, 1) ∨ U1 (d − 1) ∨ U1 ( d−f 2 + 1), is needed. In the f = 7 case use the tableaux Bp ∨ F where Bp are the c = 3 basis tableaux of shape [2d − 1, d + 1] and F are the c = 5 tableaux with r = 9. Now consider when c = 9 with r = d + f . We get s = 4d − f2 . Since s − 2d < r < s we d have 2d 3 < f < 2 and hence this case does not occur. This completes the r < s case. Hence we have proven Theorem 3. 186 Chapter 10 Two Row Partitions and the Gaussian Polynomial Let n = ab with a, b ∈ N and take ` ∈ N such that 1 ≤ ` ≤ b n2 c. Let Pba (`) be the numbers of partitions of n having at most a parts each of size less that or equal to b, that is partitions of n fitting inside a b × a rectangle. Then Pba (`) is the co-efficient of q ` in the Gaussian ¤ ¤ £ £ polynomial a+b , as in [1]. The Gaussian polynomial, a+b , is also called the Gaussian b q b q co-efficient or the generalized q-binomial coefficient. Lemma 10.0.1. Take n = ab and λ = [n − `, `]. Let K = Sb o Sa . The multiplicity of χλ in 1SKn equals Pba (`) − Pba (` − 1). Proof. Let H = Sn−` × S` and H 0 = Sn−`+1 × S`−1 . Then χ[n−`,`] = 1SHn − 1SHn0 by the determinantal formula [14]. So h1SKn , χ[n−`,`] iSn = h1SKn , 1SHn iSn − h1SKn , 1SHn0 iSn . Hence it suffices to show h1SKn , 1SHn iSn = Pba (`). Now h1SKn , 1SHn iSn is the number of orbits of K acting on the cosets of H in Sn [11]. View the numbers 1 to n in blocks of size b, that is |1, 2, . . . , b|b + 1, . . . 2b| · · · |(a − 1)b + 1, . . . ab| The copies of H in Sn correspond to the different ways S` sits in Sn , that is subsets of {1, . . . n} of size `. Given such a subset L (corresponding to a copy of H) it will be broken into a parts by intersection with the blocks above. Let µi be the size of the part of L in the ith block. Since K acts by Sb on each of the blocks, L is equivalent (under K) to a subset L0 where the first µi numbers {i · b + 1, . . . i · b + µi } are chosen from block i (starting with the 0th block). Since K also has the wreath product action by Sa acting on the blocks, L0 187 is equivalent to the subset L∗ , where the blocks are reordered so the µi ≥ µi+1 . Hence L∗ corresponds to a partition of the number ` into a parts of size at most b and every such partition corresponds to a copy of H in Sn . So every such partition is contained in some orbit of K on Sn /H, and every orbit contains some such partition. Hence it suffices to shows that no two partitions are in the same orbit. Say µ = [µ0 , . . . µa−1 ] and ν = [ν0 , . . . νa−1 ] are partitions of ` where we allow 0 ≤ µi , νi ≤ b. If µ and ν are in the same orbit, then there exists g ∈ K such that g · {i · b + j|0 ≤ i ≤ a − 1, 1 ≤ j ≤ µi } = {i · b + j|0 ≤ i ≤ a − 1, 1 ≤ j ≤ νi }. So g(ib + j) = ki,j b + ci,j . Since g moves complete blocks, we must have ki,j = ki,j 0 for all 1 ≤ j, j 0 ≤ µi . As the action is injective, we must then have ci,j 6= ci,j 0 for j 6= j 0 . Hence looking at the image, we have µi = |{ci,j }| ≤ νki . Take µ ¥ ν, µ 6= ν. There exists i such that µi > νi and µi0 = νi0 for all i0 < i. Then there are i such νi0 with νi0 ≥ µi . But if g · µ = ν then µi = |{ci,j }| ≤ νki implies there are at least i + 1 such νi0 , which is a contradiction. Hence no orbit contains two such partitions, which finishes the proof. The ideas behind this proof are due to J. Saxl stemming from discussions of his paper [13]. Since the multiplicity of irreducibles in induced characters is non-negative [16], this lemma implies the well-known unimodality of the Gaussian coefficients [1]. Now Pab (`) = ¤ £ = [ a+b Pab (`), since a+b a ]q by taking conjugate partitions. Hence this lemma shows that b q Foulkes’ Conjecture always holds for two row partitions, which is discussed in [14]. We can also interpret our results on the generalized Foulkes’ Conjecture in terms of the Gaussian coefficient. From Theorem 1 we have: Theorem 12. If n = 2b = cd, with c, d ≥ 2, then for 1 ≤ ` ≤ b n2 c, Pdc (`) − Pdc (` − 1) ≥ Pb2 (`) − Pb2 (` − 1) Similarly, Theorem 3 gives: Theorem 13. If n = 3b = cd, with c, d ≥ 3, then for 1 ≤ ` ≤ b n2 c, Pdc (`) − Pdc (` − 1) ≥ Pb3 (`) − Pb3 (` − 1) 188 Hence our results give insight into the relationship between the rates of growth of different Gaussian coefficients. 189 Chapter 11 The Alternating Character Since Foulkes’ Conjecture is based on the trivial character, it is natural to ask whether the ideas hold for the alternating character. The first question is, what we mean by the alternating character in terms of induced modules. Consider, the ‘alternating’ character of the form ((−1)Sab ↓Sa oSb ) ↑Sab , that is the usual alternating character of Sab restricted to the subgroup Sa o Sb , which is induced back up to Sab . A brief computer check of Foulkes’ Conjecture using this character shows it holds for some small values of a and b. In fact, Foulkes’ Conjecture is equivalent to the following conjecture using the alternating character. Conjecture 3 (Foulkes’ Conjecture for Alternating Characters). If a ≤ b then every irreducible character occurring in ((−1)Sab ↓Sb oSa ) ↑Sab occurs in ((−1)Sab ↓Sa oSb ) ↑Sab with multiplicity greater than or equal to its multiplicity in ((−1)Sab ↓Sb oSa ) ↑Sab . Naturally, this conjecture also generalizes to: Conjecture 4 (Generalized Foulkes’ Conjecture for Alternating Characters). Given n = ab, a ≤ b, if c, d are such that cd = n, and c, d ≥ a, then every irreducible character occurring in ((−1)Sn ↓Sb oSa ) ↑Sn occurs in ((−1)Sn ↓Sd oSc ) ↑Sn with multiplicity at least as large. Showing the equivalences of Conjecture 1 or Conjecture 2, (Foulkes’ Conjecture for trivial characters), and Conjectures 3 or 4 (Foulkes’ Conjecture for alternating characters) is straightforward. We will assume Conjecture 1 (or Conjecture 2) holds and prove the alternating character version. The same argument shows the reverse equivalence. 190 Proof. First recall that if S is a subgroup of finite index in G, F an S-module and E a G G-module over a field, then there is an isomorphism IndG S (ResS (E) ⊗ F ) ≃ E ⊗ IndS (F ). (See Chapter XVIII §7 of [16].) Note that here we have used Ind for induction and Res for restriction of modules. Also, let G = Sn , S = Sa o Sb and T = Sb o Sa or Sd o Sc as appropriate. Let E be the G-module corresponding to the character (−1) on G. Since we are working over C, we will use C to denote the trivial module over any group. Sab and χ = The characters we’re comparing are χS = IndG T S (ResS (E)) =(−1)Sab ↓Sb oSa ↑ G G IndG T (ResT (E)). Then χS ≃ IndS (ResS (E) ⊗ C) ≃ E ⊗ IndS (C) by the isomorphism mentioned above. Similarly for χT . Switching notation back to characters, we get χS ≃ G (−1)G 1G S and χT ≃ (−1)G 1T . G Since we’ve assumed Foulkes’ Conjecture on trivial characters, we have 1G S = 1T + ψ for G G some character ψ. Then χS = (−1)G (1G T + ψ) = (−1)T 1T + (−1)G ψ. Hence χS ≤ χT as desired. Since we’ve proven Theorem 1 the argument above shows: Theorem 14. If 2 ≤ b then every irreducible character occurring in ((−1)S2b ↓Sb oS2 ) ↑S2b occurs in ((−1)S2b ↓S2 oSb ) ↑S2b with multiplicity greater than or equal to its multiplicity in ((−1)S2b ↓Sb oS2 ) ↑S2b . Similarly, Theorem 2 gives: Theorem 15. Given n = 3b, 3 ≤ b, if c, d are such that cd = n, and c, d ≥ 3, then every irreducible character occurring in ((−1)Sn ↓Sb oS3 ) ↑Sn occurs in ((−1)Sn ↓Sd oSc ) ↑Sn . While Theorem 3 shows: Theorem 16. Let n = 3b = cd, with c, d ≥ 3 and let λ = [λ1 , λ2 ] be a two row partition of n. Then every irreducible character χλ occurring in ((−1)Sn ↓Sb oS3 ) ↑Sn occurs in ((−1)Sn ↓Sd oSc ) ↑Sn with multiplicity at least as large. Given the success of replacing the trivial character in Foulkes’ Conjecture with this ‘alternating’ character, it is natural to investigate if other definitions of an alternating character yield similar results. One suggestion was to try (−1)SSab − (−1)SSab for an a oAb a oSb 191 induced alternating character in place of 1SSab in Foulkes’ Conjecture. Alas, a simple a oSb computer check via GAP [9] shows Foulkes’ Conjecture for this character fails when a = 3 and b = 4. Other variations on this character, such as (−1)SAab − (−1)SSab also fail at a oSb a oSb those values. 192 Chapter 12 Discussion of General Results Theorems 1, 2, and 3 extend the current research on Foulkes’ Conjecture. Although, the proof used combinatorial techniques on Young tableaux, the results correspondingly apply to areas such as Shur functions, Rational Homotopy Theory, and other means of interpreting Foulkes’ Conjecture. In addition we may interpret the Foulkes’ Conjecture using the alternating character via these theorems, which we discussed in Chapter 11. While the construction of the tableaux themselves are cumbersome, the development of the theory illustrates new approaches to Young tableaux. These concepts could be carried forth in contexts involving tableaux other than its usage here. Although the main theorems are specific to the cases a = 2 and a = 3, some general results arise from this study. The main theoretical techniques of this paper are that of weight-set counting in Theorem 4, the application of Theorem 8, and the use of maximality to show linear independence. The theory and technique of weight-set counting developed in this paper can be implemented in general, as can the concept of maximal form. While we only used tableaux with three or fewer rows, the theoretical foundations of weight-set counting have been laid for tableaux of an arbitrary number of rows. Although the computations are impractical for a random tableau, the counting works smoothly for tableaux with suitable symmetries, particularly those tableaux in maximal form. Moreover, the technique of weight-set counting is not dependent on a filling of content [ba ]. However, for non-uniform contents, one must watch carefully the action of Sa ; the weight-set counting may need to count all rows and the definition of maximality will need adjustment. Similarly, the usage of Theorem 8 in constructing larger tableaux will also work for other 193 contents and row quantities. The use of the Lemma 3.4.9, to show weight-set disjointness by maximality, however, has only been defined for three row partitions. It should be possible to generalize it for other partitions. Tableau maximality is a very useful concept for showing weight-set disjointness and applying Theorem 8. It is vital in proving linear independence of tableaux. Linear independence through tableau maximality should allow more progress on issues such as multiplicity. The methods of proving Theorem 2 could also apply to proving Conjecture 2 with other a’s, not including multiplicities. Unfortunately, the computations are likely to be somewhat cumbersome, especially the establishments of non-zero shapes as done by Theorem 9. However, should those parameters be established through other techniques, the tableaux constructed for Theorem 2 should provide nearly all the needed shapes with three or fewer rows, thus reducing the work substantially. Moreover, the reduction procedures will also apply. Specifically, given any n = ab = cd, if the shapes having multiplicity zero in Sb o Sa are bounded, then to prove the generalized Foulkes’ Conjecture for arbitrary c, d ≥ a, we should only need to prove it for a limited number of c’s. For instance, suppose a shape had multiplicity zero only if λi − λi+1 ≤ fi . (For n = 3b, we had f1 = f2 = 4.) Assume d is even (the odd case, though more cumbersome should follows analogously). Then given a tableau, we can ‘peel off’ a column block of size d with the appropriate row length, for instance, U1 (d) and P1 (d) are the two and three row versions. We can repeat this process so long as λi − λi+1 > d + fi (and there are at least as many elements as there are rows). Hence in the end, we need only construct a tableau with λi − λi+1 ≤ d + fi . This tableau will need at P most d1 i i(d + fi ) elements, hence c will be bounded by this number. This should imply, given the fi , if the generalized Foulkes’ Conjecture is true for c up to some bound, it is true for all c. (Presumably, if these tableaux exists, we can find versions with maximal/disjoint weight sets as needed.) The existence of the fi seems probable since Theorem 2 implies f1 , f2 ≤ 4 if 3|n. It may be the case, as in Theorem 1 that the parity of λi strongly effects the multiplicity. However, since the ‘peeling off’ does not change the parity, this process should still go through. In addition to this procedure, the investigations of Theorems 1 and 2 yield some general results. Take the character 1SSab and consider the irreducible Sλ corresponding to a oSb λ = [λ1 , λ2 ]. Then Sλ always has non-zero multiplicity whenever λ1 and λ2 are even. 194 Moreover, this multiplicity is zero whenever λ2 = 1 regardless of the choice of a and b. These theorems also have implications regarding the generalized Gaussian polynomial, as discussed in Chapter 10. Finally, the techniques within the proof of Theorem 3 should extend beyond two row tableaux. Specifically, Theorem 2 can probably be strengthened to include multiplicities for all partitions. Such a result should follow the ideas of Theorem 3, though sufficient linearly independent three row tableaux for c = 4, 5, and 6 must first be established. However, weight-set maximality should be sufficient to demonstrate linear independence. In all, these results provide a strong foundation for those wishing to study the representation theory of wreath products of symmetric groups via tableaux. 195 Appendix A Association between Tableaux Spaces and Irreducibles Let n = ab and H = Sb o Sa . Recall that Wλ,a = {T |T a λ-tableau filled with 1 to a, each b times}. We will explicitly show why the multiplicity of χλ in 1SHn equals the dimension of C{qT |T ∈ Wλ,a }. View H as a subgroup of Sn , where H acts on 1, 2, . . . b|b + 1, . . . , 2b| . . . |(a − 1)b + 1, . . . , ab by Sb on each block and by Sa permuting the blocks. The elements of H are the form (π1 , . . . πa , σ) with πi ∈ Sb , σ ∈ Sa . Now Sa × Sn acts on Wλ,a with Sa acting on the numbers 1 to a and Sn acting on the positions (corresponding to labelling across the rows). Let K = {(σ −1 , (π1 , . . . πa , σ))|πi ∈ Sb , σ ∈ Sa }. So K ≤ Sa × H ≤ Sa × Sn . Let T be the λ-tableau filled across the rows with b 1’s, then b 2’s, etc. Then Sa × Sn acting on T gives Wλ,a and K fixes T . Specifically, StabSa ×Sn (T ) = K. Hence as Sa × Sn modules, Sa ×Sn a ×Sn Wλ,a ≃ 1SStab(T . ) = 1K Proposition A.0.2. Wλ,a ≃ P Sn where ϕ (µ) is the irreducible of Sa µ`a ϕSa (µ) ⊗ (ϕH (µ)) Sa indexed by µ and ϕH (µ) is the irreducible of H/(Sb × . . . × Sb ) ≃ Sa indexed by µ. Proof. Since 1×(Sb ×. . .×Sb ) ≤ K, it is in the kernel of 1K . As Sb ×. . .×Sb £H, it is in the kernel of 1SKa ×H . So we can view 1SKa ×H as an Sa × H/(Sb × . . . × Sb ) ≃ Sa × Sa module. Let Sa ×H D = {(σ −1 , σ)|σ ∈ Sa } be the image of K in Sa × H/(Sb × . . . × Sb ). Hence 1K ≃ 1SDa ×Sa P Sa ×Sa as Sa × H/(Sb × . . . × Sb ) modules. Thus we can write 1D = aµ,ν φµ ⊗ φν for µ, ν ` a and some aµ,ν , where φ is the corresponding irreducible of Sa . 196 Sa ×Sa By Frobenius reciprocity aµ,ν = (φµ ⊗ φν , 1D ) = (φµ ⊗ φν |D , 1D )D . Now φµ ⊗ P a| 1 φν |D = φµ φν . So (φµ ⊗ φν |D , 1D )D = |D| φ (σ)φν (σ) = |S σ∈S |D| · (φµ , φν )Sa . Using row a µ    1 if µ = ν . orthogonality and |Sa | = |D|, we have aµ,ν =    0 otherwise P Sa ×Sa Sa ×H So 1D = we have µ`a φµ ⊗ φµ . If we lift back to the original module 1K P 1SKa ×H = µ φSa (µ) ⊗ φH (µ) where φSa (µ) is the irreducible of Sa indexed by µ and φH (µ) Sa ×Sn Sa ×H Sa ×Sn is the irreducible of H/(Sb × · · · × Sb ) ≃ Sa indexed by µ. Since 1K = (1K ) we get X X 1SKa ×Sn = ( φSa (µ) ⊗ φH (µ))Sa ×Sn = φSa (µ) ⊗ (φH (µ))Sn . µ By this proposition we have Wλ,a ≃ µ P Sn µ`a ϕSa (µ) ⊗ (ϕH (µ)) as Sa × Sn modules. Consider the submodule on which Sa is trivial, that is, µ = (a). This corresponds to P P 1Sa ⊗ (1H )Sn . If 1SHn = ν`n mν χν , this module corresponds to ν`n 1Sa ⊗ mν χν . Now P P eλ = σ∈RT τ ∈CT ²(τ )στ is an idempotent of Sn on λ-tableau T . So the action of eλ P P on 1 ⊗ mν χν is the same as the action of qλ = π∈Sa πeλ on Wλ,a . Then qλ · Wλ,a ≃ mλ (eλ ·Sλ ) as Sn modules, as eλ ·Sν = 0 for λ 6= ν. Now Sλ is is a cyclic Sn -module generated by eλ (T ). (Correspondingly, the semi-standard tableaux which span Sλ are equivalent under the action of Sn .) Therefore dim(eλ Sλ ) = 1 and dim(qλ Wλ ) = mλ . Hence {qT |T ∈ Wλ,a }, spans a module of dimension mλ , the multiplicity of χλ in 1SSnb oSa . This proof is due to Wales, [22]. 197 Bibliography [1] George E. Andrews, The theory of partitions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1998, Reprint of the 1976 original. MR 99c:11126 [2] S. C. Black and R. J. List, A note on plethysm, European J. Combin. 10 (1989), no. 1, 111–112. MR 89m:20011 [3] Emmanuel Briand, Polynômes multisymétriques, Ph. D. dissertation, University Rennes I, Rennes, France, October 2002. [4] Michel Brion, Stable properties of plethysm: on two conjectures of Foulkes, Manuscripta Math. 80 (1993), no. 4, 347–371. MR 95c:20056 [5] C. Coker, A problem related to Foulkes’s conjecture, Graphs Combin. 9 (1993), no. 2, 117–134. MR 94g:20019 [6] Suzie C. Dent and Johannes Siemons, On a conjecture of Foulkes, J. Algebra 226 (2000), no. 1, 236–249. MR 2001f:20026 [7] William F. 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Stanley, Positivity problems and conjectures in algebraic combinatorics, Mathematics: Frontiers and Perspectives (V. Arnold, M. Atiyah, P. Lax, and B. Mazur, eds.), American Mathematical Society, Providence, RI, 2000, pp. 295–319. [20] R. M. Thrall, On symmetrized Kronecker powers and the structure of the free Lie ring, Amer. J. Math. 64 (1942), 371–388. MR 3,262d [21] Rebecca Vessenes, Foulkes’ conjecture and tableaux construction, J. Albegra (2004), forthcoming. [22] David Wales, personal communication. 199 [23] Jie Wu, Foulkes conjecture in representation theory and its relations in rational homotopy theory, http://www.math.nus.edu.sg/~matwujie/Foulkes.pdf.  Generalized Foulkes’ Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 28, 2004) ii c 2004 ° Rebecca Vessenes All Rights Reserved iii To God through whom all things, even this, are possible iv Acknowledgements I wish to thank my advisor, David Wales, for his guidance and support. I also want to thank Cherie Galvez for her help and advice, and the math department staff in general for fostering a friendly environment. In addition, I appreciate the assistance of Rowan Killip in dealing with the intricacies of LATEX and creating specific macros for my use. I enjoyed my time here at Caltech, especially the many long and varied discussions during tea; my thanks to all those who participated. Finally, I am grateful for the love and support of my husband, Ted. I also want to recognize my undergraduate alma mater, the University of Chicago, for providing me with a strong background in mathematics. I particularly wish to thank Diane Herrmann for her support and advice during those years. In addition, I wish to recognize the University of Minnesota Talented Youth Mathematics Program (UMTYMP) for giving me a strong start in mathematics during my adolescent years. v Abstract Foulkes conjectured that for n = ab and a ≤ b, every irreducible module occurring as a constituent in 1SSnb oSa occurs with greater or equal multiplicity in 1SSna oSb . We generalize part of this to say those irreducibles also occur in 1SSnd oSc , where cd = n and c, d ≥ a. We prove the generalized conjecture for a = 2 and a = 3, by explicitly constructing the corresponding tableaux. We also prove the multiplicity constraint for certain cases. For these proofs we develop a theory of construction conditions for tableaux giving rise to Sb o Sa modules and in doing so, completely classify all such tableaux for a = 2 and a = 3. vi Contents Acknowledgements iv Abstract v 1 Introduction and Statement of Main Results 1 Theorem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Theorem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Theorem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 Background 4 2.1 Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.2 Combinatorial Structures . . . . . . . . . . . . . . . . . . . . . . . . . 5 3 Theory of Tableaux Construction 9 3.1 Filling Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3.2 Showing Tableaux are Non-Zero . . . . . . . . . . . . . . . . . . . . . 12 Example: Weight-set Counting . . . . . . . . . . . . . . . . . . . . . 17 3.3 Joining Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.4 Maximal Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 4 The Tableaux of 1SS3b b oS3 31 4.1 Classification of qT 6= 0, for T filled with 1, 2, 3 . . . . . . . . . . . . 31 4.2 The Irreducibles Partitions of 1SS3b . . . . . . . . . . . . . . . . . . . b oS3 41 4.2.1 Non-Zero Partitions from Theorem 9 . . . . . . . . . . . . . . 41 4.2.2 Partition Multiplicities according to Thrall . . . . . . . . . . . 44 vii 4.3 Construction of Basis Tableaux for c = 3 . . . . . . . . . . . . . . . . 45 5 Proof of Theorem 1 52 6 Proof of Theorem 2 58 6.1 6.2 Case: d even . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Example: Tableau Reduction . . . . . . . . . . . . . . . . . . . . . . 61 Case: d odd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 7 Tableaux Construction 7.1 71 Maximality of Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . 71 Example: Maximizing a Tableau . . . . . . . . . . . . . . . . . . . . . 72 7.2 Tableaux for Two Row Partitions . . . . . . . . . . . . . . . . . . . . 75 7.3 Tableaux for c = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 7.4 Tableaux for c = 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 7.5 Tableaux for c = 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 7.6 Tableaux for c = 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 7.7 Tableaux for c = 7 and c = 8 . . . . . . . . . . . . . . . . . . . . . . . 110 8 Tableau Sufficiency 111 8.1 Sufficiency when c = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . 111 8.2 Sufficiency when c = 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 114 8.3 Sufficiency for c = 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 8.4 Sufficiency for c = 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 8.5 Sufficiency for c > 6, d even . . . . . . . . . . . . . . . . . . . . . . . 137 8.6 Sufficiency for c > 6, d odd . . . . . . . . . . . . . . . . . . . . . . . . 138 8.7 Tableaux Disjointness . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 9 Proof of Theorem 3 9.1 142 Case: s ≤ r . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 9.1.1 Basis Tableaux for c = 4, s ≤ r . . . . . . . . . . . . . . . . . 143 9.1.2 Basis Tableaux for c = 5, s ≤ r . . . . . . . . . . . . . . . . . 145 viii 9.2 9.1.3 Basis Tableaux for c = 6, s ≤ r . . . . . . . . . . . . . . . . . 146 9.1.4 Basis Tableaux for c > 6, s ≤ r . . . . . . . . . . . . . . . . . 152 Case: r < s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 9.2.1 Basis Tableaux for c = 4, r < s . . . . . . . . . . . . . . . . . 170 9.2.2 Basis Tableaux for c = 5, r < s . . . . . . . . . . . . . . . . . 172 9.2.3 Basis Tableaux for c = 6, r < s . . . . . . . . . . . . . . . . . 174 9.2.4 Basis Tableaux for c > 6, r < s . . . . . . . . . . . . . . . . . 175 10 Two Row Partitions and the Gaussian Polynomial 186 11 The Alternating Character 189 12 Discussion of General Results 192 A Association between Tableaux Spaces and Irreducibles 195 Bibliography 197 ix List of Tables 8.1 Exceptional r cases for c = 3. . . . . . . . . . . . . . . . . . . . . . . . 112 8.2 Exceptional s cases for c = 3. . . . . . . . . . . . . . . . . . . . . . . . 113 8.3 General c = 3 cases. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 8.4 Exceptional r = 0 and r = 2 cases for c = 4. . . . . . . . . . . . . . . . 116 8.5 Exceptional r = 3 cases for c = 4. . . . . . . . . . . . . . . . . . . . . . 117 8.6 Exceptional r = 4 cases for c = 4. . . . . . . . . . . . . . . . . . . . . . 118 8.7 Exceptional s cases for c = 4. . . . . . . . . . . . . . . . . . . . . . . . 120 8.8 General c = 4 cases. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 8.9 Exceptional r = 0 and r = 2 cases for c = 5. . . . . . . . . . . . . . . . 123 8.10 Exceptional r = 3 and r = 4 cases for c = 5. . . . . . . . . . . . . . . . 129 8.11 Exceptional s cases for c = 5. . . . . . . . . . . . . . . . . . . . . . . . 130 8.12 General c = 5 cases for t > 2. . . . . . . . . . . . . . . . . . . . . . . . 131 8.13 General c = 5 cases for t = 0, 1, and 2. . . . . . . . . . . . . . . . . . . 132 8.14 Exceptional r = 0 and r = 2 cases for c = 6. . . . . . . . . . . . . . . . 133 8.15 Exceptional r = 3 and r = 4 cases for c = 6. . . . . . . . . . . . . . . . 134 8.16 Exceptional s cases for c = 6. . . . . . . . . . . . . . . . . . . . . . . . 134 8.17 General c = 6 cases for odd t. . . . . . . . . . . . . . . . . . . . . . . . 135 8.18 General c = 6 cases for d ≤ 5. . . . . . . . . . . . . . . . . . . . . . . . 135 8.19 General c = 6 cases for even t. . . . . . . . . . . . . . . . . . . . . . . . 136 8.20 Exceptional r and s cases for c = 8. . . . . . . . . . . . . . . . . . . . . 139 9.1 Weights of 0 or d. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 9.2 Weights of 0 and 4. . . . . . . . . . . . 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