On the Hecke Module of GLₙ(k[[z]])\GLₙ(k((z)))/GLₙ(k((z²))) Citation Jin, Yuhui (2024) On the Hecke Module of GLₙ(k[[z]])\GLₙ(k((z)))/GLₙ(k((z²))). Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/d0bn-5e47. https://resolver.caltech.edu/CaltechTHESIS:12082023-083025167 Abstract [See Abstract in text of thesis for correct representation of mathematics] Every double coset in GLₘ(k[[z]])\GLₘ(k((z)))/GLₘ(k((z²))) is uniquely represented by a block diagonal matrix with diagonal blocks in { 1,z, (11 z \\0 zⁱ \\) (i>1) } if char(k) ≠ 2 and k is a finite field. These cosets form a (spherical) Hecke module H(G,H,K) over the (spherical) Hecke algebra H(G,K) of double cosets in K\G/H, where K=GLₘ(k[[z]]) and H=GLₘ(k((z²))) and G=GLₘ(k((z))). Similarly to Hall polynomial hλ,ν^µ from the Hecke algebra H(G,K), coefficients hλ,ν^µ arise from the Hecke module. We will provide a closed formula for hλ,ν^µ, under some restrictions over λ, ν, µ. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: Hecke Module, $\text{GL}_m(k[[z]])\backslash \text{GL}_m(k((z)))/\text{GL}_m(k((z^2)))$, symmetric elliptic difference equation Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Rains, Eric M. Thesis Committee: Mantovan, Elena (chair) Conlon, David Huang, Jia Rains, Eric M. Defense Date: 27 November 2023 Non-Caltech Author Email: yuhuijin1995 (AT) gmail.com Record Number: CaltechTHESIS:12082023-083025167 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:12082023-083025167 DOI: 10.7907/d0bn-5e47 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 16257 Collection: CaltechTHESIS Deposited By: Yuhui Jin Deposited On: 17 Jan 2024 17:35 Last Modified: 17 Jan 2024 17:35 Thesis Files PDF - Final Version See Usage Policy. 460kB Repository Staff Only: item control page

On the Hecke Module of GL𝑛 (𝑘 [[𝑧]])\GL𝑛 (𝑘 ((𝑧)))/GL𝑛 (𝑘 ((𝑧2))) Thesis by Yuhui Jin In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy CALIFORNIA INSTITUTE OF TECHNOLOGY Pasadena, California 2024 Defended November 27, 2023 ii © 2024 Yuhui Jin All rights reserved iii ACKNOWLEDGEMENTS First and foremost I would like to thank my advisor, Eric Rains, for his patient guidance and enthusiastic support. This thesis would not be possible without his help. I would like to thank my committee for their useful advice and useful conversations. I would like to thank the math Department at Caltech for the excellent supportive environment. I would like to thank Jeffrey Remmel, my undergrad advisor, for introducing me into the fascinating field of algebraic combinatorics. Finally, I would like to thank my parents for their support and encouragement. iv ABSTRACT Every double coset in GL𝑚 (𝑘 [[𝑧]])\GL𝑚 (𝑘 ((𝑧)))/GL𝑚 (𝑘 ((𝑧2 ))) is uniquely rep! 1 𝑧 resented by a block diagonal matrix with diagonal blocks in {1, 𝑧, (𝑖 > 1)} 0 𝑧𝑖 if 𝑐ℎ𝑎𝑟 (𝑘) ≠ 2 and 𝑘 is a finite field. These cosets form a (spherical) Hecke module H (𝐺, 𝐻, 𝐾) over the (spherical) Hecke algebra H (𝐺, 𝐾) of double cosets in 𝐾\𝐺/𝐻, where 𝐾 = GL𝑚 (𝑘 [[𝑧]]) and 𝐻 = GL𝑚 (𝑘 ((𝑧2 ))) and 𝐺 = GL𝑚 (𝑘 ((𝑧))). 𝜇 Similarly to Hall polynomial ℎ𝜆,𝜈 from the Hecke algebra H (𝐺, 𝐾), coefficients 𝜇 𝜇 ℎ𝜆,𝜈 arise from the Hecke module. We will provide a closed formula for ℎ𝜆,𝜈 , under some restrictions over 𝜆, 𝜈, 𝜇. v TABLE OF CONTENTS Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter I: Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Symmetric Elliptic Difference Equation . . . . . . . . . . . . . . . . 1.2 Hecke Algebra and Hall Polynomials . . . . . . . . . . . . . . . . . 1.3 Outline of the Thesis . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Future Directions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter II: symmetric coweight . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Pieri and Dual Pieri Rule . . . . . . . . . . . . . . . . . . . . . . . . Chapter III: Hecke Module . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter IV: Dual Pieri . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii iv v 1 1 2 4 4 4 6 6 13 17 43 51 1 Chapter 1 INTRODUCTION 1.1 Symmetric Elliptic Difference Equation In [5, 6, 7], Rains studies a general elliptic difference equation of the form 𝑣(𝑞 + 𝑧) = 𝐴(𝑧)𝑣(𝑧) where 𝑞 is a point of an elliptic curve and 𝐴(𝑧) is a matrix of elliptic functions with det( 𝐴(𝑧)) not identically 0. Such equations arise, for instance, in studying elliptic analogues of ordinary and 𝑞-hypergeometric special functions. At the elliptic level, these functions tend to satisfy an additional symmetry of the form 𝑣(𝑧) = 𝑣(−𝑧), which leads the equation itself to satisfy an additional symmetry, namely that 𝐴(−𝑞 − 𝑧) = 𝐴(𝑧) −1 . To understand symmetries of such equations, it is important to understand how they behave locally, i.e., over the ring of formal power series. At a typical point 𝑧0 , the equation is regular if and only if 𝐴 and 𝐴−1 are both holomorphic at 𝑧0 , since then the spaces of solutions near 𝑧0 and near 𝑞 + 𝑧0 are naturally isomorphic. However, for symmetric equations, we care about the space of symmetric solutions, which means that the condition is more subtle. (For instance, the equation 𝑣(𝑞 + 𝑧) = −𝑣(𝑧) is symmetric, but any holomorphic solution satisfying 𝑣(−𝑧) = 𝑣(𝑧) will vanish at points with 2𝑧 = −𝑞.) We thus see that to understand local singularities in complete generality, we need to better understand matrices 𝐴 satisfying this symmetry. Two such matrices give the same local behavior at 𝑧0 (satisfying 2𝑧0 = −𝑞) if and only if they are related by an invertibly holomorphic change of basis at 𝑧0 , and thus we need to understand such matrices up to (twisted) conjugation by invertibly holomorphic matrices. Instead of the difference equation, it is natural to classify the matrices 𝐴. We can rephrase the problem in an abstractly geometric way. Given an elliptic curve 𝐶𝛼 over algebraically closed field 𝑘 and a translation 𝜏𝑞 : 𝐶𝛼 → 𝐶𝛼 and 𝜂 : 𝑧 → −𝑞 − 𝑧, the problem of classifying 𝐴 becomes classifying matrices GL𝑛 (𝑘 (𝐶𝛼 )) such that 𝜂∗ 𝐴 = 𝐴−1 . By Hilbert’s theorem 90, we have the following proposition. Proposition 1.1. [5, Proposition 2.1] Let 𝐿/𝐾 be a quadratic field extension, and let 𝐴 ∈ GL𝑛 (𝐿) be a matrix such that 𝐴¯ = 𝐴−1 , where .̄ is the conjugation of 𝐿 over ¯ −1 and 𝐵 is unique up 𝑘. Then there exists a matrix 𝐵 ∈ 𝐺 𝐿 𝑛 (𝐿) such that 𝐴 = 𝐵𝐵 to right multiplication by GL𝑛 (𝐾). 2 Thus, there exist a canonical factorization 𝐴 = 𝜂∗ 𝐵−𝑡 𝐵𝑡 where 𝐵 is an injective morphism 𝐵 : 𝜋𝜂∗𝑉 → O𝐶𝑛 𝛼 , with 𝑉 a rank n vector bundle over P1 , the projective space. Take any point in 𝑝 ∈ 𝐶𝛼 , 𝐵 is a matrix over the local ring. In [5, Chapter 8] , Rains first mentioned the following without proof: if 𝑝 is fixed by 𝜂, the invariants 𝜆(𝐵; 𝑝) is determined by the equivalence relation of left multiplication by invertible matrices over the local ring and right multiplication by symmetric matrices over ! 1 𝑧 the local field. That is, 𝐵 is a direct sum of matrices 1, 𝑧, and (𝑖 > 1). In 0 𝑧𝑖 Chapter 2, we will give a proof in a different way from how Rains observed it. In [7, Chapter 13.2], Rains constructs the solution sheaf with respect to a vector bundle over 𝐶𝛼 , a system of local condiitions, and 𝑞-connections over 𝐴 for an elliptic difference equation and uses the dominant coweight to study the multiplication of 𝐴(𝑞 𝑚−1 𝑧) 𝐴(𝑞 𝑚−2 𝑧)...𝐴(𝑧), in other word, the interactions of poles and zeros for shift matrices 𝐴. In the symmetric case, we will study the interactions of 𝐴 and 𝐵 in terms of poles and zeros. 1.2 Hecke Algebra and Hall Polynomials Let 𝑘 be a finite field of 𝑞 elements with characteristic 𝑐ℎ𝑎𝑟 (𝑘) ≠ 2. Let 𝑘 [[𝑧]] be the ring of formal power series. Let 𝑘 ((𝑧)) be the the field of fractions of the ring 𝑘 [[𝑧]] of formal power series. Let 𝐺 = GL𝑛 (𝑘 ((𝑧))) and let 𝐾 = GL𝑛 (𝑘 [[𝑧]]) be the maximal compact subgroup of 𝐺. Then 𝐺/𝐾 is the affine Grassmaniann. Theorem 1.1. [1, Chapter 9] Every double coset in 𝐾\𝐺/𝐾 has a unique representation of the form diag (𝑧𝜆1 , 𝑧𝜆2 , ..., 𝑧𝜆 𝑛 ), where 𝜆1 ≥ 𝜆 2 ≥ ... ≥ 𝜆 𝑛 . We define the partition (𝜆1 , 𝜆2 , ...𝜆 𝑛 ) to be the dominant coweight of all elements in the corresponding double coset and denote it by 𝜌. This theorem is known as the Cartan decomposition. Definition 1.1. For 𝑔 ∈ 𝐺, define gcd(g) to be the fractional ideal of 𝑘 [[𝑧]] generated by the entries in 𝑔. Let ∧𝑎 be the a-th exterior power representation; the matrix entries of ∧𝑎 (𝑔) are the a by a minors of 𝑔. Then 𝑔𝑐𝑑 (∧𝑎 𝑔) is the fractional ideal generated by these minors. The fractional ideal 𝑔𝑐𝑑 (∧𝑎 𝑔) is invariant under left and right multiplication by 𝐾. In the proof of Theorem 1.1, we have the following statement. For any 𝑔 ∈ 𝐺 with 𝜌(𝑔) = (𝜆 1 , 𝜆2 , ...𝜆 𝑛 ), we have 𝑔𝑐𝑑 (𝑔) = 𝑧𝜆𝑛 and 𝑔𝑐𝑑 (∧𝑖 𝑔) = Í𝑖 𝑧 𝑗=1 𝜆 𝑛− 𝑗+1 for all i. 3 The (spherical) Hecke algebra H (𝐺, 𝐾) is the convolution algebra of all complexvalued continuous compactly supported K-bi-invariant functions on G. Let 𝑐𝜆 be the characteristic function of 𝐾𝑧𝜆 𝐾. The Hecke algebra H (𝐺, 𝐾) has a basis given by Í {𝑐𝜆 }𝑙 (𝜆)=𝑛 . In fact 𝑐 𝜇 ∗ 𝑐 𝜈 = 𝜆 𝑔𝜆𝜇𝜈 (𝑞)𝑐𝜆 , where 𝑔𝜆𝜇𝜈 (𝑞) is the Hall polynomial. Proposition 1.2. [1, Proposition 37] (special case of Satake Isomorphism) Let 𝜃 𝑟 = 𝑞 −𝑟 (𝑛−𝑟)/2 𝑐 1𝑟 for all r. The ring structure of H (𝐺, 𝐾) is a polynomial ring over 𝜃 1 , 𝜃 2 , ..., 𝜃 𝑛 , 𝜃 𝑛−1 . H (𝐺, 𝐾)  C [𝜃 1 , 𝜃 2 , ..., 𝜃 𝑛 , 𝜃 𝑛−1 ] Proposition 1.3. [3, Chapter 5, Theorem 2.7] Let 𝐺 + = 𝐺 ∩ 𝑀𝑛 (𝑘 [[𝑧]]). Then, naturally we have H (𝐺, 𝐾) = H (𝐺 + , 𝐾) [𝑐−1 1𝑛 ]. + There is a C-algebra isomorphism from H (𝐺 , 𝐾) to the ring of symmetric polynomials in n variables with coefficients in C [𝑞 −1 ], i.e., 𝑓 : 𝑐𝜆 → 𝑞 −𝑛(𝜆) 𝑃𝜆 (𝑥 1 , ...𝑥 𝑛 ; 𝑞 −1 ). Here 𝑃𝜆 is the Hall-Littlewood polynomial. In 2007, Rains and Vazirani [8] developed affine Hecke algebra techniques to prove results in terms of vanishing integrals of Macdonald and Koornwinder polynomials. However, at q = 0 (the Hall–Littlewood level), these approaches do not work, although one can obtain the results by taking the appropriate limit. In [9], Venkateswaran developed a p-adic representation theory approach dealing with this special case. The Hall polynomial was historically studied by Hall, which is an important building block in the theory of special functions. MacDonald [3, Chapter 2] provides an extensive introduction of the Hall polynomial. Since the actual explicit formula is lengthy and involves LR-sequence of type (𝜇′, 𝜈′; 𝜆′), we will only mention some results that are used in this thesis: • 𝑔𝜆(1𝑟 )𝜈 (𝑞) = 0 unless 𝜆 − 𝜈 is a vertical stripe of length r. (Pieri rule) • 𝑔𝜆(𝑟)𝜈 (𝑞) = 0 unless 𝜆 − 𝜈 is a horizontal stripe of length r. (Dual Pieri rule) The product 𝑃𝜈 𝑃 𝜇 of Hall-Littlewood polynomials is a linear combination of the Í 𝑃𝜆 ’s for all 𝜆’s with |𝜆| = |𝜈| + |𝜇| : 𝑃𝜈 (𝑥; 𝑡)𝑃 𝜇 (𝑥; 𝑡) = 𝜆 𝑔𝜆𝜇𝜈 (𝑡)𝑃𝜆 (𝑥; 𝑡). 4 1.3 Outline of the Thesis In the second chapter, we give a proof of the decomposition of 𝐾\𝐺/𝐻, where 𝐺 = GL𝑛 (𝑘 ((𝑧))), 𝐻 = GL𝑛 (𝑘 ((𝑧2 ))), and 𝐾 = GL𝑛 (𝑘 [[𝑧]]), and the Pieri rule as well as the dual Pieri rule of the interaction between dominant coweight and symmetric coweight. In the third chapter, a Hecke module will be defined and we will compute the structural constant ℎ𝜆−1𝑟 𝜈 . In the fourth chapter we will compute the structural constant ℎ𝜆−𝑟𝜈 . 1.4 Future Directions Several questions can be raised as future directions. In terms of double coset structure 𝐾\𝐺/𝐻, if we replace the maximal compact subgroup K with an Iwahori subgroup, i.e. a conjugate of the inverse map of the subgroup of all upper triangular matrices in GL𝑛 (𝑘), are there any good coset representatives and a nice Hecke module that could be possibly extended to other types? In Chapter 3, we provide a formula for ℎ𝜆−1𝑟 𝜈 . Thus, a natural question to ask is there a pure combinatorial approach to compute such ℎ𝜆−1𝑟 𝜈 , since in this part, the building block of ℎ𝜆−1𝑟 𝑎 𝑛 is obtained from induction and the simplicity of the formula itself shines a light on a possible purely combinatorial proof. Furthermore, what is the complete formula for ℎ𝜆𝜇,𝜈 and, if possible, is there any construction we can make to give a one-toone correspondence from the Hecke module structure to the ring of polynomials (similarly as the correspondence from Hall-Littlewood polynomials to the Hecke algebra)? 1.5 Notation −1 and 𝑞-factorial [𝑛]! = [𝑛] [𝑛 − 1]....[1]. The 𝑞Recall the 𝑞-integer [𝑛] = 𝑞𝑞−1 [𝑛]! 𝑛
binomial is defined as 𝑘 = [𝑛−𝑘]![𝑘]! and the 𝑞-multinomial is defined as 𝑛   𝑛 [𝑛]! = . 𝑎 1 , 𝑎 2 , ...𝑎𝑖 [𝑎 1 ]![𝑎 2 ]!...[𝑎𝑖 ]![𝑛 − 𝑎 1 − 𝑎 2 − ... − 𝑎𝑖 ]!
For the simplicity of notation in Chapters 3 and 4, we denote 𝑎1 ,𝑎2𝑛,...𝑎𝑖 = 0 if any 𝑎 𝑗 is smaller than 0 and same for the 𝑞-binomial. In this thesis, ALL terms in the

form of 𝑛𝑘 , 𝑎1 ,𝑎2𝑛,...𝑎𝑖 are 𝑞-binomial/𝑞-multinomial. In this thesis, we will use English notation for Young Diagram. Notation on partitions: A partition is a sequence 𝜆 = (𝜆1 , 𝜆2 , ...) of integers in decreasing order (we allow negative integers) and containing finitely many nonzero terms. For the sake of simplicity, we may neglect the string of zeros when writing 5 explicitly the sequence. The definition of length for 𝜆 is also different from the tradition: Here 𝑙 (𝜆) is the number of all terms in 𝜆. Furthermore, we define Í 𝑚𝑖 (𝜆) = |{ 𝑗 : 𝜆 𝑗 = 𝑖}|, and 𝑛(𝜆) = 𝑖>0 (𝑖 − 1)𝜆𝑖 . The partition 𝜆′ is the conjugate of 𝜆. Notation for matrices: Let 1 (𝑖, 𝑗) ( 𝑓 ) with 𝑖 ≠ 𝑗 be a transvectional matrix: All entries in diagonal are 1 and the only nonzero off diagonal term are at row 𝑖 and column 𝑗 with entry 𝑓 . Let 1 (𝑖,𝑖) ( 𝑓 ) be the diagonal matrix: All entries in diagonal except for row 𝑖 is 1 and the (𝑖, 𝑖)-entry is 𝑓 . Notation for Laurent series: Since we require the base field to be finite, 𝑘 ((𝑧)) is a Í local field. Given 𝑓 = 𝑖≥𝑟 𝑎𝑖 𝑧𝑖 with 𝑎𝑟 ≠ 0, the valuation of 𝑓 , 𝑣( 𝑓 ), is 𝑟. Let 𝑓𝑜𝑑𝑑 be the odd part of 𝑓 and 𝑓𝑒𝑣𝑒𝑛 be the even part of 𝑓 , that is, 𝑓𝑒𝑣𝑒𝑛 + 𝑓𝑜𝑑𝑑 = 𝑓 and 𝑓𝑒𝑣𝑒𝑛 , 𝑓𝑜𝑑𝑑 /𝑧 ∈ 𝑘 ((𝑧2 )). 6 Chapter 2 SYMMETRIC COWEIGHT 2.1 Decomposition Theorem 2.1. Every double coset in GL𝑚 (𝑘 [[𝑧]])\GL𝑚 (𝑘 ((𝑧)))/GL𝑚 (𝑘 ((𝑧2 ))) is uniquely ! represented by block diagonal matrix whose diagonal blocks are 1, 𝑧, or 1 𝑧 , where 𝑖 > 1 and the base field k is finite and has 𝑐ℎ𝑎𝑟 (𝑘) ≠ 2 . 0 𝑧𝑖 Proof. By the Iwasawa decomposition, for any 𝑔 ∈ GL𝑚 (𝑘 ((𝑧))), there exists ℎ ∈ GL𝑚 (𝑘 [[𝑧]])) such that 𝑔ℎ is upper triangular. Before we start the actual proof of the theorem, we introduce 4 tricks that are used in the actual proof. We call left multiplication by matrices in GL𝑚 (𝑘 [[𝑧]]) and right multiplication by matrices in GL𝑚 (𝑘 ((𝑧2 ))) allowed operations . ! 𝑧 𝑏 𝑓𝑒 𝑧 𝑎 Trick 1: A submatrix of the form (the column with element 𝑧 𝑏 only has 0 𝑓1 ! 𝑏 𝑧𝑎 𝑧 one nonzero entry and 𝑓𝑒 ∈ 𝑘 ((𝑧2 )) , 𝑣( 𝑓𝑒 ) = 0) can be reduced to with 0 𝑓1 allowed operations. We left multiply by 1 (1,1) (1/ 𝑓𝑒 ), right multiply by 1 (1,1) ( 𝑓𝑒 ). Note that this may change the other entries in the same row as the top row of the submatrix. Trick 2: If 𝑑 ≤ 𝑎 and for the two pairs {𝑎, 𝑑} and {𝑏, 𝑐},! elements in each pair are 𝑧𝑏 0 𝑧𝑎 of different parities, a submatrix of the form (the column with 𝑧 𝑏 only 0 𝑧𝑐 𝑧𝑑 ! 𝑧𝑏 0 0 has one nonzero entry) can be reduced to with allowed operations. 0 𝑧𝑐 𝑧𝑑 We left multiply by 1 (1,2) (−𝑧 𝑎−𝑑 ), right multiply by 1 (1,2) (𝑧 𝑎−𝑑+𝑐−𝑏 ). Note that this may change the other entries in the same row as the top row of the submatrix. ! 1 𝑓 Trick 3: A submatrix of the form (the column with the 1 only has one 0 ℎ ! ! 1 𝑧 𝑣( 𝑓odd ) 1 0 nonzero entry) can be reduced to or where 0, 𝑣( 𝑓odd ) are of 0 ℎ 0 ℎ different parity with allowed operations. Denote 𝑓 = 𝑓odd + 𝑓even . We first right 7 multiply by 1 (1,2) (− 𝑓even ). If 𝑓odd = 0, then the second submatrix is achieved. Otherwise, by Trick 1, 𝑓odd reduces to 𝑧 𝑣( 𝑓odd ) . Note that this may change the other entries in the same row as the top row of the submatrix. If the (1,1) term is 𝑧, the trick is similar. ! 1 𝑧 𝑓 Trick 4: A submatrix of the form (two columns do not have other 0 𝑧𝑎 𝑔 ! 1 𝑧 0 nonzero entries in the matrix) can be reduced to with allowed 0 𝑧 𝑎 𝑔 − 𝑓odd 𝑧 𝑎−1 operations. We right multiply by 1 (1,3) (− 𝑓even ) 1 (2,3) (− 𝑓odd /𝑧). Given the above reductions, the proof proceeds by induction on 𝑚. Suppose that the claim holds for matrices of smaller dimension. In particular, the claim applies to the upper left 𝑚 − 1 × 𝑚 − 1 submatrix of 𝑋, which allows us to assume without loss of generality that submatrix is block diagonal with blocks of the desired form. We use trick 3,4 to modify entries in column 𝑚 of 𝑋: 𝑋𝑖,𝑚 is in the form of 𝑧 𝑎 if 𝑋𝑖,𝑖 corresponds to one-dimensional blocks (trick 3). 𝑋𝑖,𝑚 = 0 if 𝑋𝑖,𝑖 corresponds to the first row of a two dimensional part (trick 4). If there exists 𝑝 such that 𝑣(𝑋 𝑝,𝑚 ) ≥ 𝑣(𝑋𝑚,𝑚 ) and 𝑋 𝑝,𝑚 is nonzero, 𝑋 𝑝,𝑚 can be reduced to 0 by row operations on row 𝑚 and 𝑝. Notice that to prove the inductive statement, it suffices to reduce one nonzero entry 𝑋𝑖,𝑚 to be 0 or completely separate one block from the matrix. (Then by induction, matrices of smaller dimension are block diagonalizable.) Therefore, one more assumption on valuations of column m entries can be added: 𝑋𝑚,𝑚 has the largest valuation among all nonzero elements in column 𝑚. We permute rows and columns in the first 𝑚 − 1 by 𝑚 submatrix such that for all nonzero 𝑋𝑖,𝑚 , 𝑣(𝑋𝑖,𝑚 ) is in increasing order and the first 𝑚 − 1 by 𝑚 − 1 submatrix is still block diagonal. There will be two cases according to the permutation: Case 1 is the first block in a twodimensional block (i.e., an entry in column m with the least valuation corresponds to a 2-dimensional block); Case 2 is the first block in a one-dimensional block (i.e., an entry in column 𝑚 with the least valuation corresponds to a one-dimensional block). Case 1: A nonzero column 𝑚 entry with the least valuation corresponds to a two-dimensional block. We will show that allowed operations can make a onedimensional block completely decomposed from the matrix. 1 𝑧 0 © ª 𝑎 𝑖 ­ 1 The following is a submatrix of {1, 2, 𝑚} rows and columns: ­0 𝑧 𝑧 𝑓1 ®® . 𝑧𝑏 ¬ «0 0 8 Step 1: Make the second entry of column 𝑚 to be the only nonzero entry by row 1 𝑧 0 © ª 𝑎 𝑖 ­ 1 operations: ­0 𝑧 𝑧 𝑓1 ®® . 𝑏+𝑎−𝑖1 𝑓 −1 «0 −𝑧 1 0 ¬ Step 2: Multiply the second row by 𝑓1−1 and make the second entry of column two in the form 𝑧𝑥 𝑓𝑒 , where 𝑓𝑒 ∈ 𝑘 ((𝑧2 )) by column operations (column m only has one nonzero element). 1 𝑧 0 1 𝑧 0 © ª © ª ′ 𝑎 𝑓 −1 𝑖 1 ® → ­0 𝑎 𝑓 𝑖1 ® , ­0 𝑧 𝑧 𝑧 𝑧 𝑒3 ­ ® ­ ® 1 𝑎+𝑏−𝑖1 𝑓 −1 𝑎+𝑏−𝑖1 𝑓 −1 0 −𝑧 0 0 −𝑧 0 « ¬ « ¬ 1 1 ′ where 𝑧 𝑎 𝑓1−1 = 𝑧𝑖1 𝑓𝑒2 +𝑧 𝑎 𝑓𝑒3 and {𝑎′, 𝑖1 } have different parity, and 𝑓𝑒3 , 𝑓𝑒2 ∈ 𝑘 ((𝑧2 )), 𝑣( 𝑓𝑒3 ) = 0. If 𝑓𝑒3 = 0, 𝑧𝑖1 is a one-dimensional block in the submatrix. −1 , first row Step 3: Turn 𝑋2,2 to the form 𝑧 𝑥 . (We multiply the second column by 𝑓𝑒3 −1 ). by 𝑓𝑒3 , first column by 𝑓𝑒3 1 𝑧 0 © ª ′ 𝑎 𝑖1 ® ­0 𝑧 𝑧 ­ ® 𝑎+𝑏−𝑖 −1 −1 1 𝑓1 𝑓𝑒3 0 ¬ «0 −𝑧 Step 4: Make the only nonzero entry in the second row be 𝑋2,𝑚 . 1 𝑧 0 1 𝑧 0 © ′ ª © ª 𝑎 −1 𝑖 ­𝑧 0 𝑧 1 ®® → ­­0 0 𝑧𝑖1 ®® ­ −1 𝑎+𝑏−𝑖 1 𝑓 −1 𝑓 −1 −𝑧 𝑎+𝑏−𝑖1 𝑓1−1 𝑓𝑒3 0¬ 0¬ « 0 «0 −𝑧 1 𝑒3 From the inequality of 𝑎′ ≥ 𝑎 ≥ 2 and the fact that {𝑎′ − 1, 𝑖1 } are of same parity, ′ trick 2 removes element 𝑧 𝑎 −1 . Thus, 𝑧𝑖1 is a one-dimensional block. We can reduce it to either 1 or 𝑧. Case 2: A nonzero column 𝑚 entry with the least valuation corresponds to a onedimensional block. If there exists at least 2 one-dimensional blocks in the 𝑚 − 1 by 𝑚 − 1 submatrix, assume that one of them is in the ℓ-th row. Then 𝑋ℓ,𝑚 reduces to 0 by trick 2. Therefore, 𝑋ℓ,ℓ is the only nonzero entry in column or row ℓ in the matrix. Thus, for this case, it remains to show that if all blocks from row 2 to 𝑚 − 1 are 2-dimensional, we can still split one block out of the matrix. (If there is no two-dimensional block, which is equivalent to m=2, the 2 by 2 matrix is a block.) We will have two major steps for this part: 9 1. We simplify our matrix into the following form: on column 𝑚 of 𝑋, 𝑋2𝑖,𝑚 = 0 (𝑋2𝑖,2𝑖 corresponds to the first row of a two-dimensional block), 𝑋2𝑖+1,𝑚 is in the form of 𝑧 𝑎 ( 𝑋2𝑖+1,2𝑖+1 corresponds to the second row of a two-dimensional block), and 𝑋1,𝑚 is still in the form of 𝑧 𝑎 . 𝑧 𝑎1,1 0 0 © ­ 0 𝑧 𝑎2,1 𝑧 𝑎2,1 ­ ­ 0 𝑧 𝑎3,1 ­ 0 ­ ­ ··· ­ 0 0 « 0 0 · · · 𝑧 𝑎1,2 ª 0 · · · 0 ®® ® 0 · · · 𝑧 𝑎3,2 ® ® ® ® 𝑎 𝑛,1 0 ··· 𝑧 ¬ 2. We completely separate one block from the matrix. Part 1: Step 1: Write 𝑋3,𝑚 = 𝑧 𝑎 𝑓1 . If 𝑓1 ∈ 𝑘 [[𝑧2 ]], then, by trick 1, we have 𝑋3,𝑚 go to 𝑧 𝑎 . If 𝑓1 ∉ 𝑘 [[𝑧2 ]], multiplying 𝑓1−1 to the third row will take 𝑋3,𝑚 to 𝑧 𝑎 and 𝑋3,3 to 𝑧 𝑘 𝑓1−1 . (The following matrix illustration is the submatrix of {1, 2, 3, 𝑚} rows and columns.) 𝑤 0 𝑧𝑖 ª 0 0 𝑧𝑖 ª ©𝑧 0 ­0 1 𝑧 0 ®® 1 𝑧 0 ®® ­ → ­ ® ® ­ 0 0 𝑧 𝑘 𝑓 −1 𝑧 𝑎 ® 0 𝑧 𝑘 𝑧 𝑎 𝑓1 ®® 1 ­ ® 𝑙 0 0 0 𝑧 0 0 𝑧𝑙 ¬ « ¬ Step 2: Row operations on row 2,3 can split 𝑋3,3 into even and odd parts to 𝑋3,2 , 𝑋3,3 ′ ′ (we have 𝑧 𝑘 𝑓1−1 = 𝑧 𝑘 𝑓𝑒1 −𝑧 𝑎 𝑓𝑒2 , with 𝑣( 𝑓𝑒1 ) = 𝑣( 𝑓𝑒2 ) = 0 and {𝑎′, 𝑎} are of different ′ ′ −1 , parity), i.e., 𝑋3,3 → −𝑧 𝑎 𝑓𝑒2 , 𝑋3,2 → −𝑧 𝑘 𝑓𝑒1 . Then, we multiply column 3 by 𝑓𝑒2 −1 . That is, 𝑓 −1 is in 𝑋 (the matrix multiply row 2 by 𝑓𝑒2 , multiply column 2 by 𝑓𝑒2 2,2 𝑒2 ′ ′ on the left). Notice 2|𝑘 − 1 + 𝑎 . Thus, by column operation on column 2,3 we eliminate 𝑋3,2 (the matrix on the right). 𝑤 ©𝑧 ­0 ­ ­ ­0 ­ «0 𝑤 ©𝑧 ­0 ­ ­ ­0 ­ «0 𝑤 0 0 𝑧𝑖 ª 0 0 𝑧𝑖 ª ©𝑧 ­ 0 1 − 𝑧 𝑘 ′−𝑎 ′ 𝑓 𝑓 −1 1 𝑧 0 ®® 𝑧 0 ®® ­ 𝑒1 𝑒2 ®→­ ® ′ ′ −1 −𝑧 𝑎 ′ 𝑧 𝑎 ® ­0 −𝑧 𝑘 −1 𝑓𝑒1 𝑓𝑒2 0 −𝑧 𝑎 𝑧 𝑎 ®® ® ­ 0 0 𝑧𝑙 ¬ 0 0 𝑧𝑙 ¬ «0 ′ ′ −1 (𝑣( 𝑓 ) = 0). Multiplying row 2 by Step 3: We write 𝑧𝑢 𝑓2 = 1 − 𝑧 𝑘 −𝑎 𝑓𝑒1 𝑓𝑒2 2 −1 𝑓𝑜2 , column operations on column 2,3 will make 𝑋2,3 /𝑧 𝑣(𝑋2,3 ) ∈ 𝑘 ((𝑧2 )). We ′ ′′ write 𝑧 𝑓2−1 = 𝑧𝑢 𝑓𝑒𝑢1 + 𝑧𝑢 𝑓𝑒𝑢2 where 2|𝑢 − 𝑢′ and 𝑢′, 𝑢′′are of different parity and 𝑣( 𝑓𝑒𝑢2 ) = 0. By column operations on columns 2,3, 𝑋2,3 would be reduced to ′′ 𝑧𝑢 𝑓𝑒𝑢2 . Next by trick 1, the term 𝑓𝑒𝑢2 and minus sign are eliminated. 10 𝑤 0 𝑤 0 0 0 𝑧𝑖 ª 0 𝑧𝑖 ª 0 𝑧𝑖 ª ©𝑧 ©𝑧 ′′ ­ 0 𝑧𝑢 𝑧𝑢 𝑓 ® ­ 0 𝑧𝑢 𝑧𝑢 ′′ 0 ® −1 0 ® 𝑧𝑢 𝑧 𝑓𝑒2 ® ­ ­ ® 𝑒𝑢2 0 ® ®→­ ®→­ ® ′ ′ ′ 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 ® ­ ® ­ ® 0 −𝑧 𝑧 ® 0 0 −𝑧 𝑧 0 0 𝑧 𝑧 ­ ® ­ ® 𝑙 𝑙 0 0 𝑧𝑙 ¬ 0 0 0 𝑧 0 0 0 𝑧 « ¬ « ¬ Now, redoing steps 1-4 for all two-dimensional blocks such that all nonzero entries in column 𝑚 are in the form of 𝑧 𝑎 would finish part 1. 𝑤 ©𝑧 ­0 ­ ­ ­0 ­ «0 Part 2. Taking any nonzero elements to be 0 will break the matrix into smaller blocks. Recall that there are two semi-block forms from step 1 above: ! ! ′′ 𝑧𝑢 𝑧𝑢 · · · 0 1 𝑧 ··· 0 and . ′ 0 𝑧𝑎 · · · 𝑧𝑎 0 𝑧 𝑘 · · · 𝑧𝑎 The former is the semi-block obtained from part 1 and the latter is the case where 𝑓1 ∈ 𝑘 [[𝑧2 ]]. The former has a restriction: For the two pairs {𝑢, 𝑢′′ } and {𝑎, 𝑎′ }, elements in each pair are of different parities. The latter has a restriction: 𝑘 > 1. ! 𝑢 𝑧 0 · · · 0 If 𝑢′′ ≥ 𝑎′, trick 2 takes the former semi-block to . ′ 0 𝑧𝑎 · · · 𝑧𝑎 ! ! ′′ ′′ 𝑧𝑢 𝑧𝑢 · · · 0 𝑧𝑢 𝑧𝑢 ··· 0 For , we rewrite it as . Similarly for ′ ′ ′′ 0 𝑧𝑎 · · · 𝑧𝑎 0 𝑧 𝑎 +𝑢−𝑢 · · · 𝑧 𝑎 ! ! 1 𝑧 ··· 0 𝑧 1 ··· 0 , if 𝑘, 𝑎 are of different parity, we rewrite it as . 0 𝑧 𝑘 · · · 𝑧𝑎 0 𝑧 𝑘−1 · · · 𝑧 𝑎 ! ′ 𝑧𝑤 𝑧𝑤 · · · 0 Now, two semi-blocks are both of the form ′ , where for the two 0 𝑧𝑏 · · · 𝑧𝑏 pairs {𝑤, 𝑤′ } and {𝑏, 𝑏′ + 1}, elements in each pair are of different parities and 𝑤 ′ < 𝑏. We claim that on the submatrix of {1, 2, 3, 𝑚} rows and columns, by those allowed operations, we can extract one two-dimensional block from the matrix and obtain 𝑎 1,1 0 0 𝑧 𝑎1,2 ª ©𝑧 ­ 0 𝑧 𝑎2,1 𝑧 𝑎2,2 0 ®® ­ ­ ® . Let condition (A′) denote 𝑎 3,1 − 𝑎 2,2 ≤ 𝑎 3,2 − 𝑎 1,2 , and 𝑎 3,1 𝑧 𝑎 3,2 ® ­ 0 0 𝑧 ­ ® 𝑎 𝑛,1 0 0 0 𝑧 « ¬ condition (B′) denote 𝑎 3,1 − 𝑎 2,2 > 𝑎 3,2 − 𝑎 1,2 . 𝑎 1,1 ©𝑧 ­ 0 ­ ′ A: ­ ­ 0 ­ « 0 0 0 𝑧 𝑎2,1 𝑧 𝑎2,2 0 𝑧 𝑎3,1 0 0 𝑎 1,1 𝑧 𝑎1,2 ª 0 0 𝑧 𝑎1,2 ©𝑧 ª ­ 0 𝑧 𝑎2,1 𝑧 𝑎2,2 𝑧 𝑎3,2 +𝑎2,2 −𝑎3,1 ® 0 ®® ® ­ ®→­ ® 𝑎 𝑎 3,2 3,1 ­ 0 ® 𝑧 ®® 0 𝑧 0 ­ ® 𝑎 𝑛,1 𝑧 𝑎 𝑛,1 ¬ 0 0 0 𝑧 « ¬ 11 𝑎 1,1 𝑧 𝑎1,1 0 0 𝑧 𝑎1,2 ª 0 0 𝑧 𝑎1,2 ª © ©𝑧 ­𝑧 𝑎3,2 +𝑎2,2 −𝑎3,1 +𝑎1,1 −𝑎1,2 𝑧 𝑎2,1 𝑧 𝑎2,2 ­ 0 𝑧 𝑎2,1 𝑧 𝑎2,2 0 ®® 0 ®® ­ ­ →­ → ® ­ ® 𝑎 3,1 𝑎 3,1 ­ ® ­ 0 ® 0 0 𝑧 0 0 𝑧 0 ­ ® ­ ® 𝑎 𝑛,1 𝑎 𝑛,1 0 0 0 𝑧 0 0 0 𝑧 « ¬ « ¬ ′ ′ Condition B is more subtle than A : First we do row operations from row 3 to 5, 7, . . . , and 𝑚 to remove all entries in 𝑋2𝑘+1,𝑚 for 𝑘 > 1 and 𝑋𝑚,𝑚 . That is, terms on column 𝑚 are transferred to column 3. In the following submatrices we rewrite 𝑎 𝑛,1 + 𝑎 3,1 − 𝑎 3,2 to be 𝑎′𝑛,1 . 0 𝑎 1,1 ©𝑧 ­ 0 ­ ­ ­ 0 ­ « 0 0 𝑧 𝑎2,1 𝑧 𝑎2,2 0 𝑧 𝑎3,1 ′ 0 𝑧 𝑎 𝑛,1 𝑎 1,1 0 𝑧 𝑎3,1 −𝑎3,2 +𝑎1,2 𝑧 𝑎1,2 ª 𝑧 𝑎1,2 ª ©𝑧 ­ 0 𝑧 𝑎2,1 𝑧 𝑎2,2 0 ®® 0 ®® ­ ® ®→­ ­ 0 0 0 𝑧 𝑎3,2 ®® 𝑧 𝑎3,2 ®® ­ ′ 𝑎 𝑛,1 0 0 𝑧 0 ¬ 0 ¬ « 𝑎 1,1 𝑧 𝑎3,1 −𝑎3,2 +𝑎1,2 +𝑎2,1 −𝑎2,2 0 𝑧 𝑎1,2 ª 0 0 𝑧 𝑎1,2 ª ©𝑧 ­ 0 𝑧 𝑎2,1 𝑧 𝑎2,2 𝑧 𝑎2,1 𝑧 𝑎2,2 0 ®® 0 ®® ­ → ® ­ ® 𝑎 3,2 ® ­ 0 0 0 𝑧 𝑎3,2 ®® 0 0 𝑧 ­ ® ′ ′ 𝑎 𝑛,1 0 𝑧 𝑎 𝑛,1 0 ¬ 0 0 𝑧 0 « ¬ Now rows 1,3 and columns 1,4 form a two-dimensional block in the matrix (column 3 still holds all the original column 𝑚 information). 𝑎 1,1 ©𝑧 ­ 0 ­ →­ ­ 0 ­ « 0 The remaining of this proof is to show such blockwise decomposition is unique. Given 𝐵(𝑧) ∈ GL𝑛 (𝑘 ((𝑧))), we write a decomposition of 𝐵 as 𝐵(𝑧) = 𝑔Λℎ, where ˜ ˜ For 𝑔 ∈ GL𝑛 (𝑘![[𝑧]]). ℎ ∈ GL𝑛 (𝑘 ((𝑧 2 ))). Denote! 𝐵(𝑧) = 𝐵−1 (−𝑧) and 𝐵ˆ = 𝐵 𝐵. −2 1 (−𝑧) 𝑖−1 1 𝑧 ˆ is ˆ = {𝑖−1, 1−𝑖}. For 𝐶 = 1, and 𝜌(𝐶) 𝐶= , the corresponding 𝐶 0 𝑧𝑖 0 (−1) 𝑖 ˆ = {0}. For 𝐶 = 𝑧, 𝐶ˆ = −1, 𝜌(𝐶) ˆ = {0}. By Cartan decomposition, 𝐶ˆ = 1, 𝜌(𝐶) ˆ is unique. 𝐵 = 𝑔Λℎ = 𝑔′Λ′ ℎ′, 𝐵ˆ = 𝑔ΛΛ̃𝑔˜ = 𝑔′Λ′Λ̃′𝑔˜ ′. By abuse of notation, 𝜌( 𝐵) ! 1 𝑧 we write 𝑧𝑖 = . For Λ = ⊕𝑖 𝑧 𝑐𝑖 , we denote 𝜎(Λ) = {𝑐 1 , ...𝑐 𝑗 }. Here a block 0 𝑧𝑖 diagonal matrix Λ has 𝑧 𝑐𝑖 on the diagonal in the order of valuation non-decreasing. 𝑛′ 𝑛′ We rewrite 𝜎(Λ) = {𝑎 1𝑛1 , 𝑎 2𝑛2 , ..., 1𝑙1 , 0𝑟1 }, 𝜎(Λ′) = {𝑏 11 , 𝑏 22 , ..., 1𝑙2 , 0𝑟2 }. Therefore ˆ = 𝜌(ΛΛ̃) = {(𝑎 1 −1) 𝑛1 , (𝑎 2 −1) 𝑛2 , ..., 0𝑙1 +𝑟1 , (1−𝑎 2 ) 𝑛2 , (1−𝑎 1 ) 𝑛1 } = 𝜌(Λ′Λ̃′). 𝜌( 𝐵) Thus we have these equalities: 𝑎𝑖 = 𝑏𝑖 , 𝑛𝑖 = 𝑛𝑖 for all i, and 𝑙1 + 𝑟 1 = 𝑙 2 + 𝑟 2 . It suffices to show 𝑙1 = 𝑙2 , 𝑟 1 = 𝑟 2 for this part. We prove this part by contradiction. Without loss of generality, 𝑙1 > 𝑙2 . We rewrite 𝑔ΛΛ̃𝑔˜ = 𝑔′Λ′Λ̃′𝑔˜ ′ as 𝑔0 ΛΛ̃ = Λ′Λ̃′𝑔0 (−𝑧). 12 We write the (𝑖, 𝑗)-th entry in the matrix 𝑔0 in the form of 𝑎𝑖, 𝑗 = 𝑎𝑖, 𝑗,𝑒 + 𝑎𝑖, 𝑗,𝑜 , where 𝑎𝑖, 𝑗,𝑜 is the odd part and 𝑎𝑖, 𝑗,𝑒 is the even part. By direct computation of the first 𝑙1 + 𝑟 1 by 𝑙1 + 𝑟 1 submatrix of LHS-RHS (i.e.,𝑔0 ΛΛ̃ − Λ′Λ̃′𝑔0 (−𝑧)), we have 𝑎 𝑎 1,2,𝑜 ··· 𝑎 1,𝑟1 +1,𝑒 ··· 𝑎 1,𝑙1 +𝑟1 ,𝑒 © 1,1,𝑜 ª ­ 𝑎 2,1,𝑜 ® 𝑎 · · · 𝑎 · · · 𝑎 2,2,𝑜 2,𝑟 +1,𝑒 2,𝑙 +𝑟 ,𝑒 1 1 1 ® ­ ­ ® ­ ... ® = 0. ­ ® ­−𝑎𝑟 +1,1,𝑒 −𝑎𝑟 +1,2,𝑒 · · · −𝑎𝑟 +1,𝑟 +1,𝑜 · · · −𝑎𝑟 +1,𝑙 +𝑟 ,𝑜 ® 2 2 1 2 1 1 ® ­ 2 ... « ¬ By direct computation on submatrix with rows {𝑙 1 +𝑟 1 , 𝑙1 +𝑟 1 +1, ..., 𝑛} and columns {1, 2, ..., 𝑙 1 + 𝑟 1 } of LHS-RHS, we have 𝑎 𝑙1 +𝑟1 +2𝑘,𝑙 = 𝑎 𝑙1 +𝑟1 +2𝑘,𝑙,𝑒 or 𝑎 𝑙1 +𝑟1 +2𝑘,𝑙,𝑜 or 0 and 𝑣(𝑎 𝑙1 +𝑟1 +2𝑘,𝑙 ) ≥ 𝑚 𝑘 for 𝑙 ≤ 𝑙1 + 𝑟 1 , 𝑘 > 0. By direct computation on submatrix with rows {1, 2, ..., 𝑙 1 + 𝑟 1 } and columns {𝑙1 + 𝑟 1 , 𝑙1 + 𝑟 1 + 1, ..., 𝑛} of LHS-RHS, we have 𝑎 𝑙,𝑙1 +𝑟1 +2𝑘+1 = 𝑎 𝑙,𝑙1 +𝑟1 +2𝑘−1,𝑒 or 𝑎 𝑙,𝑙1 +𝑟1 +2𝑘−1,𝑜 or 0 and 𝑣(𝑎 𝑙,𝑙1 +𝑟1 +2𝑘+1 ) ≥ 𝑚 𝑘 − 1 for 𝑙 ≤ 𝑙1 + 𝑟 1 , 𝑘 > 0. By direct computation on submatrix with rows {𝑙1 + 𝑟 1 , 𝑙1 + 𝑟 1 + 1, . . . , 𝑛} and columns {𝑙1 + 𝑟 1 , 𝑙1 + 𝑟 1 + 1, ..., 𝑛} of LHS-RHS, we have 𝑎 𝑙1 +𝑟1 +2𝑘 ′,𝑙1 +𝑟1 +2𝑘+1 = 𝑎 𝑙1 +𝑟1 +2𝑘 ′,𝑙1 +𝑟1 +2𝑘+1,𝑒 or 𝑎 𝑙1 +𝑟1 +2𝑘 ′,𝑙1 +𝑟1 +2𝑘+1,𝑜 or 0 where the valuation ≥ max (𝑚 𝑘 , 𝑚 𝑘 ′ ) − 1 for 𝑘 ′, 𝑘 > 0. Here is an illustration of these rules, where 𝑋 represents an element complying to the restriction. (We only need the fact that each nonzero term has valuation greater than 0.) © 𝑎 1,1,𝑒 ­ ... ­ ­ ­ 𝑎 𝑙1 +𝑟1 ,1,𝑒 ­ ­𝑎 ­ 𝑙1 +𝑟1 +1,1 ­ ­ 𝑋 ­ ­ ... ­ ­ ­ 𝑎 𝑛−1,1 ­ « 𝑋 ··· 𝑎 1,𝑙1 +𝑟1 ,𝑜 𝑋 𝑎 1,𝑙1 +𝑟1 +2 ··· 𝑋 𝑎 1,𝑛 ª ® ® ® · · · 𝑎 𝑙1 +𝑟1 ,𝑙1 +𝑟1 ,𝑜 𝑋 𝑎 𝑙1 +𝑟1 ,𝑙1 +𝑟1 +2 · · · 𝑋 𝑎 𝑙1 +𝑟1 ,𝑛 ®® · · · 𝑎 𝑙1 +𝑟1 +1,𝑙1 +𝑟1 𝑎 𝑙1 +𝑟1 +1,𝑙1 +𝑟1 +1 𝑎 𝑙1 +𝑟1 +1,𝑙1 +𝑟1 +2 · · · 𝑎 𝑙1 +𝑟1 +1,𝑛−1 𝑎 𝑙1 +𝑟1 +1,𝑛 ®® ®. ··· 𝑋 𝑋 𝑎 𝑙1 +𝑟1 +2,𝑙1 +𝑟1 +2 · · · 𝑋 𝑎 𝑙1 +𝑟1 +2,𝑛 ®® ® ® ® · · · 𝑎 𝑛−1,𝑙1 +𝑟1 𝑎 𝑛−1,𝑙1 +𝑟1 +1 𝑎 𝑛−1,𝑙1 +𝑟1 +2 · · · 𝑎 𝑛−1,𝑛−1 𝑎 𝑛−1,𝑛 ®® ··· 𝑋 𝑋 𝑎 𝑛,𝑙1 +𝑟1 +2 ··· 𝑋 𝑎 𝑛,𝑛 ¬ From Leibniz’s formula of determinant and 𝑔0 ∈ GL𝑛 (𝑘 [[𝑧]]), the least valuation of nonzero summands is 0. We write one such summand with valuation 0 by 𝜏. Notice that in row 𝑙1 + 𝑟 1 + 2𝑘 (𝑘 > 0), nonzero 𝑎 𝑙1 +𝑟1 +2𝑘,𝑙 𝑎 𝑙1 +𝑟1 +2𝑘,𝑙1 +𝑟1 +2𝑘 ′−1 where (𝑘 ′ > 0, 𝑙 ≤ 𝑙1 + 𝑟 1 ) have valuations greater than 0. Thus, we require that 𝜏 picks 𝑛−𝑙21 +𝑟1 elements in these rows from columns {𝑙1 + 𝑟 1 + 2, 𝑙1 + 𝑟 1 + 4, ..., 𝑛}. In rows ≤ 𝑙 1 + 𝑟 1 , no element in columns {𝑙1 + 𝑟 1 + 2, 𝑙1 + 𝑟 1 + 4, ..., 𝑛} is in 𝜏. From observation on column 𝑙1 + 𝑟 1 + 2𝑘 − 1(𝑘 > 0), nonzero 𝑎 𝑙1 +𝑟1 +2𝑙,𝑙1 +𝑟1 +2𝑘+1 for all 𝑙, 𝑘 have valuations greater than 0. Thus, we notice that all elements in 𝜏 which are in the first 𝑙1 + 𝑟 1 13 rows are in the first 𝑙 1 + 𝑟 1 column. That is, in the first 𝑙1 + 𝑟 1 by 𝑙 1 + 𝑟 1 submatrix of 𝑔0 (denote by 𝑔1 ), there exists a summand of Leibniz formula with a 0 valuation. 𝑎 1,2,𝑒 © 𝑎 1,1,𝑒 ­ 𝑎 𝑎 2,2,𝑒 ­ 2,1,𝑒 ­ ­ ... 𝑔1 = ­­ ­ 𝑎𝑟2 +1,1,𝑜 𝑎𝑟2 +1,2,𝑜 ­ ­ ... ­ «𝑎 𝑙1 +𝑟1 ,1,𝑜 𝑎 𝑙1 +𝑟1 ,2,𝑜 ··· ··· 𝑎 1,𝑟1 +1,𝑜 𝑎 2,𝑟1 +1,𝑜 ··· ··· 𝑎 1,𝑙1 +𝑟1 ,𝑜 ª 𝑎 2,𝑙1 +𝑟1 ,𝑜 ®® ® ® ®. · · · 𝑎𝑟2 +1,𝑟1 +1,𝑒 · · · 𝑎𝑟2 +1,𝑙1 +𝑟1 ,𝑒 ®® ® ® ® · · · 𝑎 𝑙1 +𝑟1 ,𝑟1 +1,𝑒 · · · 𝑎 𝑙1 +𝑟1 ,𝑙1 +𝑟1 ,𝑒 ¬ For nonzero 𝑎𝑖, 𝑗,𝑒 , 𝑎𝑖, 𝑗,𝑜 , we have 𝑣(𝑎𝑖, 𝑗,𝑒 ) ≥ 0, 𝑣(𝑎𝑖, 𝑗,𝑜 ) ≥ 1. Any nonzero summand will have valuation ≥ |𝑟 1 − 𝑟 2 |. The contradiction appears since 𝑟 1 < 𝑟 2 . □ We define the symmetric coweight 𝜎 on the blockwise decomposition. Recall in the proof of uniqueness in Theorem 2.1, for Λ = ⊕𝑖 𝑧 𝑐𝑖 , we have 𝜎(Λ) = {𝑐 1 , ...𝑐 𝑗 }. Therefore, for any 𝐵 ∈ GL𝑛 (𝑘 [[𝑧]])ΛGL𝑛 (𝑘 ((𝑧2 ))), we extend the definition of 𝜎(Λ) to 𝐵 such that 𝜎(𝐵) = 𝜎(Λ). 2.2 Pieri and Dual Pieri Rule Having classified matrices of this form, it is natural to ask how a small change to the matrix affects the class of the matrix. We have already considered a version of this in the proof of Theorem 2.1, where we added a single row and column to a matrix in block form and reduced it to block form. Another natural operation corresponds to a meromorphic change of basis in the difference equation, or in terms of 𝐵, left-multiplying it by a matrix 𝐴 which is not invertibly holomorphic. There are two cases in which we obtain particularly nice results on how the type can change. In this section, we define two relations 𝑎 ∼ 𝑏 and 𝑎 ∼′ 𝑏 corresponding to 𝑎𝑖 − 1 ≤ 𝑏𝑖 ≤ 𝑎𝑖 + 1 and 𝑎𝑖+1 ≤ 𝑏𝑖 ≤ 𝑎𝑖−1 for all i. In the language of Young diagram, it is equivalent to: 𝜆 ∼ 𝜇 means that 𝜆 can be obtained from 𝜇 by adding and subtracting a vertical strip and 𝜆 ∼′ 𝜇 means that 𝜆 can be obtained from 𝜇 by adding and subtracting a horizontal strip. Theorem 2.2. For any 𝐴 ∈ GL𝑛 (𝑘 ((𝑧))) with 𝜌( 𝐴) = {1𝑙 } and any B∈ GL𝑛 (𝑘 ((𝑧))) with 𝜎(𝐵) = {𝑎 1 , ..., 𝑎 𝑛 }, we write 𝜎( 𝐴𝐵) = {𝑏 1 , ..., 𝑏 𝑛 }. Then, we have 𝜎(𝐵) ∼ 𝜎( 𝐴𝐵). Theorem 2.3. For any 𝐴 ∈ GL𝑛 (𝑘 ((𝑧))) with 𝜌( 𝐴) = {𝑙} and any B∈ GL𝑛 (k((𝑧))) with 𝜎(𝐵) = {𝑎 1 , ..., 𝑎 𝑛 }, we write 𝜎( 𝐴𝐵) = {𝑏 1 , ..., 𝑏 𝑛 }. Then, we have 𝜎(𝐵) ∼′ 𝜎( 𝐴𝐵). 14 We need two lemmas to prove the above two theorems. Lemma 2.1. For any 𝐴 ∈ GL𝑛 (𝑘 ((𝑧))), we write 𝜌( 𝐴) = {𝑎 1 , . . . , 𝑎 𝑛 }. If 𝑎 𝑛 ≥ 0, 𝜎( 𝐴) = {𝑏 1 , . . . , 𝑏 𝑛 } has the property 𝑏 𝑗 ≤ 𝑎 𝑗 for all 𝑗. Proof. If 𝑎 𝑛 ≥ 0, each element in 𝐴 has nonnegative valuation. Recall in Theorem 2.1, in the algorithm of decomposing 𝑔 𝐴ℎ = Λ, ℎ can be written as a product of transvectional matrices with entries in 𝑘 [[𝑧 2 ]], diagonal matrices with entries of nonpositive valuation and permutation matrices: In Case 2 Part 1 step 2, if 𝑘 ′ − 1 < 𝑎, then, instead of making 𝑋3,2 → 0, we do similar column operations to make 𝑋3,3 → 0. In Case 2 Part 2 condition A′ and B′, we do column multiplication on column 2,3 to make 𝑋2,3 , 𝑋1,4 ∈ {1, 𝑧}. Then, all transvectional matrix (column operations) have entries in 𝑘 [[𝑧2 ]]. Thus 𝜌(𝐻) = {ℎ1 , ...ℎ𝑛 } has ℎ1 ≤ 0. 𝜎( 𝐴) = 𝜎(Λ) = 𝜌(Λ) = 𝜌(𝐺 𝐴𝐻) = 𝜌( 𝐴𝐻). From [3, Chapter 5] , 𝑏 𝑗 ≤ 𝑎 𝑗 for all 𝑗. □ Lemma 2.2. For any 𝐴 ∈ GL𝑛 (𝑘 ((𝑧))), we write 𝜌( 𝐴) = {𝑎 1 , ...𝑎 𝑛 }. If 𝑎 𝑛−1 ≥ 0 and 𝑎 𝑛 < 0 , 𝜎( 𝐴) = {𝑏 1 , ...𝑏 𝑛 } has the property 𝑏 𝑗 ≤ 𝑎 𝑗−1 for all j. Proof. Claim: There exists a 𝐷 ∈ GL𝑛 (𝑘 ((𝑧))), where the first row is (𝑧 𝑤 , 𝑧 𝑐 , 0, .., 0) (𝑤 ∈ {0, 1}; 𝑐 > 𝑤; 𝑤, 𝑐 are of different parity) or (𝑧 𝑤 , 0, 0, .., 0) and the first column is (𝑧 𝑤 , 0, 0, .., 0), with the following properties: 𝜎( 𝐴) = 𝜎(𝐷); the lower right 𝑛 − 1 by 𝑛 − 1 submatrix of 𝐷 can be written in the form of 𝐷 ′ [𝑧 𝑎 ], where 𝐷 ′ ∈ GL𝑛−1 (𝑘 ((𝑧))), 𝜌(𝐷 ′) = {𝑎 1 , 𝑎 2 , ...𝑎 𝑛−1 }, [𝑧 𝑎 ] denotes a 𝑛 − 1 by 𝑛 − 1 diagonal matrix with the first entry 𝑧 𝑎 and other entries 1 and 𝑎 ≥ 0; if 𝑤 = 1, then 𝑎 = 0. We denote 𝜌(𝐷) = {𝑑1 , ...𝑑𝑛−1 , 𝑑𝑛 }. With the claim, Lemma 2.1 implies 𝑏𝑖 ≤ 𝑑𝑖 for all 𝑖. Recall the original Dual Pieri rule for 𝜌 states: for 𝐸 and 𝐸 ′ = 𝑋 𝐸, with 𝜌(𝑋) = { 𝑓 }, 𝑓 > 0, and 𝜌(𝐸) = {𝑒 1 , ..𝑒 𝑛 }, 𝜌(𝐸 ′) = { 𝑓1 , .. 𝑓𝑛 }, we have 𝑒𝑖−1 ≥ 𝑓𝑖 ≥ 𝑒𝑖 , 𝑓𝑖 ≤ 𝑒𝑖 + 𝑓 . In other word, there is no vertical strip on the ˜ } has 𝑑˜𝑖 ≤ 𝑎𝑖−1 for 𝑖 < 𝑛. One direct diagram. Thus, 𝜌(𝐷 ′ [𝑧 𝑎 ])={ 𝑑˜1 , ... 𝑑𝑛−1 result from the claim is that 𝐷 can be decomposed in to 𝐷 ′ [𝑎] and 𝑧 𝑤 . Thus, ˜ } = 𝜌(𝐷 ′ [𝑎]) = 𝜌(𝐷 ′) = {𝑎 1 , 𝑎 2 , ...𝑎 𝑛−1 }. Inserting one if 𝑤 = 1, { 𝑑˜1 , ... 𝑑𝑛−1 additional 1 (𝑧 𝑤 part) leads to 𝑑𝑖 < 𝑎𝑖−1 for all 𝑖 ≤ 𝑛. If 𝑤 = 0, we can rewrite ˜ , 0}. Thus we have the property 𝑑𝑖 ≤ 𝑎𝑖−1 for all 𝑖 (if 𝑖 = 𝑛, 𝜌(𝐷) = { 𝑑˜1 , ... 𝑑𝑛−1 𝑑𝑛 = 0 ≤ 𝑎 𝑛−1 ). Combining with 𝑏𝑖 ≤ 𝑑𝑖 , the inequality 𝑏𝑖 ≤ 𝑎𝑖−1 is reached for all 𝑖. 15 Proof of the claim: We will present an algorithm to find such 𝐷. All matrices operations are allowed operations from Theorem 2.1. Pick 𝐴𝑖, 𝑗 to be an entry in 𝐴 with the least valuation, i.e., 𝑣( 𝐴𝑖, 𝑗 ) = 𝑎 𝑛 . Let 𝐵 = 𝑔0 (1, 𝑖) 𝐴(1, 𝑗) such that 𝐵1,1 = 𝑧 𝑎 𝑛 , 𝐵𝑖,1 = 0 for 𝑖 > 1, where (1, 𝑖), (1, 𝑗) are permutation matrices and 𝑔0 ∈ 𝐺 𝐿 𝑛 (𝑘 [[𝑧]]). (We permute rows and columns such that 𝐴𝑖, 𝑗 is in the 1,1 position and reduce it to 𝑧 𝑎 𝑛 and eliminate all other entries in column 1.) The bottom right 𝑛 − 1 by 𝑛 − 1 submatrix has 𝜌={𝑎 1 , ..𝑎 𝑛−1 }. Now, we have 𝑣(𝐵1,𝑙 ) ≥ 𝑎 𝑛 for 𝑙 > 1. By trick 3 from Theorem 2.1, 𝐵1,𝑙 = 𝑧 𝑐𝑙 𝑓𝑙 or 0, where 𝑎 𝑛 < 𝑐 𝑙 and {𝑎 𝑛 , 𝑐 𝑙 } are in different parities and 𝑓𝑙 ∈ 𝑘 [[𝑧 2 ]] ∗ for all 𝑙 > 1. If 𝐵1,𝑖 ’s are all 0, then the proof of the claim is done: 𝐷 1,2 = 0, 𝑎 = 0. If not all 𝐵1,𝑙 is 0, we pick the least 𝑐 𝑙 (assume it is 𝑐 𝑝 ) and do trick 1 to remove 𝑓𝑙 . Then we cancel all other 𝐵1,𝑙 ’s by column operations and switch column 𝑝, 2. Let 𝐶 denote the 𝑔1 𝐵ℎ1 where 𝑔1 ∈ 𝐺 𝐿 𝑛 (𝑘 [[𝑧]]), ℎ1 ∈ 𝐺 𝐿 𝑛 (𝑘 [[𝑧2 ]]), and ℎ1 is the above column operations, 𝑔1 is the above row operation. Remember that the bottom right 𝑛−1 by 𝑛−1 submatrix of 𝐶 still has 𝜌={𝑎 1 , ..𝑎 𝑛−1 }. We denote the submatrix [𝐶] 𝑛−1 . 𝐶 has the first row (𝑧 𝑎 𝑛 , 𝑧 𝑐 𝑝 , 0, .., 0). 𝑐 𝑝 > 𝑎 𝑛 and {𝑐, 𝑎 𝑛 } are of different parity. If 𝑐 𝑝 ≥ 0 and 2|𝑎 𝑛 , 𝐶 right multiplies by 1 (1,1) (𝑧 −𝑎 𝑛 ). In this case, 𝑤 = 0, 𝑐 = 𝑐 𝑝 , 𝑎 = 0. If 𝑐 𝑝 ≥ 0 and 2|𝑎 𝑛 + 1 and 𝑐 𝑝 ≠ 0, 𝐶 right multiplies by 1 (1,1) (𝑧1−𝑎 𝑛 ). In this case 𝑤 = 1, 𝑐 = 𝑐 𝑝 , 𝑎 = 0. If 𝑐 𝑝 ≥ 0 and 2|𝑎 𝑛 + 1 and 𝑐 𝑝 = 0, 𝐶 Î right multiplies by 1 (1,1) (𝑧 1−𝑎 𝑛 )(1, 2) and left multiplies by 𝑗 >1 1 ( 𝑗,1) (−𝐶 𝑗,2 ). In this case, 𝑤 = 0, 𝑐 = 1, 𝑎 = 1. If 𝑐 𝑝 < 0 and 2|𝑎 𝑛 , 𝐶 right multiplies by 1 (1,1) (𝑧−𝑎 𝑛 ) 1 (2,2) (𝑧1−𝑐 𝑝 ). In this case 𝑤 = 0, 𝑐 = 1, 𝑎 = 1 − 𝑐 𝑝 . If 𝑐 𝑝 < 0 and 2|𝑎 𝑛 + 1, 𝐶 right multiplies by 1 (1,1) (𝑧 𝑐 𝑝 +1−𝑎 𝑛 )(1, 2) 1 (1,1) (𝑧−𝑐 𝑝 ) 1 (2,2) (𝑧−𝑐 𝑝 ) and left Î multiplies by 𝑗 >1 1 ( 𝑗,1) (−𝐶 𝑗,2 𝑧−𝑐 𝑝 ). In this case, 𝑤 = 0, 𝑐 = 1, 𝑎 = 1 − 𝑐 𝑝 . □ Proof of Theorem 2.2. We write 𝑔 𝐴𝐵ℎ = Λ, 𝑔′ 𝐵ℎ′ = Λ′ with 𝑔, 𝑔′ ∈ GL𝑛 (𝑘 [[𝑧]]) and ℎ, ℎ′ ∈ GL𝑛 (𝑘 ((𝑧2 ))). Λ, Λ′ are block diagonal matrices. 𝜎(Λ′) = 𝜌(Λ′) = (𝑎 1 , 𝑎 2 , ..., 𝑎 𝑛 ). 𝜎(Λ) = 𝜎( 𝐴𝑔′−1 Λ′) = (𝑏 1 , 𝑏 2 , ..., 𝑏 𝑛 ). Also 𝜌( 𝐴𝑔′−1 = {1𝑙 }. Therefore, 𝜌( 𝐴𝑔′−1 Λ′) = (𝑐 1 , 𝑐 2 , .., 𝑐 𝑛 ) has 𝑎𝑖 ≤ 𝑐𝑖 ≤ 𝑎𝑖 + 1 for all 𝑖 ≤ 𝑛 (original Pieri rule). By Lemma 2.1, 𝑏𝑖 ≤ 𝑐𝑖 for all i. Therefore, 𝑏𝑖 ≤ 𝑎𝑖 + 1. For the other inequality, i.e., 𝑎𝑖 − 1 ≤ 𝑏𝑖 , we take 𝐴′ to be 𝐴′ = 𝑧 𝐴−1 𝑔 −1 . Now, one direct result from A is 𝜌( 𝐴′) = {1𝑛−𝑙 }. Notice that 𝐴′𝑔 𝐴𝐵ℎ = 𝐴′Λ, 𝐴′𝑔 𝐴𝐵ℎ = 𝑧𝐵ℎ. Thus, 𝜎( 𝐴′Λ) = 𝜎(𝑧Λ′) = ( 𝑎¯1 , 𝑎¯2 , ..., 𝑎¯𝑛 ). Since multiplying 𝑧 by Λ′ only switches 1 and 𝑧, we know 𝑎¯𝑖 = 𝑎𝑖 for 𝑎𝑖 > 1 and all other 𝑎¯𝑖 are less than 2. Applying the 16 result from the previous paragraph on Λ and 𝐴′Λ implies 𝑎¯𝑖 ≤ 𝑏𝑖 + 1. This equality leads to 𝑎𝑖 ≤ 𝑏𝑖 + 1 from the fact that 𝑏𝑖 ≥ 0. □ Proof of Theorem 2.3. We write 𝑔 𝐴𝐵ℎΛ and 𝑔′ 𝐵ℎ′ = Λ′ with 𝑔, 𝑔′ ∈ GL𝑛 (𝑘 [[𝑧]]) and ℎ, ℎ′ ∈ GL𝑛 (𝑘 ((𝑧2 ))),where Λ, Λ′ are block diagonal matrices. We have 𝜎(Λ′) = 𝜌(Λ′) = (𝑎 1 , 𝑎 2 , ..., 𝑎 𝑛 ) and 𝜎(Λ) = 𝜎( 𝐴𝑔′−1 Λ′) = (𝑏 1 , 𝑏 2 , ..., 𝑏 𝑛 ). Also 𝜌( 𝐴𝑔′−1 = {𝑙}. Therefore, 𝜌( 𝐴𝑔′−1 Λ′) = (𝑐 1 , 𝑐 2 , ..., 𝑐 𝑛 ) has 𝑎𝑖 ≤ 𝑐𝑖 ≤ 𝑎𝑖−1 for all 𝑖 ≤ 𝑛 (original Dual Pieri rule). By Lemma 2.2, 𝑏𝑖 ≤ 𝑐𝑖 for all i. Therefore, 𝑏𝑖 ≤ 𝑎𝑖−1 . For the other inequality, i.e., 𝑎𝑖 ≤ 𝑏𝑖−1 , we take 𝐴′ to be 𝐴′ = 𝐴−1 𝑔 −1 . Similarly to the proof of Theorem 2.2, 𝜎(Λ′) = 𝜎( 𝐴′Λ). We write 𝜌( 𝐴′Λ) = (𝑒 1 , ..., 𝑒 𝑛 ). Recall the original dual Pieri rule for 𝜌 of 𝐴′, 𝐴′Λ shows 𝑏𝑖+1 ≤ 𝑒𝑖 ≤ 𝑏𝑖 . If 𝑒 𝑛 ≥ 0, by Lemma 2.1, we have 𝑎𝑖 ≤ 𝑒𝑖 . Thus 𝑎𝑖 ≤ 𝑏𝑖 ≤ 𝑏𝑖−1 . If 𝑒 𝑛 < 0, then by the inequality on original dual Pieri rule, we have 𝑒 𝑛−1 ≥ 0. By Lemma 2.2, we have 𝑎𝑖 ≤ 𝑒𝑖−1 . Thus, both cases yield 𝑎𝑖 ≤ 𝑏𝑖−1 . □ 17 Chapter 3 HECKE MODULE In section 2.2, Pieri and Dual Pieri rules provide two nice results on how leftmultiplying 𝐵 by a matrix 𝐴 with certain type can change the symmetric coweight of 𝐵. The next natural question to ask is how spreading can the symmetric coweight be after the left multiplication. Similarly to the classical study of (spherical) Hecke algebra, we will define a Hecke module over double coset structure to describe the spreading with respect to Haar measure. Throughout this chapter, we write 𝐾 = GL𝑛 (𝑘 [[𝑧]]), 𝐻 = GL𝑛 (𝑘 ((𝑧2 ))), 𝐺 = GL𝑛 (𝑘 ((𝑧))). We define 𝜋𝜆 = diag(𝑧𝜆 𝑛 , ..., 𝑧𝜆1 ) and 𝑐𝜆 is the characteristic function of the double coset 𝐾𝜋𝜆 𝐾. We define Π 𝜆 the block diagonal matrix with valuations in nondecreasing order. Let 𝑑𝜆 be the characteristic function of the double coset 𝐾Π 𝜆 𝐻. Recall the construction of (spherical) Hecke algebra H (𝐺, 𝐾) in [3, Chapter 5], MacDonald proves that 𝑐𝜆 ’s form a C-basis of H (𝐺, 𝐾). Definition 3.1. The Hecke module H (𝐺, 𝐻, 𝐾) is the C− vector space of all complex valued continuous functions on 𝐺 that is a linear combination of 𝑑𝜆 . Equivalently, 𝑑𝜆 ’s form a C-basis of H (𝐺, 𝐻, 𝐾). We define a multiplication on H (𝐺, 𝐻, 𝐾) ∫ and H (𝐺, 𝐾): 𝑓 × 𝑔(𝑥) = 𝐺 𝑔(𝑦 −1 𝑥) 𝑓 (𝑦)𝑑𝑦 for 𝑓 ∈ H (𝐺, 𝐾), 𝑔 ∈ H (𝐺, 𝐻, 𝐾), 𝑥 ∈ 𝐺. By Proposition 3.1, the Hecke module H (𝐺, 𝐻, 𝐾) is a left H (𝐺, 𝐾)module. Unlike the definition of the (spherical) Hecke algebra, in this case, 𝐻 is not compact, thus a definition of all continuous function with compact support cannot be applied here. Any function in H (𝐺, 𝐻, 𝐾) is invariant with respect to (𝐻, 𝐾), i.e., 𝑓 (𝑘𝑥ℎ) = 𝑓 (𝑥) for all 𝑥 ∈ 𝐺, ℎ ∈ 𝐻, 𝑘 ∈ 𝐾. Proposition 3.1. H (𝐺, 𝐻, 𝐾) is a left H (𝐺, 𝐾)-module. 18 Proof. For all ℎ ∈ 𝐻, 𝑘 ∈ 𝐾, ∫ ∫ −1 𝑓 × 𝑔(𝑘𝑥ℎ) = 𝑔(𝑦 𝑘𝑥ℎ) 𝑓 (𝑦)𝑑𝑦 = 𝑔(𝑦 −1 𝑘𝑥) 𝑓 (𝑦)𝑑𝑦 ∫𝐺 ∫𝐺 = 𝑔(𝑦′−1 𝑥) 𝑓 (𝑦′)𝑑𝑦′ = 𝑔(𝑦 −1 𝑥) 𝑓 (𝑦)𝑑𝑦. 𝐺 𝐺 Given 𝑓1 , 𝑓2 ∈ H (𝐺, 𝐾) and 𝑔 ∈ H (𝐺, 𝐻, 𝐾), we have ∫ 𝑓1 × ( 𝑓2 × 𝑔)(𝑥) = ( 𝑓2 × 𝑔)(𝑦 −1 𝑥) 𝑓1 (𝑦)𝑑𝑦 ∫𝐺 ∫ = 𝑔(𝑧−1 𝑦 −1 𝑥) 𝑓2 (𝑧) 𝑓1 (𝑦)𝑑𝑦𝑑𝑧 𝐺 𝐺 ∫ ∫ = 𝑔(𝑦′−1 𝑥) 𝑓2 (𝑧′−1 ) 𝑓1 (𝑦′ 𝑧′)𝑑𝑦′ 𝑑𝑧′ . 𝐺 𝐺 The last equality is obtained by 𝑦′ = 𝑦𝑧 and 𝑧′ = 𝑧′−1 . Moreover, we have ∫ ∫ ∫ −1 ( 𝑓1 𝑓2 ) × 𝑔(𝑥) = 𝑔(𝑦 𝑥)( 𝑓1 𝑓2 )(𝑦)𝑑𝑦 = 𝑔(𝑦 −1 𝑥) 𝑓1 (𝑦𝑧) 𝑓2 (𝑧−1 )𝑑𝑧𝑑𝑦. 𝐺 𝐺 𝐺 Thus, we have 𝑓1 × ( 𝑓2 × 𝑔) = ( 𝑓1 𝑓2 ) × 𝑔. Similarly, for distributivity of 𝑓 , we have ∫ ∫ −1 (( 𝑓1 + 𝑓2 ) × 𝑔)(𝑥) = 𝑔(𝑦 𝑥)( 𝑓1 + 𝑓2 )(𝑦)𝑑𝑦 = (𝑔(𝑦 −1 𝑥) 𝑓1 (𝑦) + 𝑔(𝑦 −1 𝑥) 𝑓2 (𝑦))𝑑𝑦 𝐺 𝐺 = ( 𝑓1 × 𝑔 + 𝑓2 × 𝑔)(𝑥). Thus, we have ( 𝑓1 + 𝑓2 ) × 𝑔 = 𝑓1 × 𝑔 + 𝑓2 × 𝑔. Similarly, for distributivity of 𝑔, given 𝑔1 , 𝑔2 ∈ H (𝐺, 𝐻, 𝐾) and 𝑓 ∈ H (𝐺, 𝐾), we have ∫ ∫ −1 𝑓 × (𝑔1 + 𝑔2 )(𝑥) = (𝑔1 + 𝑔2 )(𝑦 𝑥)( 𝑓 )(𝑦)𝑑𝑦 = (𝑔1 (𝑦 −1 𝑥) 𝑓 (𝑦) + 𝑔2 (𝑦 −1 𝑥) 𝑓 (𝑦))𝑑𝑦 𝐺 𝐺 = ( 𝑓 × 𝑔‘ + 𝑓 × 𝑔2 )(𝑥). Thus, we have 𝑓 × (𝑔1 + 𝑔2 ) = 𝑓 × 𝑔1 + 𝑓 × 𝑔2 . Identity for H (𝐺, 𝐾) is 𝑐 0 , i.e., the characteristic function on 𝐾. Thus, ∫ ∫ −1 𝑐 0 × 𝑔(𝑥) = 𝑔(𝑦 𝑥)𝑐 0 (𝑦)𝑑𝑦 = 𝑔(𝑦 −1 𝑥)𝑑𝑦 = 𝑔(𝑥). 𝐺 𝐾 □ 19 Given partitions 𝜆, 𝜇 of length 𝑛, the product 𝑐 𝜇 × 𝑑𝜆 is a linear combination of 𝑑 𝜈 . Í Define ℎ𝜆𝜇𝜈 by 𝑐 𝜇 × 𝑑𝜆 = 𝜈 ℎ𝜆𝜇𝜈 𝑑 𝜈 . We have ℎ𝜆𝜇𝜈 = (𝑐 𝜇 × 𝑑𝜆 )(Π 𝜈 ) ∫ = 𝑑𝜆 (𝑦 −1 Π 𝜈 )𝑐 𝜇 (𝑦)𝑑𝑦 𝐺 ∫ = 𝑑𝜆 (𝑦 −1 Π 𝜈 )𝑑𝑦 𝐾𝜋 𝜇 𝐾 ∑︁ ∫ = 𝑑𝜆 (𝑘 𝑦𝑖−1 Π 𝜈 )𝑑𝑘 = 𝑖 ∑︁ 𝐾 𝑑𝜆 (𝑦𝑖−1 Π 𝜈 ) 𝑖 where 𝐾𝜋 𝜇 𝐾 = ∪𝑖 𝑦𝑖 𝐾, and 𝐾 has measure 1. It follows that 𝑑𝜆 (𝑦𝑖−1 Π 𝜈 ) = 1 if 𝑦𝑖−1 Π 𝜈 ∈ 𝐾Π 𝜆 𝐻 and 𝑑𝜆 (𝑦𝑖−1 Π 𝜈 ) = 0 if 𝑦𝑖−1 Π 𝜈 ∉ 𝐾Π 𝜆 𝐻. Recall that the Haar measure of 𝐾𝜋 𝜇 𝐾 is finite; a direct result is that ℎ𝜆𝜇𝜈 ∈ Z. In this chapter, we will first give a representative for 𝑦𝑖 ’s (Lemma 3.1). Then, ℎ𝜆𝜇𝜈 is equal to the number of 𝑖’s such that 𝑦𝑖−1 Π 𝜈 ∈ 𝐾Π 𝜆 𝐻. Before we start Theorem 3.1, we introduce a system of paired tuple and associated equivalence relation. The set 𝜆 denotes the set of all paired tuples of size ℓ that are associated with the partitions 𝑊ℓ,𝜈 𝜈 and 𝜆, which corresponds to symmetric coweights of elements in GL𝑛 (𝑘 ((𝑧))). Í Explicitly, we have ℓ(𝜈) = 𝑛, 𝑚 0 (𝜈) ≥ 𝑖>1 𝑚𝑖 (𝜈) and 𝑚𝑖 (𝜈) = 0 if 𝑖 < 0, similarly for 𝜆 as well. Given partitions 𝜈 and 𝜆 with ℓ(𝜈) = ℓ(𝜆) = 𝑛, let 𝑤 be a tuple of length 𝑛 consisting of 0’s and 1’s. We say 𝑤 is a paired tuple of size ℓ associated with 𝜈 and 𝜆 if the following conditions hold: 1. Í 𝑖 𝑤 𝑖 = ℓ. Here ℓ is defined as the size of 𝑤. 2. Pairing is imposed on tuple 𝑤. There is at most one pairing for any index 𝑖. If 𝜈𝑖 > 1, the indices 𝑖, 𝑛 − 𝑖 + 1 form a pair (𝑖, 𝑛 − 1 + 1). For index 𝑖 with 𝜈𝑖 = 1, it can be paired with a index 𝑗 where 𝑤 𝑗 = 0 and 𝜈 𝑗 = 0, written as (𝑖, 𝑗). Let 𝜈ˆ = ( 𝜈ˆ1 , . . . , 𝜈ˆ𝑛 ) be a tuple obtained from 𝜈 by changing each part 𝜈 according to the pairing of 𝑖. We define: 20    𝜈𝑖 + 𝑤 𝑖 − 𝑤 𝑗 , 𝜈𝑖 > 1and (𝑖, 𝑗) is a pair       2,  𝜈𝑖 = 1, and (𝑖, 𝑗) is a pair 𝜈ˆ𝑖 =   𝜈𝑖 + (−1) 𝜈𝑖 𝑤 𝑖 , 𝑖 is not paired       0, otherwise.  3. 𝜈ˆ is a reordering of 𝜆. If 𝑖 > 1, denote 𝜔 [𝑖] = (𝜔0,0 , 𝜔1,1 , 𝜔1,0 , 𝜔0,1 ), where [𝑖] [𝑖] [𝑖] [𝑖] 𝜔𝑎,𝑏 = |{(𝑟, 𝑠) : 𝑤 𝑟 = 𝑎, 𝑤 𝑠 = 𝑏, 𝑣 𝑟 = 𝑖}|. [𝑖] We also write 𝜔 [1] = (𝜔0,0 , 𝜔1,1 , 𝜔1,0 , 𝜔0,1 ), where [1] [1] [1] [1]    |{pair(𝑟, 𝑠) : 𝜈𝑟 = 1}|,       |{𝑟 : 𝑤 𝑟 = 𝜈𝑟 = 1 and 𝑟 not paired}|,  𝑎,𝑏 𝜔 [1] =   |{𝑟 : 𝑤 𝑟 = 1, 𝜈𝑟 = 0 and 𝑟 not paired }|,       0,  if (𝑎, 𝑏) = (0, 0), if (𝑎, 𝑏) = (1, 1), if (𝑎, 𝑏) = (1, 0), if (𝑎, 𝑏) = (0, 1). An equivalence relation is imposed on 𝑤 and 𝑤 ′: 𝑤 ∼ 𝑤 ′ if and only if 𝜔 [𝑖] = 𝜔′[𝑖] for all 𝑖. Next, we give a formula of ℎ𝜆𝜇𝜈 for 𝜇 comprised of −1 and 0. For 𝑎 > 1, we denote 𝑗 ,𝑎 𝑛− 𝑗−𝑘 ,(𝑎−1) 𝑘 } 𝜔 𝑖, 𝑗,𝑘 𝑖, 𝑗,𝑘 ℎ {(𝑎+1) as ℎ𝑛 and denote ℎ˜ 𝑎 [𝑎] = ℎ𝑛 𝑞 − 𝑗 𝑘−( 𝑗+𝑖) (𝑛−𝑖−𝑘) if 𝜔 [𝑎] = 2𝑖+ 𝑗+𝑘 𝑛 −1 ,{𝑎 } 𝑛2 −𝑖− 𝑗+𝑘 } {2 ,1 ¯ [1] where 𝜔 [1] = (𝑘, 𝑗, 𝑖, 0). (𝑛 − 𝑖 − 𝑗 − 𝑘, 𝑖, 𝑘, 𝑗). We denote ℎ−1 𝑖+ 𝑗+𝑘 ,{1𝑛2 ,0𝑛1 } as ℎ 1 𝑖 𝜔 ˜ ℎ. ¯ In the rest of this In Proposition 3.2 and 3.3, we will give a closed formula for ℎ, Í chapter, we write 𝑡1 = 𝑛 − 2 𝑖>1 𝑚𝑖 (𝜈). Theorem 3.1. ∑︁ ℎ𝜆−1ℓ ,𝜈 = 𝜔 𝜆 /∼ [𝑤]∈𝑊ℓ,𝜈 ∑︁ 𝑟 ([𝑤]) = ( 𝑖>1, 𝑗 ∈{0,1} 𝑗,1 𝜔 [𝑖] )𝑡 1 +( ∑︁ 𝑖>1 Ö 𝑞𝑟 ( [𝑤]) ℎ¯ 1 [1] 𝜔0,0 −𝜔1,1 )( [𝑖] [𝑖] 𝜔 ℎ˜ 𝑖 [𝑖 ] 𝜔 [𝑖 ] ,𝑖>1 ∑︁ ∑︁ ∑︁ 𝑗,𝑘 𝜔 [1] )+ 𝑛𝑖 𝑛 𝑗 + 𝑛𝑖 (𝜔0,1 −𝜔1,0 ) [ 𝑗] [ 𝑗] 1<𝑖< 𝑗 𝑗,𝑘 1<𝑖< 𝑗 ∑︁ 1 ∑︁ 0,0 1,1 2 + 𝑛𝑖 𝜔0,1 − (( 𝜔 − 𝜔 ) − ( 𝜔0,0 𝜔0,0 + 𝜔1,1 𝜔1,1 )). [𝑖] [𝑖] [𝑖] [𝑖] [𝑖] [𝑖] [𝑖] 2 𝑖>1 𝑖>1 𝑖>1 ∑︁ 21 Define A 𝑎 as the set that consists of all upper triangular matrices 𝑋 ∈ GL𝑛 (𝑘 [[𝑧]]) with the following properties: 1. 𝑋𝑖,𝑖 = 1 or 𝑧, 2. for 𝑖 such that 𝑋𝑖,𝑖 = 𝑧, we have 𝑋𝑖, 𝑗 = 0 if 𝑗 > 𝑖 and 𝑋 𝑗,𝑖 ∈ 𝑘 if 𝑗 < 𝑖, 3. for i such that 𝑋𝑖,𝑖 = 1, we have 𝑋 𝑗,𝑖 = 0 if 𝑗 < 𝑖 and 𝑋𝑖, 𝑗 ∈ 𝑘 if 𝑗 > 𝑖, 4. the number of 𝑧 on the diagonal is 𝑎. Lemma 3.1. Given 𝐴 ∈ 𝐺 with 𝜌( 𝐴) = (1𝑎 ), we have 𝐾 𝐴𝐾 = Ã 𝑋∈A 𝑎 𝐾 𝑋. Proof. Given 𝑋, 𝑌 ∈ A 𝑎 , we first show that 𝐾 𝑋 ∩ 𝐾𝑌 is empty unless 𝑋 = 𝑌 . It suffices to show 𝑋𝑌 −1 ∉ 𝐾. From direct computation, for 𝑌 with entries 𝑌𝑖, 𝑗 , 𝑌 −1 has 𝑌 −1𝑖, 𝑗 = −𝑌𝑖, 𝑗 /𝑧 if 𝑖 ≠ 𝑗 and 𝑌 −1𝑖,𝑖 = 1/𝑌𝑖,𝑖 . Since 𝑋, 𝑌 are both upper triangular, on the diagonal of 𝑋𝑌 −1 , each entry is in {1, 𝑧, 1/𝑧} and 𝑋𝑌 −1 is also upper triangular. Thus, 𝑋𝑖,𝑖 = 𝑌𝑖,𝑖 . Let 𝐼 = {𝑖 : 𝑋𝑖,𝑖 = 𝑧}. For 𝑗 ∈ 𝐼, 𝑖 ∉ 𝐼, 𝑖 < 𝑗, we have 𝑋𝑌 −1𝑖, 𝑗 = 𝑋𝑖,𝑖 (−𝑌𝑖, 𝑗 /𝑧) + 𝑋𝑖, 𝑗 (1/𝑌 𝑗, 𝑗 ) = −𝑌𝑖, 𝑗 /𝑧 + 𝑋𝑖, 𝑗 /𝑧.
Thus 𝑋𝑌 −1𝑖, 𝑗 ∈ 𝑘 [[𝑧]] iff 𝑋𝑖, 𝑗 = 𝑌𝑖, 𝑗 . Recall that the Haar measure of 𝐾 𝐴𝐾 is 𝑎𝑛 .
Also, we have 𝑎𝑛 = |A 𝑎 |. Thus, A 𝑎 is a representative of the coset. □ Lemma 3.2 and 3.3 are useful in the proof of Lemma 3.4, 3.5, 3.6. Lemma 3.2. For any 𝐴 ∈ GL𝑛 (𝑘 ((𝑧))) with 𝜌( 𝐴) = {𝜇1 , ...𝜇𝑛 } where 𝜇1 < 𝑎 and 𝜇𝑛 ≥ 0, and 𝐴′ ∈ GL𝑛 (𝑘 ((𝑧))) with 𝐴′ − 𝐴 ∈ 𝑀𝑛 (𝑧 𝑎 𝑘 [[𝑧]]), we have 𝜌( 𝐴) = 𝜌( 𝐴′). Proof. We recall a classical result on determinant from [4] : ∑︁ ∑︁ det( 𝐴 + 𝐵) = (−1) 𝑠(𝛼)+𝑠(𝛽) det( 𝐴[𝛼|𝛽]) det(𝐵(𝛼|𝛽)) 𝑟 𝛼,𝛽∈𝐼𝑟 Here 𝐴 and 𝐵 are 𝑛 by 𝑛 matrices, the outer sum on 𝑟 is over {0, ...𝑛}, 𝐼𝑟 consists of all subsets in {1, ..., 𝑛} with size 𝑟, 𝐴[𝛼|𝛽] (square brackets) is the 𝑟 by 𝑟 submatrix of 𝐴 in rows 𝛼 and columns 𝛽, and 𝐵(𝛼|𝛽) (parentheses) is the 𝑛 − 𝑟 by 𝑛 − 𝑟 submatrix of 𝐵 in rows complementary to 𝛼 and columns complementary to 𝛽. Í𝑛 Denote 𝜌𝑙 = 𝑖=𝑛−𝑙+1 𝜈𝑖 and 𝐵 = 𝐴 − 𝐴′. It suffices to show that gcd of 𝑙 by 𝑙 submatrix of 𝐴 + 𝐵 is the same as that of 𝐴. It suffices to prove the claim: for 22 any 𝑙 < 𝑛 and 𝛼, 𝛽 ∈ 𝐼𝑙 , we have 𝑣(( 𝐴 + 𝐵) [𝛼|𝛽]) ≥ 𝜌𝑙 ; pick 𝛼, 𝛽 ∈ 𝐼𝑙 such that 𝑣( 𝐴[𝛼|𝛽]) = 𝜌𝑙 ,𝑣(( 𝐴 + 𝐵) [𝛼|𝛽]) = 𝜌𝑙 . We first prove the second half of the claim. For the sake of simplifying the notation, denote 𝐶 = 𝐴[𝛼|𝛽], 𝐷 = 𝐵[𝛼|𝛽]. For 𝑟 < 𝑙 and for all 𝜙, 𝜓 ∈ 𝐼𝑟 , nonzero summand (−1) 𝑠(𝜙)+𝑠(𝜓) det(𝐶 [𝜙|𝜓]) det(𝐷 (𝜙|𝜓)) has valuation ≥ (𝑙 − 𝑟)𝑎 + 𝜌𝑟 ≥ 𝜌𝑙 . (𝜈1 < 𝑎, 𝜌𝑙 − 𝜌𝑟 ≤ (𝑙 − 𝑟)𝑎.) For 𝑟 = 𝑙 and for all 𝜙, 𝜓 ∈ 𝐼𝑟 , (−1) 𝑠(𝜙)+𝑠(𝜓) det(𝐶 [𝜙|𝜓]) det(𝐷 (𝜙|𝜓)) has valuation equal to 𝜌𝑙 . For first half of the claim, the argument is the same as the 𝑟 < 𝑙 part: For 𝑟 ≤ 𝑙 and for all 𝜙, 𝜓 ∈ 𝐼𝑟 , nonzero summand (−1) 𝑠(𝜙)+𝑠(𝜓) det(𝐶 [𝜙|𝜓]) det(𝐷 (𝜙|𝜓)) has valuation ≥ (𝑙−𝑟)𝑎+ 𝜌𝑟 ≥ 𝜌𝑙 (Note that 𝜈1 < 𝑎, 𝜌𝑙 − 𝜌𝑟 ≤ (𝑙 − 𝑟)𝑎). □ Lemma 3.3. Denote the subgroup generated by transvection matrices, i.e., all 1 (𝑖, 𝑗) ( 𝑓 ) with 𝑖 ≠ 𝑗 and 𝑣( 𝑓 ) ≥ 0, as 𝐸 ⊂ 𝐾. For any 𝑔0 ∈ 𝐺, there exists 𝑘 1 ∈ 𝐾 and 𝑒 2 ∈ 𝐸 such that 𝑘 1 𝑔0 𝑒 2 = Λ, where Λ is a diagonal matrix of the form with diagonal entries 𝑧 𝑎 and valuations of diagonal entries correspond to the dominant coweight of 𝑔0 . Proof. Let 𝑅 be a commutative ring. By [2, Theorem 4.3.9], if 𝑅 is a Euclidean domain, then 𝐸 𝑛 (𝑅) = SL𝑛 (𝑅), where 𝐸 𝑛 (𝑅) is the subgroup of SL𝑛 (𝑅) generated by transvections (also called elementary matrices). Take 𝑅 = 𝑘 [[𝑧]]. We notice that GL𝑛 (𝑘 [[𝑧]]) = (𝑘 [[𝑧]] ∗ )SL𝑛 (𝑘 [[𝑧]]). Cartan decomposition leads to 𝑘 1 𝑔0 𝑒 2 = Λ. □ In the proof of Lemma 3.4, 3.5, and 3.6, we fix matrix 𝐵 as follows: for 𝑖 with 𝜈𝑖 > 1, we have 𝐵𝑖,𝑖 = 1, 𝐵𝑖,𝑛−𝑖+1 = 𝑧, 𝐵𝑛−𝑖+1,𝑛−𝑖+1 = 𝑧 𝜈𝑖 ; if (𝑛 − 𝑡1 )/2 < 𝑖 ≤ 𝑚 0 (𝜈), then 𝐵𝑖,𝑖 = 1; if 𝑚 0 (𝜈) < 𝑖 < 𝑚 0 (𝜈) + 𝑚 1 (𝜈), then 𝐵𝑖,𝑖 = 𝑧; other entries are 0. Let 𝑐𝑖 denote column 𝑖 and 𝑟𝑖 denote row 𝑖. Lemma 3.6 is a weak version of Theorem 3.1. 1 Lemma 3.4. For any 𝑋, 𝑋 ′ ∈ A 𝑎 satisfying 𝑋𝑖, 𝑗 = 𝑋𝑖,′ 𝑗 for 0 < 𝑖, 𝑗 ≤ 𝑛−𝑡 2 or 𝑛−𝑡1 𝑛+𝑡 1 𝑛+𝑡1 ′ 2 < 𝑖, 𝑗 ≤ 2 or 2 < 𝑖, 𝑗 ≤ 𝑛, we have 𝜎(𝑋 𝐵) = 𝜎(𝑋 𝐵). Proof. It suffices to prove that the lemma holds if 𝑋 ′ satisfies the extra condition: 𝑛−𝑡1 𝑛−𝑡1 𝑛+𝑡1 𝑛+𝑡1 1 𝑋𝑖,′ 𝑗 = 0 for ( 𝑛−𝑡 2 ≥ 𝑖 and 𝑗 > 2 ) or ( 2 < 𝑖 ≤ 2 and 𝑗 > 2 ). We write 𝑌 = 𝑋 𝐵. We will do allowed row/column operations on 𝑌 to prove this lemma. This is a 6-step matrix operations procedure: 1 Step 1: For any 𝑖 ≤ 𝑛−𝑡 2 , there exist allowed operations such that 𝑋𝑖,𝑛−𝑖+1 can be reduced to 0. 23 1 For 𝑗 ≤ 𝑛−𝑡 2 and 𝑖 ≠ 𝑗, we do column operation to make 𝑌𝑖, 𝑗 = 0. Thus, 𝑛−𝑡1 1 the first 𝑛−𝑡 2 by 2 submatrix of 𝑌 is diagonal. If 𝑋𝑖,𝑖 = 1, 𝑌𝑖,𝑛−𝑖+1 = 𝑧 + 𝑋𝑖,𝑛−𝑖+1 𝐵𝑛−𝑖+1,𝑛−𝑖+1 can be reduced to 𝑧: It suffices to consider 𝑋𝑖,𝑛−𝑖+1 ≠ 0, which implies 𝑋𝑛−𝑖+1,𝑛−𝑖+1 = 𝑧. We write 𝐵𝑛−𝑖+1,𝑛−𝑖+1 as 𝑧 𝑎 . If 2|𝑎, then 𝑐 𝑛−𝑖+1 replaced by 𝑐 𝑛−𝑖+1 − 𝑋𝑖,𝑛−𝑖+1 𝐵𝑛−𝑖+1,𝑛−𝑖+1 𝑐𝑖 will make 𝑌𝑖,𝑛−𝑖+1 to 𝑧. If 2|𝑎 + 1, we replace 𝑐 𝑛−𝑖+1 by 𝑐 𝑛−𝑖+1 /(1 + 𝑋𝑖,𝑛−𝑖+1 𝐵𝑛−𝑖+1,𝑛−𝑖+1 /𝑧). Recall that on row 𝑛 − 𝑖 + 1 the only nonzero term in 𝑌 is 𝑌𝑛−𝑖+1,𝑛−𝑖+1 , thus by row operations from row 𝑛 − 𝑖 + 1 to all other rows, 𝑌𝑙,𝑛−𝑖+1 /(1 + 𝑋𝑖,𝑛−𝑖+1 𝐵𝑛−𝑖+1,𝑛−𝑖+1 /𝑧) → 𝑌𝑙,𝑛−𝑖+1 for all 𝑙 ≠ 𝑛 − 𝑖 + 1. If 𝑋𝑖,𝑖 = 𝑧, 𝑌𝑖,𝑛−𝑖+1 = 𝑧2 + 𝑋𝑖,𝑛−𝑖+1 𝐵𝑛−𝑖+1,𝑛−𝑖+1 can be reduced to 𝑧2 : similarly as 𝑋𝑖,𝑖 = 1. 𝑛−𝑡1 𝑛+𝑡 1 1 Step 2: For any 𝑖 ≤ 𝑛−𝑡 2 and 2 < 𝑗 ≤ 2 , there exist allowed operations such that 𝑋𝑖, 𝑗 can be reduced to 0. It suffices to consider 𝑋𝑖, 𝑗 ≠ 0, which implies 𝑋 𝑗, 𝑗 = 𝑧, 𝑋𝑖,𝑖 = 1. If 𝐵 𝑗, 𝑗 = 1, 𝑐 𝑗 can be replaced by 𝑐 𝑗 − 𝑋𝑖, 𝑗 𝑐𝑖 . If 𝐵 𝑗, 𝑗 = 𝑧, 𝑐 𝑗 can be replaced by 𝑐 𝑗 − 𝑋𝑖, 𝑗 𝑐 𝑛−𝑖+1 . Notice that on column 𝑛 − 𝑖 + 1, 𝑌𝑖,𝑛−𝑖+1 = 𝑧 and all other entries in 𝑌 ’s column 𝑛 − 𝑖 + 1 are 0 or of valuation 𝑧 𝑎 , 𝑧 𝑎+1 if we write 𝐵𝑛−𝑖+1,𝑛−𝑖+1 = 𝑧 𝑎 . On row 𝑗 the only nonzero term in 𝑌 is 𝑌 𝑗, 𝑗 = 𝑧2 . Thus, by row operations from row 𝑗 to all other rows, 𝑌𝑙, 𝑗 − 𝑋𝑖, 𝑗 𝑌𝑙,𝑛−𝑖+1 → 𝑌𝑙, 𝑗 for all 𝑙 ≠ 𝑖. 1 Step 3: For any 𝑖, 𝑗 ≤ 𝑛−𝑡 2 and 𝑋𝑖,𝑖 = 𝑧 and 𝑖 ≠ 𝑗 , there exist allowed operations such that 𝑋 𝑗,𝑛−𝑖+1 can be reduced to 0, i.e., 𝑌 𝑗,𝑛−𝑖+1 = 𝑋 𝑗,𝑖 𝐵𝑖,𝑛−𝑖+1 + 𝑋 𝑗,𝑛−𝑖+1 𝐵𝑛−𝑖+1,𝑛−𝑖+1 reduces to 𝑋 𝑗,𝑖 𝑧. It suffices to consider 𝑋 𝑗,𝑛−𝑖+1 ≠ 0, which implies 𝑋𝑛−𝑖+1,𝑛−𝑖+1 = 𝑧, 𝑋 𝑗, 𝑗 = 1. The only two nonzero elements in 𝑌 ’s row i are 𝑌𝑖,𝑖 = 𝑧, 𝑌𝑖,𝑛−𝑖+1 = 𝑧2 . We write 𝐵𝑛−𝑖+1,𝑛−𝑖+1 be 𝑧 𝑎 . If 2|𝑎, then 𝑐 𝑛−𝑖+1 replaced by 𝑐 𝑛−𝑖+1 − 𝑋 𝑗,𝑛−𝑖+1 𝐵𝑛−𝑖+1,𝑛−𝑖+1 𝑐 𝑗 will make 𝑌 𝑗,𝑛−𝑖+1 to 𝑋 𝑗,𝑖 𝑧. If 2|1 + 𝑎, we replace 𝑟 𝑗 by 𝑟 𝑗 − 𝑋 𝑗,𝑛−𝑖+1 𝐵𝑛−𝑖+1,𝑛−𝑖+1 /𝑧2 (𝐵𝑛−𝑖+1,𝑛−𝑖+1 has valuation ≥ 2.). Then a column operation 𝑐𝑖 = 𝑐𝑖 + 𝑐 𝑗 𝑋 𝑗,𝑛−𝑖+1 𝐵𝑛−𝑖+1,𝑛−𝑖+1 /𝑧2 will keep all other terms the same as before. 𝑛+𝑡1 1 Step 4: For any 𝑖 ≤ 𝑛−𝑡 2 , 𝑗 > 2 , 𝑋𝑛− 𝑗+1,𝑛− 𝑗+1 = 1 and 𝑖 < 𝑛 − 𝑗 + 1 , there exist allowed operations such that 𝑋𝑖, 𝑗 can be reduced to 0. 1 Starting from 𝑗 = 𝑛+𝑡 → 𝑛, for all eligible 𝑖 satisfying the above condition, 2 the following algorithm would remove 𝑋𝑖, 𝑗 . We write 𝐵 𝑗, 𝑗 be 𝑧 𝑎 . If 2|𝑎, then 𝑐 𝑗 replaced by 𝑐 𝑗 − 𝑋𝑖, 𝑗 𝐵 𝑗, 𝑗 𝑐𝑖 will make 𝑌𝑖, 𝑗 to 0. If 2|𝑎 + 1, we replace 𝑟𝑖 by 𝑟𝑖 − 𝑟 𝑛− 𝑗+1 𝑋𝑖, 𝑗 𝐵 𝑗, 𝑗 /𝑧( 𝐵 𝑗, 𝑗 has valuation ≥ 2). Recall on row 𝑛 − 𝑗 + 1 of 𝑌 , 1 𝑌𝑛− 𝑗+1,𝑛− 𝑗+1 = 1, 𝑌𝑛− 𝑗+1, 𝑗 = 𝑧; if 𝑋𝑛−𝑙+1,𝑛−𝑙+1 = 𝑧, 𝑛+𝑡 < 𝑙 < 𝑗 (𝑣(𝑌𝑛− 𝑗+1,𝑙 ) = 2 1 or ∞), then 𝑌𝑛− 𝑗+1,𝑙 = 𝑋𝑛− 𝑗+1,𝑛−𝑙+1 𝐵𝑛−𝑙+1,𝑙 ; if 𝑋𝑙,𝑙 = 𝑧, 𝑙 > 𝑗. (𝑣(𝑌𝑛− 𝑗+1,𝑙 ) ≥ 24 2 or ∞), then 𝑌𝑛− 𝑗+1,𝑙 = 𝑋𝑛− 𝑗+1,𝑙 𝐵𝑙,𝑙 . Then we need to do the next three steps to make all other entries in row 𝑖 remaining the same. 1. 𝐴 column operation 𝑐 𝑛− 𝑗+1 = 𝑐 𝑛− 𝑗+1 + 𝑐𝑖 𝑋𝑖, 𝑗 𝐵 𝑗, 𝑗 /𝑧. 1 2. If 𝑋𝑛−𝑙+1,𝑛−𝑙+1 = 𝑧, 𝑛+𝑡 2 < 𝑙 < 𝑗, we redo step 3 to eliminate terms in 𝑌𝑖,𝑙 that is caused by the 𝑟 𝑛− 𝑗+1 row operation. 3. If 𝑋𝑙,𝑙 = 𝑧, 𝑙 > 𝑗, we do row operation 𝑟𝑖 + 𝑟 𝑙 𝑋𝑖, 𝑗 𝐵 𝑗, 𝑗 𝑌𝑖,𝑙 /(𝑧𝑌𝑙,𝑙 ) (row 𝑙 has only one nonzero term 𝑌𝑙,𝑙 ). Notice that for step 4, we on purpose make current 𝑌 and original 𝑌 only differ by one element for each 𝑖, 𝑗. 𝑛+𝑡 1 1 Step 5: For any 𝑖 ≤ 𝑛−𝑡 2 , 𝑗 > 2 , 𝑋𝑛− 𝑗+1,𝑛− 𝑗+1 = 1 and 𝑖 > 𝑛 − 𝑗 + 1, there exist allowed operations such that 𝑋𝑖, 𝑗 can be reduced to 0. 1 Starting from 𝑗 = 𝑛 → 𝑛+𝑡 2 , for all eligible 𝑖 satisfying above condition, the following algorithm would remove 𝑋𝑖, 𝑗 . We write 𝐵 𝑗, 𝑗 be 𝑧 𝑎 . If 2|𝑎, then 𝑐 𝑗 replaced by 𝑐 𝑗 − 𝑋𝑖, 𝑗 𝐵 𝑗, 𝑗 𝑐𝑖 will make 𝑌𝑖, 𝑗 to 0. If 2|𝑎 + 1, we replace 𝑟𝑖 by 𝑟𝑖 − 𝑟 𝑛− 𝑗+1 𝑋𝑖, 𝑗 𝐵 𝑗, 𝑗 /𝑧( 𝐵 𝑗, 𝑗 has valuation ≥ 2). Recall that on row 𝑛 − 𝑗 + 1of 𝑌 , 1 < 𝑙 < 𝑗 (𝑣(𝑌𝑛− 𝑗+1,𝑙 ) = 𝑌𝑛− 𝑗+1,𝑛− 𝑗+1 = 1, 𝑌𝑛− 𝑗+1, 𝑗 = 𝑧; if 𝑋𝑛−𝑙+1,𝑛−𝑙+1 = 𝑧, 𝑛+𝑡 2 1 or ∞), then 𝑌𝑛− 𝑗+1,𝑙 = 𝑋𝑛− 𝑗+1,𝑛−𝑙+1 𝐵𝑛−𝑙+1,𝑙 ; if 𝑋𝑙,𝑙 = 𝑧, 𝑙 > 𝑗. (𝑣(𝑌𝑛− 𝑗+1,𝑙 ) ≥ 2 or ∞), then 𝑌𝑛− 𝑗+1,𝑙 = 𝑋𝑛− 𝑗+1,𝑙 𝐵𝑙,𝑙 . Then we need to do the next two steps to make all other entries in row 𝑖 remain the same. 1. A column operation 𝑐 𝑛− 𝑗+1 = 𝑐 𝑛− 𝑗+1 + 𝑐𝑖 𝑋𝑖, 𝑗 𝐵 𝑗, 𝑗 /𝑧. 1 2. If 𝑋𝑛−𝑙+1,𝑛−𝑙+1 = 𝑧, 𝑛+𝑡 2 < 𝑙 < 𝑗, we redo step 3 to eliminate terms in 𝑌𝑖,𝑙 that is caused by the 𝑟 𝑛− 𝑗+1 row operation. Notice that for step 5, we on purpose make current 𝑌 and original 𝑌 only differ by one element for each 𝑖, 𝑗. 𝑛+𝑡 1 1 𝑛−𝑡 1 Step 6: For 𝑛 ≥ 𝑗 > 𝑛+𝑡 2 , 2 < 𝑖 ≤ 2 , there exist allowed operations such that 𝑋𝑖, 𝑗 can be reduced to 0. It suffices to consider 𝑋𝑖, 𝑗 ≠ 0, which implies 𝑋 𝑗, 𝑗 = 𝑧, 𝑋𝑖,𝑖 = 1. If 𝐵 𝑗, 𝑗 and 𝐵𝑖,𝑖 are of the same parity, 𝑐 𝑗 can be replaced by 𝑐 𝑗 − 𝑋𝑖, 𝑗 𝐵 𝑗, 𝑗 /𝐵𝑖,𝑖 𝑐𝑖 . If 𝐵 𝑗, 𝑗 and 𝐵𝑖,𝑖 are of different parities: if 𝑋𝑛− 𝑗+1,𝑛− 𝑗+1 = 𝑧, apply step 3 again; if 𝑋𝑛− 𝑗+1,𝑛− 𝑗+1 = 1, apply step 5 again. □ 25 In the following lemma, [𝑀] 𝑛/2 denotes the 𝑛/2, 𝑛/2 submatrix of 𝑀 that consists ¯ 𝑛/2 denotes the 𝑛/2, 𝑛/2 of the {𝑛/2 + 1, 𝑛/2 + 2, ...𝑛}th rows and columns; [𝑀] ˆ 𝑛/2 submatrix of 𝑀 that consists of the {1, 2, ...𝑛/2}th rows and columns. Let [𝑀] denote a matrix 𝑌 in GL𝑛 ([[𝑧]]) such that 𝑌𝑖,𝑖 = 1, 𝑌𝑖,𝑛−𝑖+1 = 𝑧 for 𝑖 ≤ 𝑛/2 and [𝑌 ] 𝑛/2 = [𝑀] 𝑛/2 and all other entries are 0. Lemma 3.5. In this lemma, we impose a further condition on 𝐵: 𝑡1 = 0. For any 𝑋 ∈ A 𝑎 , there exist 𝑀, 𝑁 ∈ GL𝑛/2 (𝑘 ((𝑧))) such that: 𝑁 is diagonal with 𝑁𝑖,𝑖 = 1/𝑋𝑛/2−𝑖+1,𝑛/2−𝑖+1 ; 𝑋𝑖, 𝑗 = −𝑀𝑛/2+1−𝑖,𝑛/2+1− 𝑗 for 0 < 𝑖, 𝑗 ≤ 𝑛/2 and 𝑖 ≠ 𝑗; 𝑀𝑖,𝑖 = 1 for all 𝑖. Thus 𝜎(𝑋 𝐵) = 𝜌([𝑋] 𝑛/2 [𝐵] 𝑛/2 𝑀 𝑁). Proof. Lemma 3.4 shows that if 𝑋, 𝑋 ′ ∈ A 𝑎 and they follow the restrictions in Lemma 3.4, there exist 𝑘 0 ∈ 𝐾, ℎ0 ∈ 𝐻 such that 𝑘 0 𝑋 𝐵ℎ0 = 𝑋 ′ 𝐵. It suffices to show: 1. There exist columns and rows operations on 𝑋 𝐵 to a matrix 𝑌 , where 𝑌𝑖,𝑖 = 1, 𝑌𝑖,𝑛−𝑖+1 = 𝑧 for 𝑖 ≤ 𝑛/2 and [𝑌 ] 𝑛/2 = [𝑋] 𝑛/2 [𝐵] 𝑛/2 𝑀 𝑁 and all other entries are 0. 2. On [𝑌 ] 𝑛/2 , given any transvectional matrix 𝑒 3 in GL𝑛/2 (𝑘 [[𝑧]]), there exist ˆ 𝑒 3 . Thus by Lemma 3.3, row and column operations such that [𝑌ˆ ] 𝑛/2 turns to [𝑌 ] 𝑛/2 𝜎(𝑋 𝐵) = 𝜌([𝑋] 𝑛/2 [𝐵] 𝑛/2 𝑀 𝑁). For 1: we add steps 7 and 8 after the former 6 steps for the sake of proof continuity. Step 7: For 𝑖 ≤ 𝑛/2, 𝑛/2 < 𝑗 < 𝑛 − 𝑖 satisfying 𝑌𝑖,𝑖 = 1 and 𝑋𝑛− 𝑗+1,𝑛− 𝑗+1 = 𝑧 (recall 𝑌𝑖, 𝑗 = 𝑋𝑖,𝑛− 𝑗+1 𝑧), column operation on 𝑐 𝑗 by rewriting 𝑐 𝑗 to be 𝑐 𝑗 − 𝑐 𝑛−𝑖+1 𝑋𝑖,𝑛− 𝑗+1 makes 𝑌𝑖, 𝑗 = 0. Then a first half of step 1 will make 𝑌𝑖,𝑛− 𝑗+1 0. Now, [𝑌 ] 𝑛/2 = [𝑋] 𝑛/2 [𝐵] 𝑛/2 𝑀. Step 8. For all 𝑖 ≤ 𝑛/2 with 𝑌𝑖,𝑖 = 𝑧 and for all 𝑛/2 < 𝑗 < 𝑛, row operation on 𝑟 𝑗 by rewriting 𝑟 𝑗 to be 𝑟 𝑗 − 𝑟𝑖𝑌 𝑗,𝑛−𝑖+1 /𝑧2 makes 𝑌 𝑗,𝑖 = −𝑌 𝑗,𝑛−𝑖+1 /𝑧( 𝑌 𝑗,𝑛−𝑖+1 refers to the entry of 𝑌 before step 8) and 𝑌 𝑗,𝑛−𝑖+1 = 0. After each 𝑗 is proceeded, we do a column switch for 𝑐𝑖 , 𝑐 𝑛−𝑖+1 . Then rewrite 𝑐 𝑛−𝑖+1 as − 𝑐 𝑛−𝑖+1 ; 𝑟𝑖 as −𝑟𝑖 ; 𝑐𝑖 as − 𝑐𝑖 /𝑧2 . Notice that these procedures for 𝑖 are equivalent to rewrite 𝑐 𝑛−𝑖+1 as 𝑐 𝑛−𝑖+1 /𝑧. For 2, we write 𝑒 3 as a column operation that takes 𝑐𝑖 to be 𝑐𝑖 − 𝑓 𝑐 𝑗 . This is equivalent, on 𝑌 , to 𝑐 𝑛/2+𝑖 − 𝑓 𝑐 𝑛/2+ 𝑗 on bottom right submatrix of 𝑌 while preserving other parts. Let 𝑓𝑜 denote the odd part of f and 𝑓𝑒 even part of f. Similarly to step 8, for 𝑛/2 < 𝑙 < 𝑛, row operations on 𝑟 𝑙 by writing 𝑟 𝑙 to be 𝑟 𝑙 − 𝑟 𝑛/2− 𝑗+1𝑌𝑙,𝑛/2+ 𝑗 /𝑧 will make 𝑌𝑙,𝑛/2+ 𝑗 = −𝑌𝑙,𝑛/2− 𝑗+1 /𝑧 ( 𝑌 𝑗,𝑛/2− 𝑗+1 refers to the entry of 𝑌 before) and 𝑌𝑙,𝑛/2− 𝑗+1 = 0. After each 𝑙 is proceeded, we do a column operation on 𝑐 𝑛/2+𝑖 by writing 𝑐 𝑛/2+𝑖 to be 𝑐 𝑛/2+𝑖 − 𝑓𝑜 𝑐 𝑛/2− 𝑗+1 𝑧. Notice that currently 𝑌𝑛/2− 𝑗+1,𝑖+𝑛/2 = − 𝑓𝑜 𝑧, thus 𝑟 𝑛/2− 𝑗+1 + 𝑟 𝑛/2−𝑖+1 𝑓𝑜 as 𝑟 𝑛/2− 𝑗+1 and 𝑐 𝑛/2−𝑖+1 − 𝑐 𝑛/2+ 𝑗 𝑓𝑜 as 𝑐 𝑛/2−𝑖+1 will remove 26 the term 𝑌𝑛/2− 𝑗+1,𝑖+𝑛/2 . Similarly to step 8, for 𝑛/2 < 𝑙 < 𝑛, row operations 𝑟 𝑙 − 𝑟 𝑛/2− 𝑗+1𝑌𝑙,𝑛/2− 𝑗+1 on 𝑟 𝑙 will make 𝑌𝑙,𝑛/2+ 𝑗 to be the original 𝑌𝑙,𝑛/2+ 𝑗 . For 𝑓𝑒 , column operation on 𝑐 𝑛/2+𝑖 by writing 𝑐 𝑛/2+𝑖 to be 𝑐 𝑛/2+𝑖 + 𝑓𝑒 𝑐 𝑛/2+ 𝑗 makes column 𝑐 𝑛/2+𝑖 as 𝑐 𝑛/2+𝑖 − 𝑓 𝑐 𝑛/2+ 𝑗 . Notice currently 𝑌𝑛/2− 𝑗+1,𝑖+𝑛/2 = 𝑓𝑒 𝑧, row operation on 𝑟 𝑛/2− 𝑗+1 by writing 𝑟 𝑛/2− 𝑗+1 to be 𝑟 𝑛/2− 𝑗+1 − 𝑟 𝑛/2−𝑖+1 𝑓𝑒 and column operation on 𝑐 𝑛/2−𝑖+1 by writing 𝑐 𝑛/2−𝑖+1 to be 𝑐 𝑛/2−𝑖+1 + 𝑐 𝑛/2− 𝑗+1 𝑓𝑒 will remove the term 𝑌𝑛/2− 𝑗+1,𝑛/2−𝑖+1 . □ We first prove a weak version of Theorem 3.1 with the help of the above lemmas: Lemma 3.6. Theorem 3.1 holds if distinct parts of 𝜈 with 𝜈𝑖 > 1 differ by at least 5. Í 𝑗−1 Í𝑗 Proof. Define 𝐼 𝑗 = [ 𝑖=2 𝑚𝑖 (𝜈) + 1, 𝑖=2 𝑚𝑖 (𝜈)]. Let [𝐶] 𝐼 denote the submatrix of C with rows and columns in 𝐼. Let [𝐶] 𝐼,𝐽 denote the submatrix of C with rows in 𝐼 and columns in 𝐽. Claim: Following lemma 3.5, there exist 𝑔1 ∈ GL𝑛/2 (𝑘 [[𝑧]]), 𝑒 2 ∈ 𝐸 𝑛/2 such that Ê 𝑔1 [𝑋] 𝑛/2 [𝐵] 𝑛/2 𝑀 𝑁𝑒 2 = 𝑧 𝜌( [[𝑋] 𝑛/2 ] 𝐼𝑖 [[𝐵] 𝑛/2 ] 𝐼𝑖 [𝑀] 𝑛/2−𝐼𝑖 [𝑁] 𝑛/2−𝐼𝑖 ) . 𝑖 ¯ 𝑛/2 ] 𝐼 ,𝐼 do not have any In other word, if 𝑖1 ≠ 𝑖2 , all terms in [[𝑋] 𝑛/2 ] 𝐼𝑖1 ,𝐼𝑖2 , [ [𝑋] 𝑖1 𝑖2 influence on 𝜎(𝑋 𝐵). 1 𝑛+𝑡 1 We denote 𝐼1′ = [ 𝑛−𝑡 2 , 2 ], and 𝐼 ′𝑗 = [ 𝑗−1 𝑗 ∑︁ ∑︁ 𝑛 + 𝑡1 +[ 𝑚𝑖 (𝜈) + 1, 𝑚𝑖 (𝜈)]). 𝑚𝑖 (𝜈) + 1, 𝑚𝑖 (𝜈)] ∪ ( 2 𝑖=2 𝑖=2 𝑖=2 𝑖=2 𝑗−1 ∑︁ 𝑗 ∑︁ By lemma 3.4 and the claim, we have shown 𝑋, 𝑋 ′ ∈ Aℓ , 𝜎(𝑋 𝐵) = 𝜎(𝑋 ′ 𝐵) if [𝑋] 𝐼𝑖′ = [𝑋 ′] 𝐼𝑖′ for all 𝐼𝑖′. In the proof of the claim, we will show: 𝜌([[𝑋] 𝑛/2 ] 𝐼𝑖 [[𝐵] 𝑛/2 ] 𝐼𝑖 [𝑀] 𝑛/2−𝐼𝑖 [𝑁] 𝑛/2−𝐼𝑖 ) = {(𝑖 + 1) 𝑎 , 𝑖 𝑏 , (𝑖 − 1) 𝑐 }. We write the number of z and 1/z in [[𝑋] 𝑛/2 ] 𝐼𝑖 ∪ [𝑁] 𝑛/2−𝐼𝑖 to be d. Thus, 𝑑 − (𝑎 + 𝑐) ≥ 0 and 2|𝑑 − (𝑎 + 𝑐). In the proof of Proposition 3.3 , we will show on 𝐼1′ , 𝜎 = {2𝑎 , 1𝑏 , 0𝑐 } and same rule on a,b,c hold, with additional dimension requirement: 𝑧2 exists only if 1 exists in 𝜎(𝐵). These are the rules in the summation of the original theorem under [𝑤]. The remaining part of this proof is to compute 𝑟 [𝑤]. In Lemma 3.4, Step 2: For 𝑛−𝑡1 𝑛+𝑡1 1 any 𝑖 ≤ 𝑛−𝑡 2 and 2 < 𝑗 ≤ 2 , there exist allowed operations such that 𝑋𝑖, 𝑗 27 can be reduced to 0. The number of all entries satisfying above condition in 𝑋’s is Í 𝑖, 𝑗 0, 𝑗 Í 1 𝑛−𝑡 1 ( 𝑖>1, 𝑗 ∈{0,1} 𝜔 [𝑖] )( 𝑖, 𝑗 𝜔 [1] ). In Lemma 3.4, Step 6: for 𝑛 ≥ 𝑗 > 𝑛+𝑡 2 , 2 < 𝑖 ≤ 𝑛+𝑡 1 2 , there exist allowed operations such that 𝑋𝑖, 𝑗 can be reduced to 0. The number of Í Í 𝑖, 𝑗 𝑗,1 all entries satisfying the above condition in 𝑋’s is ( 𝑖>1, 𝑗 ∈{0,1} 𝜔 [𝑖] )(𝑡1 − 𝑖, 𝑗 𝜔 [1] ). 𝑛+𝑡1 1 Combining all other steps in Lemma 3.4, for any 𝑖 ≤ 𝑛−𝑡 2 and 𝑛 ≥ 𝑗 > 2 , there exist allowed operations such that 𝑋𝑖, 𝑗 can be reduced to 0. The number of all entries Í 0, 𝑗 𝑗,1 Í satisfying the above condition in 𝑋’s is ( 𝑖>1, 𝑗 ∈{0,1} 𝜔 [𝑖] )( 𝑖>1, 𝑗 ∈{0,1} 𝜔 [𝑖] ). From the claim, the number of all entries satisfying the condition in 𝑋’s is ∑︁ ∑︁ ∑︁ ∑︁ ∑︁ ∑︁ 1,𝑘 𝑘,0 ( 𝜔0,𝑘 )( 𝜔 ) + ( 𝜔 )( 𝜔 𝑘,1 ). [𝑖] [ 𝑗] [𝑖] [ 𝑗] 1< 𝑗 <𝑖, 𝑘∈{0,1} 1<𝑖< 𝑗, 𝑘 ∈{0,1} 𝑘 ∈{0,1} 𝑘∈{0,1} Í 𝜔 From the definition of ℎ˜ 𝑎 [𝑎] and Proposition 3.2, 𝑖>1 (𝜔1,0 𝜔0,1 ) need to be added [𝑖] [𝑖] in 𝑟 [𝑤]. Therefore, 𝑟 ([𝑤]) = ∑︁ 𝜔1,0 𝜔0,1 + [𝑖] [𝑖] + 𝑖>1, 𝑗 ∈{0,1} 𝑗,1 𝜔 [𝑖] ( ∑︁ 1< 𝑗 <𝑖 𝑘∈{0,1} 𝑖>1 ∑︁ ∑︁ ∑︁ ∑︁ 𝜔0,𝑘 [𝑖] ∑︁ 𝜔1,𝑘 + [ 𝑗] ∑︁ 𝑘∈{0,1} 𝑘∈{0,1} ∑︁ ∑︁ 𝜔 𝑘,0 [ 𝑗] ∑︁ 𝜔 𝑘,1 ) [𝑖] 𝑘∈{0,1} ∑︁ 0, 𝑗 0, 𝑗 𝑖, 𝑗 𝑗,1 𝑖, 𝑗 𝜔 [𝑖] + 𝜔 [𝑖] 𝜔 [1] +( 𝜔 [𝑖] )(𝑡1 − 𝜔 [1] ). 𝑖, 𝑗 𝑖, 𝑗 𝑖>1, 𝑗 ∈{0,1} 𝑖>1, 𝑗 ∈{0,1} 𝑖>1, 𝑗 ∈{0,1} Up to reordering and simplifying, we have the 𝑟 [𝑤] in the statement. Proof of claim: We are still following our steps in Lemma 3.5. However, in the following proof, we will replace [𝑋] 𝑛/2 [𝐵] 𝑛/2 𝑀 𝑁 by 𝑌 temporarily for the simplicity of notation. The row and column operations refers to the transvectional matrix in 𝐸 𝑛/2 multiplying on the left and right, plus simple row multiplication by 𝑓 with 𝑣( 𝑓 ) = 0. We start from the first nonempty set 𝐼𝑖1 . Denote 𝐶𝑖1 = [[𝑋] 𝑛/2 ] 𝐼𝑖1 [[𝐵] 𝑛/2 ] 𝐼𝑖1 [𝑀] 𝑛/2−𝐼𝑖1 [𝑁] 𝑛/2−𝐼𝑖1 . We will prove three statements: 1. 𝜌(𝐶𝑖1 ) consists of {𝑖1 , 𝑖1 − 1, 𝑖1 + 1}; 2. There exist column and row operations that make [𝑌 ] 𝐼𝑖1 diagonal and remove all other elements in the same rows or columns with statement 3 always true; 3. Each term in [𝑌 ] 𝐼𝑖 𝑗 − 𝐶𝑖 𝑗 has valuation > 𝑖 𝑗 + 1. Each nonzero term in [𝑌 ] 𝐼𝑖 𝑗 ,𝐼𝑖𝑙 has valuation ≥ max(𝑖 𝑗 , 𝑖 𝑙 ) − 1. Therefore, employing 3 statements on 𝑖’s from all nonempty sets 𝐼𝑖 with 𝑖 in increasing order will finish the proof of the claim. Statement 1: Recall that original Pieri rule guarantees that 𝜌([[𝑋] 𝑛/2 ] 𝐼𝑖1 [[𝐵] 𝑛/2 ] 𝐼𝑖1 ) consists of {𝑖 + 1, 𝑖} with the number of i+1 being the number of 𝑧’s on the diagonal of [[𝑋] 𝑛/2 ] 𝐼𝑖1 . By the construction of 𝑀, all entries in 𝑀 are in 𝑘 and 𝑑𝑒𝑡 (𝑀) ≠ 0. 28 Thus, 𝜌(𝐶𝑖1 ) consists of 𝑖1 − 1, 𝑖1 , 𝑖1 + 1. and there exists 𝑝 ≥ 0 such that the number of 𝑖1 − 1, 𝑖1 + 1 plus 2𝑝 is the number of 𝑧 or 1/𝑧 on the diagonal of [[𝑋] 𝑛/2 ] 𝐼𝑖1 , [𝑁] 𝑛/2−𝐼𝑖1 . Statement 2: Currently, each term in [𝑌 ] 𝐼𝑖1 − 𝐶𝑖1 has valuation > 𝑖 + 1 (directly from multiplication). Then, by Lemma 3.4, 𝜌([𝑌 ] 𝐼𝑖1 ) = 𝜌(𝐶𝑖1 ). By Lemma 3.5, the submatrix [𝑌 ] 𝐼𝑖1 can be diagonalized, with rows/columns operations. Then we remove all nonzero terms in column 𝐼𝑖1 under the current block. On rows in 𝐼𝑖 𝑗 , the difference in elements before and after row operation will have valuation ≥ 𝑖 𝑗 − 1 − (𝑖1 + 1). Thus on the 𝐼𝑖 𝑗 diagonal block, the difference in any element has valuation ≥ 2(𝑖 𝑗 − 1) − (𝑖1 + 1). On the off diagonal block, [𝑌 ] 𝐼𝑖 𝑗 ,𝐼𝑖𝑙 , the difference in any element has valuation ≥ 𝑖 𝑙 − 1 + 𝑖 𝑗 − 1 − (𝑖1 + 1). Then we remove all terms in rows 𝐼𝑖1 after the 𝐼𝑖1 block. Notice that with the gap of greater than 4, all above inequality satisfies the statement 3. We redo this procedure to make 𝑌 blockwise diagonal. □ Before we close the gap of Theorem 3.1, we will first provide closed formulas for ℎ with restrictions on 𝜈. Since the proof of closing the gap is pure combinatorial and lengthy, we will give a self-contained proof for the following two theorems (Proposition 3.2 is a simple case of Proposition 3.4). In the rest of this chapter, we will define B slightly different than before. Let 𝐵 denote the block diagonal matrix with valuations in nondecreasing order and 𝜎(𝐵) = 𝜈. Recall from the beginning of this chapter, this is the definition of Π 𝜈 . In this theorem, we have 𝜈 = {𝑎 𝑛 }. Equivalently, on 𝐵, it consists of 𝑧 𝑎 blocks. Proposition 3.2. (𝑛−𝑖− 𝑗−𝑘,𝑖,𝑘, 𝑗) ℎ˜ 𝑎 =   𝑛 for all a with 𝑎 > 1. 𝑖, 𝑗, 𝑘 {(𝑎+1) 𝑗 ,𝑎 𝑛− 𝑗−𝑘 ,(𝑎−1) 𝑘 } 𝑛
Proof. We will prove ℎ−1 = 𝑞 𝑛(𝑖+ 𝑗)−𝑖(𝑖+ 𝑗+𝑘) 𝑖, 𝑗,𝑘 . 2𝑖+ 𝑗+𝑘 ,{𝑎 𝑛 } {(𝑎+1) ,𝑎 ,(𝑎−1) } Let ℎ𝑛 denote ℎ−1 . We will give a detailed proof based on the 2𝑖+ 𝑗+𝑘 ,{𝑎 𝑛 } matrix decomposition of Proposition 3.4. However, instead of employing the entire paired semi-tableau construction, we only borrow the matrix decomposition techniques and the statement: All elements that must be 0s (originated from the rule in 𝑋) still remain to be 0 under some row operations. We prove by induction on 𝑛, where 𝑛 is the number of two dimensional blocks. 𝑖, 𝑗, 𝑘 correspond to ±1, +1, −1. 𝑖, 𝑗,𝑘 𝑗 𝑛− 𝑗−𝑘 𝑘 29 𝑖, 𝑗,𝑘 That is, for 𝐵 consists of 𝑛 𝑧 𝑎 block, ℎ𝑛 is the number of 𝑋 in A2𝑖+ 𝑗+𝑘 such that 𝑖, 𝑗,𝑘 𝜎(𝑋 𝐵) = {(𝑎 + 1) 𝑗 , 𝑎 𝑛− 𝑗−𝑘 , (𝑎 − 1) 𝑘 }. Assume the statement holds for all ℎ𝑛 ′ with 𝑛′ < 𝑛. We will prove the following equality: 𝑖, 𝑗,𝑘 ℎ𝑛 𝑖, 𝑗,𝑘−1 𝑛−𝑘−𝑖 𝑖, 𝑗,𝑘 = ℎ𝑛−1 + ℎ𝑛−1 𝑞 𝑖−1, 𝑗,𝑘+1 𝑛−𝑖−𝑘−1 + ℎ𝑛−1 𝑞 𝑞 (𝑞 2𝑛−𝑘− 𝑗−2𝑖 + 𝑞 𝑛−𝑖−1 (𝑞 𝑛−𝑖− 𝑗−𝑘 − 1)) 𝑖−1, 𝑗,𝑘 2𝑛−2𝑖−𝑘 (𝑞 𝑘+1 − 1) + ℎ𝑛−1 𝑖, 𝑗−2,𝑘 3𝑛−3𝑖− 𝑗−𝑘−1 + ℎ𝑛−1 𝑖, 𝑗−1,𝑘 + ℎ𝑛−1 𝑞 𝑖, 𝑗−1,𝑘−1 2𝑛−2𝑖−𝑘−1 + ℎ𝑛−1 𝑞 𝑖−1, 𝑗−1,𝑘+1 3𝑛−3𝑖−2𝑘− 𝑗−1 (𝑞 𝑛−𝑖− 𝑗−𝑘+1 − 1) + ℎ𝑛−1 𝑞 (𝑞 𝑛−𝑖− 𝑗−𝑘+1 − 1) (𝑞 𝑘+1 − 1). Following the general proof of Proposition 3.4, we embed a (2𝑛 − 2) by (2𝑛 − 2) 𝑋 ′ ∈ A into 𝑋 ∈ A2𝑖+ 𝑗+𝑘 as the first (2𝑛 − 2) by (2𝑛 − 2) submatrix. Denote ′ ′ ′ ′ 𝜎(𝑋 ′ [𝐵]) = {(𝑎 + 1) 𝑗 , 𝑎 𝑛− 𝑗 −𝑘 , (𝑎 − 1) 𝑘 }. (By abusing the notation, we write [𝐵] to be the first (2𝑛 − 2) by (2𝑛 − 2) submatrix of 𝐵.) Then on columns 2𝑛 − 1, 2𝑛 of 𝑋, the following changes in 𝜎 can be achieved: ′ ′ ′ 1. If there is no 𝑧’s in 𝑋2𝑛−1,2𝑛−1 , 𝑋2𝑛,2𝑛 , then 𝜎(𝑋 𝐵) = {(𝑎 + 1) 𝑗 , 𝑎 𝑛− 𝑗 −𝑘 , (𝑎 − ′ 1) 𝑘 } (summand 1). ′ ′ ′ 2. If there is 1 𝑧’s in 𝑋2𝑛−1,2𝑛−1 , 𝑋2𝑛,2𝑛 , then 𝜎(𝑋 𝐵) = {(𝑎 +1) 𝑗 , 𝑎 𝑛− 𝑗 −𝑘 −1 , (𝑎 − ′ ′ ′ ′ ′ 1) 𝑘 +1 } (summand 2), {(𝑎 + 1) 𝑗 +1 , 𝑎 𝑛− 𝑗 −𝑘 −1 , (𝑎 − 1) 𝑘 } (summand 3), {(𝑎 + ′ ′ ′ ′ 1) 𝑗 , 𝑎 𝑛− 𝑗 −𝑘 +1 , (𝑎 − 1) 𝑘 −1 } (summand 4). ′ ′ ′ 3. If there are 2 𝑧’s in 𝑋2𝑛−1,2𝑛−1 , 𝑋2𝑛,2𝑛 , then 𝜎 = {(𝑎 + 1) 𝑗 , 𝑎 𝑛− 𝑗 −𝑘 , (𝑎 − ′ ′ ′ ′ ′ 1) 𝑘 } (summand 5), {(𝑎 + 1) 𝑗 +1 , 𝑎 𝑛− 𝑗 −𝑘 −1 , (𝑎 − 1) 𝑘 +1 } (summand 6), {(𝑎 + ′ ′ ′ ′ ′ ′ ′ ′ 1) 𝑗 +2 , 𝑎 𝑛− 𝑗 −𝑘 −2 , (𝑎−1) 𝑘 } (summand 7), {(𝑎+1) 𝑗 +1 , 𝑎 𝑛− 𝑗 −𝑘 , (𝑎−1) 𝑘 −1 } (summand 8). Each summand is the number of 𝑋 with fixed 𝜎 times the number of all possible fillings in the last two columns given 𝜎. To simplify the notation afterward, we will reorder the 𝑋 𝐵’s 2 dimensional blocks in nondecreasing order, and make all 𝑧 𝑎 blocks with both rows affected by z in front of all the unaffected 𝑧 𝑎 blocks (𝑥 1 𝑧 𝑎−1 ’s , 𝑥2 𝑧 𝑎 ’s affected, 𝑥3 𝑧 𝑎 ’s affected, 𝑥4 𝑧 𝑎+1 ’s ). We denote 1. 𝐴1 = {𝑋2𝑟,2𝑛−1 , where 𝑟 ≤ 𝑥 1 }, 2. 𝐴2 = {𝑋2𝑟−1,2𝑛−1 , where 𝑥 1 + 𝑥 2 < 𝑟 ≤ 𝑥 3 + 𝑥 2 + 𝑥 1 }, 3. 𝐴3 = {𝑋2𝑟,2𝑛−1 , where 𝑥1 + 𝑥 2 < 𝑟 ≤ 𝑥 3 + 𝑥 2 + 𝑥 1 }, 4. 𝐴4 = {𝑋2𝑟−1,2𝑛−1 , where 𝑥 1 + 𝑥 2 + 𝑥 3 < 𝑟 ≤ 𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 1 }, 30 and similarly for 𝐵1 , 𝐵2 , 𝐵3 , 𝐵4 on 𝑋𝑟,2𝑛 . The following statements are direct results from Proposition 3.4, since these are from the same matrix operations as in case 1,2,3. Claim 1: If 𝑋2𝑛,2𝑛 = 𝑧 and 𝑋2𝑛−1,2𝑛−1 = 1, then the last 𝑧 𝑎 block becomes 𝑧 𝑎+1 . In terms of 𝐵′ 𝑠, any elements in 𝑘 can be filled in all entries in 𝐵1 , 𝐵2 , 𝐵3 , 𝐵4 (case 1 in Proposition 3.4). If 𝑋2𝑛,2𝑛 = 1 and 𝑋2𝑛−1,2𝑛−1 = 𝑧, then: Claim 2: A 𝑧 𝑎 block in [𝐵] becomes 𝑧 𝑎+1 (subcase 1 in case 2) and in terms of 𝐵′ 𝑠, any elements in 𝑘 can be filled in all entries in 𝐴1 , 𝐴2 , 𝐴4 and there exists at least one nonzero entries in 𝐴3 ; Claim 3: A 𝑧 𝑎−1 block in [𝐵] becomes 𝑧 𝑎 (subcase 1 in case 2) and in terms of 𝐵′ 𝑠, any elements in 𝑘 can be filled in all entries in 𝐴2 , 𝐴4 and there exists at least one nonzero entries in 𝐴1 and all entries in 𝐴3 are 0; Claim 4: A 𝑧 𝑎 block in [𝐵] becomes 𝑧 𝑎−1 (subcase 2 in case 2) and in terms of 𝐵′ 𝑠, any elements in 𝑘 can be filled in all entries in 𝐴4 and there exists at least one nonzero entries in 𝐴2 and all entries in 𝐴3 , 𝐴1 are 0; Claim 5: The last 𝑧 𝑎 block becomes 𝑧 𝑎−1 (subcase 3 in case 2) and in terms of 𝐵′ 𝑠, any elements in 𝑘 can be filled in all entries in 𝐴4 and all entries in 𝐴3 , 𝐴1 , 𝐴2 are 0; If 𝑋2𝑛,2𝑛 = 𝑧 and 𝑋2𝑛−1,2𝑛−1 = 𝑧, then: Claim 6: A 𝑧 𝑎 block in [𝐵] becomes 𝑧 𝑎+1 and the last 𝑧 𝑎 block becomes 𝑧 𝑎+1 (subcase 1 in case 3) and thus in terms of 𝐵′ 𝑠, any elements in 𝑘 can be filled in all entries in 𝐴1 , 𝐴2 , 𝐴4 , 𝐵1 , 𝐵2 , 𝐵3 , 𝐵4 and there exists at least one nonzero entries in 𝐴3 ; Claim 7: A 𝑧 𝑎−1 block in [𝐵] becomes 𝑧 𝑎 and the last 𝑧 𝑎 block becomes 𝑧 𝑎+1 (subcase 1 in case 3) and thus in terms of 𝐵′ 𝑠, any elements in 𝑘 can be filled in all entries in 𝐴2 , 𝐴4 , 𝐵1 , 𝐵2 , 𝐵3 , 𝐵4 and there exists at least one nonzero entries in 𝐴1 and all entries in 𝐴3 are 0; Claim 8: A 𝑧 𝑎 block in [𝐵] becomes 𝑧 𝑎−1 and the last 𝑧 𝑎 block becomes 𝑧 𝑎+1 (subcase 3 in case 3) and in terms of 𝐵′ 𝑠, any elements in 𝑘 can be filled in all entries in 𝐴4 , 𝐵1 , 𝐵2 , 𝐵3 , 𝐵4 and there exists at least one nonzero entries in 𝐴2 − 𝐵3 and all entries in 𝐴3 , 𝐴1 are 0; Claim 9: The last 𝑧 𝑎 block becomes 𝑧 𝑎 (subcase 3 in case 3) and in terms of 𝐵′ 𝑠, any elements in 𝑘 can be filled in all entries in 𝐴4 , , 𝐵1 , 𝐵2 , 𝐵3 , 𝐵4 and all entries in 31 𝐴3 , 𝐴1 , 𝐴2 − 𝐵3 are 0. Now, we explain the coefficient for each summand in the equality. Summand 1: Directly from the 𝑛 − 1 blocks with 𝑖, 𝑗, 𝑘 indices. Since there is no 𝑧’s on the two columns, there is only one possible filling. Summand 2: From the 𝑛 − 1 blocks with 𝑖, 𝑗, 𝑘 − 1 indices, the last block provides one −1 (not necessarily on that block.) Equivalently, claim 4 and 5 gives 𝐴1 , 𝐴3 = 0, and fill 𝐴2 , 𝐴4 with 𝑘. Summand 3: From the 𝑛 − 1 blocks with 𝑖, 𝑗 − 1, 𝑘 indices, the last block provides one +1 (not necessarily on that block). We distinguish two cases below. 1. 𝑋2𝑛−1,2𝑛−1 = 𝑧 : fill all𝐴3 with at least one nonzero element and fill 𝐴2 , 𝐴4 , 𝐴1 with k. This summand becomes 𝑞 𝑛−𝑖−1 (𝑞 𝑛−𝑖− 𝑗−𝑘 − 1). 2. 𝑋2𝑛,2𝑛 = 𝑧, fill 𝐵1 , 𝐵2 , 𝐵3 , 𝐵4 with 𝑘 (Claim 1). This summand becomes 𝑞 2𝑛−𝑘− 𝑗−2𝑖 . Summand 4: From the 𝑛 − 1 blocks with 𝑖 − 1, 𝑗, 𝑘 + 1 indices, the last block takes one −1 block to ±1. Equivalently, by claim 3 we can fill all entries fill all 𝐴1 with at least one nonzero element, 𝐴3 = 0, fill 𝐴2 , 𝐴4 with k. This summand becomes 𝑞 𝑛−𝑖−𝑘−1 (𝑞 𝑘+1 − 1). Summand 5: From the 𝑛 − 1 blocks with 𝑖 − 1, 𝑗, 𝑘 indices, the last block provides ±1. Equivalently, by claim 9 we can fill all entries 𝐴4 , 𝐵1 , 𝐵2 , 𝐵3 , 𝐵4 with 𝑘 and 𝐴3 , 𝐴1 , 𝐴2 − 𝐵3 with 0. This summand becomes 𝑞 2𝑛−2𝑖−𝑘 . Summand 6: From the 𝑛−1 blocks with 𝑖, 𝑗 − 1, 𝑘 − 1 indices, the last block provides +1and − 1. Equivalently, by claim 8 we can fill all entries 𝐴4 , 𝐵1 , 𝐵2 , 𝐵3 , 𝐵4 with 𝑘 and fill all𝐴2 with at least one nonzero element in 𝐴2 − 𝐵3 . This summand becomes 𝑞 2𝑛−2𝑖−𝑘−1 (𝑞 𝑛−𝑖− 𝑗−𝑘+1 − 1). Summand 7: From the 𝑛 − 1 blocks with 𝑖, 𝑗 − 2, 𝑘 indices, the last block provides +1, +1. Equivalently, by claim 6 we can fill all entries 𝐴3 with at least one nonzero element and fill 𝐴1, 𝐴2 , 𝐴4 , 𝐵1 , 𝐵2 , 𝐵3 , 𝐵4 with 𝑘. This summand becomes 𝑞 3𝑛−3𝑖− 𝑗−𝑘−1 (𝑞 𝑛−𝑖− 𝑗−𝑘+1 − 1). Summand 8: From the 𝑛 − 1 blocks with 𝑖 − 1, 𝑗 − 1, 𝑘 + 1 indices, the last block provides +1, ±1. Equivalently, by claim 7 we can fill all entries 𝐴1 with at least 32 one nonzero element and fill 𝐴2 , 𝐴4 , 𝐵1 , 𝐵2 , 𝐵3 , 𝐵4 with 𝑘. This summand becomes 𝑞 3𝑛−3𝑖−2𝑘− 𝑗−1 (𝑞 𝑘+1 − 1). Next, we group some of these summands to reduce the redundancy. [𝑛−1] 𝑖+ 𝑗+𝑘−1 𝑗+𝑛−𝑘 (2,6): 𝑞 (𝑛−1) (𝑖+ 𝑗)−𝑖(𝑖+ 𝑗+𝑘) [𝑖]![ 𝑗]![𝑘]! 𝑞 [𝑘] [𝑛−1] 𝑖+ 𝑗+𝑘−1 (3,7): 𝑞 (𝑛−1) (𝑖+ 𝑗)−𝑖(𝑖+ 𝑗+𝑘) [𝑖]![ 𝑗]![𝑘]! (𝑞 𝑛−𝑖−𝑘 + 𝑞 𝑛−𝑖− 𝑗−𝑘 − 1) [ 𝑗] [𝑛−1] 𝑖+ 𝑗+𝑘−1 (4,5,8): 𝑞 (𝑛−1) (𝑖+ 𝑗)−𝑖(𝑖+ 𝑗+𝑘) [𝑖]![ 𝑗]![𝑘]! (𝑞 𝑛−𝑖−𝑘 + 𝑞 𝑛−1 + 1) [𝑖] [𝑛−1] 𝑖+ 𝑗+𝑘−1 [𝑛 − 𝑖 − 𝑗 − 𝑘]. 1: 𝑞 (𝑛−1) (𝑖+ 𝑗)−𝑖(𝑖+ 𝑗+𝑘) [𝑖]![ 𝑗]![𝑘]! From direct computation, we have 𝑞 𝑗+𝑛−𝑘 [𝑘]+(𝑞 𝑛−𝑖−𝑘 +𝑞 𝑛−𝑖− 𝑗−𝑘 −1) [ 𝑗]+(𝑞 𝑛−𝑖−𝑘 +𝑞 𝑛−1 +1) [𝑖]+ [𝑛−𝑖− 𝑗 −𝑘] = 𝑞𝑖+ 𝑗 [𝑛]. Thus the statement is true for 𝑛. □ In the following proposition, we have 𝜈 = {1𝑛2 , 0𝑛1 }. Equivalently, on 𝐵, it consists of 2 𝑘 ,1𝑛2 +𝑖− 𝑗−𝑘 ¯ 𝜔 [1] where 𝜔 [1] = (𝑘, 𝑗, 𝑖, 0). 1-dim blocks. Recall that we define ℎ−1 𝑖+ 𝑗+𝑘 ,{1𝑛2 ,0𝑛1 } as ℎ 1 Proposition 3.3. (𝑘−1) 𝑘 𝑘 𝑛2 +𝑖− 𝑗−𝑘 ℎ2−1,1𝑖+ 𝑗+𝑘 ,{1𝑛2 ,0𝑛1 } = 𝑞 2 (𝑞 − 1) 𝑘 [𝑛1 − 𝑖] 𝑘  𝑛2 𝑘, 𝑗, 𝑛2 − 𝑘 − 𝑗   𝑛1 𝑖 Proof. Let 𝐵 denote the diagonal matrix where 𝐵𝑙,𝑙 = 1for 𝑙 ≤ 𝑛1 and 𝐵𝑙,𝑙 = 𝑧 for 𝑛1 < 𝑙 ≤ 𝑛1 + 𝑛2 . Step 1: for all 𝑙 such that 𝑋𝑙,𝑙 = 1, do column operations to remove all 𝑋𝑙,𝑟 𝐵𝑟,𝑟 . Thus, 𝑌𝑙,𝑟 = 0 if 𝑙 ≠ 𝑟, and (𝑙, 𝑟 ≤ 𝑛1 or 𝑛1 < 𝑙, 𝑟 ≤ 𝑛1 + 𝑛2 ). Step 2, for all 𝑙 such that 𝑋𝑙,𝑙 = 𝑧, and 𝑌𝑙,𝑟 ≠ 0, pick the first nonzero 𝑌𝑙,𝑟 in the 𝑟th column, then remove every 𝑌𝑙 ′,𝑟 below (i.e., 𝑙 ′ < 𝑟). Then, remove all 𝑌𝑙,𝑝 for 𝑝 > 𝑟. We call (𝑙, 𝑟) a pair. Step 3, for all 𝑙 such that 𝑋𝑙,𝑙 = 𝑧, and 𝑌𝑙,𝑟 = 0, this row becomes 1. On column 𝑙 < 𝑛1 , if there is a 𝑧 action on the column, then 1 becomes 𝑧 with weight ′ 𝑞 𝑙−1−𝑟 (𝑙) . On column 𝑛1 < 𝑙 < 𝑛1 +𝑛2 , with a 𝑧 action, either a 𝑧2 or 1 will occur. We write 𝐽 = { 𝑗1 , 𝑗2 , ... 𝑗 𝑗 } to be those columns with 𝑧 → 1, 𝐾 = {𝑘 1 , ...𝑘 𝑘 } to be those columns with 𝑧 → 𝑧 2 , {𝑖1 , ...𝑖 𝑘 } to be those columns paired with 𝑘’s, 𝐼 = {𝑖′1 , 𝑖′2 , ...𝑖𝑖′ } to be those columns with 1 → 𝑧,. A 𝑧 → 𝑧 2 occurs when there exists a 𝑙 ′ < 𝑛1 column of 1. Given 𝐽, 𝐾, for any l∈ 𝐽, 𝐾, we have 𝑟 (𝑙) = |{𝑘 𝑝 < 𝑙} ∪ { 𝑗 𝑝 < 𝑙}|. Given 𝐼, for any l∈ 𝐼, we have 𝑟 ′ (𝑙) = |{𝑖 𝑝 < 𝑙}|. 33 The total weight for 𝑘 𝑙 is (𝑞 𝑛1 −𝑖 − 𝑞 𝑙−1 )𝑞 𝑘 𝑙 −𝑛1 −1−𝑟 (𝑘 𝑙 ) . (The 𝑞 𝑙−1 term is the weight for 𝑖′ 𝑠, where 𝑧 → 𝑧2 ). The weight for 𝑗 𝑙 is 𝑞 𝑗𝑙 −𝑛1 −1−𝑟 ( 𝑗𝑙 ) 𝑞 |{𝑘 𝑙 ′ ,where 𝑘 𝑙 ′ < 𝑗𝑙 }| . Ö Ö ∑︁ Ö ′ 𝑘 𝑛2 +𝑖− 𝑗−𝑘 ′ ′ 𝑞 𝑗𝑙 −𝑛1 −1−𝑟 ( 𝑗𝑙 ) 𝑞 |{𝑘 𝑙 ′ ,where 𝑘 𝑙 ′ < 𝑗𝑙 }| (𝑞 𝑛1 −𝑖 − 𝑞 𝑙−1 )𝑞 𝑘 𝑙 −𝑛1 −1−𝑟 (𝑘 𝑙 ) ℎ2−1,1𝑖+ 𝑗+𝑘 ,{1𝑛2 ,0𝑛1 } = 𝑞𝑖𝑙 −1−𝑟 (𝑖𝑙 ) 𝐼,𝐽,𝐾 𝑙 𝑙 𝑙   ∑︁ Ö Ö 𝑛1 𝑞 𝑗𝑙 −𝑛1 −1−𝑟 ( 𝑗𝑙 ) 𝑞 |{𝑘 𝑙 ′ ,where 𝑘 𝑙 ′ < 𝑗𝑙 }| = (𝑞 𝑛1 −𝑖 − 𝑞 𝑙−1 )𝑞 𝑘 𝑙 −𝑛1 −1−𝑟 (𝑘 𝑙 ) 𝑖 𝐽,𝐾 𝑙 𝑙   Ö 𝑘−1 ∑︁ Ö 𝑛1 𝑛2 = (𝑞 𝑛1 −𝑖 − 𝑞 𝑙 ) · 𝑞 |{𝑘 𝑙 ′ ,where 𝑘 𝑙 ′ < 𝑗𝑙 }| 𝑖 𝑘 + 𝑗 𝑙=0 𝐽,𝐾 𝑙 The last summation is exactly the combinatorial description of 𝑞-binomial coefficient in terms of inversion. Thus     𝑘−1 𝑛1 𝑛2 𝑘 + 𝑗 Ö 𝑛1 −𝑖 2 𝑘 ,1𝑛2 +𝑖− 𝑗−𝑘 ℎ−1𝑖+ 𝑗+𝑘 ,{1𝑛2 ,0𝑛1 } = (𝑞 − 𝑞 𝑙 ). 𝑖 𝑘+𝑗 𝑘 𝑙=0 By rearranging, the original formula holds. □ The next step is to construct a system of paired semi-tableaux to prove the original theorem, i.e., closing the gap of 𝜈𝑖+1 −𝜈𝑖 > 4. The construction is pure combinatorial. We will define 𝐷 𝜆𝑙,𝜈 (the set of all paired diagrams which the original diagrams are 𝜈 and reduced shapes are 𝜆 with the total amount of added boxes being 𝑙) and 𝑇𝜔 (the set of all paired semi-tableau from a paired diagram 𝜔) before stating Proposition 3.4. For any Young diagram 𝜈 with 𝑚 0 (𝜈) = 𝑛/2 and 𝑚 1 (𝜈) = 0, we define an approach to add box toward 𝜈. New boxes are added to 𝜈 following the original Pieri rule, i.e. at most 1 box per row. We call one such added box diagram 𝜔 (We do not reshuffle rows in 𝜔 to make them in decreasing order). We introduce a pairing (𝑖, 𝑗) for rows 𝑖, 𝑗: for any i with 𝜈𝑖 > 1, there is a pairing for 𝜈𝑖 with 𝜈𝑛−𝑖+1 , i.e., (𝑖, 𝑛 − 𝑖 + 1). The total number of added boxes is ℓ. Let 𝑟 (𝜔) = {𝑟 1 , 𝑟 2 , ...} be the reduced shape of 𝜔: 𝑟𝑖 = 𝜔𝑖 − 𝜔 𝑗 for pair (𝑖, 𝑗). Note that 𝑟 (𝜔) is a partition. We call all such Young diagrams with added boxes paired diagrams, with the original diagram 𝜈 and reduced shape 𝜆 = 𝑟 (𝜔) and total amount of added boxes being ℓ. Let 𝐷 𝜆𝑙,𝜈 denote the set containing all such paired diagrams. Next, a definition of paired semi-tableaux based on paired diagram 𝜔 is introduced. For each ADDED box in 𝜔, numbers in {1, ..., 𝑛} will be filled in. Notice that in the original definition of Young tableaux, all boxes are filled with numbers. Rules: 34 1. Each number can appear in at most one box. 2. The number 2𝑘 + 1 can be only filled in the (𝑘 + 1)th row. 3. The number 2𝑘 can be filled in any row from the (𝑘 +1)th row to the 𝑛 − 𝑘 +1th row with the exception: 2𝑘 cannot be filled in row 𝑗 if 𝜈 𝑘 = 𝜈𝑛− 𝑗+1 and 𝜈 𝑗 = 0 and the number filled in row 𝑛 − 𝑗 + 1 is greater than 2𝑘. Before we start Proposition 3.4, a definition of weight on the paired semi-tableau 𝛾 would be necessary. We first assign a label (𝑟 (𝜔)𝑖 , 𝑛 − 𝑖) for each row of 𝛾, and a lexicographic order on labels. We create a sequence of subtableaux of {[𝛾] 1 , ..., [𝛾] 𝑛2 }, where [𝛾] 𝑖 consists of [ 𝑛2 − 𝑖 + 1, 𝑛2 + 𝑖] rows of 𝜈 with all added boxes with filled number ≥ 𝑛 − 2𝑖 + 1. Naturally, labels are assigned to rows in [𝛾] 𝑖 . Let 𝑆 (𝑎,𝑏),[𝛾] 𝑖 denote the set of all rows in [𝛾] 𝑖 with no added boxes in [𝛾] 𝑖 and labels being less than (𝑎, 𝑏). Here is the definition for weight on {1, ..., 𝑛}: 1. For 2𝑘 + 1 newly filled at [𝛾] 𝑖 (2𝑘 + 1 does not appears in [𝛾] 𝑖−1 , that is 𝑖 = 𝑛2 − 𝑘) , 𝑤𝑡 2𝑘+1 (𝛾) = 𝑞 |𝑆 (∞,𝑛), [𝜈 ] 𝑖 | . 2. For 2𝑘 newly filled at [𝛾] 𝑖 at the 𝑗th row (that is 𝑖 = 𝑛2 − 𝑘 + 1), we write the label for row j as (𝑎, 𝑛 − 𝑗). Then 𝑤𝑡2𝑘 (𝛾) = 𝑞 |𝑆 (𝑎−1,𝑛− 𝑗), [𝛾 ] 𝑖 |−1 (𝑞 − 1) if 𝑗 ≠ 𝑛 − 𝑘 − 1. 3. Under the same condition with 2, if 𝑗 = 𝑛 − 𝑘 − 1 and 2𝑘 − 1 is not in [𝛾] 𝑖 , 𝑤𝑡 2𝑘 (𝛾) = 𝑞 |𝑆 (𝑎−1,𝑛− 𝑗) , [𝛾 ] 𝑖 | . 4. Under the same condition with 2, if 𝑗 = 𝑛 − 𝑘 − 1 and 2𝑘 − 1 is in [𝛾] 𝑖 , 𝑤𝑡 2𝑘 (𝛾) = 𝑞 |𝑆 (𝑎,𝑛− 𝑗), [𝛾 ] 𝑖 | . 5. If a number 𝑘 is not filled in 𝛾, 𝑤𝑡 𝑘 (𝛾) = 1. Then, we define a weight on tableau 𝛾: 𝑤𝑡 (𝛾) = Here is an example. Î 𝑖 𝑤𝑡𝑖 (𝛾). 35 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 1 2 6 4 With pair {(1, 10), (2, 9), (3, 8), (4, 7)(5, 6)}, we write subtableaux sequence of 𝛾 {[𝛾] 1 , [𝛾] 2 , [𝛾] 3 , [𝛾] 4 , [𝛾] 5 }. 1 2 2 3 3 3 4 4 4 4 5 5 5 6 6 6 6 6 7 7 7 7 8 8 6 5 9 6 4 1 2 5 8 6 4 9 10 The reduced shape is {5, 4, 4, 3, 2}. 𝑤𝑡 6 = (𝑞 − 1)𝑞 2 . Labels in [𝛾] 3 : ((3, 7), (3, 6), (3, 5), (−3, 4), (−3, 3), (−3, 2)), 𝑤𝑡 4 = (𝑞−1). Labels in [𝛾] 4 : ((4, 8), (2, 7), (3, 6), (3, 5), (−3, 4), (−3, 3), (−2, 2), (−4, 1)), 𝑤𝑡 2 = (𝑞−1)𝑞 4 , 𝑤𝑡1 = 𝑞 6 . Labels in [𝛾] 5 : ((5, 9), (4, 8), (2, 7), (4, 6), (3, 5), (−3, 4), (−4, 3), (−2, 2), (−4, 1), (−5, 0)). The following proposition will give a combinatorial summation for ℎ𝜆ℓ𝜈 . Since in Lemma 3.4 we completely separated 1-dimensional blocks and 2-dimensional blocks, 𝜈 has the property 𝑚 0 (𝜈) = 𝑛2 and 𝑚 1 (𝜈) = 0. Again, unlike Lemma 3.4, 3.5 and 3.6, we denote B to be the block diagonal matrix with valuations in nondecreasing order and 𝜎(𝐵) = 𝜈. Recall from the beginning of this chapter, this is the definition of Π 𝜈 . 36 Proposition 3.4. ℎ𝜆−1ℓ 𝜈 = ∑︁ ∑︁ 𝑤𝑡 (𝛾). 𝜆 𝛾∈𝑇𝜔 𝜔∈𝐷 ℓ,𝜈 Proof. Recall the key property for 𝑋 ∈ Aℓ : Given any i such that 𝑋𝑖,𝑖 = 𝑧, we have 𝑋𝑖, 𝑗 = 0 for 𝑗 > 𝑖 and 𝑋 𝑗,𝑖 ∈ 𝑘 for 𝑗 < 𝑖. Given any i with 𝑋𝑖,𝑖 = 1, we have 𝑋𝑖, 𝑗 ∈ 𝑘 for 𝑗 > 𝑖 and 𝑋 𝑗,𝑖 = 0 for 𝑗 < 𝑖. We claim that each 𝑋 can be represented uniquely by a paired semi-tableau. Moreover weight of each paired semi-tableau is the number of all such 𝑋’s corresponding to the tableau. The algorithm below shows that for any 𝑖 such that 𝑋𝑖,𝑖 = 𝑧, first i by i submatrix of 𝑋 𝐵 (if 𝑖 is the second row of some two dimensional blocks of 𝐵) or first 𝑖 + 1 by 𝑖 + 1 submatrix of 𝑋 𝐵 ( if i is the first row of some two dimensional blocks of 𝐵) has one block turning from 𝑧𝑟 to 𝑧𝑟−1 or 𝑧𝑟+1 comparing to the (𝑖 − 2, 𝑖 − 2) submatrix (former) or (𝑖 − 1, 𝑖 − 1) submatrix (latter). This is equivalent to filling 𝑛 − 𝑖 + 1 in to some added box under the restriction of filling boxes. For any 𝑖 such that 𝑋𝑖,𝑖 = 𝑧, we take 𝑗 to be the number of elements in the 𝑖th column vector that are not necessarily 0. In the algorithm we will present, we call 𝑋˜ 𝑖 to be ith column vector of the matrix corresponding to 𝑌 in the 𝑋 form. Thus there are 𝑞 𝑗 choices for this column vector. In the algorithm we present below, the current column vector correspond to a matrix in GL𝑖−1 (𝑘 [[𝑧]]) times the original vector in 𝑋. Thus, the map 𝑘 𝑗 → 𝑘 𝑗 is bijective. We do not necessarily need the following statement (since the map is bijective), but the algorithm will follow the statement: All elements that must be 0s (originated from the rule in 𝑋) still remain to be 0 under some row operations. We introduce 3 tricks that are used below. We call left multiplication by matrices in GL𝑛 (𝑘 [[𝑧]]) and right multiplication by matrices in GL𝑛 (𝑘 ((𝑧2 ))) allowed operations. ! 1 𝑧 + 𝑧 𝑛1 𝑏 (the column with the 1 only has one Trick 1: A submatrix of the form 0 𝑧 𝑛1 +1 ! 1 𝑧 nonzero entry, 𝑏 ∈ 𝑘 ,𝑛1 > 1) can be reduced to with allowed operations. 0 𝑧 𝑛1 +1 If 2|𝑛1 , right multiply by 1 (1,2) (−𝑏𝑧1𝑛 ); else, left multiply by 1 (1,1) (1/(1 + 𝑏𝑧 𝑛1 −1 )) and right multiply by 1 (1,1) (1 + 𝑏𝑧 𝑛1 −1 ). Note that this may change the other entries in the same row as the top row of the submatrix. 37 ©1 𝑧 ­0 𝑧 𝑛 1 ­ Trick 2: A submatrix of the form ­ ­0 0 ­ «0 0 0 𝑓 ª 0 𝑔 ®® ® (the first two columns only do 1 𝑧 ®® 0 𝑧 𝑛2 ¬ ©1 𝑧 0 0 ª ­0 𝑧 𝑛 1 0 𝑔 ® ­ ® not have other nonzero entries, 𝑣( 𝑓 ) ≥ 0) can be reduced to ­ ® with ­0 0 1 𝑧 ® ­ ® 𝑛2 0 0 0 𝑧 « ¬ allowed operations. We left multiply by 1 (1,3) (− 𝑓odd /𝑧), right multiply by 1 (1,3) (− 𝑓odd /𝑧) 1 (1,4) (− 𝑓even ). Note that this may change the other entries in the same row as the top row of the submatrix. This trick still holds if the third row is in the form of 𝑧, 𝑧2 , given 𝑣( 𝑓 ) > 1. ©1 𝑧 ­0 𝑧 𝑎 1 ­ Trick 3: A submatrix of the form ­ ­0 0 ­ «0 0 0 𝑔ª 0 𝑓 ®® ® (the first two columns do not have 1 𝑧 ®® 0 𝑧 𝑛1 ¬ ©1 𝑧 0 0 ª ­0 𝑧 𝑎 1 0 0 ® ­ ® other nonzero entries, 𝑣( 𝑓 ) ≥ 𝑎 1 , 𝑣(𝑔) ≥ 0) can be reduced to ­ ®, ­0 0 1 𝑧 ® ­ ® 𝑛1 0 0 0 𝑧 « ¬ with allowed operations. If 2|𝑎 1 , left multiply by 1 (1,3) (− 𝑓𝑜𝑑𝑑 𝑧 −𝑎1 ) 1 (2,3) (− 𝑓𝑜𝑑𝑑 /𝑧) right multiply by1 (2,4) (− 𝑓𝑒𝑣𝑒𝑛 𝑧−𝑎1 ) 1 (2,3) (− 𝑓𝑜𝑑𝑑 𝑧−𝑎1 −1 ), then use trick 2. For 𝑎 1 being odd, vice versa. Note that this may change the other entries in the same row as the top two rows of the submatrix. Next we present an algorithm on decomposing 𝑌 = 𝑋 𝐵. We apply the algorithm in all 𝑖’s such that 𝑋𝑖,𝑖 = 𝑧, in increasing order. Then there are three cases: 𝑣(𝐵𝑖,𝑖 ) > 1; 𝑣(𝐵𝑖,𝑖 ) = 0 and 𝑣(𝐵𝑖+1,𝑖+1 ) > 1 and 𝑋𝑖+1,𝑖+1 = 1; 𝑣(𝐵𝑖,𝑖 ) = 0 and 𝑣(𝐵𝑖+1,𝑖+1 ) > 1 and 𝑋𝑖+1,𝑖+1 = 𝑧. Case 1: 𝑣(𝐵𝑖,𝑖 ) > 1. 1 𝑧 0 ··· 𝑏 1 𝑧 𝑎𝑙 ª © ­ 0 𝑧 𝑎0 0 · · · 𝑏 2 𝑧 𝑎𝑙 ®® ­ ­ ® ­· · · ® ­ ® ­ 0 · · · · · · 1 𝑧 + 𝑏𝑟 𝑧 𝑎𝑙 ® ­ ® 𝑎 +1 𝑙 𝑧 « 0 ··· ··· 0 ¬ 38 By trick 1, we remove 𝑏𝑟 𝑧 𝑎𝑙 . By trick 2, we remove 𝑏 1 𝑧 𝑎𝑙 . By trick 3, we remove 𝑏 2 𝑧 𝑎𝑙 . At the block corresponding to 𝑖, 𝑧 𝑎𝑙 → 𝑧 𝑎𝑙 +1 Thus, weight of the tableau of this 𝑋 at 𝑖 is 𝑞𝑖−𝑦 , where 𝑦 is the number of rows above 𝑖 that are permanently 0. On the tableau, this is equivalent to fill 𝑛 − 𝑖 + 1 in the added box of row 𝑛−𝑖 2 + 1. In the notation of lexicographical order, |𝑆 (∞,𝑛),[𝛾] 𝑖/2 | is the number of all rows < 𝑖 in 𝑌 |𝑆 | with possible nonzero entries. Thus, 𝑤𝑡 𝑛−𝑖+1 = 𝑞 (∞,𝑛), [𝛾 ] 𝑖/2 . Case 2: 𝑣(𝐵𝑖,𝑖 ) = 0 and 𝑣(𝐵𝑖+1,𝑖+1 ) > 1 and 𝑋𝑖+1,𝑖+1 = 1; 𝜁𝑙 =    (−𝑣(𝑌𝑙+1,𝑙+1 ), 𝑛−𝑙−1  2 ), if 𝑣(𝑌𝑙,𝑙 ) = 0 and 𝑙 ≤ 1 + 𝑖   (𝑣(𝑌𝑙,𝑙 ), 𝑙+𝑛 if 𝑣(𝑌𝑙,𝑙 ) > 0 and 𝑙 ≤ 𝑖 + 1 2 − 1),  We write 𝜁 (𝑌 ) = (𝜁1 , 𝜁2 , ...𝜁𝑖+1 ) with lexigraphical order imposed on 𝜁’s. There are three subcases based on𝑌𝑙,𝑖 : 1. There exists 𝑙 such that 𝑌𝑙,𝑖 ≠ 0 with 𝑣(𝐵𝑙,𝑙 ) > 1; 2. For any 𝑙 such that 𝑣(𝐵𝑙,𝑙 ) > 1 we have 𝑌𝑙,𝑖 = 0; There exists 𝑙 such that 𝑌𝑙,𝑖 ≠ 0 where 𝐵𝑙,𝑙 = 1, 𝑣(𝐵𝑙+1,𝑙+1 ) ≤ 𝑣(𝐵𝑖+1,1+𝑖 ). 3. For any 𝑙 such that 𝑣(𝐵𝑙,𝑙 ) > 1 or 𝐵𝑙,𝑙 = 1, 𝑣(𝐵𝑙+1,𝑙+1 ) ≤ 𝐵𝑖+1,𝑖+1 , we have 𝑌𝑙,𝑖 = 0; Subcase 1: There exists 𝑙 such that 𝑌𝑙,𝑖 ≠ 0 with 𝑣(𝐵𝑙,𝑙 ) > 1. Among all nonzero 𝑌2𝑙,𝑖 , we take 2𝑙 ′ to be the index of the largest 𝜁2𝑙 with respect to 𝜁 (𝑌 ). ©1 𝑧 ­0 𝑧 𝑎 1 ­ ­ ­0 0 ­ ­0 0 ­ ­ ­0 0 ­ «0 0 0 0 𝑏1 𝑏1 𝑧 ª 0 0 𝑏2 𝑏 2 𝑧 ®® ® 1 𝑧 𝑏 2𝑙 ′−1 𝑏 2𝑙 ′−1 𝑧 ®® 0 𝑧 𝑎𝑙 ′ 𝑏 2𝑙 ′ 𝑏 2𝑙 ′ 𝑧 ®® ® 0 0 𝑧 𝑧2 ®® 0 0 0 𝑧 𝑎 (𝑖+1)/2 ¬ Remove terms with 𝑏 2𝑙 ′−1 : left multiply by 1 (3,4) (−𝑏 2𝑙 ′−1 /𝑏 2𝑙 ′ ), then trick 1. Remove all 𝑏 2𝑚−1 ’s (i.e., in this matrix, remove 𝑏 1 ): left multiply by 1 (1,4) (−𝑏 1 /𝑏 2𝑙 ′ ), then trick 2. Remove all 𝑏 2𝑙 ’s (i.e., in this matrix, remove 𝑏 2 ): left multiply by 1 (2,4) (−𝑏 2 /𝑏 2𝑙 ′ ), then trick 3 (since we get rid of 𝑏 2𝑙 ′−1 first, row 1 and 2 are not affected in the columns 5, 6). Left multiply by 1 (5,4) (−𝑧/𝑏 2𝑙 ′ ), then trick 2 will take 𝑌4,4 to be 0. Reorder the row of 𝑧, 𝑧2 with the row of 𝑏 2𝑙 ′ : switch row 4, 5 and left multiply by 1 (4,4) (1/𝑏 2𝑙 ′ ). Notice that the current row of 𝑧 𝑎𝑙 ′ +1 is obtained by 𝑧 times that of the original row. Thus, any terms on the right will be removed to 0. Therefore, 𝑧 𝑎𝑙 ′ → 𝑧 𝑎𝑙 ′ +1 and the other entries in the corresponding row become permanent 0. On the tableau, this is equivalent to fill 𝑛 − 𝑖 + 1 in the added box of 39 row 𝑛/2 − 𝑙 ′ + 1. In the notation of lexicographical order, 𝑆 (𝑎𝑙 ′ ,𝑙 ′+ 𝑛2 −1),[𝛾] 𝑖+1/2 consists of all rows in 𝑌 above 𝑖 with possible nonzero entries with 𝜁 less than (𝑎 𝑙 ′ , 𝑙 ′ + 𝑛2 − 1) |𝑆 |−1 ′ 𝑛 and row 𝑖. Thus, 𝑤𝑡 𝑛−𝑖+1 (𝛾) = 𝑞 ( (𝑎𝑙 ′ ,𝑙 + 2 −1), [𝛾 ] (𝑖+1)/2 (𝑞 − 1). Subcase 2: For any 𝑙 such that 𝑣(𝐵𝑙,𝑙 ) > 1, we have 𝑌𝑙,𝑖 = 0. There exists 𝑙 such that 𝑌𝑙,𝑖 ≠ 0 where 𝐵𝑙,𝑙 = 1, 𝑣(𝑌𝑙+1,𝑙+1 ) ≤ 𝑣(𝐵1+𝑖,1+𝑖 ). Among all nonzero 𝑌2𝑙−1,𝑖 , we take 2𝑙 ′ −1 to be the index of the largest 𝜁2𝑙−1 with respect to 𝜁 (𝑌 ). ©1 𝑧 ­0 𝑧 𝑎 𝑙 ′ ­ ­ ­0 0 ­ ­0 0 ­ ­ ­0 0 ­ «0 0 0 0 𝑏 2𝑙 ′−1 𝑏 2𝑙 ′−1 𝑧ª 0 0 0 0 ®® ® 1 𝑧 𝑏 2𝑙−1 𝑏 2𝑙−1 𝑧 ®® 0 𝑧 𝑎𝑙 0 0 ®® ® 0 0 𝑧 𝑧2 ®® 0 0 0 𝑧 𝑎𝑖+1/2 ¬ Remove 𝑏 2𝑙 : left multiply by 1 (4,2) (−𝑧 𝑎𝑙 −𝑎𝑙 ′ 𝑏 2𝑙−1 /𝑏 2𝑙 ′−1 ) 1 (3,1) (−𝑏 2𝑙−1 /𝑏 2𝑙 ′−1 ), right multiply by1 (3,1) (𝑏 2𝑙−1 /𝑏 2𝑙 ′−1 ) 1 (4,2) (𝑏 2𝑙−1 /𝑏 2𝑙 ′−1 ). Remove the row 𝑧, 𝑧2 (we delete row 3,4): left multiply by 1 (2,2) (𝑏 2𝑙 ′−1 ) 1 (2,3) (𝑧 𝑎𝑙 ′ −2 𝑏 2𝑙 ′−1 ) 1 (4,2) (𝑧 𝑎𝑖+1/2 −𝑎𝑙 ′ /𝑏 2𝑙 ′−1 ) 1 (3,1) (𝑧/𝑏 2𝑙 ′−1 ), right multiply by 1 (1,3) (−𝑏 2𝑙 ′−1 ) 1 (2,4) (−𝑏 2𝑙 ′−1 ) 1 (4,2) (−1/𝑏 2𝑙 ′−1 ) 1 (3,1) (−1/𝑏 2𝑙 ′−1 ) 1 (4,4) (−1/𝑧 2 )(3, 4). Now, we permute rows and columns correspondingly to make 𝑧 𝑎𝑙 ′ −1 block still in row 2𝑙 ′ − 1, 2𝑙 ′. Therefore, it takes 𝑧 𝑎𝑙 ′ → 𝑧 𝑎𝑙 ′ −1 and the corresponding row becomes a permanent 0 row. On the tableau, this is equivalent to fill 𝑛 − 𝑖 + 1 in the added box of row 𝑛 ′ ′ 𝑛 2 + 𝑙 . In the notation of lexicographical order, 𝑆 (−𝑎 𝑙 ′ ,−𝑙 + 2 ),[𝛾] (𝑖+1)/2 consists of all rows above 𝑖 with possible nonzero entries with 𝜁 less than (−𝑎 𝑙 ′ , −𝑙 ′ + 𝑛2 ) and row |𝑆 |−1 ′ 𝑛 𝑖 (𝑣(𝐵 (𝑖+1)/2,(𝑖+1)/2 ) ≥ 𝑎 𝑙 ′ ). Thus, 𝑤𝑡 𝑛−𝑖+1 (𝛾) = 𝑞 (−𝑎𝑙 ′ ,−𝑙 + 2 ), [𝛾 ] 𝑖+1/2 (𝑞 − 1). Subcase 3: For any 𝑙 such that 𝑣(𝐵𝑙,𝑙 ) > 1 or 𝐵𝑙,𝑙 = 1, 𝑣(𝑌𝑙+1,𝑙+1 ) ≤ 𝑣(𝐵𝑖+1,𝑖+1 ), we have 𝑌𝑙,𝑖 = 0. Notice that this may be different from (𝑌1,𝑖 , 𝑌2,𝑖 , ..., 𝑌𝑖−1,𝑖 ) = (0, 0, .., 0). On the 2𝑙 − 1th row of 𝑌2𝑙−1,𝑖 ≠ 0 and 𝑌2𝑙,2𝑙 = 𝑧𝑌𝑖,𝑖 : (since the original 𝐵 is in increasing order, row 2𝑙 of 𝑌 is a permanent 0 row). 𝑧 𝑏 2𝑙 𝑏 2𝑙 𝑧 ª ©1 ­0 𝑧 𝑎𝑙 +1 0 0 ®® ­ ® ­ 2 ® ­0 0 𝑧 𝑧 ® ­ 𝑎 𝑖+1/2 0 0 0 𝑧 « ¬ Remove the row 𝑧, 𝑧2 : left multiply by 14,4 (−1) 14,3 (−𝑧 𝑎2𝑙 ′ −2 ) 12,4 (𝑧𝑏 2𝑙 ), right multiply by 11,3 (−𝑏 2𝑙 ) 12,4 (−𝑏 2𝑙 ) 14,4 (𝑧−2 ). Then switch columns 3, 4. 40 This takes 𝑧 𝑎𝑙 to 𝑧 𝑎𝑙 −1 and the corresponding row becomes a permanent 0 row. On the tableau, this is equivalent to fill 𝑛 − 𝑖 + 1 in the added box of row 𝑛+𝑖−1 2 . In the notation of lexicographical order, 𝑆 (−𝑎𝑖+1/2 , 𝑛−𝑖+1 ),[𝛾] 𝑖+1/2 consists of all rows ≤ 1 + 𝑖 2 in 𝑌 with possible nonzero entries with 𝜁 less than (−𝑎𝑖+1/2 , 𝑛−𝑖+1 2 ) (it is equivalent |𝑆 𝑛−𝑖+1 | to first rows with 𝑧1+𝑎𝑖+1/2 ). Thus, 𝑤𝑡 𝑛−𝑖+1 (𝛾) = 𝑞 (−𝑎𝑖+1/2 , 2 ) , [𝛾 ] 𝑖+1/2 . Moreover, it shines half light on the rule why no added box on the first row of 𝑧 𝑎𝑙 +1 . Case 3: 𝑣(𝐵𝑖,𝑖 ) = 0 and 𝑣(𝐵𝑖+1,𝑖+1 ) > 1 and 𝑋𝑖+1,𝑖+1 = 𝑧. This will be similar to case 2 combined with case 1, but with some subtle differences in subcase 3. Thus, the definition of 𝜁 applies here. 𝑋𝑙,𝑖+1 does not play a role until subcase 3. Subcase 1: There exists 𝑙 such that 𝑌𝑙,𝑖 ≠ 0 with 𝑣(𝐵𝑙,𝑙 ) > 1. It is equivalent to first do subcase 1 of case 2 then do case 1. Among all nonzero 𝑌2𝑙,𝑖 , we take 2𝑙 ′ to be the index of the largest 𝜁2𝑙 with respect to 𝜁 (𝑌 ). Therefore, 𝑧 𝑎𝑙 ′ → 𝑧 𝑎𝑙 ′ +1 ,𝑧 𝑎𝑖 → 𝑧 𝑎𝑖 +1 and the corresponding rows become permanent 0 row. On the tableau, this is equivalent to fill 𝑛 − 𝑖 + 1 in the added box of row 𝑛/2 − 𝑙 ′ + 1 and n-i in the added box of row 𝑛−𝑖+1 2 . Similarly to subcase 1 in case 2 |−1 |𝑆 ( (𝑎 ′ ,𝑙 ′ + 𝑛 −1), [𝛾 ] |𝑆 | (𝑖+1)/2 𝑙 2 (𝑞 − 1); 𝑤𝑡 𝑛−𝑖 = 𝑞 (∞,𝑛), [𝛾 ] 𝑖+1/2 , and case 1, 𝑤𝑡 𝑛−𝑖+1 (𝛾) = 𝑞 Subcase 2: For any 𝑙 such that 𝑣(𝐵𝑙,𝑙 ) > 1, we have 𝑌𝑙,𝑖 = 0. There exists 𝑙 such that 𝑌𝑙,𝑖 ≠ 0 where (𝐵𝑙,𝑙 = 1, 𝑣(𝑌𝑙+1,𝑙+1 ) < 𝑣(𝐵1+𝑖,1+𝑖 ))or(𝑣(𝑌𝑙+1,𝑙+1 ) = 𝑣(𝐵1+𝑖,1+𝑖 ) and 𝑣(𝐵𝑙+1,𝑙+1 ) = 𝑣(𝐵𝑖+1,𝑖+1 ) − 1). Among all nonzero 𝑌2𝑙−1,𝑖 , we take 2𝑙 ′ − 1 to be the index of the largest 𝜁2𝑙−1 with respect to 𝜁 (𝑌 ). (Recall in case 2’s subcase 3 had the condition 𝑣(𝑌𝑙+1,𝑙+1 ) ≤ 𝑣(𝐵1+𝑖,1+𝑖 )). This will take 𝑧 𝑎𝑙 ′ → 𝑧 𝑎𝑙 ′ −1 , 𝑧 𝑎𝑖 → 𝑧 𝑎𝑖 +1 , and the corresponding rows become permanent 0 row. On the tableau, this is equivalent to fill 𝑛 − 𝑖 + 1 in the added box of row 𝑛2 + 𝑙 ′ and 𝑛 − 𝑖 in the added box of row 𝑛−𝑖+1 2 . Similarly to subcase 2 in case |𝑆 (−𝑎 ′ ,−𝑙 ′ + 𝑛 ), [𝛾 ] |−1 |𝑆 | 𝑖+1/2 𝑙 2 2 and case 1, 𝑤𝑡 𝑛−𝑖+1 (𝛾) = 𝑞 (𝑞 − 1); 𝑤𝑡 𝑛−𝑖 = 𝑞 (∞,𝑛), [𝛾 ] 𝑖+1/2 . Subcase 3: For any 𝑙 such that 𝑣(𝐵𝑙,𝑙 ) > 1 or 𝐵𝑙,𝑙 = 1, 𝑣(𝑌𝑙+1,𝑙+1 ) ≤ 𝑣(𝐵𝑖+1,𝑖+1 ), we have 𝑌𝑙,𝑖 = 0 (except the case that 𝑣(𝑌𝑙+1,𝑙+1 ) = 𝑣(𝐵𝑙+1,𝑙+1 ) = 𝑣(𝐵𝑖+1,𝑖+1 )). 41 ©1 𝑧 ­0 𝑧 𝑐 ­ ­ ­0 0 ­ ­0 0 ­ ­ ­0 0 ­ «0 0 0 0 0 0 1 𝑧 0 𝑧𝑎 0 0 0 0 0 𝑏1 𝑧𝑎 ª ©1 𝑧 ® ­0 𝑧 𝑎 𝑎 0 𝑏2 𝑧 ® ­ ® ­ 𝑎 ® ­0 0 𝑏 𝑏𝑧 + 𝑏 3 𝑧 ® ­ → ­0 0 0 𝑏 4 𝑧 𝑎 ®® ­ ® ­ 2 ® ­0 0 𝑧 𝑧 ® ­ 0 𝑧 𝑎+1 ¬ «0 0 0 0 0 0 1 𝑧 0 𝑧𝑎 0 0 0 0 0 0 ª 0 𝑏 2 𝑧 𝑎 ®® ® ® 0 0 ® 0 (𝑏 4 − 𝑏)𝑧 𝑎 ®® ® ® 𝑧 𝑧2 ® 0 𝑧 𝑎+1 ¬ By trick 2, 𝑏 1 𝑧 𝑎 , 𝑏 3 𝑧 𝑎 go to 0. This takes 𝑧 𝑎 → 𝑧 𝑎 on col 5,6 if 𝑏 2 = 𝑏 4 − 𝑏 = 0. Also row 5, 6 are permanent 0 rows. Take 2𝑙 ′ − 1 to be the index of largest entries in 𝜁 (𝑌 ) with 𝑋˜ 2𝑙 ′,𝑖+1 − 𝑌2𝑙 ′−1,𝑖 ≠ 0 and 𝑌2𝑙 ′,2𝑙 ′ = 𝑧 𝑎 .Recall that 𝑋˜ is the corresponding X with current 𝑌 . This will take 𝑧 𝑎𝑙 ′ → 𝑧 𝑎𝑙 ′ −1 , 𝑧 𝑎𝑖 → 𝑧 𝑎𝑖 +1 and the corresponding rows become permanent 0 row. On the tableau, this is equivalent to fill 𝑛 − 𝑖 + 1 in the added box of row 𝑛−𝑖+1 𝑛 ′ 2 + 𝑙 and 𝑛 − 𝑖 in the added box of row 2 . In the notation of lexicographical order, 𝑆 (−𝑎𝑙 ′ ,−𝑙 ′+ 𝑛2 ),[𝛾] (𝑖+1)/2 consists of all rows < 𝑖 with possible nonzero entries with 𝜁 less than (−𝑎 𝑙 ′ , −𝑙 ′ + 𝑛2 ) and row i (𝑣(𝐵 (𝑖+1)/2,(𝑖+1)/2 ) = 𝑎 𝑙 ′ ). In the notation of lexicographical order, |𝑆 (∞,𝑛),[𝛾] 𝑖+1/2 | is the number of all rows above i+1 with |𝑆 ′ 𝑛 possible nonzero entries. Thus, 𝑤𝑡 𝑛−𝑖+1 (𝛾) = 𝑞 (−𝑎𝑙 ′ ,−𝑙 + 2 ), [𝛾 ] 𝑖+1/2 |𝑆 | 𝑞 (∞,𝑛), [𝛾 ] 𝑖+1/2 . |−1 (𝑞 − 1); 𝑤𝑡 𝑛−𝑖 = If no 𝑙 ′ exists, 𝑧 𝑎𝑖 → 𝑧 𝑎𝑖 and the corresponding two rows both become permanent 0 row. On the tableau, this is equivalent to fill 𝑛 − 𝑖 + 1 in the added box of row 𝑛+𝑖−1 and 𝑛 − 𝑖 in the added box of row 1+𝑛−𝑖 2 2 . Similarly to the above computation |𝑆 (−𝑎 ′ ,−𝑙 ′ + 𝑛 ), [𝛾 ] | |𝑆 (∞,𝑛), [𝛾 ] 𝑖+1/2 | 𝑖+1/2 ; 𝑤𝑡 𝑙 2 and subcase3 in case 2, 𝑤𝑡 𝑛−𝑖+1 (𝛾) = 𝑞 . 𝑛−𝑖 = 𝑞 Moreover, it shines the other half light on the rule why no added box on the first row of 𝑧 𝑎𝑙 +1 . □ Proof. [Theorem 3.1] Now we have all components to close the gap! Let 𝜈˜ denote {(𝜈1 + 5( 𝑝 − 1)) 𝑎1 , ..., (𝜈 𝑝−2 + 10) 𝑎 𝑝−2 , (𝜈 𝑝−1 + 5) 𝑎 𝑝−1 , (𝜈 𝑝 ) 𝑎 𝑝 } where 𝑎 𝜈 = {𝜈1𝑎1 , 𝜈2𝑎2 , ..., 𝜈 𝑝 𝑝 }. Given 𝜏 ∈ 𝑇𝜔 , 𝜔 ∈ 𝐷 𝜆𝑙,𝜈 , there exists corresponding 𝜏′ ∈ 𝛾 𝑇𝜔 ′ , 𝜔′ ∈ 𝐷 𝑙,𝜈˜ such that 𝜔, 𝜔′ have the same added box position and 𝜏, 𝜏′ have the ˜ Let Γ denote same numbers filled in. A key observation is that 𝛾 is not necessarily 𝜆. the set of all possible 𝛾’s. From the definition of weight, 𝑤𝑡 (𝜏) = 𝑤𝑡 (𝜏′). We call the number of pair (𝑖, 𝑗) where both rows have added boxes to be 𝑟𝑒(𝜏), the redundancy Í𝑙/2 Í Í 𝛾 of 𝜏. ℎ𝜆𝑙,𝜈 = 𝑖=0 𝜏∈𝑇𝜔 ,𝑟𝑒(𝜏)=𝑖 𝑤𝑡 (𝜏). For any 𝛾 with ℎ 𝑙, 𝜈˜ ≠ 0, all 𝜏 have 𝜔∈𝐷 𝜆 𝑙,𝜈 42 𝛾 Í Í the same redundancy, denoted by 𝑖. Thus, ℎ−1ℓ ,𝜈˜ = 𝜔∈𝐷 𝛾 𝜏∈𝑇𝜔 ,𝑟𝑒(𝜏)=𝑖 𝑤𝑡 (𝜏). Í Í Í Í𝑙, 𝜈˜ 𝛾 Therefore, ℎ𝜆−1ℓ ,𝜈 = 𝛾∈Γ 𝜔∈𝐷 𝜆 and 𝜔 ′ ∈𝐷 𝛾 𝜏∈𝑇𝜔 𝑤𝑡 (𝜏) = 𝛾∈Γ ℎ−1ℓ ,𝜈˜ . Recall the 𝑙,𝜈 𝑙, 𝜈˜ 𝜆 , each 𝑤 corresponds to one 𝜔 ∈ 𝐷 𝜆 , while the equivalent definition of [𝑤] and 𝑊𝑙,𝜈 𝑙,𝜈 relation is the rearranging order of 𝜔. □ Corollary 3.1. For any partition 𝜈 = {𝑎 1 𝑛1 , 𝑎 2 𝑛2 , ...𝑎 𝑗 𝑛 𝑗 } with distinct parts differing 𝜆 has by at least 2 and 𝑡 1 = 0, for any ℓ, 𝜆, each equivalence relation in the set 𝑊ℓ,𝜈 𝜔1,0 , 𝜔0,1 (1 ≤ 𝑖 ≤ ℓ) fixed. We denote it by Ω1,0 , Ω0,1 . For the simplicity of ℎ, we [𝑖] [𝑖] [𝑖] [𝑖] further write ∑︁ ∑︁ ∑︁ 1,0 𝑟= 𝑛𝑖 𝑛 𝑗 + 𝑛𝑖 (Ω0,1 − Ω ) + 𝑛𝑖 Ω0,1 ; [ 𝑗] [ 𝑗] [𝑖] 𝑖< 𝑗 𝑖< 𝑗 𝑀= 𝑖 ∑︁ 1 (ℓ − Ω1,0 + Ω0,1 ); [𝑖] [𝑖] 2 𝑖 𝑚𝑖 = 𝑛𝑖 − Ω1,0 − Ω0,1 for all i. [𝑎 ] [𝑎 ] 𝑖 𝑖 Then, ℎ𝜆−1ℓ ,𝜈 = 𝑞𝑟 𝑗  Ö 𝑖=1  𝑛𝑖 ∑︁ 𝑞 Ω1,0 , Ω0,1 [𝑎 𝑖 ] [𝑎 𝑖 ] 𝑒 1 +𝑒 2 +...+𝑒 𝑗 =𝑀 −1/2(( 2 𝑖 𝑚 𝑖 )−2𝑀) Í 𝑗   Ö 𝑚𝑖 2 𝑞 1/2(𝑚𝑖 +2𝑒𝑖 (𝑒𝑖 −𝑚𝑖 )) . 𝑒𝑖 𝑖=1 Proof. We have 𝑡1 = 0 if and only if there is no 1-dimensional block. Combining with the condition on gap of size 2, we know that all 𝜔1,0 , 𝜔0,1 are the same for all [𝑖] [𝑖] equivalent classes. Thus the first five summands in 𝑟 ([𝜔]) are the same among all 𝜔. Recall  𝜔 [𝑎 ¯ 𝑖] = 𝑛𝑖 𝜔0,1 , 𝜔1,1 , 𝜔1,0 [𝑎 𝑖 ] [𝑎 𝑖 ] [𝑎 𝑖 ]   = 𝑛𝑖 𝜔0,1 , 𝜔1,0 [𝑎 𝑖 ] [𝑎 𝑖 ]  𝑛 − 𝜔0,1 − 𝜔1,0  1 [𝑎 𝑖 ] 𝜔1,1 [𝑎 𝑖 ] [𝑎 𝑖 ] . Thus, summing over all 𝜔 is equivalent to summing over all 𝜔1,1 . Notice we cannot [𝑎 𝑖 ] further simplify the summand over 𝑒 1 + 𝑒 2 + ... + 𝑒 𝑗 = 𝑀 because this is not the
Í 𝑚
𝑛
(𝑚−𝑘+𝑡)𝑡 𝑞-Vandermonde formula 𝑡 𝑘−𝑡 = 𝑚+𝑛 □ 𝑡 𝑞 𝑘 . Recall in Proposition 1.2 that the ring structure of H (𝐺, 𝐾) is a polynomial ring over 𝜃 1 , 𝜃 2 , ..., 𝜃 𝑛 , 𝜃 𝑛−1 . Combining with the closed formulas of ℎ𝜆−1ℓ ,𝜈 , any ℎ𝜆𝜇,𝜈 is theoretically computable. However, due to the complexity nature of ℎ𝜆−1ℓ ,𝜈 (paired tuple structure), we only give another special case for ℎ𝜆𝜇,𝜈 in Chapter 4. 43 Chapter 4 DUAL PIERI A natural question to ask is what ℎ𝜆𝜇,𝜈 looks like under the restriction of Dual Pieri rule, i.e., the left multiplying matrix has dominant coweight {ℓ}. In this chapter, we will study ℎ𝜆{−ℓ},{𝑎 𝑛 } under the restriction ℓ < 𝑎 − 1. Theorem 2.3 states: For any 𝐴 ∈ GL𝑛 (𝑘 ((𝑧))) with 𝜌( 𝐴) = {𝑙}, any B ∈ GL𝑛 (𝑘 ((𝑧))) with 𝜎(𝐵) = {𝑎 1 , ..., 𝑎 𝑛 }, 𝜎( 𝐴𝐵) = {𝑏 1 , ..., 𝑏 𝑛 }, we have inequalities on 𝑎’s and 𝑏’s (Dual Pieri rule): 𝑎𝑖+1 ≤ 𝑏𝑖 ≤ 𝑎𝑖−1 . In Theorem 4.1, we make one simple assumption: 𝑛 ≥ 2 over 𝜈 = {𝑎 𝑛 }. Thus 𝜆 = {𝑎 + 𝑖, 𝑎 𝑛−2 , 𝑎 − 𝑗 } for 0 ≤ 𝑖, 𝑗 ≤ ℓ and 𝑖 + 𝑗 ≤ ℓ and 2|ℓ − (𝑖 + 𝑗). For other 𝜆, ℎ𝜆𝜇,𝜈 = 0. Theorem 4.2 computes ℎ𝜆{−ℓ},{𝑎 𝑛 } for 𝑛 = 1. In this chapter, we 𝑖, 𝑗 {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } rewrite 𝑝 ℓ = ℎ {−ℓ},{𝑎 𝑛 } . ℓ+𝑖+ 𝑗    𝑞 (ℓ+𝑖−2)𝑛− 2 +1 [𝑛] [𝑛 − 1] (𝑞 − 1) 2       𝑞 (ℓ+𝑖−2)𝑛−ℓ+2 [𝑛] [𝑛 − 1] (𝑞 − 1)       𝑞 (2ℓ+1)𝑛−(ℓ−1) [𝑛]     ℓ+𝑖 𝑖, 𝑗 Theorem 4.1. 𝑝 ℓ = 𝑞 (ℓ+𝑖−1)𝑛− 2 [𝑛] (𝑞 − 1)      𝑞 (ℓ−1)𝑛−(ℓ−1) [𝑛]    ℓ+ 𝑗    𝑞 (ℓ−1)𝑛− 2 [𝑛] (𝑞 − 1)     ℓ   𝑞 (ℓ−1)𝑛− 2 [𝑛] (𝑞 − 1)  𝑖, 𝑗 𝑗,𝑖 There is a symmetry on 𝑖, 𝑗: 𝑝 ℓ = 𝑞 (𝑖− 𝑗)𝑛 𝑝 ℓ . 𝑖, 𝑗 ≥ 1 and 𝑖 + 𝑗 < ℓ 𝑖, 𝑗 ≥ 1 and 𝑖 + 𝑗 = ℓ 𝑗 = 0 and 𝑖 = ℓ 𝑗 = 0 and 𝑖 < ℓ 𝑖 = 0 and 𝑗 = ℓ 𝑖 = 0 and 𝑗 < ℓ 𝑖= 𝑗 =0 Proof. From [3, Chapter 5], we recall that {𝑐𝜆 } forms a C-basis of H (𝐺, 𝐾). Here is some equalities in 𝑐𝜆 . 𝑐 {𝑘−ℓ} 𝑐 {−1𝑘 } = 𝑐 {−1𝑘−1 ,𝑘−1−ℓ} + 𝑞 𝑘 𝑐 {−1𝑘 ,𝑘−ℓ} for 𝑘 < min(ℓ, 2𝑛) 𝑐 {−1ℓ−1 } 𝑐 {−1} = [ℓ]𝑐 {−1ℓ } + 𝑐 {−1ℓ−2 ,−2} . If ℓ > 2𝑛, we notice that 𝑐 −12𝑛−1 ,2𝑛−1−ℓ = 𝑐 2𝑛−ℓ . For ℓ > 2𝑛, we have 𝑐 {−ℓ} = 𝑐 {1−ℓ} 𝑐 {−1} − 𝑞(𝑐 {2−ℓ} 𝑐 {−12 } − 𝑞 2 (𝑐 {3−ℓ} 𝑐 {−13 } − .... − 𝑞 2𝑛−1 𝑐 2𝑛−ℓ )). If ℓ ≤ 2𝑛, we have 𝑐 {−ℓ} = 𝑐 {1−ℓ} 𝑐 {−1} −𝑞(𝑐 {2−ℓ} 𝑐 {−12 } −𝑞 2 (𝑐 {3−ℓ} 𝑐 {−13 } −....−𝑞 ℓ−2 (𝑐 {−1} 𝑐 {−1ℓ−1 } −[ℓ]𝑐 {−1ℓ } ))). 44 Í From the definition of Hecke module, 𝑐 𝜇 × 𝑑𝜆 = 𝜈 ℎ𝜆𝜇𝜈 𝑑 𝜈 , and the commutativity of H (𝐺, 𝐾), we will prove the equality by induction on ℓ. The general strategy is: 𝑖, 𝑗 since each 𝑐 −𝑘 acts on 𝑑 {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } , we write explicitly the summation for 𝑝 ℓ in terms of ℎ and 𝑝. If ℓ ≤ 2𝑛, 𝑐 −ℓ ×𝑑 {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } = −[ℓ] ( ℓ−2 Ö −𝑞 𝑠 )𝑐 {−1ℓ } ×𝑑 {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } + 𝑠=1 𝑖, 𝑗 𝑝 ℓ = −[ℓ] ( ℓ−2 Ö 𝑟=1 𝑠=1 {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } −𝑞 𝑠 )ℎ {−1ℓ },{𝑎 𝑛 } ℓ−1 Ö 𝑟−1 ∑︁ ( −𝑞 𝑠 )𝑐 {−1𝑟 } 𝑐 {𝑟−ℓ} ×𝑑 {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℓ−1 Ö 𝑟−1 ∑︁ + ( −𝑞 𝑠 ) 𝑠=1 ∑︁ 𝑒, 𝑓 {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } 𝑝 ℓ−𝑟 ℎ {−1𝑟 },{𝑎+𝑒,𝑎 𝑛−2 ,𝑎− 𝑓 } . |𝑒−𝑖|,| 𝑗− 𝑓 |≤1;|𝑒−𝑖|+| 𝑓 − 𝑗 |≤𝑟;𝑒, 𝑓 ≥0 𝑟=1 𝑠=1 If ℓ > 2𝑛, 𝑖, 𝑗 𝑝ℓ = 2𝑛−1 𝑟−1 ∑︁ Ö ( 𝑟=1 𝑠=1 ∑︁ 𝑠 −𝑞 ) 𝑒, 𝑓 {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } 𝑝 ℓ−𝑟 ℎ {−1𝑟 },{𝑎+𝑒,𝑎 𝑛−2 ,𝑎− 𝑓 } +( |𝑒−𝑖|,| 𝑗− 𝑓 |≤1;|𝑒−𝑖|+| 𝑓 − 𝑗 |≤𝑟;𝑒, 𝑓 ≥0 2𝑛−1 Ö 𝑖, 𝑗 −𝑞 𝑠 ) 𝑝 ℓ−2𝑛 . 𝑠=1 For 𝑖, 𝑗 ≥ 2, 𝑖, 𝑗 𝑝ℓ = ∑︁ min ∑︁ (2𝑛,ℓ)−1 ∑︁ 𝑣∈{−1,0,1} 𝑟=1 𝑢∈{−1,0,1} {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } 𝑢+𝑖,𝑣+ 𝑗 ℎ {−1𝑟 },{𝑎+𝑢+𝑖,𝑎 𝑛−2 ,𝑎−𝑣− 𝑗 } 𝑝 ℓ−𝑟 ( 𝑟−1 Ö 𝑠=1 𝑠 −𝑞 )+( 2𝑛−1 Ö 𝑖, 𝑗 −𝑞 𝑠 ) 𝑝 ℓ−2𝑛 𝑠=1 (the last term is only nonzero when ℓ ≥ 2𝑛+𝑖+ 𝑗). To avoid extensive heavy notation, we denote 𝐴𝑟1 , 𝐴𝑟2 , ...𝐴𝑟9 the summand, in the order of the following ℎ’s. Explicitly, 𝑛−2 Î Î {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } 𝑖, 𝑗 𝑖, 𝑗 𝑠 ), 𝐴𝑟 = ℎ {𝑎+𝑖,𝑎 ,𝑎− 𝑗 } 𝑠 𝐴𝑟1 = ℎ {−12𝑟 },{𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } 𝑝 ℓ−2𝑟 ( 2𝑟−1 −𝑞 𝑝 ( 2𝑟+1 2𝑟+2 𝑠=1 𝑠=1 −𝑞 ). 2 {−1 },{𝑎+𝑖−1,𝑎 𝑛−2 ,𝑎− 𝑗+1} ℓ−2𝑟−2 Therefore, we have 1. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟 },{𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } =       𝑛 − 2 (𝑛+2−𝑟)𝑟 𝑛 − 2 −1−𝑟 2 +𝑛+𝑟𝑛 𝑛 − 2 (𝑛−𝑟) (𝑟+2) 𝑞 +2 𝑞 + 𝑞 𝑟 𝑟 −1 𝑟 −2 𝑗 ℓ−𝑖− 𝑗 𝑟 (𝐴𝑟1 has 𝑟 ∈ [1, ℓ−𝑖− 2 ] if ℓ < 2𝑛 or 𝑛 > 2 . 𝐴1 has 𝑟 ∈ [1, 𝑛−1] otherwise.); 2. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+2 },{𝑎+𝑖−1,𝑎 𝑛−2 ,𝑎− 𝑗+1} =   𝑛 − 2 𝑛(3+𝑟)−2𝑟−𝑟 2 −3 𝑞 𝑟 𝑗 ℓ−𝑖− 𝑗 𝑟 (𝐴𝑟2 has 𝑟 ∈ [0, ℓ−𝑖− 2 ] if ℓ < 2𝑛 or 𝑛 > 2 . 𝐴2 has 𝑟 ∈ [0, 𝑛−2] otherwise.); 45 3.  {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+2 },{𝑎+𝑖+1,𝑎 𝑛−2 ,𝑎− 𝑗−1} =  𝑛 − 2 𝑛(1+𝑟)−(𝑟+1) 2 ) 𝑞 𝑟 𝑗 𝑗 𝑟 (𝐴𝑟3 has 𝑟 ∈ [0, ℓ−𝑖− − 2] if ℓ < 2𝑛 or 𝑛 > ℓ−𝑖− 2 2 . 𝐴3 has 𝑟 ∈ [0, 𝑛 − 2] otherwise.); 4. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+2 },{𝑎+𝑖+1,𝑎 𝑛−2 ,𝑎− 𝑗+1} =   𝑛 − 2 𝑛(1+𝑟)−(𝑟+1) 2 )−1 𝑞 𝑟 𝑗 𝑗 𝑟 − 1] if ℓ < 2𝑛 or 𝑛 > ℓ−𝑖− (𝐴𝑟4 has 𝑟 ∈ [0, ℓ−𝑖− 2 2 . 𝐴4 has 𝑟 ∈ [0, 𝑛 − 2] otherwise.); 5. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+2 },{𝑎+𝑖−1,𝑎 𝑛−2 ,𝑎− 𝑗−1} =   𝑛 − 2 𝑛(3+𝑟)−(𝑟+1) 2 )−1 𝑞 𝑟 𝑗 𝑗 𝑟 − 1] if ℓ < 2𝑛 or 𝑛 > ℓ−𝑖− (𝐴𝑟5 has 𝑟 ∈ [0, ℓ−𝑖− 2 2 . 𝐴5 has 𝑟 ∈ [0, 𝑛 − 2] otherwise.); 6. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+1 },{𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗+1} =     𝑛 − 2 (𝑟+1)𝑛−𝑟 2 −1 𝑛 − 2 𝑛(2+𝑟)−(𝑟+1) 2 −1 𝑞 + 𝑞 𝑟 𝑟 −1 ℓ−𝑖− 𝑗 𝑗 𝑟 (𝐴𝑟6 has 𝑟 ∈ [0, ℓ−𝑖− 2 ] if ℓ < 2𝑛 or 𝑛 > 2 . 𝐴6 has 𝑟 ∈ [0, 𝑛−1] otherwise.); 7. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+1 },{𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗−1} =     𝑛 − 2 (𝑟+1)𝑛−𝑟 2 𝑛 − 2 𝑛(2+𝑟)−(𝑟+1) 2 𝑞 + 𝑞 𝑟 𝑟 −1 𝑗 𝑗 𝑟 − 1] if ℓ < 2𝑛 or 𝑛 > ℓ−𝑖− (𝐴𝑟7 has 𝑟 ∈ [0, ℓ−𝑖− 2 2 . 𝐴7 has 𝑟 ∈ [0, 𝑛 − 1] otherwise.); 8. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+1 },{𝑎+𝑖+1,𝑎 𝑛−2 ,𝑎− 𝑗 } =     𝑛 − 2 𝑟𝑛−𝑟 2 𝑛 − 2 𝑛(1+𝑟)−(𝑟+1) 2 𝑞 𝑞 + 𝑟 𝑟 −1 𝑗 𝑗 𝑟 − 1] if ℓ < 2𝑛 or 𝑛 > ℓ−𝑖− (𝐴𝑟8 has 𝑟 ∈ [0, ℓ−𝑖− 2 2 . 𝐴8 has 𝑟 ∈ [0, 𝑛 − 1] otherwise.); 9. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+1 },{𝑎+𝑖−1,𝑎 𝑛−2 ,𝑎− 𝑗 } =     𝑛 − 2 (𝑟+2)𝑛−𝑟 2 −1 𝑛 − 2 𝑛(3+𝑟)−(𝑟+1) 2 −1 𝑞 + 𝑞 𝑟 𝑟 −1 𝑗 ℓ−𝑖− 𝑗 𝑟 (𝐴𝑟9 has 𝑟 ∈ [0, ℓ−𝑖− 2 ] if ℓ < 2𝑛 or 𝑛 > 2 . 𝐴9 has 𝑟 ∈ [0, 𝑛−1] otherwise.); 46 𝑗 If 0 ≤ 𝑟 ≤ min ( ℓ−𝑖− 2 , 𝑛 − 1), 2 ℓ+𝑖+ 𝑗 𝐴𝑟2 +𝐴𝑟−1 = −[𝑛] [𝑛−1] (𝑞−1) 2 𝑞 𝑛(ℓ+𝑖−𝑟−2)−𝑟 − 2 +1 4     𝑗 𝑛 𝑛 2 𝑛(ℓ+𝑖−𝑟−1)−(𝑟+1) 2 − ℓ+𝑖+ +1 2 −[𝑛] [𝑛−1] (𝑞−1) 𝑞 𝑟 𝑟 −1 = −𝐴𝑟6 . 𝑗 − 1, 𝑛 − 2), (If 𝑟 = 𝑛 − 1, it follows that 𝐴𝑟2 = 0.) For 0 ≤ 𝑟 ≤ min ( ℓ−𝑖− 2 ℓ−𝑖− 𝑗 𝑟 𝑟−1 𝑟 𝑟 𝑟−1 𝐴3 + 𝐴5 + 𝐴7 = 0. For 1 ≤ 𝑟 ≤ min ( 2 , 𝑛 − 1), 𝐴1 + 𝐴8 + 𝐴𝑟9 = 0. We 𝑗 combine all summands under the assumption ℓ ≤ 2𝑛 or 𝑛 > ℓ−𝑖− 2 , the only left over term is 𝐴90 . We combined all the summands under the assumption ℓ > 2𝑛 and 𝑗 0 Î2𝑛−1 𝑛−1 𝑠 𝑖, 𝑗 𝑛 ≤ ℓ−𝑖− 2 , thus left over terms are 𝐴8 , 𝐴9 , ( 𝑠=1 −𝑞 ) 𝑝 ℓ−2𝑛 . From the induction, Î 𝑖−1, 𝑗 2𝑛−1 𝑖, 𝑗 0 𝑛−1 𝑠 𝑖, 𝑗 . ( 2𝑛−1 𝑠=1 −𝑞 ) 𝑝 ℓ−2𝑛 = 𝐴8 . Thus, the left over term is 𝐴9 . 𝑝 ℓ = 𝑝 ℓ−1 𝑞 For 𝑖 ≥ 2, 𝑗 = 1, 𝑖, 𝑗 𝑝ℓ = ∑︁ min ∑︁ (2𝑛,ℓ)−1 ∑︁ 𝑣∈{−1,0,1} 𝑟=1 𝑢∈{−1,0,1} {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } 𝑢+𝑖,𝑣+ 𝑗 ℎ {−1𝑟 },{𝑎+𝑢+𝑖,𝑎 𝑛−2 ,𝑎−𝑣− 𝑗 } 𝑝 ℓ−𝑟 ( 𝑟−1 Ö −𝑞 𝑠 )+( 𝑠=1 2𝑛−1 Ö 𝑠=1 We denote 𝐵𝑟1 , 𝐵𝑟2 , ...𝐵𝑟9 the summands, the same as that of 𝑖, 𝑗 ≥ 2. Therefore, we have, 1.     𝑛 − 2 (𝑛+2−𝑟)𝑟 𝑛 − 2 −1−𝑟 2 +𝑛+𝑟𝑛 𝑞 +2 𝑞 𝑟 𝑟 −1       𝑛 − 2 (𝑛−𝑟) (𝑟+2) 𝑛 − 2 𝑛𝑟−𝑟 2 −𝑟 𝑛 − 2 (𝑛−𝑟) (𝑟+1) + 𝑞 + 𝑞 + 𝑞 𝑟 −2 𝑟 − 1, 1 𝑟 − 2, 1 𝑛−2 ,𝑎−1} ℎ {𝑎+𝑖,𝑎 = 2𝑟 {−1 },{𝑎+𝑖,𝑎 𝑛−2 ,𝑎−1} 2. 𝑛−2 ,𝑎−1} ℎ {𝑎+𝑖,𝑎 = {−12𝑟+2 },{𝑎+𝑖−1,𝑎 𝑛−1 }   𝑛 − 1 𝑛(2+𝑟)−(𝑟+1) 2 𝑞 1, 𝑟 3.  {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+2 },{𝑎+𝑖+1,𝑎 𝑛−1 } = 4. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+2 },{𝑎+𝑖+1,𝑎 𝑛−2 ,𝑎− 𝑗−1} = 5. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+2 },{𝑎+𝑖−1,𝑎 𝑛−2 ,𝑎− 𝑗−1} = 6. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+1 },{𝑎+𝑖,𝑎 𝑛−1 } =    𝑛 − 1 𝑛𝑟−𝑟 2 −2𝑟 𝑞 𝑟, 1   𝑛 − 2 𝑛(𝑟+1)−(𝑟+1) 2 ) 𝑞 𝑟  𝑛 − 2 𝑛(3+𝑟)−(𝑟+1) 2 )−1 𝑞 𝑟    𝑛 − 1 𝑟𝑛−𝑟 2 𝑛 − 1 𝑛(1+𝑟)−(𝑟+1) 2 +1 𝑞 + 𝑞 𝑟, 1 1, 𝑟 − 1 𝑖, 𝑗 −𝑞 𝑠 ) 𝑝 ℓ−2𝑛 . 47 7.  {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+1 },{𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗−1} =    𝑛 − 2 (𝑟+1)𝑛−𝑟 2 𝑛 − 2 𝑛(2+𝑟)−(𝑟+1) 2 𝑞 + 𝑞 𝑟 𝑟 −1 8. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+1 },{𝑎+𝑖+1,𝑎 𝑛−2 ,𝑎− 𝑗 } =       𝑛 − 2 𝑟𝑛−𝑟 2 𝑛 − 2 𝑛(1+𝑟)−(𝑟+1) 2 𝑛 − 2 𝑛𝑟−(𝑟+1)𝑟 𝑞 + 𝑞 + 𝑞 𝑟 𝑟 −1 𝑟 − 1, 1 9. {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } ℎ {−12𝑟+1 },{𝑎+𝑖−1,𝑎 𝑛−2 ,𝑎− 𝑗 } =       𝑛 − 2 (𝑟+2)𝑛−𝑟 2 −1 𝑛 − 2 𝑛(3+𝑟)−(𝑟+1) 2 −1 𝑛 − 2 𝑛(𝑟+2)−(𝑟+1)𝑟−1 𝑞 + 𝑞 + 𝑞 . 𝑟 𝑟 −1 𝑟 − 1, 1 Before computing 𝑝𝑖,1 ℓ , one key observation on ℎ’s is the symmetry between (𝑖, 1) ,𝑎−1} ,𝑎−𝑖} = ℎ {𝑎+1,𝑎 (all five terms are the same.) and (1, 𝑖); ℎ {𝑎+𝑖,𝑎 {−12𝑟 },{𝑎+𝑖,𝑎 𝑛−2 ,𝑎−1} {−12𝑟 },{𝑎+1,𝑎 𝑛−2 ,𝑎−𝑖} 𝑛−2 𝑛−2 (𝑖−1)𝑛 𝑝 1,𝑖 . After direct computation for all nine 𝐵s, we observe that 𝑝𝑖,1 ℓ =𝑞 ℓ ℓ−𝑖−1 𝑟 𝑟−1 𝑟 If 0 ≤ 𝑟 ≤ min ( ℓ−𝑖−1 2 , 𝑛 − 2), 𝐵2 + 𝐵3 + 𝐵6 = 0. If 0 ≤ 𝑟 ≤ min ( 2 − 1, 𝑛 − 2), 𝑟 𝑟−1 + 𝐵𝑟 = 0. We 𝐵𝑟−1 + 𝐵𝑟5 + 𝐵𝑟7 = 0. If 1 ≤ 𝑟 ≤ min ( ℓ−𝑖−1 2 , 𝑛 − 1), 𝐵1 + 𝐵8 4 9 ℓ−𝑖− 𝑗 combine all summands under the assumption ℓ ≤ 2𝑛 or 𝑛 > 2 , and the only left over term is 𝐵90 . We combined all the summands under the assumption ℓ > 2𝑛 and 𝑗 𝑛−1 , 𝐵0 , ( Î2𝑛−1 −𝑞 𝑠 ) 𝑝 𝑖, 𝑗 . From the induction, 𝑛 ≤ ℓ−𝑖− , and left over terms are 𝐵 𝑠=1 8 9 ℓ−2𝑛 Î2𝑛−12 𝑠 𝑖, 𝑗 𝑛−1 ( 𝑠=1 −𝑞 ) 𝑝 ℓ−2𝑛 = 𝐵8 . Thus, the left over term is 𝐵90 . For 𝑖 ≥ 2, 𝑗 = 0, 𝑖, 𝑗 𝑝ℓ = ∑︁ min ∑︁ (2𝑛,ℓ)−1 ∑︁ 𝑣∈{−1,0,1} 𝑟=1 𝑢∈{−1,0,1} {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } 𝑢+𝑖,𝑣+ 𝑗 ℎ {−1𝑟 },{𝑎+𝑢+𝑖,𝑎 𝑛−2 ,𝑎−𝑣− 𝑗 } 𝑝 ℓ−𝑟 ( 𝑟−1 Ö −𝑞 𝑠 )+( 𝑠=1 2𝑛−1 Ö 𝑠=1 We denote 𝐶1𝑟 , 𝐶2𝑟 , ...𝐶6𝑟 the summands, the same as before. Therefore, we have 1. 𝑛−1 } ℎ {𝑎+𝑖,𝑎 = 2𝑟 {−1 },{𝑎+𝑖,𝑎 𝑛−1 }     𝑛 − 1 (𝑛+1−𝑟)𝑟 𝑛 − 1 (𝑛−𝑟) (1+𝑟) 𝑞 + 𝑞 𝑟 𝑟 −1 2. 𝑛−1 } ℎ {𝑎+𝑖,𝑎 = 2𝑟+2 {−1 },{𝑎+𝑖−1,𝑎 𝑛−2 ,𝑎−1}  3. 𝑛−1 } ℎ {𝑎+𝑖,𝑎 = {−12𝑟+2 },{𝑎+𝑖+1,𝑎 𝑛−2 ,𝑎−1} 4. 𝑛−1 } ℎ {𝑎+𝑖,𝑎 = 2𝑟+1 {−1 },{𝑎+𝑖,𝑎 𝑛−2 ,𝑎−1}   𝑛 − 2 𝑛(3+𝑟)−(𝑟+1) 2 −1 𝑞 𝑟   𝑛 − 2 𝑛(𝑟+1)−(𝑟+1) 2 𝑞 𝑟    𝑛 − 2 (𝑟+1)𝑛−𝑟 2 𝑛 − 2 𝑛(2+𝑟)−(𝑟+1) 2 𝑞 + 𝑞 𝑟 𝑟 −1 𝑖, 𝑗 −𝑞 𝑠 ) 𝑝 ℓ−2𝑛 . 48 5. 𝑛−1 } ℎ {𝑎+𝑖,𝑎 = {−12𝑟+1 },{𝑎+𝑖−1,𝑎 𝑛−1 }   𝑛 − 1 (𝑟+2)𝑛−𝑟 2 −𝑟−1 𝑞 𝑟 6. 𝑛−1 } ℎ {𝑎+𝑖,𝑎 = 2𝑟+1 {−1 },{𝑎+𝑖+1,𝑎 𝑛−1 }   𝑛 − 1 𝑟𝑛−𝑟 2 −𝑟 𝑞 . 𝑟 Before computing 𝑝𝑖,0 ℓ , one key observation on ℎ’s is the symmetry between (𝑖, 0) 𝑖𝑛 0,𝑖 and (0, 𝑖). After direct computation for all six 𝐶s, we observe that 𝑝𝑖,0 ℓ = 𝑞 𝑝ℓ . 𝑟 𝑟−1 + 𝐶 𝑟 = 0. If 1 ≤ 𝑟 ≤ min ( ℓ−𝑖 , 𝑛 − 1), If 0 ≤ 𝑟 ≤ min ( ℓ−𝑖 2 − 1, 𝑛 − 2), 𝐶2 + 𝐶3 2 4 𝑟−1 𝑟 𝑟 𝐶1 + 𝐶5 + 𝐶6 = 0. We combine all summands under the assumption ℓ ≤ 2𝑛 𝑗 0 or 𝑛 > ℓ−𝑖− 2 , and the only left over term is 𝐶5 . We combined all the sum𝑗 mands under the assumption ℓ > 2𝑛 and 𝑛 ≤ ℓ−𝑖− 2 , and left over terms are Î Î2𝑛−1 𝑠 𝑖, 𝑗 𝑠 𝑖, 𝑗 𝑛−1 𝐶6𝑛−1 , 𝐶50 , ( 2𝑛−1 𝑠=1 −𝑞 ) 𝑝 ℓ−2𝑛 . From the induction, ( 𝑠=1 −𝑞 ) 𝑝 ℓ−2𝑛 = 𝐶6 . Thus, 𝑖−1,0 2𝑛−1 the left over term is 𝐶50 . 𝑝𝑖,0 . ℓ = 𝑝 ℓ−1 𝑞 For 𝑖 = 1, 𝑗 = 0, 𝑖, 𝑗 𝑝ℓ = ∑︁ min ∑︁ (2𝑛,ℓ)−1 ∑︁ 𝑣∈{−1,0,1} 𝑟=1 𝑢∈{−1,0,1} {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } 𝑢+𝑖,𝑣+ 𝑗 ℎ {−1𝑟 },{𝑎+𝑢+𝑖,𝑎 𝑛−2 ,𝑎−𝑣− 𝑗 } 𝑝 ℓ−𝑟 − [ℓ] ( ( 𝑟−1 Ö −𝑞 𝑠 ) 𝑠=1 ℓ−2 Ö {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } −𝑞 𝑠 )ℎ {−1ℓ },{𝑎 𝑛 } +( 2𝑛−1 Ö 𝑠=1 𝑖, 𝑗 −𝑞 𝑠 ) 𝑝 ℓ−2𝑛 . 𝑠=1 Therefore, it follows that, 1. 𝑛−1 } ℎ {𝑎+1,𝑎 = 2𝑟 {−1 },{𝑎+1,𝑎 𝑛−1 }       𝑛 − 1 (𝑛+1−𝑟)𝑟 𝑛 − 1 (𝑛−𝑟) (1+𝑟) 𝑛 − 1 (𝑛−𝑟)𝑟 𝑞 + 𝑞 + 𝑞 𝑟 𝑟 −1 𝑟 − 1, 1 2. 𝑛−1 } ℎ {𝑎+1,𝑎 = {−12𝑟+2 },{𝑎 𝑛−1 ,𝑎−1}   𝑛 − 1 𝑛(2+𝑟)−(𝑟+1) 2 +1 𝑞 𝑟, 1 3. 𝑛−1 } ℎ {𝑎+1,𝑎 = 2𝑟+2 {−1 },{𝑎+2,𝑎 𝑛−2 ,𝑎−1}   𝑛 − 2 𝑛(𝑟+1)−(𝑟+1) 2 𝑞 𝑟 4. 𝑛−1 } ℎ {𝑎+1,𝑎 = {−12𝑟+1 },{𝑎+1,𝑎 𝑛−2 ,𝑎−1}       𝑛 − 2 (𝑟+1)𝑛−𝑟 2 𝑛 − 2 𝑛(2+𝑟)−(𝑟+1) 2 𝑛 − 2 (1+𝑟)𝑛−𝑟−𝑟 2 𝑞 + 𝑞 + 𝑞 𝑟 𝑟 −1 𝑟 − 1, 1 49 5. 𝑛−1 } ℎ {𝑎+1,𝑎 = {−12𝑟+1 },{𝑎 𝑛 }   𝑛 𝑞 (𝑟+1)𝑛−(𝑟+1)𝑟 𝑟, 1 6.  𝑛−1 } ℎ {𝑎+1,𝑎 = 2𝑟+1 {−1 },{𝑎+2,𝑎 𝑛−1 }  𝑛 − 1 𝑟𝑛−𝑟 2 −𝑟 𝑞 . 𝑟 Instead of computing 𝑝𝑖,1 ℓ , one key observation on ℎ’s is the symmetry between (1, 0) and (0, 1). After direct computation for all six terms, we observe that 𝑝 ℓ1,0 = 𝑞 𝑛 𝑝 ℓ0,1 . Recall the measure of 𝐾𝜋 ℓ 𝐾 (= 𝜇(𝑐 ℓ ) = 𝜇(𝑐 −ℓ )) = 𝑞 (2𝑛−1) (ℓ−1) [2𝑛], we can write 𝑖, 𝑗 (2𝑛−1) (ℓ−1) [2𝑛]. 𝑖, 𝑗 𝑝 ℓ = 𝑞 Í Í ℓ−1 We notice that 𝑖 𝑝𝑖+2𝑚−1,𝑖 = [𝑛] 2 𝑞 (𝑛−1) (ℓ−2)+( 2 +𝑚)𝑛−1 (𝑞 − 1) for 2𝑚 − 1 ≠ ℓ Í (𝑛−1) (ℓ−1)+ℓ𝑛 . Í 𝑝 𝑖−ℓ,𝑖 = 𝑝 0,ℓ = [𝑛]𝑞 (𝑛−1) (ℓ−1) . ℓ, −ℓ, 1, −1. 𝑖 𝑝𝑖+ℓ,𝑖 = 𝑝 ℓ,0 𝑖 ℓ ℓ ℓ = [𝑛]𝑞 ℓ ℓ+1 1,0 ℓ𝑛− ℓ 2 [𝑞] (𝑞 − 1) and Combining with the measure of 𝐾𝜋 𝐾, we have 𝑝 ℓ = 𝑞 ℓ−1 𝑝 ℓ0,1 = 𝑞 ℓ𝑛− 2 [𝑞] (𝑞 − 1). For 𝑖 = 𝑗 = 0, 𝑖, 𝑗 𝑝ℓ = ∑︁ min ∑︁ (2𝑛,ℓ)−1 ∑︁ 𝑣∈{−1,0,1} 𝑟=1 𝑢∈{−1,0,1} {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } 𝑢+𝑖,𝑣+ 𝑗 ℎ {−1𝑟 },{𝑎+𝑢+𝑖,𝑎 𝑛−2 ,𝑎−𝑣− 𝑗 } 𝑝 ℓ−𝑟 ( − [ℓ] ( 𝑟−1 Ö −𝑞 𝑠 ) 𝑠=1 ℓ−2 Ö {𝑎+𝑖,𝑎 𝑛−2 ,𝑎− 𝑗 } −𝑞 𝑠 )ℎ {−1ℓ },{𝑎 𝑛 } +( 𝑠=1 2𝑛−1 Ö 𝑖, 𝑗 −𝑞 𝑠 ) 𝑝 ℓ−2𝑛 . 𝑠=1 We denote 𝐷 𝑟1 , 𝐷 𝑟2 , 𝐷 𝑟3 , 𝐷 𝑟4 those summands, the same as before. It follows that 1.   𝑛 𝑛𝑟−𝑟 2 𝑞 𝑟 𝑛} ℎ {𝑎 = {−12𝑟 },{𝑎 𝑛 } 2.  𝑛} ℎ {𝑎 = {−12𝑟+2 },{𝑎+1,𝑎 𝑛−1 ,𝑎−1} 3. 𝑛} ℎ {𝑎 = {−12𝑟+1 },{𝑎+1,𝑎 𝑛−1 } 4. 𝑛} ℎ {𝑎 = {−12𝑟+1 },{𝑎 𝑛−1 ,𝑎−1}   𝑛 − 2 𝑛(1+𝑟)−(𝑟+1) 2 𝑞 𝑟   𝑛 − 1 𝑟𝑛−𝑟 2 −𝑟 𝑞 𝑟  𝑛 − 1 𝑛(𝑟+1)−𝑟 2 −𝑟 𝑞 . 𝑟 50 If 0 ≤ 𝑟 ≤ min ( 2ℓ − 1, 𝑛 − 1), 𝐷 𝑟3 = 𝐷 𝑟4 . If 1 ≤ 𝑟 ≤ min ( 2ℓ − 2, 𝑛 − 2), 𝑟 𝑟 𝑛−𝑟 [𝑟]). 𝐷 𝑟1 + 𝐷 𝑟2 = 𝐷 𝑟−1 3 + 𝐷 4 (from the equality [𝑛−𝑟] [𝑟] (𝑞 −1) + [𝑛] = 𝑞 [𝑛−𝑟] +𝑞 ℓ −2 ℓ −1 ℓ ℓ −1 Thus, we have 𝐷 42 + 2𝐷 32 + 𝐷 12 + 𝐷 12 + 𝐷 02 = 0. We combine all summands 𝑗 0 under the assumption ℓ ≤ 2𝑛 or 𝑛 > ℓ−𝑖− 2 , and the only left over term is 𝐷 4 . 0,1 𝑛 𝑝 ℓ0,0 = 𝑝 ℓ−1 𝑞 . We combined all the summands under the assumption ℓ > 2𝑛 and Î ℓ−𝑖− 𝑗 𝑠 0,0 𝑛 ≤ 2 : with 𝐷 4𝑛−2 + 2𝐷 3𝑛−1 + 𝐷 1𝑛−1 + 𝐷 02 + ( 2𝑛−1 𝑠=1 −𝑞 ) 𝑝 ℓ−2𝑛 = 0, and the left over term is 𝐷 04 . Í 𝑖, 𝑗 For 𝑖 = 𝑗 = 1, 𝑝 ℓ1,1 is obtained from: 𝑖, 𝑗 𝑝 ℓ = 𝑞 (2𝑛−1) (ℓ−1) [2𝑛]. The proof is the Í 𝑖, 𝑗 □ same as the one for 𝑖 = 1, 𝑗 = 0, by evaluating 𝑖, 𝑗 𝑝 ℓ . 𝑎+ 𝑗 We now analyze ℎ {−ℓ},{𝑎} , i.e., 𝑛 = 1. 𝑗+ℓ    𝑞 2 −1 (𝑞 − 1)     𝑎+ 𝑗 Theorem 4.2. ℎ {−ℓ},{𝑎} = 1      𝑞ℓ  𝑗 = ℓ − 2, ℓ − 4, ..., 2 − ℓ 𝑗 = −ℓ 𝑗 =ℓ Ã Proof. Claim: For all 𝐴 ∈ 𝐺 𝐿 2 (𝑘 [[𝑧]]) with 𝜌( 𝐴) = 𝑎, we have 𝐾 𝐴𝐾 = 𝑋∈A 𝐾 𝑋 where A consists of upper triangular matrices 𝑋 with following properties: 1. 𝑋1,1 = 𝑧𝑡 , 𝑋2,2 = 𝑧𝑟 2. 𝑋1,2 = 𝑏 0 + 𝑏 1 𝑧 + 𝑏 2 𝑧 2 + ... + 𝑏𝑟−1 𝑧𝑟−1 where 𝑏𝑖 ∈ 𝑘 3. if 𝑡 ≠ 0, then 𝑏 0 ≠ 0 4. 𝑡 + 𝑟 = ℓ The proof of the claim follows the same strategy of Lemma 3.1. From trick 1 of Proposition 3.4 in Chapter! 3, we have the following: each 𝑋 ∈ ℓ− 𝑗 ℓ+ 𝑗 1 𝑧 is in the form of 𝑧 𝑎+ 𝑗 . The number A with 𝑋2,2 = 𝑧 2 , 𝑋1,1 = 𝑧 2 , 𝑋 𝑎 0 𝑧 𝑗+ℓ of X in A with the above condition is 𝑞 2 −1 (𝑞 − 1) if 𝑗 ≠ −ℓ, ℓ, and ℎ𝑎+ℓ = {−ℓ},{𝑎} 𝑞 ℓ , ℎ𝑎−ℓ = 1. □ {−ℓ},{𝑎} 51 BIBLIOGRAPHY [1] Daniel Bump. “Hecke Algebras”. 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