On the Link Floer Homology of L-space Link Citation Dawra, Nakul (2015) On the Link Floer Homology of L-space Link. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/Z9CZ353T. https://resolver.caltech.edu/CaltechTHESIS:05222015-133207861 Abstract We will prove that, for a 2 or 3 component L-space link, HFL - is completely determined by the multi-variable Alexander polynomial of all the sub-links of L, as well as the pairwise linking numbers of all the components of L. We will also give some restrictions on the multi-variable Alexander polynomial of an L-space link. Finally, we use the methods in this paper to prove a conjecture of Yajing Liu classifying all 2-bridge L-space links. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: Heegaard Floer Homology, Low-dimensional Topology Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Awards: Apostol Award For Excellence In Teaching In Mathematics, 2013 Thesis Availability: Public (worldwide access) Research Advisor(s): Ni, Yi Thesis Committee: Ni, Yi (chair) Marcolli, Matilde Graber, Thomas B. Vafaee, Faramarz Defense Date: 19 May 2015 Record Number: CaltechTHESIS:05222015-133207861 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:05222015-133207861 DOI: 10.7907/Z9CZ353T Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 8879 Collection: CaltechTHESIS Deposited By: Nakul Dawra Deposited On: 26 May 2015 21:22 Last Modified: 05 Jul 2022 19:06 Thesis Files Preview PDF - Final Version See Usage Policy. 539kB Repository Staff Only: item control page

On the link Floer homology of L-space links Thesis by Nakul Dawra In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2015 (Defended May 19, 2015) ii c 2015 Nakul Dawra All Rights Reserved iii Acknowledgements I would like to thank my advisor Yi Ni for sharing his knowledge and providing invaluable support. I owe so much to your gentle and patient guidance. Thank you for being such a wonderful mentor and friend, and helping me keep my Legandrians and Lagrangians straight. Thanks to Matilde Marcolli, Tom Graber and Faramarz Vafee for participating in my defense. Thanks to Yajing Liu, whose work served as a key source of inspiration for this project and who also read my early drafts. Thanks again to Faramarz Vafaee for offering me significant guidance while helping me edit my work. I was partially funded through this project by the Troesh family through the Troesh Family Fellowship, which made focusing on research much easier. My graduate life was made more enjoyable by the presence of Foo Yee Yeo, Daiqi Linghu, Scott Geraedts, and my wonderful office mate Liling Gu. My roommates Andrew Friedman, Stuart Friedel, and Lee and Sadie Spivak added endless color and spontaneity to my life. I am incredibly fortunate to have met Ross Elliot, my most stalwart and trusted friend. Nothing in my life would be possible without the foundation my parents have built for me and I can’t thank them enough for all they have done. I would like to thank my sister for being a wonderful person and an unyielding source of courage and inspiration. Thanks to James Jensen for the myriad bits of help and pleasant company. Lastly, I must thank Emily Jensen for standing by me through all of this. I could not have done it without you. iv Abstract We will prove that, for a 2 or 3 component L-space link, HF L− is completely determined by the multi-variable Alexander polynomial of all the sub-links of L, as well as the pairwise linking numbers of all the components of L. We will also give some restrictions on the multi-variable Alexander polynomial of an L-space link. Finally, we use the methods in this paper to prove a conjecture of Yajing Liu classifying all 2-bridge L-space links. v Contents Acknowledgements iii Abstract iv 1 Introduction 1 2 Homological Preliminaries 6 3 The Chain Complex 10 4 Proof of the Main Theorems 16 5 Application to 2-Bridge Links 21 Bibliography 35 1 Chapter 1 Introduction In [18] and [23], knot and link Floer homology were defined as a part of Ozsváth and Szabó’s Heegaard Floer theory (introduced in [20]). These give rise to graded homology groups which are invariants of isotopy classes of knots and links embedded in S 3 . Carefully examining these groups has yielded a wealth of topological insights (see [14], [15], [16], [25], [26] and [30]). The Euler characteristic of link (knot) Floer homology is the multi-variate (single variable) Alexander polynomiala . Throughout this paper, we will work over the field F = Z/2Z, and L = L1 t L2 t . . . t Ll will always be an l component link inside S 3 unless otherwise specified. We will focus on links all of whose large positive surgeries yield L-spaces. L-spaces are rational homology spheres whose Heegaard Floer homology is the simplest possible. d (Y )) ≥ More specifically, recall that for any rational homology 3-sphere Y we must have dim(HF |H1 (Y )|, and so we define an L-space as: d (Y )) = |H1 (Y )|. Definition 1.1 Y a QHS 3 is an L-space if dim(HF Lens spaces are the simplest examples of L-spaces. Further examples include any connected sums of 3-manifolds with elliptic geometry [21], as well as double branched covers of quasi-alternating links [22]. It was shown in Theorem 1.4 of [20] that such manifolds do not admit co-orientable C 2 -taut foliations. We will define an L-space link as follows: Definition 1.2 L ⊂ S 3 is an L-space link if the 3-manifolds Sn3 1 ,...,nl (L) obtained by surgery on L are all L-spaces when all of the ni are sufficiently large.b . L-space links were first studied in [6], where it was shown that any link arising as the embedded link of a complex plane curve singularity (i.e. algebraic link) is an L-space link (note that this a This is “almost true”, we will make it precise in Definition 1.4. b Note that this definition does not depend on the orientation of the components of L. 2 includes all torus links). The general study of properties and examples of L-space links was initiated in [10] see also [5]. L-space knots were first examined in [21]. In that paper it was shown that for an L-space knot, the knot Floer homology is completely determined by its Euler Characteristic (i.e. the Alexander polynomial). In this paper, we give a generalization of this statement to 2 and 3 component L-space links inside S 3 . First, we recall some standard facts and notation. Definition 1.3 Let H(L)i denote the affine lattice over Z given by lk(Li , L\Li )/2 + Z. We define: H(L) := l M H(L)i . i=1 We can think of every element of H(L) as an element of the set of relative Spinc structures of L ⊂ S 3 via the identification H(L) → Spinc (S 3 , L) given in section 8.1 of [23]. Note that H(L) is an affine lattice over H1 (S 3 − L) ∼ = Zl . \ Both HF L− and HF L for a link L inside S 3 split into direct summands indexed by pairs (d, s), where d ∈ Z (the homological grading) and s ∈ H(L). We will write these summands as HF L− d (L, s) \ and HF Ld (L, s). Now, if s = (s1 , s2 , . . . , sl ) ∈ H(L), we denote by us the monomial us11 . . . usl l . Definition 1.4 In this paper, we define the symmetric multi-variable Alexander polynomial ∆L (u1 , u2 , . . . , ul ) for L so that the following equalityc holds: X \ χ(HF L∗ (L, s))us = s∈H(L) l   Y 1/2 −1/2 ui − ui ∆L (u1 , u2 , . . . , ul ). i=1 Theorem 1.5 Let L ⊂ S 3 be a 2 or 3 component L-space link and let s ∈ H(L). Then HF L− (L, s) is completely determined by the symmetric multi-variable Alexander polynomials ±∆M for every sub-link M ⊂ L, as well as the pairwise linking numbers of components of L. In [21], it was shown that being an L-space knot forces strong restrictions on the Alexander polynomial, and we will generalize this to links. Our restrictions will depend on the Alexander polynomial of the link L, as well as the Alexander polynomial of all its sub-links after a shift depending on various linking numbers. Definition 1.6 Given a proper subset S = {i1 , i2 , . . . , ik } ( {1, . . . , l}, we let {j1 , j2 , . . . , jl−k } = {1, . . . , l}\S where ja < jb when a < b. Let LS ⊂ L be the sub-link Li1 t Li2 t . . . t Lik . The c In proposition 9.1 of [23], the above equality was only shown to hold up to sign. So our sign convention for ∆ L here may not be standard, but it will make the statement of some of our Theorems easier. For our main Theorem, we only need to know ∆L up to sign. 3 polynomial PLLS is defined as follows: When S = ∅ we have, ! l Y P∅L = 1/2 ui ∆L (u1 , . . . , ul ); i=1 When l − k > 1 we have, ! l−k Y PLLS (uj1 , uj2 , . . . , ujl−k ) = 1/2+lk(Lji ,LS )/2 uji ∆L\LS (uj1 , . . . , uj(l−k) ); p=1 And finally when l − k = 1 we have, lk(Lj1 ,LS ) PLLS (uj1 ) = uj1 2   X   ∆L\LS (uj1 ). u−i j 1 i≥0 Now, fix some s = (s1 , s2 , . . . , sl ) ∈ H(L) and r ∈ {1, . . . , l} so that r 6∈ S. Then, define R s0 ≥s (PLLS ) s0r =sr s0 s0j l−k to be the sum of all the coefficients of monomials uj11 . . . ujl−k of PLLS that satisfy s0r = sr and s0jp ≥ sjp for jp 6= r. Example 1.7 Consider the 2-bridge link L = b(20, −3) (see Section 5 for definitions and notation). Then; 1/2 3/2 ∆L (u1 , u2 ) =u1 u2 3/2 1/2 + u1 u2 1/2 1/2 − u1 u2 1/2 −1/2 + u1 u2 −1/2 1/2 −3/2 −1/2 −1/2 −3/2 3/2 3/2 u2 + u1 u2 + u1 u2 − u1 u2 + u1 −1/2 −1/2 −3/2 −3/2 u2 − u1 u2 . − u1 P∅L (u1 , u2 ) =u1 u22 + u21 u2 + u1 + u2 + 1 1 1 + − u21 u22 − u1 u2 − 1 − . u1 u2 u1 u2 L = L1 t L2 is a 2 component link with both components unknots. The linking number of the 2 components is 2 so;  PLL1 (u2 ) = u2   X i≥0   and PLL (u1 ) = u1  u−i 2 2  X i≥0 . u−i 1 4 Theorem 1.8 If L is an L-space link, then for any s ∈ H(L) and r ∈ {1, 2, . . . , l}: X (−1)l−1−|S| R s0 ≥s (PLLS ) = 0 or 1. s0r =sr S⊂{1,...,l} r6∈S Remark 1.9 When l = 1, this says that the coefficients of P∅L are all 1 or 0, which follows from the work in [21]. Given any 2 variable polynomial F (u1 , u2 ), we define F |(i,j) , where i = 1 or 2, to be the polynomial obtained from F by discarding all monomials where the exponent of ui is not equal to j. Then the above Theorem, when restricted to the l = 2 case, reads as follows: Corollary 1.10 Suppose that L = L1 t L2 is an L-space link. Then the nonzero coefficients of P∅L are all ±1. The nonzero coefficients of P∅L |(r,s0r ) for r = 1 or 2 and any s0r ∈ H(L)r , alternate in s0 sign. The first nonzero coefficient of P∅L |(r,s0r ) is −1 if the coefficient of urr in PLL3−r is 0; and the s0 first nonzero coefficient of P∅L |(r,s0r ) is 1 if the coefficient of urr in PLL3−r is 1. Proof: As in Theorem 1.8, fix s0 = (s01 , s02 ). Suppose without loss of generality that r = 1. We denote by as1 ,s2 the coefficient of us11 us22 in P∅L (u1 , u2 ), and as1 the coefficient of us11 in PLL2 (u1 ). Then according to Theorem 1.8: as01 − X as01 ,s2 = 0 or 1. (1.1) as01 ,s2 = 0 or 1. (1.2) s2 ≥s02 Similarly; as01 − X s2 ≥s02 +1 Subtracting 1.1 from 1.2 gives as01 ,s02 = −1, 0 or 1. We have thus shown that all the coefficients of P∅L or −1, 0 or −1. We know that as01 must be either 1 or 0 (see Remark 1.9). Combining this with P P equation 1.1 gives that s2 ≥s0 as01 ,s2 = 0 or 1 if as01 = 1, and s2 ≥s0 as01 ,s2 = 0 or −1 if as01 = 0. 2 2 The rest of the corollary now immediately follows.  Part of the above corollary was already shown directly in Theorem 1.15 of [10]. Additionally in [10], it was shown that when q and k are odd positive integers b(qk − 1, −k) is an L-space link. This was conjectured to be a complete list of 2-bridge L-space links, which is correct. Theorem 1.11 If L is a 2-bridge L-space link, then, after possibly reversing the orientation of one of the components, L is equivalent to b(qk − 1, −k) for some positive odd integers q and k. The organization of this paper is as follows. Section 2 consists of some homological algebra needed to compute HF L− (L) from its Euler characteristic when L is a 2 or 3 component L-space 5 link. Section 3 generalizes the arguments in [23] to work on links. In Section 4 Theorem 1.5 is proved, as well the the restrictions on the Alexander polynomials of L-spaces. In Section 5 we prove the classification of 2 bridge L-space links. 6 Chapter 2 Homological Preliminaries Definition 2.1 Let En = {0, 1, 2}n ⊂ Rn where n ≥ 1. We will denote (0, 0, . . . , 0), (1, 1, . . . 1) and (2, 2, . . . , 2) by 0,1 and 2 respectively. For any ε ∈ En , we denote by εj the jth coordinate of ε and by ej the jth elementary coordinate vector. We define an n-dimensional short exact cube of chain complexes, C (or short exact cube for short), as follows: 1 For every ε ∈ En there is a chain complex C ε over F. 2 Suppose that ε0 , ε and ε00 are in En and only differ in the jth coordinate with ε0j = 0, εj = 1 and ε00j = 2. Then there is a short exact sequence 0 / C ε0 iε0 ε / Cε jεε00 / C ε00 / 0. 3 The diagram made by all of the complexes C ε and maps iε0 ε , jεε00 is commutative. We will denote C (2,2,...,2) as C for short. We define the cube of inclusions, C I , to be the subdiagram consisting of all the chain complexes C ε with ε ∈ {0, 1}n and the corresponding inclusion maps. We call a short exact cube basic if the following additional properties hold: 4 For ε ∈ {0, 1}n , H∗ (C ε ) ∼ = F[U ] where multiplication by U drops homological grading by 2. We do not specify what the top grading for F[U ] is, but we do require that it is even. 5 All of the maps (iε0 ε )∗ , induced by homology in the cube of inclusions are either isomorphisms in all degrees, or (iε0 ε )∗ is injective in all degrees and the top degree supported in H∗ (Cε ) is 2 higher than the top degree supported in H∗ (Cε0 ). In other words U H∗ (C ε ) ∼ = H∗ (C ε0 ) When the top grading for F[U ] is d, we will write it as F(d) [U ]. Similarly, F(d) will be used to denote F supported in degree d. Given an n dimensional basic short exact cube C, if we restrict to the commutative diagram coming from the subset of En with jth coordinate i where i = 0, 1 or 2, this can be thought of as 7 an n − 1 dimensional short exact cube of chain complexes which we will denote by ji C. For any j ∈ {1, 2, . . . , n}, j2 C is the same as C; and j0 C and j1 C are basic. Lemma 2.2 Suppose C is a basic short exact cube of chain complexes. Also let ε ∈ En have some coordinate equal to 2. Then, H∗ (C ε ) is finite dimensional. Proof: In the n = 1 case, H∗ (C) is either F or 0 by property 5 of basic short exact cubes. Thus, for any n-dimensional basic short exact cube C, the homologies of the complexes in j21 C I are only either F or 0 for any j1 . From here we can conclude that the homologies of the complexes in j22 j21 C I are finite and continuing with this argument proves the claim.  Definition 2.3 If C is a basic short exact cube, then we define the hypercube graph of C, HC(C), as a directed graph with labeled edges as follows: • The vertices correspond to the elements of the set {0, 1}n . • There is a directed edge from ε0 to ε if the two agree in all coordinates except the jth for some 1 ≤ j ≤ n and ε0j = 0, εj = 1. We will denote the edge from ε0 to ε by eε0 ε . • An edge eε0 ε is labeled with 0 if (iε0 ε )∗ is an isomorphism in all degrees and 1 otherwise. We will denote the label of an edge e by lC (eε0 ε ) or l(eε0 ε ) when C is clear from context. g We will denote by HC(C) the subgraph of HC(C) induced by all the vertices except the origin and g we will refer to HC(C) as the hypercube subgraph of C. Remark 2.4 Note that, since C I is a commutative diagram, for any two directed paths between vertices the sum of the edge labels must be the same in HC(C). If we are given a directed hypercube graph G (directed as in definition 2.3) with edge labels 0 and 1 that satisfies the property that the sum of the edge labels along any two directed paths between vertices is the same, we can easily construct a basic short exact cube with G as its hypercube graph. Also note that χ(H∗ (C)) is completely determined by HC(C). Lemma 2.5 Suppose that C is a basic short exact cube. There are only two mutually exclusive possibilities: g 0 ) = HC(C) g 1 If C 0 is another basic short exact cube then HC(C ⇒ HC(C 0 ) = HC(C). g 2 Either all of the edges in HC(C)\HC(C) (i.e. all the edges emerging from 0 )are labeled with 0 or they are all labeled with 1. 8 Proof: Note first that, if possibility 1 is satisfied, possibility 2 cannot also be satisfied since if all the edges emerging from 0 are labeled with i (where i is 0 or 1) then we can get another valid labeling by simply replacing all the i’s emerging from the origin with (1 − i)s (see Remark 2.4). g 0 ) = HC(C), g Suppose that C and C 0 satisfy HC(C but HC(C 0 ) 6= HC(C). Then there must be some ε0 connected to the origin such that the edge from 0 to ε0 is labeled differently in HC(C 0 ) and HC(C). Assume without loss of generality that lC (e0ε0 ) = 1 and lC 0 (e0ε0 ) = 0. Consider any other vertex ε connected to the origin and consider lC (e0ε ). We claim that lC (e0ε ) must be 1. To see this, consider the square subgraph induced by the vertices 0, ε, ε0 and δ = ε + ε0 . If lC (e0ε ) = 0 then since lC (e0ε0 ) = 1 this forces lC (eεδ ) = 1 = lC 0 (eεδ ) and lC (eε0 δ ) = 0 = lC 0 (eε0 δ ) (see the Remark 2.4). However this is impossible because we know lC 0 (e0ε0 ) = 0 and if 0 = lC 0 (eε0 δ ), 1 = lC 0 (eεδ ) there is no label that works for e0ε0 . So we get that in C every edge emerging from the origin must be labeled 1 if one of them is. By the same argument, we can show that every edge emerging from the origin must be labeled 0 if one of them is. This proves that the two cases stated in the Lemma are exhaustive and mutually exclusive.  g g Lemma 2.6 Suppose that A and B are two basic short exact cubes satisfying HC(A) = HC(B), g g every edge in HC(A)\HC(A) is labeled with 0, and every edge in HC(B)\HC(B) is labeled with 1. Then, χ(H∗ (A)) = χ(H∗ (B)) + (−1)n . Proof: We will prove this inductively. For the n = 1 case using the fact that both H∗ (A0 ) and H∗ (B 0 ) have even top grading we directly compute that χ(H∗ (A)) = 0 and χ(H∗ (B)) = 1. Now we can proceed with the induction. Note that we have: χ(A) = χ(11 A) − χ(10 A) and χ(B) = χ(11 B) − χ(10 B). g and also χ(11 A) = χ(11 B) since they are both completely determined by the hypercube subgraph HC χ(10 A) = χ(10 B) + (−1)n−1 by induction.  Lemma 2.7 Suppose that C is a 1, 2 or 3-dimensional basic cube of chain complexes. Then we can compute H∗ (C) as a graded vector space if we know H∗ (C ε ) for any C ε in the cube of inclusions C I , as well as all the maps (iε0 ε )∗ induced by homology in the cube of inclusions C I . Proof: When n = 1, we have a short exact sequence: 0 / C0 i01 / C1 j12 /C / 0. 9 Thus if (i01 )∗ is an isomorphism, we get that H∗ (C) ∼ = 0; and if not, then H∗ (C) ∼ = F. For the n = 2 case we show all possibilities for HC in Figure 2.1. (1) (2) (3) (4) (5) (6) Figure 2.1: All possible hypercube graphs in the n = 2 case ((0,0) is on the bottom left). The dotted lines denote edges labeled with 0 and the solid lines are edges labeled with 1 If we assume that H∗ (C 0 ) ∼ = F(0) [U ], then H∗ (C) is 0, 0, 0, F(3) , F(2) , F(4) ⊕ F(3) for the 6 possibilities shown in Figure 2.1, respectively. In the n = 3 case we only need to consider those HC which do not have a facet equal to (1), (2) or (3) in Figure 2.1 , as otherwise we would have for some j = 1, 2 or 3, H∗ (j0 C) = 0 or H∗ (j1 C) = 0. This would allow us to compute H∗ (C) from the long exact sequence for the short exact sequence: 0 / jC 0 / jC 1 /C / 0. We show all the possibilities for HC when n = 3 and none of the facets are as (1), (2) or (3) of Figure 2.1 in Figure 2.2. (1) (2) (3) (4) (5) Figure 2.2: Some hypercube graphs in the n = 3 case. Once again the (0, 0, 0) is on the bottom left and the dotted lines denote edges labeled with 0 and the solid lines are edges labeled with 1 If we assume that H∗ (C 0 ) ∼ = F(0) [U ], then H∗ (C) is F2(3) , F(4) ⊕ F2(3) , F2(4) , F2(5) ⊕ F(4) , F(6) ⊕ F2(5) ⊕ F(4) for the five cases shown, respectively.  Remark 2.8 The above Lemma does not hold when n ≥ 4. Consider the basic 4 dimensional short exact cube C where every edge of HC(C) is labeled with 1 and H∗ (C 0 ) ∼ = F(0) [U ]. For j j j j any j1 , j2 ∈ {1, 2, 3, 4} we have H∗ (j21 j22 C 00 ) ∼ = F4 ⊕ F3 , H∗ (21 22 C 10 ) ∼ = H∗ (21 22 C 01 ) ∼ = F6 ⊕ F5 and j j H∗ (j21 j22 C 11 ) ∼ = F8 ⊕ F7 . It follows that all maps on homology in the cube of inclusions for 21 22 C are j trivial. So for all j the map from H∗ (j0 C) ∼ = F(6) ⊕ F2(5) ⊕ F(4) to H∗ (1 C) ∼ = F(8) ⊕ F2(7) ⊕ F(6) may be of rank 0 or 1 without violating commutativity. Thus H∗ (C) may be either F(8) ⊕ F3(7) ⊕ F3(6) ⊕ F(5) or F(8) ⊕ F2(7) ⊕ F2(6) ⊕ F(5) . See also Theorem 1.5.1.d in [7]. 10 Chapter 3 The Chain Complex For a complete overview of Heegaard Floer homology, admissible multi-pointed Heegaard diagrams for knots and links, the definition of L-spaces and their relationship with the Heegaard Floer complex, see [20],[19], [18], [23], [21], [24], [25] and [12]. Suppose that L ⊂ S 3 is an oriented l component link. In this paper, we define a multi-pointed Heegaard diagram H = (Σg , α, β,w,z) for L with the following propertiesa : • Σg is a closed oriented surface of genus g. • α = (α1 , . . . , αg+m−1 ) is a collection of disjoint simple closed curves which span a g-dimensional lattice of H1 (Σ, Z), and the same goes for β = (β1 , . . . , βg+m−1 ). Thus, α and β specify handlebodies Uα and Uβ . We require that Uα ∪Σ Uβ = S 3 . • z = (z1 , z2 , . . . , zl ) and w = (w1 , w2 , . . . , wm ) are both collections of basepoints in Σ where l ≤ m. We will call wl+1 , wl+2 , . . . wm free basepoints. m • If {Ai }m i=1 and {Bi }i=1 are the connected components of Σ\ S g+m−1 αi i=1  and Σ\ S g+m−1 βi i=1  respectively then wi ∈ Ai ∩Bi for any 1 ≤ i ≤ m; and there is some permutation σ of {1, . . . , l} such that zi ∈ Ai ∩ Bσ(i) when 1 ≤ i ≤ l. • The diagram as defined so far specifies the link L ⊂ S 3 . • We require that all of the α and β curves intersect transversely and that every non-trivial periodic domain have both positive and negative local multiplicities (see section 3.4 of [23]). Also recall that for every intersection point x∈ Tα ∩ Tβ there is a Maslov grading M (x) and an Alexander multigrading Ai (x) ∈ H(L)i . Definition 3.1 Suppose we have a multi-pointed Heegaard diagram H = (Σg , α, β,w,z) for the pair L as above. We define the complex CF − (H) to be free over F with generators [x, i1 , j1 , . . . , il a This is identical to the definition given in [23] except we want to allow “spare” basepoints that will arise in the proof of the main theorem. , 11 , jl , il+1 , . . . , im ] where ik ∈ Z≤0 , and jk ∈ Q satisfying jk − ik = Ak (x). The differential is, as usual, given by counting holomorphic disks: ∂[x, i1 , j1 , . . . , il , jl , il+1 , . . . , im ] = X X [y, i1 −nw1 (φ), j1 −nz1 (φ), . . . , il −nwl (φ), jl −nzl (φ), il+1 −nwl+1 (φ), . . . , im −nwm (φ)]. y∈Tα ∩Tβ φ∈π2 (x,y) µ(φ)=1 In the notation of [11], this differential and Heegaard diagram correspond to the maximally colored case. The complex CF − is also an F[U1 , U2 , . . . , Um ]-module. The action of Uk for 1 ≤ k ≤ l is given by: Uk [x, i1 , j1 , . . . , ik , jk , . . . il , jl , il+1 . . . , im ] = [x, i1 , j1 , . . . , ik − 1, jk − 1, . . . il , jl , il+1 , . . . , im ]; and for l < k < m is given by; Uk [x, i1 , j1 , . . . , il , jl , il+1 , . . . , ik , . . . , im ] = [x, i1 , j1 , . . . , il , jl , il+1 . . . , ik − 1, . . . , im ]. We define the Maslov grading of [x, i1 , j1 , . . . , ik , jk , . . . il , jl , il+1 . . . , im ] by setting it equal to M (x) when all the ik are 0 and letting the action of each Ui drop the Maslov grading by 2. Note that both as a complex and F[U1 , U2 , . . . , Um ]-module CF − is isomorphic to CF − as defined in [23] via the isomorphism induced by [x, i1 , j1 , . . . , il , jl , il+1 , . . . , im ] 7→ U1−i1 . . . Ul−im x. And so it follows that CF − is a chain complex with homology HF − (S 3 ). Definition 3.2 Suppose that we have a Heegaard diagram H for L ⊂ S 3 as above. Fix some s = (s1 , . . . , sl ) ∈ H(L). Now suppose that we restrict CF − (H) to only those generators [x, i1 , j1 , . . . , il , jl , il+1 , . . . , im ] which satisfy Ak (x) = jk and force the differential to only count holomorphic disks φ with nzk (φ) = 0 when 1 ≤ k ≤ l. Then this quotient complex of CF − (H) will be denoted by CF L− (H,s). Note that CF L− inherits an F[U1 , . . . Um ] module action from CF − . Theorem 3.3 If CF L− (H, s) is as above, then its homology is HF L− (S 3 , L, s) Proof: If the diagram H has no free points, then CF L− (H,s) is the same as the complex computing HF L− in [23]. so we only need to show what happens in the case when there are free basepoints in H. Suppose that H0 is another Heegaard diagram that only has l-pairs of basepoints (one pair 12 for each link component) and no others. Then we claim that H can be obtained from H0 via the following moves: 1 a 3-manifold isotopy 2 α and β curve isotopy 3 α and β handleslide 4 index one/two stabilization We may also need the inverses of moves 1-4 5 free index zero/three stabilization, but we do not need the inverse of 5. We follow the argument from proposition 4.13 of [11] which relies on [12] Lemma 2.4 . Basically, we can apply moves 1 − 4 to H0 to obtain a Heegaard diagram that differs from a diagram with exactly l pairs of basepoints (one pair for each component) by index zero/three stabilizations only. Then we can apply moves 1 − 4 again to obtain the diagram H. Now we know that moves 1 − 4 and their inverses give chain homotopy equivalences for the complexes CF L− by the arguments given in [20] and proposition 3.9 of [23], so we will focus on move 5. Suppose that H1 and H2 are two Heegaard diagrams for L, and H2 is obtained from H1 by a free index zero/three stabilization. Then H2 has an extra free basepoint wr that H1 does not have. By the argument of Lemma 6.1 in [23], we see that the complex CF L− (H2 , s) is just the mapping cone CF L− (H1 , s)[Ur ] / CF L− (H1 , s)[Ur ], Ur −Uk where k is an index corresponding to some w basepoint in H1 . Now, k may correspond to a free basepoint, or it may correspond to some link component (in which case the action of Uk is trivial); but in either case, the homology of this mapping cone is the same as the homology of CF L− (H1 , s). So we see that all of the above 5 Heegaard moves induce quasi-isomorphisms of chain complexes, and this gives the desired result.  Definition 3.4 Fix a Heegaard diagram H for L. For a given s ∈ H(L) and ε ∈ El , we define the − complex A− s,ε (H) to be the quotient complex of CF (H) generated by those [x, i1 , j1 , . . . , il , jl , il+1 , . . . , im ] that satisfy • max{ik , jk − (sk − 1)} ≤ 0 if εk = 0 • max{ik , jk − sk } ≤ 0 if εk = 1 • ik ≤ 0 and jk = sk if εk = 2. 13 − − By A− s (H) we mean As,1 (H). We will write As,ε when the choice of diagram is clear from context. − The complex A− s,1 (H) inherits an F[U1 , . . . Um ] action from CF (H). When H is clear from context we will omit H from the notation. Remark 3.5 If we complete A− s (H) with respect to the maximal ideal (U1 , . . . , Um ), there is an − isomorphism between the completed version of A− s (H) and A (H, s) as defined in section 4.2 of [11], given by: −max{i1 ,j1 −s1 } [x, i1 , j1 , . . . , il , jl , il+1 , . . . , im ] 7→ U1 −max{i2 ,j2 −s2 } U2 −max{il ,jl −sl } . . . Ul U −il+1 . . . U −im x. We can use the proofs in section 4.3 and 4.4 of [11] to show that the homology of the complex A− s (H) does not depend on the choice of a Heegaard diagram. For this reason we will sometimes − − write H∗ (A− s (H)) as H∗ (As (L)). In this paper we could have just used the complexes As to get the same results about link Floer homology. The choice to use the notation here has been made to make the analogy with the work in [18] and [21] more clear. Theorem 3.6 Suppose that L ⊂ S 3 is an L-space link and s ∈ H(L). Then, as F[U1 , U2 , . . . , Ul ]modules, ∼ H∗ (A− s ) = F[U ]. where all of the Ui have the same action as U on the right hand side. Proof: We can use the proof of Theoremb 10.1 in [11] to see that for any s ∈ H(L), H∗ (A− s ) is isomorphic (as a module) to HF − (Y, s), where Y is some L space obtained by large positive surgery on L and s is a Spinc structure over Y .  Remark 3.7 The above property characterizes L-space links. See also proposition 1.11 of [10]. Suppose that, for a fixed s ∈ H(L), we have ε0 , ε and ε00 in El so that they only differ in the jth coordinate with ε0j = 0, εj = 1 and ε00j = 2. Then, for a given Heegaard diagram H of L, there is a short exact sequence: 0 / A− 0 (H) s,ε i ε0 ε / A− s,ε (H) jεε00 / A− 00 (H) s,ε / 0. So, we can define a short exact cube of chain complexes A− (H, s) by setting A− (H, s)ε = A− s,ε (H). Note also that A− (H, s) is just CF L− (H, s). Theorem 3.8 For any s ∈ H(L), A− (H, s) is a basic short exact cube when L is an L-space link. b As was mentioned in Remark 3.5, the only difference between the complex in that paper and this one is that it is defined over F[[U1 , U2 , . . . , Um ]] as opposed to F[U1 , U2 , . . . , Um ]. However the proof of Theorem 10.1 in [11] does not rely on F[[U1 , U2 , . . . , Um ]] in any way. See also the proof of Theorem 4.1 in [18]. 14 Proof: We want to show properties 4 and 5 in definition 2.1. Note that, by Theorem 3.6, we ∼ already know that for all ε ∈ {0, 1}n we have H∗ (A− s,ε ) = F[U ]. First, we will examine all maps induced on homology in the cube of inclusions. Suppose that ε0 and ε are in {0, 1}l and differ only in the jth coordinate with ε0j = 0 and εj = 1. Also define ε00 to agree in all coordinates with ε except the jth and ε00j = 2. Now, following the proof of Lemma 3.1 in [21], we define X to be the set of generators [x, i1 , j1 , . . . , il , jl , il+1 , . . . , lm ] of CF − that satisfy: 1 max{ik , jk − (sk − 1)} ≤ 0 if ε00k = 0 2 max{ik , jk − sk } ≤ 0 if ε00k = 1 3 ik ≤ 0 and jk = sk if ε00k = 2, i.e. when k = j. We define a set Y similarly, except 3 is replaced with; 3 ik = 0 and jk < sk if ε00k = 2, i.e. when k = j. Note that X naturally generates a sub-complex of a quotient complex of CF − , which we will denote by C{X} = A− s,ε00 . Similarly, there are complexes C{Uj X}, C{Y }, C{X ∪ Y }, C{Uj X ∪ Y } and − C{X ∪ Uj X ∪ Y }, all of which inherit differentials from CF − . Since C{X ∪ Y } = A− s,ε /Uj (As,ε ) d of some L-space obtained by some large surgery on L (see section 11.2 of [11]). its homology is HF Therefore H∗ (C{X ∪ Y }) ∼ = F. Similarly H∗ (C{Uj X ∪ Y }) ∼ = F. Now we have two short exact sequences of complexes: 0 / C{Y } i1 / C{X ∪ Y } / C{Uj X} i2 / C{Uj X ∪ Y } j1 / C{X} /0 and 0 j2 / C{Y } / 0. We will denote the connecting homomorphims for these two complexes by δ1 and δ2 , respectively. First note that δ2 ◦ δ1 = 0 (this follows from the fact the differential ∂ on the quotient complex C{X ∪ Uj X ∪ Y } satisfies ∂ 2 = 0). Now it follows from the exact same argument as in Lemma 3.1 in [21] that either H∗ (C{X}) = H∗ (A− s,ε00 ) is 0 and H∗ (C{Y }) is F, or H∗ (C{X}) is F and − H∗ (C{Y }) is 0 . If H∗ (C{X}) = 0 then the map iε0 ε : A− s,ε0 → As,ε clearly induces an isomorphism on homology. If H∗ (A− s,ε00 ) is F supported in some degree k then it follows from the first short exact − sequence that H∗ (C{X ∪ Y }) = H∗ (A− s,ε /Uj (As,ε )) is also F supported in degree k. Then, from the − second short exact sequence it follows that H∗ (C{Uj X}) ∼ = H∗ (C{Uj X ∪ Y }) = H∗ (A− s,ε0 /Uj (As,ε0 )) is F supported in degree k − 2. So we now have that the top grading in H∗ (A− s,ε0 ) is two less than the top grading in H∗ (A− s,ε ), and we have now completely verified property 5 in the definition of a basic short exact cube. 15 The only thing that is left to check in property 4 is that for any ε ∈ {0, 1}l , H∗ (A− (H, s)ε ) ∼ = F[U ] ∼ has even top degree. For any sufficiently large (s1 , s2 , . . . , sl ) = s ∈ H(L) we have H∗ (A− s ) = HF − (S 3 ) = F(0) [U ]. For any s0 ≤ s, we can decrease the sj by one over finitely many steps to get from s to s0 . By property 5 we know that each of these steps will either preserve the top degree or drop it by 2. The result now follows.  Corollary 3.9 For an L-space link L ⊂ S 3 with Heegaard diagram H, HC(A− (H, s)) depends only on L and s. 3 Proof: The top gradings of all the H∗ (A− s,ε ) are invariants of L ⊂ S and s. The maps induced by homology in A− (L, s)I are completely determined by these gradings since we have shown that A− (H, s) is a basic short exact cube.  Here is another fact that we will use often: Lemma 3.10 Fix some s ∈ H(L) where L in S 3 is an arbitrary link (i.e. not necessarily an L-space \ link). Then, if HF L− (L, s+ε) is trivial for every ε ∈ {0, 1}l , ε 6= 0 we get HF L− (L, s) ∼ L(L, s). = HF Proof: First fix a Heegaard diagram H for L ⊂ S 3 . We define an l-dimensional short exact cube C s as follows: for ε ∈ {0, 1}l and s ∈ H(L) we define C s,ε to be a quotient complex of CF − (H) generated by those [x, i1 , j1 , . . . , il , jl , il+1 , . . . , im ] that satisfy the following: • ik = 0 and jk < sk if εk = 0 • ik = 0 and jk ≤ sk if εk = 1 • ik = 0 and jk = sk if εk = 2. Then the inclusion and quotient maps of C s are defined naturally from CF − (H). \ L(L, s) and for ε ∈ {0, 1}l we have By definition, H∗ (C s ) ∼ = HF 1−ε H∗ (C s,ε ) ∼ = U11−ε1 U21−ε2 . . . Ul l HF L− (L, s + 1 − ε). And so H∗ (C ε ) is only nonzero when ε = 1. So it follows by taking iterated quotients that, \ L(L, s). HF L− (L, s) ∼ = HF = H∗ (C 1 ) ∼ = H∗ (C) ∼  16 Chapter 4 Proof of the Main Theorems Remark 4.1 Suppose that M = Li1 t Li2 t . . . Lik is a sub-link of L = L1 t L2 . . . t Ll with the inherited orientation. Fix some Heegaard diagram H for L. Now choose any s = (s1 , . . . sl ) ∈ H(L) so that all sj for Lj 6∈ M are sufficiently large (for instance larger than max{Aj (x)} for every generator x in some fixed diagram H for L ⊂ S 3 ). Then it is easy to see that for some r∈ H(M ) − 0 0 and any ε ∈ El the complex A− s,ε (H) is the same as Ar,ε0 (H ) where ε ∈ El−k is obtained from ε by deleting εi1 , . . . , εik and reordering and H0 is obtained by deleting zi1 , . . . , zik and reordering. The explicit value for r can be computed by the formula in section 4.5 of [11] (see also section 3.7 of [23]). So r = (r1 , . . . , rk ) ∈ H(M ) is given by rj = sij − lk(Lij , L\M )/2. The next Lemma was observed in [10] Lemma 1.10. Lemma 4.2 Every sub-link of an L-space link is an L-space link. Proof: Suppose that M ⊂ L is some sub-link. It suffices to show that, for any r ∈ H(M ), we − − ∼ ∼ have H∗ (A− r (M )) = F[U ]. This is true because H∗ (Ar (M )) = H∗ (As (L)) for some s ∈ H as shown above.  Lemma 4.3 X χ(HF L− (L, s))us = P∅L (u1 , . . . , ul ) s∈H Proof: It was shown in [23] proposition 9.1 that when l > 1 X \ χ(HF L(L, s))us = ± l Y −1 ui − ui 2 1 2 ! ∆L i=1 s∈H and we have chosen sign conventions so that X s∈H \ χ(HF L(L, s))us = l Y i=1 1 2 −1 ui − ui 2 ! ∆L . 17 and so for l > 1 it follows that   X − χ(HF L (L, s))u s  =  s∈H Y  ua1 1 . . . ual l    i=1 √ ∆L ! i = = 1 − l Y u2 − u 2 ! i=1 (a1 ,...,al )∈Zl≤0 1 \ χ(HF L(L, s))u l −1  Y 12 ui − ui 2 ua1 1 . . . ual l  Y s s∈H (a1 ,...,al )∈Zl≤0  =  ! X i 1 − u−1 i ∆L u1 u2 . . . ul ∆L = P∅L . When l = 1, it was shown in [18] that: X \ χ(HF L(L, s))us = ±∆L (u1 ); s∈H and so the result follows by the same argument as above.  Definition 4.4 Suppose we are given a Heegaard diagram H for an L-space link L ⊂ S 3 . Define a directed labeled graph T(H) as follows: • The vertices correspond to the elements of H(L). • There is a directed edge from s = (s1 , . . . , sl ) to s0 = (s01 , . . . , s0l ) if for some i we have s0i = si +1 and s0j = sj for every j 6= i. We will call this edge ess0 . • If s and s0 , are as above then define ε ∈ El so that εj = 1 if j 6= i and εi = 0. Then the label of edge ess0 is the same as the label of the edge between ε and 1 in HC(A− (L, s0 )). Just as in corollary 3.9, the graph T(H) is an invariant of L ⊂ S 3 . So we will simply say T(L). We will denote by js T(L) the subgraph of T(L) that is obtained by restricting to the hyperplane with jth coordinate equal to s. Definition 4.5 Suppose that L ⊂ S 3 is an L-space link. Then we recursively define m(L) ∈ H(L) as follows. If L has only one component let m(L) be the degree of ∆L . In general;  m(L)i = max {degui (P∅L )} ∪     lk(Li , Lj ) lk(Li , Lj ) j < i ∪ m(L\Lj )i + j>i m(L\Lj )i−1 + 2 2 where by degui (P∅L ) we mean the maximal degree of ui in any monomial of P∅L . 18 Proposition 4.6 For an L-space link L ⊂ S 3 suppose that s ≥ m(L)j . Then js T(L) is completely determined by T(L\Lj ) and all the edges from js T(L) to s+1j T(L) must be labeled with 0. Proof: First note that T(L\Lj ) only makes sense in light of Lemma 4.2 from which it follows that L\Lj is an L-space link. Pick m = (m1 , . . . , ml ) ∈ H(L) so that for any 1 ≤ i ≤ l, mi > Ai (x) for every generator x. Then, we claim that whenever si > mi , sii T(L) is completely determined by T(L\Li ) and all the edges from sii T(L) to si +1i T(L) must be labeled with 0. We prove this − claim when i = l. Since sl > ml , the inclusion between A− (s1 ,...,sl ) and A(s1 ,...,sl +1) induces an isomorphism on homology. So the edge between (s1 , . . . , sl ) and (s1 , . . . , sl + 1) is labeled with 0. Following Remark 4.1 we get that the edge between (s1 , . . . , si , . . . , sl ) and (s1 , . . . , si + 1, . . . , sl )   ,Ll ) has the same label as the edge between s1 − lk(L21 ,Ll ) , . . . , si − lk(L2i ,Ll ) , . . . sl−1 − lk(Ll−1 and 2   ,Ll ) s1 − lk(L21 ,Ll ) , . . . , si − lk(L2i ,Ll ) + 1, . . . , sl−1 − lk(Ll−1 in T(L\Ll ) and so this proves the claim. 2 Now we are ready to prove the proposition. We will prove this by induction on l. If mj − 1 ≥ m(L)j , for some fixed j, the edge between (s1 , . . . , mj −1, . . . , sl ) and (s1 , . . . , mj , . . . , sl ) is labeled zero if si ≥ mi for every i 6= j (by induction). g − (L, (s1 , . . . , mj , . . . , sl ))). One valid (in the sense of Remark Notice that this determines HC(A 2.4) labeling of the remaining edges in HC(A− (L, (s1 , . . . , mj , . . . , sl ))) is given by setting all the j edges between HC(A− (L, (s1 , . . . , mj , . . . , sl ))) ∩ mj −1 T(L) and HC(A− (L, (s1 , . . . , mj , . . . , sl ))) ∩ j mj T(L) to be zero and letting an edge between s1 and s2 in HC(A− (L, (s1 , . . . , mj , . . . , sl ))) ∩ j − 0 0 mj −1 T(L) have the same labeling as the edge between s1 and s2 in HC(A (L, (s1 , . . . , mj , . . . , sl )))∩ 0 0 j mj T(L) where s1 and s2 are the same as s1 and s2 after adding one to the jth coordinate. Since mj −1 > deguj P∅L we must have χ(H∗ (A− (L, (s1 , . . . , mj , . . . , sl )))) = 0 and so the labeling for HC(A− (L, (s1 , . . . , mj , . . . , sl ))) described above is the correct one since it yields the correct j Euler characteristic (see Remark 2.6 and Lemma 2.5). We can similarly fill in all of mj −1 T(L) and j all the edges between mj −1 T(L) and mjj T(L) are labeled 0. Repeating this process by inductively decreasing the jth coordinate proves the claim.  Lemma 4.7 For a 2 or 3 component L-space link, T(L) completely determines HF L− (L, s) for every s ∈ H(L). Proof: Note that T(L) determines all the hypercube graphs of A− (L, s) for any s ∈ H(L). Thus, by Lemma 2.7 and Remark 2.4 we get that T(L) determines all the HF L− (L, s) upto an even shift in absolute grading. To fix the grading note that we can pick s ∈ H(L) so that any edge emerging − 3 ∼ from s0 ≥ s is 0 since for s sufficiently large H∗ (A− s (L)) = HF (S ) = F(0) [U ]. This fixes the 19 grading as required.  Lemma 4.8 For an L-space link L, the graph T(L) is determined by the polynomials ±∆M and the linking numbers lk(Li , M ) where M is any sublink of L. Proof: We will prove this by inducting on l. First suppose that l = 1. Then ±∆L completely P determines ±P∅L = s∈Z as (u1 )s . The only possibilities for |as | are either 1 or 0. If |as | = 1 then this forces the edge between s − 1 and s to be labeled with 1. If as = 0 then this forces the edge between s − 1 and s to be labeled with 0. This proves the case when l = 1. By proposition 4.6, we see that the subgraph of T(L) that is induced by all the vertices s = (s1 , . . . , sl ) satisfying si ≥ m(L)i for some i, is completely determined by the relevant polynomials and linking numbers. g − (L, m(L))) is contained inside the part of the For the rest of T(L) note that every edge of HC(A graph whose labels we have already determined. By Lemma 2.5, this either completely determines HC(A− (L, m(L))), or all the edges emerging from (m(L)1 −1, . . . , m(L)l −1) are labeled with a 0 or g − (L, m)), they are all labeled with 1. If HC(A− (L, m(L))) is not completely determined by HC(A then we can use Lemma 2.6 to see that the absolute values of the coefficients of ∆L are enough to determine if all the edges emerging from (m(L)1 − 1, . . . , m(L)l − 1) are labeled with a 0 or 1. Thus, g − (L, (m1 , . . . , mi − 1, . . . , ml )) for any i and so we can proceed as we now have computed HC(A before to inductively fill out all of T(L). This proves the Lemma.  Proof: [Proof of Theorem 1.5] This follows immediately from the previous two Lemmas  Lemma 4.9 Let S = {i1 , . . . , ik } ( {1, . . . , l} and suppose that {j1 , . . . , jl−k } = {1, . . . , l}\S where ja < jb when a < b. Pick s ∈ H(L) so the sip ≥ m(L)ip . Then if asj1 ,sj2 ,...,sjl−k is the coefficient s sj l−k in PLLS , we have asj1 ,sj2 ,...,sjl−k = χ(H∗ (A− of uj1j1 . . . ujl−k s,ε (L))), where ε ∈ El satisfies εr = 2 if r = jp for some p and εr = 1 otherwise. Proof: This follows from Remark 4.1 and Lemma 4.3.  Proof: [Proof of Theorem 1.8] We will assume WLOG that r = 1. Then let S = {i1 , . . . , ik } ⊂ {2, . . . , l} and {j1 , . . . , jl−k−1 } = {2, . . . , l}\S with ja < jb if a < b. s = (s1 , . . . , sl ) ∈ H (L) is arbitrary. Fix (m1 , . . . , ml ) ∈ H (L) so that mi > m (L)i + 1. Then we have the following:
R s0 ≥s PLLS = s01 =s1 X s0 =(s01 ,...,s0l )∈H s01 =s1 ,s0ip =mip mjp ≥s0jp ≥sjp    χ H ∗ A− , 0 s ,ρ 20 where ρ ∈ El is fixed and satisfies ρk = 2 if k = jp for some p, and ρk = 1 otherwise. This follows by the previous Lemma. We get that the above quantity is equal to: X X (−1) number of 0’s in ε    χ H ∗ A− . 0 s ,ε (4.1) s0 =(s01 ,...,s0l )∈H ε∈El ,ε1 =2,εip =1 εjp =1 or 0 s01 =s1 ,s0ip =mip mjp ≥s0jp ≥sjp Note that if ε ∈ El with ε1 = 2, εi = 0 or 1 if i 6= 1 we get: − A− s0 ,ε = As00 ,(2,1,...,1) . where s00 is given by s001 = s01 and s00k =s0k + εk − 1. So all of the terms in (4.1) that correspond to s0 with s0i 6= si or mi will cancel out. This leaves, X (−1) number of 0’s in ν s0
   χ H ∗ A− (L) , 0 0 s ,ν(s ) (4.2) s0 =(s01 ,...,s0l )∈H s01 =s1 ,s0ip =mip s0jp =sjp or mjp where here ν (s0 )1 = 2, ν (s0 )ip = 1 and ν (s0 )jp = 1 if s0jp = mjp , and ν (s0 )jp = 0 otherwise. Given S ⊂ {2, . . . , l}, we define s(S) by setting s(S)1 = s1 , s(S)k = mp if p ∈ S, and s(S)k = sp − 1 otherwise. Then we can rewrite (4.2) as X (−1) l−1−|S|−|S 0 |    χ H∗ A− . 0 s(S∪S ),(2,1,...,1) (4.3) S 0 ⊂{2,...,l}\S Thus, we finally get: X S⊂{2,...,l} (−1) l−1−|S| R s0 ≥s PLLS
= s01 =s1 X X (−1) −|S 0 |    χ H ∗ A− s(S∪S 0 ),(2,1,...,1) S⊂{2,...,l} S 0 ⊂{2,...,l}\S = X X (−1) −|S\A|    χ H∗ A− s(S),(2,1,...,1) S⊂{2,...,l} A⊂S    = χ H ∗ A− . s(∅),(2,1,...,1) Now (4.4) must be either 1 or 0 by Theorem 3.8. (4.4)  21 Chapter 5 Application to 2-Bridge Links We would like to use the recursive formula for the multivariate Alexander polynomial of a 2-bridge link given in [8], so we will use the conventions from that paper. A circle labeled k or −k will represent a braid with k crossings as in Figure 5.1 Suppose we are given a collection of nonzero k = ... = -k ... Figure 5.1: integers a1 , . . . , an . Then we can define α and β via 1 α = a1 + β (5.1) 1 a2 + .. .+ 1 an where α > 0, g.c.d(α, β) = 1, and α > |β| > 0. Now, if α is even we can use (a1 , . . . , an ) to construct an oriented link C(a1 , . . . , an ) as shown in Figure 5.2. Links of this form are called 2-bridge links, and we have the following classification from [3] and page 144 of [28] (see also chapter 12 in [1]): Theorem 5.1 If L = C(a1 , . . . , an ) and L0 = C(b1 , . . . , bm ) are two 2 bridge links where we define α and β from a1 , . . . , an , as in equation 5.1, and similarly α0 and β 0 from b1 , . . . , bm . Then L and L0 are equivalent iff α0 = α and β 0 ≡ β ±1 mod 2α. If β 0 ≡ β + α mod 2α or β 0 β ≡ 1 + α mod 2α, then L and L0 are equivalent after reversing the orientation of one of the components. We will denote the 2-bridge link determined by α and β as above by b(α, β). To use the formulas 22 a1 a3 n = 2k a2k-1 a5 ... -a2 -a4 -a2k a2k-1 a2k+1 n = 2k+1 ... -a2k Figure 5.2: Diagram for constructing 2-bridge link given a sequence of non-zero integers. in [8], we need an expansion of α β of the following form: 1 α = 2p1 + β . 1 2q1 + 1 2p2 + 1 2q2 + .. .+ 1 2pn We will denote b(α, β) = C(2p1 , 2q1 , . . . , 2pn−1 , 2qn−1 , 2pn ) by D(p1 , q1 , p2 , q2 , . . . , pn ) for convenience. We define two variable polynomials Fr (u1 , u2 ) for r ∈ Z:  r−1 X    (u1 u2 )i      i=0 Fr (u1 , u2 ) = 0   −1  X     − (u1 u2 )i  i=r if r > 0 if r = 0 if r < 0. 23 ± Now let us define polynomials ∆k ∈ Z[u± 1 , u2 ] for 0 ≤ k ≤ n recursively as follows: ∆0 = 0 ∆1 = Fp1 ∆k = (qk−1 (u1 − 1)(u2 − 1)Fpk + 1)∆k−1 + (u1 u2 )pk−1 Also set lk = Pk ˜ i=1 pi and lk = Fp k (∆k−1 − ∆k−2 ). Fpk−1 (5.2) Pk i=1 |pk |. Then by Theorems 1, 3 and corollary 1 of [8] we have; Theorem 5.2 If L = D(p1 , q1 , p2 , q2 , . . . , pk ), then: (u1 u2 ) 1−lk 2 ∆k (u1 , u2 ) = ±∆L (u1 , u2 ). l̃k The minimal degree of u1 (or u2 ) in any monomial of ∆k is lk − and the maximal degree of u1 (or 2 l̃k u2 ) in any monomial of ∆k is lk + − 1. 2 Define q(k) = Qk−1 i=1 qi and F (k) = Qk i=1 Fpi where, as usual, the empty product is 1. Also recall that the linking number of D(p1 , q1 , p2 , q2 , . . . , pn ) is −ln . ± Given any P ∈ Z[u± 1 , u2 ] where P = r s r,s∈Z ar,s u1 u2 , P we define P [i] to be the polynomial j+i j j∈Z aj+i,j u1 u2 . If P [i] 6= 0 we say that P is supported on the diagonal i. Note that if Q ∈ P ± [i] Z[u± = P [i] + Q[i] and (P Q)[i] = a+b=i P [a] Q[b] . Thus, it follows that if P [0] 1 , u2 ], then (P + Q) P divides Q, then (Q/P [0] )[k] = Q[k] /P [0] . Using equation (5.2), we get the following identity: ∆[k] n = X (qn−1 (u1 − 1)(u2 − 1)Fpn + 1) [i] [j] ∆n−1 +  pn−1 (u1 u2 ) i+j=k Fpn Fpn−1  (∆n−1 − ∆n−2 )[k] . (5.3) This can then be expanded to: [k+1] [k−1] [k] ∆[k] n = (qn−1 (−u2 )Fpn )∆n−1 + (qn−1 (−u1 )Fpn )∆n−1 + (qn−1 (u1 u2 + 1)Fpn + 1)∆n−1   Fpn (∆n−1 − ∆n−2 )[k] . (5.4) + (u1 u2 )pn−1 Fpn−1 Lemma 5.3 If t > n − 1 then ∆n is not supported on the diagonal t. Also: ∆n[n−1] = q(n)(−u1 )n−1 F (n). [0] [t] Proof: First note that ∆1 = ∆1 = Fp1 . Now the claim that ∆n = 0 when t > n − 1 can be easily [n−1] seen by induction via equation (5.4). We will prove that ∆n = q(n)(−u1 )n−1 F (n) for n > 1 by 24 induction on n using equation (5.4): n−2 Y ∆[n−1] = (qn−1 (−u1 )Fpn ) n ! qi n−2 (−u1 ) i=1 n−1 Y ! Fpi i=1 = q(n)(−u1 )n−1 F (n).  Lemma 5.4 For n ≥ 2: ∆[n−2] = P1 + P2 + P3 n where: P1 = (n − 1)(u1 u2 + 1)q(n)F (n)(−u1 )n−2 P2 = P3 = n X q(n) F (n) q Fp i i=2 i−1 n−1 X (u1 u2 )pi i=1 (−u1 )n−2 and q(n) F (n) (−u1 )n−2 . qi Fpi (5.5) Proof: When n = 2, we directly compute that: ∆2 = q1 (u1 − 1)(u2 − 1)Fp2 Fp1 + Fp1 + (u1 u2 )p1 Fp2 . [n−2] For n > 2, we can recursively compute ∆n : [n−2] [n−3] ∆[n−2] = (qn−1 (u1 − 1)(u2 − 1)Fpn + 1)[0] ∆n−1 + (qn−1 (u1 − 1)(u2 − 1)Fpn + 1)[1] ∆n−1 n Fpn [n−2] (∆ ) Fpn−1 n−1 q(n) F (n) [n−3] = (qn−1 (u1 u2 + 1)Fpn + 1) (−u1 )n−2 + qn−1 (−u1 )Fpn ∆n−1 qn−1 Fpn q(n) F (n) + (u1 u2 )pn−1 (−u1 )n−2 . qn−1 Fpn−1 + (u1 u2 )pn−1 The result now follows by induction. Lemma 5.5 Let ∆n = P i j i,j aij u1 u2 .  Suppose that all the nonzero aij are ±1. Suppose also that for fixed i0 (or j 0 ) the nonzero ai0 j (or aj 0 i ) alternate in sign. Then we must have |qi | = 1 for every 1 ≤ i ≤ n − 1. For the pi , one of the following two possibilities holds: • For i 6= 1 all pi are equal. For i 6= 1, pi = ±1 and pi = −qi−1 • For i 6= n all pi are equal. For i 6= n, pi = ±1 and pi = −qi . 25 Proof: First note that when n = 1, the Lemma is vacuously true. So from now on we will assume [n−1] that n ≥ 2. If ∆n has all coefficients ±1 or 0, then so does ∆n = q(n)F (n)(−u1 )n−1 . For this to happen |q(n)| must be 1 which implies that qi = ±1 for every 1 ≤ i ≤ n − 1. F (n) has coefficients ±1 if for all but possibly one i, we have pi = ±1. [n−2] Now we focus on ∆n . There are four cases: Case 1 (There is some k ∈ {1, 2, . . . , n} such that pk > 1) Suppose that r of the pi are −1 (and so except for pk , the rest are 1.) First, we get that: F (n) = (−1)r pX k −1 (u1 u2 )i−r . i=0 Now, since all the nonzero coefficients of ∆n are by assumption ±1, the same must be true for ∆[n−2] n q(n)(−u1 u2 )−r (−u1 )n−2 . ∆[n−2] We will compute the coefficient of u1 u2 in q(n)(−u1 u2n)−r (−u1 )n−2 . Now recall that ∆[n−2] = P1 + P2 + P3 n i where P1 , P2 and P3 are as defined in equation (5.5). Set Pi0 := q(n)(−u1 u2P)−r (−u1 )n−2 for i = 1, 2 or 3. Then, P10 = (n − 1)(u1 u2 + 1) pX k −1 (u1 u2 )i , i=0 and so the coefficient of (u1 u2 ) in P10 is 2(n − 1). Similarly, P20 = qk−1 + X (qj−1 ) pj =1 2≤j≤n pX k −1 X (u1 u2 )i + (−qj−1 ) pj =−1 2≤j≤n i=0 pk X (u1 u2 )i . i=1 So the coefficient of (u1 u2 ) in P20 is X pj qj−1 , 2≤j≤n j6=k and similarly the coefficient of (u1 u2 ) in P30 is X pj qj . 1≤j≤n−1 j6=k ∆[n−2] So finally, the coefficient of u1 u2 in q(n)(−u1 u2n)−r (−u1 )n−2 is 2(n − 1) + X 2≤j≤n j6=k pj qj−1 + X 1≤j≤n−1 j6=k pj qj , (5.6) 26 which must be 1, −1 or 0. Notice first that, if 2 ≤ k ≤ n − 1 then the sum X pj qj−1 + 2≤j≤n j6=k X pj qj 1≤j≤n−1 j6=k is bounded above in absolute value by 2(n − 2), which makes it impossible for equation (5.6) to be equal to 1, −1 or 0. So, we get that k must be 1 or n. If k is 1 then equation (5.6) becomes 2(n − 1) + pn qn−1 + n−1 X pj (qj + qj−1 ). j=2 Notice that the above quantity has smallest possible value 1 and this only occurs if all of the qi are equal and have opposite sign as all the pi+1 , which proves the claim in this case. When k = n the argument is similar. Case 2 (There is some k ∈ {1, 2, . . . , n} such that pk < −1) The argument is the same as [n−2] in the previous case, except we divide ∆n by q(n)(−u1 u2 )−r (−u1 )n−2 and examine the coefficient of (u1 u2 )−1 . Case 3 (All of the pi are ±1 and n ≥ 3) We will start by showing that all the qi are equal. [n−1] Suppose as in the previous cases that the number of pi that are −1 is r. In this case ∆n is the monomial (−1)n−1+r q(n)u1n−1−r u−r 2 6= 0. This has the maximal possible degree for u1 and minimal possible degree for u2 by Theorem 5.2. [n−2] This immediately forces ∆n [n−3] to have at most 2 nonzero coefficients, and ∆n to have at most [n−2] is of the form 3 nonzero coefficients. So ∆n n−1−r 1−r an−2−r,−r un−2−r u−r u2 1 2 + an−1−r,1−r u1 Using the symmetry of the Alexander polynomial under the involution ui 7→ u−1 i , as well as the symmetry given by exchanging u1 and u2 (there is an isotopy of S 3 exchanging the two components of a 2 bridge link which is easy to see using the Schubert normal form [28]); we can conclude that an−2−r,−r = an−1−r,1−r . Suppose that an−2−r,−r = an−1−r,1−r 6= 0. Then since we have required the signs of ai,j to be alternating for fixed i (and j), this forces one of the following possibilities for [n−3] ∆n n−2−r 1−r ∆[n−3] = ±(un−3−r u−r u2 + un−1−r u2−r ) or ± (u1n−2−r u1−r ) or 0. n 1 2 + u1 1 2 2 n−1−r 2−r We have ruled out ±(un−3−r u−r u2 ) due to Theorem 3 (see also definition 2(iv)) in [8]. 1 2 + u1 27 [n−3] In all the possibilities for ∆n , we have ∆[n−3] (−1, 1) = ±1 or 0. n Fpn (−1, 1) is always 1 since we have assumed pn = ±1. From this we conclude ∆[n−1] (−1, 1) = q(n) and ∆n[n−2] (−1, 1) = 0. n [n−3] Using this in the recursive formula for ∆n given in equation (5.4), we get [n−4] ∆[n−3] (−1, 1) = −q(n) + q(n − 2) + qn−1 ∆n−1 (−1, 1). n [0] We manually compute ∆3 = 1 − 2q1 q2 . So this gives the formula ∆[n−3] (−1, 1) = n n−2 X q(n) − (n − 1)q(n). qq i=1 i i+1 If the above sum is to equal ±1 (note that it cannot be 0), we must have n−2 X 1 = n − 2, qq i=1 i i+1 and this can only happen if all the qi are equal. ∆[n−2] Now suppose that an−2−r,−r = an−1−r,1−r = 0. The constant term of q(n)(−u1 u2n)−r (−u1 )n−2 is: (n − 1) + X X qi−1 + 2≤i≤n pi =1 (−qi ), (5.7) 1≤i≤n−1 pi =−1 which by our assumption must be 0. We can rewrite (5.7) as; (n − 1) + qn−1 pn + qn−1 q1 p1 − q1 + + 2 2 X 2≤i≤n−1 qi−1 pi + qi pi + qi−1 − qi , 2 (5.8) qi−1 pi + qi pi . 2 (5.9) which simplifies to (n − 1) + qn−1 pn + q1 p1 + 2 X 2≤i≤n−1 Note that qn−1 pn + q1 p1 + 2 X 2≤i≤n−1 qi−1 pi + qi pi 2 (5.10) has a maximum absolute value of n − 1 which can only happen if all the qi are equal (and have opposite sign as all the pi ). 28 So we have shown in all cases that all the qi are equal. This allows us to rewrite equation 5.9 (which ∆[n−2] is the constant term of q(n)(−u1 u2n)−r (−u1 )n−2 ) as; (n − 1) + n−1 X  q1 pi + q1 i=2 p1 + pn 2  . (5.11) n We must have (5.11) equal to ±1 or 0. First note that we cannot have q1 p1 +p = 1 since 2 Pn−1 i=2 q1 pi is n n = −1 or 0. If q1 p1 +p =0 bounded above in absolute value by n − 2. So we must have that q1 p1 +p 2 2 Pn−1 then i=2 q1 pi must be −n + 2 which implies that all the pi for 2 ≤ i ≤ n − 1 have the opposite sign n as q1 and since q1 p1 +p = 0 we get that one of p1 and pn must also have the opposite sign as q1 which 2 Pn−1 n proves the claim in this case. If we assume that q1 p1 +p = −1 then we need i=2 q1 pi ≤ 3 − n. 2 Pn−1 Pn−1 However i=2 q1 pi = 3 − n is impossible since changing the pi always changes the sum i=2 q1 pi Pn−1 by a multiple of 2. Thus we once again have that i=2 q1 pi = 2 − n. This along with the fact that n q1 p1 +p = −1 implies that all of the pi have the opposite sign as q1 . 2 Case 4 (n = 2 and all the pi are ±1) The only tuples (p1 , q1 , p2 ) that do not satisfy the condition given in the Lemma are (1, 1, 1) and (−1, −1, −1), and we can manually compute ∆2 in both these cases to check that they do not satisfy that all of the nonzero coefficients are ±1. In particular for (1, 1, 1) we have ∆2 = 2 − u1 − u2 + 2u1 u2 and for (−1, −1, −1) we have ∆2 = − u22u2 + u21u2 + u11u2 − u12u2 1 2 1 2  Now, if an oriented 2-bridge link L is an L-space link, it must satisfy the conditions of the Lemma 5.5 by corollary 1.10 and so if L = D(p1 , q1 , . . . , pn−1 , qn−1 , pn ), then we have narrowed things down to the following 8 possibilities where w > 0 is an integer, q := 2w + 1, q 0 := 2w − 1 and k := 2n − 1. L = D(−1, 1, . . . , −1, 1, w) = b(qk − 1, q − (qk − 1)) or = D(−1, 1, . . . , −1, 1, −w) = b(q 0 k + 1, q 0 − (q 0 k + 1)) or = D(1, −1, . . . , 1, −1, w) = b(q 0 k + 1, q 0 k + 1 − q 0 ) or = D(1, −1, . . . , 1, −1, −w) = b(qk − 1, qk − 1 − q) or = D(w, −1, 1, . . . , −1, 1) = b(q 0 k + 1, k) or = D(−w, −1, 1, . . . , −1, 1) = b(qk − 1, −k) or = D(w, 1, −1, . . . , 1, −1) = b(qk − 1, k) or = D(−w, 1, −1, . . . , 1, −1) = b(q 0 k + 1, −k). We can further reduce these 8 possibilities down to 4 by noting b(qk−1, ±k) = b(qk−1, ±(q−(qk−1))) which can be seen by rotating the diagram given by 5.2 by 180◦ , and similarly b(q 0 k + 1, ±k) = 29 b(q 0 k + 1, ±(q 0 k + 1 − q 0 )). Now we compute the signatures of these four possibilities. Lemma 5.6 When q,q 0 and k are odd positive integers and q 6= 1 if k = 1; σ(b(qk − 1, ±k)) = ±(q − 2) (5.12) σ(b(q 0 k + 1, ±k)) = ±q 0 . (5.13) 0 = q 0 + k1 , we can use Figure Proof: First we compute the signature of b(q 0 k + 1, k). Since q k+1 k 5.2 to give a diagram D for b(qk − 1, k). Now we will use the Gordon-Litherland formula for knot signature(see [4]) on D. Since the surface given by a checkerboard coloring of D is orientable, the signature of the link is simply the signature of the Goeritz matrix for D (see the end of the first page in [4]). We denote by An (p) the n × n matrix with A11 = p, Aii = 2 when 2 ≤ i ≤ n, Aij = −1 when |j − i| = 1 and 0 everywhere else. A Goeritz matrix for D is given by A q (1 + k).  We claim that if p > 1, An (p) has signature n. This is easy to see inductively; let B(p) =  1 0 1 p 1 , In denote   B(p) 0 . Then the n × n identity matrix and Bn (p) =  0 In−2  p Bn (p)An (p)Bn (p)T =  0  0 An−1 (2 − 1/p)  so σ(An (p)) = 1 + σ(An (2 − 1/p)) and the claim follows. So the signature of b(q 0 k + 1, k) is q 0 . Since b(q 0 k + 1, −k) is the mirror image of b(q 0 k + 1, k), the signature of b(q 0 k + 1, −k) is −q 0 . Now we consider b(qk − 1, k) where k > 1 (k = 1 has already been covered above). qk−1 = q − k1 . k In this case a Goeritz matrix is Aq (1 − k) and  1−k Bq (1 − k)Aq (1 − k)Bq (1 − k)T =  0 0 Aq−1 (2 − 1/(1 − k))  . Now 1 − k < 0 and 2 − 1/(1 − k) > 1, so σ(An (1 − k)) = −1 + σ(Aq−1 (2 − 1/(1 − k))) = q − 2. Since b(qk − 1, −k) is the mirror image of b(qk − 1, k), σ(b(qk − 1, −k)) = −q + 2 as desired.  Proposition 5.7 If L is an L-space link of the form b(qk − 1, k) = D(−1, 1, . . . , −1, 1, w) then L = b(2, 1) Proof: Let us assume that L = b(qk − 1, k) is an L-space link. Now if s < n, it is easy to see by induction that 1 ∆s (u1 , u2 ) = − s s u1 u2 s−1 X i=0 ! ui1 u2s−1−i . 30 So by equation (5.2) we get (u1 − 1)(u2 − 1) ∆n (u1 , u2 ) = ! w−1 X (u1 u2 ) i ! 1 − n−1 n−1 u1 u2 !! +1 i=0 (u1 u2 )−1 −(u1 u2 )−1 ! w−1 X i (u1 u2 ) i=0 1 − n−1 n−1 u1 u2 n−2 X ui1 un−2−i 2 n−2 X !! ui1 u2n−2−i n−3 X 1 − n−2 n−2 u1 u2 − i=0 + i=0 !!! ui1 u2n−3−i . i=0 This simplifies to X ∆n (u1 , u2 ) = X ui+j+1−n u2i−j − 1 0≤i≤w−1 0≤j≤n−1 ui+j+1−n ui−j−1 . 1 2 0≤i≤w 0≤j≤n−2 Now note that L = L1 t L2 , where both L1 and L2 are unknots and lk(L1 , L2 ) = −ln = −w + n − 1, so we get: PLL1 (u2 ) = (u2 ) n−w−1 2 ∞ ∞ X n−w−1 X (u2 )−i and PLL2 (u1 ) = (u1 ) 2 (u1 )−i . i=0 i=0 Finally, by Theorem 5.2 we also get P∅L = ±(u1 u2 ) n−w+1 2 ∆n (u1 , u2 ). Expanding this then gives ±P∅L = (u1 u2 ) n−w+1 2 ∆n (u1 , u2 ) = i+j+ 3−n−w 2 X u1 i−j+ n−w+1 2 u2 − 0≤i≤w−1 0≤j≤n−1 X i+j+ 3−n−w 2 u1 i−j+ n−w−1 2 u2 0≤i≤w 0≤j≤n−2 If n = 1, we get: ±P∅L = X i+1 −w 2 u1 i+1 −w 2 u2 . 0≤i≤w−1 We can then fix the sign for P∅L using corollary 1.10 to get P∅L = − X i+1 −w 2 u1 i+1 −w 2 u2 . 0≤i≤w−1 Then, using the method given in the proof of Theorem 1.5, we can compute T(L). In this case m(L) = (w/2, w/2). The edge between (s1 , w/2 − 1) and (s1 , w/2) is labeled with 0 whenever s1 ≥ w/2. Similarly, the edge between (w/2 − 1, s2 ) and (w/2, s2 ) is labeled 0 whenever s2 ≥ w/2. w/2 w/2 u2 in P∅L is −1, which forces both edges from (w/2 − 1, w/2 − 1) to be The coefficient of u1 labeled with 1. This along with Lemma 3.10 allows us to compute   w w  ∼ \ HF L L, , = F(1) . 2 2 (5.14) . 31 \ Now, recall that when L is alternating, HF L(L, s) is completely determined by its Euler characteristic and σ(L), using Theorem 1.3 in [23]. Specifically, if s = (s1 , s2 ) and as is the coefficient of us −1 L in (1 − u−1 1 )(1 − u2 )P∅ then |a | \ HF L(L, s) ∼ = Fs s+s + σ−1 . 1 2 2 Therefore   w w  ∼ \ HF L L, , = F(2w−1) 2 2 (5.15) by Lemma 5.6. Combining equations (5.14) and (5.15) gives w = 1, which along with n = 1, gives that L = b(2, 1). If n 6= 1, the leading coefficient of P∅L |(1,j) and P∅L |(1,j+1) have opposite sign iff j = w−n+1 , or 2 in other words there is a sign change in the leading coefficients of P∅L |(1,j) at j = w−n+1 . Also note 2 and uj1 otherwise. Combining these facts using corollary 1.10, we that in PLL2 |(1,j) = 0 if j > n−w−1 2 must have w = n − 1. When w = n − 1 we fix the sign of P∅L using corollary 1.10 to get P∅L = X i+j+ 3−n−w 2 u1 i−j+ n−w+1 2 u2 − 0≤i≤n−2 0≤j≤n−1 i+j+ 3−n−w 2 X u1 i−j+ n−w−1 2 u2 . 0≤i≤n−1 0≤j≤n−2 We now know enough to compute T(L). We will compute the part of T(L) inside the region bounded by s1 + s2 ≥ n − 2, s1 ≥ 0 and s2 ≥ 0. This is shown in Figure 5.3. Using this and Lemma 3.10 we compute \ HF L(L, (1, n − 1)) ∼ = F(1) . (5.16) Once again, using Theorem 1.3 in [23]: if s = (s1 , s2 ) and as is the coefficient of us in (1 − u−1 1 )(1 − L u−1 2 )P∅ then, |a | \ HF L(L, s) ∼ = Fs s+s + σ−1 . 1 2 2 and therefore \ HF L(L, (1, n − 1)) ∼ = F(2n−2) . (5.17) Combining this with equation (5.16) gives a contradiction, since n is an integer.  Proposition 5.8 Suppose L = b(q 0 k + 1, k) = D(1, −1, . . . , 1, −1, w) is an L-space link, then q 0 = 1. Proof: We follow the same proof as the previous proposition. First note that, in this case lk(L1 , L2 ) = −ln = −w − n + 1; and so PLL1 (u2 ) = (u2 ) −w−n+1 2 ∞ ∞ X −w−n+1 X (u2 )−i and PLL2 (u1 ) = (u1 ) 2 (u1 )−i . i=0 i=0 32 s1 = 0 (1, n-1) (n-1,1) s1 + s2 = n-2 s2 = 0 Figure 5.3: Part of T(L), for b(k 2 − 1, k) assuming it is an L-space link. Edges labeled with 1 are drawn in black and edges labeled with 0 are not shown. We can compute ∆n (u1 , u2 ) = X u1i+j ui−j+n−1 − 2 0≤i≤w−1 0≤j≤n−1 X i−j+n−2 ui+j , 1 u2 1≤i≤w−1 0≤j≤n−2 which gives P∅L = − X 0≤i≤w−1 0≤j≤n−1 i+j+ −w−n+3 2 u1 i−j+ −w+n+1 2 u2 + X i+j+ −w−n+3 2 u1 i−j+ −w+n−1 2 u2 , 1≤i≤w−1 0≤j≤n−2 where the signs are fixed by corollary 1.10. Using this, we compute T(L) inside the region bounded and s2 ≥ w−n+1 and it is shown in Figure 5.4. by s1 + s2 ≥ w − 2,s1 ≥ w−n+1 2 2    w−n+1 w+n−1 ∼ \ HF L L, , = F(1) 2 2 (5.18) 33 s1 = w-n+1 2 , w+n-1 ( w-n+1 2 2 ) , w-n+1 ( w+n-1 2 2 ) s2 = w-n+1 2 s1 + s2 = w-2 Figure 5.4: Part of T(L) for b(q 0 k + 1, k), assuming it is an L-space link. We can do this computation again using the fact that L is alternating and to get    w−n+1 w+n−1 ∼ \ HF L L, , = F(2w−1) . 2 2 (5.19) combining equation 5.18 and equation 5.19 then gives w = 1 which implies q 0 = 1 as desired  Proposition 5.9 If L = b(q 0 k + 1, −k) = D(−1, 1, . . . , −1, 1, −w) is an L-space link, then k = 1. Proof: Here lk(L1 , L2 ) = −ln = w + n − 1, and so PLL1 (u2 ) = (u2 ) w+n−1 2 ∞ X (u2 )−i and PLL2 (u1 ) = (u1 ) i=0 w+n−1 2 ∞ X (u1 )−i i=0 and P∅L = − X 1−w≤i≤−1 0≤j≤n−2 i+j+ w−n+3 2 u1 i−j+ w+n−1 2 u2 + X i+j+ w−n+3 2 u1 i−j+ w+n+1 2 u2 −w≤i≤−1 0≤j≤n−1 where we have fixed signs for P∅L , as in the previous two propositions using corollary 1.10. Note that
both edges going to w+n−1 , w+n−1 must be labeled with 1 because they are determined by PLLi 2 2

since m(L) = w+n−1 , w+n−1 . Also notice that when n > 1, the point w+n−1 , w+n−1 is outside 2 2 2 2
of the Newton polytope for P∅L . Thus both edges from w+n−3 , w+n−3 are also labeled with 1. So 2 2 34 we get    w+n−1 w+n−1 ∼ \ HF L L, , = F(0) ⊕ F(−1) , 2 2 \ which is a contradiction because for an alternating link L we know HF L(L, s) is only supported in one degree. Thus, we must have n = 1, which forces k = 1 as well.  Proof: [proof of Theorem 1.11] Combining the previous three propositions (also Lemma 5.5) shows that, if b(α, β) is an L-space link, then it is either b(qk − 1, −k) for q and k odd positive integers, or of the form b(k + 1, k) where k is odd. Note that reversing the orientation of one of the components of b(k + 1, k) gives b(k + 1, −1), which proves the Theorem.  35 Bibliography [1] Gerhard Burde and Heiner Zieschang. Knots, volume 5 of de Gruyter Studies in Mathematics. Walter De Gruyter & Co, Berlin, 1985. [2] Jae C. Cha and Charles Livingston. Linkinfo: Table of knot invariants. http://www.indiana.edu/ linkinfo, October 2014. [3] John H. Conway. An enumeration of knots and links, and some of their algebraic properties, volume 346.H of Computational problems in abstract algebra. Pergamon Press, Oxford and New York, Oxford and New York, 1969. [4] Cameron Mc Gordon and Rick A. Litherland. On the signature of a link. Invent. math., 47(1):53–69, 1978. [5] Eugene Gorsky and Jennifer Hom. Cable links and L-space surgeries. preprint, 2015. arXiv:1502.05425. [6] Eugene Gorsky and András Némethi. Links of plane curve singularities are L-space links. preprint, 2014. arXiv:1403.3143v2. [7] Eugene Gorsky and Andrs Nmethi. Lattice and Heegaard-Floer homologies of algebraic links. preprint, 2015. arXiv:1301.7636v4. [8] Taizo Kanenobu. Alexander polynomials of two-bridge links. J. Austral. Math. Soc. (Series A), 36:59–68, 1984. [9] Yajing Liu. Heegaard Floer homology of surgeries on two-bridge links. preprint, 2014. arXiv:1402.5727v2. [10] Yajing Liu. L-space surgeries on links. preprint, 2014. arXiv:1409.0075v2. [11] Ciprian Manolescu and Peter Ozsváth. Heegaard Floer homology and integer surgeries on links. preprint, 2010. arXiv:1011.1317. [12] Ciprian Manolescu, Peter Ozsváth, and Sucharit Sarkar. Knot Floer homology and integer surgeries. Ann. of Math. (2), 169:633–660, 2009. 36 [13] Kunio Murasugi. Knot theory & its applications, volume 38 of Modern Birkhuser Classics. Birkhuser Boston Inc., Boston, MA, 2008. Translated by Bohdan Kurpita. [14] Yi Ni. Knot Floer homology detects fibred knots. Invent. Math., 170(3):577–608, 2007. [15] Peter Ozsváth and Zoltán Szabó. Knot Floer homology and the four-ball genus. Geom. Topol., 7:615–639, 2003. [16] Peter Ozsváth and Zoltán Szabó. Holomorphic disks and genus bounds. Geom. Topol., 8:311– 344, 2004. [17] Peter Ozsváth and Zoltán Szabó. Holomorphic disks and genus bounds. Geom. Topol., 8:311– 334, 2004. [18] Peter Ozsváth and Zoltán Szabó. Holomorphic disks and knot invariants. Adv. Math., 186(1):58– 116, 2004. [19] Peter Ozsváth and Zoltán Szabó. Holomorphic disks and three-manifold invariants: properties and applications. Ann. of Math. (2), 159(3):1159–1245, 2004. [20] Peter Ozsváth and Zoltán Szabó. Holomorphic disks and topological invariants for closed threemanifolds. Ann. of Math. (2), 159(3):1027–1158, 2004. [21] Peter Ozsváth and Zoltán Szabó. On knot Floer homology and lens space surgeries. Topology, 44(6):1281–1300, 2005. [22] Peter Ozsváth and Zoltán Szabó. On the Heegaard Floer homology of branched double-covers. Adv. Math., 194(1):1–33, 2005. [23] Peter Ozsváth and Zoltán Szabó. Holomorphic disks, link invariants and the multi-variable Alexander polynomial. Algebr. Geom. Topol., 8(2):615–692, 2008. [24] Peter Ozsváth and Zoltán Szabó. Knot Floer homology and integer surgeries. Algebr. Geom. Topol., 8(1):101–153, 2008. [25] Peter Ozsváth and Zoltán Szabó. Link Floer homology and the Thurston norm. J. Amer. Math. Soc., 21(3):671–709, 2008. [26] Peter Ozsváth and Zoltán Szabó. Knot Floer homology and rational surgeries. Geom. Topol., 11(1):1–68, 2011. [27] Dale Rolfsen. Knots and Links, volume 346.H. AMS Chelsea Publishing, Providence, Rhode Island, 2003. [28] Horst Schubert. Knoten mit zwei brücken. Mathematische Zeitschrift, 65:133–170, 1956. 37 [29] Charles A. Weibel. An introduction to homological algebra, volume 38 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1994. [30] Zhongtao Wu. Cosmetic surgery in L-space homology spheres. Geom. Topol., 15(2):1157–1168, 2011.