Nontrivial Alexander Polynomials of Knots and Links

In this paper we present a sequence of link invariants, defined from twisted Alexander polynomials, and discuss their effectiveness in distinguishing knots. In particular, we recast and extend by geometric means a recent result of Silver and Williams on the nontriviality of twisted Alexander polynomials for nontrivial knots. Furthermore we prove that these invariants decide if a genus one knot is fibered. Finally we also show that these invariants distinguish all mutants with up to 12 crossings. 1. Definition of the invariant and main results Let L ⊂ S3 be an oriented m-component link, and denote by X(L) = S3 νL its exterior. Let R = Z or R = Fp; given a representation α : π1(X(L)) → GL(R, k) we can consider the associated multivariable twisted Alexander polynomial ΔL ∈ R[t±1 1 , . . . , t±1 m ] (where t1, . . . , tm correspond to a basis of H1(X(L)) determined by the meridians to each link component), well-defined up to units. In Subsection 2.1 we recall the details of the definition. Let now α : π1(X) → Sk be a homomorphism into the symmetric group. Using the action of Sk on R by permutation of the coordinates, we get a representation π1(X) → Sk → GL(R, k) that we will denote by α as well. Consider now the set of representations of π1(X(L)) in the symmetric group, modulo conjugation: Rk(L) = {α : π1(X(L)) → Sk}/ ∼ where two representations are equivalent if they are the same up to conjugation by an element in Sk. Given α : π1(X(L)) → Sk the polynomial ΔL depends only on the equivalence class [α] of α in Rk(L). We now define the invariant ΔL = ∏ [α]∈Rk(L) ΔL ∈ R[t±1 1 , . . . , t±1 m ]. We will illustrate the effectiveness of this invariant by discussing some of the topological information that it carries, and by using explicit calculations we show its ability to distinguish many examples of inequivalent mutant knots. Our first result relates the link invariants ΔL with epimorphisms of the link group onto finite groups, which will lead to a useful topological interpretation (cf. Lemma 2.3). Precisely, consider an epimorphism γ : π1(X(L)) → G, where G is a finite group of order k = |G|. Using the left action of G on its group ring we can define a representation, denoted with the same symbol, γ : π1(X(L)) → AutR(R[G]) ∼= GL(R, k). We have the following proposition. Proposition 1.1. Let γ : π1(X(L)) → G be a homomorphism to a finite group G of order k. Then ΔγL divides Δ k L. This relation is crucial in proving the following theorem, which shows that the sequence ΔL detects the unknot and the Hopf link. Received 19 May 2006; revised 1 February 2007; published online 19 June 2007. 2000 Mathematics Subject Classification 57M27 (primary). S. Vidussi was partially supported by NSF grant #0629956. NONTRIVIAL ALEXANDER POLYNOMIALS OF KNOTS AND LINKS 615 Theorem 1.2. Let L ⊂ S3 be an oriented link which is neither the unknot nor the Hopf link (with either orientation). Then there exists a k such that ΔL = ±1 ∈ Z[t±1 1 , . . . , t±1 m ]. In fact we will show that if L is neither the unknot nor the Hopf link, then there exists an epimorphism γ : π1(X(L)) → G to a finite group such that ΔγL = 1. (This result is nontrivial when m = 1 or 2.) For the case of knots this provides a different approach to a recent result by Silver and Williams [21]. The proof is based on the relation between twisted Alexander polynomials and covers of the link exterior, using ideas from previous papers by the authors [7, 8], combined with information on the topology of those covers arising from the work in [3, 15, 18]. If K is a fibered knot, its ordinary Alexander polynomial is monic. The following result, combining results from [6] and [8], generalizes that assertion to ΔK and shows that, at least in some cases, the converse holds true. Theorem 1.3. Let K ⊂ S3 be a fibered knot, then ΔK ∈ Z[t±1] is monic for any k. Conversely, if ΔK is monic for all k and if K is a genus one knot, then K is fibered. Note that the converse also holds for knots whose exterior has fundamental group that satisfies suitable subgroup separability properties. We refer the interested reader to [8] for details (the results in [8] are only stated for closed 3-manifolds, but they also hold for 3manifolds with toroidal boundary). For a knot K the calculation of ΔK can be done using the program KnotTwister [5]. Our computations in Section 4 confirm that the ΔK are very strong knot invariants. For example computing ΔK ∈ F13[t] distinguishes all pairs and triples of mutants with up to 12 crossings (cf. Section 4 for the definition of mutants). In Section 4 we also show that ΔK is not determined by either the HOMFLY polynomial, Khovanov homology or knot Floer homology. The paper is organized as follows. In Section 2 we give a precise definition of twisted Alexander polynomials and discuss some basic properties. In particular we give a proof of Proposition 1.1. In Section 3 we give the proofs of Theorems 1.2 and 1.3. We conclude the paper in Section 4 with several examples and questions. 2. Twisted Alexander polynomials and finite covers 2.1. Twisted Alexander modules and their polynomials In this section we give the precise definition of the (twisted) Alexander polynomials. Twisted Alexander polynomials were introduced, for the case of knots, in 1990 by Lin [17], and further generalized to links by Wada [24]. We follow the approach taken by Cha [2, 7]. For the remainder of this section let N be a 3-manifold (by which we always mean a compact, connected and oriented 3-manifold) and denote by H := H1(N)/TorH1(N) the maximal free abelian quotient of π1(N). Furthermore let F be a free abelian group and let R be Z or the field Fp := Z/pZ where p is a prime number. Now let φ ∈ Hom(H,F ) be a non-trivial homomorphism. Through the homomorphism φ, π1(N) acts on F by translations. Furthermore let α : π1(N) → GL(R, k) be a representation. We write R[F ] = R ⊗R R[F ]. We get a representation α⊗ φ : π1(N) → Aut(R[F ]) g → ∑ i ai ⊗ fi → ∑ i α(g)(ai)⊗ (fi + φ(g)). We can therefore view R[F ] as a left Z[π1(N)]-module. Note that this module structure commutes with the natural R[F ]-multiplication on R[F ]. 616 STEFAN FRIEDL AND STEFANO VIDUSSI Let Ñ be the universal cover of N . Note that π1(N) acts on the left on Ñ as the group of deck transformation. The chain groups C∗(Ñ) are in a natural way right Z[π1(N)]-modules, with the right action on C∗(Ñ) defined via σ · g := g−1σ, for σ ∈ C∗(Ñ). We can form by tensor product the chain complex C∗(Ñ) ⊗ Z[π1(N)] R [F ]. Now define H1(N ;R[F ]) := H1(C∗(Ñ) ⊗ Z[π1(N)] R [F ]), which inherits the structure of R[F ]-module. The R[F ]-module H1(N ;R[F ]) is a finitely presented and finitely related R[F ]-module since R[F ] is Noetherian. Therefore H1(N ;R[F ]) has a free R[F ]-resolution R[F ] S −→ R[F ] → H1(N ;R[F ]) → 0 of finite R[F ]-modules, where we can always assume that r s. Definition 2.1. The twisted Alexander polynomial of (N,α, φ) is defined to be the order of the R[F ]-module H1(N ;R[F ]), that is, the greatest common divisor of the s× s minors of the s× r matrix S. It is denoted by ΔN,φ ∈ R[F ]. Note that this definition only makes sense since R[F ] is a unique factorization domain (UFD). It is well known that ΔN,φ is well defined only up to multiplication by a unit in R[F ] and its definition is independent of the choice of the resolution. When φ is the identity map on H, we will simply write ΔN . Also, we will write ΔN,φ in the case that α : π1(N) → GL(Z, 1) is the trivial representation. If N = X(L) is the exterior of an oriented ordered link L = L1 ∪ · · · ∪ Lm, then we write ΔL for the twisted Alexander polynomial of X(L). Also, we can identify H with the free abelian group generated by t1, . . . , tm and we can view the corresponding twisted Alexander polynomial ΔL as an element in R[t ±1 1 , . . . , t ±1 m ]. 2.2. Twisted Alexander polynomials and homomorphisms to finite groups Let N be a 3-manifold and let γ : π1(N) → G be an epimorphism onto a finite group G of order k = |G|. We get the induced regular representation γ : π1(N) → G → Aut(R[G]) where g ∈ G acts on R[G] by left multiplication. Since R[G] ∼= R|G| we can identify AutR(R[G]) = GL(R, k). It is easy to see that the isomorphism type of the R[H]-module H1(N ;R[H]) does not depend on the identification AutR(R[G]) = GL(R, k). The following lemma clearly implies Proposition 1.1. Lemma 2.2. Let γ : π1(N) → G be an epimorphism onto a finite group G of order k. Then there exists a homomorphism α : π1(N) → Sk such that the corresponding representation α : π1(N) → Sk → GL(R, k) is given by the regular representation γ : π1(N) → G → GL(R, k). Proof. Denote the elements of G by g1, . . . , gk. Since γ defines an action on the set G = {g1, . . . , gk} via left multiplication we get an induced map α : π1(N) → Sk. Clearly the corresponding representation α : π1(N) → Sk → GL(R, k) is isomorphic to the regular representation γ : π1(N) → G → GL(R, k). 2.3. Twisted Alexander polynomials and finite covers For the remainder of this section let γ : π1(N) → G be an epimorphism onto a finite group G of order k, and take R = Z. Denote the induced G-cover of N by π : NG → N . Also, denote by NONTRIVIAL ALEXANDER POLYNOMIALS OF KNOTS AND LINKS 617 HG the maximal free abelian quotient of π1(NG): the map π∗ : HG → H is easily seen to have maximal rank; hence in particular b1(NG) b1(N). Given any homomorphism φ : H → F to a free abelian group F we can consider the induced homomorphism φG := π∗φ : HG → F . In particular, when φ is the identity map on H, we have φG = π∗ : HG → H. We can now formulate the relationship between the twisted Alexander polynomials of N and the untwisted Alexander polynomial of NG. Lemma 2.3 [7]. Let γ : π1(N) → G be an epimorphism onto a finite group G and π : NG → N the induced G-cover. Then ΔγN = ΔNG,π∗ ∈ Z[H]. Finally, we need to rewrite the Alexander polynomial ΔNG,π∗ in terms of the full Alexander polynomial of NG; their relation is the following. Proposition 2.4 [7, 22]. Let N be a 3-manifold with non-empty toroidal boundary, and let NG be the 3-manifold defined as above. Furthermore let ΔNG ∈ Z[HG] be the (ordinary multivariable) Alexander polynomial. Then we have the following equality in Z[H]. If b1(NG)> 1, then ΔNG,π∗ = ⎧⎨ ⎩ π∗(ΔNG) if b1(N) > 1, (tdivπ∗ − 1)π∗(ΔNG) if b1(N) = 1, Imπ∗ = 〈tdiv ∗〉, t ∈ H indivisible. (1) If b1(NG) = 1, then b1(N) = 1 and ΔNG,π∗ = π∗(ΔNG). (2) 3. Proof of Theorems 1.2 and 1.3 3.1. Proof of Theorem 1.2 The topological ingredient of the proof is a result on the virtual Betti number of link exteriors. This result can be deduced quite directly from [3, Theorem 1.3], but it is perhaps appropriate, in order to illustrate its nature, to break down the proof to emphasize the role of the JSJ decomposition of a link exterior. We start with the following results. Theorem 3.1 [15, 18]. Let N be an irreducible 3-manifold containing an essential torus or annulus S; up to a lift to a finite cover, we can assume that S is non-separating. Then S is either the fiber of a fibration over S1, or the virtual Betti number vb1(N) of N is infinite. Remark. Recall that having virtual Betti number vb1(N) infinite means that N admits finite covers of arbitrarily large Betti number. A priori, the covers do not have to be regular; however, to any finite cover N̂ with fundamental group π̂ we can canonically associate a finite regular cover N̄ determined by the subgroup π̄ := ⋂ p∈π1(N) pπ̂p −1. This subgroup is clearly a normal subgroup of both π̂ and π1(N). Also, since π̂ ⊂ π1(N) is of finite index we see easily that π̄ is in fact the intersection of finitely many subgroups of π1(N) of finite index. Therefore π̄ ⊂ π̂ ⊂ π1(N) is of finite index as well, and N̄ is a finite cover. From standard arguments, we have b1(N̄) b1(N̂) b1(N), so we can assume that N admits finite regular covers of arbitrarily large Betti number. The set of left cosets π1(N)/π̄ is a finite group, that we denote by G, hence π̄ is the kernel of an epimorphism γ : π1(N) → G, so that N̄ = NG. 618 STEFAN FRIEDL AND STEFANO VIDUSSI Theorem 3.2 [3, Theorem 2.7]. Let N be an irreducible 3-manifold with non-empty incompressible boundary all of whose components are tori. Suppose that the interior of N has a complete hyperbolic structure of finite volume. Then vb1(N) = ∞. The topological ingredient in the proof of Theorem 1.2 is then the following observation. Lemma 3.3. Let L = L1 ∪ · · · ∪ Lm ⊂ S3 be an oriented link which is neither the unknot nor the Hopf link. Then vb1(X(L)) = ∞. Proof. First note that if L is a split link, that is, if X(L) = S3 νL is reducible, then π1(X(L)) maps onto a free group with two generators, which implies that vb1(X(L)) = ∞ (cf. for example [15]). We can therefore now assume that L is not a split link. In particular no component of L bounds a disk in the complement of the components. By Dehn’s lemma this implies that the boundary of X(L) is incompressible. As X(L) is irreducible and has boundary, X(L) is Haken, hence it admits a geometric decomposition along a (possibly empty) family of essential tori T . We will break the argument into subcases. First assume that T is non-empty. Clearly X(L) cannot be covered by a torus bundle over S1 since X(L) has boundary. It therefore follows from Theorem 3.1 that vb1(X(L)) = ∞. Now assume that T is empty. By Thurston’s geometrization of Haken manifolds we deduce that either X(L) is Seifert-fibered or the interior of X(L) has a complete hyperbolic structure of finite volume. In the hyperbolic case, Theorem 3.2 asserts that vb1(X(L)) = ∞. We are left with the Seifert-fibered case. The classification of Seifert links (see [4, Chapter II]) shows that L is the link obtained by removing m fibers, regular or singular, from the (p, q)Seifert fibration of S3, where (p, q) are coprime integers or (0,±1). Depending on the type of the orbifold quotient (see Jaco [13, Chapter VIII]), X(L) either contains essential tori or is special. In the former case, Theorem 3.1 implies that vb1(X(L)) = ∞ right away. If X(L) is special, checking case-by-case, L is either: a (nontrivial) (p, q)-torus knot, obtained by removing a regular fiber; the union of the unknot and its (p, q)-cable, obtained by removing a regular fiber and the fiber with multiplicity p (whose exterior is the p/q-cable space); or one of a family of 3-component links obtained by removing a regular fiber and the two singular fibers. In the last two cases, we can identify an essential, non-separating cabling annulus joining a regular and a singular fiber of the Seifert fibration. With the exception of the Hopf link with either orientations (corresponding to q = ±1) these annuli do not fiber X(L) by [4, Theorem 11.2]. For a (p, q)-torus knot traced on a torus T , the annulus X(L) ∩ T is the only essential annulus, and it is separating, so we pass to some finite cover. However, this cover cannot be an annulus bundle over S1 (T 2 × I or the twisted I-bundle over a Klein bottle), as by [11, Theorems 10.5 and 10.6] the only manifolds covered by T 2 × I are T 2 × I itself and the twisted I-bundle over a Klein bottle, which does not embed in S3. It follows that, with the exception of the Hopf link with either orientation (for whom X(L) = T 2 × I), all these links have vb1(X(L)) = ∞ by Theorem 3.1. The following theorem, together with Proposition 1.1, immediately implies Theorem 1.2. Theorem 3.4. Let L = L1 ∪ · · · ∪ Lm ⊂ S3 be an oriented link which is neither the unknot nor the Hopf link. Then there exists an epimorphism γ : π1(X(L)) → G onto a finite group G such that ΔγL = ±1. Proof. Since L is neither the unknot nor the Hopf link, Lemma 3.3 implies that there exists a cover X(L)G with b1(X(L)G) > 2. As X(L)G has non-empty boundary all of whose NONTRIVIAL ALEXANDER POLYNOMIALS OF KNOTS AND LINKS 619 components are tori, [23, Corollary II.4.4] implies that the sum of the coefficients of ΔX(L)G is zero. Hence, by Proposition 2.4 the sum of the coefficients of ΔX(L)G,π∗ is zero as well, hence, by Lemma 2.3, ΔγX(L) cannot be ±1. When L is the unknot or the Hopf link, X(L) is homeomorphic to S1 ×D2 and T 2 × I respectively. In particular, the maximal (free) abelian cover X̂(L) is contractible. Given any representation π1(X(L)) → GL(R, k), we have H1(X(L);R[H1(X(L))]) ∼= H1(X̂(L);R) = 0, where the first isomorphism follows from the Eckmann–Shapiro lemma. As the corresponding twisted Alexander module is trivial, ΔL = 1 for all k. This implies that the sequence Δ k L detects the unknot and the Hopf link. 3.2. Proof of Theorem 1.3 Let K ⊂ S3 be a fibered knot; it is shown in [6] that ΔK is monic for any representation α : π1(X(K)) → GL(Z, k) (cf. also [2] and [10]). This clearly implies that ΔK is monic for all k. Now let K be a genus one knot such that ΔK is monic for all k. We denote by NK the zero framed surgery along K. Gabai [9] showed that K is fibered if and only if NK is fibered. Clearly, NK has vanishing Thurston norm. Under this hypothesis we show in [8] that if NK is not fibered, then there exists an epimorphism β : π1(NK) → G onto a finite group G such that ΔβNK = 0. Now consider the homomorphism γ : π1(X(K)) → π1(NK) → G. Since π1(X(K)) → π1(NK) is an epimorphism, it follows from the five-term exact sequence (cf. [1, Chapter VII, Corollary 6.4]) that H1(NK ;Rk[t±1]) is a quotient of H1(X(K);R[t]), hence there exists a polynomial p(t) ∈ Z[t±1] such that ΔγK = p(t)Δ β NK (cf. also [14]). In particular ΔγX(K) = 0. Then Theorem 1.3 follows from Proposition 1.1.