Twisted Alexander Polynomials of Hyperbolic Knots Nathan M

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Experimental Mathematics Publication details, including instructions for authors and subscription information: http://www.tandfonline.com/loi/uexm20 Twisted Alexander Polynomials of Hyperbolic Knots Nathan M. Dunfield a , Stefan Friedl b & Nicholas Jackson c a Department of Mathematics, University of Illinois, 1409 W. Green Street, Urbana, IL, 61801, USA b Mathematisches Institut, Raum 020, Weyertal 86-90, Universität zu Köln, Köln, 50931, Germany c Mathematics Institute, Zeeman Building, University of Warwick, Coventry, CV4 7AL, UK Version of record first published: 17 Dec 2012.

To cite this article: Nathan M. Dunfield , Stefan Friedl & Nicholas Jackson (2012): Twisted Alexander Polynomials of Hyperbolic Knots, Experimental Mathematics, 21:4, 329-352 To link to this article: http://dx.doi.org/10.1080/10586458.2012.669268

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Twisted Alexander Polynomials of Hyperbolic Knots

Nathan M. Dunﬁeld, Stefan Friedl, and Nicholas Jackson

CONTENTS We study a twisted Alexander polynomial naturally associated 1. Introduction to a hyperbolic knot in an integer homology 3-sphere via a lift of 2. Twisted Invariants of 3-Manifolds the holonomy representation to SL(2, C ). It is an unambiguous 3. Twisted Torsion of Cyclic Covers symmetric Laurent polynomial whose coefficients lie in the trace 4. Torsion Polynomials of Hyperbolic Knots field of the knot. It contains information about genus, fibering, 5. Example: The Conway and Kinoshita–Terasaka Knots and chirality, and moreover, is powerful enough to sometimes 6. Knots with at Most 15 Crossings detect mutation. 7. Twisted Torsion and the Character Variety of a Knot We calculated this invariant numerically for all 313 209 hyper- 8. Character Variety Examples bolic knots in S3 with at most 15 crossings, and found that in Acknowledgments all cases it gave a sharp bound on the genus of the knot and References determined both fibering and chirality.

We also study how such twisted Alexander polynomials vary

as one moves around in an irreducible component X 0 of the SL(2, C )-character variety of the knot group. We show how to understand all of these polynomials at once in terms of a poly-

nomial whose coefﬁcients lie in the function ﬁeld of X 0.Weuse this to help explain some of the patterns observed for knots in S3, and explore a potential relationship between this universal polynomial and the Culler–Shalen theory of surfaces associated to ideal points.

1. INTRODUCTION A fundamental invariant of a knot K in an integral Downloaded by [87.153.35.205] at 11:13 07 January 2013 homology 3-sphere Y is its Alexander polynomial ΔK . While ΔK contains information about genus and fibering, it is determined by the maximal metabelian quotient of the fundamental group of the complement M = Y \ K, and so this topological information has clear lim- its. In 1990, Lin introduced the twisted Alexander polynomial associated to K and a representation α : π1 (M) → GL(n, F ), where F is a field. These twisted Alexander polynomials also contain information about genus and fibering and have been studied by many authors (see the survey [Friedl and Vidussi 10]). Much of this work has focused on those α that factor through a finite quotient of π1 (M), which is closely related to studying the ordinary 2000 AMS Subject Classification: 57M25, 57M27, 57M50 Alexander polynomial in finite covers of M. In contrast, Keywords: hyperbolic knots, twisted Alexander polynomial, we study here a twisted Alexander polynomial associated twisted torsion, genus bounds, character varieties to a representation coming from hyperbolic geometry.

329 330 Experimental Mathematics, Vol. 21 (2012), No. 4

∗ Suppose that K is hyperbolic, i.e., the complement (d) If K denotes the mirror image of K, then TK ∗ (t)= M has a complete hyperbolic metric of finite volume, TK (t), where the coefficients of the latter polynomial and consider the associated holonomy representation are the complex conjugates of those of TK . → + H 3 + H 3 ∼ C α : π1 (M) Isom ( ). Since Isom ( ) = PSL(2, ), (e) If K is amphichiral, then TK is a real polynomial. there are two simple ways to get a linear representa- (f) The values T (1) and T (−1) are mutation- tion so that we can consider the twisted Alexander K K invariant. polynomial: compose α with the adjoint representation to get π1 (M) → Aut(sl2 C ) ≤ SL(3, C ), or alterna- T tively lift α to a representation π1 (M) → SL(2, C ). Moreover, K both determines and is determined by The former approach is the focus of the recent paper a sequence of C -valued torsions of finite cyclic covers of [Dubois and Yamaguchi 09]; the latter method is what X. Specifically, let Xm be the m-fold cyclic cover coming ±1 Z we use here to define an invariant TK (t) ∈ C [t ], called from the free abelianization of H1 (X; ). For the restric- the hyperbolic torsion polynomial. tion αm of α to π1 (Xm ), we consider the correspond- C The hyperbolic torsion polynomial TK is surprisingly ing -valued torsion τ(Xm ,αm ). A standard argument little studied. To our knowledge, it has previously been shows that TK determines all the τ(Xm ,αm ) (see Theo- looked at only for 2-bridge knots, in [Morifuji 08, Kim rem 3.1). More interestingly, the converse holds: the tor- and Morifuji 10, Hirasawa and Murasugi 08, Silver and sions τ(Xm ,αm ) determine TK (see Theorem 4.3). This Williams 09]. Here we show that it contains a great deal of latter result follows from work in [Fried 86] (see also topological information. In fact, we show that TK deter- [Hillar 05]) and [Menal-Ferrer and Porti 10]. mines genus and fibering for all 313 209 hyperbolic knots in S3 with at most 15 crossings, and we conjecture this Remark 1.2. Conjecturally, the torsions τ(Xm ,αm )can to be the case for all knots in S3 . be expressed as analytic torsions and as Ruelle zeta functions defined using the lengths of prime geodesics. [Ray and Singer 71, Cheeger 77, Cheeger 79, Müller 78, 1.1. Basic Properties Park 09] for details and background material. We hope More broadly, we consider here knots in Z2 -homology that this point of view will be helpful in the further study 3-spheres. The ambient manifold Y containing the knot of TK . K will always be oriented, not just be orientable, and The torsions τ(Xm ,αm ) are interesting invariants TK depends on that orientation. Following Turaev, we in their own right. For example, it is shown in T formulate K as a Reidemeister–Milnor torsion, since [Menal-Ferrer and Porti 10] that τ(Xm ,αm ) is nonzero this reduces its ambiguity; in that setting, we work for every m. Furthermore, J. Porti showed us in a pri- with the compact core of M, namely the knot exterior vate communication (2009) that τ(X1 ,α1 )=τ(X, α)= \ X := Y int(N(K)) (see Section 2 for details). By fixing TK (1) is not obviously related to hyperbolic vol- certain conventions for lifting the holonomy representa- ume. More precisely, using a variation on [Porti 97, → C → C

Downloaded by [87.153.35.205] at 11:13 07 January 2013 tion α : π1 (X) PSL(2, )toα : π1 (X) SL(2, ), we Théorème 4.17], one can show that there exists a se- construct in Section 4 a well-defined symmetric polyno- quence of knots Kn whose volumes converge to a T ∈ C ±1 T mial K [t ]. The first theorem summarizes its basic positive real number, but the numbers K n (1) con- properties: verge to zero. See [Porti 97, Menal-Ferrer and Porti 10, Menal-Ferrer and Porti 11] for further results. Theorem 1.1. Let K be a hyperbolic knot in an oriented Z2 -homology 3-sphere. Then TK has the following properties: 1.2. Topological Information: Genus and Fibering ±1 We define x(K) to be the Thurston norm of a gener- (a) TK is an unambiguous element of C [t ] that satisfies −1 Z ∼ Z TK (t )=TK (t). It does not depend on an orienta- ator of H2 (X, ∂X; ) = ;ifK is null-homologous in tion of K. Y , then x(K)=2· genus(K) − 1, where genus(K)isthe least genus of all Seifert surfaces bounding K. Also, we (b) The coefficients of TK lie in the trace field of K.IfK say that K is fibered if X fibers over the circle. has integral traces, the coefficients of TK are algebraic integers. A key property of the ordinary Alexander polynomial ΔK is that (c) TK (ξ) is nonzero for every root of unity ξ. In particular, TK =0 . x(K) ≥ deg(ΔK ) − 1. Dunfield et al.: Twisted Alexander Polynomials of Hyperbolic Knots 331

When K is fibered, this is an equality, and moreover, alently its genus. Moreover, the knot K is fibered if and the leading coefficient of ΔK is 1 (here we normalize ΔK only if TK is monic. so that the leading coefficient is positive). As with any twisted Alexander/torsion polynomial, we get similar in- We have not done extensive experiments for knots in manifolds other than S3 , but so far, we have not encoun- formation out of TK : tered any examples in which TK does not contain perfect genus and fibering information. Theorem 1.3. Let K be a knot in an oriented Z2 -homology sphere. Then 1.4. Topological Information: Chirality and Mutation 1 T x(K) ≥ deg(TK ). When K is amphichiral, K is a real polynomial (The- 2 orem 1.1(e)). This turns out to be an excellent way to 3 If K is fibered, this is an equality and TK is monic, i.e., detect chirality. Indeed, among hyperbolic knots in S has leading coefficient 1. with at most 15 crossings, the 353 knots for which TK is real are exactly the amphichiral knots (Section 6.3). Theorem 1.3 is an immediate consequence of the def- Also, hyperbolic invariants often do not detect muta- initions below and of [Friedl and Kim 06, Theorem 1.1] tion, for example the volume [Ruberman 87], the invari- (for the genus bound) and of [Goda et al. 05] (for the ant trace field [Maclachlan and Reid 03, Corollary 5.6.2], fibered case); see also [Cha 03, Kitano and Morifuji 05, and the birationality type of the geometric component of Pajitnov 07, Kitayama 09, Friedl and Kim 06] and the character variety [Cooper and Long 96, Tillmann 00, [Friedl and Vidussi 10, Theorem 6.2]. Tillmann 04]. The ordinary Alexander polynomial ΔK is also mutation-invariant for knots in S3 . However, x(K) 1.3. Experimental Results can change under mutation, and given that x(K) determines the degree of T for all 15 crossing knots, it follows The calculations in [Cha 03, Goda et al. 05, Friedl and K that T can change under mutation; we discuss many Kim 06]gave evidence that when one can freely choose K such examples in Section 6.4. However, sometimes muta- the representation α, the twisted torsion polynomial is tion does preserve T , and we do not know of any exam- very successful at detecting both x(K) and nonfibered K ples of two knots with the same T that are not mutants. knots. Moreover, [Friedl and Vidussi 01] shows that K collectively, the twisted torsion polynomials of represen- 1.5. Adjoint Torsion Polynomial tations coming from homomorphisms to finite groups As we mentioned earlier, there is another natural determine whether a knot is fibered. However, it is not way to obtain a torsion polynomial from the holon- known whether all twisted torsion polynomials together omy α : π (M) → PSL(2, C ), namely by considering the always detect x(K). 1 adjoint representation of PSL(2, C ) on its Lie alge- Instead of considering many different representations bra. The corresponding torsion polynomial was stud-

Downloaded by [87.153.35.205] at 11:13 07 January 2013 as in the work just discussed, we focus here on a single, al- ied in [Dubois and Yamaguchi 09], partly building on beit canonical, representation. Despite this, we find that [Porti 97]. We refer to this invariant here as the adjoint T alone is a very powerful invariant. In Section 6, we K torsion polynomial and denote it by T adj. We also numer- describe computations showing that the bound on x(K) K ically calculated this invariant for all knots with at most is sharp for all 313 209 hyperbolic knots with at most 15 15 crossings. In contrast to what we found for T ,the crossings; in contrast, the bound from Δ is not sharp K K degree of T adj was not determined by the genus for 8 252 for 2.8% of these knots. Moreover, among such knots, K of these knots. Moreover, we found 12 knots for which the T was monic only when the knot was fibered, whereas K genus bound from T adj was not sharp even after account- 4.0% of these knots have monic Δ but are not fibered. K K ing for the fact that x(K) is necessarily an odd integer. (Here we computed T numerically to a precision of 250 K The differing behaviors of these two polynomials seems decimal places; see Section 6.7 for details.) very mysterious to us; understanding what is behind it Given all this data, we are compelled to propose the might shed light on Conjecture 1.4. See Sections 5.1 and following conjecture, even though on its face it feels 6.6 for the details on what we found for T adj. quite implausible, given the general squishy nature of K Alexander-type polynomials. 1.6. Character Varieties Conjecture 1.4. For a hyperbolic knot K in S3 , the hyper- So far, we have focused on the twisted torsion polyno- T bolic torsion polynomial K determines x(K),orequiv- mial of (a lift of) the holonomy representation of the 332 Experimental Mathematics, Vol. 21 (2012), No. 4

T X 0 hyperbolic structure on M. However, this representa- several explicit examples of K in Section 8 and use tion is always part of a complex curve of representations these to explore a possible avenue for bringing the Culler– π1 (M) → SL(2, C ), and it is natural to study how the Shalen theory of surfaces associated to ideal points of torsion polynomial changes as we vary the representa- X(K) to bear on Conjecture 1.7. tion. In Sections 7 and 8, we describe how to understand all of these torsion polynomials at once, and use this to 1.7. Other Remarks and Open Problems help explain some of the patterns observed in Section 6. For the special case of 2-bridge knots, how the torsion For simplicity, we have restricted ourselves here to the 3 polynomial varies with the representation had previously study of knots, especially those in S . However, we been studied in [Morifuji 08, Kim and Morifuji 10], and expect that many of the results and conjectures are valid our results here extend some of that work to more general for more general 3-manifolds. In the broader settings, knots. the appropriate question is whether the twisted torsion Consider the character variety polynomial detects the Thurston norm and fibered classes (see [Friedl and Kim 06, Friedl and Kim 08, C C X(K):=Hom π1 (M), SL(2, ) // SL(2, ), Friedl and Vidussi 08] for more details). which is an affine algebraic variety over C .Weshowin We conclude this introduction with a few questions Section 7 that each χ ∈ X(K) has an associated torsion andopenproblems: polynomial T χ . These T χ vary in an understandable way, K K T in terms of a polynomial with coefficients in the ring of 1. Does K determine the volume of the complement of K? Some calculational evidence is given regular functions C [X0 ]: in [Friedl and Jackson 11] and in Section 6.4 in this paper. Theorem 1.5. Let X0 be an irreducible component of 3 X(K) that contains the character of an irreducible repre- 2. If two knots in S have the same TK , are they ± T X 0 ∈ C 1 sentation. There is a unique K [X0 ][t ] such that necessarily mutants? See Section 6.4 for more on χ ∈ T T X 0 for all χ X0 , one has K (t)= K (χ)(t). Moreover, this. T X 0 is itself the torsion polynomial of a certain repre- K 3. Does the invariant T contain information about sentation π (M) → SL(2, F ), and thus has all the usual K 1 symmetries of the knot besides information on properties (symmetry, genus bound, etc.). chirality?

Corollary 1.6. Let K be a knot in an integral homology 4. Does there exist a hyperbolic knot with TK (1) = 3-sphere. Then the following hold: 1? { ∈ | T χ } T (a) The set χ X(K) deg( K )=2x(K) is Zariski 5. If K is a real polynomial, is K necessarily am- open. phichiral? χ

Downloaded by [87.153.35.205] at 11:13 07 January 2013 { ∈ |T } (b) The set χ X(K) K is monic is Zariski 6. For an amphichiral knot, is the top coefficient of closed. TK always positive? 7. For fibered knots, why is the second coefficient It is natural to focus on the component X0 of X(K) of TK so often real? This coefficient is the sum of that contains the (lift of) the holonomy representation of the eigenvalues of the monodromy acting on the the hyperbolic structure, which we call the distinguished twisted homology of the fiber. See Section 6.5 component. In this case, X0 is an algebraic curve, and we for more. show that the following conjecture is implied by Conjec- 8. Why is |T (−1)| > |T (1)| for 99.99% of the ture 1.4. K K knots considered in Section 6.5?

Conjecture 1.7. Let K be a hyperbolic knot in S3 ,and X0 the distinguished component of its character variety. 2. TWISTED INVARIANTS OF 3-MANIFOLDS T X 0 T X 0 Then 2x(K) = deg( K ),and K is monic if and only if K is fibered. In this section, we review torsions of twisted homology groups and explain how they are used to define At the very least, Conjecture 1.7 is true for many 2- the twisted torsion polynomial of a knot together with bridge knots, as we discuss in Section 7.2. We also give a representation of its fundamental group to SL(2, C ). Dunfield et al.: Twisted Alexander Polynomials of Hyperbolic Knots 333

∈ · We then summarize the basic properties of these torsion then there are unique gj π such thatc ˜j = gj c˜j .We polynomials, including how to calculate them. say that these two Euler lifts are equivalent if (−1)dim(cj ) 2.1. Torsion of Based Chain Complexes gj j Let C∗ be a finite chain complex over a field F . Sup- represents the trivial element in H (X; Z). An equiva- pose that each chain group Ci is equipped with an or- 1 lence class of Euler lifts is called an Euler structure on dered basis ci and that each homology group Hi(C∗) X. The set of Euler structures on X, denoted by Eul(X), is also equipped with an ordered basis hi. Then there × Z is an associated torsion invariant τ(C∗,c∗,h∗) ∈ F := admits a canonical free transitive action by H1 (X; ); F \{0}, as described in the various excellent exposi- see [Turaev 90, Turaev 01, Turaev 02, Friedl et al. 11] for tions [Milnor 66, Turaev 01, Turaev 02, Nicolaescu 03]. more on these Euler structures. Finally, a homology ori- R We will follow the convention of Turaev, which is the mul- entation for X is just an orientation of the -vector space R tiplicative inverse of Milnor’s invariant. If the complex C∗ H∗(X; ). is acyclic, then we will write τ(C∗,c∗):=τ(C∗,c∗, ∅). Now we can define the torsion τ(X, α, e, ω) associated to X, a representation α, an Euler structure e,and α 2.2. Twisted Homology a homology orientation ω.IfH∗ (X; V ) = 0, we define τ(X, α, e, ω) := 0, and so now assume that Hα (X; V )= For the rest of this section, fix a finite CW complex X ∗ → 0. Up to sign, the torsion we seek will be that of the and set π := π1 (X). Consider a representation α : π α twisted cellular chain complex C∗ (X; V ) with respect GL(V ), where V is a finite-dimensional vector space over { } F Z to the following ordered basis. Let vk be any ba- . We can thus view V as a left [π]-module. To define { } α sis of V ,and c˜j any Euler lift representing e.Or- the twisted homology groups H∗ (X; V ), consider the uni- α der the basis {c˜j ⊗ vk } for C∗ (X; V ) lexicographically, versal cover X of X. Regarding π as the group of deck i.e.,c ˜j ⊗ vk < c˜j ⊗ vk if either j

n ±1 ∼ ±1 n for V . Similar elementary arguments prove the following R ⊗R R[t ] = R[t ] given by lemma. Here −ω denotes the opposite homology orienta- g · (v ⊗ p):= α(g) · v ⊗ tφ(g)p tion to ω, and note that (det ◦α):π → F factors through Z n ±1 H1 (X; ). for g ∈ π and v ⊗ p ∈ R ⊗R R[t ]. Put differently, we get a representation α ⊗ φ : π → GL(n, R[t±1 ]). We de- Lemma 2.1. If β is conjugate to α, then given h ∈ note by Q(t) the field of fractions of R[t±1 ]. The repre- Z ∈{− } ± H1 (X; ) and 1, 1 , one has sentation α ⊗ φ allows us to view R[t 1 ]n and Q(t)n as −1 Z s ∈ c τ(X, β, h · e, · ω)= dim(V ) · (det ◦α)(h) ·τ(X, α, e, ω). left [π]-modules. Given Spin (X), we define

τ(K, α, s):=τ(XK ,α⊗ φ, s,ωK ) ∈ Q(t) 2.4. Twisted Torsion of 3-Manifolds to be the twisted torsion polynomial of the oriented knot Let N be a 3-manifold whose boundary is empty or K corresponding to the representation α and the Spinc - consists of tori. We first recall some facts about Spinc - structure s. Calling τ(K, α, s) a polynomial even though structures on N; see [Turaev 02, Section XI.1] for de- it is defined as a rational function is reasonable given tails. The set Spinc (N) of such structures admits a free Theorem 2.4 below. and transitive action by H1 (N; Z). Moreover, there ex- c ists a map c1 : Spin (N) → H1 (N; Z) that has the prop- Remark 2.2. The study of twisted polynomial invari- erty that c1 (h · s)=2h + c1 (s) for every h ∈ H1 (N; Z) ants of knots was introduced in [Lin 01]. The invari- and s ∈ Spinc (N). ant τ(K, α, s) can be viewed as a refined version of Now consider a triangulation X of N. By [Turaev 02, the twisted Alexander polynomial of a knot and of Section XI], there exists a canonical bijection Wada’s invariant. We refer to [Wada 94, Kitano 96, Spinc (N) → Eul(X) that is equivariant with respect to Friedl and Vidussi 10] for more on twisted invariants of the actions by H1 (N; Z)=H1 (X; Z). Given a represen- knots and 3-manifolds. c tation α : π1 (N) → GL(V ), an element s ∈ Spin (N), and a homology orientation ω for N, we define 2.6. The SL(2, F ) Torsion Polynomial of a Knot τ(N,α,s,ω):=τ(X, α, e, ω), Our focus in this paper is on 2-dimensional representations, and we now give a variant of τ(K, α, s) that does where e is the Euler structure on X corresponding to s. not depend on s or the orientation of K. Specifically, for It follows from [Turaev 86, Turaev 90] that τ(N,α,s,ω) an unoriented knot K in a Q HS and a representation is independent of the choice of triangulation and hence α : π → SL(2, F ), we define is well defined. See also [Friedl et al. 11] for more details s T α φ(c1 ( )) · s about τ(N,α,s,ω). K := t τ(K, α, ) ( 2–1) for every s ∈ Spinc (X) and choice of oriented meridian μ 2.5. Twisted Torsion Polynomial of a Knot

Downloaded by [87.153.35.205] at 11:13 07 January 2013 and prove the following result. Let K be a knot in a rational homology 3-sphere Y . \ Throughout, we write XK := Y int(N(K)) for the knot → F T α Theorem 2.3. For α : π SL(2, ), the invariant K is exterior, which is a compact manifold with torus bound- a well-defined element of F (t) that is symmetric, i.e., ary. We define an orientation of K to be a choice of T α −1 T α K (t )= K (t). oriented meridian μK ;ifY is oriented, instead of just orientable, this is equivalent to the usual notion. Sup- T α ∈ F We will call K (t)thetwisted torsion polynomial pose now that K is oriented. Let πK := π1 (XK ) and take associated to K and α. φK : πK → Z to be the unique epimorphism such that φ(μK ) > 0. There is a canonical homology orientation ωK Proof. That the right-hand side of (2–1) is independent for XK as follows: take a point as a basis for H0 (XK ; R ) of the choice of s follows easily from Lemma 2.1 using 2φ(g) and take {μK } as the basis for H1 (XK ; R ). We will drop the observation that det ((α ⊗ φ)(g)) = t for g ∈ π K from such notation if the knot is understood from the and the properties of c1 given in Section 2.4. context. The choice of meridian μ affects the right-hand side of For a representation α : π → GL(n, R) over a com- (2–1) in two ways: it is used to define the homology ori- mutative domain R, we define a torsion polynomial entation ω and the homomorphism φ : π → Z. The first as follows. Consider the left Z[π]-module structure on doesn’t matter, by Lemma 2.1, but switching φ to −φ is Dunfield et al.: Twisted Alexander Polynomials of Hyperbolic Knots 335

equivalent to replacing t with t−1 . Hence being indepen- For part (b), conjugate α so that it is upper diagonal: dent of the choice of meridian is equivalent to the ﬁnal T α a(g) b(g) claim that K is symmetric. α(g)= for all g ∈ π. −1 Now any SL(2, F )-representation preserves the bilin- 0 a(g) F 2 → ear form on given by (v,w) det(vw). Using this The diagonal part of α is the representation β given by observation, it is shown in [Friedl et al. 11, Theorem 7.3], generalizing [Hillman et al. 10, Corollary 3.4] and build- a(g)0 g → . ing on work in [Turaev 86, Turaev 90], that in our con- 0 a(g)−1 text, we have It is easy to see, for instance using (2–2), that T α = T β . − s K K s 1 2φ(c1 ( )) · s τ(K, α, ) t = t τ(K, α, ), Finally, part (c) follows from a straightforward calcula- T α tion with (2–2); see, e.g., [Turaev 01, Turaev 02]. which establishes the symmetry K and hence the theorem. 2.7. Calculation of Torsion Polynomials Using T α While in general, K is a rational function, it is fre- Fox Calculus quently a Laurent polynomial or can be computed in Suppose we are given a knot K in a Q HS and a rep- terms of the ordinary Alexander polynomial ΔK . resentation α : π1 (XK ) → SL(2, F ). In this section, we give a simple method for computing T α . As usual, we Q → K Theorem 2.4. Let K be a knot in a HS and let α : πK write π := π (X )andφ = φ . We can extend the group SL(2, F ) be a representation. 1 K K homomorphism α ⊗ φ : π → GL(2, F [t±1 ]) to a ring ho- T α F ±1 Z → F ±1 (a) If α is irreducible, then K lies in [t ]. momorphism [π] M(2, [t ]), which we also denote ⊗ × Z (b) If α is reducible, then T α = T β ,whereβ is the di- by α φ. Given a k l matrix A =(aij )over [π], K K ⊗ × agonal part of α, i.e., a diagonal representation such we denote by (α φ)(A) the 2k 2l matrix obtained that tr β(g) =tr α(g) for all g ∈ π. from A by replacing each entry aij by the 2 × 2 matrix (α ⊗ φ)(a ). (c) If α is reducible and factors through ij Now let F = x1 ,...,xn be the free group on n gener- H1 (XK ; Z)/(torsion), ators. By [Fox 53, Fox 54, Crowell and Fox 63] and also then [Harvey 05, Section 6], there exists for each xi a Fox derivative · −1 T α ΔK (zt) ΔK (z t) K (t)= − − , ∂ t − (z + z 1 )+t 1 : F → Z[F ] ∂xi −1 where z,z are the eigenvalues of α(μK ),andΔK is the symmetrized Alexander polynomial. with the following two properties: ∂xj ∂(uv) ∂u ∂v

Downloaded by [87.153.35.205] at 11:13 07 January 2013 = δ and = + u for all u, v ∈ F . ∂x ij ∂x ∂x ∂x When the ambient manifold Y is a ZHS, then the i i i i torsion polynomial of any reducible representation α to We also need the involution of Z[F ] that sends g ∈ F − SL(2, F ) can be computed by combining (b) and (c); to g 1 and respects addition (this is not an algebra au- Z T α tomorphism, since it induces an antihomomorphism for when H1 (Y ; ) is finite but nontrivial, then K is the product of the torsion polynomials of two 1-dimensional multiplication). We denote the image of a ∈ Z[F ] under Z representations, but it may not be directly related to ΔK . this map by a,andifA is a matrix over [F ], then A denotes the result of applying this map to each entry. Proof. Part (a) is due to [Kitano and Morifuji 05] and T α The following allows for the efficient calculation of K , is seen as follows. Since α is irreducible, there is a since π always has such a presentation (e.g., if K is a knot ∈ g [π,π] such that α(g) does not have trace 2 (see, in S3 , one can use a Wirtinger presentation). e.g., [Culler and Shalen 83, Lemma 1.5.1] or the first part of the proof of [Kitano and Morifuji 05, Theorem 1.1]). Proposition 2.5. Let K beaknotinaQ HS,and Then take a presentation of π for which g is a genera- x1 ,...,xn | r1 ,...,rn−1 a presentation of πK of tor and apply Proposition 2.5 below with xi = g; since deficiency one. Let A be the n × (n − 1) matrix with φ(g)=0andtr α(g) = 2, the denominator in (2–2) lies ∂rj entries aij = ∂x . Fix a generator xi and consider the F × T α F ±1 i in , and hence K is in [t ]. matrix Ai obtained from A by deleting the ith row. Then 336 Experimental Mathematics, Vol. 21 (2012), No. 4

there exists an l ∈ Z such that for every even-dimensional equivalent to a triangulation. By standard results (see, representation α : π → GL(V ), one has e.g., [Turaev 01, Section 8]), we can now use X to calculate the torsion of XK . det (α ⊗ φ) Ai T α (t)=tl · ( 2–2) Consider the Euler structure e for X that is given by K det (α ⊗ φ) x − 1 i picking an arbitrary lift of the vertex of X to the universal whenever the denominator is nonzero. cover X and then taking the lift of each xi that starts at this base point. Reading out the words rj in x1 ,...,xn The same formula also holds, up to a sign, when starting at the base point gives a canonical lift for the 2- dim(V ) is odd. An easy way to ensure a nonzero denom- cells corresponding to the relators. With respect to this inator in (2–2) is to choose an xi such that φ(xi) =0; basing, the chain complex C∗(X) is isomorphic to the then det (α ⊗ φ) xi − 1 is essentially the characteris- chain complex −1 tic polynomial of α(xi ) and hence nonzero. 0 → Z[π]n−1 −∂→2 Z[π]n −∂→1 Z[π] → 0. Remark 2.6. Wada’s invariant (see [Wada 94]) is deﬁned The bases of C2 (X)andC1 (X) are abusively denoted by to be {rj } and {xi}, and the basis of C0 (X) is the lifted base det (α ⊗ φ) At i . point b.Thus det (α ⊗ φ) x − 1 i ∂r ∂ (r )= j x = a x and ∂ (x )=(x − 1)b. In [Friedl and Vidussi 10, p. 53], it is erroneously claimed 2 j ∂x i ij i 1 i i i i i that up to multiplication by a power of t, the tor- { } sion polynomial T α agrees with Wada’s invariant. Since Now ﬁx a basis vk for V . If we then view ele- K ∈ there seems to be some confusion in the literature re- ments v V as vertical vectors and β(g) as a matrix, Z · garding the precise relationship between twisted torsion the left [π]-module structure on V is given by g v = β ⊗ and Wada’s invariant, we discuss it in detail in Sec- β(g)v. Thus in the complex C∗ (X; V )=C∗(X) Z[π ] V , we have tion 2.8. In that section, we will also see that for representations into SL(2, F ), Wada’s invariant does in fact ∂2 (rj ⊗ vk )= (aij xi ⊗ vk )= (xi · aij ⊗ vk ) T α agree with K (t). In particular, the invariant studied in i i [Kim and Morifuji 10] agrees with T α (t). K = (xi ⊗ aij · vk )= (xi ⊗ β(aij )vk ) . i i Proposition 2.5 is an immediate consequence of the Thus with the basis-ordering conventions of Section 2.3, following. β the twisted chain complex C∗ (X; V ) is given by

− − Proposition 2.7. Let K, π, A be as above. For each gen- → n−1 −β−−(A→) n −(−−−−−−−−−−−−−β (x1 1),...,β(xn 1))→ → c 0 V V V 0, ( 2–4) erator xi, there is an s ∈ Spin (XK ) such that for every

Downloaded by [87.153.35.205] at 11:13 07 January 2013 even-dimensional representation β : π → GL(V ), one has where as usual, matrices act on the left of vertical vectors. − det β(A ) From now on, we assume that det (β(xi 1)) =0, s i τ(XK ,β, )= ( 2–3) since otherwise, there is nothing to prove. First, consider det (β(xi − 1)) the case that det β(Ai ) = 0. We claim in this case that β whenever the denominator is nonzero. C∗(X; V ) is not acyclic, and thus (2–3) holds by the definition of τ. Consider any v ∈ V n−1 that is in the kernel of 2 The homology orientation ω is suppressed in (2–3) be- Ai ; because β(xi − 1) is nonsingular, the fact that ∂ =0 cause by Lemma 2.1, it doesn’t affect τ, since dim(V )is β forces v to be in the kernel of A. Hence H2 (X; V ) =0, even. as needed. When instead, det β(A ) = 0, then both boundary Proof of Proposition 2.7. Let X be the canonical 2- i maps in (2–4) have full rank, and hence the complex is complex corresponding to the presentation of π, i.e., X acyclic. Following [Turaev 01, Section 2.2], we can use a has one cell of dimension zero, n cells of dimension one, suitable matrix τ-chain to compute the desired torsion. and n − 1 cells of dimension two. Since the Whitehead Specifically, [Turaev 01, Theorem 2.2] gives group of π is trivial [Waldhausen 78], it follows that X is simple-homotopy equivalent to every other CW de- det β(Ai ) τ(X, β, e)= . ( 2–5) composition of XK ; in particular, it is simple-homotopy det (β(xi − 1)) Dunfield et al.: Twisted Alexander Polynomials of Hyperbolic Knots 337

Here we are using that dim(V ) is even, which forces the where here At denotes the transpose of A,andsoAt is sign discussed in [Turaev 01, Remark 2.4] to be positive. an (n − 1) × n matrix over Z[π]. As in the proof of The- Also, the convention of [Turaev 01] is to record a basis as orem 2.7, in the generic case, [Turaev 01, Theorem 2.2] the rows of a matrix, whereas we use the columns; this is gives that irrelevant, since the determinant is transpose-invariant. t c det β(Ai ) Given (2–5), if we take s be the Spin -structure corre- τ V ⊗Z C∗(X) = . β [π ] det (β(x − 1)) sponding to e, we have established (2–3). i Up to the sign of the denominator, this is precisely the 2.8. Connection to Wada’s Invariant formula for Wada’s invariant given in [Wada 94]. It is important to note here that β(At ) is not neces- We now explain why the formula (2–2) differs from the sarily the same as β(A) t , and hence Wada’s invariant one used to define Wada’s invariant [Wada 94], and how may differ from our τ(X, β). However, note that there Wada’s invariant also arises as a torsion of a suitable exists a canonical isomorphism chain complex. To start, suppose we have a representa- → F op op tion β : π GL(d, ), where as usual, π is the funda- Vβ ⊗Z[π ] C∗(X) → C∗(X) ⊗Z[π ] Vβ mental group of a knot exterior. The representation β v ⊗ σ → σ ⊗ v makes V := F d into both a left and a right Z[π]-module. that moreover, respects the ordered bases. Thus these The left module β V is defined by g · v := β(g)v, where v ∈ V is viewed as a column vector, and the right mod- chain complexes have the same torsion invariant. It is op · ∈ easy to see that the left module (Vβ ) is isomorphic to ule Vβ is defined by v g := vβ(g), where now v V is ∗ ∗ → F viewed as a row vector. β V , where β : π GL(d, ) is the representation given ∗ −1 t Z op by β (g):= β(g) . Thus Wada’s invariant for β is our Given a left [π]-module W , we denote by W the ∗ right Z[π]-module given by w · f := f · w. Similarly, we torsion τ(X, β ). ⊗ can define a left module W op for a given right Z[π]- Our focus in this paper is on β of the form α φ, → F → Z module W . In Section 2, we started with the left modules where α : π SL(2, )andφ : π is the usual epimorphism. Note that α∗ is conjugate to α (see, e.g., C∗(X)andβ V and used the chain complex [Hillman et al. 10]), and hence (α ⊗ φ)∗ is conjugate to β d op ∗ α C∗ (X,F ):=C∗(X) ⊗Z[π ] β V α ⊗ (−φ). Since we argued in Section 2.6 that T is independent of the choice of φ, it follows that in this case, in defining the torsion. our T α is exactly Wada’s invariant for α. One could instead consider the chain complex ⊗ Vβ Z[π ] C∗(X). 3. TWISTED TORSION OF CYCLIC COVERS

Here, if {vk } is a basis for V and {c˜j } is a Z[π]-basis As usual, let K be a knot in a Q HS with exterior X for C∗(X), then we endow Vβ ⊗Z[π ] C∗(X) with the basis and fundamental group π. For an irreducible representa- Downloaded by [87.153.35.205] at 11:13 07 January 2013 {vk ⊗ c˜j } ordered reverse lexicographically, i.e., vk ⊗ c˜j < tion α : π → SL(2, C ), in this section we relate the tor-

⊗ T α C vk c˜j if either j

is the following (see [Dubois and Yamaguchi 09, Corol- where the last two equalities follow from (2–1) and the lary 27] for a related result). fact that ζ =1.

Theorem 3.1. Let K be a knot in a Q HS with exterior X Proof of Lemma 3.3. The idea is that for suitable and fundamental group π.Letα : π → SL(2, C ) be an ir- choices, one gets an isomorphism reducible representation. Then for every m ∈ N , we have α m γm ⊗α C∗ (Xm ; V ) → C∗ X; C [Zm ] ⊗C V T α K (ζ)=τ(Xm ,αm ). ∈µ as based chain complexes over C , and hence their torsions ζ m are the same. Fix a triangulation for X with an ordering cj of its Note here that since α is irreducible, the torsion poly- cells, as well as an Euler lift c → c˜ of the cells to the T α C ±1 T α j j nomial K is in [t ] by Theorem 2.4(a), and so K (ξ)is universal cover X.Letφm : π → Zm be the epimorphism well deﬁned for every ξ ∈ C ×. Our proof of Theorem 3.1 whose kernel is πm = π1 (Xm ), and ﬁx g ∈ π, where g = is inspired by an argument from [Turaev 86, Section 1.9]. φm (g) generates Zm . Combining Theorem 3.1 with a (generalization of) a re- Consider the triangulation of Xm that is pulled back sult of [Fried 86], we will prove the following. from that of X, and let cj be the cell in Xm that is the image ofc ˜ under X → X . Then each cell of X has a If τ(X ,α ) is nonzero for every m ∈ N , j m m Theorem 3.2. m m k · { − } T α ∈ C unique expression as g cj for k in 0,...,k 1 , where then the τ(Xm ,αm ) determine K (t) (t). k here g acts on Xm as a deck transformation. k k We order these cells so that g · c

Downloaded by [87.153.35.205] at 11:13 07 January 2013 = f c˜ ⊗ (h · v) , Proof of Theorem 3.1. Using Lemma 3.3 and the fact that γm and ∈µ λζ are conjugate representations of ∈ · C Z ζ m wherewehaveusedh πm to see that h 1=1in [ m ]. π,wehave Clearly, f is a chain map of complexes of C -vector spaces, and it is an isomorphism, since it sends the elements τ(X ,α )=τ(X, α ⊗ γ )= τ(X, α ⊗ λ , s). m m m ζ { k · ⊗ } { ⊗ ζ ∈µ of the basis (g c˜j ) v to those of the basis c˜k m −k −k −k −k (g ⊗ g · v )}.Nowchoose{vk, = g ⊗ g · v } as Note here that while the other terms do not depend on our basis for C [Z ] ⊗C V and order them by v

only issue is that we need to ensure the nonvanish- 4. TORSION POLYNOMIALS OF HYPERBOLIC ing of the denominators in (2–3) for both α ⊗ φ and KNOTS α ⊗ λ . Since α is irreducible, we can choose g ∈ [π,π] ξ Let K be a hyperbolic knot in an oriented Z -homology so that tr(α(g)) = 2 (see, e.g., [Culler and Shalen 83, 2 sphere Y . In this section, we define the hyperbolic torsion Lemma 1.5.1]). Notice then that polynomial TK associated to a certain preferred lift to SL(2, C ) of the holonomy representation of its hyperbolic ⊗ −1 − ⊗ −1 − −1 − α φ g 1 = α λξ g 1 = α g 1 , structure. and since tr(α(g)) = 2, we have det α(g−1 − 1) =0. 4.1. The Discrete and Faithful SL(2, C ) Hence if we take a suitable presentation of π where g is a Representations generator, then we can apply Proposition 2.7 with xi = g As usual, we write π = πK := π1 (XK ), and let μ ∈ π be to both α ⊗ φ and α ⊗ λξ and so prove the lemma. a meridian for K. The orientation of μ, or equivalently of K, will not matter in this section, but fix one so that φ : π → Z is determined. ∼ We turn now to the proof of Theorem 3.2, which From now on, assume that M = Y \ K = int(X)has says that typically the torsions τ(Xm ,αm ) collectively a complete hyperbolic structure. The manifold M inher- T α determine K (by Theorem 3.1, the hypothesis that its an orientation from Y , and so its universal cover M T α τ(Xm ,αm ) = 0 for all m is equivalent to no root of K can be identified with H 3 by an orientation-preserving C being a root of unity). A polynomial p in [t] of de- isometry. This identification is unique up to the action of d gree d is palindromic if p(t)=t p(1/t), or equivalently Isom+ (H 3 )=PSL(2, C ), and the action of π on M = H 3 ≤ ≤ if its coefficients satisfy ak = ad−k for 0 k d.Fora gives the holonomy representation α : π → PSL(2, C ), ∈ C ∈ N polynomial p [t]andm , we denote by rm (p)the which is unique up to conjugation. resultant of tm − 1andp, i.e., Remark 4.1. By Mostow–Prasad rigidity, the complete r (p)=Res(p, tm − 1) = (−1)mdRes(tm − 1,p) n hyperbolic structure on M is unique. Thus α is deter- =(−1)md p(ζ), mined, up to conjugacy, solely by the knot K and the ∈µ ζ m orientation of the ambient manifold Y . A subtle point is that there are actually two conjugacy classes of discrete where here d is the degree of p. The following theorem faithful representations πK → PSL(2, C ); the other one was proved in [Fried 86] for p ∈ R [t] and generalized in corresponds to reversing the orientation of Y (not K)or [Hillar 05] to the case of C [t]. equivalently complex-conjugating the entries of the image matrices. Theorem 3.5. Suppose p and q are palindromic polyno- Downloaded by [87.153.35.205] at 11:13 07 January 2013 mials in C [t].Ifrm (p)=rm (q) =0 for all m ∈ N , then To define a torsion polynomial, we want a rep- p = q. resentation into SL(2, C ) rather than PSL(2, C ). Thurston proved that α always lifts to a representation Theorem 3.2 now follows easily from Theorems 3.1 and α : π → SL(2, C ); see [Thurston 97] and [Shalen 02, T α 3.5 and the symmetry of K shown in Theorem 2.3. Section 1.6] for details. In fact, there are exactly two such lifts, the other being g → (−1)φ(g) α(g); the Remark 3.6. We just saw that under mild assumptions, point is that any other lift has the form g → (g)α(g) the torsions τ(Xm ,αm ) of cyclic covers determine the for some homomorphism : π →{±1}, i.e., some el- C T a 1 Z Z (t)-valued torsion polynomial K .Itwouldbevery ement of H (M ; 2 )= 2 .Nowα(μ) is parabolic, interesting if one could directly read off the degree and so tr α(μ) = ±2. Since Y is a Z2 HS, we know T α and the top coefficient of K from the τ(Xm ,αm ). See that φ(μ) is odd; hence there is a lift α such that [Hillar and Levine 07] for some of what is known about tr α(μ) = 2; arbitrarily, we focus on that lift and call recovering a palindromic polynomial p from the sequence it the distinguished representation. This representation rm (p); in particular, when p is monic and of even degree is determined, up to conjugacy, solely by K (sans d, Sturmfels and Zworski conjecture that one needs to orientation). We explain below the simple change that know rm (p)onlyform ≤ d/2 + 1 to recover p. results if we instead required the trace to be −2. 340 Experimental Mathematics, Vol. 21 (2012), No. 4

4.2. The Hyperbolic Torsion Polynomial lies in SL(2, F K ), where F K is the trace field; hence T ∈ F [t±1 ], proving the first part of (b). To see the For a hyperbolic knot K in an oriented Z2 HS, we define K K the hyperbolic torsion polynomial to be other part, first using [Maclachlan and Reid 03, Theo- rem 5.2.4], we can conjugate α so that α(π) ⊂ SL(2, OK ), T T α K (t):= K (t), where here OK is the ring of algebraic integers in some K where α : π → SL(2, C ) is the distinguished representa- number field (which might be a proper extension of F T α tion. Before proving Theorem 1.1, which summarizes K ). We now compute K by applying Proposition 2.5 to a presentation of π in which μ is our preferred genera- basic properties of TK (t), we give a few definitions. F tor. Since α(μ) is parabolic with trace 2, the denominator The trace field K of K is the field obtained by adjoining to Q the elements tr α(g) for all g ∈ π; this in (2–2) is Q is a finite extension of and an important number- p(t) := det (α ⊗ φ)(μ−1 − 1) =(tk − 1)2 , theoretic invariant of the hyperbolic structure on M;see [Maclachlan and Reid 03] for more. We say that K has where k = −φ(μ) = 0. Thus by (2–2), we know that integral traces if every tr α(g) is an algebraic integer ·Tα O ±1 ∈ Z ±1 p(t) K is in K [t ]. Then since p(t) [t ] is monic, (this is necessarily the case if M does not contain a closed T α the leading coefficient of K must be integral. An easy essential surface; see, e.g., [Maclachlan and Reid 03, The- inductive argument now shows that all the other coeffi- ∗ orem 5.2.2]). Also, we denote by K the result of switch- cients are also integral, proving part (b). ∗ ing the orientation of the ambient manifold Y ; we call K The proof of (c) uses Theorem 3.1, and we handle all the mirror image of K. We call K amphichiral if Y has mth roots of unity at once. In the notation of Section 3, an orientation-reversing self-homeomorphism that takes we have K to itself; equivalently, K = K∗ in the category of knots in oriented 3-manifolds. TK (ζ)=τ(Xm ,αm ). (4−1) ∈µ ζ m Theorem 4.2. Let K be a hyperbolic knot in an oriented By [Menal-Ferrer and Porti 10, Theorem 0.4], Z2 -homology 3-sphere. Then TK has the following properties: which builds on [Raghunathan 65], we have that α 2 H∗ m (Xm ; C ) = 0, or equivalently, τ(Xm ,αm )is T C ±1 (a) K is an unambiguous element of [t ] that satisfies nonzero. Thus by (4–1), we must have TK (ζ) =0 for T −1 T K (t )= K (t). It does not depend on an orienta- every mth root of unity, establishing part (c). tion of K. For (d), the distinguished representation for the mir- ∗ (b) The coefficients of TK lie in the trace field of K.IfK ror knot K is α : π → SL(2, C ), where each α(g)isthe has integral traces, the coefficients of TK are algebraic matrix that is the complex conjugate of α(g). Since our integers. choice of orientation for the meridian μ was arbitrary, we T T ∗ (c) TK (ξ) is nonzero for every root of unity ξ. can use the same φ in calculating both K and K .Thus Downloaded by [87.153.35.205] at 11:13 07 January 2013 ∗ we have (d) If K denotes the mirror image of K, then TK ∗ (t)= T K (t), where the coefficients of the latter polynomial φ(c1 (s)) φ(c1 (s)) TK ∗ (t)=t τ(X, α ⊗ φ)=t τ(Xk ,α⊗ φ) are the complex conjugates of those of TK . = TK (t), (e) If K is amphichiral, then TK is a real polynomial. proving (d). Next, claim (e) follows immediately (f) The values TK (1) and TK (−1) are mutation- invariant. from (d). Finally, claim (f) is a recent result of [Menal-Ferrer and Porti 11].

For the special case of 2-bridge knots and ξ = ±1, As in Section 3, we now consider the C -valued tor- assertion (c) is also a consequence of [Hirasawa and sions τ(X ,α ) of ﬁnite cyclic covers of X . Somewhat Murasugi 08] and [Silver and Williams 09]. m m K surprisingly, these determine TK : Proof. Since the distinguished representation α is irre- Z ducible, part (a) follows from Theorems 2.3 and 2.4(a). Theorem 4.3. Let K be a hyperbolic knot in a 2 HS with → C Next, since M has a cusp, by [Neumann and Reid 92, distinguished representation α : πK SL(2, ).Then T ∈ C Lemma 2.6], we can conjugate α so that its image K is determined by the torsions τ(Xm ,αm ) . Dunﬁeld et al.: Twisted Alexander Polynomials of Hyperbolic Knots 341

Proof. As discussed in the proof of Theorem 1.1(c), every our exploration of TK , and we devote this section to ex- τ(Xm ,αm ) in nonzero, so the result is immediate from amining them in detail. Theorem 3.2. The Conway knot C is the mirror of the knot 11n34 in the numbering of [Hoste and Thistlethwaite 99, Hoste et Remark 4.4. In choosing our distinguished representa- al. 98]. The program Snap [Goodman and Neumann 12, tion, we arbitrarily chose the lift α : π → SL(2, C ), where Coulson et al. 00] finds that the trace field F of the hy- tr(α(μ)) = 2. As discussed, the other lift β is given by perbolic structure on the exterior of C is the extension → − φ(g) ∈ g ( 1) α(g). Note that given g π,wehave of Q obtained by adjoining the root θ ≈ 0.1233737 − (β ⊗ φ)(g) (t)=β(g) · tφ(g) = α(g) · (−1)φ(g) · tφ(g) 0.5213097i of φ(g) = α(g) · (−t) = (α ⊗ φ)(g) (−t). p(x)=x11 − x10 +3x9 − 4x8 +5x7 − 8x6 +8x5 − 5x4 3 2 T β T α − +6x − 5x +2x − 1. It follows from Proposition 2.5 that K (t)= K ( t)= T − K ( t). Put differently, using β instead of α would sim- Snap also finds the explicit holonomy representation ply replace t by −t. πC → SL(2, F ), and one can directly apply Proposi- tion 2.5 to compute TC .Ifweset Remark 4.5. When Y is not a Z2 HS, the choice of lift α of 1 the holonomy representation can have a more dramatic η = 20θ10 +9θ9 +28θ8 +3θ7 + θ6 +19θ5 +10θ4 effect on T α . For example, consider the manifold m130 in 53 +47θ3 +6θ +1 , the notation of [Callahan et al. 99, Culler et al. 12]. This 2 10 manifold is a twice-punctured genus-1 surface bundle then η,θ,θ ,...,θ is an integral basis for OF ,and Z Z ⊕ Z over the circle, and since H1 (M; )= 8 , there are we obtain the result shown in Figure 1. four distinct lifts of the holonomy representation. Two of The Kinoshita–Terasaka knot is the mirror of 11n42. these lifts give T α = t2 + t−2 − 2i, and the other two K Its trace field is the same as for the Conway knot (since give √ [F : Q ] is odd, the trace field is also the invariant trace T α 2 −2 − − 1 −1 − K = t + t + 8 8i t + t 6i field, which is mutation-invariant), and one obtains the result shown in Figure 2. for the two distinct square roots of −8 − 8i. In particular, From the above, we see that T is not invariant under the fields generated by the coefficients are different; only K mutation. Since C and KT have genera 3 and 2 respec- the latter two give the whole trace field. tively and deg(TC ) = 10 and deg(TKT ) = 6, we see that Conjecture 1.4 holds for both knots. Also note that the 5. EXAMPLE: THE CONWAY AND KINOSHITA– coefficients of these polynomials are not real, certifying TERASAKA KNOTS the fact that both knots are chiral.

The Conway and Kinoshita–Terasaka knots are a famous Remark 5.1. It was shown in [Friedl and Kim 06, Sec- Downloaded by [87.153.35.205] at 11:13 07 January 2013 pair of mutant knots both of which have trivial Alexander tion 5] that twisted Alexander polynomials correspond- polynomial. Despite their close relationship, they have ing to representations over finite fields detect the genus of different genera. Thus they are a natural place to start all knots with at most twelve crossings. For example, for

T − 10 − 9 − 8 − 7 − 6 − 5 − 4 − 3 − 2 − 5 −5 C (t)= 79θ 35θ 111θ 11θ 4θ 71θ 38θ 187θ 2θ 24θ + 206η t +t 10 9 8 7 6 5 4 3 2 − 4 −4 + 257θ + 114θ + 361θ +36θ +13θ + 232θ + 124θ + 608θ +6θ +78θ 671η t + t − 10 − 9 − 8 − 7 − 6 − 5 − 4 − 3 − 2 − 3 −3 + 372θ 165θ 523θ 51θ 21θ 334θ 183θ 877θ 11θ 111θ + 972η t + t 10 9 8 7 6 5 4 3 2 − 2 −2 + 373θ + 162θ + 528θ +40θ +33θ + 312θ + 200θ + 866θ +24θ +99θ 968η t + t − 10 − 9 − 8 7 − 6 − 5 − 4 − 3 − 2 − 1 −1 + 303θ 115θ 445θ +14θ 75θ 152θ 227θ 649θ 73θ 29θ + 749η t + t 10 9 8 − 7 6 − 5 4 3 2 − − + 116θ +14θ + 200θ 88θ + 116θ 122θ + 204θ+ 146θ + 124 θ 78θ 220η ≈ 5 −5 − − 4 −4 − 3 −3 4.89524 + 0.09920i t +t + 15.68571 0.29761i t + t + 23.10363 0.07842i t + t + −26.94164 + 4.84509i t2 + t−2 + 38.38349 − 24.49426i t1 + t−1 + −43.32401 + 44.08061i .

FIGURE 1. The hyperbolic torsion polynomial of the Conway knot. 342 Experimental Mathematics, Vol. 21 (2012), No. 4

T − 10 − 9 − 8 − 7 − 6 − 5 − 4 − 3 − 2 − 3 −3 KT (t)= 55θ 24θ 78θ 6θ 5θ 45θ 29θ 128θ 5θ 15θ + 142η t + t 10 9 8 7 6 5 4 3 2 − 2 −2 + 293θ + 126θ + 416θ +28θ +29θ + 236θ + 160θ + 678θ +24θ +75θ 756η t +t − 10 − 9 − 8 − 7 − 6 − 5 − 4 − 3 − 2 − 1 −1 + 699θ 291θ 1001θ 42θ 95θ 512θ 419θ 1585θ 81θ 149θ + 1785η t + t 10 9 8 7 6 5 4 3 2 − + 790θ + 314θ + 1146θ +8 θ + 150θ + 494θ + 532θ + 1738θ + 136θ + 126θ 1986η ≈ − 3 −3 − 2 −2 − 1 −1 4.41793 0.37603i t +t + 22.94164 + 4.84509i t + t + 61.96443 24.09744i t + t + −82.69542 + 43.48539i .

FIGURE 2. The hyperbolic torsion polynomial of the Kinoshita-Terasaka knot.

11 2 the Conway knot, there is a representation α : π1 (XC ) → + −4.2204 − 60.5497i t − t F GL(4, 13) such that the corresponding torsion polyno- + 52.0953 + 134.5013i t10 − t3 mial T α ∈ F [t±1 ] has degree 14, and hence C 13 − − 9 − 4 + 147.7856 46.07448i t t 1 8 5 x(C) ≥ deg (T α )=3.5. + 897.2087 + 62.3265i t − t 4 C + −2465.8556 − 1308.0110i t7 − t6 , In particular, this shows that x(C) = 5, since x(C)= 2 genus(C) − 1 is an odd integer. and for the Kinoshita-Terasaka knot we obtained The calculation using the discrete and faithful SL(2, C ) representation is arguably more satisfactory, T adj ≈ − 7 − KT (t) 0.7378 + 12.4047i t 1 since it gives the equality − 6 − + 29.9408 56.5548i t t 1 + −655.7823 − 173.0400i t5 − t2 x(C)= deg(TC ) 2 + 2056.7509 + 1678.4875i t4 − t3 . on the nose, and not just after rounding up to odd integers. Interestingly, we have not found an example Since dim sl(2, C ) = 3, it follows from [Friedl and Kim for which this rounding trick applies to TK ;atleast 06, Theorem 1.1] that for knots with at most 15 crossings, one always has 1 x(K) = deg(T )/2 (see Section 6). x(K) ≥ deg T adj(t) (5−1) K 3 K

5.1. The Adjoint Representation and hence For an oriented hyperbolic knot K with distinguished 13 7 x(C) ≥ and x(KT ) ≥ . (5−2) representation α : π1 (XK ) → SL(2, C ), we now consider 3 3 the adjoint representation

Downloaded by [87.153.35.205] at 11:13 07 January 2013 ≥ Thus using that x(K) is an integer, we get x(C) 5and ≥ αadj : π1 (XK ) → Aut sl(2, C ) x(KT ) 3, which are sharp. Intriguingly, in contrast to − T g → A → α(g)Aα(g) 1 the case of K , one does not have equality in (5–1) and (5–2) for these two knots. Below, in Section 6.6, we de- associated to α. It is well known that this representa- T adj scribe some knots for which K fails to give a sharp tion is also faithful and irreducible. Using sign-refined bound on x(K) even after one takes into account that torsion and the orientation on K, one gets an invari- x(K) is an odd integer. T adj ∈ C ±1 ant K [t ] that is well defined up to multiplication by an element of the form tk . We refer to [Dubois and Yamaguchi 09] for details on this construc- 6. KNOTSWITHATMOST15CROSSINGS T adj tion and for further information on K ; one thing that − There are 313 231 prime knots with 15 or fewer cross- is shown there is that T adj(t)=−T adj(t 1 )uptoapower K K ings [Hoste et al. 98], of which all but 22 are hyperbolic. of t,andsoT adj has odd degree. K For each of these hyperbolic knots, we computed a high- For the Conway knot we calculate that T precision numerical approximation to K (see Section 6.7 T adj(t) ≈ −0.2788 + 16.4072i t13 − 1 C for details), and this section is devoted to describing the + −3.9858 − 20.1706i t12 − t various properties and patterns we found. Dunfield et al.: Twisted Alexander Polynomials of Hyperbolic Knots 343

6.1. Genus 6.3. Chirality

The genus bound from TK given in Theorem 1.3 is sharp For hyperbolic knots with at most 15 crossings, we found for all 313 209 hyperbolic knots with 15 or fewer cross- that a knot is amphichiral if and only if TK has real coef- ings; that is, x(K) = deg(TK )/2 for all these knots. In ficients. In particular, there are 353 such knots with TK contrast, the ordinary Alexander polynomial fails to de- real, and SnapPy [Culler et al. 12] easily confirms that tect the genus for 8 834 of these knots, which is 2.8% of they are all amphichiral. (This matches the count of am- the total. phichiral knots from [Hoste et al. 98, Table A1].) We showed that the genus bound from TK is sharp In contrast, the numbers TK (1) and TK (−1) do not using the following techniques to give upper bounds on always detect chirality. For example, the chiral knot the genus. First, for the alternating knots (36% of the 10153 = 10n10 has TK (1) = 4, and 10157 = 10n42 has total), the genus is simply determined by the Alexan- TK (−1) = 576. Moreover, the knot 14a506 has both der polynomial [Murasugi 58, Crowell 59]. For the non- TK (1) and TK (−1) real. (This last claim was checked to alternating knots, we first did 0-surgery on the knot K the higher precision of 10 000 decimal places.) to get a closed 3-manifold N; by [Gabai 87], the genus

of K is the same as that of the simplest homologically 6.4. Knots with the Same TK nontrivial surface in N. We then applied the method of While we saw in Section 5 that TK is not mutation- [Dunfield and Ramakrishnan 10, Section 6.7] to a trian- invariant, there are still pairs of knots with the same gulation of N to quickly find a homologically nontrivial TK . In particular, among knots with at most 15 cross- surface. Since this surface need not be of minimal genus, ings, there are 2 739 groups of more than one knot that when necessary we randomized the triangulation of N share the same TK , namely 2 700 pairs and 39 triples. until we found a surface whose genus matched the lower Here we do not distinguish between a knot and its mir- bound from TK . ror image, and having the same TK means that the coefficients agree to 5 000 decimal places. Stoimenow found that there are 34 349 groups of mutant knots among those 6.2. Fibering with at most 15 crossings, involving some 77 680 distinct We also found that T gives a sharp obstruction to fiber- K knots [Stoimenow 12]. Thus there are many examples in ing for all hyperbolic knots with at most 15 crossings. which mutation changes TK . However, all of the examples In particular, the 118 252 hyperbolic knots for which we found of knots with the same TK are in fact mutants. T is monic are all fibered. In contrast, while the or- K As mentioned, it is shown in [Menal-Ferrer and dinary Alexander polynomial always certifies nonfiber- Porti 11] that the evaluations TK (1) and TK (−1) are ing for alternating knots [Murasugi 63, Gabai 86], among mutation-invariant. We found 38 pairs of nonmutant the 201 702 nonalternating knots there are 7 972, or knots with the same TK (1) and the same TK (−1). 4.0%, whose Alexander polynomials are monic but do Suggestively, several of these pairs (including the five not fiber. Downloaded by [87.153.35.205] at 11:13 07 January 2013 pairs shown in [Dunfield et al. 10, Figure 3.9]; see also To confirm fibering when T is monic, we K [Stoimenow and Tanaka 09, Tables 2 and 3]) are known used a slight generalization of the method of to be genus-2 mutants. We also found a triple of mutu- [Dunfield and Ramakrishnan 10, Section 6.11]. Again by ally nonmutant knots {10a121, 12a1202, 12n706} where [Gabai 87], it is equivalent to show that the 0-surgery TK (+1) = −4, and a similar sextet N is fibered. Starting with the minimal-genus surface S found as above, we split N open along S, and tried to sim- {10n10, 12n881, 13n592, 13n2126, 15n9378, 15n22014} plify a presentation for the fundamental group of N \ S T until it was obviously that of a surface group. If it was, where K (+1) = 4; however, within these groups, the T − then it followed that N \ S = S × I and N was fibered. value K ( 1) did not agree. The difference with [Dunfield and Ramakrishnan 10] is that we allowed S to be a general normal surface instead 6.5. Other Patterns of the restricted class of [Dunfield and Ramakrishnan 10, We found two other intriguing patterns that we are un- Figure 6.13]. We handled this by splitting the manifold able to explain. The first is that the second-highest coef- open along S and triangulating the result using Regina ficient of TK is often real for fibered knots. In particular, [Burton 09]. this is the case for 53.1% (62 763 of 118 252) of the fibered 344 Experimental Mathematics, Vol. 21 (2012), No. 4

knots in this sample. In contrast, the second coefficient 6.7. Computational Details is real for only 0.2% (364 of 194 957) of nonfibered knots. The complete software used for these computations, as (Arguably, the right comparison is with the leading co- well as a table of TK for all these knots, is available T efficient of K for nonfibered knots; even fewer (0.05%) at http://dunfield.info/torsion. The software runs within of these are real.) For fibered knots, the twisted homol- Sage,1 and makes use of SnapPy [Culler et al. 12] and ogy of the universal cyclic cover can be identified with t3m [Culler and Dunfield 10]. It finds very high precision that of the fiber; hence the action of a generator of the solutions to the gluing equations, in the manner of Snap deck group on this homology of the cover can be thought [Goodman and Neumann 12, Coulson et al. 00], and ex- of as the action of the monodromy of the bundle on the tracts from this a high-precision approximation to the twisted homology of the fiber. The second coefficient of distinguished representation. Except as noted above, we T K is then just the sum of the eigenvalues of this mon- did all computations with 250 decimal places of preci- odromy, but it is unclear why this should often be a real sion. Even at this accuracy, TK is fast to compute for number. these knots, taking only a couple of seconds each on a |T − | |T | The second observation is that K ( 1) > K (1) for late-2010 high-end desktop computer. However, to save all but 22 (less than 0.01%) of these knots. The excep- space, only 40 digits were saved in the final table. tions are nonalternating, and all but one (15n151121) is To guard against error, two of the authors indepen- fibered. dently wrote programs that computed TK , and the out- puts of these programs were then compared for all non- 6.6. Adjoint Polynomial alternating knots with 14 crossings. T adj As discussed in Section 5.1, a torsion polynomial K constructed by composing the holonomy representation 7. TWISTED TORSION AND THE CHARACTER with the adjoint representation of PSL(2, C ) on its Lie VARIETY OF A KNOT algebra was studied in [Dubois and Yamaguchi 09]. We Z also numerically calculated this invariant for all knots As usual, consider a hyperbolic knot K in a 2 HS, with at most 15 crossings. In contrast to what we found and let π := π1 (XK ). So far, we have focused on the for TK , there was not always an equality in the bound of torsion polynomial of the distinguished representation (5−1) and (5−2). In fact, some 8 252 of these knots had α : π → SL(2, C ) coming from the hyperbolic structure. x(K) > 1 deg(T adj). All such knots were nonalternating, However, this representation is always part of a complex 3 K → C and were among the 8 834 knots for which ΔK fails to give curve of representations π SL(2, ), and it is natural a sharp bound on x(K). However, using the trick from to ask whether there is additional topological information Section 5.1 that x(K) is an odd integer, we showed that in the torsion polynomials of these other representations. T adj In this section, we describe how to understand all of these the bound on x(K)from K is effectively sharp in all T adj torsion polynomials at once, and use this to help explain but 12 cases. The 12 knots for which K fails to deter-

Downloaded by [87.153.35.205] at 11:13 07 January 2013 mine the genus are as follows: there are seven knots where some of the patterns observed in Section 6. For the spe- T adj cial case of 2-bridge knots, it had previously been stud- x(K) = 7 (i.e., genus 4) but deg( K ) = 15, namely ied in [Morifuji 08, Kim and Morifuji 10] how the torsion {15n75595, 15n75615, 15n75858, 15n75883, 15n75948, polynomial varies with the representation, and we extend 15n99458, 15n112466}, here some of those results to more general knots. To state our results, we must first review some and five knots with x(K) = 9 (i.e., genus 5) but basics about character varieties; throughout, see the T adj deg( K ) = 21, namely classic paper [Culler and Shalen 83] or the survey [Shalen 02] for details. Consider the representation {15n59545, 15n62671, 15n68947, 15n109077, 15n85615}. variety R(K):=Hom π,SL(2, C ) , which is an affine algebraic variety over C . The group SL(2, C ) acts on T adj In these 12 cases, we computed K to the higher R(K) by conjugating each representation; the algebro- accuracy of 10 000 decimal places. geometric quotient X(K):=R(K)// SL(2, C ) is called T adj Intriguingly, the polynomial K did better at pro- the character variety. More concretely, X(K) is the set viding an obstruction to fibering; just as for TK ,itwas monic only for those knots in the sample that are actually fibered. 1 Available at http://www.sagemath.org. Dunfield et al.: Twisted Alexander Polynomials of Hyperbolic Knots 345

∈ T X 0 of characters of representations α R(K), i.e., functions We give several explicit examples of K later, in Sec- χα : π → C of the form χα (g)=tr(α(g)) for g ∈ π. tion 8. The following result is immediate from Theo- When α is irreducible, the preimage of χα under the rem 1.5. projection R(K) → X(K) is just all conjugates of α, but distinct conjugacy classes of reducible representations Corollary 7.3. Let X0 be an irreducible component of can sometimes have the same character. Still, it makes X(K)irr.Then χ sense to call a character irreducible or reducible depend- ∈ T ≤ T X 0 (a) For all χ X0 , we have deg ( K ) deg K with ing on which kind of representations it comes from. equality on a nonempty Zariski-open subset. The character variety X(K) is also an affine algebraic χ (b) If T X 0 is monic, then T is monic for all χ ∈ X . variety over C ; its coordinate ring C [X(K)], which con- K K 0 Otherwise, T χ is monic only on a proper Zariski- sists of all regular functions on X(K), is simply the sub- K closed subset. ring C [R(K)]SL(2,C ) of regular functions on R(K) that are invariant under conjugation. We start by showing T χ In particular, when X is a curve, the genus bound that it makes sense to define a torsion polynomial K 0 ∈ T χ T α and fibering obstruction given by T χ are the same for all for χ X(K)via K := K for every α with χα = χ. K ∈ T χ χ X0 except on a finite set where K provides weaker Lemma 7.1. If α, β ∈ R(K) have the same character, then information. We can also repackage Corollary 7.3 as a T α T β K = K . uniform statement on all of X(K).

Proof. As discussed, if α is irreducible, then β must be Corollary 7.4. Let K be a knot in an integral homology conjugate to α; hence they have the same torsion poly- 3-sphere. Then nomial. If instead α is reducible, then Theorem 2.4(b) { ∈ | T χ } (a) The set χ X(K) deg( K )=2x(K) is Zariski T α shows that K depends only on the diagonal part of α, open. which can be recovered from its character. Since β must { ∈ |Tχ } (b) The set χ X(K) K is monic is Zariski also be reducible and has the same character as α,we closed. T α T β again get K = K .

Proof. It suﬃces to consider the intersections of these

An irreducible component X0 of X(K) has dimC (X0 ) sets with each irreducible component X0 of X(M). If X0 ≥ 1, since the exterior of K has boundary a torus. There consists solely of reducible characters, the result is im- are two possibilities for X0 : either it consists entirely of mediate from Theorem 2.4(c). Otherwise, it follows from T X 0 ≤ reducible characters, or it contains an irreducible char- Corollary 7.3 combined with the fact that deg( K ) acter. In the latter case, it turns out that irreducible 2x(K). characters are Zariski open in X0 , and every character Downloaded by [87.153.35.205] at 11:13 07 January 2013 We now turn to the proof of Theorem 1.5. in X0 is that of a representation with nonabelian image. Since the torsion polynomials of reducible represen- Proof of Theorem 1.5. By [Culler and Shalen 83, Propo- tations are boring (see Theorem 2.4 and the discussion sition 1.4.4], there is an irreducible component R0 of immediately following), we focus on those components R(K) such that the projection R0 → X(K) surjects onto containing an irreducible character. We denote the union X0 . Consider the tautological representation of all such components by X(K)irr; equivalently, X(K)irr ρ : π → SL 2, C [R ] , is the Zariski closure of the set of irreducible characters. taut 0 T χ ∈ It is natural to ask how K varies as a function of χ. which sends g π to the matrix ρtaut(g) of regular func- We have obtained the following result. tions on R0 so that

ρtaut(g)(α)=α(g) for all α ∈ R0 . Theorem 7.2. Let X0 be an irreducible component of ± irr T X 0 ∈ C 1 C X(K) .Thereisaunique K [X0 ][t ] such that Since R0 is irreducible, [R0 ] is an integral domain. Thus χ ∈ T T X 0 for all χ X0 , one has K (t)= K (χ)(t). Moreover, we can consider its ﬁeld of fractions, i.e., the ﬁeld of ra- T X 0 C C K is itself the torsion polynomial of a certain represen- tional functions (R0 ). Working over (R0 ), there is an → F T taut C ±1 tation π SL(2, ) and thus has all the usual properties associated torsion polynomial K that is in (R0 )[t ], (symmetry, genus bound, etc.). since ρtaut is irreducible. From Lemma 2.5, it is clear that 346 Experimental Mathematics, Vol. 21 (2012), No. 4

∈ T α T taut C ±1 Z for every α R0 ,wehave K (t)= K (α)(t)in [t ]. Proposition 7.7. Let K be a knot in a HS whose distin- T taut Hence the coefficients of K have well-defined values guished representation is sufficiently generic and whose at every point α ∈ R0 , and so lie in C [R0 ]. Now, since distinguished component of X(M) contains a reducible the torsion polynomial is invariant under conjugation, character. Then deg(TK ) ≥ 2 deg(ΔK ) − 2,andifΔK is T taut C C SL(2,C ) T each coefficient of K lies in [X0 ]= [R0 ] ,and nonmonic, then so is K . T taut C ±1 hence K descends to an element of [X0 ][t ], which T X 0 is the K we seek. 7.2. 2-Bridge Knots For 2-bridge knots in S3 , the torsion polyno- 7.1. The Distinguished Component mial as a function on X(M)irr was studied in

It is natural to focus on the component X0 of X(M) that [Kim and Morifuji 10]. Since 2-bridge knots are alter- contains the distinguished representation. In this case, nating, the ordinary Alexander polynomial ΔK deter- X0 is an algebraic curve, and we refer to it as the distin- mines the genus and whether K fibers [Murasugi 63, guished component. By Corollary 7.3, the following con- Crowell 59, Murasugi 63, Gabai 86]. However, as men- T X 0 jecture that K detects both the genus and fibering of tioned, there seems to be no a priori reason that the same T X 0 K is implied by Conjecture 1.4. must be true for K . We now sketch what is known about this special case, starting with two results from Conjecture 7.5. Let K be a hyperbolic knot in S3 ,and [Kim and Morifuji 10]. X0 the distinguished component of its character variety. T X 0 T X 0 Then 2x(K) = deg( K ),and K is monic if and only Theorem 7.8. [Kim and Morifuji 10, Theorem 4.2] Let K if K is fibered. be a hyperbolic 2-bridge knot. Then there exists a com- ⊂ irr T X 0 ponent X0 X(K) such that 2x(K) = deg( K ) and T X 0 As we explain in Section 7.2, this conjecture is true for K is monic if and only if K is fibered. In particular, if many 2-bridge knots. X(K)irr is irreducible, then Conjecture 1.7 holds for K. One pattern in Section 6 is that TK never gave worse topological information than the ordinary Alexander Proof. Since ΔK detects the genus, it is nonconstant and Z polynomial ΔK . In certain circumstances, Corollary 7.3 so has a nontrivial root. For every knot in a HS, a → allows us to relate ΔK to TK , as we now discuss. First, root of ΔK gives rise to a reducible representation πK T X 0 C we can sometimes show that K must contain at least SL(2, ) with nonabelian image. In the case of 2-bridge as much topological information as ΔK . knots, the character of any such representation belongs irr to a component X0 ⊂ X(M) (see [Hilden et al. 95, Re-

Lemma 7.6. Let K be a knot in a ZHS.SupposeX0 is a mark 1.9 and Corollary 2.9], originating in [Burde 90, component of X(K)irr that contains a reducible charac- Proposition 2.3] and the comment following it), and T X 0 ≥ − ter. Then deg( K ) 2 deg(ΔK ) 2,andifΔK is non- Lemma 7.6 now finishes the proof. T X 0 Downloaded by [87.153.35.205] at 11:13 07 January 2013 monic, then so is K . Theorem 7.9. [Kim and Morifuji 10, Lemma 4.8 and The- Proof. Let α be a reducible representation whose charac- orem 4.9] Let K = K(p, q) be a hyperbolic 2-bridge knot, ter lies in X0 . By Theorem 2.4(c), the torsion polynomial and let c be the leading coefficient of ΔK . Suppose there T α − ± K has degree 2 deg(ΔK ) 2, and its leading coefficient exists a prime divisor of p such that if c = 1, then ±1 {− } is the square of that of ΔK ∈ Z[t ]. The result now fol- the reduction of c modulo is not in 1, 0, 1 .Then lows from Corollary 7.3. Conjecture 1.7 holds for K.

irr Now consider the distinguished representation α and Proof sketch. Let X0 be any component of X(K) . distinguished component X0 ⊂ X(K). We say that α First, one shows that X0 contains a character χ such T T X 0 T is sufficiently generic if deg( K ) = deg( K )and K is that tr(μK ) = 0. In [Kim and Morifuji 10, Lemma 4.6], T X 0 monic only if K is. Corollary 7.3 suggests that most this is shown using the particular structure of πK ,andit knots will have sufficiently generic distinguished repre- also follows from the following more general fact. sentations; however, because the distinguished character takes on only values that are algebraic numbers, there Claim 7.10. Let K beaknotinS3 whose exterior con- seems to be no a priori reason why this must always be tains no closed essential surface. If X0 is a component the case. Regardless, our intuition is that the hypothesis of X(M), then given a ∈ C , there is χ ∈ X0 such that of this next proposition holds quite often. tr(μK )=a. Dunfield et al.: Twisted Alexander Polynomials of Hyperbolic Knots 347

Two-bridge knots satisfy the hypothesis of Claim 7.10 manifolds of minimal volume. The manifold is m003 in by [Hatcher and Thurston 85], and the proof of the claim the SnapPea census [Callahan et al. 99, Culler et al. 12], is straightforward from the Culler–Shalen theory of sur- and is the once-punctured torus bundle over the circle −21 Z faces associated to ideal points of X0 . Specifically, on with monodromy 1 −1 . Its homology is H1 (M; )= the smooth projective model of X0 , the rational function Z ⊕ Z/5Z, and it is, for instance, the complement of a tr(μK ) takes on the value a somewhere, and if this were null-homologous knot in L(5, 1). After randomizing the at an ideal point, the associated essential surface would triangulation a bit, SnapPy gives the following presenta- have either to be closed or have meridian boundary; the tion: latter situation also implies the existence of a closed es- 3 −2 π := π1 (M)= a, b bab aba =1 . sential surface by [Culler et al. 87, Theorem 2.0.3]. The representation corresponding to a χ such that We will view X(π) as a subvariety of X a, b , where { } ∼ C 3 tr(μK ) = 0 is irreducible but has metabelian image, and a, b is the free group on a, b .Now X a, b = , T χ where the coordinates are (x, y, z)= tr(a), tr(b), tr(ab) ; in the 2-bridge case one can use this to calculate K explicitly. In particular, in [Kim and Morifuji 10], the this is because the trace of every word w ∈a, b can authors find that, provided there exists a prime as be expressed in terms of these coordinates using the T χ fundamental relation tr(UV)=tr(U)tr(V ) − tr(UV −1 ) in the hypothesis, the polynomial K is nonmonic and deg T χ =2x(K). We then apply Corollary 7.3 to see that for U, V ∈ SL(2, C ). Since π is defined by the single K − Conjecture 1.7 holds. relator R = bab3 aba 2 , the character variety X(π)is cut out by the polynomials corresponding to tr(R)=2, Another interesting class of characters in X(M)irr tr [a, R] =2,andtr [b, R] = 2. Using Gröbner bases comprises those of representations for which μK is in SAGE to decompose X(π) into irreducible compo- parabolic (e.g., the distinguished representation); such nents over Q , we found a unique component X0 that parabolic representations must occur on every compo- contains an irreducible character, i.e., contains a point nent X0 by Claim 7.10. For the 3 830 nonfibered 2-bridge such that tr [a, b] = 2. Explicitly, the ideal of X0 is ≤ T χ knots with q

Downloaded by [87.153.35.205] at 11:13 07 January 2013 ∼ terized by v ∈ C \{0}, and hence C (R ) = C (v), and we 8. CHARACTER VARIETY EXAMPLES 0 have an explicit π → GL (2, C (v)), which is the restric- As with many things related to the character variety, tion of the tautological representation. Using Lemma 2.5, T X 0 while K is a very natural concept, actually computing we find that the torsion polynomial of this representation it can be difficult. Here, we content ourselves with finding is T X 0 4 2 K for three of the simplest examples. In each case, 2 v + v +1 t − + t−1 . there is only one natural choice for X0 , and moreover, it v3 + v √ is isomorphic to C \{finite set}.ThusX0 is rational, and Substituting in v = ± 1 − −4u2 +1 2u to eliminate C (X0 ) is just rational functions in one variable, which v gives the final answer makes it easy to express the answer. We do one fibered 2 example and two that are nonfibered; in all cases, the 2 u − 1 − T X 0 (t)=t + + t 1 . simplest Seifert surface has genus 1. K u

8.1. Example: m003 8.2. Example: m006 We start with the sibling M of the ﬁgure-8 complement, The census manifold M = m006 can also be described which is one of the two orientable cusped hyperbolic 3- as 5/2 surgery on one component of the Whitehead link 348 Experimental Mathematics, Vol. 21 (2012), No. 4

L. (Here our conventions are such that +1 surgery on where either component of L gives the trefoil knot, whereas −1 surgery gives the figure-8 knot.) Thus M is, for in- u2 +4 u2 +4 8u f(u)= , , , stance, the complement of a null-homologous knot in 4u u2 − 4 u2 − 4 the lens space L(5, 2), and again H1 (M; Z)=Z ⊕ Z/5Z. Using spun-normal surfaces, it is easy to check via and then calculate [Culler and Dunfield 10] that there is a Seifert surface in M that has genus 1 with one boundary component. 4 (u +2) − SnapPy gives the presentation T X 0 1 K (t)= t + t 16u2 π := π (M)= a, b b2 abab2 a−2 =1 . (u +2) u4 +4u3 − 8u2 +16u +16 1 + . − 2 −1 8(u 2)u Changing the generators to anew = a and bnew = ab rewrites this as π = a, b a3 bab3 ab =1 . 8.4. The Role of Ideal Points Using the same setup as in the previous example, we find A key part of the Culler–Shalen theory is the associa- a single component X0 containing an irreducible charac- 2 tion of an essential surface in the manifold M to each ter. The ideal of X0 is x − y,y z − z − 1 , and hence we ideal point of a curve X0 ⊂ X(M). The details can be can bijectively parameterize X0 by found, for example, in [Shalen 02], but in brief, consider 2 −1 f : C \{1, −1}→X , where f(u)= u, u, (u − 1) . the smooth projective model X0 with its rational map 0 − X0 → X0 .NowX0 is a smooth Riemann surface, and (8 1) the finitely many points where X0 → X0 is undefined Considering the curve of representations given by are called the ideal points of X0 . To each such point x,

there is an associated nontrivial action of π := π1 (M) v 1 v−1 0 ρ(a)= and ρ(b)= , on a simplicial tree Tx . One then constructs a surface −1 3−2u 2 0 v u 2 −1 v S in M dual to this action, which can be taken to be − essential (i.e., incompressible, boundary incompressible, where v + v 1 = u, and again directly applying and not boundary parallel). Since minimal complexity Lemma 2.5 and eliminating v gives Seifert surfaces often arise from an ideal point of some 2 3 2u − 1 − 2u T X 0 1 X0 , a very natural idea is thus to try to use such an K (t)= t + t + . 2 − 2 − T X 0 u 1 u 1 ideal point x to say something about K .Moreover, irr provided that X0 ⊂ X(M) , a surface associated to an 8.3. m037 ideal point is never a fiber or semifiber, which suggests Downloaded by [87.153.35.205] at 11:13 07 January 2013 T X 0 that one might hope to prove nonmonotonicity of K by The census manifold M = m037 has H1 (M; Z)=Z ⊕ examining T X 0 (x). Thus, we now compute what happens Z/8Z, and so is not a knot in a Z/2-homology sphere. K to T X 0 at such ideal points in our two nonfibered exam- However, this makes no difference in this character va- K ples m006 and m037. (Aside: It is known that even for riety context. Again, using spun-normal surfaces, one knots in S3 , not all boundary slopes need arise from ideal easily checks that there is a Seifert surface in M that points [Chesebro and Tillmann 07], so it is probably too has genus 1 with one boundary component. Now π = − much to expect that there is always an ideal point that a, b a3 ba2 ba3 b 2 =1 , and this time, there are two gives a Seifert surface.) components of X(π) containing irreducible characters. However, one of these consists entirely of metabelian representations that factor through the epimorphism π → 8.5. Ideal Points of m006 Z/2 ∗ Z/2= c, d c2 = d2 =1 given by a → c and − b → d. Focusing on the other component X , it turns out If we view the parameterization (8 1) above as a rational 0 1 C → that the ideal is xz − 2y,4y2 − z2 − 4 , and so we can map from P ( ) X0 , we have ideal points corresponding to u ∈{−1, 1, ∞}. To calculate the boundary slopes parameterize X0 by of the surfaces associated to each of these, we consider f : C \{−2, 0, 2}→X0 , the trace functions of SnapPy’s preferred basis μ, λ for Dunfield et al.: Twisted Alexander Polynomials of Hyperbolic Knots 349

π1 (∂M). In our presentation for π, we calculate and 2 2 T X 0 − tr(μ)=tr(a bab)=x z − z − 1 K (u = 2)(t)=0. u u4 − u2 − 1 = − , (u − 1)2 (u +1)2 8.7. General Picture for Ideal Points tr(λ)=tr(a3 ba)=−x3 − xz +2x Based on the preceding examples and a heuristic calcula- u u4 − 3 u2 +3 tion for tunnel-number-one manifolds, we conjecture the − = − , following: (u 1)(u +1) u2 − 2 u4 − u2 +1 4 − 2 − tr(μλ)=x 2x z +1= − . Conjecture 8.1. Let K be a knot in a rational homology (u 1)(u +1) irr 3-sphere, and X0 a component of X(K) .Supposex is Now consider an ideal point x with associated surface S, an ideal point of X0 that gives a Seifert surface (hence T X 0 and pick a simple closed curve γ on π1 (∂M). Then the K is nonfibered). Then the leading coefficient of K has number of times γ intersects ∂S is twice the order of the a finite value at x. pole of tr(γ)atx (here if tr(γ) has a zero of order m at x, this counts as a pole of order 0, not one of order −m). Unfortunately, Conjecture 8.1 does not seem particu- The above formulas thus show that the points u =1and larly promising as an attack on Conjecture 1.7 for dis- u = −1 give surfaces with boundary slope μλ2 , whereas tinguishing fibered versus nonfibered cases. Moreover, in u = ∞ gives one with boundary slope μ3 λ−1 . The latter terms of looking at such Seifert ideal points to show that T X 0 − is the homological longitude, and since there is only one K determines the genus, the second ideal point u = 2 T X 0 spun-normal surface with that boundary slope and each in Section 8.6 where K vanishes is not a promising sign. choice of spinning direction, it follows that the surface However, in trying to use an ideal point x of X0 that ∞ T X 0 associated to ξ = must be the minimal-genus Seifert gives a Seifert surface to understand K ,itmaybe T X 0 surface. Thus we are interested in wrong to focus on just the value of K at x. After − all, there is no representation of π corresponding to x. T X 0 ∞ 1 0 K (u = )(t)=2 t + t + (simple pole)t . Rather, as in the construction of the surface associated to x, perhaps one should view x as giving a valuation on C (R0 ), where R0 is a component of R(M) surjecting 8.6. Ideal Points of m037 onto X0 . This time, we have four ideal points corresponding to If we unwind the definition of the associated surface u = −2, 2, 0, ∞. We obtain and its properties, we are left with the following abstract situation. There is a field F with an additive valuation u8 − 48u6 +96u4 − 768u2 + 256 tr(μ)=tr(a2 ba3 )= , v : F × → Z with a representation ρ : π → SL(2, F ) such 64(u − 2)(u +2)u3 ∈ ≥| |

Downloaded by [87.153.35.205] at 11:13 07 January 2013 that for each γ π,wehavev (tr(γ)) φ(γ) , where φ : tr(λ)=tr(a−1 ba3 b) π → Z is the usual free abelianization homomorphism. 1 = − This alone is not enough, because even in the fibered − 2 2 4 256 (u 2) (u +2) u case, one always has such a setup by looking at an ideal × 12 − 10 8 − 6 4 u 72u + 1264u 12032u + 20224u point of a component of X(M) consisting of reducible − 18432u2 + 4096 , representations. Thus it seems that the key to such an ap- u8 − 16u6 + 352u4 − 256u2 + 256 proach must be to exploit the fact that since X0 contains tr(μλ)=− . ∈ 4(u − 2)3 (u +2)3 u an irreducible character, there is a γ π with φ(γ)=0, yet v (tr(γ)) is arbitrarily large. Hence {2, −2} give surfaces with boundary slope μ2 λ−1 , and {0, ∞} give surfaces with boundary slope μ4 λ3 .In fact, the homological longitude is μ2 λ−1 , and again using ACKNOWLEDGMENTS spun-normal surfaces, one easily checks that surfaces as- We thank Jérôme Dubois, Taehee Kim, Takahiro Kitayama, sociated to {2, −2} are the minimal-genus Seifert surface. Vladimir Markovic, Jinsung Park, Joan Porti, Dan Silver, Alexander Stoimenow, Susan Williams, and Alexandru Za- Thus, we care about harescu for interesting conversations and helpful sugges- tions. We are particularly grateful to Joan Porti for sharing − T X 0 1 0 K (u = 2)(t)=4 t + t + (simple pole)t his expertise and early drafts of [Menal-Ferrer and Porti 10, 350 Experimental Mathematics, Vol. 21 (2012), No. 4

Menal-Ferrer and Porti 11] with us. We also thank the referee [Culler et al. 12] M. Culler, N. M. Dunﬁeld, and J. R. Weeks. for helpful comments. Dunﬁeld was partially supported by US “SnapPy: A Computer Program for Studying the Geom- NSF grants DMS-0707136 and DMS-1105476. etry and Topology of 3-Manifolds.” Available at http: //snappy.computop.org/, 2012.

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Downloaded by [87.153.35.205] at 11:13 07 January 2013 Nathan M. Dunﬁeld, Department of Mathematics, University of Illinois, 1409 W. Green Street, Urbana, IL 61801, USA (nathan@dunﬁeld.info)

Stefan Friedl, Mathematisches Institut, Raum 020, Weyertal 86-90, Universität zu Köln, 50931 Köln, Germany ([email protected])

Nicholas Jackson, Mathematics Institute, Zeeman Building, University of Warwick, Coventry, CV4 7AL, UK ([email protected])