Finiteness of the image of the Reidemeister torsion of a splice

ANNALES MATHÉMATIQUES

BLAISE PASCAL

Teruaki Kitano & Yuta Nozaki

Finiteness of the image of the Reidemeister torsion of a splice

<http://ambp.centre-mersenne.org/item?id=AMBP_2020__27_1_19_0>

Cet article est mis à disposition selon les termes de la licence Creative Commons attribution 4.0.

https://creativecommons.org/licenses/4.0/

L’accès aux articles de la revue « Annales mathématiques Blaise Pascal » (http://ambp.centre-mersenne.org/), implique l’accord avec les conditions générales d’utilisation (http://ambp.centre-mersenne.org/legal/).

Publication éditée par le laboratoire de mathématiques Blaise Pascal de l’université Clermont Auvergne, UMR 6620 du CNRS

Clermont-Ferrand — France

Publication membre du

Finiteness of the image of the Reidemeister torsion of a splice

Teruaki Kitano Yuta Nozaki

Abstract

The setRT(M)of values of the SL(2,C)-Reidemeister torsion of a 3-manifoldMcan be both finite and infinite. We prove thatRT(M)is a finite set ifMis the splice of two certain knots in the 3-sphere.

The proof is based on an observation on the character varieties andA-polynomials of knots.

1.

Introduction

LetKbe the figure-eight knot andE(K)the exterior of an open tubular neighborhood of K in the 3-sphere S³. The first author [13] computed the SL(2,C)-Reidemeister torsionτ_ρ(E(K))for any acyclic irreducible representationρ:π₁(E(K)) →SL(2,C). As a consequence, for the doubleM =E(K) ∪_idE(K)ofE(K), the setRT(M)of values of the SL(2,C)-Reidemeister torsion τ_ρ(M) is the set of all complex numbersC. In contrast, his computation also shows thatRT(Σ(K,K))is a finite set. Here, for knotsK₁ andK₂ inS³, letΣ(K₁,K₂)denote the closed 3-manifoldE(K₁) ∪_h E(K₂), wherehis an orientation-reversing homeomorphism∂E(K₁) →∂E(K₂)interchanging meridians and preferred longitudes of the knots. We callΣ(K₁,K₂)thespliceofE(K₁)andE(K₂) (or simply thespliceofK1 andK2). By definition, a splice is an integral homology 3-sphere. Recently, Zentner [20] showed that the fundamental group of any integral homology 3-sphereMadmits an irreducibleSL(2,C)-representation, and therefore, it is worth studyingRT(M).

The purpose of this paper is to generalize the above result on splices to a certain class of knots. We focus on the character varietyX(E(K))andA-polynomialA_K(L,M) ∈Z[L,M] of a knotKand prove the following main theorem and its corollary.

Theorem 1.1. Suppose that knotsK1andK2inS³satisfy the following conditions:

• for any irreducible componentC ⊂ X(E(K_i)) (i =1,2), eitherdimC =0, or dimC=1and its image under the mapX(E(K_i)) → X(∂E(K_i))is not a point.

• gcd(AK1(L,M),A_K2(M,L))=1.

Then RT(Σ(K₁,K₂))is a finite set.

Keywords:Reidemeister torsion,A-polynomial, character variety, splice, bending, Riley polynomial.

2020Mathematics Subject Classification:57M27, 57M25, 20C99, 14M99.

Corollary 1.2. For any2-bridge knotsK₁andK₂, the set RT(Σ(K₁,K₂))is finite.

Curtis [5, 6] defined an SL(2,C)-Casson invariant λ_SL(2,C)(M) for any homology 3-sphere M. Roughly speaking, this invariant counts the number of isolated points of X(M). It is known thatλ_SL(2,C)(Σ(K₁,K₂))is vanishing for anyK₁,K₂ by Boden and Curtis [2]. By definition, this implies that there are no isolated points in X(Σ(K₁,K₂)) and any connected component of X(Σ(K₁,K2))has a positive dimension. However by the main theoremRT(Σ(K₁,K₂))is a finite set for any knots with the above conditions.

In fact, we concretely describe X(Σ(K,K))for the cases whereKis the trefoil knot or figure-eight knot in Section 4.

Recently Abouzaid and Manolescu defined anSL(2,C)-Floer homology and also a full Casson invariant by taking its Euler characteristic in [1]. That is a problem to study a relation with our Reidemeister torsion for a splice.

Acknowledgments

The authors wish to express their thanks to Takahiro Kitayama and Luisa Paoluzzi for useful discussions. They would also like to thank the referee for their careful reading of the manuscript and for various comments. In particular, the geometric description of 2-parameter representations in Example 3.8 is suggested by the referee.

This study was supported in part by JSPS KAKENHI 16K05161. The second author started this study when he visited to QGM, Aarhus University. He would like to thank the institution for the warm hospitality. His visit to QGM was supported by JSPS Overseas Challenge Program for Young Researchers and he was also supported by Iwanami Fujukai Foundation.

2.

Character variety,

2.1.

Representation variety and character variety

LetΓbe a finitely generated group. We define theSL(2,C)-representation varietyR(Γ)of Γto be the affine algebraic set Hom(Γ,SL(2,C))overC. Considering the GIT quotient of R(Γ)by the action ofSL(2,C)by conjugation, one obtains theSL(2,C)-character variety X(Γ):=R(Γ)//SL(2,C)ofΓ(see [9, Section 2] for instance). The character varietyX(Γ) is again an affine algebraic set and not necessarily irreducible. LetR^irr(Γ)denote the subset of irreducible representations andX^irr(Γ)the image ofR^irr(Γ)under the projection R(Γ) X(Γ). It is known that the induced mapR^irr(Γ)/SL(2,C) → X^irr(Γ)is bijective.

We focus on the case Γ = π₁(M) for a connected compact manifold M and call R(M):=R(π₁(M))(resp.X(M):=X(π₁(M))) the representation variety (resp. character

variety) ofM. For instance, the character variety of a torusT²is described explicitly as follows: Letλ,µbe generators ofπ₁(T²)=Z²andρ∈R(T²). Sinceλandµcommute, there exists a representationρ⁰such thatρ⁰is conjugate toρand bothρ⁰(λ)andρ⁰(µ) are upper triangular. Considering the (1,1)-entries of these matrices, one can define the mapθ: R(T²) → (C^×)²/∼byθ(ρ)=(ρ⁰(λ)₁₁, ρ⁰(µ)₁₁), where(L,M) ∼ (L⁰,M⁰)if L=L⁰, M =M⁰orL⁻¹ =L⁰, M⁻¹ =M⁰.

It is easy to see that this map gives an identificationθ: X(T²) → (C^×)²/∼.

The character variety of the complementE(K)of a knotKis complicated in general.

However, it is well known that ifK is a 2-bridge knot thenX(E(K))does not have an irreducible component of dimension larger than one. More generally, if a 3-manifoldM contains no irreducible closed surface and∂M T², then dimC=1 for every irreducible componentCofX(M)(see [3, Section 2.4]).

2.2. A-polynomial of knots

We briefly review the A-polynomial introduced by Cooper, Culler, Gillet, Long, and Shalen [3] (see also [4]) and a relation with the boundary slopes of knots. For an oriented knotK, letr: X(E(K)) →X(∂E(K))denote the regular map between affine algebraic sets induced by the inclusion and letπ:(C^×)²→ (C^×)²/∼be the natural projection. Here one takesλ, µ ∈ π₁(E(K))as a pair of a longitudeλand a meridianµ. We takeλto be homologically trivial inH₁(E(K);Z). By using theseλand µone can also identify π₁(∂E(K))withZ².

For any[ρ] ∈ X(E(K))one can take[ρ⁰]=[r(ρ)]. To define theA-polynomial of a knot, we writeLforρ⁰(λ)₁₁andMforρ⁰(µ)₁₁as above.

Then, the Zariski closure ofπ⁻¹(θ◦r(X(E(K)))) ⊂C²is an affine algebraic set whose irreducible components are curvesC₁, . . . ,C_n and some points. Since codimC_j = 1, the ideal I(C_j)is known to be principal, namelyI(C_j)=(f_j)for some f_j ∈ C[L,M]. It is known that there is c ∈ C such that c f₁(L,M) · · · f_n(L,M) ∈ Z[L,M] and its coefficients have no common divisor. TheA-polynomialAK(L,M)ofKis now defined byA_K(L,M)=c f₁(L,M) · · ·f_n(L,M)up to sign, and it is independent of the choice of an orientation ofK.

Remark 2.1. SinceAK(L,M)has the factorL−1 coming from abelian representations ofπ₁(E(K)), theA-polynomial is sometimes defined to beA_K(L,M)/(L−1). This is not essential in our main theorem due to Lemma 2.2.

Lemma 2.2. Ifθ◦r(ρ)=(L,1), thenL=1. In particular, theA-polynomialAK(L,M) does not have the factorM−1.

Proof. It follows fromr(ρ)=(L,1)thatρ(µ)is equal to the identity matrixI₂or ^{1 1}_{0 1} up to conjugate. In the case ρ(µ)=I2,ρis trivial. In the latter case,ρ(λ)is of the form

1u 0 1

for someu∈C, and henceL=1.

We next see a relation between theA-polynomial and boundary slopes ofK. The rest of this subsection is devoted to proving Corollary 2.6 which is used in Corollary 1.2, not in Theorem 1.1. Here, p/q ∈Q∪ {∞}is called aboundary slopeofK if there exists a properly embedded incompressible surfaceSinE(K)such that∂Sis parallel copies of a simple closed curve of slope p/q, namely the homology class of each boundary component ofSequalspµ+qλ∈H₁(E(K))up to sign. We denote byBS(K)the set of boundary slopes ofK.

For a polynomial f(L,M)=Í

i,jai jLⁱM^j ∈Z[L,M], theNewton polygonN(f)of f is defined byN(f)=Conv({(i,j) ∈Z² |ai j ,0}), where Conv(T)denotes the convex hull of a subsetTinR².

We write bySS(P) ⊂Q∪ {∞}the set of slopes of the sides of a polygonP. Note that SS(N(f))=∅if and only if f is a monomial. The setSS(N(A_K))is closely related to BS(K).

Theorem 2.3([3, Theorem 3.4]). The inclusion SS(N(A_K)) ⊂ BS(K)holds for every knotK.

Let us review some facts about the Minkowski sum. For subsetsT andUofR², the Minkowski sumT+Uis defined byT+U={t+u∈R² |t∈T, u∈U}. One can see that Conv(T+U)=Conv(T)+Conv(U), and henceN(fg)=N(f)+N(g). The following proposition is well known and plays a key role in the next lemma.

Proposition 2.4(see [7, Section 15.1] for example). LetPandQbe convex polygons.

Then SS(P+Q)=SS(P) ∪SS(Q).

For a subset S ofQ∪ {∞}, we denote byS⁻¹ the set{s⁻¹ ∈ Q∪ {∞} | s ∈ S}, where we use the convention 0· ∞=1. Also, for a polynomial f ∈Z[L,M], we define

f^T ∈Z[L,M]by f^T(L,M)= f(M,L).

Lemma 2.5. Let f₁,f₂∈Z[L,M]. If SS(N(f₁)) ∩SS(N(f₂))⁻¹ =∅, thengcd(f₁,f₂^T)is a monomial.

Proof. Let g = gcd(f₁,f₂^T). Then g | f₁ andg^T | f₂. By Proposition 2.4, we have SS(N(g)) ⊂SS(N(f₁))andSS(N(g^T)) ⊂SS(N(f₂)). SinceSS(N(g))=SS(N(g^T))⁻¹, the assumption implies thatSS(N(g))=∅, namelygis a monomial.

Corollary 2.6. IfK₁andK₂be any2-bridge knots, then it holds thatgcd(AK1,A^T_K

2)=1.

Proof. By [8, Theorem 1(b)], BS(K_i) ⊂ 2Zholds. It follows from Theorem 2.3 that SS(N(A_K1)) ∩SS(N(A_K2))⁻¹ =∅, and hence gcd(AK1,A^T_K

2)is a monomial by Lemma 2.5.

Here, in general, theA-polynomial of a knotKis divided by neitherLnorMby definition.

Therefore, the monomial must be 1.

2.3.

The

For precise definitions of a Reidemeister torsion, please see Johnson [10], Kitano [12, 13]

and Milnor [14, 15] as references.

Let M be a 3-manifold and let ρ ∈ R(M) be an acyclic representation. That is, C∗(M;C²ρ)is an acyclic chain complex with twisted coefficients.

Then one gets a nonzero complex numberτ_ρ(M) ∈C^×for an acyclic chain complex C∗(M;C²ρ). We call it theSL(2,C)-Reidemeister torsionofMforρ.

Remark 2.7. Throughout this paper, we setτ_ρ(M)=0 ifρis not acyclic. Thenτ_ρ(M) can be regarded as a function onR(M)and also onX(M).

One can use the well-known multiplicativity of the Reidemeister torsion to compute it as below.

Proposition 2.8. LetM be a3-manifold decomposed intoM1andM2by an embedded torusT². Letρ: π₁(M) →SL(2,C)be a representation. Suppose thatρis acyclic on π₁(T²). Then it holds thatρis acyclic onπ₁(M)if and only if it is acyclic on bothπ₁(M₁) andπ₁(M₂). Further in this case it holds that

τ_ρ(M)=τ_ρ(M₁)τ_ρ(M₂).

One needs the acyclicity of representations to use the above. First we mention the following lemma.

Lemma 2.9. Letρbe a representationπ₁(T²) →SL(2,C). Then it holds thatρis acyclic if and only if ρis not parabolic. Here ρis said to beparaboliciftrρ(x) =2for any x∈π1(T²).

Proof. First note that for a basis{x,y}ofπ₁(T²)the chain complexC∗(T²;C²ρ)is given by 0→C²

∂2

−−→C²⊕C²

∂1

−−→C²→0, where

∂₂ =

−(ρ(y) −I₂) ρ(x) −I₂ , ∂₁=

ρ(x) −I₂ ρ(y) −I₂

.

We here show thatρis not parabolic if and only ifH0(T²;C²ρ)=0. Ifρis not parabolic, then there is a basis {x,y} such that det(ρ(x) − I₂) , 0, and thus H₀(T²;C²ρ) = 0.

Conversely, ifρis parabolic, thenρ(x)andρ(y)are simultaneously of the form ¹_{0 1}^∗ by taking conjugate, and thereforeH0(T²;C²ρ),0.

Next, if H₀(T²;C²ρ) = 0, then ρ is acyclic. Indeed, by Kronecker duality (or the universal coefficient theorem) and Poincaré duality, H₂(T²;C²ρ) H₀(T²;C²_ρˇ), where ˇρ(γ) := ^tρ(γ)⁻¹. When ρis parabolic, so is ˇρ. It follows from χ(T²) =0 that

H₁(T²;C²ρ)=0.

3.

Proof of the main theorem

Recall thatΣ(K₁,K₂)denotes the splice. The following lemma is shown in [2, Proof of Corollary 3.3]. We give a proof to be self-contained.

Lemma 3.1. Ifρis irreducible onπ₁(Σ(K₁,K₂)), then the restrictions ofρonπ₁(E(K₁)) andπ₁(E(K₂))are also irreducible.

Proof. Assume that ρis reducible on π₁(E(K₁)). Then we may take ρas an upper triangular representation on it. Since the longitudeλ₁ofK₁belongs to the commutator subgroup[π₁(E(K₁)), π₁(E(K₁))], then one can see thatL₁is an upper triangular parabolic matrix asL1=ρ(λ₁)= ¹_{0 1}^α

.

Ifα=0, thenL₁ is the identity and henceX₂ =L₁is also the identity matrix. This means thatρmust be trivial onπ₁(E(K₂))and this is a contradiction.

Therefore we may assumeα,0. SinceX₁commutes withL₁, thenX₁is also an upper triangular matrix as X₁ = ^±₀¹_±^β₁ (β,0). Hence the imageρ(π₁(E(K₁)))is an upper triangular subgroup. Since this is an abelian subgroup inSL(2,C), thenL1must be also

the identity. This is a contradiction.

Remark 3.2. By the above arguments, it can be seen that there exists no reducible representation except the trivial representation.

Next we can see the following.

Proposition 3.3. Ifρ: π₁(Σ(K₁,K₂)) →SL(2,C)be an acyclic representation, then its restrictionρ|_π1(T2)is also acyclic.

Proof. Assume thatρ|_π1(T2)is not acyclic. Consider the homology long exact sequence for

0→C∗(T²;C²ρ) →C∗(E(K₁);C²ρ) ⊕C∗(E(K₂);C²ρ) →C∗(Σ(K₁,K₂);C²ρ) →0.

Here we simply writeρfor each ofρ|_π1(T2),ρ|_π1(E(K1)), andρ|_π1(E(K2)).

SinceC∗(Σ(K₁,K₂);C²ρ)is acyclic, we have the exact sequences 0→H₂(T²;C²ρ) →H₂(E(K₁);C²ρ) ⊕H₂(E(K₂);C²ρ) →0, 0→H₁(T²;C²ρ) →H₁(E(K₁);C²ρ) ⊕H₁(E(K₂);C²ρ) →0, 0→H₀(T²;C²ρ) →H₀(E(K₁);C²ρ) ⊕H₀(E(K₂);C²ρ) →0.

Since ρis not acyclic onπ₁(T²),ρis parabolic on it by Lemma 2.9. If it is trivial on π₁(T²), it should be trivial onπ₁(Σ(K₁,K₂)). Then it is not acyclic onΣ(K₁,K₂). For any non-trivial parabolic representationρonπ₁(T²), it is easy to see

H2(T²;C²ρ)H0(T²;C²ρ)C, H1(T²;C²ρ)C²

by the proof of Lemma 2.9. Ifρis irreducible, then bothρ|_π1(E(K1))andρ|_π1(E(K2))are irreducible by Lemma 3.1. Then it holds that H₀(E(K₁);C²ρ)and H₀(E(K₂);C²ρ) are vanishing. ThereforeH₀(T²;C²ρ)is vanishing in this case by the above exact sequences. It is contradiction.

Next assume thatρis reducible. Now we may assume that the image ofρbelongs to the upper triangular subgroup. It is easily seen that the images of the longitudes are trivial I₂or ¹_{0 1}^∗

since the longitudes belong to the commutator subgroup. Therefore the image of each meridian is in the upper triangular parabolic subgroup by the definition of a splice, and thusρis abelian. This contradicts the fact that the abelianization ofπ₁(Σ(K₁,K₂))is

trivial.

Lemma 3.4. Let f: X→Ybe a non-constant regular map between affine algebraic sets XandY. IfXis irreducible anddimX=1, then f⁻¹({y})is a finite(possibly empty)set for anyy∈Y.

Proof. The inverse image f⁻¹({y})is a closed subset ofX, namely f⁻¹({y})is a finite union of irreducible algebraic sets. Since they are proper algebraic subsets ofX, they are

of dimension zero.

The next lemma follows from Lemma 3.4 or Bézout’s theorem.

Lemma 3.5. Let f,g ∈ C[L,M]. Then{f =g =0} ⊂C² is a finite set if and only if gcd(f,g)=1.

Using the above lemmas and propositions, we prove the main theorem.

Proof of Theorem 1.1. First note that gcd_Z[L,M](f,g)=gcd_C[L,M](f,g)holds for f,g∈ Z[L,M]up to multiplication by elements ofC^×. By Lemma 3.5, the intersection

{(L,M) ∈C² | A_K1(L,M)=A^T_K2(L,M)=0}

of the algebraic curves defined by A_K1 and A^T_K

2 is a finite set A. Let us prove that the image ofX(Σ(K₁,K2)) →X(E(K_i))is a finite setXifori=1,2. Then Propositions 2.8 and 3.3 complete the proof.

By the definition of theA-polynomial,θ◦ri(X_i) ⊂ A. It follows from Lemma 3.4 and the second condition in Theorem 1.1 thatr_i⁻¹(θ⁻¹(A))is a finite set. Thus,X_i is also a

finite set.

We next prove Corollary 1.2. LetKbe a 2-bridge knot. Take and fix a presentation of π₁(E(K))and writeφ(s,t)to its Riley polynomial (see Section 4). Then the following lemma is a consequence of [19, Lemma 2].

Lemma 3.6. The coefficient of the leading term ofφ(s,t) ∈Z[s^±1,t]with respect totis a monomial ofs.

Proof of Corollary 1.2. It suffices to check that any pair of 2-bridge knotsK₁ andK₂ satisfies the conditions in Theorem 1.1. First, Corollary 2.6 implies gcd(AK1(L,M),

A_K2(M,L))=1. LetCbe an irreducible component ofX(E(K_i)).

IfCconsists of reducible representations, then dimC=1 andr_i(C) ⊂X(∂E(K_i))is not a point. Otherwise,Cis described by an irreducible factor of the Riley polynomial ofK_i, and hence dimC=1. Assume thatr_i(C)is a pointθ⁻¹(L,M). Then the function trρ(µ)is the constant M +M⁻¹ onC, and thus s−M | φ(s,t). Since M , 0, this

contradicts Lemma 3.6.

We put the following problem.

Problem 3.7. When RT(Σ(K₁,K2))is an infinite set? Or is it always a finite set?

Here we give an observation when dimC>1 in Theorem 1.1.

Example 3.8. LetKbe the Montesinos knotM(1/3,1/3,1/3,1/3,1/2)(see Figure 3.1).

Thenπ₁(E(K))has the presentation

*

µ₁, . . . , µ₅

µ_iµ⁻¹_i+1µ⁻¹_i µ_i+1µ⁻¹_i =µ_i+1µ⁻¹_i+2µ_i+1µ⁻¹_i+2µ⁻¹_i+1µ_i+2µ⁻¹_i+1 (i=1,2,3) µ₄µ⁻¹₅ µ⁻¹₄ µ₅µ⁻¹₄ =µ₅µ⁻¹₁ µ₅µ₁µ⁻¹₅

+ .

Note that µ₁is conjugate toµ⁻¹₂ ,µ₃,µ⁻¹₄ andµ₅. It follows from [18, Theorem 1] that there is an irreducible component of X(E(K))with dim ≥ 2. In fact, we construct a 2-parameter familyCof representations by a bending (see Section 4) along the sphereS intersectingKat 4 points illustrated in Figure 3.1.

For s ∈ C^×\ {1}, we first define the representation ρ_s: π₁(E(K)) → SL(2,C)by ρ_s(µ_j)= _s2−1+s^s−1−20s

if j=1,3,5 and ρ_s(µ_j)= ₀^s_s¹⁻¹

if j=2,4. Note thatρ_sfactors throughπ₁(E(¯3₁))=hx,y|xyx=yxyi, where ¯31denotes the right-handed trefoil knot.

That is,ρ_scomes from ¯ρ_s:π₁(E(¯31)) →SL(2,C)by ¯ρ_s(x)=ρ_s(µ_j)if j =1,3,5 and ρ¯_s(y)=ρ_s(µ_j)⁻¹if j =2,4. Here we haveρ_s(µ⁰

1)=ρ_s(µ₁)sinceρ_s(µ₁)=ρ_s(µ₅). By finding the tangle enclosed by the dotted circle drawn in Figure 3.1 which is a part of ¯31, we can see that the restriction ofρ_stoπ₁(S\K)is invariant under conjugation by

P_u =©

­

­

«

s²−1+s⁻² u

1/2 s2−1+s−2

u 1/2

− s2−1+s−2

u −1/2

s−s⁻¹

0

s²−1+s⁻² u

⁻1/2

ª

®

®

®

¬ ,

whereu∈C^×(see Lemma 4.1). Therefore, one obtains representationsρ_s,u: π₁(E(K)) → SL(2,C)by

ρ_s,u(µ_j)=

(ρ_s(µ_j) if j=3, P_uρ_s(µ_j)P⁻¹_u if j=1,2,4,5.

By the above bending construction, the setC={ρ_s,u}is still a 2-parameter family in X(E(K)). Now one can also check it directly

τ_ρs,u(E(K))=144s⁻⁴(s−1)⁸

−s⁻¹(s−1)² =−144(trρ_s,u(µ₂) −2)³, and henceτ_ρs,u(E(K))depends only on trρ_s,u(µ₂)=s+s⁻¹.

On the other hand, to be independent ofu, it can be explained by the generalized multiplicativity of the Reidemeister torsion to the decompositionE(K)=M1∪_S0M2along the surfaceS₀=S∩E(K)with 4 boundary components. AlthoughM₁,M₂andS₀are not acyclic, after fixing suitable bases ofH₁(M₁;C²ρs,u),H₁(M₂;C²ρs,u)andH₁(S₀;C²ρs,u), we obtainτ_ρs,u(E(K))=τ_ρs,u(M₁)τ_ρs,u(M₂)/τ_ρs,u(S₀). By the construction ofρ_s,u, we see that the value of the right-hand side is independent ofu.

Let RTC be the subset ofRT(Σ(K,K)) consisting of τ_ρ(Σ(K,K))’s where the re- striction of ρ to each E(K) belongs to C. Then RT_C is a finite set even though C ⊂ X(Σ(K,K))is 2-dimensional. Indeed, one can check that trρ_s,u(λ) =s²⁴+s⁻²⁴, and thus there are finitely many solutions (s₁,s₂) of trρ_s1,u1(µ) = trρ_s2,u2(λ) and trρ_s1,u1(λ) = trρ_s2,u2(µ). We conclude that there are finitely many possibilities of the valueτ_ρ(Σ(K,K))=τ_ρs

1,u1(E(K))τ_ρs

2,u2(E(K)).

Problem 3.9. Can we relax the assumption “eitherdimC =0, ordimC =1and its image under the mapX(E(K_i)) →X(∂E(K_i))is not a point” in Theorem 1.1?

Figure 3.1. The Montesinos knotK =M(1/3,1/3,1/3,1/3,1/2). 4.

Computational observation

4.1.

Concrete description of

In this section we observe some examples of splices. We use Mathematica to compute matrices. Recall that X^irr(M)is identified with R^irr(M)/SL(2,C). The construction of deformations of a representation used in this section is called abending constructionor simply abending. See [11, 17] as a reference.

Here we compute X^irr(Σ(K,K))for the trefoil knot and the figure-eight knotK by using a presentation of a twist knot. Let J(2,2q)be a twist knot whereqis a nonzero integer. Please see [16] as a reference for twist knots.

A presentation ofπ₁(E(J(2,2q)))is given as

π₁(E(J(2,2q)))=hx,y| z^qx=yz^qi, z=[y,x⁻¹]

We take a representationρ: hx,yi →SL(2,C)from the free grouphx,yiinSL(2,C)by the correspondence

ρ(x)= s 1 0 1/s

, ρ(y)= s 0

−t 1/s

(s,t∈C^×).

We use a small letter for a group element and its capital letter for the image of a small letter, likeXforρ(x). Forρ(z^q)=Z^q, we put the matrixZ^q = ^zz¹¹21^zz¹²22

.

We define the Riley polynomial to beφ_q(s,t)=z₁₁+(1/s−s)z₁₂. It can be checked that the previous representation gives an irreducible representation ofπ₁(E(J(2,2q)))in SL(2,C)if and only if(s,t)satisfiesφ_q(s,t)=0.

It is seen that any[ρ] ∈X(E(J(2,2q)))can be parametrized by ξ=trρ(x)=trρ(y)=s+1/s,

trρ(xy)=s²+1/s²−t=(s+1/s)²−t−2=ξ²−t−2, and then byξandt.

Here we take other wordsez =[x,y⁻¹]and λ =ez^qz^q. This x gives a meridian of J(2,2q)and thisλdoes the corresponding longitude forx. Hereλis homologically trivial.

Thereforehx, λiis the free abelian group of rank 2 andρ(x)=Xcommutes withρ(λ)=L.

We can find another matrix which commutes withXandLby direct computations.

Lemma 4.1. Any matrix Awhich commutes withX = ^{s c}_{0 1/s}²

(s,±1, c,0)has a form of

A= a ^a−1/a_s−1/sc²

0 1/a

!

for somea∈C^×.

Now we consider two copiesK1,K2ofJ(2,2q)and π₁(E(K₁))=

x₁,y₁ |z^q₁x₁ =y₁z₁^q, z₁ =[y₁,x₁⁻¹], π₁(E(K₂))=

x2,y₂ | z₂^qx2=y₂z^q₂, z2=[y₂,x₂⁻¹].

Further

π₁(Σ(K₁,K₂))=π₁(E(K₁)) ∗_π1(T2)π₁(E(K₂))

=

x₁,y₁,x₂,y₂ |z₁^qx₁=y₁z^q₁, z₂^qx₂=y₂z^q₂, x₁ =λ₂, λ₁=x₂. We consider an irreducible representationρ:π₁(Σ(K₁,K2)) →SL(2,C). Up to conju- gate, we can set that

X₁=ρ(x₁)=

s1 1 0 1/s1

, Y₁=ρ(y₁)=

s1 0

−t₁ 1/s1

.

First note that we treat cases ofs₁,±1. Further we may assume thatX₂is conjugate to ^s_{0 1/s}^{2 1}₂

, andY₂to _−t^s2_21/s⁰₂

simultaneously, as X₂=H

s₂ 1 0 1/s2

H⁻¹, Y₂=H

s₂ 0

−t₂ 1/s2

H⁻¹ for someH∈SL(2,C).

Here we require the conditions

X₁ =L₂, L₁=X₂

to get a representation on π₁(Σ(K₁,K2)). It can be seen that L1 is an upper triangular matrix and thenX₂is also an upper triangular matrix. By taking

H= c 0

0 1/c

(c,0),

one has

X₂=H

s2 1 0 1/s₂

H⁻¹=

s2 c² 0 1/s₂

,

Y₂=H

s₂ 0

−t₂ 1/s2

H⁻¹ =

s₂ 0

−t₂/c² 1/s2

.

HereL₂is also an upper triangular matrix andL₂=X₁. Now any[ρ]=[ρ₁∗ρ₂] ∈ X(Σ(K₁,K2))is corresponding to(X₁,Y₁,X2,Y₂)=(X₁,Y1,L1,Y2)of the above forms. For a∈C^×we defineA_aby

A_a= a _s^a−1/a

1−1/s1

0 1/a

!

and now consider deformations[ρ_a]=[(A_aρ₁A⁻¹_a ) ∗ρ₂]of[ρ]=[ρ₁∗ρ₂]as (A_aX₁A⁻¹_a,A_aY₁A⁻¹_a ,X₂,Y₂)=(X₁,A_aY₁A⁻¹_a ,X₂,Y₂).

Lemma 4.2. It holds thatAaL1A⁻_a¹=L1.

Proof. We prove AaL1A⁻¹_a =L1. We may assumes ,±1. Here we take eigenvectors u₁= ¹₀,u₂∈C²forX₁such thatX₁u₁ =s₁u₁,X₁u₂=s⁻¹₁ u₂. SinceX₁L₁=L₁X₁, one has

X1L1u2=L1X1u2

=L₁s₁⁻¹u₂

=s⁻¹₁ L1u2.

Hence there exists a nonzero constantγsuch thatL₁u₂ =γu₂. This means thatL₁has alsou₁,u₂ ∈C²as eigenvectors. By similar arguments for A_aX₁=X₁A_a, one seesA_a hasu1,u2 ∈C²as eigenvectors. Therefore it is seen that X1,L1,Aaare simultaneously diagonalizable and in particular A_aL₁A⁻¹_a =L₁. By the above lemma, one can see A_aρ₁A⁻¹_a =ρ₁on the subgroupπ₁(T²)generated by {x₁,l₁}={x₂,l₂}and thenρ_a =(A_aρ₁A⁻¹_a ) ∗ρ₂gives an irreducible representation of π₁(Σ(K₁,K₂)).

Further ifa,1, thenAaY1A⁻¹_a ,Y1. This impliesρ_a,ρ∈R(Σ(K₁,K2)).It can be seen by the following computations. First one sees that

tr(ρ₁∗ρ₂(y₁x₂))=tr(Y1X₂)

=s₁s₂+ 1 s1s2

−c²t₁.

On the other hand, one sees that tr

(A_aρ₁A⁻_a¹) ∗ρ₂)(y₁x₂)

=tr(A_aY₁A⁻_a¹X₂)

=s₁s₂+ 1 s₁s₂ +

((s₂−_s¹

2) (s₁−_s¹

1) 1

a² −1

−c² a² )

t₁. Therefore we can find one character

[ρ] 7→trρ(y₁x2)

which is not constant on X(Σ(K₁,K₂)) and we know X(Σ(K₁,K₂)) has at least one dimension near[ρ].

Proposition 4.3. X(Σ(K₁,K2))has just one dimension near[ρ].

Proof. Take and fix any[ρ] =[ρ₁∗ρ₂] ∈ X(Σ(K₁,K₂)). It is seen that the character varietyX(Σ(K₁,K₂))has at least one dimension near[ρ]by a bending construction.

Consider another one parameter family

{[ρ_u]}_u ={[ρ_1,u∗ρ_2,u]}_u ⊂X(Σ(K₁,K₂))

such that[ρ₀]=[ρ]. Here recall there exist only finitely many quadruples{(s₁,t₁,s₂,t₂)’s}

for this fixed[ρ]=[ρ₁∗ρ₂] ∈X(Σ(K₁,K₂))by the proof of the main theorem. Then we may assume that[ρ_u]=[ρ_1,u∗ρ₂]and[ρ_1,u]=[ρ₁] ∈ X(K₁)for anyu. Hence one gets [ρ_u] =[ρ_1,u∗ρ₂]=[(Bρ₁B⁻¹) ∗ρ₂], whereB ∈ SL(2,C). BecauseBmust commute withX₁andL₁, thenBhas a similar form asA_ain Lemma 4.1. Therefore this is a bending

construction and the dimension of deformations is one.

4.2. q =1

Case

Here we putq=1. ThisJ(2,2)is the trefoil knot. We write again X =

s 1 0 1/s

, Y =

s 0

−t 1/s

and

Z =[Y,X⁻¹]=

1−s²t ¹_s −s(1+t)

−^t

s +st(1+t) 1+(2− ¹

s²)t+t²

,

eZ =[X,Y⁻¹]= 1− −2+s²

t+t^{2 −1+s}_s^2−t

t(1−s²+t)

s 1− ^t

s²

! ,

Z X−Y Z=

0 −1+1/s²+s²−t s(−1+1/s²+s²−t) 0

.

The condition that(s,t)gives a representation is−1+1/s²+s²−t=0. On the other hand,

φ₁(s,t)=w₁₁+(1/s−s)w₁₂

=1−s²t+(1/s−s)(1/s−s(1+t))

=1−s²t+1/s²−1−t−1+s²+s²t

=−1+1/s²+s²−t

=ξ²−3−t, wherem=s+1/s.

Hence in the case of the trefoil knot, one sees t=ξ²−3 andX^irr(E(J(2,2)))is given by

{(ξ,t) ∈C² |t=ξ²−3, t,0}.

Ift=0, then the corresponding representation is not irreducible.

Remark 4.4. Ifs=1, that is,ξ=2, then the chain complex is not acyclic. In the other cases,τ_ρ(E(J(2,2)))=2.

Compute L=Z Ze

= 1−t²+s⁴t²−t³+^t(1+t)_s2 −s²t(1+t+t²) (1+s²)t(1+t+s⁴(1+t)−s²(3+3t+t²))

s³ t²(1+s⁶−s²t−s⁴t)

s³ 1−t²+_s^t24−t³+s²t(1+t) −^t(1+t+t²)

s²

! .

By puttingt=1− (1/s²+s²), one obtains

L= −s⁶ −^{(1+s2+s4)(1+s6)}

s⁵

0 −1/s⁶

!

and

tr(L)=−s⁶−1/s⁶=−T₆(m).

HereT₆(x)=x⁶−6x⁴+9x²−2 is the normalized Chebyshev polynomial of degree 6.

Remark thatT₆(x)has the propertyT₆(2 cosθ)=2 cos 6θ.

By relationsx₁ =λ₂, λ₁=x₂, one has

tr(X1)=tr(L2), tr(L1)=tr(X2).

By puttingξ₁ =s1+1/s1, ξ₂ =s2+1/s2, one obtains ξ₁=−T₆(ξ₂), −T₆(ξ₁)=ξ₂.

Hence we obtain only one equation

ξ=−T₆(−T₆(ξ))=−T₆(T₆(ξ)).

This equationξ+T₆(T₆(ξ))=0 is a polynomial equation of degree 36 with distinct 36 roots as follows:

−2=2 cosπ, 2 coskπ

35 (k=1,3, . . . ,33), 2 coskπ

37 (k=1,3, . . . ,35).

It is seen that they are the roots as

T₆(T₆(−2))=T₆(T₆(2 cosπ))

=2 cos 36π=2, T₆

T₆

2 coskπ

35 =2 cos36kπ

35 =−2 coskπ 35, T₆

T₆

2 coskπ

37 =2 cos36kπ

37 =−2 coskπ 37.

Further one easily sees thatξ =−2 does not give a representation on the splice. The roots 2 cos^kπ₃₅(k=1,3, . . . ,33)are corresponding to the conditions³⁶₁ =s₁coming from matrix equationsL₁ =X₂andL₂ =X₁.

It can be seen that there exists a k such that tr(ρ(x₁)) =2 cos^kπ₃₅ and tr(ρ(x₂)) =

−T₆(2 cos^kπ₃₅)for any[ρ] ∈ X^irr(Σ(K₁,K₂)). On the other hand, the roots 2 cos^kπ₃₇(k = 1,3, . . . ,35)are corresponding to the conditions₁³⁶=s₁⁻¹coming from equationsL1 =X₂⁻¹ andL₂=X₁. They give representations of the splicing of 31and its mirror image, not 31.

Take[ρ]=[ρ₁∗ρ₂] ∈ X^irr(Σ(K₁,K2))and identify it with(X₁,Y1,X2,Y2). Consider A_a= a _s^a−1/a

1−1/s1

0 1/a

! ,

where a ∈ C^×, s1,s2 ∈ C^× are satisfying s1 +1/s1 = ξ₁,s2+1/s2 = ξ₂ and ξ₁ =

−T₆(T₆(ξ₁)), ξ₂=−T₆(ξ₁). In this case, one gets tr

(A_aρ₁A⁻¹_a ) ∗ (ρ₂)(y₁x₂)

=tr(X2A_aY₁A⁻¹_a )

=s₁s₂+ 1 s1s2

−c²

a²(s²₁+1/s²₁−1)+(1−a²)(s₂−1/s2)

(s₁−1/s1) (s²₁+1/s₁²−1).

Herecis determined byX₂=L₁, namely

c²=−(1+s₁²+s⁴₁)(1+s₁⁴) s⁵₁ .

4.3. q =−1

Case

We putq=−1. ThisJ(2,−2)is the figure-eight knot. In this case the Riley polynomial φ₋₁(s,t)is given by

φ₋₁(s,t)=t²− (s²+1/s²−3)t−s²−1/s²+3

=t²− (ξ²−5)t−ξ²+5, whereξ=s+1/s.

Then the irreducible representation part ofX^irr(E(J(2,4)))is {(ξ,t) ∈C² |t²− (ξ²−5)t−ξ²+5=0, t ,0}.

Under the same notations, one obtains

ξ₁ =ξ₂⁴−5ξ₂²+2, ξ₁⁴−5ξ₁²+2=ξ₂. Hence we obtain only one equation

ξ=ξ¹⁶−20ξ¹⁴+158ξ¹²−620ξ¹⁰+1244ξ⁸−1190ξ⁶+487ξ⁴−60ξ²−2 and 16 rootsν₀ =−2, ν_i ,±1(i=1, . . . ,15). For anyν_i(i,0), we can take a bending construction to do deformations inX^irr(Σ(J(2,−2),J(2,−2))).

Remark 4.5. In this case,tis a root oft²− (ν_i²−5)t−ν_i²+5=0.

References

[1] Mohammed Abouzaid and Ciprian Manolescu. A sheaf-theoretic model for SL(2,C) Floer homology. to appear inJ. Eur. Math. Soc.

[2] Hans U. Boden and Cynthia L. Curtis. Splicing and the SL2(C)Casson invariant.

Proc. Am. Math. Soc., 136(7):2615–2623, 2008.

[3] Daryl Cooper, Marc Culler, Henri Gillet, Darren D. Long, and Peter B. Shalen. Plane curves associated to character varieties of 3-manifolds.Invent. Math., 118(1):47–84, 1994.

[4] Daryl Cooper and Darren D. Long. Remarks on theA-polynomial of a knot.J. Knot Theory Ramifications, 5(5):609–628, 1996.

[5] Cynthia L. Curtis. An intersection theory count of the SL2(C)-representations of the fundamental group of a 3-manifold.Topology, 40(4):773–787, 2001.

[6] Cynthia L. Curtis. Erratum to: “An intersection theory count of the SL2(C)- representations of the fundamental group of a 3-manifold” [Topology40(4):773–787, 2001].Topology, 42(4):929, 2003.

[7] Branko Grünbaum.Convex polytopes, volume 16 ofPure and Applied Mathematics.

Interscience Publishers, 1967. With the cooperation of Victor Klee, M. A. Perles and G. C. Shephard.

[8] Allen Hatcher and William Thurston. Incompressible surfaces in 2-bridge knot complements.Invent. Math., 79(2):225–246, 1985.

[9] Michael Heusener. SLn(C)-representation spaces of knot groups.RIMS Kokyuroku, 1991:1–26, 2016.

[10] Dennis Johnson. A geometric form of Casson’s invariant and its connection to Reidemeister torsion. unpublished lecture notes.

[11] Dennis Johnson and John J. Millson. Deformation spaces associated to compact hyperbolic manifolds. InDiscrete groups in geometry and analysis (New Haven, Conn., 1984), volume 67 ofProgress in Mathematics, pages 48–106. Birkhäuser, 1987.

[12] Teruaki Kitano. Reidemeister torsion of Seifert fibered spaces for SL(2;C)- representations.Tokyo J. Math., 17(1):59–75, 1994.

[13] Teruaki Kitano. Reidemeister torsion of the figure-eight knot exterior for SL(2;C)- representations.Osaka J. Math., 31(3):523–532, 1994.

[14] John Milnor. Two complexes which are homeomorphic but combinatorially distinct.

Ann. Math., 74:575–590, 1961.

[15] John Milnor. A duality theorem for Reidemeister torsion.Ann. Math., 76:137–147, 1962.

[16] Takayuki Morifuji. Twisted Alexander polynomials of twist knots for nonabelian representations.Bull. Sci. Math., 132(5):439–453, 2008.

[17] Kimihiko Motegi. Haken manifolds and representations of their fundamental groups in SL(2,C).Topology Appl., 29(3):207–212, 1988.

[18] Luisa Paoluzzi and Joan Porti. Non-standard components of the character variety for a family of Montesinos knots.Proc. Lond. Math. Soc., 107(3):655–679, 2013.

[19] Robert Riley. Nonabelian representations of 2-bridge knot groups.Q. J. Math., Oxf.

II. Ser., 35(138):191–208, 1984.

[20] Raphael Zentner. Integer homology 3-spheres admit irreducible representations in SL(2,C).Duke Math. J., 167(9):1643–1712, 2018.

Teruaki Kitano

Department of Information Systems Science, Faculty of Science and Engineering, Soka University Tangi-cho 1-236, Hachioji, Tokyo 192-8577, Japan kitano@soka.ac.jp

Yuta Nozaki

Organization for the Strategic Coordination of Research and Intellectual Properties, Meiji University 4-21-1 Nakano, Nakano-ku, Tokyo, 164-8525, Japan Current address:

Graduate School of Advanced Science and Engineering, Hiroshima University 1-3-1 Kagamiyama, Higashi-Hiroshima City, Hiroshima, 739-8526, Japan

nozakiy@hiroshima-u.ac.jp