Reidemeister torsion and Seifert surgery

# Reidemeister torsion for linear representations and Seifert surgery on knots

## Abstract.

We study an invariant of a -manifold which consists of Reidemeister torsion for linear representations which pass through a finite group. We show a Dehn surgery formula on this invariant and compute that of a Seifert manifold over . As a consequence we obtain a necessary condition for a result of Dehn surgery along a knot to be Seifert fibered, which can be applied even in a case where abelian Reidemeister torsion gives no information.

###### Key words and phrases:
Reidemeister torsion, Dehn surgery, Seifert fibered space
###### 2000 Mathematics Subject Classification:
Primary 57M27, Secondary 55R55, 57Q10

## 1. Introduction

Let be a knot in a homology -sphere and the complement of an open tubular neighborhood of . We denote by the result of -surgery along for an irreducible fraction . The aim of the paper is to give a necessary condition for to be a certain closed -manifold, in particular a Seifert manifold, using Reidemeister torsion for linear representations.

It is known that the Alexander polynomial of has useful information on Dehn surgery. In [1] and [2] Kadokami used abelian Reidemeister torsion to provide obstructions to lens surgery and Seifert surgery in terms of . In [9], [10] and [6] Ozsváth-Szabó and Kronheimer-Mrowka-Ozsváth-Szabó gave other obstructions for to lens surgery and Seifert surgery in terms of the Heegaard Floer homology of , the knot Floer homology of and the Monopole Floer homology of , which deduce those in terms of . It is of interest to investigate information on Dehn surgery that Reidemeister torsion for linear representations has. Reidemeister torsion of coincides with a twisted Alexander invariant of up to multiplication of units. See [3], [4], [7] and [12] for the definition of twisted Alexander invariants and the relation with Reidemeister torsion.

We fix orientations of and the ambient homology sphere. Let be a closed connected -manifold with and a linear representation over a field of a finite group . All homology groups and cohomology groups are with respect to integral coefficients unless specifically noted. First we define an invariant of for , where acts on by

 g′⋅(g,h):=(g′gg′−1,g′hg′−1)

for and , and an invariant of for a surjection , where is a primitive -root of (Definition 3.3). These invariants are sets which consist of Reidemeister torsion of and respectively for representations which pass through surjectively. The pair corresponds with the images of longitudinal and meridional elements by the representations. It is worth pointing out that for , if we know all surjective homomorphisms from to , is combinatorially computable from a presentation of as Reidemeister torsion is. We establish a Dehn surgery formula which computes from with (Theorem 3.4). Therefore by this formula we obtain a necessary condition for to be homeomorphic to if we have . Next we compute the invariant for a Seifert manifold over (Theorem 4.4). Note that every Seifert manifold which is a result of Dehn surgery along a knot has or as its base space. Finally as an application we consider the Kinoshita-Terasaka knot , whose Alexander polynomial is . We show that for any integer , is not homeomorphic to any Seifert manifold over with three singular fibers. In this case we can check that abelian Reidemeister torsion gives no information.

This paper is organized as follows. In the next section we give a brief exposition of fundamental facts about Reidemeister torsion. In Section we develop a key lemma of Reidemeister torsion on gluing a solid torus along a torus boundary. Furthermore we define the invariants and and describe a Dehn surgery formula on these invariants. Section is devoted to computations of for Seifert manifolds over . In the last section we apply these results to the Kinoshita-Terasaka knot.

## 2. Reidemeister torsion

We first review the definition of Reidemeister torsion. See [8] and [11] for more details.

For given bases and of a vector space, we denote by the determinant of the base change matrix from to .

Let be a commutative field and an acyclic chain complex of finite dimensional vector spaces over . For a basis of for , choosing a lift of in and combining it with , we obtain a basis of .

###### Definition 2.1.

For a given basis of , we choose a basis of and define

 τ(C∗,c):=m∏i=0[bibi−1/ci](−1)i+1 ∈F∗.

It can be easily checked that does not depend on the choices of and .

The torsion has the following multiplicative property. Let

 0→C′∗→C∗→C′′∗→0

be a short exact sequence of acyclic chain complexes and , and bases of , and respectively. Choosing a lift of in and combining it with the image of in , we obtain a basis of .

###### Theorem 2.2.

([8, Theorem 3. 1], [11, Theorem 1. 5]) If for all , then

 τ(C∗,c)=τ(C′∗,c′)τ(C′′∗,c′′).

Let be a connected finite CW-complex and a linear representation over a commutative ring . We regard as a left -module by

 γ⋅v:=ρ(γ)v,

where and . Then we define the twisted homology group and the twisted cohomology group of associated to as follows:

 Hρi(X;Rn) :=Hi(C∗(˜X)⊗Z[π1X]Rn), Hiρ(X;Rn) :=Hi(HomZ[π1X](C∗(˜X),Rn)),

where is the universal covering of .

###### Definition 2.3.

For a representation with , we define the Reidemeister torsion of associated to as follows. We choose a lift in for each cell of and a basis of . Then

 τρ(X):=[τ(Cρ∗(X;Fn),~c)] ∈F∗/(±1)nImdet∘ρ,

where

 ~c:=⟨~e1⊗f1,…,~e1⊗fn,…,~edimC∗(X)⊗f1,…,~edimC∗(X)⊗fn⟩.

For a representation with , we set .

It is known that does not depend on the choices of and and is a simple homotopy invariant.

###### Remark 2.4.

For a link exterior of , given a presentation of the link group, Reidemeister torsion can be computed efficiently using Fox calculus (cf. e.g. [3], [4]).

## 3. A surgery formula

### 3.1. A gluing lemma

In this subsection we discuss a gluing lemma (Proposition 3.1) which we need to establish a surgery theorem (Theorem 3.4) and to compute Reidemeister torsion of Seifert manifolds (Lemma 4.3).

Let be a compact connected orientable 3-manifold whose boundary consists of tori and a 3-manifold obtained by gluing a solid torus to along a component of . We take a generator and a representation . Let us denote by and the homomorphisms and induced by the inclusion maps respectively.

###### Proposition 3.1.

If there exists such that , then

 τρ∘π(E)=[det(ρ∘i(ν)−I)]τρ(M).

To prove this proposition we begin by collecting the following computations.

###### Lemma 3.2.

(i) The following conditions are equivalent.
(a) vanishes.
(b) vanishes.
(c) .
(ii) If satisfies one of the conditions in (i), then

 τρ∘i(Z) =[det(ρ∘i(ν)−I)−1], τρ∘i(∂Z) =[1].
###### Proof.

We only consider the case of . The proof for the case of is very similar. Taking the natural cell structure on with one 0-cell, two 1-cells and one 2-cell, one can identify with

 0→Fn∂2−→F2n∂1−→Fn→0,

where

 ∂1=(ρ(ν−1)−I0) and ∂2=(0ρ(ν−1)−I).

Therefore vanishes if and only if and for appropriate choices of bases and ,

 τρ∘i(∂Z)=[det(ρ∘i(ν−1)−I)det(ρ∘i(ν−1)−I)]=[1].

We define a representation of to be

 ρ†(γ):=ρ(γ−1)T,

where . Then we have an isomorphism

 (3.1) C∗ρ†(M;Fn)≅Hom(Cρ∗(M;Fn),F)

defined by

 ψ↦(c⊗v↦ψ(c)Tv),

where , and .

###### Proof of Proposition 3.1.

We first prove that (a) vanishes if and only if (b) vanishes and (c) . By Lemma 3.2(i) and the Mayer-Vietoris long exact sequence we check at once that two of the conditions (a), (b) and (c) deduce the other one. Therefore it suffices to show that (a) deduce (c).

Let us assume that (a) holds and that . From the proof of Lemma 3.2 one can see that . By the Mayer-Vietoris long exact sequence we obtain . If , then collapses onto a 2-dimensional subcomplex, which contradicts it. If is closed, then by Poincaré duality, (3.1) and the universal coefficient theorem we have

 Hρ†0(M;Fn) ≅H3ρ†(M;Fn) ≅H3(Hom(Cρ∗(M;Fn),F)) ≅Hom(Hρ3(M;Fn),F)≠0.

However, there exists such that , and so , a contradiction.

Next we assume that vanishes. It follows from the above argument that is defined. By Lemma 3.2(i) and are also defined. Considering the exact sequence

 0→Cρ∘i∗(∂Z;Fn)→Cρ∘π∗(E;Fn)⊕Cρ∘i∗(Z;Fn)→Cρ∗(M;Fn)→0,

by the multiplicative property of torsion (Theorem 2.2) we obtain

 τρ∘π(E)τρ∘i(Z)=τρ(M)τρ∘i(∂Z).

Combining it with Lemma 3.2 (ii), we completes the proof. ∎

### 3.2. Description of the formula

Fix a finite group . For a group , we denote by the set of conjugacy classes of surjective homomorphisms from to . Let be an oriented smooth knot in an oriented homology -sphere. We take a longitude-meridian pair , which is compatible with the orientations of and the ambient space and define the abelianization map which maps to .

###### Definition 3.3.

Let be a representation.
(i)For , we define to be the set of for such that , where is a representation which maps to .
(ii)For a closed connected 3-manifold with and a surjection , where is a primitive -root of , we define to be the set of for , where is defined as .

###### Theorem 3.4.

We take integers and such that . Let be a surjection which maps the image to . If for any such that and is not empty, and , then

 TφK(p/q),β={τ|t=ζ[det(ζrφ(gshr)−I)] ; τ∈TφK([g,h]) with gqhp=1}.

This theorem easily follows from Proposition 3.1 and the following lemma.

###### Lemma 3.5.

Let be a surjection which maps to and a representation. If , then

 τα′⊗ρ(EK)=τα⊗ρ(EK)|t=ζ.
###### Proof.

Choose a triangulation of and maximal trees and in the -skeleton and in the dual -skeleton respectively. Collapsing and all the -cells along , we have a -dimensional CW-complex which is simple homotopic to . Let us denote the number of -cells of by , then it follows from that there are -cells. We can arrange the chain complex of the form

 0→C2(˜W)∂2−→C1(˜W)∂1−→C0(˜W)→0,

where

 ∂1=(γ1−1…γm−1)

and is a generator set of . If necessary, attaching one -cell and one -cell along the word of in , we can assume that . Let be the result of deleting st row of the matrix of .

First we assume that vanishes. Then and deduce , and so , where is the -dimensional matrix with entries in which is the result that linearly operates all the entries of and is defined similarly. This gives . Since , we obtain . Considering

 2∑i=0(−1)idimHα⊗ρi(EK;F(t)n)=nχ(EK)=0,

we can see that vanishes. In this case we have

 τα⊗ρ(EK)|t=ζ =[det(α⊗ρ(A))det(tρ(μ)−I)∣∣∣t=ζ] =[det(α′⊗ρ(A))det(ζρ(μ)−I)] =τα′⊗ρ(EK)|t=ζ≠0.

Now assume that vanishes and that . Then , and so the same argument as above shows that vanishes. These prove the lemma. ∎

## 4. Torsion of Seifert manifolds

In this section we compute the invariant for a Seifert manifold over .

Let be the link in represented in Figure 1 and the exterior of an open tubular neighborhood of . We denote by the 3-manifold which has a surgery description shown in Figure 1 and take integers and such that for . We assume that and that for .

From the diagram we have presentations of and as follows:

 (4.1) π1EL =⟨x,y1,y2,…,ym | [x,yi]=1 for i=1,…,m⟩, (4.2) π1M(p1/q1,…,pm/qm) =⟨x,y1,y2,…,ym | y1…ym=1,[x,yi]=xqiypii=1 for i=1,…,m⟩.

We fix a finite group . The group acts on by

 g′⋅(g,h1,…,hm):=(g′gg′−1,g′h1g′−1,…,g′hmg′−1)

for and .

###### Definition 4.1.

We define to be the set of such that

 ⟨g,h1,…,hm⟩=G, g∈Z(G), h1…hm=1  and  gqihpii=1 for i=1,…,m,

where is the center of .

###### Lemma 4.2.

The map which maps to is bijective.

The proof is straightforward from (4.2).

###### Lemma 4.3.

Let be a representation. If , then

 τρ(M(p1/q1,…,pm/qm))=[det(ρ(x)−I)m−2∏midet(ρ(xsiyrii)−I)].
###### Proof.

Let be the natural surjection. From (4.1) we can directly compute that

 τρ∘π(EL)=[det(ρ(x)−I)m−2]

(Remark 2.4). The details are left to the reader. Now we use Proposition 3.1 repetitiously, and the lemma follows. ∎

Now we easily obtain the next theorem as a corollary of Lemma 4.2 and Lemma 4.3.

###### Theorem 4.4.

Let be a representation and a surjection, which maps to and to for . If for any , , then

 TφM(p1/q1,…,pm/qm),β={[det(ζaφ(g)−I)m−2∏mi=1det(ζasi+biriφ(gsihrii)−I)] ; [g,h1,…,hm]∈SG(p1/q1,…,pm/qm)}.
###### Remark 4.5.

In [5] Kitano gave a formula which computes for a general Seifert manifold and an irreducible representation such that vanishes.

## 5. Application

Let be the Kinoshita-Terasaka knot illustrated in Figure 2. It is well known that . As an application we show that is not homeomorphic to for any integer and any pair .

For example, let us consider , whose st homology group is . We set . Since we can compute that

 τα′(EKT)=[1]

for any surjection (Remark 2.4), it follows from Proposition 3.1 that

 τβ(KT(6/q))=[1]

for any surjection . Furthermore Lemma 4.3 yields

 τβ′(M(3/2,−3,−5))=[1]

for any surjection , hence abelian Reidemeister torsion gives no information in this case.

First we have the following data on . By direct computations we obtain

 (5.1) S(π1KT(6/q),A4) =∅, (5.2) ♯S(π1KT(6/q),A5) =2,

where is the alternating group on letters. Let be the representation induced by the natural action of the symmetric group on . Then we computes that

 TφKT([g,h])=⎧⎨⎩{[(t2+t+1)(5t6+5t5−5t4−9t3−5t2+5t+5)(t−1)4]}if [g,h]=[1,(3,4,5)],∅otherwise

(Remark 2.4). By Theorem 3.4 we have

 (5.3) TφKT(6/q),β={[29]}

for any surjection .

Second we have the following lemma on .

###### Lemma 5.1.

Let be a surjection, which maps to . If , then for any ,

 |τ|=AB1B2B3,

where

 A =1,9,16, Bi =1,2,4,9,16for i=1,2,3.
###### Proof.

By Theorem 4.4 there exist and for such that

 τ=[(ζa−1)4∏3i=1det(ζciφ(h′i)−I)].

Note that . The possible values of are , , and these of are , , , , , which proves the lemma. ∎

Now let us suppose that is homeomorphic to . Since we have

 (5.4) |q1p2p3+p1q2p3+p1p2q3|=6.

From (5.1) and (5.2) we have

 SA4(p1/q1,p2/q2,p3/q3) =∅, ♯SA5(p1/q1,p2/q2,p3/q3) =2.

By direct computations these are equivalent to the conditions that () we cannot realize that

 2∣p1,3∣p2,3∣p3

by permuting the indices and that only one of the following holds:

 (i) after possible permuting the indices,2∣p1,3∣p2,5∣p3, (ii) after possible permuting the indices,2∣p1,5∣p2,5∣p3, (iii) after possible permuting the indices,3∣p1,3∣p2,5∣p3, (iv) after possible permuting the indices,5∣p1,5∣p2,5∣p3.

In the case (i) we have from (5.4). If , then (iii) also holds. If , then (0) does not hold. In the case (ii) we have from (5.4), and (iv) also holds. In the case (iv) (5.4) does not hold. Therefore we only have to consider the case (iii).

Let us assume (iii). If , then (i) also holds, hence . Since

 ζaq1+b1p1=ζaq2+b1p2=ζb1+b2+b3=1,

where is an integer such that for , if , then for all , and cannot be surjective. Therefore and , in consequence, the assumption of Lemma 5.1 is satisfied. Comparing (5.3) and Lemma 5.1, we have a contradiction, and we obtain the desired conclusion.

Acknowledgement. The author would like to express his gratitude to Toshitake Kohno for his encouragement and helpful suggestions. He also would like to thank Hiroshi Goda, Teruhisa Kadokami, Takayuki Morifuji, Masakazu Teragaito and Yuichi Yamada for fruitful discussions and advices. This research is supported by JSPS Research Fellowships for Young Scientists.

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