CHARACTER VARIETIES IN $\mathrm{SL}_{2}$ AND KAUFFMAN SKEIN ALGEBRAS (Topology, Geometry and Algebra of low-dimensional manifolds)

CHARACTER VARIETIES IN $SL_{2}$ AND KAUFFMAN SKEIN

ALGEBRAS

JULIEN MARCH\’E

1. INTRODUCTION

These notes collect

some

general facts about character varieties of finitely generated

groups

in$SL_{2}$

.

We stress that

one can

study character varieties with the point of view of

skein modules: using

a

theoremof K. Saito,

we recover

some

standard results ofcharacter varieties, including

a

construction of

a

so-called tautological representation. This allowsto

give

a

global construction ofthe Reidemeister torsion, which should be useful for further

study such as its singularities or differential equations it should satisfy.

The main motivation of the author is to understand the relation between character

varieties and topological quantumfieldtheory (TQFT)with gaugegroup $SU_{2}$

.

Thistheory

-in the [1] version-makes fundamental use of the Kauffman bracket skein module, an

object intimately related to character varieties. Moreover the non-abelian Reidemeister

torsionplaysa fundamental role in the Wittenasymptotic expansion conjecture, governing

the asymptotics of quantuminvariants of 3-manifolds. However, these notes do not deal

with TQFT, and strictly speaking do not contain any

new

result. Let us describe and

comment its content.

(1) The traditional definition of character varieties

uses an

algebraic quotient, al-though it is not always presented that way. On the contrary, the skein algebra is

given bygenerators and relations. These twopoints ofview arein fact equivalent.

This

was

previously shown by Bullock up to nilpotent elements (see [3]) and by

Przytycki and Sikoraingeneral (see [10]) usingwork ofBrumfiel and Hilden ([2]).

Indeed, the proof is in a short article of Procesi (see [9]) and follows from the

fundamental theorems of invariant theory. We explain this in Section 2.

(2) A theorem of K. Saito (with unpublished proof) allows to

recover a

representa-tion from its character in a very general situation. We present this theorem and

advertise it by applying it in different situations in Section 3.1.

- Given a character $\chi$ with values in a field $k$, in which extension of

$k$ lives a

representation with character $\chi$?

- Compare the tangent space of the character variety with the twisted coho-mology of the

group

with values inthe adjoint representation.

- Definetautological representations with values inthefield of functions of (an

irreducible component of) the character variety.

-Study points of the character variety in valuation rings, related to Culler-Shalen theory.

$\langle$3) Using the tautological representation gives

a

convenient framework for studying

global aspectsoftheReidemeister torsion. We show in what

sense

theReidemeister Received October 1st, 2015.

torsion may be seen

as

a rational volume form

on

the character variety of

a

3-manifold with boundary and give examples in Section 4.

Acknowldegements: These working notesgrewslowly and benefited from many

con-versations. It is my pleasure to thank L. Benard, R. Detcherry, A. Ducros, E. Falbel, L.

Funar, M. Heusener, T. Q. T. Le, M. Maculan, C. Peskine, J. Porti, R. Santharoubane, M.

Wolff for their help and interest. I also thank the organizers of the conference Topology, Geometry andAlgebra

_of

Low-Dimensional

_Manifolds

in Numazu (Japan), June 2015 for welcoming me and publishing these notes.

2. Two DEFINITIONS OF CHARACTER VARIETIES

In all the article, $k$ will denote a field with characteristic O. Most results of Section3

hold for any field with characteristic different from 2. We stayed in characteristic $0$ in

view of

our

applications.

2.1. Algebraic quotient. Let $\Gamma$ be a finitely generated group. We define its

represen-tation variety into $SL_{2}$ and we denote by $Hom(\Gamma, SL_{2})$ the spectrum of the algebra $A(\Gamma)=k[X_{i,j}^{\gamma}, i,j\in\{1, 2\}, \gamma\in\Gamma]/(\det(X^{\gamma})-1,$$X^{\gamma\delta}-X^{\gamma}X^{\delta}$

with $\gamma,$$\delta\in\Gamma)$

In this formula, $X^{\gamma}$ stands for the matrix with entries

$X_{i_{J}}^{\gamma},,$ $i,j\in\{1$,2$\}$. In particular

the last equation above is a collectionof 4 equations. The

name

representation variety is

justified bythe following universal property which holds for any $k$-algebra $R$:

$Hom_{k-alg}(A(\Gamma), R)=Hom(\Gamma, SL_{2}(R))$

Let $SL_{2}(k)$ act on the space $Hom(\Gamma, SL_{2})$ by conjugation. This action is algebraic

as

it

comes

from the action $(g.P)(X^{\gamma})=P(g^{-1}X^{\gamma}g)$ where

we

have$g\in SL_{2}(k)$ and$P\in A(\Gamma)$

.

Definition 2.1. We define the character variety of$\Gamma$

and denote by$X(\Gamma)$ the spectrum

of the algebra$A(\Gamma)^{SL_{2}}$ of invariants.

In other words, $X(\Gamma)$ isthe quotient of$Hom(r, SL_{2})$ in the

sense

ofgeometricinvariant

theory. Standard arguments from this theory gives the following theorem:

Theorem 2.2.

_If

$k$ is algebraically closed, there is a bijection between thefollowing sets:

- The $k$-points

of

$X(\Gamma)$ (or equivalently $Hom_{k-alg}(A(\Gamma)^{SL_{2}},$ $k)$)

- The closed orbits

of

$SL_{2}(k)$ acting on $Hom(\Gamma, SL_{2}(k))$

- The conjugacy classes

of

semi-simple representations

of

$\Gamma$ into _{$SL_{2}(k)$}

- The characters

of

representations in$Hom(\Gamma, SL_{2}(k))$

.

Recallthatby characterof

a

representation$p$ : $\Gammaarrow SL_{2}(k)$

we mean

themap$\chi_{\rho}$ :

$\Gammaarrow k$

given by $\chi_{\rho}(\gamma)=Tr\rho(\gamma)$. This theoremwill be made moreprecise in thesequel using the

point of view of skein algebras.

Remark 2.3. We point out the fact that the algebra $A(\Gamma)$ may have nilpotent elements,

as the skein algebra. A big part of the literatureon charactervarieties uses the reduction

2.2. The skein algebra.

Definition 2.4. We define the skein character variety $X_{8}(\Gamma)$ as the spectrum of the

algebra

$B(\Gamma)=k[Y_{\gamma}, \gamma\in\Gamma]/(Y_{1}-2, Y_{\alpha\beta}+Y_{\alpha\beta^{-1}}-Y_{\alpha}Y_{\beta}$ with $\alpha, \beta\in\Gamma)$

One

can

show that $B(r)$ _is a _{finitely generated} $k$-algebra (see [4], Proposition 1.4.1).

Moreover, any representation $\rho$ : $\Gammaarrow SL_{2}(k)$ gives rise to

an

algebra morphism $\chi_{\rho}$ :

$B(\Gamma)arrow k$ by the formula $\chi_{\rho}(Y_{\gamma})=Trp(\gamma)$

.

This is a consequence of the famous trace

relation:

$rr_{r}(AB)+Tr(AB^{-1})=Tr(A)Tr(B) \forall A, B\in SL_{2}(k)$

The character of the $tauto\log_{\grave{1}}ca1$ representation $\rho$ : $\Gammaarrow SL_{2}(\mathcal{A}(\Gamma))$ defined by $\rho(\gamma)=$

$X^{\gamma}$ is a map $\Phi$ : $B(r)arrow A(\Gamma\rangle^{SL_{2}}$. The following theorem

as

been proved by Przytyscki

and Sikora,

see

[10]. In pract\’ice, it follows from [9] Theorem 2.6, see also [2] and [3].

Theorem 2.5. Forany

_field

$k$

of

characteristic $0$ and any finitely generatedgroup$\Gamma$, the

map $\Phi$ : _{$B(r)arrow A(\Gamma\rangle^{SL_{2}}$} is

an

isomorphism.

Thestatement of Theorem 2.6 in [9] is much

rrxore

general and deals with algebraswith

trace satisfying the Cayley-Hamilton identity. We derive our statement from his below.

Proof

Let $k[\Gamma]$ be the group algebra of $\Gamma$

. We denote by $[\gamma]$ the generator associated to

$\gamma\in r$andset$\mathfrak{R}([\gamma])=[\gamma]+[\gamma^{-1}]$ that

we

extend by$k$-linearity. We define $H(\Gamma)=k[\Gamma]/I$

where $I$ is the twesided ideal generated by the elements_$Tx(x)y$ -y&(x$\rangle$, for any

$x,$ $y\in$

$k[\Gamma]$

.

Thetrace factors to a $k$-linear endomorphismof $H(I^{\gamma})$

.

By direct computation, one

checks that the following identities hold for any $x,$$y\in H(\Gamma\rangle_{\backslash }.$

(i) $Tr(x)y=y$Tr(x) (ii) $\ulcorner fr(xy)=rfr(yx)$

(iii) $Tr(Tx(x)y)=\ulcorner fr(x)Tr(y)$

(iv) $x^{2}-r \ (x\rangle x+\frac{1}{2}(Tr(x)^{2}-Tr(x^{2}))=0.$

The last equation iscalled the Cayley-Hamilton identity of order 2. The map$j$ : $H\langle\Gamma$) $arrow$

$M_{2}(A(\Gamma))$ defined by $j([\gamma])=X^{\gamma}$ is

an

algebra morphism preserving the trace. It is

universal in the

sense

that for any $k$-algebra $B$ and morphism $j’$ : $H(\Gamma)arrow M_{2}(B)$, there

is aunique algebra morphism $\varphi$ :$A(\Gamma)arrow B$ such that the following diagramcommutes.

$H(\Gamma)arrow^{j}M_{2}(A(r^{t}))$

$\backslash ^{j’} \downarrow M_{2}(\varphi)$

$M_{2}(B)$

Denote by $G$ the group _{$GL_{2}(k)$}

.

The universal property implies that if

we

compose $j$

with

a

conjugation $\pi_{g}$ with $g\in G$, there is an automorphism $\varphi_{g}$ of $A(\Gamma)$ such that

$7r_{g}oj=M_{2}(\varphi_{g})oj$. The action of $G$

on

$A(\Gamma)$ is the one described in Subsection 2.1: the

formula $\rho(g)=\pi_{g}\circ M_{2}(\varphi_{g})^{-\lambda}$ defines an action of$G$ on $M_{2}(A(\Gamma))$ fixing $X^{\gamma}$ for all

$\gamma$ and

Theorem 2.6 of [9] says thefollowing:

To endthe proof,

we

observe that $\Phi$ is the restriction of_$j$ tothecenter. More precisely, the map $B(\Gamma)arrow H(\Gamma)$ defined by $Y_{\gamma}\mapsto Tr([\gamma])$ is injective and its image by $j$ is the

center of $M_{2}(A(\Gamma))^{G}$, that is $A(\Gamma)^{G}Id$

.

We end the proof by observing the equality

$A(\Gamma)^{G}=A(\Gamma)^{SL_{2}}.$ $\square$

Thanks to this theorem,

we can remove

the ‘s’ in $X_{s}(\Gamma)$

.

2.3. Irreducible, reducible and central characters. Given $\alpha,\beta\in\Gamma$,

we

define the

following element of$B(\Gamma)$:

$\Delta_{\alpha,\beta}=Y_{\alpha}^{2}+Y_{\beta}^{2}+Y_{\alpha\beta}^{2}-Y_{\alpha}Y_{\beta}Y_{\alpha\beta}-4.$

The equation $\Delta_{\alpha,\beta}\neq 0$ defines an open subset $U_{\alpha,\beta}$ of $X(\Gamma)$ and

we

set

$X^{irr}( \Gamma)=\bigcup_{\alpha,\beta\in I}, U_{\alpha,\beta}.$

We begin our study with the following lemma, a slightly different version of Lemma

1.2.1 in [4]. We will say that

a

representation $\rho$ : $\Gammaarrow SL_{2}(k)$ is absolutely irreducible if

it is irreducible in

an

algebraic closure of$k.$

Lemma 2.7. Given

a

_field

$k$ and a representation

$\rho$ : $\Gammaarrow SL_{2}(k)$, $\rho$ is absolutely

irreducible

_if

and only

_if

there exists $\alpha,$$\beta\in\Gamma$ such that $rb(\rho(\alpha\beta\alpha^{-1}\beta^{-1}))\neq 2.$

Proof.

We first recall Burnside irreducibility criterion which states that $p$ is absolutely

irreducible ifand only if$Span\{\rho(\gamma), \gamma\in\Gamma\}=M_{2}(k)$.

Let $M$ be the matrix defined by $M_{ij}=Tr(p(\gamma_{i}\gamma_{j}))$ for $i,j\in\{1$,

.

.

.

,4$\}$ where _{$\gamma_{1}=$}

$1,\gamma_{2}=\alpha,$$\gamma_{3}=\beta$ and $\gamma_{4}=\alpha\beta$

.

A simple computation shows that $\det M=-\chi_{\rho}(\Delta_{\alpha,\beta})^{2}$

where we have $\chi_{\rho}(\Delta_{\alpha,\beta})=Tr\rho(\alpha\beta\alpha^{-1}\beta^{-1})-2$. Hence if there exists $\alpha,$$\beta$ such that $\chi_{\rho}(\Delta_{\alpha,\beta})\neq 0$, then $(\rho(\gamma_{i}))_{i=1..4}$ is

a

basis of $M_{2}(k)$ and by Burnside criterion, $\rho$ is

abso-lutely irreducible.

Conversely, suppose that for all $\alpha,$$\beta\in\Gamma$ one has $\chi_{\rho}(\Delta_{\alpha,\beta})=$ O. Then there exists

a

non-zero

subspace $F_{\alpha,\beta}$ of $k^{2}$ fixed by the commutator $\rho([\alpha,$$\beta$ Suppose that there

exists $\gamma,$

$\delta$ such that _{$F_{\alpha,\beta}\cap F_{\gamma,\delta}=\{O\}$} then in

a

basis made of these two lines,

one can

write $\rho([\alpha, \beta])=(\begin{array}{ll}1 x0 1\end{array})$ and $\rho([\gamma, \delta])=(\begin{array}{ll}1 0y 1\end{array})$ with $x$ and $y$

non zero.

We compute

then $Tr([[\alpha, \beta], [\gamma, \delta =2+(xy)^{2}$

.

The hypothesis implies that $x$

or

$y$ is

zero

which is

impossible. Thisfinallyimplies that $\bigcap_{\alpha},{}_{\beta}F_{\alpha,\beta}\neq\{0\}$. But this subset is -invariant which

implies that $\rho$ is reducible.

$\square$

Consider the closed subset $X^{cen}(\Gamma)$ of $X(\Gamma)$ defined by the equations $Y_{\gamma}^{2}=4$ for all

$\gamma\in\Gamma.$ A $k$-point of$X^{cen}(\Gamma)$ has the form $\varphi(Y_{\gamma})=2\epsilon(\gamma)$ for

some

$\epsilon\in H^{1}(\Gamma, \mathbb{Z}/2\mathbb{Z})$

.

This

is the character ofthe central representation $\rho$ : $\gamma\mapsto\epsilon(\gamma)id.$

Theclosedsubset$X(\Gamma)\backslash X^{irr}(\Gamma)$ is denoted by$X^{red}(\Gamma)$: letusdescribe itmoreprecisely.

Let $Hom(\Gamma, G_{m})$ be the spectrum of the algebra

$C(\Gamma)=k[Z_{\gamma}, \gamma\in\Gamma]/(Z_{\gamma}Z_{\delta}-Z_{\gamma\delta}, \gamma, \delta\in\Gamma)$

We have for any $k$-algebra $R,$ $Hom_{k-alg}(C(\Gamma), R)\simeq Hom(\Gamma, R^{x})$, which explains the

notation.

Let $\sigma$ be the automorphism of $C(\Gamma)$ defined by $\sigma(Z_{\gamma})=Z_{\gamma^{-1}}$ and $\pi$ : $B(\Gamma)arrow C(\Gamma)$,

Proposition 2.8. The map $7r:B(\Gamma)arrow C(\Gamma)$ is

a

morphism

of

algebras whose image is

the $\sigma$-invariant part. $It_{\mathcal{S}}$

kernel is the radical

_of

the ideal generated by $\Delta_{\alpha,\beta}.$

In other words, a $k$-point

$\varphi$

of

$X^{red}$

lifts

to a $k$-point

of

$Hom(\Gamma, G_{m})/\sigma$

.

Hence, there

exists

an

extension $\hat{k}$

of

$k$ (at most quadratic) and a morphism $\psi$ : $\Gammaarrow\hat{k}^{X}$ such that

$\varphi(Y_{\gamma}\rangle=\psi(\gamma)+\psi(\gamma)^{-1}.$

Proof

Weprove easily that $\pi$ isahomomorphism whose imageis $C(\Gamma)^{\sigma}$. Let $k$ be

a

field

and $\varphi$ : $B(r)arrow k$ a morphism satisfying $\varphi(\Delta_{\alpha,\beta})=0$ for all $a,$$\beta\in\Gamma$

.

We write $\varphi(\alpha)=$

$\varphi(Y_{\alpha})$ for short: by assumption

one

has $\varphi([\alpha, \beta])=2$

.

Moreover, if $\varphi\langle\alpha$) _{$=\varphi(\beta)=2$}

then $\varphi(\Delta_{\alpha,\beta})=0$ implies $(\varphi(\alpha\beta)-2)^{2}=0$ and hence $\varphi(\alpha\beta)=2$

.

We conclude that

$\varphi(\alpha)=2$ for any $\alpha\in[\Gamma, \Gamma]$. Moreover, if $\alpha\in\Gamma$ and $\beta\in[\Gamma, \Gamma]$, then $\varphi(cx\beta)$ satisfies $(\varphi(\alpha)-\varphi(\alpha\beta))^{2}=0$and hence $\varphi(\alpha\beta)=\varphi(\alpha)$

.

This proves that$\varphi$ factors through

a

map

$\varphi^{ab}$ : _{$B(\Gamma^{ab})arrow k$} where $\Gamma^{ab}=\Gamma/[\Gamma, \Gamma]$

.

Moreover, we clearly have $C(\Gamma^{ab})=C(\Gamma)$ and

the map $B(\Gamma^{at\}})arrow c(r^{a\})})^{\sigma}$ is invertible. This proves the first part of the lemma. The

second part follows. $\square$

Remark

2.9.

We don’t know if the $\Delta_{\alpha,\beta}s$ generate the kernel of$7\ulcorner$

.

Moreover, it is well known that the deformationsofreducible representations into irreducible

ones

are related to the Alexander module. It would be interesting to have

an

explicit description of the (co-)normal bundle of$X^{red}(I)$ in $X(\Gamma)$

.

More precisely, definethe ideal $I$bythefollowing

exact sequence:

$0arrow Iarrow B(f^{\backslash })arrow C\langle\Gamma)^{\sigma}arrow 0$

Then there should be a relation between the $c(r)$-module $I/I^{2}\otimes_{C(\Gamma)^{\sigma}}C(\Gamma)$ and the

module $H^{1}く\Gamma,$$c(r)$) where the action of$\gamma\in\Gamma$ is by multiplication by $Z_{\gamma}.$

2.4. Functorial properties. If $\varphi$ :

$\Gammaarrow\Gamma’$ is agroup homomorphism, there is anatural

map $\varphi_{*}:B(\Gamma)arrow B(\Gamma’)$ mapping $Y_{\gamma}$ to $Y_{\varphi(\gamma\rangle}$

.

There is also

a

natural map $\varphi_{*}:A(r)arrow$ $A(\Gamma‘)$ which makes the followingdiagram commute:

$B(\Gamma)arrow^{\varphi*}\mathcal{B}(\Gamma’)$

$\downarrow\Phi \}\Phi$

$A(\Gamma)\underline{\varphi_{*}}A(\Gamma’)$

$2.4.\lambda$

.

Double quotients. Let

us

show an example which arises when computing the

fun-damental group ofa 3 manifold presented by

a

Heegard splitting.

Proposition 2.10. Let$\Gamma,$$\Gamma_{\lambda},$$\Gamma_{2}$ be threefinitely generated groups and$\varphi_{i}:\Gammaarrow r_{i}$ be two

surjective morphisms. Denote by$\Gamma’$ the

amalgamatedproduct$r’=\Gamma_{1}\star\Gamma_{2}\Gamma^{\cdot}$

Then $x(r)$ is isomorphic to the

_fiber

product $X(I_{1}^{\gamma})\cross X(\Gamma_{2})$.

$x(r)$

Moregeometrically, ifwedenote by $L_{i}$ the images of$X(\Gamma_{i})$ in$x(r)$, then $X(\Gamma’)$ isthe

Proof.

Given any morphisms $\varphi_{i}$ : $\Gammaarrow\Gamma_{i}$ and any $k$-algebra $R$,

one

has the following

natural isomorphisms:

$Hom_{k-alg}(A(\Gamma’), R) = Hom(\Gamma’, SL_{2}(R))$

$= \{\rho_{i}:\Gamma_{i}arrow SL_{2}(R), \rho_{1}\circ\varphi_{1}=\rho_{2}0\varphi_{2}\}$

$= Hom_{k-alg}(A(\Gamma_{1})\bigotimes_{A(\Gamma)}A(\Gamma_{2}), R)$

Hence we have the isomorphism $A( \Gamma’)=A(\Gamma_{1})\bigotimes_{A(\Gamma)}A(\Gamma_{2})$

.

We notice that the invariant

part of this tensorproduct isnot thetensorproduct of the invariant parts (think about

a

wedge of two circles). However, this holds when$A(\Gamma_{i})$ is aquotient of$A(\Gamma)$, that iswhen

the morphisms $\varphi_{i}$ are surjective.

Denotingby$I_{i}$the kernelof the map$A(\Gamma)arrow A(\Gamma_{i})$, the result follows from the following

standard fact in invariant theory: $A(\Gamma’)=A(\Gamma)/(I_{1}+I_{2})$ and $(I_{1}+I_{2})^{SL_{2}}=I_{1}^{SL_{2}}+I_{2}^{SL_{2}}.$

On the other hand,

one

shows easily that the map $B(\Gamma_{1})\otimes_{B(\Gamma)}B(\Gamma_{2})arrow B(\Gamma’)$ is

an

isomorphism, see for instance [10]. $\square$

2.4.2. Semi-directproducts. Thefollowing situation appears for fundamental groups of

3-manifolds which fiber over the circle. Let $\Gamma$

be afinitely generated group and $\varphi\in Aut(\Gamma)$

an automorphism. Set $\Gamma’=\Gamma x_{\varphi}\mathbb{Z}$ so that we have the following (split) exact sequence.

Let $t=s(1)$

.

$0arrow\Gammaarrow^{\alpha}\Gamma^{\prime_{arrow}}\mathbb{Z}-0\wedge s$

Lemma 2.11. Suppose that $k$ is algebraically closed and consider only closed points

of

the character varieties. Then the map $\alpha^{*}:X(\Gamma’)arrow X(\Gamma)\omega$restricted to $X^{irr}(\Gamma)$ is a

$\mathbb{Z}/2\mathbb{Z}$ principal bundle over $F$, where $F=\{x\in X^{irr}(\Gamma), \varphi^{*}x=x\}.$

Proof.

Let $\rho’$ : _{$\Gamma’arrow SL_{2}(K)$} be a representation which restricts to an irreducible repre-sentation $\rho$ : $\Gammaarrow SL_{2}(K)$. This representation satisfies $\rho’(t)\rho(\gamma)\rho’(t)^{-1}=\rho(\varphi(\gamma))$ hence

the character$\chi_{\rho}$ is fixed by

$\varphi^{*}$. Suppose

we

have another representation$\rho’$ : $\Gamma’arrow SL_{2}(K)$

withthe

same

restriction, then $\rho"(t)\rho’(t)^{-1}$ commuteswith the imageof$\rho$

.

Hence$\rho"(t)=$ $\pm\rho’(t)$

.

There

are

precisely two preimages for $\rho$ and the lemmais proved.

$\square$

3. APPLICATIONS OF SAITO’S THEOREM

We know from the isomorphism between $X_{s}(\Gamma)$ and $X(\Gamma)$ that if $k$ is algebraically

closed, any $k$-point of$X_{s}(\Gamma)$ is thecharacter ofarepresentation $\rho:\Gammaarrow SL_{2}(k)$

.

However

this fact is non-trivial to prove directly: the following theorem of [11] allows it and has

further applications.

Theorem 3.1. Let $R$ be a $k$-algebra and

$\varphi$ : $B(\Gamma)arrow R$ be a morphism

of

$k$-algebras.

Suppose that there exists $\alpha,$$\beta\in\Gamma$ such that $\varphi(\Delta_{\alpha,\beta})$ is invertible and $A,$$B\in SL_{2}(R)$

such that Tr$A=\varphi(Y_{\alpha})$,$Tr(B)=\varphi(Y_{\beta})$,$Tx(AB)=\varphi(Y_{\alpha\beta})$. Then, there is a unique

representation$\rho$ : $\Gammaarrow SL_{2}(R)$ such that$\rho(\alpha)=A,$$\rho(\beta)=B$ and

for

all$\gamma\in\Gamma,$$Tr(\rho(\gamma))=$

$\varphi(Y_{\gamma})$.

Proof.

(Sketch) Define $\gamma_{i}$ and $M$

as

in Lemma 2.7. As $\det M=-\varphi(\triangle_{\alpha,\beta})^{2}$, the matrix $M$ isinvertibleover$R$. Givenany$\gamma\in\Gamma$, wedenote by $C_{\gamma}$ the vectorofcoefficients of$\rho(\gamma)$

in the basis $(\rho(\gamma_{\mathfrak{i}}))_{i=1..4}$ and by $T_{\gamma}$ the vector $(\varphi(Y_{\gamma\gamma_{i}}))_{i=1..4}$

.

We then have $MC_{\gamma}=T_{\gamma}$

this actually defines a representation in $SL_{2}$ whose traces are prescribed by $\varphi$

.

This is a

non-trivial consequenceof the trace relations, we refer to [11] for a proof. $\square$

3.1.

Points of$X(\Gamma)$ over arbitrary fields.

Definition 3.2. Let $k$ be a field and $p:\Gammaarrow SL_{2}(\hat{k})$ be a representation where $\hat{k}$

is

an

extension of $k$ and such that Tr$p(\gamma)\in k$ for all $\gamma\in\Gamma$

.

We set

$M(\rho)=Span_{k}\{\rho(\gamma),\gamma\in\Gamma\}.$

It is

a

sub-k-algebra of $M_{2}\langle\hat{k}\rangle$

.

We will say that two representations

$\rho_{i}$ : $rarrow SL_{2}(\hat{k}_{\eta}\cdot)$

where $i=1$,2

are

equivalent if there is

an

isomorphism of$k$-algebras$\sigma$ : $M(\rho_{1})arrow M(\rho_{2})$

such that $p_{2}=\sigma 0\rho_{1}.$

We have the following proposition:

Proposition 3.3. Let $k$ be a

fietd.

There is

a

functorial

isomorphism between $k$-points

of

$X^{irr}(\Gamma\rangle\wedge$ and equivalence classes

of

absolutely irreducible representations $p:\Gammaarrow SL_{2}(\hat{k})$ where $k$ is a

finite

extension

of

$k$ and

for

all

$\gamma$ in $\Gamma,$ $Tr(p(\gamma))\in k.$

Proof of

Proposition 3.3. Given$\rho:\Gammaarrow SL_{2}(\hat{k})$ asin the statement,

we

define amorphism

$\varphi\in Hom_{alg}(B(\Gamma), k)$ by setting $\varphi(Y_{\gamma})=Tr(p(\gamma))$

.

By Lemma 2.7, there exists $\alpha,$$\beta\in\Gamma$

such that $\chi_{\rho}(\triangle_{\alpha,\beta})\neq 0$

.

Hence, $\varphi$ indeed corresponds to a $k$-point of$X^{irr}(\Gamma)$

.

Onthe otherhand,

a

$k$-pointof$X^{irr}(\Gamma)$ correspondsto

a

morphism$\varphi$in$Hom_{alg}(B(\Gamma), k)$

such that there exists $\alpha,$$\beta\in r$ with $\varphi(\triangle_{\alpha,\beta}\rangle\neq 0.$

We define the matrices $A=$ $(^{\varphi(Y_{\alpha})}1$ $-0^{1})$ and $B=(_{u}^{0}$ $\varphi(Y_{\beta})-1/u)$ where $u$ belongs to

anextension $\hat{k}$

of$k$ such that theequation$u^{2}+u\varphi(Y_{\alpha\beta})+1=0$ holds $(\hat{k}$can

bechosen at

most quadratic). By Theorem 3.1, there is a unique representation $\rho$ : $\Gammaarrow SL_{2}(\hat{k})$ such

that $p(\alpha)=A,$$\rho(\beta)=B$ and $Tr(\rho(\gamma))=\varphi(Y_{\gamma})$ for all $\gamma\in\Gamma.$

In ordertoshow that this representation doesnot depend on the choiceof$\alpha$and $\beta$, we

observe that ifwe make

an

other choice for $A’,$$B’\in M_{2}(\hat{k}’)$ whichdefines arepresentation

$\rho’$ : $\Gammaarrow SL_{2}(\hat{k}^{;})$, the map defined by _{$\sigma(A)=A’$} and _{$\sigma(B)=B’$} extends to a

k-isomorphismfrom$M(\rho)$ to$M(p’\rangle$ showingthatthetworepresentations

are

equivalent. $\square$

Thealgebra $M(\rho)$ associated toanabsolutely irreducible representation$\rho$ issimple and

central and hence defines

a

class in the Brauer group $Br(k)$

.

_Moreover, $\dim_{k}M(\rho)=4.$

It follows that $M(p)$ is a quaternion algebra and in particular, its square is $0$ in $Br(K)$

.

The set of$k$-points of$X^{irr}(\Gamma)$ has then a partition into Brauer classes. Weget finally the

following proposition:

Proposition 3.4. Let $k$ be a

field

with $Br(k)=0$ (this occurs

_for

instance

_if

$ki\mathcal{S}$

alge-braically closed or

_of

transcendence degree

one

over

an

algebraically closed field). Then

$k$-points in $X^{\dot{I}X}(\Gamma)$ correspond bijectively to conjugacy classes

of

absolutely irreducible

representations $\rho\prime\Gammaarrow SL_{2}(k)$

.

Proof

Anirreduciblepoint $\varphi$ : $B(\Gamma)arrow k$corresponds tothe character of

a

representation $\rho:\Gammaarrow SL_{2}(\hat{k})$for

some

extension$\hat{k}$

of$k$

.

As theBrauergroupis trivial, thereis

an

algebra

isomorphism $\sigma$ : $M_{2}(k)arrow M(\rho)$

.

Tensoring by

$\hat{k}$

, we find that $\sigma\otimes\lambda$ : $M_{2}(\hat{k})arrow M(\rho)\otimes\hat{k}$

is a $\hat{k}$

-linear automorphism of $M_{2}(\hat{k})$ and hence a conjugation by some $g\in GL_{2}(\hat{k})$ by

in $k$

.

As two such representation $\rho$ and $\rho’$

are

equivalent, it

means

that there is

an

automorphism$\sigma$of$M_{2}(k)$ suchthat$\rho’=\sigma 0\rho$and this automorphismisaconjugation. $\square$

Example 3.5. Let $F_{2}$ be the free

group

on

the generators $\alpha$and $\beta$

.

It iswell-known that

the map $\Psi$ : _{$\mathbb{Q}[x, y, z]arrow B(F_{2})$} defined by _{$\Psi(x)=Y_{\alpha},$ $\Psi(y)=Y_{\beta}$} and _{$\Psi(z)=Y_{\alpha\beta}$} is

an

isomorphism. Hence,

one

has

a

rational point of$X(F_{2})$ by sending$x,$ $y,$$z$ to 1. However,

there is

no

representation $\rho$ : $F_{2}arrow SL_{2}(\mathbb{Q})$ with this character. However, one

can

find

a

solution in

some

(non uniquely defined) quadratic extension of $\mathbb{Q}.$

3.2. Tangent spaces. Let $k$ be a field and

$\rho$ : $\Gammaarrow SL_{2}(k)$ be an absolutely irreducible

representation. The character $\chi_{\rho}$ may be

seen

as a

$k$-point of $X^{irr}(\Gamma)$

.

We derive from

Saito’s Theorem

a

simple proof of the following well-known result:

Proposition 3.6. There is a natural isomorphism$T_{\chi_{\rho}}X_{k}(\Gamma)\simeq H^{1}(\Gamma, Ad_{\rho})$ where$Ad_{\rho}$ is

the adjoint representation

_of

$\Gamma$ on_{$sl_{2}(k)$} induced by

$\rho.$

Proof.

It is well-known that

a

tangent vector at $\chi_{\rho}$ corresponds to an algebra morphism

$\varphi_{\epsilon}$ : $B(\Gamma)arrow k[\epsilon]/(\epsilon^{2})$ such that $\varphi_{0}=\chi_{\rho}.$

As$\rho$is absolutelyirreducible,there exists $\alpha,$$\beta\in\Gamma$such that $\varphi(\Delta_{\alpha,\beta})\neq 0$

.

Considerthe

map $\pi$ : $SL_{2}(k)\cross SL_{2}(k)arrow k^{3}$ definedby $\pi(A, B)=(Tr(A), Tx(B), Tx(AB))$

.

Lemma 3.7

below shows that this mapisasubmersion precisely

on

the preimageof the open set of$k^{3}$ defined by $\Delta_{\alpha,\beta}\neq 0$

.

This proves that we

can

find two matrices $A_{\epsilon},$$B_{e}\in SL_{2}(k[\epsilon]/(\epsilon^{2}))$

such that $Tx(A_{\epsilon})=\varphi_{\epsilon}(Y_{\alpha})$,$Tr(B_{\epsilon})=\varphi_{\epsilon}(Y_{\beta})$ and $b(A_{\epsilon}B_{\epsilon})=\varphi_{\epsilon}(Y_{\alpha\beta})$. By Theorem 3.1,

thereis

a

unique representation $\rho_{\epsilon}$ : $\Gammaarrow SL_{2}(k[\epsilon]/(\epsilon^{2}))$ such that$\rho_{\epsilon}(\alpha)=A_{\epsilon},$ $\rho_{e}(\beta)=B_{\epsilon}$

and $\chi_{\rho_{\mathcal{E}}}=\varphi_{\epsilon}.$

We define the cocyle $\psi\in Z^{1}(\Gamma, Ad_{\rho})$ by the formula

(1) $\psi(\gamma)=\frac{d}{d\epsilon}|_{\epsilon=0}\rho(\gamma)^{-1}\rho_{\epsilon}(\gamma)$.

The map $\varphi_{\epsilon}\mapsto\psi$ is

a

linear map$T_{\chi_{\rho}}X_{k}(\Gamma)arrow H^{1}(\Gamma, Ad_{\rho})$

.

Weconstruct the inverse map

by sending the cocycle $\psi$ to the character ofthe representation $\rho_{\epsilon}(\gamma)=\rho(\gamma)(1+\epsilon\psi(\gamma))$

.

Ifwehad chosen other matrices$A_{\epsilon}’,$$B_{\epsilon}’$ defining

a

representation $\rho_{\epsilon}’$, then

as

in the proof

of Proposition 3.3, we would have found an automorphism $\sigma$ of $M_{2}(k[\epsilon](\epsilon^{2}))$ such that $\rho’=\sigma 0\rho$

.

Moreover$\rho_{\epsilon}$ and

$\rho_{\epsilon}’$ coincide up tofirst orderso thatone can write$\sigma=Id+\epsilon D$

for some derivation $D$ of _{$M_{2}(k)$}. All such derivations have the form _{$D(X)=[\xi, X]$} for

some

$\xi\in sl_{2}(k)$

.

A computation shows that $\psi’=\psi+d\xi$ and the map $\varphi_{\epsilon}arrow\psi$ is well-defined. Reciprocally, changing $\psi$ to $\psi+d\xi$ amounts in conjugating$\rho_{\epsilon}$ by

$e^{\epsilon\xi}$

, which does not change the character, hence the theorem is proved. $\square$

Lemma 3.7. The map $\pi$ : $SL_{2}(k)^{2}arrow k^{3}$

defined

by the

formula

$\pi(A, B)=(Tr(A), Tr(B), \ulcorner fr\langle AB))$

is a submersion at any pair$(A, B)$ such that Tr$[A, B]\neq 2.$

Proof.

Let $\xi,$

$\eta$ be in $sl_{2}(k)$ corresponding to

a

tangent vector $(A\xi, B\eta)\in T_{(A},{}_{B)}SL_{2}(k)^{2}.$

If $D_{(A,B)}\pi$ is not surjective, there exists $u,$$v,$$w\in k$ such that $uTr(A\xi)+vTx(B\eta)+$

$wTr(A\xi B+AB\eta)=0$ for all $\xi,$$\eta\in sl_{2}(k)$. This is possible if and only if there exists

$\lambda,$_{$\mu\in k$} such that $uA+wBA=\lambda Id$ and $vB+wAB=\mu Id$

.

Hence, there is a linear

relation in the family $(Id, A, B, AB)$_{, which is equivalent to} the equality $\mathfrak{t}R([A, B])=2,$

One has to be careful that the isomorphism between $T_{\chi_{p}}X(\Gamma\rangle and H^{1}(f, Ad_{\rho})$ does

not longer hold if$\rho$ is not absolutely irreducible. However there is still

a

natural map

: $H^{1}(\Gamma, Ad_{\rho})arrow T_{\chi_{p}}X(\Gamma)$ sending $\psi$ to the character of$\rho(1+\epsilon\psi)$

.

Luckily, the skein

interpretation of the character variety still allows to compute the tangent space at these points. Let

us

consider

an

extreme case, that is the tangent space at the character ofthe

trivial representation. One has $H^{1}(\Gamma, Ad_{\rho})=H^{\lambda}(\Gamma, k)\otimes sl_{2}(k)$ and the map vanishes.

Proposition3.8. Foranyfinitelygeneratedgroup$r$ and

field

$k$, thetangentspace$T_{1}X(\Gamma)$

at the character

_of

the trivial representation $\dot{u}$ isomorphic to the space

of functions

$\psi$ :

$\Gammaarrow k$ satisfying the following quadratic

functional

equation (see [6]):

$\forall\gamma, \delta\in\Gamma, \psi(\gamma\delta)+\psi(\gamma^{-1}\delta)=2\psi(\gamma)+2\psi(\delta)$

Proof.

Itis sufficient to write down the conditionsfor $\varphi_{\epsilon}$

:

$Y_{\gamma}\mapsto 2+\epsilon\psi(\gamma)$ to be

an

algebra

morphism from $B(\Gamma)$ to $k[\epsilon]/(\epsilon^{2})$

.

$\square$

Observethat the map $\mathscr{S}H^{1}(\Gamma, k)arrow\tau_{1}x(r)$ mapping$\lambda\mu$to the map$\psi$ : $\gamma\mapsto\lambda(\gamma\rangle\mu(\gamma)$

is injective but not necessarilysurjective.

3.3.

Tautological representations. Let$\Gamma$

beafinitelygeneratedgroupand$k$beafield.

Consider

an

irreducible component $Y$ of$X(\Gamma)$

.

We will say that $Y$ isof irreducibletype

if it contains the character of an absolutely irreducible representation. More formally,

this component corresponds toa minimal prime $\mathfrak{p}\subseteq B(\Gamma\rangle$: we will write $k|Y$] $=B(\Gamma)/\mathfrak{p}.$

The component $Y$ will be of irreducible type if and only if the tautological character

$\chi_{Y}$ : $B(f^{I})arrow k[Y]$ is irreducible.

From Proposition 3.3, there exists

an

extension $K$ (at most quadratic) of the field

$k(Y)$ _and a _{representation} $p_{Y}$ : $rarrow SL_{2}(K)$ such that $\chi_{\rho Y}=\chi_{Y}$

.

We will call such a

representation atautological representation.

Remark 3.9. The Brauer group of

a

transcendance degree

one

extension of

an

alge-braically closed field vanishes. Hence, if $k$ is algebraically closed and $\dim(Y)=1$,

one

can choose $K=\mathbb{C}(Y)$. This

case

will appear very oftenif $\Gamma$ is the fundamental group of

a

knot complement.

Remark 3.10. If the component $Y$ is of reducible type, the tautological representation

still exists and

can

be constructed directly. Moreover, it will be convenient to replace

$k(Y)$ _by

an

extension for more natural formulas. For $instance_{\}}$ if Iab $=\mathbb{Z}$and $\varphi$ :

$\Gammaarrow \mathbb{Z}$ is the abelianization morphism, then

one

component of$X(\Gamma)$ will be ofreducible type.

Moreover, it will be isomorphic to $A^{1}$ via the map

$\varphi_{*}$ : $B(\Gamma)arrow B(\mathbb{Z})\simeq k[Y_{1}].$ $A$

tautological representation will be given by

$\rho_{Y}:\Gammaarrow SL_{2}(k\langle t)) , \rho_{Y}(\gamma)=(\begin{array}{ll}t^{\varphi(\gamma)} 00 t^{-\varphi(\gamma)}\end{array})$

and the extension $k(Y_{1})\subset k(t)$ corresponds to thesubstitution $Y_{1}=t+i^{-1}.$

3.4. Valuation rings and Culler-Shalen theory. We give in this section an example ofresult ofCuller-Shalentheory (thoughnot explicitlystated in theirarticles). The proof

will follow standard lines for what concerns $Culle\triangleright$Shalen theory. We add it in order to

convincethe readerthat thetechniquesof thepreceding sectionarewell-adaptedto these questions.

Let $M$be

a

3-manifold by which

we mean

acompactoriented and connected topological

- $S$ is not

a

2-sphere,

- the map $\pi_{1}(S)arrow\pi_{1}(M)$ induced by the inclusion is injective, $-S$ _{is not boundary parallel.}

We will say that $M$ is small if it does not contain any incompressible surface without

boundary.

Given a topological space $Y$ with finitely many connected components $(Y_{i})_{i}\in I$ each

of them having a finitely generated fundamental group, we set $X(Y)= \prod_{i\in I}X(\pi_{1}(Y_{i}))$

.

Proposition 3.11. Let $M$ be a small

3-manifold.

Then the restriction map$r:X(M)arrow$

$X(\partial M)$ is proper.

Proof.

Fix

a

base point $y$ in $M$ and let $I=\pi_{0}(\partial M)$

.

We choose a base point $y_{i}$ for each component $\partial_{i}M$ of $\partial M$. Set _{$\Gamma=\pi_{1}(M, y)$} and $\Gamma_{i}=\pi_{1}(\partial_{i}M, y_{i})$. As $X(\partial M)=$

$\prod_{i\in I}X(\Gamma_{i})$, it is sufficient to define the map $r_{i}:X(\Gamma)arrow X(\Gamma_{i})$ for any $i\in I$

.

This map

is the

one

induced by the inclusion $\partial_{i}M\subset M$

on

fundamental groups.

By the valuative criterion ofproperness (see Thm4.7in [7]), $r$isproperif and only iffor

any discrete valuation $k$-algebra $R$ with fraction field $K$ fitting in the following diagram

there is

a

morphism $Spec(R)arrow X(M)$ which makes the diagram commute.

(2) $Spec(K)arrow^{\varphi}X(M)$

$Spec(R)\downarrowarrow X(\partial M)\prime|r$

We will show that if such

a

morphism does not exist, then there is

an

incompressible

surface by standard Culler-Shalen arguments. First, by the preceding section and

re-marking that $K$ has transcendence degree 1 over $k$, the morphism $\varphi$ corresponds to a

representation $\rho$ : $\Gammaarrow SL_{2}(K)$. Let $T$ be the Bass-Serre tree on which $SL_{2}(K)$ acts.

Through $\rho$, the group $\Gamma$

acts on $T$

.

If this action is trivial- i.e. it fixes a vertex of

T-the representation $\rho$ is conjugate to

a

representation in $SL_{2}(R)$

.

For any $\gamma\in\Gamma$,

we

have

Tr$\rho(\gamma)\in R$

.

This shows that the map $B(\Gamma)arrow R$ defined by $Y_{\gamma}\mapsto Tr\rho(\gamma)$ corresponds

to a map$Spec(R)arrow X(M)$ which yieldsacontradiction. Hence, the action isnon-trivial

and is dual to a non-empty essential surface $\Sigma\subset M.$

For all $i\in I$, denote by $\rho_{i}:\Gamma_{i}arrow SL_{2}(K)$ the representation induced by $\rho$

.

From the

diagram (2), we get that for all $\gamma\in\Gamma_{i}$, Tr$\rho_{i}(\gamma)\in R.$

Let us prove that $\gamma$ fixes a vertex in$T$

.

If$\rho_{i}(\gamma)=\pm 1$, then it is trivial. In the opposite

case, one can find a vector $v\in K^{2}$ such that $v$ and $\rho_{i}(\gamma)v$ form

a

basis. The matrix of

$\rho_{i}(\gamma\rangle$ in that basis is $(\begin{array}{ll}0 -11 b(\rho_{\iota’}(\gamma))\end{array})$. This proves that $\rho_{i}(\gamma)$ is conjugate to an element

in $SL_{2}(R)$ and hence fixes a vertex in $T$. A standard lemma (Corollaire 3 p90 in [12])

shows that ifevery $\gamma\in\Gamma_{i}$ fixes a vertex in $T$, then $\Gamma_{i}$ itself fixes a vertex in$T.$

The relative construction ofdual surfaces shows that one can find an essential surface

$\Sigma$

dual to the action without boundary, see Corollary 6.0.1 in [13]. Any component of $\Sigma$

is an incompressible surface. This is not possible

as

$M$ is small. $\square$

4. THE REIDEMEISTER TORSION AS A RATIONAL VOLUME FORM

In this section, we construct the Reidemeister torsion of a 3-manifold with boundary

an element of$\Lambda^{d}\Omega_{k(Y)/k}^{1}=\Omega_{k(Y)/k}^{d}$ where $d=\dim Y$

.

Some assumptions

are

necessary for

this construction to work, and these

are

given in Subsection 4.2. We will end this section with examples.

4.1. The adjoint representation. Let be

a

finitely generated group, $k$ a field and

$\varphi$ : $B(\Gamma)arrow k$ an irreducible character. Then, we

now

from Proposition 3.3 that there

exists

an

extension $\hat{k}$

of $k$ and a representation

$\rho$ : $\Gammaarrow SL_{2}(\hat{k})$ such that $\chi_{\rho}=\varphi.$

Moreover, the subalgebra $M(\rho)=Span\{\rho(\gamma), \gamma\in r\}$ _satisfies $M(\rho)\otimes_{k}\hat{k}\simeq M_{2}(\hat{k})$ and

depends only

on

$\varphi$ up to isomorphism.

Recallthe splitting$M_{2}(\hat{k})=kId\oplus s1_{2}(\hat{k})$ and for any$X\in M_{2}(\hat{k})$, denote by_{$X_{0}=X-$}

$\frac{1}{I^{2}}\mathfrak{B}(X)Idthep$$etM(\rho)_{0}=S.pan_{k}\{p(\gamma)_{0}, \gamma\in\Gamma\}tisa3-$rojection o$nthes$econd factor.Then s

dinxensional k$-$vector space o$nw$hich $\Gamma$ acts b

$yc$onjugation W$ewi11$ denote b$y$

$Ad_{\varphi}$ this $\Gamma$

-module and call it the adjoint representation. The relation between$Ad_{\varphi}$ and

$Ad_{\rho}$ is the following:

$Ad_{p}=Ad_{\varphi}\otimes_{k}\hat{k}.$

The representation $Ad_{\varphi}$ is

more

natural

as

it depends only

on

the character.

The Killing form is $an$ invariant non-degenerate pairing on $M(\rho)_{0}$ and there is

an

invariant volume form $\epsilon\in\Lambda^{3}(M\langle\rho)_{0})^{*}$ given by $\epsilon(\zeta, \eta, \theta)=Tr(\zeta[\eta,$$\theta$

Proposition 4.1. Let $Y$ be an irreducible component

of

$B(\Gamma)$

of

irreducible type

corre-sponding to a minimal prime ideal$\mathfrak{p}$. Let $Ad_{y}$ be the adjoint representation associatedto

the tautological character $\chi_{Y}$

.

There is an isomorphism between $\Omega_{B(\Gamma)/k}^{1}\otimes_{B(\Gamma)}k(Y)$ and

$H_{i}(\Gamma, Ad_{Y})$. This provides the following exact sequence:

$\mathfrak{p}/\mathfrak{p}^{2_{\otimes_{k[Y]}}}k(Y)arrow H_{1}(\Gamma, Ad_{Y})arrow\Omega_{k(Y)/k}^{1}arrow 0.$

Proof.

Thisfollows by duality from Proposition3.6with thetwist that $Ad_{Y}$ isnot exactly $Ad_{\rho_{Y}}$ which takes values in $SL_{2}(\hat{K})$ for

some

extension

$\hat{K}$

of $k(Y)$

.

Instead of adapting

the previous proof, we describe directly this dual picture.

Recall that $H_{1}(\Gamma, Ad_{Y})$ is thefirst homology ofacomplex$C_{*}(\Gamma, Ad_{Y})$ where$C_{k}(\Gamma, Ad_{Y})$

is generated by elements of the form $\xi\otimes[\gamma_{1}, ..., \gamma_{k}]$ with $\xi\in Ad_{Y}$ and $\gamma_{1}$,

.

. . ,$\gamma_{k}\in\Gamma.$

We have the formulas $\partial\xi\otimes[\gamma]=\rho(\gamma)^{-1}\xi\rho(\gamma)-\xi$ and $\partial\xi\otimes[\gamma, \delta]=\rho(\gamma)^{-1}\xi p(\gamma)\otimes[\delta]-$ $\xi\otimes[\gamma\delta]+\xi\otimes[\gamma]$

.

This allows to check directly that the map $dY_{\gamma}\mapsto p(\gamma)_{0}\otimes[\gamma]$ induces

a well-defined map of $B(\Gamma)$-modules $\Omega_{B(r)/k}^{1}arrow H_{1}(\Gamma, Ad_{Y})$ and hence

a

map

$\Psi:\Omega_{B(\Gamma)/k}^{1}\otimes k(Y)arrow H_{1}(r, Ad_{y})$

.

Reciprocally, let $\Lambda=k(Y)\oplus\epsilon(\Omega_{B(\Gamma)/k}^{1}\otimes k(Y))$ be the $k(Y)$-algebra with $e^{2}=0$

.

_The

map $\varphi$ : $B(\Gamma)arrow$ A given by$Y_{\gamma}arrow Y_{\gamma}+\epsilon dY_{\gamma}$ isan algebra morphism. By Lemma 3.7, one

can

fin\‘a $A_{\epsilon},$$B_{\epsilon}\in SL_{2}(\Lambda)$ whose traces

are

prescribed and by Saito’s theorem (using the

irreducibility of $\chi_{y}$),

one

finds

a

representation $\rho_{\epsilon}$ : $\Gammaarrow SI_{J}2(\Lambda)$ satisfying $\chi_{p_{\’{e}}}=\varphi.$ $A$

simplecheckshowsthatthemap$\xi\otimes\gamma\mapsto\frac{d}{\ } Tr(\xi\rho(\gamma)^{-1}p_{\epsilon}(\gamma))$ induces precisely the inverse

of $\Psi$

.

The exact sequence of the proposition follows from standard facts ofcommutative

algebra. $\square$

4.2. Characters of 3-manifolds. Let $M$ be a connected 3-manifold whose boundary

is

non

empty and has the following decomposition into connected components: $\partial M=$ $\bigcup_{i\in I}\partial_{i}M$

.

Let $g_{i}$ be the genus of $\partial_{i}M$ :

we

suppose that $g_{i}>0$ for all $i\in I$ and set

Denote by $\Gamma$ the fundamental group of _$M$

and let $Y$ be

an

irreducible component of

X.

($\Gamma$)

.

We denoteby$\chi_{y}$ : $B(\Gamma)arrow k[Y]$ the tautological character andby$Ad_{Y}$ the adjoint

representation. The followingtechnical lemma will be useful in the sequel.

Lemma4.2. Let$\alpha,\beta$ be two elements

of

$\Gamma$ suchthat

$\Delta_{\alpha,\beta}$ isnon

zero

in$k[Y]$_, Then, there

is

a

$k(Y)$-basis

of

$Ad_{Y}$ such that the adjoint representation has

coeficients

in $k[Y][ \frac{1}{\Delta_{\alpha,\beta}}].$

Proof.

Set $R=k[Y|[\Delta_{\alpha,\beta}^{-1}]$ and consider the composition $\chi_{Y}$ : $B(\Gamma)arrow k[Y]arrow R$

.

One

can find an extension $\hat{R}$

with $A,$ $B\in SL_{2}(\hat{R})$ and by Saito’s theorem,

a

unique

repre-sentation $\rho_{\alpha,\beta}$ :

$\Gammaarrow SL_{2}(\hat{R})$ with character

$\chi_{Y}$ satisfying $p(\alpha)=A$ and $p(\beta)=B$

.

By

construction, $p(\gamma)$ lies in _{$Span_{R}\{1, A, B, AB\}$}

.

The basis $\mathcal{A}_{0},$$B_{0},$ $(AB)_{0}$ of $M(\rho_{\alpha,\beta})_{0}$

sat-isfies the assumption of the lemma and one has $M(\rho)_{0}\simeq M(\rho_{\alpha,\beta})_{0}\simeq Ad_{Y}$. This proves

thelemma. $\square$

4.2.1. Computing $H^{1}.$

Definition 4.3. Recall that

we

say that

a

representation $\rho$ : $\Gammaarrow SL_{2}(k)$ belongs to $Y$

if its character $\chi_{\rho}$ : $B(\Gamma)arrow k$ factorizes through $k[Y]$

.

Equivalently,

we

will say that $Y$

contains the representation $\rho$. For instance,

a

representation is of irreducible type if it

contains

an

irreducible character.

A representation $\rho$ : $\Gammaarrow SL_{2}(k)$ is said regular if $H^{1}(\Gamma, Ad_{\rho})$ has dimension $d$

.

An

irreduciblecomponent$Y$of$X(\Gamma)$ will besaid regularif it containsaregular representation.

Proposition 4.4. An irreducible component $Y$

of

$X(\Gamma)$

of

irreducible type $\dot{u}$ regular

if

and only

_if

$\dim H^{1}(\Gamma, Ad_{Y})=d.$

Proof.

We start with a standard argument of Poincar\’e duality, taken from [8]. Let $\rho$ :

$\Gammaarrow SL_{2}(k)$ be any representation and $k$ be any field. From Poincar\’e duality and using

the trace to identify $Ad_{\rho}$ and $Ad_{\rho}^{*}$, we get for any integer $l$ an isomorphism of $k$-vector

spaces:

$PD$ : $H^{l}(M, Ad_{\rho})\simeq H^{3-l}(M, \partial M, Ad_{\rho})^{*}$

The exact sequence of the pair $(M, \partial M)$ andPoincar\’e duality together give the following

commutative diagram:

$H^{1}(M, Ad_{\rho})arrow^{\alpha}H^{1}(\partial M, Ad_{\rho})arrow^{\beta}H^{2}(M, \partial M, Ad_{\rho})$

$\{PD |PD \downarrow PD$

$H^{2}(M, \partial M, Ad_{\rho})^{*}arrow H^{1}(\partial M, Ad_{\rho})^{*}\beta^{*}arrow^{\alpha^{*}}H^{1}(M, Ad_{\rho})^{*}$

This gives rank$(\alpha)=rank(\beta)=$ dimker$\beta$, dimker_$\beta+$ rank_{$\beta=\dim H^{1}(\partial M, Ad)$} and

finally rank$( \alpha)=\frac{1}{2}\dim H^{1}(\partial M, Ad_{\rho})$. Hence, $\dim H^{1}(M, Ad_{\rho})\geq\frac{1}{2}\dim H^{1}(\partial M, Ad_{p})$

.

Moreover,$\chi(H^{*}(\partial_{i}M, Ad_{\rho}))=3\chi(\partial_{i}M)=6-6g_{i}$ and$H^{0}(\partial_{i}M, Ad_{\rho})$ and$H^{2}(\partial_{i}M, Ad_{\rho})$

have the

same

dimension. Hence, $\dim H^{1}(M, Ad_{\rho})\geq\sum_{i}(\dim H^{0}(\partial_{i}M)+3g_{i}-3)$

.

If $g_{i}=1,$ $\rho$ restricted to $\partial_{\’{i}}M$ isabelian and hence, $\dim H^{0}(\partial_{i}M, Ad_{\rho})\geq 1$. This proves the

following inequality whatever be $\rho$:

(3) $\dim H^{1}(M, Ad_{\rho})\geq d.$

Let$\rho$ : $\Gammaarrow SL_{2}(k)$ bearegular representation and $\alpha,$$\beta\in\Gamma$ besuchthat $\chi_{\rho}(\Delta_{\alpha,\beta})\neq 0.$

has coefficients in $R=k[Y][ \frac{1}{\Delta_{\alpha,\beta}}]$

.

Let

us

call $Ad_{Y}^{R}$ the free $R$-module such that $Ad_{Y}=$ $Ad_{Y}^{R}\otimes_{R}k(Y)$

.

Then $C^{*}(M, Ad_{Y}^{R})$ is a finite complex of free $R$-modules. By standard semi-continuity

arguments (See [7], Chap. 12), the function $p_{i}(x)=\dim_{k(x)}H^{i}(M, Ad_{Y}^{R}\otimes k(x)\rangle$ is upper

semi continuous

on

$Spec(R)$ _{for any} $i\in N$

.

Comparing the character $\chi_{p}$ with the generic

point, we getthe inequality $( \lim_{k(Y)}H^{1}(M, Ad_{Y})\leq\dim_{k}H^{1}(M, Ad_{\rho})$.

It follows that if$Y$ contains

a

regular representation, then $H^{1}(M, Ad_{p})$ has dimension

$d$

.

Reciprocally, if _{$H^{1}(M, Ad_{Y})$} has dimension $d$, there is

an

open set in $Spec(R)$, and

hence

some

closed points $\chi_{p}$ for which $H^{1}(M, Ad_{\rho})$ has dimension less than

$d$ by upper

serni continuity, hence equal to $d$ by the inequality (3). $\square$

Corollary 4.5. Let $Y$ be an iweducible component

of

$X(\Gamma)$ which contains a regular

representation and such that$\mathfrak{p}/\mathfrak{p}^{2}\otimes k(Y)=0$ where$p$ is the minimalprime ideat

of

$B(\Gamma)$

associated to Y.

Then

$Y$ has dimension$d.$

Proof.

$\mathfrak{R}om$ Proposition 4.1 and Proposition 4.4,

we

get the sequence of inequalities

$d=\dim H^{1}(M, Ad_{Y})=\dim\Omega_{k(Y)/k}^{1}=tr.deg_{k}k(Y)$

.

$\square$

Iftheassumptionsofthecorollary

are

verified, wewill say that$Y$isaregular component

of$X(\Gamma)$

.

If $M$ is a hyperbolic manifold with finite volume and $\rho$ : $\pi_{1}\langle M$) $arrow SL_{2}(k)$ is

a

lift of the holonomy representation, then the component of$X(\Gamma\rangle$ containing _$p$ is regular

as

$p$ is regular and $\dim Y=d$

.

We suppose from

now

on

that $Y$ is

a

regular component

of$X(\Gamma)$

.

4.2.2. Computing $H^{2}$. Let $I_{0}\subseteq I$ be the subset parametrizing the toric components of

$\partial M$ and let $Y$ be a regular component of$X(\Gamma)$

.

Then, from the equality in (3), we get

the isomorphisms $H^{0}(\partial_{i}M,Ad_{Y})=0$ if $g_{i}>1$ and $H^{0}(\partial_{i}M, Ad_{Y})$ is

one

dimensional if

$g_{i}=1$

.

Pick $\xi_{i}$

a

generator of $H^{0}(\partial_{i}M)$ for$i\in I_{0}.$

Lemma 4.6. The map $H^{2}(M, Ad_{Y})arrow K^{Io}$ mapping $\eta$ to the family $(\langle r_{i}^{*}\eta, \xi_{i}\rangle)_{i\in I_{0}}$ is an

isomorphism, where $r_{i}$ : $\partial_{i}Marrow M$ is the $inclu\mathcal{S}ion$ map.

Proof.

This map is part of the exact sequence of the pair $(M, \partial M)$

.

Poincar\’e duality

gives $H^{2}(\partial M, Ad_{Y})=\oplus_{i}H^{0}(\partial_{i}M, Ad_{Y})^{*}=K^{Io}$ where a basis is given by evaluation

on $\xi_{i}$

.

The map is surjective

as

the group $H^{3}(M, \partial M;Ady)=H^{0}(M, Ad_{Y})^{*}=0$ as

$Ad_{Y}$ is

an

irreducible representation. From

a

computation ofEuler characteristic, we get $\dim H^{2}(M,Ad_{Y})=|I_{0}|$ and the result follows. $\square$

4.3. Reidemeister Torsion. For a vector space $E$ ofdimension $n$

over

$k$,

we

denote by

$\det(E)$ the space $\Lambda^{n}E$ and if $n=1$,

we

set $E^{-1}=E^{*}$

.

Given

a

finite complex of finite

dimensional vector spaces $C^{*}=C^{0}arrow\cdots C^{k}$_,

we

set $\det C^{*}=\otimes_{i=0}^{k}(\det C^{i})^{(-1)^{i}}$ The

cohomology of $C^{*}$ is viewed

as

a complex with trivial differentials. There is a natural

(Euler) isomorphism $\det C^{*}\simeq\det H^{*}$ between the determinant of

a

complex and the

determinant of its cohomology whichis well-defined up to sign.

Picking

a

cellular decomposition of $M$ with $0$, 1 and 2-cells, we obtain a complex

$C^{*}(M,Ad_{Y})$ with a preferred volume element: we associate to each cell the volume

Through the isomorphism $\det C^{*}(M,Ad_{Y})\simeq\det H^{*}(M, Ad_{Y})$,

we

get

an

element $T(M)\in\det H^{*}(M, Ad_{Y})=\det\Omega_{k(Y)/k}^{1}\otimes\det H^{2}(M, Ad_{Y})$

.

Given a system of

genera-tors $\xi=(\xi_{i})_{i\in I_{0}}$ of$H^{0}(\partial M, Ad_{Y})$,

we

define

$T(M, \xi)=\langle T(M) , \bigotimes_{t\in I_{0}}\xi_{i}\rangle\in\Omega_{k(Y)/k}^{d}.$

In practice, we will evaluate it on preferred generators $\xi_{i}\in H^{0}(\partial_{i}M, Ad_{Y})$ such that

$b(\xi_{i}^{2})=2$ or on elements of the form $\xi_{i}=\rho(\gamma_{i})_{0}$ for

some

$\gamma_{i}\in\pi_{1}(\partial_{i}M)$

.

4.3.1. The handlebody. Let $M$ be handlebody of genus 2. Let $\alpha,$$\beta$ be the generators of

$F_{2}=\pi_{1}(M)$. One has $B(\Gamma)=k[x, y, z]$ with $x=Y_{\alpha},$$y=Y_{\beta}$ and $z=Y_{\gamma}$

.

Moreover

a

tautological representationis given by

$\rho(\alpha)=(\begin{array}{ll}x -1l 0\end{array}), \rho(\beta)=(\begin{array}{ll}0 -u^{-1}u y\end{array})$

with $K=k(x, y, z,u)/(u^{2}+uz+1)$

.

The handlebodycollapses to

a

cellular complex with

one

-cell and two 1-cells and the twisted complex is given by $C^{0}=sl_{2}(K)$,$C^{1}=sl_{2}(K)^{2}$

and

$d\xi=(\rho(\alpha)^{-1}\xi\rho(\alpha)-\xi, \rho(\beta)^{-1}\xi\rho(\beta)-\xi)$

Let

us

write$T(M)=fdx\wedge dy\wedge dz$

.

_{Then, in order to compute} $f$, itsuffices to evaluate it

onthe vector fields $\partial_{x},$$\partial_{y},$$\partial_{z}$ which correspondto the twisted cocycles $\psi_{x}=\rho^{-1}\partial_{x}\rho,$_{$\psi_{y}=$}

$\rho^{-1}\partial_{y}\rho,$$\psi_{z}=\rho^{-1}\partial_{z}\rho.$

Finally, $f$ is the determinant ofthe matrix of $d$ and the three cocycles in a basis of

$sl_{2}(K)$ of volume 1. This gives _$f=2$ and $T(M)=2dx\wedge dy\wedge dz.$

4.3.2. The 3-chain link complement. Let $p:S^{3}arrow S^{2}$ be the Hopf fibration and $\Sigma\subset S^{2}$

be the complement of three disjoint discs. Then $M=p^{-1}(\Sigma)$ isthe complement of

a

link

in $S^{3}$

with 3 components. Clearly, the Hopf bundle restricted to $\Sigma$

istrivial and we have

indeed $M\simeq\Sigma\cross S^{1}.$

In particular, $\pi_{1}(M)=F_{2}\cross \mathbb{Z}$

.

If $\rho$ : $\Gammaarrow SL_{2}(k)$ is irreducible, it has to map the

generator of$\mathbb{Z}$

to $\pm 1$

.

This shows that $X(\Gamma)$ has two irreducible components $Y_{+}$ and $Y_{-}$

ofirreducible type, each isomorphic to$X(F_{2})=A^{3}$

.

_Let $\alpha,$$\beta$ be the generators of $F_{2}$ and

$t$ the generator of$\mathbb{Z}$

.

The tautological

representation $\rho$ : $\Gammaarrow SL_{2}(K)$ corresponding to

$Y_{\pm}$ is given by the

same

formula

ae

above with in addition

$\rho(t)=\pm Id.$

The manifold $M$ collapses to the product ofa _{wedge of two circles with} a _{circle. This}

cell complex _{has respectively 1,3,2 cells of dimension 0,1,2. We have}

$d^{0}\xi=(\rho(\alpha)^{-1}\xi\rho(\alpha)-\xi, \rho(\beta)^{-1}\xi\rho(\beta)-\xi, 0)$ and

$d^{1}(\zeta, \eta, \theta)=(\rho\langle\alpha)^{-1}\theta p(\alpha)-\theta, \rho(\beta)^{-1}\theta\rho(\beta)-\theta)$

Without surprise, this complex is the tensor product ofthe complex of thehandlebody

with the cellular complex of the circle. In particular, it splits into two independent

complexes. Its torsion will be $\frac{2}{g}dx\wedge dy\wedge dz$ where $g$ is the torsion ofthe acyclic complex

Using $\xi_{\alpha}=p(\alpha)_{0},\xi_{\beta}=\rho(\beta)_{0}$ and $\xi_{\alpha\beta}=\rho(\alpha\beta)_{0}$,

we

get the formula $g=4$ and hence

$T(M,$$\xi\rangle=\frac{1}{2}dx\wedge dy\wedge dz$. In the normalization Tr$\xi_{i}^{2}=2$, weget

$T(M)= \frac{dx\wedge dy\wedge dz}{2\sqrt{(x^{2}-4)(y^{2}-4)(z^{2}-4)}}=\frac{1}{2}\frac{du}{u}\wedge\frac{dv}{v}\wedge\frac{dw}{w}$

where we have set $x=u+u^{-1},$ $y=v+v^{-1}$ _and $z=w+w^{-1}.$

4.3.3.

_Manifolds

fibering over the circle. The following example is an adaptation of [5].

Let $\Sigma$

be compact oriented surface with boundary and $\varphi$ : $\Sigmaarrow\Sigma$ be

a

homeomorphism

preserving the orientationandfixing the boundary pointwise. The suspension$M$ isdetined

as

$M=\Sigma\cross[0, 1]/\sim$where ($x, 1)\sim(\varphi(x), 0)$. Itsboundaryis$\partial\Sigma\cross S^{1}$

and its fundamental group is

$\pi_{1}(M)=\pi_{1}(\Sigma)\rangle\triangleleft \mathbb{Z}$

where the action of $\mathbb{Z}$

is given by the action $\varphi$ on $\pi_{1}(M)$

.

More precisely, picking an

element $t\in\pi_{1}(M\rangle iping to 1, we$have $t\gamma t^{-1}=\varphi_{*}(7)$ where$\varphi_{*}:\pi_{1}(\Sigma)arrow 7r_{1}(\Sigma)$ is the

map induced by $\varphi.$

Recall from Subsection 2.4.2 that the map $r:X(M)arrow X(\Sigma)$ corestricted to $X^{irr}(\Sigma)$

isa $\{\pm 1\}$-principal covering on its image, the (irreducible) fixed point set of$\varphi^{*}$ actingon

$X^{irr}(\Sigma)$. We choose

an

irreducible component$Y$of_$X(M)$ofirreducibletype: it maps to

an

irreduciblecomponent $Z=r(Y)$ of$X(\Sigma)$ of irreducible type. Moreover, the map $Yarrow Z$

is a covering of order at most 2. The adjoint representation $Ad_{Y}$ is clearly independent

on

the representation of $\pi_{1}(M)$ chosen, hence the Reidemeister torsion actually lives

on

$Z.$

Consider the long exact sequence of the pair $(M, \Sigma)$ (Wang sequence), together with

the corresponding sequence on theboundary. It gives the following diagram (weremoved

the coefficients $Ad_{Y}$ from thenotation).

$0 H^{1}(M)-H^{1}(\Sigma)arrow^{\alpha}H^{1}(\Sigma)arrow H^{2}(M)arrow 0$

$H^{0}(\partial\Sigma)-H^{1}(\partial M)\uparrowarrow H^{1}(\partial\Sigma)\}arrow H^{1}(\partial\Sigma)\downarrowarrow^{\sim}H^{2}(\partial M)|\simarrow 0$

Multiplicativity properties imply that the torsion of the first line is the element $T(M)\in$

$\det\Omega_{k(Y)/k}^{1}\otimes\det H^{2}(M)$. Choosinggenerators $\xi$of$H^{0}(\partial M)$, gives the preferred generator

of$\det H^{2}(M)$ and hence the torsion $T(M, \xi)$

as

a

volume form

on

$Z.$

It remains to interpret the map $\alpha=\varphi^{*}$ –id, which

we

do in the following lemma,

proved in the

same

way

as

Proposition 4.1.

Lemma 4.7. The natural map $\Omega_{B(\Sigma)/k}^{1}\otimes_{k[Z]}k(Y)arrow H_{1}(\Sigma_{\}}Ad_{Y})$ is an isomorphism. $In$

geometric terms, $H_{1}(\Sigma, Ad_{Y})$ is the space

of

rational sections

of

the cotangent bundle

of

$X(\Sigma)$, pulled-back to $Y.$

In particular, themap$\varphi^{*}$ issimply the derivative of the action of$\varphi$

on

$X(\Sigma)$

.

Restricted

to $Z$, it is an endomorphism of the restriction ofthe cotangent space of$X(\Sigma)$ to $Z.$

For any component $\gamma_{i}$ of $\partial\Sigma,$ $i\in\gamma r_{0}(\partial\Sigma)$, we consider the generator $\xi_{i}=\rho_{Y}(\gamma_{i})_{0}.$

Plugging it into the sequence, we have the following interpretation of $T(M,\xi)$ and its

Proposition 4.8.

On

any irreducible component

_of

the

_fixed

point set

_of

$\varphi^{*}$ on$X(\Sigma)$

we

have

$T(M, \xi)=\frac{1}{2}\frac{\wedge dY_{\gamma_{i}}}{\det(D\varphi-1)|_{kerdY_{\gamma_{i}}}}$ and$T(M)= \frac{1}{2}\frac{\wedge u_{\iota’}^{-1}du_{i}}{\det(D\varphi-1)|_{kerdu_{l}}}$

where $u_{i}+u_{i}^{-1}=Y_{\gamma_{i}}.$

In particular, we

recover

the example of the magic manifold, where $\varphi$ isthe identity.

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