Character Varieties in SL 2 and Kauffman Skein Algebras

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2. Two definitions of character varieties In all the article, k will denote a field with characteristic 0. Most results of Section 3 hold for any field with characteristic different from 2. We stayed in characteristic 0 in view of our applications. 2.1. Algebraic quotient. Let Γ be a finitely generated group. We define its representation variety into SL2 and we denote by Hom(Γ, SL2) the spectrum of the algebra γ γ γδ γ δ A(Γ) = k[Xi,j, i, j ∈ {1, 2}, γ ∈ Γ]/(det(X ) − 1, X − X X with γ, δ ∈ Γ) γ γ In this formula, X stands for the matrix with entries Xi,j, i, j ∈ {1, 2}. In particular the last equation above is a collection of 4 equations. The name representation variety is justified by the following universal property which holds for any k-algebra R:

Homk−alg(A(Γ),R) = Hom(Γ, SL2(R)) Let SL2(k) act on the space Hom(Γ, SL2) by conjugation. This action is algebraic as it comes γ −1 γ from the action (g.P )(X )= P (g X g) where we have g ∈ SL2(k) and P ∈ A(Γ). Deﬁnition 2.1. We deﬁne the character variety of Γ and denote by X(Γ) the spectrum of the algebra A(Γ)SL2 of invariants.

In other words, X(Γ) is the quotient of Hom(Γ, SL2) in the sense of geometric invariant theory. Standard arguments from this theory gives the following theorem: Theorem 2.2. If k is algebraically closed, there is a bijection between the following sets: SL2 - The k-points of X(Γ) (or equivalently Homk−alg(A(Γ) , k)) - The closed orbits of SL2(k) acting on Hom(Γ, SL2(k)) - The conjugacy classes of semi-simple representations of Γ into SL2(k) - The characters of representations in Hom(Γ, SL2(k)).

Recall that by character of a representation ρ : Γ → SL2(k) we mean the map χρ : Γ → k given by χρ(γ) = Tr ρ(γ). This theorem will be made more precise in the sequel using the point of view of skein algebras. Remark 2.3. We point out the fact that the algebra A(Γ) may have nilpotent elements, as the skein algebra. A big part of the literature on character varieties uses the reduction of these algebras: this is not the case of these notes. 2.2. The skein algebra.

Deﬁnition 2.4. We deﬁne the skein character variety Xs(Γ) as the spectrum of the algebra

B(Γ) = k[Yγ, γ ∈ Γ]/(Y1 − 2,Yαβ + Yαβ−1 − YαYβ with α, β ∈ Γ) One can show that B(Γ) is a ﬁnitely generated k-algebra (see [CS83], Proposition 1.4.1). Moreover, any representation ρ :Γ → SL2(k) gives rise to an algebra morphism χρ : B(Γ) → k by the formula χρ(Yγ ) = Tr ρ(γ). This is a consequence of the famous trace relation: −1 Tr(AB) + Tr(AB ) = Tr(A) Tr(B) ∀A, B ∈ SL2(k)

2 γ The character of the tautological representation ρ :Γ → SL2(A(Γ)) defined by ρ(γ)= X is a map Φ : B(Γ) → A(Γ)SL2 . The following theorem as been proved by Przytyscki and Sikora, see [PS00]. In practice, it follows from [P87] Theorem 2.6, see also [BH91] and [Bul97]. Theorem 2.5. For any field k of characteristic 0 and any finitely generated group Γ, the map Φ : B(Γ) → A(Γ)SL2 is an isomorphism. The statement of Theorem 2.6 in [P87] is much more general and deals with algebras with trace satisfying the Cayley-Hamilton identity. We derive our statement from his below. Proof. Let k[Γ] be the group algebra of Γ. We denote by [γ] the generator associated to γ ∈ Γ and set Tr([γ]) = [γ] + [γ−1] that we extend by k-linearity. We define H(Γ) = k[Γ]/I where I is the two-sided ideal generated by the elements Tr(x)y − y Tr(x), for any x,y ∈ k[Γ]. The trace factors to a k-linear endomorphism of H(Γ). By direct computation, one checks that the following identities hold for any x,y ∈ H(Γ): (i) Tr(x)y = y Tr(x) (ii) Tr(xy) = Tr(yx) (iii) Tr(Tr(x)y) = Tr(x) Tr(y) 2 1 2 2 (iv) x − Tr(x)x + 2 (Tr(x) − Tr(x )) = 0. The last equation is called the Cayley-Hamilton identity of order 2. The map j : H(Γ) → γ M2(A(Γ)) defined by j([γ]) = X is an algebra morphism preserving the trace. It is universal in ′ the sense that for any k-algebra B and morphism j : H(Γ) → M2(B), there is a unique algebra morphism ϕ : A(Γ) → B such that the following diagram commutes.

j H(Γ) / M (A(Γ)) ❑ 2 ❑❑ ′ ❑❑❑j ❑❑❑ M2(ϕ) ❑% M2(B)

Denote by G the group GL2(k). The universal property implies that if we compose j with a conjugation πg with g ∈ G, there is an automorphism ϕg of A(Γ) such that πg ◦ j = M2(ϕg) ◦ j. −1 The action of G on A(Γ) is the one described in Subsection 2.1: the formula ρ(g)= πg ◦M2(ϕg) γ deﬁnes an action of G on M2(A(Γ)) ﬁxing X for all γ and Theorem 2.6 of [P87] says the following:

GL2 Theorem 2.6 ([P87], 2.6). The map j : H(Γ) → M2(A(Γ)) is an isomorphism. To end the proof, we observe that Φ is the restriction of j to the center. More precisely, the map B(Γ) → H(Γ) deﬁned by Yγ 7→ Tr([γ]) is injective and its image by j is the center of G G G SL2 M2(A(Γ)) , that is A(Γ) Id. We end the proof by observing the equality A(Γ) = A(Γ) .

Thanks to this theorem, we can remove the ’s’ in Xs(Γ). 2.3. Irreducible, reducible and central characters. Given α, β ∈ Γ, we deﬁne the following element of B(Γ): 2 2 2 ∆α,β = Yα + Yβ + Yαβ − YαYβYαβ − 4.

The equation ∆α,β 6= 0 deﬁnes an open subset Uα,β of X(Γ) and we set irr X (Γ) = Uα,β. α,β[∈Γ We begin our study with the following lemma, a slightly diﬀerent version of Lemma 1.2.1 in [CS83]. We will say that a representation ρ : Γ → SL2(k) is absolutely irreducible if it is irreducible in an algebraic closure of k.

3 Lemma 2.7. Given a field k and a representation ρ :Γ → SL2(k), ρ is absolutely irreducible if and only if there exists α, β ∈ Γ such that Tr(ρ(αβα−1β−1)) 6= 2. Proof. We first recall Burnside irreducibility criterion which states that ρ is absolutely irreducible if and only if Span{ρ(γ), γ ∈ Γ} = M2(k). Let M be the matrix defined by Mij = Tr(ρ(γiγj)) for i, j ∈ {1,..., 4} where γ1 = 1, γ2 = 2 α, γ3 = β and γ4 = αβ. A simple computation shows that det M = −χρ(∆α,β) where we −1 −1 have χρ(∆α,β) = Tr ρ(αβα β ) − 2. Hence if there exists α, β such that χρ(∆α,β) 6= 0, then (ρ(γi))i=1..4 is a basis of M2(k) and by Burnside criterion, ρ is absolutely irreducible. Conversely, suppose that for all α, β ∈ Γ one has χρ(∆α,β) = 0. Then there exists a non-zero 2 subspace Fα,β of k fixed by the commutator ρ([α, β]). Suppose that there exists γ, δ such that 1 x F ∩ F = {0} then in a basis made of these two lines, one can write ρ([α, β]) = and α,β γ,δ 0 1 1 0 ρ([γ, δ]) = with x and y non zero. We compute then Tr([[α, β], [γ, δ]]) = 2 + (xy)2. The y 1 hypothesis implies that x or y is zero which is impossible. This finally implies that α,β Fα,β 6= {0}. But this subset is ρ-invariant which implies that ρ is reducible. T cen 2 Consider the closed subset X (Γ) of X(Γ) defined by the equations Yγ = 4 for all γ ∈ Γ. A cen 1 k-point of X (Γ) has the form ϕ(Yγ ) = 2ε(γ) for some ε ∈ H (Γ, Z/2Z). This is the character of the central representation ρ : γ 7→ ε(γ)id. The closed subset X(Γ) \ Xirr(Γ) is denoted by Xred(Γ): let us describe it more precisely. Let Hom(Γ, Gm) be the spectrum of the algebra

C(Γ) = k[Zγ , γ ∈ Γ]/(Zγ Zδ − Zγδ,γ,δ ∈ Γ) × We have for any k-algebra R, Homk−alg(C(Γ),R) ≃ Hom(Γ,R ), which explains the notation. Let σ be the automorphism of C(Γ) defined by σ(Zγ ) = Zγ−1 and π : B(Γ) → C(Γ), the morphism defined by π(Yγ )= Zγ + Zγ−1 . Proposition 2.8. The map π : B(Γ) → C(Γ) is a morphism of algebras whose image is the σ-invariant part. Its kernel is the radical of the ideal generated by ∆α,β. red In other words, a k-point ϕ of X lifts to a k-point of Hom(Γ, Gm)/σ. Hence, there exists × an extension kˆ of k (at most quadratic) and a morphism ψ : Γ → kˆ such that ϕ(Yγ ) = ψ(γ)+ ψ(γ)−1. Proof. We prove easily that π is a homomorphism whose image is C(Γ)σ. Let k be a field and ϕ : B(Γ) → k a morphism satisfying ϕ(∆α,β) = 0 for all α, β ∈ Γ. We write ϕ(α) = ϕ(Yα) for short: by assumption one has ϕ([α, β]) = 2. Moreover, if ϕ(α) = ϕ(β) = 2 then ϕ(∆α,β) = 0 implies (ϕ(αβ) − 2)2 = 0 and hence ϕ(αβ) = 2. We conclude that ϕ(α) = 2 for any α ∈ [Γ, Γ]. Moreover, if α ∈ Γ and β ∈ [Γ, Γ], then ϕ(αβ) satisfies (ϕ(α) − ϕ(αβ))2 = 0 and hence ϕ(αβ)= ϕ(α). This proves that ϕ factors through a map ϕab : B(Γab) → k where Γab = Γ/[Γ, Γ]. Moreover, we clearly have C(Γab) = C(Γ) and the map B(Γab) → C(Γab)σ is invertible. This proves the first part of the lemma. The second part follows.

Remark 2.9. We don’t know if the ∆α,βs generate the kernel of π. Moreover, it is well known that the deformations of reducible representations into irreducible ones are related to the Alexan- der module. It would be interesting to have an explicit description of the (co-)normal bundle of Xred(Γ) in X(Γ). More precisely, deﬁne the ideal I by the following exact sequence: 0 / I / B(Γ) / C(Γ)σ / 0

2 Then there should be a relation between the C(Γ)-module I/I ⊗C(Γ)σ C(Γ) and the module 1 H (Γ,C(Γ)) where the action of γ ∈ Γ is by multiplication by Zγ.

4 2.4. Functorial properties. If ϕ : Γ → Γ′ is a group homomorphism, there is a natural map ′ ′ ϕ∗ : B(Γ) → B(Γ ) mapping Yγ to Yϕ(γ). There is also a natural map ϕ∗ : A(Γ) → A(Γ ) which makes the following diagram commute:

ϕ∗ B(Γ) / B(Γ′)

Φ Φ ϕ∗ A(Γ) / A(Γ′)

2.4.1. Double quotients. Let us show an example which arises when computing the fundamental group of a 3-manifold presented by a Heegard splitting.

Proposition 2.10. Let Γ, Γ1, Γ2 be three ﬁnitely generated groups and ϕi : Γ → Γi be two ′ ′ surjective morphisms. Denote by Γ the amalgamated product Γ =Γ1 ⋆ Γ2. Γ ′ Then X(Γ ) is isomorphic to the ﬁber product X(Γ1) × X(Γ2). X(Γ)

′ More geometrically, if we denote by Li the images of X(Γi) in X(Γ), then X(Γ ) is the (schematic) intersection L1 ∩ L2.

Proof. Given any morphisms ϕi : Γ → Γi and any k-algebra R, one has the following natural isomorphisms: ′ ′ Homk−alg(A(Γ ),R) = Hom(Γ , SL2(R))

= {ρi :Γi → SL2(R), ρ1 ◦ ϕ1 = ρ2 ◦ ϕ2}

= Homk−alg(A(Γ1) ⊗ A(Γ2),R) A(Γ) ′ Hence we have the isomorphism A(Γ )= A(Γ1) ⊗ A(Γ2). We notice that the invariant part of A(Γ) this tensor product is not the tensor product of the invariant parts (think about a wedge of two circles). However, this holds when A(Γi) is a quotient of A(Γ), that is when the morphisms ϕi are surjective. Denoting by Ii the kernel of the map A(Γ) → A(Γi), the result follows from the following ′ SL2 SL2 SL2 standard fact in invariant theory: A(Γ )= A(Γ)/(I1 + I2) and (I1 + I2) = I1 + I2 . ′ On the other hand, one shows easily that the map B(Γ1) ⊗B(Γ) B(Γ2) → B(Γ ) is an isomorphism, see for instance [PS00].

2.4.2. Semi-direct products. The following situation appears for fundamental groups of 3-manifolds which ﬁber over the circle. Let Γ be a ﬁnitely generated group and ϕ ∈ Aut(Γ) an automorphism. ′ Set Γ =Γ ⋊ϕ Z so that we have the following (split) exact sequence. Let t = s(1).

z s 0 / Γ α / Γ′ / Z / 0 Lemma 2.11. Suppose that k is algebraically closed and consider only closed points of the character varieties. Then the map α∗ : X(Γ′) → X(Γ) corestricted to Xirr(Γ) is a Z/2Z principal bundle over F , where F = {x ∈ Xirr(Γ), ϕ∗x = x}.

′ ′ Proof. Let ρ :Γ → SL2(K) be a representation which restricts to an irreducible representation ′ ′ −1 ρ : Γ → SL2(K). This representation satisﬁes ρ (t)ρ(γ)ρ (t) = ρ(ϕ(γ)) hence the character ∗ ′ ′ χρ is ﬁxed by ϕ . Suppose we have another representation ρ : Γ → SL2(K) with the same restriction, then ρ′′(t)ρ′(t)−1 commutes with the image of ρ. Hence ρ′′(t) = ±ρ′(t). There are precisely two preimages for ρ and the lemma is proved.

5 3. Applications of Saito’s Theorem

We know from the isomorphism between Xs(Γ) and X(Γ) that if k is algebraically closed, any k-point of Xs(Γ) is the character of a representation ρ : Γ → SL2(k). However this fact is non-trivial to prove directly: the following theorem of [S93] allows it and has further applications. Theorem 3.1. Let R be a k-algebra and ϕ : B(Γ) → R be a morphism of k-algebras. Suppose that there exists α, β ∈ Γ such that ϕ(∆α,β) is invertible and A, B ∈ SL2(R) such that Tr A = ϕ(Yα), Tr(B)= ϕ(Yβ), Tr(AB)= ϕ(Yαβ). Then, there is a unique representation ρ :Γ → SL2(R) such that ρ(α)= A, ρ(β)= B and for all γ ∈ Γ, Tr(ρ(γ)) = ϕ(Yγ ). 2 Proof. (Sketch) Define γi and M as in Lemma 2.7 . As det M = −ϕ(∆α,β) , the matrix M is invertible over R. Given any γ ∈ Γ, we denote by Cγ the vector of coefficients of ρ(γ) in the basis (ρ(γi))i=1..4 and by Tγ the vector (ϕ(Yγγi ))i=1..4. We then have MCγ = Tγ which we solve −1 by setting Cγ = M Tγ and define this way ρ(γ). It remains to show that this actually defines a representation in SL2 whose traces are prescribed by ϕ. This is a non-trivial consequence of the trace relations, we refer to [S93] for a proof.

3.1. Points of X(Γ) over arbitrary ﬁelds.

Deﬁnition 3.2. Let k be a ﬁeld and ρ :Γ → SL2(kˆ) be a representation where kˆ is an extension of k and such that Tr ρ(γ) ∈ k for all γ ∈ Γ. We set

M(ρ) = Spank{ρ(γ), γ ∈ Γ}.

It is a sub-k-algebra of M2(kˆ). We will say that two representations ρi : Γ → SL2(kî) where i = 1, 2 are equivalent if there is an isomorphism of k-algebras σ : M(ρ1) → M(ρ2) such that ρ2 = σ ◦ ρ1. We have the following proposition: Proposition 3.3. Let k be a field. There is a functorial isomorphism between k-points of irr X (Γ) and equivalence classes of absolutely irreducible representations ρ :Γ → SL2(kˆ) where kˆ is a finite extension of k and for all γ in Γ, Tr(ρ(γ)) ∈ k.

Proof of Proposition 3.3. Given ρ : Γ → SL2(kˆ) as in the statement, we define a morphism ϕ ∈ Homalg(B(Γ), k) by setting ϕ(Yγ ) = Tr(ρ(γ)). By Lemma 2.7, there exists α, β ∈ Γ such irr that χρ(∆α,β) 6= 0. Hence, ϕ indeed corresponds to a k-point of X (Γ). irr On the other hand, a k-point of X (Γ) corresponds to a morphism ϕ in Homalg(B(Γ), k) such that there exists α, β ∈ Γ with ϕ(∆α,β) 6= 0. ϕ(Y ) −1 0 −1/u We define the matrices A = α and B = where u belongs to an 1 0 u ϕ(Yβ) 2 extension kˆ of k such that the equation u + uϕ(Yαβ)+1 = 0 holds (kˆ can be chosen at most quadratic). By Theorem 3.1, there is a unique representation ρ : Γ → SL2(kˆ) such that ρ(α)= A, ρ(β)= B and Tr(ρ(γ)) = ϕ(Yγ ) for all γ ∈ Γ. In order to show that this representation does not depend on the choice of α and β, we ′ ′ ′ observe that if we make an other choice for A ,B ∈ M2(kˆ ) which defines a representation ′ ′ ′ ′ ρ : Γ → SL2(kˆ ), the map defined by σ(A) = A and σ(B) = B extends to a k-isomorphism from M(ρ) to M(ρ′) showing that the two representations are equivalent.

The algebra M(ρ) associated to an absolutely irreducible representation ρ is simple and central and hence deﬁnes a class in the Brauer group Br(k). Moreover, dimk M(ρ) = 4. It follows that M(ρ) is a quaternion algebra and in particular, its square is 0 in Br(K). The set of k-points of Xirr(Γ) has then a partition into Brauer classes. We get ﬁnally the following proposition:

6 Proposition 3.4. Let k be a field with Br(k) = 0 (this occurs for instance if k is algebraically closed or of transcendence degree one over an algebraically closed field). Then k-points in Xirr(Γ) correspond bijectively to conjugacy classes of absolutely irreducible representations ρ : Γ → SL2(k). Proof. An irreducible point ϕ : B(Γ) → k corresponds to the character of a representation ρ : Γ → SL2(kˆ) for some extension kˆ of k. As the Brauer group is trivial, there is an algebra isomorphism σ : M2(k) → M(ρ). Tensoring by kˆ, we find that σ ⊗ 1 : M2(kˆ) → M(ρ) ⊗ kˆ is a kˆ-linear automorphism of M2(kˆ) and hence a conjugation by some g ∈ GL2(kˆ) by Skolem- −1 Noether theorem. Finally, the representation gρg :Γ → SL2(kˆ) takes its values in k. As two ′ such representation ρ and ρ are equivalent, it means that there is an automorphism σ of M2(k) such that ρ′ = σ ◦ ρ and this automorphism is a conjugation.

Example 3.5. Let F2 be the free group on the generators α and β. It is well-known that the map Ψ : Q[x,y,z] → B(F2) defined by Ψ(x) = Yα, Ψ(y) = Yβ and Ψ(z) = Yαβ is an isomorphism. Hence, one has a rational point of X(F2) by sending x,y,z to 1. However, there is no representation ρ : F2 → SL2(Q) with this character. However, one can find a solution in some (non uniquely defined) quadratic extension of Q.

3.2. Tangent spaces. Let k be a ﬁeld and ρ : Γ → SL2(k) be an absolutely irreducible rep- irr resentation. The character χρ may be seen as a k-point of X (Γ). We derive from Saito’s Theorem a simple proof of the following well-known result:

1 Proposition 3.6. There is a natural isomorphism Tχρ Xk(Γ) ≃ H (Γ, Adρ) where Adρ is the adjoint representation of Γ on sl2(k) induced by ρ.

Proof. It is well-known that a tangent vector at χρ corresponds to an algebra morphism ϕε : 2 B(Γ) → k[ε]/(ε ) such that ϕ0 = χρ. As ρ is absolutely irreducible, there exists α, β ∈ Γ such that ϕ(∆α,β) 6= 0. Consider the 3 map π : SL2(k) × SL2(k) → k defined by π(A, B) = (Tr(A), Tr(B), Tr(AB)). Lemma 3.7 below shows that this map is a submersion precisely on the preimage of the open set of k3 defined 2 by ∆α,β 6= 0. This proves that we can find two matrices Aε,Bε ∈ SL2(k[ε]/(ε )) such that Tr(Aε)= ϕε(Yα), Tr(Bε)= ϕε(Yβ) and Tr(AεBε)= ϕε(Yαβ). By Theorem 3.1, there is a unique 2 representation ρε :Γ → SL2(k[ε]/(ε )) such that ρε(α)= Aε, ρε(β)= Bε and χρε = ϕε. 1 We define the cocyle ψ ∈ Z (Γ, Adρ) by the formula

d −1 (1) ψ(γ)= ρ(γ) ρε(γ). dε ε=0

1 The map ϕε 7→ ψ is a linear map Tχρ Xk(Γ) → H (Γ, Adρ). We construct the inverse map by sending the cocycle ψ to the character of the representation ρε(γ)= ρ(γ)(1 + εψ(γ)). ′ ′ ′ If we had chosen other matrices Aε,Bε defining a representation ρε, then as in the proof of 2 ′ Proposition 3.3, we would have found an automorphism σ of M2(k[ε](ε )) such that ρ = σ ◦ ρ. ′ Moreover ρε and ρε coincide up to first order so that one can write σ = Id+ εD for some derivation D of M2(k). All such derivations have the form D(X) = [ξ, X] for some ξ ∈ sl2(k). ′ A computation shows that ψ = ψ + dξ and the map ϕε → ψ is well-defined. Reciprocally, εξ changing ψ to ψ + dξ amounts in conjugating ρε by e , which does not change the character, hence the theorem is proved.

2 3 Lemma 3.7. The map π : SL2(k) → k deﬁned by the formula π(A, B) = (Tr(A), Tr(B), Tr(AB)) is a submersion at any pair (A, B) such that Tr[A, B] 6= 2.

7 2 Proof. Let ξ, η be in sl2(k) corresponding to a tangent vector (Aξ, Bη) ∈ T(A,B)SL2(k) . If D(A,B)π is not surjective, there exists u, v, w ∈ k such that u Tr(Aξ)+ v Tr(Bη)+ w Tr(AξB + ABη) = 0 for all ξ, η ∈ sl2(k). This is possible if and only if there exists λ, µ ∈ k such that uA + wBA = λId and vB + wAB = µId. Hence, there is a linear relation in the family (Id,A,B,AB), which is equivalent to the equality Tr([A, B]) = 2, see Lemma 2.7. 1 One has to be careful that the isomorphism between Tχρ X(Γ) and H (Γ, Adρ) does not longer 1 hold if ρ is not absolutely irreducible. However there is still a natural map Ξ : H (Γ, Adρ) →

Tχρ X(Γ) sending ψ to the character of ρ(1 + εψ). 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 of the trivial representation. One has 1 1 H (Γ, Adρ)= H (Γ, k) ⊗ sl2(k) and the map Ξ vanishes.

Proposition 3.8. For any ﬁnitely generated group Γ and ﬁeld k, the tangent space T1X(Γ) at the character of the trivial representation is isomorphic to the space of functions ψ : Γ → k satisfying the following quadratic functional equation (see [FS04]): ∀γ, δ ∈ Γ, ψ(γδ)+ ψ(γ−1δ) = 2ψ(γ) + 2ψ(δ)

Proof. It is sufficient to write down the conditions for ϕε : Yγ 7→ 2+ εψ(γ) to be an algebra morphism from B(Γ) to k[ε]/(ε2). 2 1 Observe that the map S H (Γ, k) → T1X(Γ) mapping λµ to the map ψ : γ 7→ λ(γ)µ(γ) is injective but not necessarily surjective. 3.3. Tautological representations. Let Γ be a finitely generated group and k be a field. Consider an irreducible component Y of X(Γ). We will say that Y is of irreducible type if it contains the character of an absolutely irreducible representation. More formally, this component corresponds to a minimal prime p ⊂ B(Γ): we will write k[Y ]= B(Γ)/p. The component Y will be of irreducible type if and only if the tautological character χY : B(Γ) → k[Y ] is irreducible. From Proposition 3.3, there exists an extension K (at most quadratic) of the field k(Y ) and a representation ρY : Γ → SL2(K) such that χρY = χY . We will call such a representation a tautological representation. Remark 3.9. The Brauer group of a transcendance degree one extension of an algebraically closed field vanishes. Hence, if k is algebraically closed and dim(Y ) = 1, one can choose K = C(Y ). This case will appear very often if Γ 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 Γab = Z and ϕ :Γ → Z is the abelianization morphism, then one component of X(Γ) will be of reducible type. Moreover, it will be isomorphic 1 to A via the map ϕ∗ : B(Γ) → B(Z) ≃ k[Y1]. A tautological representation will be given by tϕ(γ) 0 ρ :Γ → SL (k(t)), ρ (γ)= Y 2 Y 0 t−ϕ(γ) −1 and the extension k(Y1) ⊂ k(t) corresponds to the substitution Y1 = t + t . 3.4. Valuation rings and Culler-Shalen theory. We give in this section an example of result of Culler-Shalen theory (though not explicitly stated in their articles). The proof will follow standard lines for what concerns Culler-Shalen theory. We add it in order to convince the reader that the techniques of the preceding section are well-adapted to these questions. Let M be a 3-manifold by which we mean a compact oriented and connected topological 3-manifold. An oriented and connected surface S ⊂ M is said incompressible if - S is not a 2-sphere,

8 - the map π1(S) → π1(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 (Yi)i ∈ I each of them having a ﬁnitely generated fundamental group, we set X(Y )= X(π1(Yi)). iQ∈I Proposition 3.11. Let M be a small 3-manifold. Then the restriction map r : X(M) → X(∂M) is proper.

Proof. Fix a base point y in M and let I = π0(∂M). We choose a base point yi for each component ∂iM of ∂M. SetΓ= π1(M,y) and Γi = π1(∂iM,yi). As X(∂M) = i∈I X(Γi), it is sufficient to define the map ri : X(Γ) → X(Γi) for any i ∈ I. This map is the oneQ induced by the inclusion ∂iM ⊂ M on fundamental groups. By the valuative criterion of properness (see Thm 4.7 in [Ha77]), r is proper if and only if for any discrete valuation k-algebra R with fraction field K fitting in the following diagram there is a morphism Spec(R) → X(M) which makes the diagram commute.

ϕ (2) Spec(K) / X(M) 9 r Spec(R) / X(∂M)

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 remarking that K has transcendence degree 1 over k, the morphism ϕ corresponds to a representation ρ :Γ → SL2(K). Let T be the Bass-Serre tree on which SL2(K) acts. Through ρ, the group Γ acts on T . If this action is trivial - i.e. it fixes a vertex of T - the representation ρ is conjugate to a representation in SL2(R). For any γ ∈ Γ, we have Tr ρ(γ) ∈ R. This shows that the map B(Γ) → R defined by Yγ 7→ Tr ρ(γ) corresponds to a map Spec(R) → X(M) which yields a contradiction. Hence, the action is non-trivial and is dual to a non-empty essential surface Σ ⊂ M. For all i ∈ I, denote by ρi :Γi → SL2(K) the representation induced by ρ. From the diagram (2), we get that for all γ ∈ Γi, Tr ρi(γ) ∈ R. Let us prove that γ fixes a vertex in T . If ρi(γ)= ±1, then it is trivial. In the opposite case, 2 one can find a vector v ∈ K such that v and ρi(γ)v form a basis. The matrix of ρi(γ) in that 0 −1 basis is . This proves that ρi(γ) is conjugate to an element in SL2(R) and hence 1 Tr(ρi(γ)) fixes a vertex in T . A standard lemma (Corollaire 3 p90 in [Ser83]) shows that if every γ ∈ Γi fixes a vertex in T , then Γi itself fixes a vertex in T . The relative construction of dual surfaces shows that one can find an essential surface Σ dual to the action without boundary, see Corollary 6.0.1 in [Sh02]. Any component of Σ is an incompressible surface. This is not possible as M is small.

4. The Reidemeister torsion as a rational volume form In this section, we construct the Reidemeister torsion of a 3-manifold with boundary on an irreducible component Y ⊂ X(Γ). It will be a rational volume form on Y , that is an element of d 1 d Λ Ωk(Y )/k = Ωk(Y )/k 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 Γ be a ﬁnitely generated group, k a ﬁeld and ϕ : B(Γ) → k an irreducible character. Then, we now from Proposition 3.3 that there exists an

9 extension kˆ of k and a representation ρ : Γ → SL2(kˆ) such that χρ = ϕ. Moreover, the sub- algebra M(ρ) = Span{ρ(γ), γ ∈ Γ} satisﬁes M(ρ) ⊗k kˆ ≃ M2(kˆ) and depends only on ϕ up to isomorphism. Recall the splitting M2(kˆ) = kˆId ⊕ sl2(kˆ) and for any X ∈ M2(kˆ), denote by X0 = X − 1 2 Tr(X)Id the projection on the second factor. Then set M(ρ)0 = Spank{ρ(γ)0, γ ∈ Γ}. It is a 3-dimensional k-vector space on which Γ acts by conjugation. We will denote by Adϕ this Γ-module and call it the adjoint representation. The relation between Adϕ and Adρ is the following: Adρ = Adϕ ⊗kk.ˆ The representation Adϕ is more natural as it depends only on the character. The Killing form is an invariant non-degenerate pairing on M(ρ)0 and there is an invariant 3 ∗ volume form ε ∈ Λ (M(ρ)0) given by ε(ζ,η,θ) = Tr(ζ[η,θ]). Proposition 4.1. Let Y be an irreducible component of B(Γ) of irreducible type corresponding to a minimal prime ideal p. Let AdY be the adjoint representation associated to the tautological 1 character χY . There is an isomorphism between ΩB(Γ)/k ⊗B(Γ) k(Y ) and H1(Γ, AdY ). This provides the following exact sequence: 2 1 p/p ⊗k[Y ] k(Y ) → H1(Γ, AdY ) → Ωk(Y )/k → 0.

Proof. This follows by duality from Proposition 3.6 with the twist that AdY is not exactly AdρY which takes values in SL2(Kˆ ) for some extension Kˆ of k(Y ). Instead of adapting the previous proof, we describe directly this dual picture. Recall that H1(Γ, AdY ) is the ﬁrst homology of a complex C∗(Γ, AdY ) where Ck(Γ, AdY ) is generated by elements of the form ξ ⊗ [γ1, . . . , γk] with ξ ∈ AdY and γ1, . . . , γk ∈ Γ. We have the formulas ∂ξ ⊗ [γ] = ρ(γ)−1ξρ(γ) − ξ and ∂ξ ⊗ [γ, δ] = ρ(γ)−1ξρ(γ) ⊗ [δ] − ξ ⊗ [γδ]+ξ ⊗[γ]. This allows to check directly that the map dYγ 7→ ρ(γ)0 ⊗[γ] induces a well-deﬁned 1 map of B(Γ)-modules ΩB(Γ)/k → H1(Γ, AdY ) and hence a map 1 Ψ:ΩB(Γ)/k ⊗ k(Y ) → H1(Γ, AdY ).

1 2 Reciprocally, let Λ = k(Y ) ⊕ ε ΩB(Γ)/k ⊗ k(Y ) be the k(Y )-algebra with ε = 0. The map ϕ : B(Γ) → Λ given by Yγ → Yγ + εdYγ is an algebra morphism. By Lemma 3.7, one can ﬁnd Aε,Bε ∈ SL2(Λ) whose traces are prescribed and by Saito’s theorem (using the irreducibility of

χY ), one ﬁnds a representation ρε :Γ → SL2(Λ) satisfying χρε = ϕ. A simple check shows that d −1 the map ξ ⊗ γ 7→ dε Tr(ξρ(γ) ρε(γ)) induces precisely the inverse of Ψ. The exact sequence of the proposition follows from standard facts of commutative algebra. 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: ∂M = i∈I ∂iM. Let gi be the genus of ∂iM : we suppose that gi > 0 for all i ∈ I and set d = i max(1S , 3gi − 3). Denote by Γ the fundamental group of M and let Y be an irreducible componentP of X(Γ). We denote by χY : B(Γ) → k[Y ] the tautological character and by AdY the adjoint representation. The following technical lemma will be useful in the sequel.

Lemma 4.2. Let α, β be two elements of Γ such that ∆α,β is non zero in k[Y ]. Then, there is a k(Y )-basis of Ad such that the adjoint representation has coefficients in k[Y ][ 1 ]. Y ∆α,β −1 Proof. Set R = k[Y ][∆α,β] and consider the composition χY : B(Γ) → k[Y ] → R. One can find an extension Rˆ with A, B ∈ SL2(Rˆ) and by Saito’s theorem, a unique representation ρα,β : Γ → SL2(Rˆ) with character χY satisfying ρ(α) = A and ρ(β) = B. By construction, ρ(γ) lies in SpanR{1,A,B,AB}. The basis A0,B0, (AB)0 of M(ρα,β)0 satisfies the assumption of the lemma and one has M(ρ)0 ≃ M(ρα,β)0 ≃ AdY . This proves the lemma.

10 4.2.1. Computing H1.

Definition 4.3. Recall that we say that a representation ρ : Γ → SL2(k) belongs to Y if its character χρ : B(Γ) → k factorizes through k[Y ]. Equivalently, we will say that Y contains the representation ρ. For instance, a representation is of irreducible type if it contains an irreducible character. 1 A representation ρ :Γ → SL2(k) is said regular if H (Γ, Adρ) has dimension d. An irreducible component Y of X(Γ) will be said regular if it contains a regular representation. Proposition 4.4. An irreducible component Y of X(Γ) of irreducible type is regular if and only 1 if dim H (Γ, AdY )= d. Proof. We start with a standard argument of Poincaréduality, taken from [HK98]. Let ρ :Γ → SL2(k) be any representation and k be any field. From Poincaréduality and using the trace to ∗ identify Adρ and Adρ, we get for any integer l an isomorphism of k-vector spaces: l 3−l ∗ P D : H (M, Adρ) ≃ H (M,∂M, Adρ) The exact sequence of the pair (M,∂M) and Poincaréduality together give the following commutative diagram:

1 α / 1 β / 2 H (M, Adρ) / H (∂M, Adρ) / H (M,∂M, Adρ)

PD PD PD ∗ ∗ 2 ∗ β / 1 ∗ α / 1 ∗ H (M,∂M, Adρ) / H (∂M, Adρ) / H (M, Adρ)

1 This gives rank(α) = rank(β) = dim ker β, dim ker β + rank β = dim H (∂M, Adρ) and ﬁnally 1 1 1 1 1 rank(α)= 2 dim H (∂M, Adρ). Hence, dim H (M, Adρ) ≥ 2 dim H (∂M, Adρ). ∗ 0 2 Moreover, χ(H (∂iM, Adρ)) = 3χ(∂iM) = 6 − 6gi and H (∂iM, Adρ) and H (∂iM, Adρ) 1 0 have the same dimension. Hence, dim H (M, Adρ) ≥ i(dim H (∂iM) + 3gi − 3). If gi = 1, 0 ρ restricted to ∂iM is abelian and hence, dim H (∂iM,PAdρ) ≥ 1. This proves the following inequality whatever be ρ:

1 (3) dim H (M, Adρ) ≥ d.

Let ρ : Γ → SL2(k) be a regular representation and α, β ∈ Γ be such that χρ(∆α,β) 6= 0. Then, by Lemma 4.2, one can find a k(Y )-basis of AdY such that the adjoint representation has coefficients in R = k[Y ][ 1 ]. Let us call AdR the free R-module such that Ad = AdR ⊗ k(Y ). ∆α,β Y Y Y R ∗ R Then C (M, AdY ) is a finite complex of free R-modules. By standard semi-continuity ar- i R guments (See [Ha77], Chap. 12), the function pi(x) = dimk(x) H (M, AdY ⊗k(x)) is upper semi-continuous on Spec(R) for any i ∈ N. Comparing the character χρ with the generic point, 1 1 we get the inequality dimk(Y ) H (M, AdY ) ≤ dimk H (M, Adρ). 1 It follows that if Y contains a regular representation, then H (M, Adρ) has dimension d. 1 Reciprocally, if H (M, AdY ) has dimension d, there is an open set in Spec(R), and hence some 1 closed points χρ for which H (M, Adρ) has dimension less than d by upper semi-continuity, hence equal to d by the inequality (3). Corollary 4.5. Let Y be an irreducible component of X(Γ) which contains a regular representation and such that p/p2 ⊗ k(Y ) = 0 where p is the minimal prime ideal of B(Γ) associated to Y . Then Y has dimension d. Proof. From Proposition 4.1 and Proposition 4.4, we get the sequence of inequalities d = 1 1 dim H (M, AdY ) = dimΩk(Y )/k = tr.degkk(Y ).

11 If the assumptions of the corollary are veriﬁed, we will say that Y is a regular component of X(Γ). If M is a hyperbolic manifold with ﬁnite volume and ρ : π1(M) → SL2(k) is a lift of the holonomy representation, then the component of X(Γ) containing ρ is regular as ρ is regular and dim Y = d. We suppose from now on that Y is a regular component of X(Γ).

2 4.2.2. Computing H . Let I0 ⊂ I be the subset parametrizing the toric components of ∂M and let Y be a regular component of X(Γ). Then, from the equality in (3), we get the isomorphisms 0 0 H (∂iM, AdY )=0if gi > 1 and H (∂iM, AdY ) is one dimensional if gi = 1. Pick ξi a generator 0 of H (∂iM) for i ∈ I0.

2 I0 ∗ Lemma 4.6. The map H (M, AdY ) → K mapping η to the family (hri η,ξii)i∈I0 is an isomorphism, where ri : ∂iM → M is the inclusion map. Proof. This map is part of the exact sequence of the pair (M,∂M). Poincar´eduality gives 2 0 ∗ I0 H (∂M, AdY ) = i H (∂iM, AdY ) = K where a basis is given by evaluation on ξi. The 3 0 ∗ map is surjective asL the group H (M,∂M; AdY ) = H (M, AdY ) = 0 as AdY is an irreducible 2 representation. From a computation of Euler characteristic, we get dim H (M, AdY )= |I0| and the result follows.

4.3. Reidemeister Torsion. For a vector space E of dimension n over k, we denote by det(E) the space ΛnE and if n = 1, we set E−1 = E∗. Given a finite complex of finite dimensional ∗ 0 k ∗ k i (−1)i vector spaces C = C → ··· C , we set det C = i=0(det C ) . The cohomology of C∗ is viewed as a complex with trivial differentials. ThereN is a natural (Euler) isomorphism det C∗ ≃ det H∗ between the determinant of a complex and the determinant of its cohomology which is well-defined up to sign. ∗ Picking a cellular decomposition of M with 0, 1 and 2-cells, we obtain a complex C (M, AdY ) 3 with a preferred volume element: we associate to each cell the volume element in Λ AdY dual to the trilinear form ε. ∗ ∗ Through the isomorphism det C (M, AdY ) ≃ det H (M, AdY ), we get an element T (M) ∈ ∗ 1 2 det H (M, AdY ) = detΩk(Y )/k ⊗ det H (M, AdY ). Given a system of generators ξ = (ξi)i∈I0 of 0 H (∂M, AdY ), we define d T (M,ξ)= hT (M), ξii∈ Ωk(Y )/k. Oi∈I0 0 2 In practice, we will evaluate it on preferred generators ξi ∈ H (∂iM, AdY ) such that Tr(ξi ) = 2 or on elements of the form ξi = ρ(γi)0 for some γi ∈ π1(∂iM).

4.3.1. The handlebody. Let M be handlebody of genus 2. Let α, β be the generators of F2 = π1(M). One has B(Γ) = k[x,y,z] with x = Yα,y = Yβ and z = Yγ. Moreover a tautological representation is given by x −1 0 −u−1 ρ(α)= , ρ(β)= 1 0 u y with K = k(x, y, z, u)/(u2 + uz + 1). The handlebody collapses to a cellular complex with one 0 1 2 0-cell and two 1-cells and the twisted complex is given by C = sl2(K),C = sl2(K) and

dξ = (ρ(α)−1ξρ(α) − ξ, ρ(β)−1ξρ(β) − ξ) Let us write T (M)= fdx∧dy ∧dz. Then, in order to compute f, it suﬃces to evaluate it on the −1 −1 vector ﬁelds ∂x, ∂y, ∂z which correspond to the twisted cocycles ψx = ρ ∂xρ, ψy = ρ ∂yρ, ψz = −1 ρ ∂zρ. Finally, f is the determinant of the matrix of d and the three cocycles in a basis of sl2(K) of volume 1. This gives f = 2 and T (M) = 2dx ∧ dy ∧ dz.

12 4.3.2. The magic manifold. Let p : S3 → S2 be the Hopf ﬁbration and Σ ⊂ S2 be the complement of three disjoint discs. Then M = p−1(Σ) is the complement of a link in S3 with 3 components called the magic manifold. Clearly, the Hopf bundle restricted to Σ is trivial and we have indeed M ≃ Σ × S1. In particular, π1(M)= F2 × Z. If ρ :Γ → SL2(k) is irreducible, it has to map the generator of Z to ±1. This shows that X(Γ) has two irreducible components Y+ and Y− of irreducible 3 type, each isomorphic to X(F2) = A . Let α, β be the generators of F2 and t the generator of Z. The tautological representation ρ : Γ → SL2(K) corresponding to Y± is given by the same formula as above with in addition ρ(t)= ±Id. The manifold M collapses to the product of a wedge of two circles with a circle. This cell complex has respectively 1,3,2 cells of dimension 0,1,2. We have d0ξ = (ρ(α)−1ξρ(α) − ξ, ρ(β)−1ξρ(β) − ξ, 0) and

d1(ζ,η,θ) = (ρ(α)−1θρ(α) − θ, ρ(β)−1θρ(β) − θ) Without surprise, this complex is the tensor product of the complex of the handlebody with the cellular complex of the circle. In particular, it splits into two independent complexes. Its 2 torsion will be g dx ∧ dy ∧ dz where g is the torsion of the acyclic complex

h·,ξ ⊗ξ ⊗ξ i / d / 2 α β αβ / 3 / 0 / SL2(K) / SL2(K) / K / 0

Using ξα = ρ(α)0,ξβ = ρ(β)0 and ξαβ = ρ(αβ)0, we get the formula g = 4 and hence T (M,ξ)= 1 2 2 dx ∧ dy ∧ dz. In the normalization Tr ξi = 2, we get dx ∧ dy ∧ dz 1 du dv dw T (M)= = ∧ ∧ 2 (x2 − 4)(y2 − 4)(z2 − 4) 2 u v w where we have set x = u + u−1p,y = v + v−1 and z = w + w−1. 4.3.3. Manifolds fibering over the circle. The following example is an adaptation of [D06]. Let Σ be compact oriented surface with boundary and ϕ :Σ → Σ be a homeomorphism preserving the orientation and fixing the boundary pointwise. The suspension M is defined as M =Σ×[0, 1]/∼ where (x, 1) ∼ (ϕ(x), 0). Its boundary is ∂Σ × S1 and its fundamental group is

π1(M)= π1(Σ) ⋊Z where the action of Z is given by the action ϕ on π1(M). More precisely, picking an element −1 t ∈ π1(M) mapping to 1, we have tγt = ϕ∗(γ) where ϕ∗ : π1(Σ) → π1(Σ) is the map induced by ϕ. Recall from Subsection 2.4.2 that the map r : X(M) → X(Σ) corestricted to Xirr(Σ) is a {±1}-principal covering on its image, the (irreducible) ﬁxed point set of ϕ∗ acting on Xirr(Σ). We choose an irreducible component Y of X(M) of irreducible type: it maps to an irreducible component Z = r(Y ) of X(Σ) of irreducible type. Moreover, the map Y → Z is a covering of order at most 2. The adjoint representation AdY is clearly independent on the representation of π1(M) chosen, hence the Reidemeister torsion actually lives on Z. Consider the long exact sequence of the pair (M, Σ) (Wang sequence), together with the corresponding sequence on the boundary. It gives the following diagram (we removed the coeﬃcients AdY from the notation).

0 / H1(M) / H1(Σ) α / H1(Σ) / H2(M) / 0 _ ∼ H0(∂Σ) / H1(∂M) / H1(∂Σ) / H1(∂Σ) ∼ / H2(∂M) / 0

13 Multiplicativity properties imply that the torsion of the first line is the element T (M) ∈ 1 2 0 detΩk(Y )/k ⊗ det H (M). Choosing generators ξ of H (∂M), gives the preferred generator of det H2(M) and hence the torsion T (M,ξ) as a volume form on Z. It remains to interpret the map α = ϕ∗ − id, which we do in the following lemma, proved in the same way as Proposition 4.1. 1 Lemma 4.7. The natural map ΩB(Σ)/k ⊗k[Z] k(Y ) → H1(Σ, AdY ) is an isomorphism. In geometric terms, H1(Σ, AdY ) is the space of rational sections of the cotangent bundle of X(Σ), pulled-back to Y . In particular, the map ϕ∗ is simply the derivative of the action of ϕ on X(Σ). Restricted to Z, it is an endomorphism of the restriction of the cotangent space of X(Σ) to Z. For any component γi of ∂Σ, i ∈ π0(∂Σ), we consider the generator ξi = ρY (γi)0. Plugging it into the sequence, we have the following interpretation of T (M,ξ) and its normalized version Proposition 4.8. On any irreducible component of the fixed point set of ϕ∗ on X(Σ) we have 1 dY 1 u−1du T (M,ξ)= γi and T (M)= i i 2 det(DϕV− 1)| 2 det(DϕV − 1)| ker dYγi ker dui −1 where ui + ui = Yγi . In particular, we recover the example of the magic manifold, where ϕ is the identity. References [BHMV95] C. Blanchet, N. Habegger, G. Masbaum and P. Vogel. Topological quantum field theories derived from the Kauffman bracket. Topology, 34, (1995), 883-927. [BH91] G. W. Brumfiel and H.M. Hilden. SL2 Representations of Finitely Presented Groups. Contemporary Mathematics, 187, American Mathematical Society, Providence. [Bul97] Doug Bullock. Rings of SL2(C)-characters and the Kauffman bracket skein module. Comment. Math. Helv., 72, no. 4, (1997), 521-542. [CS83] M. Culler and P. B. Shalen. Varieties of group representations and splittings of 3-manifolds. Ann. of Math. 117 (1983), 109-146. [D06] J. Dubois. Non abelian twisted Reidemeister torsion for fibered knots. Canadian Math.Bull. 49 (2006), 55-71. [FS04] P. P. Friis and H. Stetkær. On the quadratic functional equation on groups. [Ha77] R. Hartshorne. Algebraic Geometry Graduate Text in Mathematics, Springer (1977). [HK98] C. D. Hodgson and S. P. Kerckhoff. Rigidity of hyperbolic cone-manifolds and hyperbolic Dehn surgery. J. Differential Geometry 48 (1998), 1-60. [P87] C. Procesi. A formal inverse to the Cayley-Hamilton Theorem. Journal of algebra, 107, (1987), 63-74. [PS00] J. H. Przytycki and A. S. Sikora. On skein algebras and SL2(C)-character varieties. Topology, 39, (2000), 115-148. [S93] K. Saito Representation variety of a finitely generated group into SL2 or GL2. Kyoto University, RIMS, 1993. [Ser83] J. P. Serre. Arbres, amalgames, SL2. Astérisque, 46, SociétéMathématique de France (1983). [Sh02] P. B. Shalen. Representations of 3-manifold groups. Handbook of geometric topology, 955-1044, North- Holland, Amsterdam, 2002.

Institut de Mathematiques,´ Universite´ Pierre et Marie Curie, 75252 Paris cedex´ 05, France E-mail address: [email protected]