# Character variety of the free group

A classical result of Fricke--Klein--Vogt from the late 1800s implies that the character variety $$\chi_\mathbb{C}$$ associated to the free group $$F_2$$ and the algebraic group $$\mathrm{SL}_2(\mathbb{C})$$ is isomorphic to $$\mathbb{C}^3$$.

Question: What happens if we change $$\mathbb{C}$$ to a different (commutative unital) ring $$R$$? I.e., can one explicitly describe the character variety $$\chi_R$$ associated to $$F_2$$ and $$\mathrm{SL}_2$$ over an arbitrary ring $$R$$? For which rings $$R$$, do we have $$\chi_R\simeq R^3$$?

I suspect that $$\chi_\mathbb{R}$$ is not isomorphic to $$\mathbb{R}^3$$ while $$\chi_{\mathbb{F}_q}\simeq \mathbb{F}_q^3$$.

• $R$ is commutative and unital? Otherwise what is $SL_2$? Sep 29, 2021 at 10:29
• Yes. Thanks for the clarification Mark. Sep 29, 2021 at 11:28

Let $$\pi$$ be a free group of rank 2. The character variety of $$SL_{2,k}$$-representations of $$\pi$$ is always isomorphic to affine 3 space $$\mathbb{A}^3_k$$ for any ring $$k$$.

Let $$A[\pi] = A[\pi]_k$$ denote the coordinate ring of $$SL_{2,k}\times SL_{2,k}$$, which we identify with $$Hom(\pi, SL_{2,k})$$ by picking a basis of $$\pi$$. Thus $$A[\pi] \cong k[\{a_i,b_i,c_i,d_i\}_{i=1,2}]/(a_id_i-b_ic_i-1)_{i=1,2}$$ Let $$X_i$$ be the matrix variable $$[[a_i,b_i],[c_i,d_i]]$$. I claim that if we let $$GL_{2,k}$$ act on $$A[\pi]$$ by simultaneous conjugation on matrices, then $$A[\pi]^{GL_{2,k}}$$ is a polynomial ring in 3 variables. $$\newcommand{\tr}{\operatorname{tr}}$$ $$\newcommand{\bZ}{\mathbb{Z}}$$

This is treated in Brumfiel-Hilden's book "SL(2)-representations of finitely presented groups" (combine their Propositions 9.1(ii) and Proposition 3.5). Specifically 9.1(ii) says that for any ring $$k$$, the ring of invariants $$A[\pi]^{GL_{2,k}}$$ is isomorphic to a certain ring $$TH[\pi]$$ (a subalgebra generated by "traces"), and Proposition 3.5 says that $$TH[\pi]$$ is in fact a polynomial ring over $$k$$ in the variables $$A = \tr(X_1)$$, $$B = \tr(X_2)$$, and $$C = \tr(X_1X_2)$$.

However, there is an issue with their statement - they actually claim that $$k[A,B,C] = TH[\pi] = A[\pi]^{GL_2(k)}$$ (taking invariants by $$GL_2(k)$$ instead of $$GL_{2,k}$$). This cannot be true, since when $$k$$ is a finite field, $$A[\pi]^{GL_2(k)}$$ is the invariant ring of a 6-dimensional algebra by a finite group, which must also be 6-dimensional, whereas $$k[A,B,C]$$ is visibly 3-dimensional. However, this is the only issue in their exposition. If you replace all instances of $$GL_2(k)$$ by $$GL_{2,k}$$, then their proof is fine.

I recently encountered this error when I tried to use their results in a paper of mine. Here are two ways to address their gap:

1. When $$k = \bZ$$ or $$k$$ is an infinite field, $$GL_2(k)$$ is Zariski-dense in $$GL_{2,k}$$, and hence in these cases their argument essentially works as is. You can bootstrap from these cases to the general case as follows. Since you know the case for $$k = \mathbb{Z}$$, you have $$\bZ[A,B,C] = A[\pi]_\bZ^{GL_{2,\bZ}}$$, so it suffices to check that taking invariants commutes with base change to any ring $$k$$. For any ring $$k$$, the universal coefficients theorem (Jantzen, I Proposition 4.18) gives an exact sequence $$0\longrightarrow A[\pi]_\bZ^{GL_{2,\bZ}}\otimes_\bZ k\longrightarrow A[\pi]_k^{GL_{2,k}}\longrightarrow Tor_1^\bZ(H^1(GL_{2,\bZ},A[\pi]_\bZ),k)\longrightarrow 0$$ Thus it suffices to show that $$H^1(GL_{2,\bZ},A[\pi]_\bZ)$$ is $$\bZ$$-flat (in fact it is 0, but we don't need this). For this, it suffices to check vanishing of the Tor group when $$k = \mathbb{F}_p$$ (for all $$p$$), and since $$\overline{\mathbb{F}_p}$$ is faithfully flat over $$\mathbb{F}_p$$, it suffices to check this when $$k = \overline{\mathbb{F}_p}$$, but since we are assuming Brumfiel-Hilden's result over infinite fields, the exact sequence above gives us this Tor vanishing for $$k = \overline{\mathbb{F}_p}$$, as desired.

2. Another approach is to put together pieces of their argument to make your own direct argument. Firstly, Brumfiel-Hilden's arguments for Proposition 3.5 are correct. Specifically, you can check that for any ring $$k$$, the map $$k[x,y,z] \rightarrow A[\pi]^{GL_{2,k}}$$ gives an isomorphism onto the subalgebra $$T[\pi]\subset A[\pi]$$ generated by traces $$\tr(X_{i_1}X_{i_2}\cdots X_{i_r})$$ where each $$i_j\in\{1,2\}$$ and $$r\ge 1$$ (injectivity is Prop 3.2, surjectivity is prop 1.7). The next piece you need is the correct statement of "The first fundamental theorem of invariant theory". Namely, you need to replace $$GL_2(k)$$-invariants in their Proposition 9.2 (first part) with $$GL_{2,k}$$-invariants. This is due to Donkin (Invariants of several matrices) when $$k = \bZ$$ or $$k$$ is an algebraically closed field, but again the same argument as above using the universal coefficient theorem extends this to all rings. This theorem says that if $$A$$ denotes the polynomial ring on 8 variables representing two 2x2 matrices, then $$A^{GL_{2,k}}$$ ($$GL_2$$ acting by simultaneous conjugation) is generated as a $$k$$-subalgebra by traces and determinants of arbitrary products of the universal matrices. Thus, to complete the argument it suffices to check that the surjective map $$A\rightarrow A[\pi]$$ induces a surjection on $$GL_{2,k}$$-invariants (note the determinants map to $$1\in A[\pi]$$). For this one can use a piece of Brumfiel-Hilden's argument (specifically in $$\S9$$, p97-98, "surjectivity of $$\rho_n$$"), which is correct and holds over arbitrary rings $$k$$.

• Regarding the general issue of being invariant for the group $SL_2(k)$ versus being invariant for the group scheme $SL_2$ over $k$, see mathoverflow.net/questions/374993/… Sep 29, 2021 at 16:51

I think it depends on what you want.

The $$\mathrm{SL}_2(\mathbb{C})$$-character variety of a free group $$F_r$$ admits a model over $$\mathbb{Z}$$ (maybe its better to say $$\mathbb{Z}[1/2]$$); see Rank 1 character varieties of finitely presented groups. And that model gives $$\mathbb{A}^3$$ for $$r=2$$. So the $$R$$-points are $$R^3$$. However, if you start from $$\mathrm{SL}_2(R)\times \mathrm{SL}_2(R)\cong \mathrm{Hom}(F_2,\mathrm{SL}_2(R))$$ and consider the action of $$\mathrm{SL}_2(R)$$ by conjugation, I think it depends on what "quotient" you consider.

By Seshadri’s extension of GIT to arbitrary base, there exists a scheme $$\mathrm{Spec}\left(R[\mathrm{Hom}(F_r, \mathrm{SL}_2(R))]^{\mathrm{SL}_2(R)}\right)$$ for $$R = \mathbb{Z}[1/2]$$. Then since $$R ֒\hookrightarrow \mathbb{C}$$ is a flat morphism we conclude: $$R[\mathrm{Hom}(F_r, \mathrm{SL}_2(R))]^{\mathrm{SL}_2(R)}\otimes_R \mathbb{C} \cong \mathbb{C}[\mathrm{Hom}(F_r, \mathrm{SL}_2(\mathbb{C}))]^{\mathrm{SL}_2(\mathbb{C})}.$$

On the other hand, if you start with $$R=\mathbb{R}$$ and define the quotient as the "polystable quotient" then you do not get $$\mathbb{R}^3$$. In fact for the $$r=1$$ case you don't get $$\mathbb{R}$$ either, as you would expect if you considered the $$\mathbb{R}$$-points of the GIT quotient $$\mathrm{Hom}(F_1,\mathrm{SL}_2)/\!/\mathrm{SL}_2\cong \mathbb{A}^1$$.

See the examples in Section 6 here (both the $$r=1$$ and $$r=2$$ cases are done):

Topology of Moduli Spaces of Free Group Representations in Real Reductive Groups

• Thanks Sean. I meant GIT quotient. I had a quick look at the Rank 1 character varieties paper. Where precisely do you discuss integral models? Sep 30, 2021 at 4:20
• Theorem 3.1 gives a model over $\mathbb{Z}[1/2]$. Sep 30, 2021 at 10:28
• Thank you very much Sean. But this theorem 3.1 is stated for complex coefficients. Are you saying the statement and the proof works for any ring R (in which 2 is invertible)? In the proof, there is an appeal to a result of Weyl. It seems like one has to at least make sure this goes through over general rings. Oct 1, 2021 at 4:03
• In the first part of my answer, I mentioned a model for the complex character variety and noted what that means for its $R$-points (with respect to that model). The second part starts with a specific $R$ and uses GIT to obtain a similar result. The third part mentions there are "other quotients" that give different answers. I hope that helps. If you have further questions, please send me an email. Good luck! Oct 1, 2021 at 11:07
• Thanks Sean. Yes, probably my question was not precise enough, hence the confusion. Thankfully it seems we are all on the same page now. Oct 10, 2021 at 23:12