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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 $\chi_{\mathbb{F}_q}\simeq \mathbb{F}_q^3$.

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  • $\begingroup$ $R$ is commutative and unital? Otherwise what is $SL_2$? $\endgroup$
    – markvs
    Sep 29, 2021 at 10:29
  • $\begingroup$ Yes. Thanks for the clarification Mark. $\endgroup$
    – Dr. Evil
    Sep 29, 2021 at 11:28

2 Answers 2

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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 $A[2]$ denotes the polynomial ring on 8 variables representing two 2x2 matrices, then $A[2]^{GL_{2,k}}$ ($GL_2$ 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 $A[2]\rightarrow A[\pi]$ 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 $k$.

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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

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  • $\begingroup$ 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? $\endgroup$
    – Dr. Evil
    Sep 30, 2021 at 4:20
  • $\begingroup$ Theorem 3.1 gives a model over $\mathbb{Z}[1/2]$. $\endgroup$ Sep 30, 2021 at 10:28
  • $\begingroup$ 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. $\endgroup$
    – Dr. Evil
    Oct 1, 2021 at 4:03
  • $\begingroup$ 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! $\endgroup$ Oct 1, 2021 at 11:07
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    $\begingroup$ Thanks Sean. Yes, probably my question was not precise enough, hence the confusion. Thankfully it seems we are all on the same page now. $\endgroup$
    – Dr. Evil
    Oct 10, 2021 at 23:12

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