# Character varieties of finitely generated groups

Consider the following situation: $\Gamma_0\leq\Gamma$ are both finitely generated groups and $\Gamma_0$ has finite index in $\Gamma$. The restriction gives a well defined map between the character varieties of these groups: $$\mathrm{Res}:M(\Gamma,\mathrm{GL}_N(\mathbb{C}))\longrightarrow M(\Gamma_0,\mathrm{GL}_N(\mathbb{C}))$$ where $M(\Gamma,\mathrm{GL}_N(\mathbb{C}))$ is the GIT quotient of the variety $\mathrm{Hom}(\Gamma,\mathrm{GL}_N(\mathbb{C}))$ under the action of $\mathrm{GL}_N(\mathbb{C})$ by conjugation. Is it always true that the map $\mathrm{Res}$ is proper ? If not, can you provide a (simple) counterexample ?

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Free groups of different ranks? – Mark Sapir Jul 10 '12 at 8:51
By proper do you mean that preimage of every compact is compact? Then the answer is yes. – Misha Jul 10 '12 at 9:25
If a group acts on a Euclidean building and a finite-index subgroup fixes a point then the original group also fixes a point. To see this note that a group fixes a point iff it has a bounded orbit. The reduction of your question to buildings goes through compactification of char. varieties by group actions on buildings worked out by Ann Parreau in her thesis. – Misha Jul 10 '12 at 9:44
Misha - this should be an answer! – HJRW Jul 10 '12 at 10:05
Dear Benoit, Perhaps I'm confused (I'm not used to working with character varieties in precisely your set-up, but rather in the related, but technically slightly different, contexts), but won't the map even be finite (in the sense of algebraic geometry), i.e. proper with finite fibres. Concretely, given an $n$-dimensional rep'n of $\Gamma_0$, won't there be at most finitely many non-isomorphic ways to extend it to an $n$-dimensional representation of $\Gamma$ if the index $[\Gamma:\Gamma_0]$ is finite? Regards, – Emerton Jul 10 '12 at 15:00

Claim. The restriction map is always proper, where the target group $G$ is the group of $K$-points of a reductive group over a local field $K$, e.g. $G=GL_N({\mathbb C})$.

Proof. First, some generalities, details for which you can find, for instance, here. Let $X$ be the symmetric space or a locally compact Euclidean building corresponding to $G$. For each representation $\rho$ to $G$ of a group $\Lambda$ with generators $\gamma_1,...,\gamma_k$, define the min-max displacement $$d_\rho:=\inf_{x\in X} d_\rho(x), d_\rho(x):=\max_j \rho(\gamma_j)(x).$$ Let $x_\rho\in X$ denote a point for which $d_\rho(x)-d_\rho\le 1$. It is not hard to check that the number $d_\rho$ depends only on the projection $[\rho]$ of $\rho$ to $M(\Lambda, G)$. Furthermore, a sequence $[\rho_i]$ is precompact in $M(\Lambda, G)$ iff the sequence $(d_{\rho_i})$ is bounded. Suppose that $[\rho_i]$ is not precompact and the sequence $(d_{\rho_i})$ diverges to infinity. Then you take the asymptotic cone $Cone(X)$ of $X$ centered at the points $x_{\rho_i}$ with scaling factors $(d_{\rho_i})^{-1}$. The result is an isometric action $\rho$ of $\Lambda$ on $Cone(X)$ without a common fixed point. The key fact is that $Cone(X)$ is a Euclidean building, by a theorem of Kleiner and Leeb. By the Cartan-Tits theorem, the action $\rho$ has no bounded orbits.

Now, we can prove the claim. Suppose to the contrary, that there exists a sequence of representations $\rho_i$ of $\Gamma$ whose projections to $M=M(\Gamma, G)$ diverge, while their restrictions $\rho'_i$ to $\Gamma_0$ project to a relatively compact sequence in $M(\Gamma_0,G)$. Then the above construction yields an action $\rho$ of $\Gamma$ on $Cone(X)$.

Note that the restricted actions $\rho_i'$ have bounded $d_{\rho'_i}$. There are two cases to consider:

1. The distance between points $x_{\rho_i'}$ and $x_{\rho_i}$ is $O(d_{\rho_i})$. Then the sequence $(x_{\rho_i'})$ represents a point $x$ in $Cone(X)$. This point is necessarily fixed by $\Gamma_0$. Thus, since $|\Gamma:\Gamma_0|<\infty$, the $\Gamma$-orbit of $x$ is bounded, which is a contradiction.

2. $d_{\rho_i}=o(d(x_{\rho_i'},x_{\rho_i}))$. Then geodesic segments between $x_{\rho_i'},x_{\rho_i}$ represent a geodesic ray in $Cone(X)$. The point at infinity $\xi$ represented by this ray is fixed by $\Gamma_0$ and, moreover, every element of $\Gamma_0$ acts as a unipotent isometry of $Cone(X)$. Hence, $\Gamma_0$ has a common fixed point in $X$. Now, the argument is the same as in Case 1. QED.

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Many thanks for your precise answer Misha. Bests – Benoît Jul 10 '12 at 15:28