# on the zariski closure and cocompactness of discrete groups

Assume $\Gamma$ is a discrete subgroup of some $GL_n$, and let $G$ be its Zariski closure. Let $H$ an algebraic cocompact normal subgroup of $G$. Do we have that $H\cap \Gamma$ is of finite index in $\Gamma$. I'm not sure this is true (actually i think it should be false), i'm far from being an expert in this field, and i don't have any couterexample yet.

Many thanks

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Can you define a Zariski discrete subgroup, which is not finite? –  Marc Palm Jul 17 '13 at 15:17
thanks for your answer. "Can you define a Zariski discrete subgroup, which is not finite?", i can't :) . i'm not sure i really get your counterexample. in this case $\overline{\Gamma}^{Zar}$ should be the lower unipotent matrices with coefficients in $\mathbb{R}$ (am i right ? ), and $H$ is not a cocompact subgroup ? –  user37253 Jul 17 '13 at 15:24
Assuming your field is real or complex numbers, the statement is true. The point is that modulo a compact normal subgroup of $G$, the group $H$ has finite index in $G$. –  Misha Jul 17 '13 at 15:28
Sorry I confused H being cocompact in GL(n) versus in Gamma:( –  Marc Palm Jul 17 '13 at 15:32
@MarcPalm: That would be the only reasonable (and quite common) interpretation. –  Misha Jul 17 '13 at 15:40

This is not true. There are irreducible (arithmetic) lattices $\Gamma=G({\mathbb Z})$ in $$SL _2({\mathbb R})\times SU(2)=G({\mathbb R})$$ such that $\Gamma$ projects densely into $SU(2)$ and its projection to $SL_2({\mathbb R})$ is a cocompact lattice. $G$ is an algebraic $\mathbb Q$-subgroup of $GL_n$ and the Zariski closure of $\Gamma$ is not $H= SL_2({\mathbb R})$ but it is $G({\mathbb R})$. $H$ is an algebraic co-compact normal subgroup of $G({\mathbb R})$ but its intersection with $\Gamma$ is only the trivial element.
Yes. $G({\mathbb Z})$ is Zariski dense in $G({\mathbb R})$. However, $G$ is ${\mathbb Q$ simple, but not absolutely simple (it is not simple even over $\mathbb R$). –  Venkataramana Jul 18 '13 at 7:40
this is not a specific lattice; there are many $\mathbb Z$ structures on $G({\mathbb R})$. –  Venkataramana Jul 19 '13 at 15:30