# Zariski density of conjugates of subgroups by arithmetic subgroups?

Let $G$ be a linear algebraic $\mathbb{Q}$-group, which is assumed to be connected, $\mathbb{Q}$-simple, and of adjoint type, such that the Lie group $G(\mathbb{R})$ has no compact factor defined over $\mathbb{Q}$. Let $\Gamma\subset G(\mathbb{Q})$ be a congruence subgroup. It is known, from the theory of Margulis, that $\Gamma\subset G(\mathbb{R})$ is Zariski dense. For convenience assume that $\Gamma\subset G(\mathbb{R})^+\cap G(\mathbb{Q})$ and that $\Gamma$ is torsion free. Note also that in this case, if one takes $X$ to be the non-compact symmetric domain associated to $G(\mathbb{R})^+$, then the quotient $X/\Gamma$ is a localy symmetric manifold of negative curvature (a typical example of hyperbolic manifold}.

I'd like to consider conjugates of linear $\mathbb{Q}$-subgroups of $G$ under $\Gamma$. More restrictively, let me take $H\subset G$ a connected semi-simple $\mathbb{Q}$-group such that $H(\mathbb{R})$ again has no compact factors defined over $\mathbb{Q}$. Then

(1) is the union $\bigcup_{g\in\Gamma}gHg^{-1}$ Zariski dense in $G$?

(2) if $\Gamma'$ is a finitely generated subgroup of $\Gamma$, and $H'$ be the Zariski closure of the subgroup of $G(\mathbb{Q})$ generated by $\bigcup_{g\in \Gamma'}gH(\mathbb{Q}) g^{-1}$, then how far is $\Gamma'$ from being an arithmetic subgroup of $H'$?

Thanks!

• Maybe you should assume in your question that $G$ is $\mathbb{Q}$-simple. Otherwise you can take $H$ to be a normal $\mathbb{Q}$-subgroup of $G$ and the answer to question (1) is trivially "no". Commented Mar 14, 2011 at 18:13

Part (2) is also hopeless: If you look at the proof of the Tits' alternative, you see that every lattice $\Gamma$ in a semisimple Lie group will contain a "Schottky subgroup", which is a free subgroup of finite rank $\Gamma'$, Zariski dense in $G({\mathbb R})$, but not a lattice. Infinite index subgroups of lattices which are Zariski dense are called thin subgroups, it is an active research area in the last few years. What is not well-understood is how to construct non-free thin subgroups (they do exist sometimes, of course): There are several general construction of non-free thin subgroups in rank 1 case, but higher rank is a different story (especially, if you look at lattices in $SL(3,{\mathbb R})$).
Unless I am missing something, the answer to (1) is NO, in general. For example, take $H$ to be the image of $SL_2$ in its adjoint representation, which lands in $G=SL_3$. Then, every element of the conjugate set in (1) has the property that $det (g-1) =0$, which is not true for $g\in SL_3$.
(2) needs a reformulation. You can take $\Gamma '$ to be trivial. Then $H'=H$ and then $\Gamma '$ is not going to be an arithmetic subgroup of $H'$.