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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!

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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". –  Mikhail Borovoi Mar 14 '11 at 18:13
    
thanks! Corrected as suggested. –  turtle Mar 15 '11 at 7:25
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2 Answers

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 $H'$.

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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 $SL(3,{\mathbb R})$).

Given what we know in rank 1 case, the "reasonable" class of thin subgroups consists of groups which one should call "geometrically finite." In rank 1 case, they are similar to lattices in the sense that they act with finite covolume on certain convex subsets of the symmetric space. In higher rank, there is not even a good general definition (the one using convex subsets is known to be wrong), the best we have so far are the "Anosov actions" of Guichard and Wienhard. Free groups obtained via the ping-poing argument as in Tits' proof satisfy Anosov property.

Incidentally, Zariski density result for lattices is due to Borel, not Margulis.

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