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Assume that $G$ is a semi-simple linear algebraic group defined over $\mathbb{Q}$, which is $\mathbb{Q}$-simple, and that $G(\mathbb(R)$ is non-compact, without $\mathbb{R}$-factors of rank 1. Then by Margulis's works, the arithmetic subgroups of $G(\mathbb{R})$ are the same as discrete lattice in $G(\mathbb{R})$. Here a lattice is a discrete subgroup $\Gamma$ such that the quotient $\Gamma\backslash G(\mathbb{R})$ is of finite volume with respect to the measure deduced from the left Haar measure. In particular, an arithmetic subgroup in $G(\mathbb{R})$ is Zariski dense in $G_\mathbb{R}$.

Conversely, a discrete subgroup $\Gamma$ in $G(\mathbb{R})$ is given, such that $\Gamma$ is also dense in $G_\mathbb{R}$ for the Zariski topology, what condition should one impose to make it arithmetic? Shall I assume $\Gamma$ to be finitely generated, or stable under certain actions such as $Aut(\mathbb{R/Q})$? I feel that such kind of results are more or less available in the literature, bu I'm far from an expert in this field.

Many thanks!

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  • $\begingroup$ The group Aut($\mathbf{R}$) is trivial. $\endgroup$
    – BCnrd
    Commented Nov 17, 2010 at 14:46
  • $\begingroup$ here by arithmetic subgroup I mean a subgroup of $G(\mathbb{R})$ that is commensurable with some congruence subgroup, the latter is given in term of the $\mathbb{Q}$-structure for $G$, namely $\Gamma$ is commensurable with some $\Gamma'=K\cap G(\mathbb{Q})$ with $K$ a compact open subgroup of $G(\mathbb{A}_f)$, $\mathbb{A}_f$ being the ring of finite adeles of $\mathbb{Q}$. Since $G(\mathbb{Q})$ is often too large (cf. the real approaximation), $\Gamma$ is not that large, rather it contains a cofinite subgroup of $G(\mathbb{Z})$, provided the latter being reasonably defined. $\endgroup$
    – genshin
    Commented Nov 17, 2010 at 15:17
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    $\begingroup$ Suppose $G = {\rm{SO}}(q)$ for non-deg. quad. space $(V,q)$ over $\mathbf{Q}$ that is pos.-def. at the real place, and let $G' = {\rm{SO}}(q')$ be another, where $(V,q)$ and $(V',q')$ are not $\mathbf{Q}$-isomorphic but $\dim V = \dim V'$, so $G'_{\mathbf{R}} \simeq G_{\mathbf{R}}$. Then $G'$ is a different $\mathbf{Q}$-structure on $G_{\mathbf{R}}$, and provides lots of discrete subgroups of $G(\mathbf{R}) = G'(\mathbf{R})$ that aren't arithmetic relative to the $\mathbf{Q}$-structure $G$ yet pass all "algebraic" tests. Margulis' result includes existence of suitable $\mathbf{Q}$-structure. $\endgroup$
    – BCnrd
    Commented Nov 17, 2010 at 15:41

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Arbitrary Zariski-dense subgroups in a semisimple group can be very small from a real-analytic point of view. It seems that algebra cannot distinguish between "small" and "large" Zariski-dense subgroups, so most criteria to distinguish between the two have a strong non-algebraic flavour. (Of course one can also characterize arithmetic groups algebraically, but this has even less to do with the line of argument you seem to suggest.) From a dynamical point of view, the key difference between lattices and arbitrary Zariski-dense subgroups is that the former act transitively on the product of the Furstenberg boundary of the ambient Lie group with itself ("double ergodicity"). This is a sort of "largeness" property. There are various ways to capture this property, the most systematic way seems to me the concept of a generalized Weyl group due to Bader and Furman.

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an arithmetic subgroup is a lattice. This is in characteristic zero due to Borel and Raghunathan (MR0147566) and in positive characteristic due to Harder and Behr.

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  • $\begingroup$ The OP is well-aware of this, since it is stated in the first paragraph of the question (in a more precise form). The question being posed (I think) is to give "algebraic" and/or "group-theoretic" conditions (not involving measure) that imply arithmeticity for a discrete subgroup of $G(\mathbf{R})$ that is Zariski-dense in $G_{\mathbf{R}}$. $\endgroup$
    – BCnrd
    Commented Nov 17, 2010 at 15:14
  • $\begingroup$ I wonder if an alternative description is possible, so that discrete Zariski dense subgroups $\Gamma$ would be close to being arithmetic. Of course I should not directly require $\Gamma$ to be also a lattice, which would be reduced back to Margulis' theorem. $\endgroup$
    – genshin
    Commented Nov 17, 2010 at 15:20
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    $\begingroup$ Well, there is also a characterization which is totally algebraic. It is in a paper by Lubotzky and Venkataramana: A group theoretical characterisation of S-arithmetic groups in higher rank semi-simple groups. This does not use even the Zariski density, since you do not assume your group to be linear. $\endgroup$
    – Keivan Karai
    Commented Nov 17, 2010 at 15:33
  • $\begingroup$ Plus, I do not think that Zariski dense subgroups are "close" in any sense to being arithmetic. Venkataramana has a very nice paper in which he constructs a variety of Zariski dense subgroups. $\endgroup$
    – Keivan Karai
    Commented Nov 17, 2010 at 15:35
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    $\begingroup$ Dear Keivan: great, the constructions in that paper of Venkataramana are contained in arithmetic groups relative to a specified $\mathbf{Q}$-structure. So that seems to settle things in the negative (even if it is assumed that $\Gamma$ is commensurable with $\Gamma \cap G(\mathbf{Q})$). $\endgroup$
    – BCnrd
    Commented Nov 17, 2010 at 16:16

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