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SETUP: Let $G$ be a connected Lie group, and $H\subset G$ be a FINITELY GENERATED dense subgroup.

I am interested in knowing what kind of information one can infer on the complexity of $H$.

I am especially interested in the case in which $G$ is simply connected, non compact, and non diffeomorphic to $\mathbb{R}^n$. After some research online, the only result I found in this direction is in "On dense free subgroups of Lie groups", by Breuillard, E. and Gelander, T.. Here the authors prove that if $G$ is not solvable, and $H\subset G$ is finitely generated and dense, then it contains a free group of rank $r=2\dim G$.

Does anyone have other references of result in this direction? I hope to find results of the type "such a group $H$ needs to be at least this complicated".

In the case I am interested in, $H$ is the fundamental group of a compact manifold, so I have an "upper bound" on the complexity of "H". Now I want a "lower bound", if this makes any sense.

Thank you in advance!

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    $\begingroup$ Discrete dense subgroups of Lie groups can be a lot of things. For example, the braid group $B_n$ is dense in the unitary group of rank $n \choose 2$ matrices via the Lawrence-Krammer representation and work of Stoimenow. But then there's a lot of subgroups of $B_n$ that would be dense as well, presumably including infinite-rank free groups and other groups. $\endgroup$
    – Ryan Budney
    Commented Dec 17, 2012 at 19:32
  • $\begingroup$ A discrete subgroup usually means a subgroup that is discrete in the induced topology. So if $\dim(G)>0$ there is no dense discrete subgroup. You probably mean a countable or finitely generated dense subgroup of $G$. $\endgroup$
    – YCor
    Commented Dec 17, 2012 at 19:50
  • $\begingroup$ @Ryan, Yves: I edited my question, I hope it makes more sense now... $\endgroup$
    – CuriousUser
    Commented Dec 17, 2012 at 19:54
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    $\begingroup$ Marco: The result you quoted is due to Tits and is usually called "Tits' alternative". Every finitely presented group is the fundamental group of a compact manifold, so the restriction you have does not say much. What kind if complexity are you talking about? For instance, do free groups have high complexity? $\endgroup$
    – Misha
    Commented Dec 17, 2012 at 20:38
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    $\begingroup$ @Marco & Misha: the Breuillard-Gelander result also contains the requirement that the dense free subgroup is also dense (actually for some G it cannot be taken to be of rank 2, unlike in the Tits alternative) $\endgroup$
    – YCor
    Commented Dec 17, 2012 at 22:35

2 Answers 2

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You might also be interested in a theorem of Breuillard, Gelander, Souto and Storm. They prove that a connected, semisimple Lie group contains a dense copy of any finitely generated, fully residually free group (a 'limit group' in Sela's terminology). The reference is

Breuillard, Gelander, Souto, Storm, Dense embeddings of surface groups. Geom. Topol. 10 (2006), 1373--1389.

The class of limit groups is large but very well understood, and includes fundamental groups of most surfaces.

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I recently completed a preprint with Michael Larsen in this direction. Here is a link to the paper:

https://arxiv.org/abs/1312.7294

Here is the abstract: When does Borel's theorem on free subgroups of semisimple groups generalize to other groups? We initiate a systematic study of this question and find positive and negative answers for it. In particular, we fully classify fundamental groups of surfaces and von Dyck groups that satisfy Borel's theorem. Further, as a byproduct of this theory, we make headway on a question of Breuillard, Green, Guralnick, and Tao concerning double word maps.

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    $\begingroup$ For completeness: what you define as "satisfy Borel's theorem" is the following property for a group $\Gamma$: for every semisimple connected complex group $G$ and proper subvariety $V$ of $G$ and $\gamma\in\Gamma$ of infinite order, there exists a homomorphism $f:\Gamma\to G(\mathbf{C})$ such that $f(\gamma)\notin V$. Borel's theorem is that free groups satisfy this property (hence in its original statement there's no allusion to torsion: the condition is $\gamma\neq 1$). $\endgroup$
    – YCor
    Commented Nov 15, 2018 at 14:09

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