Subgroups of $SL_3(\mathbb{Z})$ that are finitely generated, Zariski-dense, infinite index, and torsion-free My question stems from Misha's answer of a MathOverflow question. Misha supplied the following question in his answer:

Open question: Does there exist a finitely generated Zariski-dense torsion-free subgroup of $SL_3(\mathbb{Z})$ that is not a subgroup of some surface group?

I also find this question interesting. An answer to this question demonstrates how the well-known result "infinite-index subgroups of surface groups are free" might extend to higher-rank arithmetic groups. In light of this, I'd like to learn more about this question. So here is a broader version of this question, which I hope is easier:

My question: Let $\Delta \leq SL_3(\mathbb{Z})$ be an infinite-index, Zariski dense, and finitely generated subgroup that is torsion-free. What properties does $\Delta$ necessarily have?

Thank you for your time!
 A: Here is what's known about this question:


*

*The problem is hard and requires new ideas. The situation in the $SL(4,Z)$ case is very different and the analogy is misleading. 

*$\Gamma=SL(3,Z)$ contains no singular semisimple elements (of infinite order). This implies that there cannot be "nontrivial" semisimple RAAGs in $\Gamma$. (I call RAAG trivial if it is a free product of free abelian groups of rank $\le 2$.) One (at least, mine) motivation for looking at nontrivial RAAGs is Serre's question going back to 1977 on coherence of $SL(3,Z)$, which is still open. 

*If $\Lambda<\Gamma$ is Zariski dense and infinite index, then it cannot contain a lattice in the 3-dimensional Heisenberg group. (This follows from a more general theorem of Benoist and Oh.) This result suggests (but does not quite prove) that $\Gamma$ contains no nontrivial RAAGs. 

*It is still an interesting problem to embed RAAGs discretely in $SL(3,R)$: Such embeddings are known only for "trivial RAAGs". However, even proving existence of a discrete embedding of $Z^2 *Z$ in $SL(3,R)$ (actually, it embeds in a subgroup $SL(3, Z(1/p))$) is nontrivial. It was done independently few years ago by my student, James Forehand and Grisha Soifer. The proof uses "supersingular embeddings" of $Z^2$ in $SL(3,R)$, i.e., a semisimple embedding where both generators and their product map to singular elements. (This is a bit strange at the first glance, since Tit's ping-pong works best when elements are very proximal, which in $SL_3$ means regular.) These embedding results came as the outcome of unsuccessful attempts to play ping-pong with abelian subgroups of $SL(3,Z)$. 

*The following papers on this topic are known to be wrong: 
S. Wang, Representations of surface groups and right-angled Artin groups
in higher rank, Algebr. Geom. Topol 7 (2007), 1099-1117.
(the entire proof I think is hopeless)  
and 
G. Soifer, Free subgroups of linear groups. 
Pure Appl. Math. Q. 3 (2007), no. 4, part 1, 987-1003. 
more precisely, the proof that $Z^2 *Z$ embeds in $SL(3,Z)$ in this paper is (hopelessly) wrong. 
Edit. Lastly, and just for the record (I am pretty sure that OP knows this): Zariski dense subgroups of $\Gamma$ of course have all the standard properties of subgroups of general linear lattices, e.g., they  are residually finite, have virtual cohomological dimension $\le 3$, contain free nonabelian subgroups, etc. Some further information about their finite quotients can be derived from the general theory of "thin groups". I talked to Lena Fuchs about this problem a year ago, it appears that the "thin groups theory" does not provide much further insight into the problem. 
