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!

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    $\begingroup$ It's very likely that $SL_3(\mathbf{Z})$ contains free products such as $\mathbf{Z}^2\ast\mathbf{Z}$ (necessarily Zariski dense). Anyway the construction of such groups probably requires a little effort. $\endgroup$
    – YCor
    Apr 17, 2014 at 20:18
  • $\begingroup$ That's neat. Do you have any feeling on whether there might be an example that is not a right-angled Artin group? $\endgroup$ Apr 17, 2014 at 20:25
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    $\begingroup$ I don't know, there are plenty of possible variations. Still this sounds likely: I'd guess that there exists a Zariski dense subgroup of infinite index containing the integral Heisenberg group, and thus not isomorphic to a subgroup of a RAAG. $\endgroup$
    – YCor
    Apr 17, 2014 at 22:12
  • $\begingroup$ See the edit to my answer. $\endgroup$
    – Igor Rivin
    Apr 18, 2014 at 3:15
  • $\begingroup$ And another edit... $\endgroup$
    – Igor Rivin
    Apr 18, 2014 at 3:32

2 Answers 2


Here is what's known about this question:

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

  2. $\Gamma=SL(3,Z)$ 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.

  3. 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.

  4. 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 $SL(3,Z)$.

  5. 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)


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.

  • $\begingroup$ @Misha: is it known at least if there's a Zariski-dense representation of $Z^2*Z$ into $SL_3(Z)$ whose restriction to $Z^2$ is faithful, and with image of infinite index? $\endgroup$
    – YCor
    Apr 18, 2014 at 22:20
  • $\begingroup$ @YvesCornulier: No, even this is unknown. Proving this would be a big step forward (this would answer my question mentioned by OP in the very beginning). $\endgroup$
    – Misha
    Apr 18, 2014 at 22:31
  • $\begingroup$ @Misha: thanks for the reference to Benoist-Oh (math.u-psud.fr/~benoist/prepubli/09sl3.pdf), t's very interesting. Btw in our case, in their notation, $F_1$ is "reducible", and in this case they give no proof, but refer to an earlier paper of Oh "Discrete subgroups generated by lattices in opposite horospherical subgroups, CRAS 323 (1996)" gauss.math.yale.edu/~ho2/cras.pdf $\endgroup$
    – YCor
    Apr 18, 2014 at 22:33
  • $\begingroup$ @YvesCornulier: Thanks, I lost track of what Oh knew before her work with Benoist, she covered almost all cases in her PhD thesis, the remaining cases were done with Benoist. $\endgroup$
    – Misha
    Apr 18, 2014 at 22:35
  • $\begingroup$ Actually, the subject of discrete subgroups with opposite unipotents was started in the (now classical) paper of T.N.Venkataramana in the late '80s. I am not quite certain what the value added in Oh's thesis is, maybe @YvesCornulier knows... $\endgroup$
    – Igor Rivin
    Apr 19, 2014 at 18:57

MR0548436 (81k:20064) Reviewed Margulis, G. A.; Soĭfer, G. A. Nonfree maximal subgroups of infinite index of the group SLn(Z). (Russian) Uspekhi Mat. Nauk 34 (1979), no. 4(208), 203–204. 20H05 (22E40) More links PDF Clipboard Journal Article Make Link

In previous work [Dokl. Akad. Nauk SSSR 234 (1977), no. 6, 1261–1264; MR0466412 (57 #6292)], the authors showed among other things that for n≥3, SL(n,Z) has a maximal subgroup of infinite index. Independently, V. P. Platonov and G. Prasad asked whether such a subgroup must possess a free subgroup of finite index. Here the authors provide a negative answer: at least when n≥4, SL(n,Z) has a maximal subgroup of infinite index containing a free abelian group of rank 2. Besides their own previous techniques, the authors make use of the methods of J. Tits [J. Algebra 20 (1972), 250–270; MR0286898 (44 #4105)]. {English translation: Russian Math. Surveys 34 (1979), no. 4, 178–179.} Reviewed by James E. Humphreys

EDIT to address Khalid's and Yves' comments: First, it is a result of yours truly that a RANDOM (two-generator) subgroup of a Zariski-dense subgroup is Zariski dense, and result of R. Aoun that such a random subgroup is free (he proves it in slightly less generality, but the result is true, essentially by ping-pong combined with the work of Guivarc'h and Goldsheid. Note that free is NOT what Khalid wants. However, if you read Margulis-Soifer (not the paper I cited, but the one in Journal of Algebra, which actually contains complete proofs), they construct a finitely generated Zariski-dense subgroup which contains a $\mathbb{Z} \oplus \mathbb{Z}.$ They then take the maximal subgroup containing it, but you don't care about maximality, so don't do that part. Unfortunately, the Margulis-Soifer example is constructed only for $n\geq 4,$ not $n=3.$

However, progress marched on, and Venkataramana (who, I believe is a frequent contributor here), in this paper:

@article {MR892908, AUTHOR = {Venkataramana, T. N.}, TITLE = {Zariski dense subgroups of arithmetic groups}, JOURNAL = {J. Algebra}, FJOURNAL = {Journal of Algebra}, VOLUME = {108}, YEAR = {1987}, NUMBER = {2}, PAGES = {325--339}, ISSN = {0021-8693}, CODEN = {JALGA4}, MRCLASS = {20G30 (20H05 22E40)}, MRNUMBER = {892908 (88i:20068)}, MRREVIEWER = {Alexander Lubotzky}, DOI = {10.1016/0021-8693(87)90106-2}, URL = {http://dx.doi.org/10.1016/0021-8693(87)90106-2}, }

provided examples of (finitely generated, infinite index) zariski-dense subgroups which contain unipotent subgroups, thus bearing out Yves' conjecture (more or less) about the Heisenberg group.

  • $\begingroup$ Thanks for the reference. Can one draw finitely generated examples from Margulis and Soĭfer's work? It seems to me that most (all?) infinite-index maximal subgroups of, say, $\text{SL}_4(\mathbb{Z})$ are not finitely generated. $\endgroup$ Apr 17, 2014 at 21:38
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    $\begingroup$ @Khalid: any Zariski dense subgroup therein contains a finitely generated Zariski dense subgroup. $\endgroup$
    – YCor
    Apr 17, 2014 at 22:10
  • $\begingroup$ @YvesCornulier see the edit. $\endgroup$
    – Igor Rivin
    Apr 18, 2014 at 3:15
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    $\begingroup$ @Igor: your edit is confusing to me: I don't think Venkataramana claims to embed the Heisenberg group (or a finite index subgroup therein) into an infinite index Zariski-dense subgroup of $SL_3(\mathbf{Z})$, but only provides such a subgroup with a (power of a) given unipotent element. He even claims to get a free pair of commuting unipotent elements only in $SL_n(\mathbf{Z})$ for $n\ge 4$. $\endgroup$
    – YCor
    Apr 18, 2014 at 8:53
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    $\begingroup$ I am sorry; I saw this (very interesting post) just now. What Yves Cornulier says is correct; if a finite index subgroup of the Heisenberg group embeds in a Zariski dense subgroup $\Gamma$ of $SL_3({\mathbb Z}$, then $\Gamma$ does have finite index in $SL_3({\mathbb Z})$. I proved this ( a long time ago) but it is an easy consequence of a result of Tits, on unipotent generators of arithmetic groups. $\endgroup$ Sep 23, 2014 at 6:14

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