Let $\Gamma$ be a torsion free Zariski-dense discrete subgroup of $\operatorname{SL}_3(\mathbb{R})$. Then one can show that the cohomological dimension of $\Gamma$ is less than or equal to 5. The equality case is if and only if $\Gamma$ is a cocompact lattice of $\operatorname{SL}_3(\mathbb{R})$. The question is what properties can one derive if we assume that the cohomology of $\Gamma$ is “big” enough, namely, that $\operatorname{cd} \Gamma\geq \operatorname{vcd}\operatorname{SL}_3(\mathbb{Z})=3$. Can one expect that $\Gamma$ is a lattice?
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$\begingroup$ Please use a high-level tag like gr.group-theory. I added this tag now. Regarding high-level tags, see meta.mathoverflow.net/q/1075 $\endgroup$GH from MO– GH from MO2025-11-22 20:49:08 +00:00Commented Nov 22 at 20:49
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2$\begingroup$ Well, there's an obvious non-lattice of cd = 3: subgroup of (strictly) upper triangular matrices in $SL(3, \mathbb Z)$. $\endgroup$Denis T– Denis T2025-11-22 20:56:46 +00:00Commented Nov 22 at 20:56
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$\begingroup$ Thanks @DenisT, Indeed, I forgot something in the statement! Namely, being Zariski-dense. $\endgroup$M. Han– M. Han2025-11-22 21:41:56 +00:00Commented Nov 22 at 21:41
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5$\begingroup$ The answer is "we do not have a clue," compare the last paragraph of my answer here. $\endgroup$Misha– Misha2025-11-22 22:37:01 +00:00Commented Nov 22 at 22:37
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1$\begingroup$ See also my answer here. To appreciate how hard such questions are, note that 11 years since that post, there was no progress on these problems (even though various people tried and proved some results under extra geometric assumptions). $\endgroup$Misha– Misha2025-11-23 01:27:33 +00:00Commented Nov 23 at 1:27
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1 Answer
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To summarize our current knowledge (for simplicity, restricting to finitely generated subgroups):
All known Zariski dense finitely generated torsion-free discrete subgroups of $G=SL_3(\mathbb R)$ are either lattices in $G$ or are surface groups or free products of free abelian groups of ranks 1 and 2. For instance, we do not know if there exists a discrete subgroup isomorphic to the free product of $\mathbb Z$ and the fundamental group of a closed hyperbolic surface. In particular, even list of known subgroups (up to an isomorphism) of cohomological dimension 2 is very limited. In particular, there are no known examples of virtual cohomological dimensions $3$ and $4$ which are discrete, Zariski dense and not lattices.
There are no known tools to prove nonexistence of finitely generated Zariski dense non-lattice discrete subgroups of virtual cohomological dimensions $3, 4$. We do not know enough to even make a well-informed conjecture.
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$\begingroup$ Thank you @Misha for the extended comment. I have two remarks: why considered the cohomology up to 7, I thought that the cohomology is bounded by the dimension of the symmetric space, which is 5. On the other hand, we know that $\operatorname{vcd}\operatorname{SL}_3(\mathbb{Z})=3$, hence at least we have the existence of group of cohomological dimension 3 at least. $\endgroup$M. Han– M. Han2025-11-23 12:29:04 +00:00Commented Nov 23 at 12:29
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1$\begingroup$ @M.Han: Oh, that's because my head was not clear when I was typing. It is corrected now. $\endgroup$Misha– Misha2025-11-23 14:09:35 +00:00Commented Nov 23 at 14:09
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$\begingroup$ I see now, thanks again! $\endgroup$M. Han– M. Han2025-11-23 15:45:06 +00:00Commented Nov 23 at 15:45