Let $G$ be a semisimple Lie group with finite center, $K$ a maximal compact subgroup, and $d=\dim(G/K)$. Let $\Gamma$ be a non-cocompact discrete subgroup of $G$. [Edit: assume that $\Gamma$ is virtually torsion-free (which is automatic if $G$ is linear and $\Gamma$ is finitely generated).]

Is it true that the virtual cohomological dimension (vcd) of $\Gamma$ is $<d$?

The virtual cohomological dimension in this case is the cohomological dimension of some/every torsion-free finite-index subgroup. I guess it's also the rational cohomological dimension.


  • if $\Gamma$ is cocompact the vcd equals $d$;
  • in general, the vcd is $\le d$;
  • if $\Gamma$ is a non-cocompact lattice, then the vcd is $<d$ (at least in the arithmetic case, where it's related to the $\mathbf{Q}$-rank, cf work of Borel-Serre, and also in the rank-1 case; I think the general case follows).

I'd also be interested by variants of this question, where vcd is replaced by the asymptotic dimension, or by Roe's coarse cohomological dimension.

  • $\begingroup$ In case $\Gamma$ is not virtually torsion-free, its vcd is infinite so one has to assume this, as emphasized by Misha in his answer. (When $G$ is non-linear, in some cases it can happen that $\Gamma$ is finitely generated but not virtually torsion-free.) $\endgroup$ – YCor Apr 11 at 8:34

First of all, since you are not assuming finite generation, you should at least assume that $\Gamma$ is virtually torsion-free. (Otherwise, you need to work rationally and reprove Whitehead's lemma, in the setting of orbifolds/orbicomplexes; see below.)

With this extra assumption, here is a proof:

WLOG, $\Gamma$ is torsion-free; let $X=G/K$ be the associated symmetric space so $M=\Gamma\backslash X$ is an open $d$-dimensional manifold. Whitehead proved in Lemma 2.1 in

"The immersion of an open 3-manifold in euclidean space", Proc. London Math. Soc 11 1961, 81-90.

that every open triangulated $d$-dimensional manifold $M$ deformation retracts to its $d-1$-dimensional skeleton $Y\subset M$.

See also Andy Putman's write-up of the proof here.

Thus, in your case, there exists a $d-1$-dimensional $Y=K(\Gamma,1)$. Hence, $cd(\Gamma)\le d-1$.

Edit: If $X$ is an irreducible symmetric space of rank $\ge 2$, I do not know of any examples of discrete subgroups $\Gamma< G$ (which are not non-uniform lattices) of vcd equal to $d-1$. (Most likely, such subgroups simply do not exist.) In contrast, such examples abound in the rank 1 case (more precisely, for real and complex hyperbolic spaces).

  • $\begingroup$ Excellent, thanks! I'm happy with the case when $\Gamma$ is finitely generated, so I accept the answer. $\endgroup$ – YCor Apr 10 at 20:13
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    $\begingroup$ In addition, it works for an arbitrary virtually connected Lie group $G$. $\endgroup$ – YCor Apr 10 at 20:52

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