# “Dimension” of discrete subgroups of infinite covolume in Lie groups

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 $$?

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.

Remarks:

• 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 $$ (at least in the arithmetic case, where it's related to the $$\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.

• 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.) – YCor Apr 11 '19 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 $$Y\subset M$$.

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).
• Excellent, thanks! I'm happy with the case when $\Gamma$ is finitely generated, so I accept the answer. – YCor Apr 10 '19 at 20:13
• In addition, it works for an arbitrary virtually connected Lie group $G$. – YCor Apr 10 '19 at 20:52