# Finite models for torsion-free lattices

Let $$G$$ be a real, connected, semisimple Lie group and $$\Gamma < G$$ a torsion-free lattice. Then does there exist a finite $$CW$$-model for $$B\Gamma$$?

I know this to be true in many instances (e.g. when $$\Gamma$$ is uniform, when $$G=\mathrm{SO}_0(n,1)$$ or when $$\mathrm{rank}_{\mathbb R}(G) \geq 2$$ and $$\Gamma$$ is irreducible). In these instances, there always exist certain "canonical" finite CW-models for $$B\Gamma$$.

However, I am unaware of the situation for general such lattices, hence the question.

• I guess many (all?) cases were done in the 70's by Borel-Serre (at least when $G$ has finite center): there's a model that's a compact manifold (with boundary). I'd thought it was known in general, but I'm not sure. – YCor Feb 11 at 12:10
• Let $\Gamma$ be such a lattice, and assume $G$ has trivial center (if $G$ has finite center this is no restriction). Then there is a unique product decomposition $G=G_1\times \dots G_n$ such that $\Gamma$ meets $G_i$ in an irreducible lattice. Passing to the overgroup of finite index generated by projections, we can reduce to the irreducible case. Then rank $\ge 2$ is fine (probably by arithmeticity + Borel-Serre), remains rank 1. This case is certainly OK too (removing a disjoint family of horoballs and modding out directly yields a manifold with boundary). – YCor Feb 11 at 13:09
• There is a Ph.D. thesis, I don't remember now exactly by whom, whose main result is the existence of compactifications in the rank $1$ case (basically generalizing the situation for hyperbolic lattices in the way that you described). The Lie groups I am interested in are all the real points of an algebraic group, so they all have finite center anyways. – H1ghfiv3 Feb 11 at 13:16
• I think your argument shows only that each $\Gamma$ virtually has a finite CW-model, i.e that there exists a finite-index subgroup $\Gamma' < \Gamma$ possessing a finite CW-model. – H1ghfiv3 Feb 11 at 13:45
• "I think your argument shows only that each $\Gamma$ virtually..." No, I was careful on that point: I project the lattices on factors, so I embed $\Gamma$ into a larger torsion-free lattice which is a direct product (I said "overgroup"). – YCor Feb 11 at 16:17

I'll try to summarize YCor's comments into an answer (using big guns): Let $$G$$ be the real points of an algebraic group (a restriction by the OP in the comments) and assume $$\Gamma$$ irreducible.
Then Raghunathan shows that the answer is "yes" if $$\Gamma$$ is arithmetic. Margulis (Discrete subgroups of semisimple Lie groups) says that $$\Gamma$$ will be arithmetic if $$G$$ has rank at least $$2$$ (this is the sum over the ranks of the almost-factors). This leaves the case where $$G$$ is a single rank-1 factor. In that case $$\Gamma\backslash G/K$$ is a finite volume hyperbolic manifold from which one can cut off the cusps, see Theorem 12.7.2 in Ratcliffe, "Foundations of hyperbolic manifolds".
• The reduction on the irreducible case that YCor employed might be correct, I am not yet convinced (but that doesn't mean anything, I am not an expert in this area and all of what I've been asking might be very obvious to experts on Lie groups). However, the question on whether a general torsion-free discrete group $\Gamma$ has a finite CW-model for $B\Gamma$ whenever this virtually is the case is highly non-trivial, and as far as I know still open. A statement conforming this would indeed be of some profoundness (at least to me) – H1ghfiv3 Feb 12 at 18:30
• @H1ghfiv3 My attempt was to embed $\Gamma$ as subgroup of finite index in a product, so I didn't try to get finiteness of $B\Gamma$ from a finite index subgroup, but from a finite index overgroup. The problem with my approach is that the overgroup can have torsion even if the small group doesn't. – YCor Feb 14 at 16:45