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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.

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    $\begingroup$ 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. $\endgroup$ – YCor Feb 11 at 12:10
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    $\begingroup$ 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). $\endgroup$ – YCor Feb 11 at 13:09
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    $\begingroup$ 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. $\endgroup$ – H1ghfiv3 Feb 11 at 13:16
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    $\begingroup$ 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. $\endgroup$ – H1ghfiv3 Feb 11 at 13:45
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    $\begingroup$ "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"). $\endgroup$ – YCor Feb 11 at 16:17
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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".

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  • $\begingroup$ I probably should have clarified that the irreducible case was clear to me, precisley because of the arguments that you employ. What I currently fail to understand is YCor's reasoning for general (reducible) lattices. Thanks anyways! $\endgroup$ – H1ghfiv3 Feb 12 at 14:26
  • $\begingroup$ Your original question was a "reference request", I think these are the references. The reduction to the irreducible case may be technical but I don't think anything profound happens (no new "references"). $\endgroup$ – Stefan Witzel Feb 12 at 16:21
  • $\begingroup$ 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) $\endgroup$ – H1ghfiv3 Feb 12 at 18:30
  • $\begingroup$ Oh yes, I didn't mean to claim anything in that direction. What you can get on abstract grounds is finite domination as I mentioned above. I didn't want to think about the details of YCor's reduction, but I probably should. $\endgroup$ – Stefan Witzel Feb 12 at 21:24
  • $\begingroup$ @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. $\endgroup$ – YCor Feb 14 at 16:45

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