# 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
Commented Feb 11, 2020 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
Commented Feb 11, 2020 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. Commented Feb 11, 2020 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. Commented Feb 11, 2020 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
Commented Feb 11, 2020 at 16:17

In fact, more is true and you do not need separate arguments for rank 1 and higher rank.

The following is Theorem 13.1(i) in the book of Ballmann, Gromov and Schroeder "Manifolds of nonpositive curvature":

Suppose that $$(M,g)$$ is a complete real-analytic Riemannian manifold of nonpositive curvature and finite volume. Then $$M$$ is tame: It is diffeomorphic to the interior of a compact manifold with boundary $$M'$$.

Moreover, the proof shows that $$M'$$ can be realized as a submanifold (with boundary) of $$M$$.

Applying this to the locally-symmetric space $$(M,g)=X/\Gamma$$, where $$\Gamma$$ is a torsion-free lattice in the isometry group of a nonpositively curved symmetric space $$X$$, after triangulating $$M'$$, we obtain a finite model for $$\Gamma$$.

• Just one question: Do they also show that locally symmetric spaces of non-positive curvature are analytic ? Commented Feb 19, 2020 at 7:42
• @H1ghfiv3 I think this will come from the fact that your groups are real points of algebraic groups. Commented Feb 19, 2020 at 7:52
• @MoisheKohan Neat! You don't happen to know of any generalization to other lattices in locally compact CAT(0)-groups (like algebraic groups over local fields)? I'm asking because in certain cases there is an answer similar to the one I sketched, but it uses heavy machinery (reduction theory) and it is an open problem to understand how much of that one could do purely geometrically. Commented Feb 19, 2020 at 8:00
• @H1ghfiv3 There are many arguments, for instance, use an equivariant isometric embedding in the symmetric space of $SL(n, R)$, which would have to be real analytic. Or use the fact that the metric on the symmetric space is obtained by applying action of it's isometry group (which is real analytic) to a K- invariant metric on the tangent soace at the point corresponding to K in G/K. Commented Feb 19, 2020 at 9:06
• @StefanWitzel That I do not know, Gromov's argument is differential-geometric. Commented Feb 19, 2020 at 9:08

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".

Edit: I'll also try to summarize the discussion about the reduction to the irreducible case. Suppose $$G = G_1 \times \ldots \times G_k$$ and the image of $$\Gamma$$ under the projection to $$G_i$$ is an irreducible lattice $$\Gamma_i$$. Let $$\Gamma'' = \Gamma_1 \times \ldots \times \Gamma_k$$. Now the above discussion shows that each $$\Gamma_i$$ acts properly and cocompactly on a contractible CW-complex $$X_i$$. Hence $$\Gamma''$$ acts properly and cocompactly on $$X = X_1 \times \ldots \times X_k$$. This action restricts to a proper and cocompact action of $$\Gamma$$. If $$\Gamma$$ happens to be torsion-free, the action is free and $$\Gamma \backslash X$$ is a finite model for $$B\Gamma$$.

• 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! Commented Feb 12, 2020 at 14:26
• 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"). Commented Feb 12, 2020 at 16:21
• 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) Commented Feb 12, 2020 at 18:30
• 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. Commented Feb 12, 2020 at 21:24
• @H1ghfiv3 Well, in general of course it's not, apply induction. Commented Feb 19, 2020 at 10:43