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Suppose $G$ is a residually-finite group and $H < G$ a torsion-free subgroup of finite index.

What characterizes such $G$ such that $BH$ is homotopic to a finite complex?

I believe Serre showed that if $G$ is arithmetic, the result holds. But I am sure since then much more must be known.

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Here's a result that gives some idea of how hard it is to characterise linear (let alone residually finite) groups of type $F$ (ie with a $K(G,1)$ that's a finite complex).

Theorem: There is a sequence of finite subsets $S_i\subseteq GL_{n_i}(\mathbb{Z})$ with the property that:

  1. for every $i$, either $G_i=\langle S_i\rangle$ is of type $F$ or $G_i$ is not finitely presentable (in particular not of type $F$);
  2. the set of $i$ such that $G_i$ is of type $F$ is recursively enumerable but not recursive.

So there is no algorithm to determine whether or not $G_i$ is of type $F$.

I can give details of the proof if anyone's interested. Basically, it's an easy application of the Haglund--Wise version of the Rips Construction.


The first ingredient is a sequence of finite presentations for groups $(Q_i)$, with the property that the set $\{i\mid Q_i\cong 1\}$ is recursively enumerable but not recursive. We also want to set things up so the non-trivial $Q_i$ are infinite. Such sequences are quite well known, see for instance Chuck Miller's survey article.

The second ingredient is provided by Haglund and Wise, in one of many variations of a famous construction of Rips. For any finite presentation for a group $Q$, Haglund and Wise construct a short exact sequence

$1\to G\to \Gamma\to Q\to 1$

with the following properties:

  1. $\Gamma$ is the fundamental group of a `virtually special', non-positively curved square complex $X$; and
  2. $G$ is finitely generated.

Non-positive curvature is a local condition on $X$ which ensures that its universal cover is contractible; in particular, $\Gamma$ is of type $F$. Being `special' is a condition on the hyperplanes of $X$. All you need to know is that it ensures that $\Gamma$ is (virtually) a subgroup of a right-angled Coxeter group, from which it follows that $\Gamma$ is a subgroup of $GL_{n}(\mathbb{Z})$ for some $n$.

Everything in this construction is completely explicit. Given a presentation for $Q$, one can write down a presentation for $\Gamma$ and the generators $S$ for $G$. Furthermore, you can also write down an explicit embedding $\Gamma\hookrightarrow GL_n(\mathbb{Z})$.

Finally, we apply this construction to the $Q_i$. If $Q_i\cong 1$ then we have $G_i\cong\Gamma_i$, so in particular it's of type $F$. On the other hand, a result of Bieri ensures that if $Q_i$ is infinite then $G_i$ isn't finitely presentable. (This uses the fact that the $\Gamma_i$ are of cohomological dimension two.)

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Oh, that's interesting. I'd love to see details on this. –  Romeo Oct 14 '10 at 2:49
I must sleep now, but I'll try to find time tomorrow. –  HJRW Oct 14 '10 at 2:58
Also, if you have any comments on my comment after Andy's above, I'd be very interested to hear them. –  Romeo Oct 14 '10 at 2:59
Nice, thanks for that. –  Romeo Oct 14 '10 at 23:15
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There are plenty of torsion-free residually finite groups with no finite dimensional Eilenberg-Mac Lane space.

You can even take $G = H$ to be finitely presented: Stallings showed if $F_2$ is the free group of rank two generated by $a$ and $b$, then the kernel of the map

$F_2 \times F_2 \times F_2 \to \mathbb{Z}$

given by sending each copy of $a$ and $b$ to the generator of $\mathbb{Z}$ is finitely presented but has infinitely generated third homology.

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Right; thanks for the example. I know the result is false in general, but I'm not clear on what properties G must have to guarantee that BH is finite (this may not even be known). But the geometric group theory literature is vast.... –  Romeo Oct 13 '10 at 23:51
Yeah, I would guess that there's no reasonable characterization. –  Richard Kent Oct 14 '10 at 1:06
I concur with Richard -- though a number of classes of groups are known to have the property in question (arithmetic groups, mapping class group, automorphism groups of free groups, etc.), once you leave the world of well-behaved groups like that all hell breaks loose. There's no hope for any kind of general criterion. Maybe the most useful thing for you to do would be to read the section on finiteness properties in Brown's book on the cohomology of groups and then follow it up with Bestvina-Brady's paper "Morse Theory and Finiteness Properties of Groups". –  Andy Putman Oct 14 '10 at 2:28
Can we pin these down by requiring that they act on some nice (eg, highly-connected) space in a nice (eg, finite stabilizers) way? Or can someone shoot this down by giving an example of a group acting on a contractible space X with finite stabilizers such that BH isn't finite? I've looked at Brown's book and some old papers of Serre, but I'm not sure these quite address this. I'll check out the B-B paper, though, thanks. –  Romeo Oct 14 '10 at 2:56
If $G$ acts cocompactly on a contractible CW complex $X$ (in a reasonable way) and all the stabilizers have compact classifying spaces, then $G$ has a compact classifying space. The idea is to construct a classifying space for $G$ out of $X/G$ and classifying spaces for the stabilizers using a homotopy colimit type construction. This is a "spaces" version of what Brown does in his book in the sections on equivariant homology. You can't relax contractibility to high connectivity -- the group action tells you nothing about the homology of $G$ above the range of connectivity of $X$. –  Andy Putman Oct 14 '10 at 3:12
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