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Oct 14, 2010 at 14:59 comment added Andy Putman Isn't that the definition of having a compact classifying space?
Oct 14, 2010 at 4:50 comment added Romeo Ah, ok, this is nice, thanks. Does it go the other way, viz. if we have such a G, can we always cook-up such an X?
Oct 14, 2010 at 3:12 comment added Andy Putman 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 $X$.
Oct 14, 2010 at 2:56 comment added Romeo 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.
Oct 14, 2010 at 2:28 comment added Andy Putman 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".
Oct 14, 2010 at 1:06 comment added Autumn Kent Yeah, I would guess that there's no reasonable characterization.
Oct 13, 2010 at 23:51 comment added Romeo 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....
Oct 13, 2010 at 23:37 history answered Autumn Kent CC BY-SA 2.5