Timeline for cohomological dimension of a group acting on a product
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 18, 2012 at 13:05 | vote | accept | Andrew Zimmer | ||
| Jul 16, 2012 at 22:15 | answer | added | Ralph | timeline score: 6 | |
| Jul 16, 2012 at 17:25 | comment | added | Andrew Zimmer | @Tom, no you were right originally. When I added some background I also added the words "torsion free." I was careless when I first posted the question. | |
| Jul 16, 2012 at 17:00 | comment | added | Tom Goodwillie | Oh, sorry, I think I overlooked the words "torsion free". | |
| Jul 16, 2012 at 15:31 | answer | added | Ian Agol | timeline score: 2 | |
| Jul 16, 2012 at 14:47 | history | edited | BS. |
added tag
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| Jul 16, 2012 at 9:15 | comment | added | YCor | @Chris: have you got a nontrivial (= with no finite index cyclic subgroup) example of a discrete group acting properly on $M\times\mathbf{R}$ with $M$ compact manifold? @Tom: if I don't miss anything, for $n=0$ it's clear that the conclusion is that $G$ is finite (properness means that for any compact $K$, the set of $g$ such that $gK\cap K$ is nonempty is finite; then take $K=M$) i.e. trivial if $G$ is assumed torsion-free. | |
| Jul 16, 2012 at 0:38 | comment | added | Andrew Zimmer | Thanks for your comments. I have modified the question to give some background. | |
| Jul 16, 2012 at 0:37 | history | edited | Andrew Zimmer | CC BY-SA 3.0 |
added motivation
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| Jul 16, 2012 at 0:04 | comment | added | Chris Gerig | Even with $n=1$, for $cd(G)=1$ iff $G$ is free. You shouldn't expect anything unless your space is a $K(G,1)$-complex, for then $cd(G)\le$ geometric dimension of $G$ (minimal dimension of such a complex). | |
| Jul 15, 2012 at 23:57 | history | edited | Andrew Zimmer | CC BY-SA 3.0 |
added 2 characters in body
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| Jul 15, 2012 at 23:02 | comment | added | Tom Goodwillie | Consider the case when $n=0$. | |
| Jul 15, 2012 at 22:48 | history | asked | Andrew Zimmer | CC BY-SA 3.0 |