Timeline for Why is BG infinite dimensional for G finite ?
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13 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Nov 1, 2011 at 22:42 | answer | added | user18951 | timeline score: 16 | |
| Nov 1, 2011 at 19:35 | answer | added | Tom Goodwillie | timeline score: 31 | |
| Nov 1, 2011 at 19:23 | comment | added | Guillaume Brunerie | @Hiro Lee: I assume that bounded means that there is only a finite number of non trivial homotopy groups and that all homotopy groups are finite. | |
| Nov 1, 2011 at 19:17 | comment | added | Hiro Lee Tanaka | @Aaron, Guillaume: Are you excluding obvious K(G,1)s, like hyperbolic manifolds, the torus, or the circle? I'm assuming you are implicitly using a simply-connected assumption. | |
| Nov 1, 2011 at 19:10 | answer | added | Ian Agol | timeline score: 72 | |
| Nov 1, 2011 at 19:10 | comment | added | Guillaume Brunerie | @Aaron Do you have a reference or an explanation for this more general statement? | |
| Nov 1, 2011 at 19:03 | comment | added | Dylan Wilson | (Also the reason I didn't post this as an answer is because, while "topological", it certainly does not avoid homological algebra!! Spectral sequences make an appearance...) | |
| Nov 1, 2011 at 19:01 | comment | added | Dylan Wilson | (BTW the above is from notes for a course being taught by Steve Mitchell, you should check this out for lots more fun group cohomology stuff: math.washington.edu/~mitchell/Quillen/quillen.html) | |
| Nov 1, 2011 at 18:59 | comment | added | Dylan Wilson | The homological proof seems to be the easiest... But maybe this would work: First show this for BA, where A is cyclic (you could use an explicit model for BA to do this...). Then show that the inclusion of any subgroup H in G gives rise to a map of rings H^*G--->H^*H that displays the latter as a finitely generated module over the former. (To prove this, look hard at the fibration U(n)/G ---> BG--->BU(n) arising from a faithful unitary representation of G, say via G---> S_n ---> U(n)). Then combine these two facts! | |
| Nov 1, 2011 at 18:47 | comment | added | Mark Grant | I rather like Johannes Ebert's answer to the linked question. It is purely topological (using the theory of characteristic classes). | |
| Nov 1, 2011 at 18:23 | comment | added | Aaron Mazel-Gee | Well, there's the more general statement that any space with bounded homotopy groups is homologically infinite-dimensional. | |
| Nov 1, 2011 at 18:18 | history | asked | tj_ | CC BY-SA 3.0 |