Timeline for Can both G and BG be finite CW complexes?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 11, 2013 at 5:59 | comment | added | Jesper Grodal | @DavidSpeyer it depends on what you mean by "be" ;) See answers below. | |
| Oct 10, 2013 at 23:11 | answer | added | Jesper Grodal | timeline score: 13 | |
| Oct 3, 2013 at 6:12 | vote | accept | Dan Petersen | ||
| Oct 3, 2013 at 1:46 | answer | added | Tom Goodwillie | timeline score: 31 | |
| Oct 3, 2013 at 0:37 | answer | added | John Klein | timeline score: 9 | |
| Oct 3, 2013 at 0:20 | comment | added | Tom Goodwillie | Yes, Chris. First reduce to the case when $G$ is connected, replacing $BG$ by its (finite) universal covering space. Now if $BG$ is not contractible then its top homology (with coefficients in a suitable field) is in some positive dimension $d$. The top homology of $G$ is in some positive dimension $e$. The spectral sequence shows that the contractible space $EG$ has nontrivial $H_{d+e}$, contradiction. | |
| Oct 2, 2013 at 22:54 | answer | added | David E Speyer | timeline score: 21 | |
| Oct 2, 2013 at 22:47 | comment | added | David E Speyer | Can a topological group be a finite CW complex and not a Lie group? | |
| Oct 2, 2013 at 21:34 | comment | added | Chris Schommer-Pries | My guess is that this not possible. In fact I think if you look at the Serre spectral sequence for the fibration you mention you can see that if the cohomology of G is bounded, then the cohomology of BG must be unbounded (hence it can't be finite complex). | |
| Oct 2, 2013 at 21:20 | history | asked | Dan Petersen | CC BY-SA 3.0 |