Timeline for Euler characteristic and universal cover
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 15, 2012 at 14:46 | comment | added | Mark Grant | @Marco: 1) The not-so-helpful answer is "for groups of type FL over the field $k$". This seems to include finite groups (by Fernado's comment) and also any group which admits a finite classifying space. I'm not sure about finitely presented, which corresponds to haveing a classifying space with finite $2$-skeleton. 2) Yes, that's correct. | |
| Oct 15, 2012 at 13:45 | comment | added | CuriousUser | Mark, thanks a lot for this! Two questions though: 1) For which groups does this work? from the other comments, i seem to understand that it holds for $\pi$ finitely presented, is this right? 2) I seem to understand that for the definition of $\chi(\tilde{M})$ you used the "usual" cohomology groups (as opposed to the compactly supported ones), is this correct? again, thank so much! | |
| Oct 15, 2012 at 7:55 | comment | added | Mark Grant | @Will: Note also that the condition is satisfied whenever $\pi$ admits a finite (in the sense of finitely many cells) classifying space $B\pi$. For then we can take $F$ to be the cellular chains on $E\pi$. | |
| Oct 14, 2012 at 23:37 | comment | added | Ralph | @Fernando: Yes, of course. Thanks for clarifying. | |
| Oct 14, 2012 at 23:11 | comment | added | Fernando Muro | @Ralph, over a field of characteristic zero (one of Mark's assumptions) all finite groups are FP. | |
| Oct 14, 2012 at 22:52 | comment | added | Ralph | @Mark: To my understanding, your proof works for $\pi$ of type $FP$ while by the comments the statement also holds if $\pi$ is finite. I wonder, if these two cases can be unified to the case that $\pi$ has a finite index subgroup of type FP ? | |
| Oct 14, 2012 at 22:42 | comment | added | Ralph | @Will: In group theoretical notion, the condition says that $\pi$ is of type $FP$. | |
| Oct 14, 2012 at 22:22 | comment | added | Will Sawin | What does this condition mean in terms of ordinary group theory or topology? | |
| Oct 14, 2012 at 17:54 | comment | added | John Klein | @Mark: your're right, the fibration doesn't need to be orientable, and in fact is might not be. | |
| Oct 14, 2012 at 16:18 | comment | added | Mark Grant | @John: Yes, I think that's right. Although I don't really see why your fibration is orientable? (In particular, couldn't $\pi$ act non-trivially on the homology of $\tilde M$?) Meanwhile, it seems orientability is not really necessary for this argument; see mathoverflow.net/questions/80326/… | |
| Oct 14, 2012 at 16:01 | comment | added | John Klein | @Mark: Isn't your argument a variant/generalization of the following: suppose that $\tilde M$ (although not necessarily compact!) has the homotopy type of a finitely dominated space. Consider the fibration $\tilde M \to M \to B\pi$. Then if $B\pi$ is finitely dominated, the Euler characteristic of $M$ is a product of the Euler characteristic of $\tilde M$ with the Euler characteristic of $B\pi$ (because the fibration is orientable). | |
| Oct 14, 2012 at 13:36 | history | edited | Mark Grant | CC BY-SA 3.0 |
sounded more confident
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| Oct 14, 2012 at 12:12 | history | answered | Mark Grant | CC BY-SA 3.0 |