Timeline for Euler characteristic and universal cover
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 15, 2012 at 13:48 | comment | added | CuriousUser | Hi Johannes! actually, my problem is that $\tilde{M}$ retracts to a lie group, but is not one. in particular, i don't see how i can make $\pi(M)$ act on the retraction, in a free fashion | |
| Oct 14, 2012 at 15:46 | comment | added | Johannes Ebert | If $\tilde{M}$ is a Lie group and the fundamental group $\pi$ a subgroup, then there is a nonvanishing $\pi$-equivariant vector field on $\tilde{M}$ that descends to a nonvanishing vector field on $M$, whence $\chi(M)=0$. What is the case you are interested in? | |
| Oct 14, 2012 at 12:12 | answer | added | Mark Grant | timeline score: 9 | |
| Oct 9, 2012 at 13:31 | answer | added | Liviu Nicolaescu | timeline score: 2 | |
| Oct 9, 2012 at 13:30 | history | edited | CuriousUser | CC BY-SA 3.0 |
made definition more precise
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| Oct 9, 2012 at 13:11 | comment | added | CuriousUser | In the case I am interested in, $\tilde{M}$ turns out to be a vector bundle over a Lie group, so it has compact topology. To me the Euler characteristic is $\chi(M)=\sum_i (-1)^{i}\dim H_{i}(M,k)$, where $k$ is some field, say $\mathbb{R}$ or $\mathbb{Q}$. I don't know how much the question might change, but one could use compactly supported homology groups. I am very flexible at the moment. | |
| Oct 9, 2012 at 12:53 | comment | added | Oscar Randal-Williams | If $\pi_1(M)$ is infinite, then $\widetilde{M}$ is not compact: in what sense of Euler characteristic do you mean in this case? | |
| Oct 9, 2012 at 12:40 | comment | added | HJRW | And, as you say, it is obvious for finite-sheeted covers. | |
| Oct 9, 2012 at 12:40 | comment | added | HJRW | Sorry, I just realised I miscalculated the Euler characteristic! My mistake. I'm going to delete the comment, since it was silly. | |
| Oct 9, 2012 at 12:38 | comment | added | CuriousUser | HW, I am not sure what you are asking. In your example $\chi(\tilde{M})$ is not 0... if $M_1\to M_2$ is a $n$-sheeted covering then $\chi(M_1)=n\cdot\chi(M_2)$, so $\chi(M_1)=0$ iff $\chi(M_2)=0$. Probably I didn't understand your comment though. | |
| Oct 9, 2012 at 12:03 | history | asked | CuriousUser | CC BY-SA 3.0 |