Timeline for On the cohomology of a finite covering map
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 4, 2021 at 9:05 | comment | added | HJRW | I think the bit in Hatcher you're referring to is the discussion of the transfer map. | |
| May 4, 2021 at 7:04 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
typos
|
| Mar 3, 2011 at 18:42 | history | edited | David Sprehn | CC BY-SA 2.5 |
added 13 characters in body
|
| Mar 3, 2011 at 18:38 | comment | added | David Sprehn | Yes, that's a much quicker way to see it than the chain-level argument I had in mind. Of course spectral sequence for $X\to X/G\to BG$ will not collapse without some assumptions... if you want a totally general answer (i.e. with integer coefficients) there will be no way to avoid doing a spectral sequence. I pointed out in my answer (because I thought the questioner was thinking of it) that the spectral sequence for $G\to X\to X/G$ does collapse, but without giving any useful information -- sorry that wasn't clear. | |
| Mar 2, 2011 at 6:36 | comment | added | Pierre | Second, the last statement, about coefficients for which the order of the group is invertible, follows immediately from the spectral sequence: in this case it has only one column, which is $H^0(G, H^*(X)) = (H^*(X))^G$. | |
| Mar 2, 2011 at 6:24 | comment | added | Pierre | A couple of comments/corrections. First the spectral sequence of $X \to X_{hG} \to BG$ need not collapse for all coefficients, even when the action is free so that the Borel construction on $X$ (which I wrote $X_{hG}$) is just $X/G$. Think of $\mathbb{Z}/2$ acting on $S^1$ by the antipodal map, which induces the identity on cohomology; the quotient is still $S^1$. On the $E_2$ page you have two copies of the cohomology of the group, which is very big, while the target $S^1$ has a cohomology in degrees 0 and 1 only. Plenty of differentials all over the place. (continued) | |
| Mar 2, 2011 at 5:11 | comment | added | David Sprehn | Right, perfect! Of course, in this case (free action) the Borel construction is homotopy-equivalent to the orbit space $X/G$. A small quibble: it's not a principle bundle (the fiber is $X$, not $G$) but merely a fiber bundle with structure group $G$. Of course, it's still a fibration so we get a Serre spectral sequence as desired. | |
| Mar 2, 2011 at 4:52 | comment | added | Chris Gerig | Thats the principal bundle $X\rightarrow X_G\rightarrow BG$ where $X_G$ is the Borel construction | |
| Mar 2, 2011 at 4:39 | comment | added | David Sprehn | Also: I have a vague memory that there's another sequence you can use, in case you want the more general case (I haven't checked this!): If memory serves, there is a different fibration, $X\to X/G\to BG$. The latter map is the classifying map of the principle bundle $X\to X/G$. But studying this would give a relationship involving the group cohomology of $G$ (maybe that's what you meant in the question though....) | |
| Mar 2, 2011 at 4:35 | history | answered | David Sprehn | CC BY-SA 2.5 |