Timeline for What is the equivariant cohomology of a group acting on itself by conjugation?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2019 at 12:43 | comment | added | user404153 | What is the statement for disconnected, eg. finite, groups? | |
| Feb 19, 2017 at 2:05 | comment | added | Anton Mellit | But now I'm confused because the same would show that $G/G$ (right action) is formal. Which is nonsense. | |
| Feb 19, 2017 at 1:38 | comment | added | Anton Mellit | As I understand, the main idea of this proof is that by Leray-Hirsch it is enough to show that every cohomology class of $G$ can be represented by a $G_{ad}$- invariant differential form. This can be achieved, perhaps in a more conceptual way as follows: any cohomology class on $G$ can be represented by a regular form. Then we average this form over a maximal compact subgroup $K$ to get a regular form invariant under $K$, since $K$ is Zariski dense in $G$, it must be invariant under $G$. | |
| Dec 31, 2011 at 3:33 | comment | added | user2529 | Is the equivariant homology of $G$ acting on G by conjugation still the tensor product of homology of $BG$ with homology of $G$? Thanks. | |
| Dec 28, 2011 at 16:51 | comment | added | David Ben-Zvi | It doesn't hold for example for the left translation action - as for any free action, the equivariant cohomology in that case is the cohomology of the quotient, i.e., of a point, so very far from a tensor product of cohomology of $G$ and that of $BG$. | |
| Dec 28, 2011 at 5:43 | comment | added | Gao 2Man | Does this decomposition into tensor product hold for arbitrary $G$ action on $G$, or only for the conjugation action? | |
| Apr 9, 2010 at 20:19 | comment | added | HJRW | Nice to see MO bringing UT math department together! | |
| Apr 9, 2010 at 17:15 | vote | accept | Tim Perutz | ||
| Apr 9, 2010 at 16:35 | history | answered | David Ben-Zvi | CC BY-SA 2.5 |