Timeline for What is the equivariant cohomology of a group acting on itself by conjugation?
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Apr 9, 2014 at 21:46 | answer | added | jdc | timeline score: 6 | |
| Apr 11, 2010 at 13:32 | answer | added | Craig Westerland | timeline score: 6 | |
| Apr 9, 2010 at 17:15 | vote | accept | Tim Perutz | ||
| Apr 9, 2010 at 17:02 | history | edited | Tim Perutz | CC BY-SA 2.5 |
Reverted, given answer just in from David Ben-Zvi.
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| Apr 9, 2010 at 16:35 | answer | added | David Ben-Zvi | timeline score: 43 | |
| Apr 8, 2010 at 3:53 | answer | added | David Ben-Zvi | timeline score: 7 | |
| Apr 8, 2010 at 0:46 | history | edited | Tim Perutz | CC BY-SA 2.5 |
Corrected statement about E_2
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| Apr 8, 2010 at 0:46 | answer | added | Dan Ramras | timeline score: 25 | |
| Apr 8, 2010 at 0:45 | comment | added | Tim Perutz | @Paul. No, that was carelessness; I will correct it. | |
| Apr 8, 2010 at 0:42 | comment | added | Paul | One other thought, your fibration has a section (e.g. the identity gauge transformation) and that might help collapse the spectral sequence. | |
| Apr 8, 2010 at 0:30 | comment | added | Paul | I don't know, but are you assuming field coefficients? Or is there another reason why $E_2^{p,q}$ is a tensor product? | |
| Apr 7, 2010 at 23:11 | history | asked | Tim Perutz | CC BY-SA 2.5 |