Timeline for Elementary question: Intuition for equivariant cohomology
Current License: CC BY-SA 3.0
12 events
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Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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Aug 8, 2016 at 6:49 | vote | accept | HLC | ||
Aug 2, 2016 at 16:52 | vote | accept | HLC | ||
Aug 2, 2016 at 16:52 | |||||
Aug 2, 2016 at 13:32 | answer | added | Allen Knutson | timeline score: 1 | |
Aug 2, 2016 at 2:34 | review | Close votes | |||
Aug 2, 2016 at 11:31 | |||||
Aug 2, 2016 at 2:09 | comment | added | mme | The module structure comes from the map $X_G = (X \times EG)/G \to EG/G = BG$. When $G$ acts freely on $X$, $X_G = X/G \times EG$. When $X$ is a finite-dimensional manifold, $X/G$ has finite-dimensional cohomology, so the $t_i$ must be torsion in its cohomology ring. It is not strictly true that the non-torsion part of $H^*_G(M)$ is contributed by the $G$-fixed part, but rather by non-free $G$-orbits; though in the case of $T^n$, the non-free $G$-orbits have trivial rational cohomology ring if they're not fixed; for a more complicated case see $SU(2)$ acting on $S^2$. Keyword: localization. | |
Aug 2, 2016 at 1:57 | history | edited | HLC | CC BY-SA 3.0 |
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Aug 2, 2016 at 1:47 | comment | added | Michael Albanese | Don't ask the question on both sites at the same time. Pick one and ask there. | |
Aug 2, 2016 at 1:46 | comment | added | HLC | @MichaelAlbanese I have edited my post to mention that. If this is not ok, I can delete the MSE post. | |
Aug 2, 2016 at 1:45 | history | edited | HLC | CC BY-SA 3.0 |
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Aug 2, 2016 at 1:40 | comment | added | Michael Albanese | Crossposted on MSE. Please don't do that. | |
Aug 2, 2016 at 1:22 | history | asked | HLC | CC BY-SA 3.0 |