Timeline for Is a Lie group equivariantly formal under conjugation by a maximal torus?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Apr 4, 2014 at 1:57 | answer | added | jdc | timeline score: 0 | |
| Jul 7, 2013 at 2:50 | vote | accept | jdc | ||
| Jul 5, 2013 at 19:04 | answer | added | Tom Baird | timeline score: 5 | |
| May 22, 2013 at 11:48 | comment | added | Allen Knutson | No, I think it's the same -- in both cases the only interesting map is a certain $G/T$-bundle. I like your description a lot more, though! | |
| May 21, 2013 at 19:02 | comment | added | Dave Anderson | @jdc, as you saw, $G$ is equivariantly formal for the adjoint action, so $H_G^*(G) \to H^*(G)$ is surjective. But this factors through the pullback $H_G^*(G) \to H_T^*(G)$, so the map $H_T^*(G) \to H^*(G)$ is surjective, as well. (@Allen, is the map you describe different from this change-of-groups homomorphism?) | |
| May 21, 2013 at 1:51 | comment | added | Allen Knutson | Let's see, $H^*_{T_{Ad}}(G) = H^*_{T_{Ad}\times G}(G\times G) = H^*_{G_{Ad}}(G\times^T G) \from H^*_{G_{Ad}}(G)$ where the last is pullback along the multiplication map $G\times^T G \to G$. So I suspect you could use the representative from that last cohomology group. | |
| May 21, 2013 at 0:23 | history | edited | jdc | CC BY-SA 3.0 |
Added "?" in title
|
| May 20, 2013 at 20:31 | history | asked | jdc | CC BY-SA 3.0 |