Timeline for Equivariant singular cohomology
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 25, 2011 at 4:35 | answer | added | Peter May | timeline score: 11 | |
| Feb 24, 2011 at 22:44 | answer | added | Chris Gerig | timeline score: 2 | |
| Feb 8, 2010 at 12:01 | vote | accept | Kevin H. Lin | ||
| Dec 24, 2009 at 13:36 | comment | added | Chris Schommer-Pries | Those adjunctions can only exist if G is discrete. It's not clear from the question if we are assuming this or if G was allowed to be, say, a compact Lie group. | |
| Dec 23, 2009 at 12:32 | comment | added | user2146 | Is something like the following possible? I guess there are adjunctions $GTop\rightleftarrows GsSet \rightleftarrows GsAb\rightleftarrows Ch(GAb)$. I would denote the composition with $N\circ\mathbb{Z}\circ Sing$. Then perhaps one can define $H^n(X,\mathbb{Z})= H^n(N\circ\mathbb{Z}\circ Sing(X))$. | |
| Dec 23, 2009 at 5:34 | comment | added | shenghao | or hypercohomology of the space BG with coeff in ... | |
| Dec 23, 2009 at 3:57 | answer | added | Ben Webster♦ | timeline score: 5 | |
| Dec 23, 2009 at 3:15 | comment | added | Kevin H. Lin | Ah, ok, so this is, umm, hyper-group-cohomology? | |
| Dec 23, 2009 at 3:02 | answer | added | Kevin McGerty | timeline score: 12 | |
| Dec 23, 2009 at 2:52 | comment | added | Mariano Suárez-Álvarez | If $G$ acts on $M$, then $G$ acts on $M$'s de Rham complex $\Omega^\bullet(M)$. Now take $\mathbb{H}^\bullet(G,\Omega^\bullet(M))$. This gives you an equivariant theory. You can do it replacing $\Omega^\*(M)$ by $S^\*(M)$, the singular complex, of course. Since hypercohomology sees only the quasi-isomorphism type of its argument, you get isomorphisms between what you get from de Rham and what you get from $S^\*(M)$, &c. | |
| Dec 23, 2009 at 2:47 | comment | added | Kevin H. Lin | Do you really mean just the de Rham complex? There's no $G$−stuff in the de Rham complex, and shouldn't equivariant stuff involve $G$−stuff? | |
| Dec 23, 2009 at 2:46 | history | edited | Kevin H. Lin | CC BY-SA 2.5 |
added 156 characters in body; added 1 characters in body
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| Dec 23, 2009 at 2:35 | comment | added | Mariano Suárez-Álvarez | Isn't Cartan's equivariant cohomology the hypercohomology of the de Rham complex? | |
| Dec 23, 2009 at 2:30 | history | asked | Kevin H. Lin | CC BY-SA 2.5 |