Timeline for For G a Lie group, can I make sense of G/G as a derived manifold in a nice way?
Current License: CC BY-SA 3.0
11 events
when toggle format | what | by | license | comment | |
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Oct 2, 2013 at 17:56 | vote | accept | Qiaochu Yuan | ||
Oct 2, 2013 at 11:51 | answer | added | Ben Webster♦ | timeline score: 5 | |
Oct 2, 2013 at 0:36 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
added 338 characters in body
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Oct 1, 2013 at 22:52 | answer | added | Sam Gunningham | timeline score: 9 | |
Oct 1, 2013 at 18:36 | answer | added | A non. | timeline score: 5 | |
Oct 1, 2013 at 18:04 | comment | added | Qiaochu Yuan | @Jason: the nicest thing would be to have $\Omega^{\bullet}(M/G) \cong \Omega^{\bullet}(M)^G$ but this doesn't hold in the above example. The quotienting procedure is maybe not so important; what I really want is to write down some smooth object which has de Rham algebra $\Lambda^{\bullet}(\mathfrak{g}^{\ast})$, vector fields $\mathfrak{g}$, etc. | |
Oct 1, 2013 at 17:24 | comment | added | user36931 | Well one can work with G/G as an ordinary stack and then the natural version of de Rham cohomology will be the Cartan model for equivariant cohomology, you can get back to the de Rham algebra of G by taking derived tensor over H^*(BG) with H^*(pt). Don't see any riches down that path... | |
Oct 1, 2013 at 17:11 | comment | added | Jason Starr | Okay, got it. So you are observing that the de Rham algebra on $G$ is (left) $G$-equivariantly isomorphic to the pullback of the Chevalley-Eilenberg algebra under projection to a point. In what sense is that a problem with quotients that needs to be "fixed"? | |
Oct 1, 2013 at 16:59 | comment | added | Qiaochu Yuan | @Jason: I don't think I am. I want left-invariant differential forms on $G$, not conjugation-invariant differential forms. | |
Oct 1, 2013 at 16:56 | comment | added | Jason Starr | Are you sure you aren't confusing the left regular action with the conjugation action? | |
Oct 1, 2013 at 16:47 | history | asked | Qiaochu Yuan | CC BY-SA 3.0 |