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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
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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