Timeline for Poincare dual in equivariant (co)homology?
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2013 at 22:56 | answer | added | Stefan Waner | timeline score: 5 | |
| Jan 25, 2012 at 19:18 | vote | accept | Guangbo Xu | ||
| Jul 21, 2011 at 11:01 | answer | added | Haggai Tene | timeline score: 6 | |
| Oct 21, 2010 at 1:24 | comment | added | Dave Anderson | @stankewicz: I wish I could say "revised and completely error-free notes", but that should be happening soon! | |
| Oct 21, 2010 at 1:20 | comment | added | Dave Anderson | @Guangbo: in your last sentence, are you asking for a "geometric" interpretation for a class in $H_G^*X$ of degree greater than $\dim X$? If so, that's also an interesting question, and a rather different answer from the one I gave below can be found in arxiv.org/abs/0910.2316 . | |
| Oct 19, 2010 at 14:58 | comment | added | stankewicz | (I should say Fulton's notes by Dave Anderson) | |
| Oct 19, 2010 at 14:32 | comment | added | stankewicz | I'm sorry but my only reference is a set of lecture notes based partly on Fulton's notes on equivariant cohomology: math.washington.edu/~dandersn/eilenberg the statement I know is just that $H^{BM,G}_i(X) = H^{2\dim_\mathbf{C}(X) - i}_G(X)$ where $X$ is a nonsingular variety. | |
| Oct 19, 2010 at 14:14 | answer | added | Dave Anderson | timeline score: 15 | |
| Oct 19, 2010 at 8:18 | answer | added | Mark Grant | timeline score: 3 | |
| Oct 19, 2010 at 0:52 | comment | added | Guangbo Xu | @ stankewicz Thank you. It seems that the Borel-Moore homology is kind of homology with compact support. But what kind of duality is it? Can you provide some reference? | |
| Oct 18, 2010 at 19:21 | comment | added | stankewicz | The method I've seen is to use Borel-Moore homology instead of regular homology so that some form of duality is thrust upon you. | |
| Oct 18, 2010 at 13:02 | comment | added | Guangbo Xu | @ Tom Goodwillie What do you mean more than one thing? Yes, for $M\times_G EG$ is not good in that sense. | |
| Oct 18, 2010 at 2:37 | comment | added | Tom Goodwillie | The expressions "equivariant (co)homology" is used to mean more than one thing. The (co)homology of $M\times _GEG$ is not very well suited to duality statements, I believe. | |
| Oct 17, 2010 at 22:25 | history | edited | Guangbo Xu | CC BY-SA 2.5 |
edited title
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| Oct 17, 2010 at 22:20 | history | asked | Guangbo Xu | CC BY-SA 2.5 |