Timeline for Differential forms of a Lie group giving cohomology of the Lie group
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2019 at 8:50 | vote | accept | Praphulla Koushik | ||
| Jul 24, 2019 at 6:32 | history | edited | Praphulla Koushik | CC BY-SA 4.0 |
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| Jul 24, 2019 at 1:28 | comment | added | Praphulla Koushik | @FernandoMuro thank you. I will see that... | |
| Jul 23, 2019 at 18:26 | comment | added | Fernando Muro | See the motivation section in en.m.wikipedia.org/wiki/Lie_algebra_cohomology | |
| Jul 23, 2019 at 18:05 | comment | added | Praphulla Koushik | @FernandoMuro I mean manifold cohomology only... I do not know why it seems like this is not what I want.. I am looking for some easy way to compute using the Lie group structure.. I was thinking if I can take some subcomplex and still get cohomology of manifold... I will read about Chevalley-Eilenberg complex. Thanks,. It looks like you see something is not ok in my question.. Please let me know what it is.. I can explain or correct myself.. | |
| Jul 23, 2019 at 18:00 | comment | added | Praphulla Koushik | I see one downvote.. if there is anything I can improve in the question, do let me know.. I may be wrong.. | |
| Jul 23, 2019 at 17:59 | comment | added | Fernando Muro | @praphullakoushik if you mean the manifold cohomology you have the trivial answer: take the whole complex. I guess this is not what you want. As pointed above, under some hypotheses (simply connected, etc.) the Chevalley-Eilenberg complex of its Lie algebra also computes the same cohomology and it has finite dimension. | |
| Jul 23, 2019 at 17:28 | comment | added | Praphulla Koushik | @MikeMiller I did not think of the possible confusion... If I say cohomology of the Lie group $G$ it is not clear if I am saying about cohomology of the underlying manifold or the cohomology of topological space $BG$.. Here I mean not $BG$. Are there other possible interpretation of “cohomology of Lie group G” | |
| Jul 23, 2019 at 17:13 | comment | added | Praphulla Koushik | @FernandoMuro I mean the cohomology of underlying manifold.. I am looking for some easy way to compute the cohomology.. sorry for the confusion... is it ok now? | |
| Jul 23, 2019 at 17:09 | comment | added | Fernando Muro | Could you clarify what notion of Lie group cohomology you're talking about? | |
| Jul 23, 2019 at 15:25 | comment | added | mme | Wait, I am confused. When you say "cohomology of this Lie group", do you intend to mean the cohomology of the classifying space BG? You can't possibly extract this from a subcomplex of the de Rham complex because $H^*(BG)$ is nonzero in arbitrarily large degrees (and in particular past $\dim G$). If what you mean is that you want a nice complex computing cohomology of the underlying manifold then Gael's complex does what you want. | |
| Jul 23, 2019 at 14:46 | answer | added | Gael Meigniez | timeline score: 7 | |
| Jul 23, 2019 at 14:13 | comment | added | Praphulla Koushik | @YCor if it is already written somewhere, can you please suggest reference.. I will read :) | |
| Jul 23, 2019 at 14:02 | comment | added | YCor | When $G$ is compact a suitable subcomplex consists of those $G$-invariant forms. When $G$ is arbitrary (connected), maybe the subcomplex of $K$-invariant forms, for $K$ maximal compact subgroup of $G$, works, but I haven't checked. | |
| Jul 23, 2019 at 14:00 | history | edited | YCor | CC BY-SA 4.0 |
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| Jul 23, 2019 at 13:58 | history | asked | Praphulla Koushik | CC BY-SA 4.0 |