Timeline for Cohomology of the quotient of a Lie group by a finite subgroup
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Apr 23, 2022 at 7:30 | history | edited | YCor | CC BY-SA 4.0 |
formatting
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| S Apr 23, 2022 at 6:53 | history | suggested | Daniel Asimov | CC BY-SA 4.0 |
Corrected title of question
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| Apr 22, 2022 at 21:58 | review | Suggested edits | |||
| S Apr 23, 2022 at 6:53 | |||||
| Jun 27, 2011 at 7:13 | vote | accept | Alexander Lytchak | ||
| S Jun 24, 2011 at 10:14 | vote | accept | Alexander Lytchak | ||
| Jun 27, 2011 at 7:13 | |||||
| Jun 24, 2011 at 2:17 | answer | added | Craig Westerland | timeline score: 8 | |
| Jun 23, 2011 at 7:02 | comment | added | Ulrich Pennig | @Chris: A fibration sequence in spaces would not give you a long exact sequence in cohomology. You have to treat this with the Leray-Serre spectral sequence. | |
| Jun 22, 2011 at 21:38 | comment | added | Chris Gerig | Am I missing something? We have the short exact sequence $\Gamma\hookrightarrow G\rightarrow G/\Gamma$, and applying $H^3(-;\mathbb{Z})$ we see (from the long exact cohomology sequence) that the projection is surjective on $H^3$ when the inclusion is trivial on $H^3$ (due to exactness). But $H^3(\Gamma)=0$, so this is always the case. I am no pro on Lie groups, so I guess perhaps there is no long exact cohomology sequence on general Lie groups? Or I have to assume $G$ compact, etc. | |
| Jun 22, 2011 at 19:38 | vote | accept | Alexander Lytchak | ||
| S Jun 24, 2011 at 10:14 | |||||
| Jun 22, 2011 at 18:54 | answer | added | Konrad Waldorf | timeline score: 4 | |
| Jun 22, 2011 at 17:57 | comment | added | Alexander Lytchak | @Konrad: This sounds intersting and would help me to understand the situation at least a little bit. Is it difficult to see? Or is there a reference for this claim? | |
| Jun 22, 2011 at 17:48 | comment | added | Konrad Waldorf | For $G=SU(3)$ and $\Gamma=Z(G)$, the map $is$ surjective, for example. More general, for $SU(n)$ the map is surjective if and only if $n$ is odd. | |
| Jun 22, 2011 at 17:45 | comment | added | Konrad Waldorf | @Alexander: Right, I didn't notice "prime". Well, I would know how to use it: for central subgroups the question can be reformulated in terms of levels of orbifold WZW models, and these are all known. | |
| Jun 22, 2011 at 17:20 | comment | added | Alexander Lytchak | In fact, I would not know how to use that $\Gamma$ is in the center. For istance, I do not know if the map on $H^3$ is a surjection, if $G=SU(3)$ and $\Gamma$ the center of $G$. | |
| Jun 22, 2011 at 17:15 | comment | added | Alexander Lytchak | No, I would not like to assume that $\Gamma$ is in the center. If my Lie group is $Spin$, as in the question, and the order of the group not divisible by $2$, it cannot be in the center anyway. | |
| Jun 22, 2011 at 16:43 | comment | added | Konrad Waldorf | Do you assume that $\Gamma$ is a subgroup of the center? | |
| Jun 22, 2011 at 11:41 | comment | added | Jim Conant | "Finite quotient" usually means that $G/\Gamma$ is finite not that $\Gamma$ is finite as in your question. | |
| Jun 22, 2011 at 11:16 | history | asked | Alexander Lytchak | CC BY-SA 3.0 |