Timeline for $SO(3)$ 2-cocycle trivialized to a 2-coboundary in $SU(2)$?
Current License: CC BY-SA 3.0
13 events
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Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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Dec 2, 2016 at 3:01 | vote | accept | miss-tery | ||
Nov 13, 2016 at 23:38 | comment | added | Ehud Meir | Sorry if that was not clear enough. It works only for $n=2$, I added a more detailed answer. | |
Nov 13, 2016 at 23:37 | answer | added | Ehud Meir | timeline score: 3 | |
Nov 13, 2016 at 21:20 | comment | added | miss-tery | @Ehud Meir, even $H^2$ case alone counts as an aswer already. | |
Nov 12, 2016 at 22:26 | comment | added | miss-tery | yes, I am asking the case that the nontrivial element in $H^2(SO(3),R/Z)$. | |
Nov 12, 2016 at 20:53 | comment | added | miss-tery | Does your example only work in $H^2$ or does it work for $H^n$ for other $n$? If it is clear, in either cases, you can write it as an answer even if it is trivial to you, it is non trivial to me still! Thanks. | |
Nov 12, 2016 at 12:34 | comment | added | Ehud Meir | Isn't it the case that the extension $$1\to \mathbb{Z}/2\to SU(2)\to SO(3)\to 1$$ corresponds to the nontrivial element in $H^2(SO(3),\mathbb{R}/\mathbb{Z})$? If so, it is more or less tautological that the cocycle trivializes, and it is also quite clear how to write the cochain $\beta$. | |
Nov 4, 2016 at 15:00 | history | edited | miss-tery | CC BY-SA 3.0 |
added 128 characters in body
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Nov 4, 2016 at 14:02 | comment | added | miss-tery | thanks, I mean the "group cohomology of the underlying group." | |
Nov 4, 2016 at 12:53 | comment | added | David Roberts♦ | The question you link to seems to consider arbitrary discrete groups, whereas you have Lie groups. This is kinda different! | |
Nov 4, 2016 at 12:52 | comment | added | David Roberts♦ | What sort of cohomology are you taking? Do you mean continuous (or smooth) cohomology, or the group cohomology of the underlying group? | |
Nov 4, 2016 at 2:23 | history | asked | miss-tery | CC BY-SA 3.0 |