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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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
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