Timeline for Two ways of getting a cohomology class from an extension of a discrete group by $\mathbb C^*$
Current License: CC BY-SA 3.0
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Apr 4, 2013 at 3:40 | comment | added | Angelo | Dear Chris, yes, I think I can show that one of the classes vanishes if and only if the other vanishes. Also, the two classes have the same orders; this gives a strong indication that they should be coincide up to sign. | |
Apr 4, 2013 at 0:14 | comment | added | Chris Gerig | Addendum: This fibration I think is a principal bundle, and so existence of a section would mean the bundle is trivial, which should mean that the cohomology class in $H^2$ is zero, so that its image in $H^3$ is also zero, giving agreement here. | |
Apr 4, 2013 at 0:11 | comment | added | Chris Gerig | Also, your extension $\alpha$ corresponds to the $B\mathbb{C}^*$-fibration that you write, and so we want an explicit description, in terms of the underlying groups, of the failure of a set-theoretic section to be the desired map. Ideally this will have either a cocyle-description (and then check it with $\delta\alpha$) or a crossed module extension description (and then compare extensions). On afterthought, these comments are probably recasting your question into a harder one. | |
Apr 4, 2013 at 0:06 | comment | added | Chris Gerig | Just in case you haven't already looked in this direction (I got stuck): $H^3(A,B)$ corresponds to crossed module extensions. So your extension $\alpha\in H^2(G,\mathbb{C}^*)$ maps under the connecting homomorphism to a particular $\mathbb{Z}\to N\to E\to G$, which ideally you can read off from the definitions (with the help of MacLane's paper on this notion). | |
Apr 3, 2013 at 18:51 | comment | added | Angelo | To Fernando: you are right; but I am assuming that $\overline G$ is locally connected, or, equivalently, that is a Lie group, so $G$ is discrete. I edited the question to reflect this. | |
Apr 3, 2013 at 18:49 | history | edited | Angelo | CC BY-SA 3.0 |
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Apr 3, 2013 at 15:48 | comment | added | Fernando Muro | There's a difference though, in your first construction you obtain a class in $H^3(B'G,\mathbb Z)$, where $B'G$ denotes the classifying space of $G$ as a discrete group, while in your second construction you get a class in $H^3(BG,\mathbb Z)$ where $BG$ is the classifying space of the topological group $G$. Of course, you have a canonical map $B'G\rightarrow BG$ and your question may be whether the first class coincides with the pull-back along this map of the second class. | |
Apr 3, 2013 at 9:50 | history | asked | Angelo | CC BY-SA 3.0 |