Timeline for Is it true that there are exactly two conjugacy classes of order two elements in Out(R)?
Current License: CC BY-SA 3.0
6 events
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| Aug 9, 2012 at 21:55 | comment | added | André Henriques | $Out(R)$ is typically taken to be a Borel group (group object in the category of Borel spaces), so that's one kind of cohomology that you can take. You can also say that $Out(R)=Aut(R)/PU(R)$ where both $Aut(R)$ and $PU(R)$ are topological group (one with dense image in the other), and so $Out(R)$ is a sheaf of groups on the site of topological spaces. Then you can take a notion of group cohomolgoy that is appropriate for that context. Concerning the $H^3(G,S^1)$ that shows up in my question, there I want any cohomology that ends up being the same as $H^4(BG,\mathbb Z)$. | |
| Aug 8, 2012 at 11:55 | comment | added | David Roberts♦ | ... depending on the sort of topological 2-group you are classifying. | |
| Aug 8, 2012 at 11:53 | comment | added | David Roberts♦ | So what cohomology do you have in mind when classifying topological 2-groups? Several answers are possible... | |
| Aug 8, 2012 at 6:29 | comment | added | André Henriques | Topological 2-group. Take Aut(M) as the space of objects, with the topology of pointwise convergence on the predual (wich makes it a Polish group), and its action by unitaries of M, with the ultraweak topology (also a Polish group). | |
| Aug 8, 2012 at 5:31 | comment | added | David Roberts♦ | Is $S^1$ there being considered with the discrete topology, or as a topological group? Going back a step, are you considering your 2-group as being a topological 2-group? | |
| Aug 6, 2012 at 17:54 | history | asked | André Henriques | CC BY-SA 3.0 |