Timeline for divisible 2nd cohomolgy group $H^2(G,\mathbb{Z}G)$
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2015 at 15:47 | comment | added | ThiKu | You would have to find a space $M$ with $\pi_1M$ als desired and whose universal covering is (weakly contractible but) not homeomorphic to $R^n, n\not=2$. This doesn't look like an easy problem but perhaps I'm just unimaginative. | |
| Jul 24, 2015 at 14:48 | comment | added | Jiang | @Thiku, yes, it is a result of Borel-Serre. Do you know any example such that the 2nd cohomology group is finitely generated? Of course, G is assumed to have (T) and this would solve my question. | |
| Jul 24, 2015 at 14:37 | comment | added | Fernando Muro | @ThiKu your M must be aspherical. | |
| Jul 24, 2015 at 14:00 | comment | added | ThiKu | The first examples of a (torsionfree) Kazhdan Group are perhaps (torsionfree, finite-index subgroups of) higher rank lattices like $SL(n,Z), n\ge 3$. In this case however $\widetilde{M}$ is homeomorphic to $R^N$, so $H^2(G,ZG)=0$. | |
| Jul 24, 2015 at 13:24 | comment | added | Jiang | @ThiKu, thanks, I learned this interpretation from Brown's GTM book, but do not know how to use it. | |
| Jul 24, 2015 at 13:19 | comment | added | ThiKu | where the right hand side means cohomology with Compact Support of the universal covering. Not sure whether that helps, though. | |
| Jul 24, 2015 at 13:18 | comment | added | ThiKu | There is a topological Interpretation of that cohomology: when G is the fundamental group of a compact space M, then $$H^*(G,ZG)=H_c^*(\widetilde{M})$$ | |
| Jul 24, 2015 at 12:59 | comment | added | Jiang | @YCor, I do not have a precise description of the class of $\mathcal{G}$, but you can think this is the class of the group which satisfies $\mathbb{T}$-valued co cycle super-rigidity for its Bernoulli shift action. So, if possible, I expect an example from (T) groups. | |
| Jul 24, 2015 at 12:55 | history | edited | Jiang | CC BY-SA 3.0 |
Remove confusion
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| Jul 24, 2015 at 12:19 | comment | added | YCor | Also "infinite non-amenable group $G_1\times G_2$" precisely means "infinite non-amenable group", which most likely is not what you mean. | |
| Jul 24, 2015 at 12:18 | comment | added | YCor | Could you define precisely the class $\mathcal{G}$? or your question does not make sense. | |
| Jul 24, 2015 at 12:06 | history | asked | Jiang | CC BY-SA 3.0 |