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17h
revised symmetric measurable 2-cocycles on compact abelian groups vanish?
Fix typo
18h
comment symmetric measurable 2-cocycles on compact abelian groups vanish?
thanks, but I am still worried that the 2-cocycle relation holds almost everywhere, not everywhere in my problem, is it appropriate to have a pure algebraic argument as above to show this?
19h
asked symmetric measurable 2-cocycles on compact abelian groups vanish?
Jul
24
comment divisible 2nd cohomolgy group $H^2(G,\mathbb{Z}G)$
@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
comment divisible 2nd cohomolgy group $H^2(G,\mathbb{Z}G)$
@ThiKu, thanks, I learned this interpretation from Brown's GTM book, but do not know how to use it.
Jul
24
comment divisible 2nd cohomolgy group $H^2(G,\mathbb{Z}G)$
@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
revised divisible 2nd cohomolgy group $H^2(G,\mathbb{Z}G)$
Remove confusion
Jul
24
asked divisible 2nd cohomolgy group $H^2(G,\mathbb{Z}G)$
Apr
18
awarded  Self-Learner
Apr
18
awarded  Yearling
Apr
18
accepted $\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $l^1(G)$ is a flat $\mathbb{Z}G$-(right) module?
Apr
18
answered $\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $l^1(G)$ is a flat $\mathbb{Z}G$-(right) module?
Nov
22
accepted amenable + without $BS(m,n)$+finite $K(G,1)$implies virtually cyclic?
Nov
18
comment torsion free for the 2nd cohomology group?
@YCor, if my proof is correct, then this is a ``direct" consequence of Popa's cocycle superrigidity result for Bernoulli shift of property (T) group (and of course it holds for a more wider class of groups) plus taking advantage of the principal algebraic action setting, although this is not what my primary goal...
Nov
18
comment torsion free for the 2nd cohomology group?
@YCor, thanks for mentioning this group, I learned it from the book on Kazhdan's property (T). And I asked this question because I find a proof that my question has a positive answer but I am not sure whether this is known or not...
Nov
18
comment amenable + without $BS(m,n)$+finite $K(G,1)$implies virtually cyclic?
@QiaochuYuan, I am not sure it is suitable to be posted in MO, that's why I first asked it here.
Nov
18
comment amenable + without $BS(m,n)$+finite $K(G,1)$implies virtually cyclic?
@QiaochuYuan, yes.
Nov
17
asked amenable + without $BS(m,n)$+finite $K(G,1)$implies virtually cyclic?
Nov
11
asked torsion free for the 2nd cohomology group?
Sep
17
comment vanishing higher cohomology group for property T group?
@BenWieland, I checked the references you mentioned, but I do not know why the group mentioned by YCor is of type FP. Could you please explain this?