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17h

revised 
symmetric measurable 2cocycles on compact abelian groups vanish?
Fix typo 
18h

comment 
symmetric measurable 2cocycles on compact abelian groups vanish?
thanks, but I am still worried that the 2cocycle 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 2cocycles 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 BorelSerre. 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 superrigidity 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  SelfLearner 
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? 