586 reputation
416
bio website
location
age
visits member for 4 years, 2 months
seen 13 hours ago

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?
Sep
14
comment vanishing higher cohomology group for property T group?
@BenWieland, thanks, I would check that.
Sep
14
comment vanishing higher cohomology group for property T group?
@BenWieland, could you give me a reference why the group mentioned by YCor has nontrivial $H^2(G, \mathbb{Z}G)$?
Sep
14
comment vanishing higher cohomology group for property T group?
@YCor, thanks for the reference!
Sep
14
comment vanishing higher cohomology group for property T group?
@YCor, in Ben's answer above, -3 paragraph, maybe there is some misunderstanding?
Sep
14
comment vanishing higher cohomology group for property T group?
@YCor, could you give me the reference on the property T group $G$ you mentioned such that $H^2(G;\mathbb{Z}G)$ is nontrivial? Thanks.
Aug
21
asked Reference on calculation of 2nd cohomology group
Aug
11
accepted vanishing higher cohomology group for property T group?
Aug
9
comment vanishing higher cohomology group for property T group?
@IgorBelegradek, thanks a lot for clarification!
Aug
9
comment vanishing higher cohomology group for property T group?
@IgorBelegradek, I am not able to download the paper you mentioned right now, if I understand your remark correctly, you mean there exists a property $T$ group $G$ with $H^2(G, \mathbb{Z})\neq 0$?
Aug
8
comment vanishing higher cohomology group for property T group?
@YCor, you mean for n=2, $G=SL_3(\mathbb{Z})$, the cohomology group vanishes?
Aug
8
asked vanishing higher cohomology group for property T group?
Jul
3
asked seek another proof of a result in Fourier analysis