bio  website  

location  
age  
visits  member for  4 years, 2 months 
seen  yesterday  
stats  profile views  788 
2d

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 