bio | website | |
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location | ||
age | ||
visits | member for | 4 years, 8 months |
seen | Apr 18 at 0:21 | |
stats | profile views | 844 |
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? |
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! |