Reputation
678
Next privilege 1,000 Rep.
See votes, expandable usercard
Badges
5 17
Impact
~12k people reached

  • 0 posts edited
  • 0 helpful flags
  • 87 votes cast
Sep
30
comment torsion free for the 2nd cohomology group?
@YCor, I have decided to write down the proof, you can find it as corollary 4.1 in arxiv.org/abs/1509.08278. Thanks!
Sep
30
accepted symmetric measurable 2-cocycles on compact abelian groups vanish?
Sep
19
comment Infinite number of non-isomorphic von Neumann algebras with property Gamma?
@JonBannon, Sometimes, big theorems suffer from the same fate as nuclear bombs: once invented, never used. I like this theorem very much. One reason is that it involves three fundamental objects in II$_1$ factors: the hyperfinite II$_1$ factor $R$, the free group factor $L(F_n)$ and tensor product. And many questions on $R\otimes L(F_n)$ is unanswered, e.g., I think it is not known whether there exists a MASA $A$ in $R\otimes L(F_n)$ such that $A$ is a maximal injective subalgebra. I tend to believe no.
Sep
18
comment Infinite number of non-isomorphic von Neumann algebras with property Gamma?
You can apply Ozawa and Popa's theorem proved in the paper arxiv.org/abs/math/0302240.
Aug
28
comment symmetric measurable 2-cocycles on compact abelian groups vanish?
thanks, it seems that theorem 10 in ams.org/mathscinet-getitem?mr=414775 is the statement you are talking above?
Aug
27
revised symmetric measurable 2-cocycles on compact abelian groups vanish?
Fix typo
Aug
27
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?
Aug
27
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...