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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... |