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seen Aug 24 at 13:59

Aug
20
awarded  Disciplined
Aug
12
comment a measurable function on a LCA group coincide with an mulitplicative character almost everywhere
Hi, could you tell me where you have met this fact? I asked an essentially same question recently, and I want to understand this stuff better.
Aug
12
asked Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
Aug
11
comment Existence of proper I.C.C. subgroup
my fault, I should mention that $G$ is inner amenable group if there exists a finite additive probability measure on the power set of $G$ which is invariant under conjugate action of $G$.(Recall for amenable group, we require that measure is invariant under left action of $G$).
Aug
11
comment induction exercise doubt
Hint: When do induction, you can not change the assumption, in your case, the assumption, $x_1x_2... x_{n+1}=1$ is the assumption, but you could translate your assumption into different form. Try set $x_i'=x_i for 1\leq i\leq n-1, x_n'=x_nx_{n+1}$.
Aug
11
comment What is the closed form for this sequence, powers of $4$?
try oeis.org, but it seems does not work...
Aug
11
comment Proof of Cauchy's Lemma in the case that G is abelian
Do reduction on order of $G$. As you did, take $U$, when $r>0$, focus on $U$, when $r=0$, consider $G/U$.
Aug
11
comment Existence of proper I.C.C. subgroup
Thanks for your quick answer, do you know any I.C.C. non amenable inner amenable groups satisfy this property?
Aug
11
accepted Existence of proper I.C.C. subgroup
Aug
11
comment Existence of proper I.C.C. subgroup
I do not understand why "g is contained in two separate cyclic subgroups". I think the following argument also works. Clearly, we can pick some $h_i$ with $h_i\not\in \langle g\rangle$ then $\langle h_i, g\rangle=T$ so $T$ is abelian, a contradition.
Aug
11
revised Question on amenable direct summand
add more assumption
Aug
11
comment Compact operators on $L^2(G)$ as a reduced cross product of $C_0(G)$ and $G$.
Why $\mathbb{C}1 \otimes C^*(\pi(c_0(G)), \{\lambda_g : g \in G\}) \cong \mathcal{K}(\ell^2(G))$? And a first feeling is that $\lambda_g$ is not a compact operator.
Aug
11
asked Existence of proper I.C.C. subgroup
Aug
11
comment Almost everywhere measurable homomorphism
Could you give any reference for this result?
Aug
4
asked Question on extension of cocycles
Aug
1
comment II$_1$-factors with finite commutant and trivial intersection generate $B(H)$?
Hope Jesse would help clarify this. I am thinking whether the other extreme happens, I.e., $\mathcal{A}'\cap B(H*H)=\mathbb{C}$...
Jul
30
comment II$_1$-factors with finite commutant and trivial intersection generate $B(H)$?
I suspect the following might be a counterexample. Take $A, N'$ to be II$_1$ factors in $B(H), B(K)$ respectively. Then consider $M=(A*A)\otimes N$ as a Von Neumann algebra of $B(H*H\otimes K)$. then take a haar unitary $u$ in the second copy of $A$, set $\mathcal{A}=A, \mathcal{B}=uAu^*$, here $A$ denotes the first copy.
Jul
27
comment Why is the natural map from maximal to reduced C star algebra surjective?
I see, the second fact you mentioned is also used to show $||.||_{max}$ is the largest norm on algebraic tensor of two C star algebras, thanks a lot!
Jul
27
accepted Why is the natural map from maximal to reduced C star algebra surjective?
Jul
26
comment Question on amenable direct summand
Oops, sorry, I should have mentioned $A\subset B\subset N$.