bio  website  math.vanderbilt.edu/~peters10 

location  Vanderbilt University  
age  
visits  member for  5 years, 2 months 
seen  7 hours ago  
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10h

comment 
A relative property gamma and $L(\mathbb F_2)$
This is basically the only case where you have a negative answer. Ozawa showed that if $\mathcal M$ is a II$_1$ subfactor of $L(\mathbb F_2)$, then either $\mathcal M \cong \mathcal R$ or else $\mathcal M' \cap L(\mathbb F_2)^{\mathcal U}$ is a direct sum of matrix algebras: ams.org/mathscinetgetitem?mr=2079600 
Jul 7 
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Tomita Takesaki theory and boundeness of $S$
Yes, my answer only deals with a cyclic and separating unit vector $\xi$, and the corresponding state $x \mapsto \langle x \xi, \xi \rangle$. 
Jul 6 
answered  Tomita Takesaki theory and boundeness of $S$ 
May 31 
awarded  Yearling 
May 28 
answered  Free actions of nonamenable groups 
Apr 21 
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Nonergodic Dye Theorem for orbit equivalent automorphisms
You'll need to assume that the powers of $S$ also act freely so that the orbits are infinite. But more to the point, I don't remember Dye assuming ergodicity, have you checked his original paper? 
Apr 17 
awarded  Enlightened 
Apr 17 
awarded  Nice Answer 
Mar 15 
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Question about projections in von Neumann algebras
Suppose $\mathbb M_n(\mathbb C) \subset M$, and let $e_{i j} \in M$ denote the matrix with $1$ in the $ij$th position and $0$ elsewhere. Set $N = e_{1 1} M e_{1 1}$. Then show that the map $\theta: M \to \mathbb M_{n}(N)$ given by $\theta(x)_{ij} = e_{1 i} x e_{j 1}$ is a $*$isomorphism with inverse $\theta^{1}( ( a_{i j} )_{ij} ) = \sum_{i j = 1}^n e_{i 1} a_{i j} e_{1 j}$. Considering diagonal matrices then gives an isomorphism $\mathbb M_n(\mathbb C)' \cap M \cong N$. 
Mar 11 
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Question about projections in von Neumann algebras
The $z$ will not be the same as the $z$ in Takesaki's book, sorry for the confusion. Theorem V.1.41 in Takesaki shows that either $W^*(p,q)$ is not a factor, in which case $W^*(p, q)' \cap M \not= \mathbb C$. Or else $W^*(p, q)$ is $\mathbb C$ or $\mathbb M_2(\mathbb C)$, and then we again have $W^*(p, q)' \cap M \not= \mathbb C$ since $M \not\cong \mathbb C$ and $M \not\cong \mathbb M_2(\mathbb C)$ (this is an easy exercise to show). If $z$ is any nontrivial projection in $W^*(p, q)' \cap M$ then either $zp \not= 0$, or else $z^\perp p \not= 0$. 
Mar 10 
answered  Question about projections in von Neumann algebras 
Mar 9 
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${\rm II}_1$factors with finite commutant: $\mathcal{A} \cap \mathcal{B} = \mathbb{C} \Rightarrow \mathcal{A}' \cap \mathcal{B}'$ hyperfinite?
The answer to that question is yes, since there exist von Neumann algebras (even II$_1$ factors) which have no nontrivial outer automorphisms. Do you want to assume that certain groups have properly outer faithful actions? If so do you want some uniqueness properties? Then I'm guessing that the answer will be no. 
Mar 9 
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${\rm II}_1$factors with finite commutant: $\mathcal{A} \cap \mathcal{B} = \mathbb{C} \Rightarrow \mathcal{A}' \cap \mathcal{B}'$ hyperfinite?
I'm not sure how to make your question precise, could you explain more what you are looking for? 
Mar 6 
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${\rm II}_1$factors with finite commutant: $\mathcal{A} \cap \mathcal{B} = \mathbb{C} \Rightarrow \mathcal{A}' \cap \mathcal{B}'$ hyperfinite?
It's similar, $(\mathcal R_\infty^\Gamma \subset \mathcal R_\infty)$ does not encode $\Gamma$ either. 
Feb 16 
answered  Embedding the group von Neumann algebra into an injective von Neumann algebra on the same Hilbert space 
Feb 13 
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Connes' correspondences of two $L^\infty$algebras
Yes, but that is what is unclear to me. How are you associating a finiteadditive probability measure on $X \times Y$ to a binormal functional on $N \otimes_{max} M^o$? I think this is where the confusion lies. 
Feb 12 
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Connes' correspondences of two $L^\infty$algebras
I'm not so sure about your last paragraph. Are you sure this is the association given by the Riesz representation theorem? 
Feb 11 
revised 
Characterization of amenable actions
added 514 characters in body 
Feb 10 
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Is the center of the automorphism group of a von Neumann algebra M trivial whenever M is a factor?
Here is a way to find other examples. If $f \in L^\infty(X, \mu)$ is a real function then $e^{if} = 1$ if and only if $f(x)/2\pi \in \mathbb Z$, a.e. $x \in X$. Taking different masas in $\mathcal B(\mathcal H)$ you can then find $x, y \in \mathcal B(\mathcal H)$, selfadjoint, which don't commute, yet $e^{ix} = e^{iy} = 1$. 
Feb 9 
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Is the center of the automorphism group of a von Neumann algebra M trivial whenever M is a factor?
By the way, $e^{ix} = e^{iy}$ doesn't imply $e^{itx} = e^{ity}$ for all $t$. 