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bio website math.vanderbilt.edu/~peters10
location Vanderbilt University
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visits member for 5 years, 2 months
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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/mathscinet-getitem?mr=2079600
Jul
7
comment 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 non-amenable groups
Apr
21
comment Non-ergodic 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
comment 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
comment 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 non-trivial 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
comment ${\rm II}_1$-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 non-trivial 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
comment ${\rm II}_1$-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
comment ${\rm II}_1$-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
comment Connes' correspondences of two $L^\infty$-algebras
Yes, but that is what is unclear to me. How are you associating a finite-additive probability measure on $X \times Y$ to a bi-normal functional on $N \otimes_{max} M^o$? I think this is where the confusion lies.
Feb
12
comment 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
comment 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 $x, y \in \mathcal B(\mathcal H)$, self-adjoint, which don't commute, yet $e^{ix} = e^{iy} = 1$.
Feb
9
comment 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 $t$.