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bio website math.vanderbilt.edu/~peters10
location Vanderbilt University
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visits member for 4 years, 11 months
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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 $t$.
Feb
9
comment Is the center of the automorphism group of a von Neumann algebra M trivial whenever M is a factor?
Here is an alternate way to finish the argument (for von Neumann algebras): Automorphisms preserve the spectrum, and so since $\phi(u) = \lambda_u u$ for any unitary it follows that $\phi$ must fix all unitaries whose spectrum has no rotational symmetry. By the spectral theorem such unitaries are dense in the set of all unitaries.
Nov
17
awarded  Enlightened
Nov
17
awarded  Nice Answer
Oct
31
comment Faithful and weakly-mixing representations of Property (T) groups in relation to left regular rep
Any non-amenable group without property (T) has such a representation. By Theorem 1 in [Bekka, Valette: Kazhdan's property (T) and amenable representations. Math. Z. 212, 293-299 (1993)] any group without property (T) has a weak mixing amenable representation, which cannot be weakly contained in the left-regular representation if the group is non-amenable.
Aug
7
comment Infinite amenable group subfactors
How about $R \overline \otimes (\otimes_{\gamma \in \Gamma} \mathbb M_2(\mathbb C) )$, where $\Gamma$ acts trivially on the first copy of $R$?