bio | website | math.vanderbilt.edu/~peters10 |
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location | Vanderbilt University | |
age | ||
visits | member for | 5 years, 1 month |
seen | 8 hours ago | |
stats | profile views | 2,264 |
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 |
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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 |
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 |
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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 $t$. |
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 |