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Sep
6 |
revised |
The converse of von Neumann's mean ergodic theorem
Fixed the definition of $F_n$. |
Sep
6 |
answered | The converse of von Neumann's mean ergodic theorem |
Sep
6 |
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The converse of von Neumann's mean ergodic theorem
For discrete groups, if $H_{\Gamma'}$ is always equal to $H_\Gamma$ then $\Gamma' = \Gamma$. You can just consider the representation $\ell^2(\Gamma/\Gamma')$. |
Aug
6 |
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Rank–nullity theorem for finite von Neumann algebras
You still need that $T$ is contained in $M$. In Michael's example the operator $T$ is not contained in $M$. |
Aug
5 |
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Rank–nullity theorem for finite von Neumann algebras
You just need a dimension function which assigns equal dimensions to equivalent projections. You don't need factoriality. In the abelian case $M = L^\infty(X, \mu)$ this just says $\mu( E \cup F ) = \mu(E) + \mu(F)$ for disjoint measurable sets $E$ and $F$. |
Aug
4 |
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Rank–nullity theorem for finite von Neumann algebras
With the extra assumption this is just as easily seen to be true since then the projections onto the ranges of $\tilde Tp$ and $(\tilde T p)^*$ are equivalent in $M$ (just consider the partial isometry in the polar decomposition of $\tilde T p$). |
Jul
30 |
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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 |
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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 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 |
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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. |