Jesse Peterson
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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 comment 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 comment 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 comment 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 comment 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 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.