bio | website | math.vanderbilt.edu/~peters10 |
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location | Vanderbilt University | |
age | ||
visits | member for | 4 years, 3 months |
seen | 2 days ago | |
stats | profile views | 2,109 |
Aug 7 |
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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 $R$? |
Aug 6 |
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Infinite amenable group subfactors
I don't believe it is correct. Why should uniqueness up to outer conjugacy imply $\mathcal R^\Gamma = \mathbb C$? |
Aug 6 |
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Infinite amenable group subfactors
You should be more precise about what you mean by "only one manner". Why do you conclude that $\mathcal R^\Gamma = \mathbb C$? |
Aug 5 |
awarded | oa.operator-algebras |
Aug 5 |
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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?
$(\mathcal R^\Gamma \subset \mathcal R)$ does not encode $\Gamma$ if it is infinite. |
Aug 4 |
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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?
The situation is similar, for the shift action $\mathcal R^\Gamma = \mathbb C$ if and only if $| \Gamma | = \infty$. |
Aug 4 |
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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?
@SébastienPalcoux: Yes, $M \cong L(\mathbb F_\infty)$. The reason that $M^\Gamma = \mathbb C$ is not because the action of $\Gamma$ on itself is transitive, but rather because the action has no finite orbits. |
Aug 4 |
answered | ${\rm II}_1$-factors with finite commutant: $\mathcal{A} \cap \mathcal{B} = \mathbb{C} \Rightarrow \mathcal{A}' \cap \mathcal{B}'$ hyperfinite? |
Aug 1 |
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How well do we know relative commutants in $L(\mathbb{F}_\infty)$?
@SébastienPalcoux: If $N \subset M$ is an inclusion of ${\rm II}_1$ factors, then $N' \cap \mathcal B(L^2(M))$ is anti-isomorphic to the basic construction $\langle M, N \rangle = (JNJ)' \cap \mathcal B(L^2(M))$. This is always a semi-finite factor, and is finite if and only if $N$ is a finite index subfactor of $M$. In the case of free products, $N \subset N * B$ is finite index only in the case $B = \mathbb C$. |
Jul 30 |
awarded | Enlightened |
Jul 30 |
awarded | Nice Answer |
Jul 24 |
answered | Multiplicative domains and conditional expectations |
May 31 |
awarded | Yearling |
May 19 |
answered | von Neumann algebras generated by commutators |
Apr 25 |
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Existence of orthogonal projections generating Von Neumann algebras
For a counter-example take $H = \mathbb C^2$. |
Jan 31 |
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The category of subfactors extending the category of groups?
You could define a ``morphism'' from $(N_1 \subset M_1)$ to $(N_2 \subset M_2)$ to be a group homomorphism from the normalizer group $\mathcal N_{M_1}(N_1) / \mathcal U(N_1)$ to $\mathcal N_{M_2}(N_2) / \mathcal U(N_2)$. But I don't think you'll get much insight from this perspective. |
Jan 31 |
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The category of subfactors extending the category of groups?
Are you taking specific actions or are you looking for something which holds for arbitrary actions? |
Jan 31 |
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The category of subfactors extending the category of groups?
How are you having $G$ and $G'$ act on $R$? |
Dec 20 |
answered | Do syndetic sets on amenable semigroups have positive upper density? |
Nov 13 |
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Sets $E$ in $\mathbb{Z}$ such that any $l^2$ function with support on $E$ comes from Fourier of a continuous function
One characterization of an amenable group $\Gamma$ is that $\| \lambda(f) \|_\infty = \| f \|_1$ for any non-negative valued function $f \in \ell^1\Gamma$. So no amenable groups will have such an infinite subset. Which then begs the question: What about non-amenable groups which do not contain a non-cyclic free group? |