3,132 reputation
1221
bio website math.vanderbilt.edu/~peters10
location Vanderbilt University
age
visits member for 4 years, 6 months
seen Nov 20 at 21:49

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 $R$?
Aug
6
comment 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
comment 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
comment ${\rm II}_1$-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
comment ${\rm II}_1$-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
comment ${\rm II}_1$-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
comment 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
comment Existence of orthogonal projections generating Von Neumann algebras
For a counter-example take $H = \mathbb C^2$.
Jan
31
comment 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
comment 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?