bio | website | math.vanderbilt.edu/~peters10 |
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location | Vanderbilt University | |
age | ||
visits | member for | 3 years, 10 months |
seen | yesterday | |
stats | profile views | 1,982 |
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? |
Jan 31 |
comment |
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 |
comment |
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? |
Oct 14 |
awarded | Constituent |
Oct 14 |
awarded | Caucus |
Sep 26 |
answered | Relative amenability of subgroups |
Sep 8 |
comment |
A non-hyperfinite type III factor from an action of the free group on the circle
I don't follow your logic in part (d), why does $a.x = \gamma^n.x$? |
Sep 6 |
answered | Infinite finitely generated non-amenable groups |
Aug 9 |
comment |
Relative commutants of abelian von Neumann algebras
@Jiang: Yes, that's correct. This example is far from typical though. |
Jul 28 |
comment |
Is the fundamental group of $II_{1}$ factors invariant under a relation?
@SébastienPalcoux: This is because if $u \in L(\mathbb F_\infty) \subset M * L(\mathbb F_\infty)$ is a unitary with trace $0$ then $M$ is in free position from $u M u^*$, hence $M * M \cong W^*( M, u M u^* ) \subset M * L(\mathbb F_\infty) \cong M$. |
Jul 27 |
revised |
Is the fundamental group of $II_{1}$ factors invariant under a relation?
removed open problem tag. |
Jul 27 |
answered | Is the fundamental group of $II_{1}$ factors invariant under a relation? |
Jul 2 |
awarded | Enlightened |
Jul 2 |
awarded | Nice Answer |
Jun 25 |
awarded | Revival |
May 31 |
awarded | Yearling |
May 28 |
answered | Characterization of amenable actions |
May 28 |
comment |
Characterization of amenable actions
@Martin: Thanks, I thought this was the case, but I wanted to make sure. |