Linked Questions

22
votes
4answers
5k views

What's a noncommutative set?

This issue is for logicians and operator algebraists (but also for anyone who is interested). Let's start by short reminders on von Neumann algebra (for more details, see [J], [T], [W]): Let $H$ ...
3
votes
3answers
717 views

Are the finite dimensional von Neumann algebras, singly generated?

Let $\mathcal{M}$ be a finite dimensional von Neumann algebra, then : $$\mathcal{M} \simeq \bigoplus_i M_{n_i}(\mathbb{C})$$ Question : Is it singly generated (as von Neumann algebra)? how ? ...
18
votes
2answers
2k views

Non weakly-group-theoretical integral fusion category

Is there an integral fusion category of rank $7$, FPdim $210$ and type $(1,5,5,5,6,7,7)$ with the following fusion rules (or the little $\color{purple}{\text{variation}}$ below)? $$\small{\begin{...
8
votes
3answers
377 views

What are the intermediate subfactors of the tensor product of two maximal subfactors?

Let $(N_1 \subset M_1)$ and $(N_2 \subset M_2)$ be two maximal subfactors. Their tensor product, the subfactor $(N_1 \otimes N_2 \subset M_1 \otimes M_2)$, admits four obvious intermediate ...
5
votes
3answers
560 views

Jordan-Hölder theorem for subfactors?

All the subfactors $(N\subset M)$ are irreducible and finite index inclusions of II$_1$ factors. First recall that in this paper, D. Bisch characterizes the Jones projections $e_K$ of the ...
3
votes
1answer
595 views

Abelian subfactors, a relevant concept?

Through the questions below, this post asks whether the concept of abelian subfactor is relevant. Remark : here abelian qualifies an inclusion of II$_1$ factors $(N \subset M)$, $N$ is not an abelian ...
2
votes
1answer
155 views

Are every finitely generated planar algebras, also singly generated?

Let $\mathcal{P}$ be a finitely generated planar algebra. Question : Is it also singly generated ? I ask this question, because, on one hand I've read on this paper of V. Jones and D. Bisch : "...
4
votes
0answers
300 views

Are the homogeneous single chain subfactors, Dedekind?

Background: See here and there. Recall that a subfactor is Dedekind if all its intermediate subfactors are normal. A subfactor $(N \subset M)$ is Homogeneous Single Chain (HSC) if its lattice ...
3
votes
0answers
292 views

What's the ratio of inclusions of finite groups with a distributive lattice?

Definition: Two inclusions of finite groups are equivalent, $(A \subset B) \sim (C \subset D)$, if: $(A/A_B \subset B/A_B) \simeq (C/C_D \subset D/C_D)$ with $A_B$ the normal core of $A$ in $B$. ...
6
votes
0answers
221 views

Existence of a Kac algebra for a given fusion ring in a particular class

A $n$-dimensional Kac algebra (i.e., a Hopf C*-algebra), admits finitely many irreducible representations, whose cardinal $r$ is called its rank, the increasing sequence $(d_{1},d_{2},d_{3}, ..., d_{r}...
0
votes
1answer
265 views

Chinese remainder theorem for cyclic subfactor planar algebras

This post was inspired by an exchange with the indian woman mathematician Ajit Iqbal Singh. The chinese remainder theorem can be stated as follows: Let $n_1, \dots, n_r \ge 2$ be positive integers ...
4
votes
0answers
238 views

An embedding theorem for a fusion ring planar algebra?

We first recall the embedding theorem for finite depth subfactor planar algebras: The planar algebra generated by a (finite depth) subfactor, is embeddable into the planar algebra generated by its ...
4
votes
0answers
135 views

Is there a maximal finite depth infinite index irreducible subfactor?

A subfactor $N \subset M $ is irreducible if $N' \cap M = \mathbb{C} $. It's maximal if it admits no non-trivial intermediate subfactors $N \subset P \subset M $. It's cyclic if its lattice of ...
2
votes
0answers
146 views

Planar algebraic translation of a subfactor property

Let $N \subset M$ be an irreducible finite depth and finite index subfactor. $M$ is a completely reducible (algebraic) $N$-$N$ bimodule, it decomposes into irreducibles as follows : $$M=\bigoplus_{...