Linked Questions

23 votes
4 answers
6k 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 $H$ ...
Sebastien Palcoux's user avatar
22 votes
2 answers
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)? $$\scriptsize{\begin{...
Sebastien Palcoux's user avatar
9 votes
3 answers
406 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 ...
Sebastien Palcoux's user avatar
7 votes
3 answers
691 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 ...
Sebastien Palcoux's user avatar
6 votes
0 answers
236 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}...
Sebastien Palcoux's user avatar
5 votes
0 answers
305 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 ...
Sebastien Palcoux's user avatar
4 votes
1 answer
697 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 ...
Sebastien Palcoux's user avatar
4 votes
0 answers
249 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 ...
Sebastien Palcoux's user avatar
4 votes
0 answers
148 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 ...
Sebastien Palcoux's user avatar
3 votes
3 answers
1k 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 ? ...
Sebastien Palcoux's user avatar
3 votes
0 answers
299 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 $B$. ...
Sebastien Palcoux's user avatar
2 votes
1 answer
177 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 : "...
Sebastien Palcoux's user avatar
2 votes
0 answers
149 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_{...
Sebastien Palcoux's user avatar
0 votes
1 answer
308 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 ...
Sebastien Palcoux's user avatar