Linked Questions
14 questions linked to/from The cyclic subfactors theory: a quantum arithmetic?
22
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4
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6k
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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
3
answers
1k
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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 ?
...
22
votes
2
answers
2k
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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{...
7
votes
3
answers
698
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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 ...
9
votes
3
answers
409
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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 ...
4
votes
1
answer
710
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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
1
answer
180
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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 :
"...
5
votes
0
answers
305
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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 ...
0
votes
1
answer
311
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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 ...
6
votes
0
answers
239
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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}...
3
votes
0
answers
302
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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$.
...
4
votes
0
answers
252
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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
0
answers
151
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
0
answers
149
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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_{...