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5 votes
0 answers
161 views

Are the integer index finite depth irreducible subfactors Kac-coideal?

Is every integer index finite depth irreducible subfactors planar algebra, the intermediate of an irreducible finite index depth $2$ subfactors planar algebra? In other words, of the following form (...
Sebastien Palcoux's user avatar
8 votes
0 answers
306 views

Are there only finitely many maximal irreducible amenable subfactors at fixed finite index?

A subfactor $N \subset M $ is maximal if it admits no non-trivial intermediate subfactors $N \subset P \subset M $. Question: Are there only finitely many maximal irreducible amenable subfactors at ...
Sebastien Palcoux's user avatar
3 votes
0 answers
200 views

What are the first non-maximal non-group-subgroup simple irreducible subfactors?

Definition: For an irreducible (finite index) subfactor $(\mathcal{N} \subset \mathcal{M})$, an intermediate $(\mathcal{N} \subset \mathcal{P} \subset \mathcal{M})$ is normal if the biprojections $e_{\...
Sebastien Palcoux's user avatar
3 votes
1 answer
335 views

What's the natural equivalence of subfactors in general?

Let $A$ be a factor and $\mathcal{C}_{A}$ be the category of all the subfactors $(M \subset N)$ such that $M$ and $N$ are isomorphic to $A$. The most famous of them is perhaps $\mathcal{C}_{R}$ with $...
Sebastien Palcoux's user avatar
2 votes
0 answers
101 views

Are there infinitely many amenable Hadamard-Petrescu subfactors?

The complex Hadamard matrices of dimension $n$ are used to build index $n$ subfactors through the commuting square construction. For more details, see the paper Subfactors and Hadamard Matrices by W....
Sebastien Palcoux's user avatar
1 vote
1 answer
181 views

Existence of a third intermediate if there are two intermediate subfactors of index 2

Let $(N \subset M)$ be an irreducible finite index unital inclusion of hyperfinite ${\rm II}_1$ factors. Let $K_1$ and $K_2$ be two distinct intermediate subfactors $N \subset K_i \subset M$, such ...
Sebastien Palcoux's user avatar