All Questions
14 questions
1
vote
0
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91
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Subfactors with integer Jones index
Is there any integer (Jones) index subfactor which is not extremal?
- 393
1
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0
answers
111
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Inclusion of finite dimensional C*-algebras and relative commutants of subfactors
Given a subfactor $N\subset M$ with finite Jones index, the inclusion of relative commutants $N^{\prime}\cap M\subset N^{\prime}\cap M_1$ (here, $M_1$ is the basic construction of $N\subset M$) is a ...
- 393
2
votes
0
answers
118
views
Depth of the reduced subfactor
Suppose $N\subset M$ is a finite depth subfactor with $[M:N]<\infty$. Consider the reduced subfactor $pNp\subset pMp$ for some projection $p\in N$. How to calculate the depth of $pNp\subset pMp$ in ...
- 393
0
votes
1
answer
122
views
On index 2 and square of subfactors without extra intermediate
Let $N \subsetneq K_i \subsetneq M$, $i=1,2$, be a square of irreducible finite index unital inclusion of hyperfinite ${\rm II}_1$ factors, such that there is no extra intermediate, with $K_1 \not \...
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 ...
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....
0
votes
1
answer
122
views
Is there an irreducible subfactor with an infinite homogeneous single chain lattice?
We know that we can build an irreducible subfactor realizing a finite single chain lattice containing any finite index irreducible maximal subfactors, by using the free composition (see here).
Now ...
1
vote
0
answers
308
views
Is a finite depth-index irreducible subfactor, intermediate of a depth ≤ 3 one?
Let $(N \subset M)$ be a finite depth-index irreducible subfactor.
Main question: Is $(N \subset M)$ the intermediate of a finite index depth $\le 3$ irreducible subfactor?
(In others words, is ...
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 (...
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_{\...
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_{...
3
votes
0
answers
304
views
Is the fundamental group of a maximal subfactor always $\mathbb{R}_{+}^{*}$?
The fundamental group $\mathcal{F}(N \subset M)$ of a unital inclusion of II$_{1}$ factors $N \subset M$ is defined as : $\mathcal{F}(N \subset M) =\{t >0 \ | \ (N \subset M)^{t} \simeq (N \...
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 $...
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 ...