# 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 there $P \subset N$ or $Q \supset M$ such that $(P \subset M)$ or $(N \subset Q)$ is a finite index depth $\le 3$ irreducible subfactor?)

Remark: By the prop. 9.1.1 p 37 here, an irreducible finite depth-index subfactor is always an intermediate subfactor of a depth $2$ reducible subfactor. But, this post asks if it is always the intermediate of a finite index depth $\le 3$ irreducible subfactor.

Remark: the first examples to look at should be the $A_n$-subfactors (of depth $n-1$).
It's ok for $n=2,3,4$; for $n=5$ also: it's isomorphic to $(R^{S_3} \subset R^{S_2})$, intermediate of $(R^{S_3} \subset R)$.
Question: Is $A_6$ a counter-example? If yes, why? Else, is it true for all $n \ge 2$ ?

Definition: A finite index depth $d$ irreducible subfactor $(N \subset M)$ is called top if it can't be the intermediate of a strictly smaller depth subfactor (in others words, such a subfactor is top if $\forall P \subset N$ and $\forall Q \supset M$ such that $(P \subset M)$ and $(N \subset Q)$ are irreducible, then their depths are $\ge d$).

Examples: A group-subgroup subfactor is top iff it's a group subfactor.
The non-integer index depth $3$ irreducible subfactors are top.

The question Are the integer index finite depth irreducible subfactors Kac-coideal? is reformulable by: Are the integer index finite depth irreducible top subfactors, depth $2$?

The main question is also reformulable by: Is a finite depth-index irreducible top subfactor, depth $\le 3$?

Optional part: top-simple reduction

If the main question admits a negative answer:
Question: Is there an irreducible finite index top subfactor of arbitrary large depth $d$?
If yes, is there one for all $d \ge 2$?

Remark: The classification of the subfactors reduces to the classification of the top subfactors, because all the others come as intermediates.

Let $(N \subset M)$ be a finite depth-index irreducible top subfactor, and $(N \subset P \subset M)$ an intermediate.
Question: $P$ normal $\Rightarrow$ depth$(N \subset P)$, depth$(P \subset M) \le$ depth$(N \subset M)$?
Remark: For the depth $2$ case, it's true and the converse also. Is it also an equivalence at depth $d$?

Question: Does the Jordan-Hölder theorem work for the finite depth-index irreducible top subfactors?
Remark: this anwser shows that, without the top assumption, Jordan-Hölder theorem does not work, counter-examples are given the $(A_n \subset S_{n+1})$ group-subgroup subfactors (for $n \ge 3$).

If the two previous questions admit positive answers then the classification of the depth $\le d$ finite index irreducible subfactors reduces to the top simple ones.

• I guess, an irreducible finite index finite depth subfactor is always an intermediate subfactor of a depth 2 reducible subfactor. – Marcel Bischoff Sep 4 '14 at 19:28
• What let you think that every finite depth subfactor is intermediate of an irreducible depth $\leq 3$ guy?? – Marcel Bischoff Dec 18 '14 at 22:41
• Ok, I think for $A_n$ subfactors this is clear, still not sure how this should work in general... – Marcel Bischoff Dec 18 '14 at 22:49
• What's your argument for $A_n$, $n\ge 6$? – Sebastien Palcoux Dec 19 '14 at 0:18
• What let me know is for example the paper Quantum subgroups of the Haagerup fusion categories of Grossman-Snyder in which the Haagerup subfactor and its Izumi generalizations, appear at intermediate of depth $3$ irreducible subfactors. In general, such irreducible depth $3$ "reduction" seems possible via this quantum subgroups way. – Sebastien Palcoux Dec 19 '14 at 0:29