
> 
> All the [subfactors](http://en.wikipedia.org/wiki/Subfactor) $(N\subset M)$ are irreducible and finite index inclusions of II$\_1$ factors.
> 
> 
> 


First recall that in [this paper](http://projecteuclid.org/euclid.pjm/1102622455), D. Bisch characterizes the Jones projections $e\_K$ of the intermediate subfactors $(N \subset K \subset M)$ as projections $p \in N' \cap M\_1$ such that $p \geq e\_N$ and $\mathscr{F}(p) = \lambda q$,  

with $q$ a projection, $\lambda \in \mathbb{C}$ and $ \mathscr{F} : N' \cap M\_1 \to M' \cap M\_2$ the Ocneanu Fourier transform.


In [this paper](https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-50/issue-2/Normal-intermediate-subfactors/10.2969/jmsj/05020469.full) of T. Teruya, an intermediate subfactor
$(N \subset K \subset M)$ is defined as **normal** if:


* $e\_K \in Z(N' \cap M\_1 )$
* $\mathscr{F}(e\_K) \in Z(M' \cap M\_2)$


with $Z(X)$ the center of X, and $\mathscr{F}$ as above ($\mathscr{F}(x)=[M : N]^{-3/2}E\_{M'}^{N'}(xe\_{M}e\_{N})$).


Teruya proves that for the depth $2$ case (Kac algebra), the normal intermediate subfactors gives exactly the normal Kac subalgebras (in particular, the normal subgroups for the group subfactors).


**Remark** : If $N' \cap M\_1 $ and $M' \cap M\_2$ are abelian, then every intermediate subfactor is normal.


**Example**: Let $(A \subset B)$ and $(C \subset D)$ be $2$-supertransitive subfactors, $N= A \otimes B$ and $M= C \otimes D$, then $P=A \otimes D$ and $Q=B \otimes C$ are normal intermediate subfactors of $(N \subset M) $ because $N' \cap M\_1 $ and $M' \cap M\_2$ are $\mathbb{C}^4$, and so abelian (see [Watatani](http://www.sciencedirect.com/science/article/pii/S002212369690110X) prop 5.1 p329).  

This result is true without the $2$-supertransitivity assumption if $\mathscr{F}\_{(N\subset M)} = \mathscr{F}\_{(A\subset B)} \otimes \mathscr{F}\_{(C\subset D)}$.


A subfactor is **simple** if it has no non-trivial normal intermediate subfactor.  

A group subfactor is simple iff the group is simple. A maximal subfactor is *a fortiori* simple.



> 
> **Question**: Let $(N \subset M)$ be a subfactor, and let $$ N=K\_1 \subset K\_2 \subset \dots \subset K\_r = M $$ be a normal chain such that each subfactor
> $(K\_i \subset K\_{i+1})$ is simple, and $K\_i \neq K\_{i+1}$ for $0<i<r$.
> Then any other normal chain of $(N \subset M)$ having the same properties is
> equivalent to this one (i.e. the sequence of subfactors in our two
> chains are the same up to isomorphisms, and a permutation of the
> indices) ?
> 
> 
> 




---


The rest of the post is dedicated to a reformulation of the question.


Through [this comment](https://mathoverflow.net/questions/156103/jordan-holder-theorem-for-planar-algebras#comment399695_156103), Benjamin Steinberg shows me that the Jordan-Hölder theorem is a generic property of [modular lattices](http://en.wikipedia.org/wiki/Modular_lattice) (i.e. lattices checking : $x ≤ b \Rightarrow x ∨ (a ∧ b) = (x ∨ a) ∧ b$, $\forall a$).


In the lattice theory framework, the subfactors $(A \subset B)$ and the isomorphisms, are replaced by intervals $[a,b]$ and projectivities (two intervals $[a, b]$ and $[c, d]$ are **perspective** if $b ∨ c = d$ and $b ∧ c = a$ or vice versa. **Projectivity** is the transitive closure of perspectivity). There is a well-known Jordan-Hölder theorem for modular lattices (also semimodular, see [this paper](http://citeseerx.ist.psu.edu/viewdoc/download?rep=rep1&type=pdf&doi=10.1.1.215.8386) of Grätzer-Nation).


So we would need, firstly to prove that the set of normal intermediate subfactors is a lattice and is modular, and secondly that projective intervals in such lattices give isomorphism of subfactors.


In [this paper](http://www.sciencedirect.com/science/article/pii/S002212369690110X), Y. Watatani introduced the notion of quasi-normal intermediate subfactors (by using two commuting squares) and proved modular identites ([W](http://www.sciencedirect.com/science/article/pii/S002212369690110X) thm3.9 p323).  

But Teruya proved that a normal intermediate subfactor is quasi-normal ([T](https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-50/issue-2/Normal-intermediate-subfactors/10.2969/jmsj/05020469.full) thm3.4 p377).


So it rests to prove that we have a lattice and the second point about projectivity and isomorphism.  

This is like the [second isomorphism theorem for groups](http://en.wikipedia.org/wiki/Isomorphism_theorem#Second_isomorphism_theorem), and it's the content of the reformulation :



> 
> **Reformulation**: Let $(N \subset M)$ be a subfactor, let $P$, $Q$ be normal intermediate subfactors, then:  
> 
> Are $P \wedge Q$ and $P \vee Q$ normal, and $(P \wedge Q \subset Q)$ isomorphic to $(P \subset P \vee Q)$ ?
> 
> 
> 


**Remark** : $P \wedge Q = P \cap Q$ and $P \vee Q = PQ=QP$.  

The latticeness part seems reduced to know if $\mathscr{F}(e\_P.e\_Q) \in Z(M' \cap M\_2)$.



Jordan-Hölder theorem for subfactors?
The Jordan-Hölder property is true for large classes of subfactors: see class $\mathcal{C}$ and beyond, above,  

but it's false in general, counter-examples are given by $(A\_n \subset S\_{n+1})$, see [this answer](https://mathoverflow.net/questions/159167/a-second-isomorphism-theorem-for-the-inclusions-of-groups/160253#160253): 



> 
> **Theorem:** $\forall n \ge 3$, $A\_{n+1}$ and $S\_n$ are *normal intermediate subgroups* of the inclusion $(A\_n \subset S\_{n+1})$, but $(A\_n \subset S\_n \subset S\_{n+1} )$ and $(A\_n \subset A\_{n+1} \subset S\_{n+1} )$ are not equivalent. 
> 
> 
> 


Because there are non-equivalent inclusions of groups which are isomorphic as group-subgroup subfactors (see [here](https://mathoverflow.net/questions/136021/is-there-a-purely-group-theoretic-reformulation-of-an-equivalence-of-subgroups?rq=1)), we need to add that $(R^{S\_{n+1}} \subset R^{S\_{n}}) \not\simeq (R^{A\_{n+1}} \subset R^{A\_{n}})$. 


In fact, more generally we have the following properties (see M Izumi [here](https://www.youtube.com/watch?v=I52MOU9F-sg&index=5&list=LLhntpxxSKIETTxsQMN8nywg), 23:30 and 27:50):  

If $(R^{B} \subset R^{A}) \simeq (R^{D} \subset R^{C})$ then $Rep(A/B\_A) \simeq Rep(C/D\_C)$, with $A\_B$ the [normal core](http://groupprops.subwiki.org/wiki/Normal_core).  

If moreover the inclusions are maximal then $(A \subset B) \sim (C \subset D)$ 


