Jordan-Hölder theorem for subfactors?

Question

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 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 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 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) ?

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

Through this comment, Benjamin Steinberg shows me that the Jordan-Hölder theorem is a generic property of modular lattices (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 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, Y. Watatani introduced the notion of quasi-normal intermediate subfactors (by using two commuting squares) and proved modular identites (W thm3.9 p323).
But Teruya proved that a normal intermediate subfactor is quasi-normal (T 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, 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 $\mathscr{F}(e_P.e_Q) \in Z(M' \cap M_2)$.

Answer 1

A proof for the class $\mathcal{C}$ group-subgroup subfactors:

First of all, by the Galois correspondence, an intermediate subfactor $R^G \subset P \subset R^H$ is given by an intermediate subgroup $H \subset K \subset G$, with $P=R^K$.

Next, by Teruya prop.3.3 p476, $P$ is a normal intermediate subfactor iff $\forall g \in G$ : $KgH=HgK$ $(\star)$ Let's call such an intermediate subgroup $K$ a normal intermediate subgroup.

Examples: If $H=\{ e \}$ then $K$ is a normal intermediate subgroup iff $K$ is a normal subgroup of $G$.
$H_i$ and $G_i$ are obviously normal intermediate subgroups of the inclusion $(H_i \subset G_i)$, and
$H_1 \times G_2$ and $G_1 \times H_2$ are normal intermediate subgroups of $(H_1 \times H_2 \subset G_1 \times G_2)$.

An inclusion of groups is called simple if it admits no non-trivial normal intermediate subgroups.
Examples: the maximal inclusions $(H \subset G)$ are simple and $(\{ e\} \subset G)$ is simple iff $G$ is simple.

For the rest of the answer, $K$ and $L$ are normal intermediate subgroups of the inclusion $(H \subset G)$.

Lemma: $K \vee L = KL=LK$
Proof: Let $k \in K$ and $l \in L$, then $kl = kle = hlk'$ by $(\star)$, but $l'=hl \in L$, so $KL \subset LK$.
Idem, $LK \subset KL$, so $KL=LK$. Then $K \vee L := \langle K , L \rangle = KL=LK$. $\square$

Lemma: The set of normal intermediate subgroups of $(H \subset G)$ is a lattice for $\wedge$ and $ \vee $.
Proof: $K \wedge L=K \cap L$, $K \vee L = KL=LK$.
Let $g \in G$, $(K \cap L)gH \subset HgK$ and $HgL$ by $(\star)$, so $(K \cap L)gH \subset Hg(K \cap L)$.
Idem $Hg(K \cap L) \subset (K \cap L)gH$. Then $Hg(K \cap L) = (K \cap L)gH$.
Now, $(KL)gH = KHgL = KgL = KgHL = Hg(KL)$.
Conclusion, $K \wedge L$ and $K \vee L$ are also normal intermediate subgroups. $\square$

Definition: Let $\sim$ be the equivalence relation on inclusions of finite groups, generated by :
$(H \subset G) \sim (\phi(H) \subset \phi(G))$, with $ \phi: G \to L$ a finite group morphism and with $ker(\phi) \subset H$.
Remark : $(H_1 \subset G_1) \sim (H_2 \subset G_2) \Rightarrow (R^{G_1} \subset R^{H_1}) \simeq (R^{G_2} \subset R^{H_2})$, see here.

Let $\Omega = L / (K \cap L)$ and $\Omega' = KL / K$.
Let $\pi: L \to S_{\Omega}$ such that $\pi(l).l'(K \cap L) = ll'(K \cap L)$. Idem let $\pi': KL \to S_{\Omega'}$.
$ker(\pi) = \{ l' \in L : \forall l \in L, l^{-1}l'l \in K \cap L \} = core_L(K \cap L) \subset K \cap L $. $ker(\pi') = \{ g' \in KL : \forall g \in KL, g^{-1}g'g \in K \} = core_{KL}(K) \subset K $.

Definition: $(H \subset G)$ is of class $\mathcal{C}$ if for all $K$, $L$ and $\forall k \in K$, $k.core_{KL}(K) \cap L \neq \emptyset$.
For the rest of the answer, the inclusion $(H \subset G)$ is supposed to be of class $\mathcal{C}$.

Lemma: $(\pi(K \cap L) \subset \pi(L)) \simeq (\pi'(K) \subset \pi'(KL))$
Proof: Let $\phi : \pi(L) \to \pi'(KL)$ such that $\phi(\pi(l)) = \pi'(l)$.
$\phi$ is well-defined because $ker(\pi) \subset ker(\pi')$, because if $l' \in ker(\pi)$ and $g \in KL$,
then $g=lk$ and $g^{-1}l'g = k^{-1}l^{-1}l'lk \subset k^{-1}(K \cap L)k \subset K$.
$\phi$ is injective because $ker(\pi') \cap L = ker(\pi)$, because $ker(\pi) \subset ker(\pi') \cap L$ and $ker(\pi') \cap L \subset ker(\pi)$ because if $l' \in ker(\pi') \cap L$ and $l \in L$, then $l^{-1}l'l \in K$ and $L$.
$\phi$ is surjective thanks to class $\mathcal{C}$ assumption. In particular $\phi(\pi(K \cap L)) = \pi'(K)$.

Corollary: $(K \wedge L \subset L) \sim (K \subset K \vee L)$.
Proof : $(K \cap L \subset L) \sim (\pi(K \cap L) \subset \pi(L)) \simeq (\pi'(K) \subset \pi'(KL)) \sim (K \subset KL)$. $\square $

Theorem: The Jordan-Hölder theorem is true for the class $\mathcal{C}$ group-subgroup subfactors.
Proof: It's a consequence of the results above and what it's explained in the second part of the post. $\square$

Problem: Are all the inclusions $(H \subset G)$ of class $\mathcal{C}$ (see here)?
Examples: If $(H_i \subset G_i)$ is a maximal inclusion, then it is obviously of class $\mathcal{C}$, and $(H_1 \times H_2 \subset G_1 \times G_2)$ is also of class $\mathcal{C}$ (see here).

Answer 2

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:

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

Answer 3

Beyond the class $\mathcal{C}$:

The inclusion $(D_{10} \subset A_6)$ is not of class $\mathcal{C}$ (see this answer), nevertheless, it checks Jordan-Hölder: in fact, $(R^{A_6} \subset R^{D_{10}})$ has exactly two intermediate subfactors $R^K$ and $R^L$ (with $K \simeq L \simeq A_5$), each are normal intermediate subfactors and, $(R^{L} \subset R^{D_{10}}) \simeq (R^{K} \subset R^{D_{10}})$ and $(R^{A_6} \subset R^{L}) \simeq (R^{A_6} \subset R^{K})$ because the natural isomorphism $L \simeq K$ (exchanging $(123)$ with $(14)(56)$, see here) extends into an automorphism of $A_6$.