# Existence of homogeneous single chain compositions of a given maximal subfactor?

All the subfactors here are irreducible inclusion of hyperfinite II$_1$ factors.

A subfactor $(N \subset M)$ is Homogeneous Single Chain ($HSC$) if its lattice of intermediate subfactors is a single chain: $N=P_0 \subset P_1 \subset \dots \subset P_{n-1} \subset P_n= M$, with $(P_i \subset P_{i+1}) \simeq (P_0 \subset P_1)$.

We qualify such a subfactor by the notation $HSC^n_{(P_0 \subset P_1)}$, because it's a subfactor whose lattice is a single chain of length $n$ and whose intermediate maximal inclusions are all isomorphic to $(P_0 \subset P_1)$.

Question: Is there a $HSC^n_{(P \subset Q)}$ subfactor for all maximal subfactor $(P \subset Q)$ and for all $n>1$ ?

Warning: The subfactors given by the basic construction are not irreducible.

Group-subgroup formulation:

Definition : Let $(A \subset B)$ and $(C \subset D)$ be inclusions of groups, then $(A \subset B) \sim (C \subset D)$ if $(A/A_B \subset B/A_B) \simeq (C/C_D \subset D/C_D)$, with $A_B$ the normal core of $A$ in $B$.

Remark : In the finite group case, if $(A \subset B) \sim (C \subset D)$ then $(R^B \subset R^A) \simeq (R^D \subset R^C)$, but the converse is false, see here ($R$ is the hyperfinite II$_1$ factor).

Definition: An inclusion of groups is $HSC^n_{(A \subset B)}$ if its lattice of intermediate subgroups is a single chain of length $n$ and if its intermediate maximal inclusions of groups are all $\sim$ to $(A \subset B)$.

Question: Is there a $HSC^n_{(A \subset B)}$ inclusion for all maximal inclusion $(A \subset B)$ and for all $n>1$ ?

Remark: An inclusion of group $(A \subset B)$ is given (up to $\sim$) by a transitive permutation group $G$ of degree $d = [A:B]$ (i.e. $G$ is the image of the coset representation and $(A \subset B) \sim (G_1 \subset G$)).
Idem a maximal inclusion of groups is given (up to $\sim$) by a primitive permutation group.

Notation: A transitive group $G$ is $HSC^n_{M}$ (with $M$ a primitive group) if $(G_1 \subset G)$ is $HSC^n_{(M_1 \subset M)}$.

Experimental answer: Yes, if we restrict to the transitive groups of degree $\le 30$.
More precisely, let $G_{d,r}$ be the $r$-th transitive group of degree $d$ given by the GAP Data Library :

gap> TransitiveGroup(5,4);
A5

Remark: $G_{d,1}$ is the cyclic group $\mathbb{Z}/d\mathbb{Z}$

The primitive groups of deg. $\le 5$: $G_{2,1}$ ; $G_{3,1}$, $G_{3,2}$ ; $G_{4,4}$, $G_{4,5}$ ; $G_{5,1}$, $G_{5,2}$,$G_{5,3}$, $G_{5,4}$, $G_{5,5}$.

The following tables gives the number of $HSC^n_{G_{d,r}}$ transitive groups of degree $\le 30$ related to the primitive group $G_{d,r}$, with $n>1$, and the first examples.

$\begin{array}{c|c} &G_{2,1}&G_{3,1}&G_{3,2}&G_{4,4}&G_{4,5}&G_{5,1}&G_{5,2}&G_{5,3}&G_{5,4}&G_{5,5} \newline \hline 2 &2&4&9&18&30&8&15&31&1&5 \newline \hline 3 &15&152&476 \newline \hline 4 &597 \end{array}$

$\begin{array}{c|c} &G_{2,1}&G_{3,1}&G_{3,2}&G_{4,4}&G_{4,5}&G_{5,1}&G_{5,2}&G_{5,3}&G_{5,4}&G_{5,5} \newline \hline 2 &G_{4,1}&G_{9,1}&G_{9,3}&G_{16,63}&G_{16,195}&G_{25,1}&G_{25,4}&G_{25,8}&G_{25,192}&G_{25,196} \newline \hline 3 &G_{8,1}&G_{27,1}&G_{27,8} \newline \hline 4 &G_{16,1} \end{array}$

These tables show the existence of all the $HSC^n_{G_{d,r}}$ transitive groups of degree $\le 30$.

Remark: There is one and only one $HSC^2_{G_{5,4}}$ transitive group: $G_{25,192}$, of order $46656000000$, whereas $G_{5,4} \simeq A_5$ is order $60$.

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Yes by free compositions:

Theorem: Let $N \subset M$ be an irreducible finite index subfactor with $P$ an intermediate subfactor ($N \subset P \subset M$) such that $N \subset M$ is a free composition of $N \subset P$ and $P \subset M$.
If $L$ is another intermediate subfactor $N \subset L \subset M$, then $N \subset L \subset P$ or $P \subset L \subset M$.

I've just written a proof in this short note.
This result was already in this recent paper of Zhengwei Liu (thm 2.11 p 9) with a different proof.
(Note that the two last paragraphs of my proof can be avoided by using Liu's thm 4.1 p18)

Now, thanks to this paper of Sante Gnerre, there always exists a finite index irreducible subfactor realizing a free composition of two given finite index irreducible subfactors.

Conclusion: Given a finite index irreducible maximal subfactor $(P\subset Q)$, there exists a $HSC_{(P\subset Q)}^n$ subfactor for all $n$, it suffices to take free compositions of $(P\subset Q)$.

Remark: A free compostion of group-subgroup subfactors is not group-subgroup in general, so it's still open at this level, and harder (see the last remark of the section "group-subgroup formulation" above).

Remark: For non-free compositions, a sufficient condition for the theorem is true (i.e. "no extra intermediate") is : $\alpha \xi_i \overline{\alpha}$ irreducible for all $i \neq 0$ (see notation here) [observation of Zhengwei and me].

Remark-questions: a free composition (with a non-trivial subfactor) is not a "finite depth" subfactor, so we can also ask, given a finite depth finite index irreducible maximal subfactor $(P\subset Q)$:
- Is there a finite depth $HSC_{(P\subset Q)}^n$ subfactor for all $n$? (or equivalently, a finite depth "quotient" of the free composition subfactor (planar algebra) with "no extra intermediate", see the remark above).
- If $(P\subset Q)$ is depth $r$, what's the minimal depth of a $HSC_{(P\subset Q)}^n$ subfactor?

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