# Abelian subfactors, a relevant concept?

Through the questions below, this post asks whether the concept of abelian subfactor is relevant.
Remark : here abelian qualifies an inclusion of II$$_1$$ factors $$(N \subset M)$$, $$N$$ is not an abelian algebra.

First some useful reminders about groups and lattices :

Definitions: A lattice $$(L, \wedge, \vee)$$ is :

• Distributive if $$a∨(b∧c) = (a∨b) ∧ (a∨c)$$
• Modular if $$a ≤ c \Rightarrow a ∨ (b ∧ c) = (a ∨ b) ∧ c$$

$$(\forall a,b,c \in L)$$

Remark: Distributivity $$\Rightarrow$$ Modularity

Let $$G$$ be a finite group and let $$\mathcal{L}(G)$$ be its lattice of subgroups, and $$\mathcal{N}(G)$$, of normal subgroups.

Theorems : A finite group $$G$$ is

• Cyclic iff $$\mathcal{L}(G)$$ is distributive (Ore 1938)
• Abelian iff $$\mathcal{L}(G \times G)$$ is modular (Lukacs-Palfy 1986)

(see here thm2.3 p431 and thm6.5 p449)

Remark : Of course, a cyclic group is abelian, and a direct product of abelian groups is abelian.
Theorem : Every finite abelian groups is a direct product of finite cyclic groups.

Theorem : $$\mathcal{N}(G)$$ is modular.
Definition : $$G$$ is Dedekind if all its subgroups are normal. The abelian groups are Dedekind.
A non-abelian Dedekind group is called Hamiltonian (for example the quaternion group $$Q_8$$).
Remark : $$G$$ abelian implies $$\mathcal{L}(G)$$ modular, but the converse is false (see $$Q_8$$).

All the subfactors $$(N\subset M)$$ are irreducible and finite index inclusions of II$$_1$$ factors.

Let $$(N\subset M)$$ be a subfactor and $$\mathcal{L}(N\subset M)$$ its lattice of intermediate subfactors.
Galois correspondence for group subfactors: $$\mathcal{L}(R^G\subset R)$$ $$\leftrightarrow$$ $$\mathcal{L}(G)$$ $$\leftrightarrow$$ $$\mathcal{L}(R \subset R \rtimes G)$$.
Recall also that $$(R^G \otimes R^H\subset R \otimes R) \simeq (R^{G \times H}\subset R)$$

Definitions : A subfactor $$(N\subset M)$$ is

• Cyclic if $$\mathcal{L}(N\subset M)$$ is distributive.
• Abelian if $$\mathcal{L}(N \otimes N \subset M \otimes M)$$ is modular.

Remark : here abelian qualifies the inclusion of factors $$(N \subset M)$$, $$N$$ is not an abelian algebra.
Remark: $$(R^G\subset R)$$ is cyclic (resp. abelian) iff $$G$$ is cyclic (resp. abelian).

Question 1a : Are the cyclic subfactors abelian ?

Examples: If $$(N\subset M)$$ is $$2$$-supertransitive, then it is maximal, so cyclic. If also $$[M:N]>2$$ then $$\mathcal{L}(N \otimes N \subset M \otimes M)$$ is distributive (W prop5.1 p329), so modular, and then $$(N\subset M)$$ is abelian.
All the maximal group-subgroup subfactors $$(R^G\subset R^H)$$ are abelian (see the corollary here).

Let $$(\otimes_{i \in I} A_i \subset \otimes_{i \in I} B_i)$$ be the tensor product of the subfactors $$(A_i \subset B_i)_{i \in I}$$, with $$I$$ finite.

Question 1b : Is a tensor product of abelian subfactors also abelian ?
Question 1c : Is every abelian subfactor a tensor product of cyclic subfactors ?

In this paper, T. Teruya introduced the notion of normal intermediate subfactors, generalizing exactly the notion of normal subgroups (see the post Jordan-Hölder theorem for subfactors for more details).

Definitions : A subfactor $$(N\subset M)$$ is

• Dedekind if all its intermediate subfactors are normal.
• Hamiltonian if it is Dedekind and non-abelian.

Remark : If $$(N\subset M)$$ is Dedekind then $$\mathcal{L}(N\subset M)$$ is modular (W thm3.9 p323, T thm3.4 p377).

Question 2 : Are the abelian subfactors Dedekind ?

Remark : Positive answers for questions 1a, 2 and Jordan-Hölder, would solve the question 1 here.

Problem : Find Hamiltonian subfactors not coming from group theory.

Definition : A subfactor is basically abelian if $$(N' \cap M_1)$$ and $$(M' \cap M_2)$$ are abelian algebras. Remark : A group subfactor is abelian iff it is basically abelian.

Question 3 : Is a subfactor abelian iff it is basically abelian ?

Remark : the implication $$(\Leftarrow)$$ is clear if the relative commutants deal with the tensor product.
If the implication $$(\Rightarrow)$$ and the question 1a are true, then there is no non-trivial maximal Kac algebra ! (the original motivation for this post).

Question 1a: Yes, a cyclic subfactor is abelian, if it admits no depth $2$ intermediate inclusions, because then, thanks to the corollary here, its tensor square is also cyclic.

Remark : In general, the question seems reduced to know if non-trivial maximal Kac algebras exist, and if the lattice of left coideals of their tensor square is modular.

Question 2: No, $(R^{A_6} \subset R^{D_8})$ is a counterexample. $(D_8 \subset A_6)$ admits exactly two non-trivial intermediate subgroups ($2^2:S_3$, see here) which are of order $24$ and isomorphic to $S_4$ (see here).
So $(R^{A_6} \subset R^{D_8})$ is cyclic, and admits no depth $2$ intermediate inclusion ($A_6$ is simple and $D_8$ is not a normal subgroup of $S_4$), so thanks to the answer of Q1a, it is abelian.
If $(R^{A_6} \subset R^{D_8})$ is Dedekind, then the two copies of $S_4$, I call $K$ and $L$, would be normal intermediate subgroups (see here), but then $A_6=KL$ and so by the product formula we would have $\vert A_6 \vert . \vert D_8 \vert = \vert K \vert . \vert L \vert$, unfortunately $360*8=2880 \neq 576 = 24^2$, contradiction.
So, $A_6 \neq KL$, so $K$ or $L$ are not a normal intermediate subgroups.
Conclusion, $(R^{A_6} \subset R^{D_8})$ is abelian (and cyclic) but not Dedekind.

Remark: It could be relevant to add the assumption Dedekind for being an abelian subfactor.

Question 3: No, there are group-subgroup subfactors counterexamples:

An irreducible subfactor is basically abelian iff the edges between vertices of depth $1$ and $2$ have multiplicity one, in its principal and dual principal graphs.

Thanks to the computation of the principal and dual principal graphs of the group-subgroup subfactors (see this book of Jones-Sunder, prop. A.4.4 p141), a group-subgroup subfactor $(R^G \subset R^H)$ is basically abelian iff for all irreducible complex representations $V$ of $G$ then $dim(V^H) \le 1$ with $V^H$ the subspace of vectors invariant under the action of $H$ (note that the dual 2-box space is always abelian).

Now, the group-subgroup maximal subfactors $(R^G \subset R^H)$ are abelian, and thanks to the previous paragraph and the answers of Jack Schmidt here and there, some of them are not basically abelian:

A counter-example of group-subgroup maximal subfactor with a depth $1$-$2$ edge of mult. $>1$ on its principal graph given by $(D_{12} \subset L_2(11))$ of index $55$ (the first ?).

Remark : the question is still open if we restrict to the depth $2$ case, but this negative answer in general is a nice encouragement for the existence of non-trivial maximal Kac algebras (see here).