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).