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).
