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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).

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1 Answer 1

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 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 $G$ and $W$ of $H$, and for all $g \in G$, then $dim(V^H) \le 1$ and $dim(W^K) \le 1$, with $K = H \cap g^{-1}Hg$ and $V^H$ the subspace of vectors invariant under the action of $H$.

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:

The smallest counterexample seems to be given by $(S_3 \subset A_5)$ of index $10$: the first group-subgroup maximal subfactor with a depth $1$-$2$ edge of mult. $>1$ on its principal or dual principal graph. There are also counterexamples with such edges on its principal and dual principal graphs, for example, the one 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).


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.

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