Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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).

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.