Through the questions below, this post asks whether the concept of **abelian [subfactor](http://en.wikipedia.org/wiki/Subfactor)** is relevant.  

**Remark** : here *abelian* qualifies an inclusion of II$\_1$ [factors](http://en.wikipedia.org/wiki/Von_Neumann_algebra#Factors) $(N \subset M)$, $N$ is not an abelian algebra.




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> 
> First some useful reminders about groups and lattices :
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**Definitions**: A [lattice](http://en.wikipedia.org/wiki/Lattice_%28order%29) $(L, \wedge, \vee)$ is :


* [**Distributive**](http://en.wikipedia.org/wiki/Distributive_lattice) if $a∨(b∧c) = (a∨b) ∧ (a∨c)$
* [**Modular**](http://en.wikipedia.org/wiki/Modular_lattice) 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](http://www.math.hawaii.edu/%7Ewilliamdemeo/latticetheory/Palfy-GroupsAndLattices-GStA-2001.pdf) 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**](https://en.wikipedia.org/wiki/Hamiltonian_group) 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 $Q\_8$).




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> 
> All the [subfactors](http://en.wikipedia.org/wiki/Subfactor) $(N\subset M)$ are irreducible and finite index inclusions of II$\_1$ factors.
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Let $(N\subset M)$ be a subfactor and $\mathcal{L}(N\subset M)$ its lattice of intermediate subfactors.  

**[Galois correspondence](http://projecteuclid.org/euclid.pja/1195524026)** 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](https://mathoverflow.net/questions/133050/the-cyclic-subfactors-theory-a-quantum-arithmetic)** 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 ?
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**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](http://www.sciencedirect.com/science/article/pii/S002212369690110X) 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](https://mathoverflow.net/questions/157775/products-of-maximal-inclusions-of-finite-groups-with-a-non-obvious-intermediate/157860#157860)).


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 ?
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In [this paper](https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-50/issue-2/Normal-intermediate-subfactors/10.2969/jmsj/05020469.full), T. Teruya introduced the notion of normal intermediate subfactors, generalizing exactly the notion of normal subgroups (see the post [Jordan-Hölder theorem for subfactors](https://mathoverflow.net/questions/156311/jordan-holder-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](http://www.sciencedirect.com/science/article/pii/S002212369690110X) thm3.9 p323, [T](https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-50/issue-2/Normal-intermediate-subfactors/10.2969/jmsj/05020469.full) thm3.4 p377).



> 
> **Question 2** : Are the abelian subfactors Dedekind ?
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**Remark** : Positive answers for questions 1a, 2 and [Jordan-Hölder](https://mathoverflow.net/questions/156311/jordan-holder-theorem-for-subfactors), would solve the question 1 [here](https://mathoverflow.net/questions/133050/the-cyclic-subfactors-theory-a-quantum-arithmetic).


**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 ?
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**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](https://mathoverflow.net/questions/132866/non-weakly-group-theoretical-integral-fusion-categories)).



Abelian subfactors, a relevant concept?
**Question 1a**: Yes, a cyclic subfactor is abelian, **if** it admits no depth $2$ intermediate inclusions, because then, thanks to the corollary [here](https://mathoverflow.net/questions/135335/what-are-the-intermediate-subfactors-of-the-tensor-product-of-two-maximal-subfac/158350#158350), 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. 




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**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](http://homepages.ulb.ac.be/~tconnor/atlaslat/alt6.pdf)) which are of order $24$ and isomorphic to $S\_4$ (see [here](http://groupprops.subwiki.org/wiki/Subgroup_structure_of_alternating_group:A6)).  

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](https://mathoverflow.net/questions/158662/normal-intermediate-subgroup-and-normal-core)), but then $A\_6=KL$ and so by the [product formula](http://groupprops.subwiki.org/wiki/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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**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](http://www.cambridge.org/us/knowledge/isbn/item1155032/?site_locale=en_US) 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](https://mathoverflow.net/users/3710/jack-schmidt) [here](https://math.stackexchange.com/questions/684172/maximal-subgroup-and-representations-principal-part) and [there](https://math.stackexchange.com/questions/684103/maximal-subgroup-and-representations-dual-part), 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](https://mathoverflow.net/questions/132866/non-weakly-group-theoretical-integral-fusion-categories)). 


