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

Context: First recall some results:
- Actions of finite groups on the hyperfinite type $II_{1}$ factor $R$ (Jones 1980).
- A Galois correspondence for depth 2 irreducible subfactors (Izumi-Longo-Popa 1998).
- A finite group is cyclic if and only if its subgroups lattice is distributive (Ore 1938).
- Any irreducible subfactor of finite index has only finitely many intermediate subfactors (Watatani 1993).

It's then natural to give the following definition :
A cyclic subfactor is a subfactor admitting a distributive intermediate subfactors lattice.

Examples :

  • Let $G$ be a finite group. By Jones, there is a unique irreducible subfactor $(R^{G} \subset R)$, and by the Galois correspondence, its intermediate subfactors lattice is given by the subgroups lattice of $G$. Conclusion, by Ore, $(R^{G} \subset R)$ is cyclic if and only if $G$ is a cyclic group : $\mathbb{Z}_{n}$ with $n\in \mathbb{N}$.

  • Every maximal subfactor (i.e. no non-trivial intermediate subfactors) is cyclic. The $2$-supertransitive subfactors are maximal : for example, the Haagerup subfactor (for more examples see here).

  • Every irreducible depth $2$ subfactor is of the form $(R^{\mathbb{A}} \subset R)$ with $\mathbb{A}$ a Kac algebra (for ex. $\mathbb{C}G$). By Galois correspondence, the lattice for $(R^{\mathbb{A}} \subset R)$ is given by the left coideal subalgebras of $\mathbb{A}$. It's unknown, whether or not, an irreducible depth $2$ cyclic subfactor is always trivial (i.e. the first example, with $\mathbb{A} = \mathbb{C}\mathbb{Z}_{n}$). There are attempts of counter-examples (see link1, link2 and link3).

Motivation: Thanks to the first example, I perceive the cyclic subfactors as a quantum generalization of the natural numbers. Here is a slogan promoting the idea that they are central in the subfactor theory :
The prime numbers are to the natural numbers what the maximal subfactors are to the cyclic subfactors,
and the cyclic groups are to the groups what the cyclic subfactors are to the subfactors.

Questions: Here are some questions asking if it's a good "quantum arithmetic" :
(The subfactors are supposed to be finite index and irreducible)

1. As well as an integer $n$ decomposes uniquely (up to permutation) into a product of prime numbers: $n = \prod p_{i}^{r_{i}}$, a cyclic subfactor $N \subset M$ decomposes uniquely (up to permutation and isomorphism) into a chain of maximal intermediate subfactors $N = P_{1} \subset P_{2} \subset ... \subset P_{r} \subset M$ ?

2. Any intermediate subfactor of a cyclic subfactor is cyclic ? (obvious)

3. Two cyclic subfactors of max. depth m compose into at most n+1 cyclic subfactors of depth m+n ?

4. How generalize the multiplication and the addition of $\mathbb{N}$ onto the cyclic subfactors ? (hard)


Clarification of question 1

Let $N \subset M$ be a cyclic subfactor. Let two chains of maximal intermediate subfactors $$N = P_{1} \subset P_{2} \subset ... \subset P_{r} \subset P_{r+1} = M $$ $$N = Q_{1} \subset Q_{2} \subset ... \subset Q_{s} \subset Q_{s+1} = M $$ then $r=s$ and $\exists \sigma \in S_{r}$ such that $(P_i \subset P_{i+1}) \simeq (Q_{\sigma(i)} \subset Q_{\sigma(i)+1})$ ?

More, $\forall \sigma \in S_{r}$, $\exists (R_i)$ such that $(P_i \subset P_{i+1}) \simeq (R_{\sigma(i)} \subset R_{\sigma(i)+1})$ and $$N = R_{1} \subset R_{2} \subset ... \subset R_{r} \subset R_{r+1} = M $$ is a chain of maximal intermediate subfactors ?


Examples for question 3

Let the subfactor $(R^{\mathbb{Z}_2} \subset R)$, it's a depth $2$ irreducible maximal (and so cyclic) subfactor. It composes with itself into $n-1$ cyclic subfactors of depth $n$. These subfactors have principal graph $D_{n+2}^{(1)}$.

Remark : About the composition of subfactors, there is this paper of Izumi-Morrison-Penneys (2013).


Acknowledgment: Thank you to Vaughan Jones who encouraged me to develop this theory by saying :
<< Your cyclic idea might have potential... see what you can prove about such subfactors >>. Thank you also to André Henriquez and Dave Penneys: I adjusted the questions 1 and 3 after their remarks.

share|improve this question
4  
@Sebastien: You should maybe try to find of a precise formulation of your question 1 (without any reference to prime factorization of integers) and ask that on its own: what do you mean exactly by "decomposes uniquely". Concerning 3, could you explain why you say "fixed depth": do you have a counterexample when you don't impose fixed depth? If yes: what is it (knowing it will help others think about your question). –  André Henriques Jun 7 '13 at 19:01
    
@AndréHenriques : I've added a clarification of question 1, and examples for question 3. I hope it's ok. –  Sébastien Palcoux Nov 18 '13 at 18:54

1 Answer 1

up vote 0 down vote accepted

The cyclic subfactors theory defines a too large class$^{\star}$ of subfactors for being an good quantum arithmetic (i.e. quantum generalization of the natural numbers theory).
Nowadays, my better attempt for this is the natural subfactors theory (see the optional part here).

$^{\star}$Up to equivalence, more than $70$% of the inclusions of groups of index $≤30$ are distributive lattice! Nevertheless the percentage is decreasing for the following indices (see this post).


Question 1: No, $(R^{A_7} \subset R^{S_4})$ is a counterexample.
Thanks to the subgroups lattice of $A_7$ (see here), there is a subgroup of $A_7$ called $2^2:S_3$ isomorphic to $S_4$, such that $(S_4 \subset A_7)$ has exactly two non-trivial intermediate subgroups: $K=L_2(7)$ and $L=A_4:S_3$, so that $(R^{A_7} \subset R^{S_4})$ is cyclic.
Now if we consider the two maximal chains $(S_4 \subset K \subset A_7)$ and $(S_4 \subset L \subset A_7)$, then the intermediate indices are $(7,15)$ and $(3,35)$, so these chains can't be equivalent.

See the questions: Jordan-Hölder theorem for subfactors? and Abelian subfactors, a relevant concept?


Question 3: No, the maximal subfactor $(R^{\mathbb{Z}_3 \rtimes \mathbb{Z}_2} \subset R^{\mathbb{Z}_2})$ gives counter-examples.
It is depth $4$ and it admits several compositions with itself which are also cyclic subfactors of depth $4$,
for example $(R^{(\mathbb{Z}_3 \rtimes \mathbb{Z}_2) \times (\mathbb{Z}_3 \rtimes \mathbb{Z}_2)} \subset R^{\mathbb{Z}_2 \times \mathbb{Z}_2})$ and $(R^{\mathbb{Z}_9 \rtimes \mathbb{Z}_2} \subset R^{\mathbb{Z}_2})$.
Their intermediate subfactors lattices are as $\mathcal{L}_6$ and $\mathcal{L}_4$, so they are not isomorphic.
($\mathcal{L}_n$ is the lattice of divisors of $n$).

share|improve this answer
    
About question 1, there are two parts ($\exists \sigma$... and $\forall \sigma$...). For the second part, there is a counterexample simplier than $(R^{A_7} \subset R^{S_4})$: just take $(R^{S_4} \subset R^{\mathbb{Z}_4})$ because its lattice is an inhomogeneous single chain. –  Sébastien Palcoux May 7 at 18:41

Your Answer

 
discard

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.