**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.