The **fundamental group** $\mathcal{F}(N \subset M)$ of an inclusion of $II_{1}$ factors $N \subset M$ is defined as : $\mathcal{F}(N \subset M) =\{t >0 \ | \ (N \subset M)^{t} \simeq (N \subset M) \} $ (see here)

**Examples**:

- It's $\mathbb{R}_{+}^{*}$ for every finite index finite depth irreducible subfactor.

- It's trivial for uncountably many subfactors of the form $R^{\mathbb{Z}_{2}} \subset R⋊\mathbb{Z}_{3}$ (Bisch-Nicoara-Popa)

A subfactor $N \subset M $ is **maximal** if it admits no non-trivial intermediate subfactors $N \subset P \subset M $.

**Examples**:

- Every $2$-supertransitive subfactors: $A_{n}$-subfactor, Haagerup subfactor...

- Every group-subgroup subfactor $(R^{G} \subset R^{H})$ such that $(H \subset G)$ is a maximal subgroup

(i.e. $\pi_{H}(G)$ is a primitive permutation group with $\pi_{H} : G \to S_{X}$ canonical for $X = G/H$).

Is the fundamental group of a maximal subfactor always $\mathbb{R}_{+}^{*}$ ?

**Remark** : For being relevant, we need to restrict to factors with full fundamental group. And for being reasonable, we can start by inclusion of hyperfinite $II_{1}$ factors. Next, we could enlarge to every factors with full fundamental group (for example $L(\mathbb{F}_{\infty})$).

quantum generalizationof a prime number, because $R^{G} \subset R$ is maximal iff $G = \mathbb{Z}/p\mathbb{Z}$ with $p$ a prime number. – Sebastien Palcoux Jul 12 '13 at 8:21stable subfactorin the BNP: << i.e. splits a common copy of the hyperfinite $II_{1}$ factor $R$ >>, but I don't understand what does it mean. – Sebastien Palcoux Jul 12 '13 at 8:26