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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 $L(\mathbb{F}_{\infty})$).

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I highly doubt this is known. I don't have much intuition about what being a maximal subfactor gives you, but I think the only results about fundamental groups of subfactors is what you give above. (Well almost, by Popa's work any amenable subfactor, not just finite depth, has full fundamental group, and as mention in the BNP article you cite it is true for any stable subfactor). As far as results where it is not all of $\mathbb{R}^+$, I think that the BNP examples are all that is known. –  Owen Sizemore Jul 11 '13 at 22:01
    
Intuitively, a maximal subfactor is a kind of quantum generalization of a prime number, because $R^{G} \subset R$ is maximal iff $G = \mathbb{Z}/p\mathbb{Z}$ with $p$ a prime number. –  Sébastien Palcoux Jul 12 '13 at 8:21
    
I had read about stable subfactor in the BNP: << i.e. splits a common copy of the hyperfinite $II_{1}$ factor $R$ >>, but I don't understand what does it mean. –  Sébastien Palcoux Jul 12 '13 at 8:26
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Stable means the that $P\subset Q\simeq P\otimes\mathcal{R}\subset Q\otimes \mathcal{R}$. –  Owen Sizemore Jul 13 '13 at 4:48
    
Thank you again ! –  Sébastien Palcoux Jul 13 '13 at 9:31
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