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

Question: are there only finitely many maximal subfactors of a fixed finite index (up to isomorphism)?

(Need to add "finite depth" and "irreducible" ?)

Bonus question: let $\alpha$ the index of a finite depth irreducible subfactor.
Does there exist a maximal finite depth irreducible subfactor of index $\alpha$ ?

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Presumably, you are imagining fixing the isomorphism class of $N$ and $M$ as von Neumann algebras (otherwise, even for index=1, there are infinitely many). Another way of formalizing the question so that it has a chance of have a positive answer, is to ask for isomorphism classes of planar algebras. –  André Henriques Jun 8 '13 at 18:00
    
What about subfactors of index 4? I think subfactors of affine type $D_n$ are all maximal. –  Victor Ostrik Jun 8 '13 at 18:12
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Dave Penneys says to me that each the $D_{n}^{(1)}$ subfactors decompose into intermediate subfactors isomorphic to $R^{\mathbb{Z}_{2}} \subset R$ (with $R$ the hyperfinite $II_{1}$ factor), but I have not verified this information, I will try to find a reference. –  Sébastien Palcoux Jun 8 '13 at 18:32
    
If anyone know a reference containing the explicit intermediate subfactors lattice for all the $D^{(1)}_{n}$ subfactors, please post it here. –  Sébastien Palcoux Jun 10 '13 at 12:54
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It says: "If $\Gamma$ is $D^{(1)}_n$ $n > 4$ or $D^{(1)}_\infty$, then $N\subset M$ has the only one nontrivial intermediate subfactor." in the proof they use that the Galois group is $\mathbb{Z}_2$. –  Marcel Bischoff Jun 16 '13 at 15:53
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