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Given a finite index inclusion, $N\subset M$, of $II_1$ factors we can construct two towers of finite dimensional algebras known as the $\textit{standard invariant}$. For low index, this has allowed for a complete calculation of subfactors. However, at index 6 this breaks downs because there is an uncountable family of non-isomorphic factors with the same standard invariant.

Recent research in this direction has centered on Jones' planar algebra formulism. I have seen many talks by Vaughan where he says that things are hopeless at index 6 and cites the result above.

However, if one doesn't care above classifying subfactors and only wants to classify standard invariants (planar algebras) the above problem does not come up. So my questions is

How hard it is to classify (subfactor) planar algebras at index 6? And why?

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up vote 11 down vote accepted

At the very least there's one Bisch-Haagerup subfactor standard invariant for every quotient group of $\mathbb{Z}/2 \ast \mathbb{Z}/3 \cong \mathrm{PSL}_2(\mathbb{Z})$. That's already way too much to try to classify. Even if you restrict to finite depth, there's way too many finite quotients of the modular group.

On the other hand, if you're very optimistic you could hope that Bisch-Haagerups are the only wild part of the classification of standard invariants at index 6. I don't know any evidence for such a conjecture or counterexamples.

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are these Bisch-Haagerup subfactor standard invariant for quotient groups of $PSL_2(\mathbb{Z})$ pairwise non-isomorphic? – André Henriques Jul 7 '13 at 16:59
I don't think they're all pairwise non-isomoprhic, but most of them are. For example, you can recover the category of representations of the quotient group Rep(G) from the standard invariant. – Noah Snyder Jul 8 '13 at 1:13
  • If a subfactor $N \subset M$ at index 6 is composed, i.e., it admits a non-trivial intermediate subfactor $N \subset P \subset M$, then the two components are necessarily of indices $2$ and $3$. There exists one subfactor at index $2$ given by the graph $A_{3}$, and two at index $3$ given by $A_{5}$ and $D_{4}$.
    Then, the standard invariant of a composed subfactor at index $6$ is necessarily given by a quotient of $A_{3} \star A_{5}$ or $A_{3} \star D_{4}$ (see here and here), the second is given by $\mathbb{Z}_{2} \star \mathbb{Z}_{3}$ (see Noah's answer). In this case, there exists many quotients with property (T), related to a continuous family of non-isomorphic subfactors classified by cocycles (Bisch -Nicoara-Popa).
    Conclusion: the composed subfactors are completely classified by these quotients and cocycles.

  • So now, it remains to classify all the maximal subfactors at index 6, here are some examples:
    $R^{S_{6}} \subset R^{S_{5}} $ , $R^{A_{6}} \subset R^{A_{5}} $, ... (with $S_{n}$ and $A_{n}$ the symmetric and alternating groups).

List all the group-subgroup maximal subfactors at index $6$.

1. Is there a non group-subgroup maximal subfactor at index $6$ (other than $A_{\infty}$) ?
2. Are there only finitely many maximal subfactors of a fixed finite index ?

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The term "exercise" is usually reserved for problems that you know how to solve, and whose difficulty is such that the reader also has a reasonable chance of solving on his/her own. Is that the case of your "exercise"? – André Henriques Jul 7 '13 at 12:47
Ok André, I replace exercise by problem. – Sebastien Palcoux Jul 7 '13 at 13:00
I thought it suffices to check the maximal subgroups of index 6 (up to isomorphism) on an atlas of (finite) groups, but I don't know if there exists a theorem about upper-bound. – Sebastien Palcoux Jul 7 '13 at 13:47
I have posted a generalization of this problem here. – Sebastien Palcoux Jul 9 '13 at 9:55
Thanks to a @BS. answer here, the classification of group-subgroup maximal subfactors at index $n$ reduced to the knowledge of primitive permutation groups (already classified for all n<4096). There is (up to equiv.) 5, 4, 7, 7, 11 such objects for n= 5, 6, 7, 8, 9. – Sebastien Palcoux Jul 9 '13 at 15:03

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