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?



Are subfactor planar algebras hard to classify at index 6?
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.


