# Invertibility of the planar algebra-subfactor correspondence

In Jones's paper "Planar Algebras I", Theorem 4.2.1 establishes that an extremal finite index subfactor admits a spherical C*-planar algebra structure, and Theorem 4.3.1 establishes that spherical C*-planar algebra arise from subfactors. It seems unlikely, due to the key ingredient of the proof of Theorem 4.3.1, that these processes are inverses of one another.

Question: Is there a class of subfactors for which one can associate planar algebras in a reversible way? (I.e. for which there is a known inverse for the way one passes from the subfactor to the planar algebra, and vice versa?)

(In particular I am wondering if this is now possible in light of the work of Jones, Shlyakhtenko and Guionnet.)

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Sorry, Dmitri. I wasn't aware of this. –  Jon Bannon Jan 22 '11 at 1:56

Yes, strongly amenable subfactors of the hyperfinite $II_1$-factor are completely classified by their standard invariant. The finite depth case was done by Popa's Classification of subfactors: the reduction to commuting squares (MR1055708), and the infinite depth case was finished by Popa's Classification of amenable subfactors of type $II$ (MR1278111). The reconstruction theorem in this case reproduces a hyperfinite $II_1$-subfactor.

The Guionnet-Jones-Shlyakhtenko construction reproduces an inclusion of interpolated free-group factors (arXiv:0911.4728), and there is a specific formula for which factors you get. So you need to start with the right factors (modulo the free-group factor isomorphism problem...).

EDIT: Noah's answer makes a really important point. I should point out that at index 6, Bisch, Nicoara, and Popa constructed an uncountable family of (non-amenable) subfactors of the hyperfinite $II_1$-factor with the same standard invariant with property (T) (MR2314611). As they say in the abstract:

We exploit the fact that property (T) groups have uncountably many non-cocycle conjugate cocycle actions on the hyperfinite $II_1$ factor.

For a discrete group, if you're amenable and you have property (T), then you're finite. For a subfactor, if you're amenable and you have property (T), then you're finite depth. So once you're in the infinite depth property (T) setting, there's no hope for a bijective correspondence between subfactors and planar algebras.

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Thanks Dave and Noah! I'll accept this first answer. –  Jon Bannon Jan 23 '11 at 14:51