Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.)

share|improve this question
    
@Jon: You shouldn't link directly to the PDF version of an arXiv paper. See tea.mathoverflow.net/discussion/157/… for a discussion about this. Here is the correct link: arxiv.org/abs/1009.0939 –  Dmitri Pavlov Jan 21 '11 at 23:53
    
Sorry, Dmitri. I wasn't aware of this. –  Jon Bannon Jan 22 '11 at 1:56
add comment

2 Answers 2

up vote 6 down vote accepted

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.

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

Dave's already answered the question quite well, but let me try a different take.

It's informative to look at the simpler case of group subfactors M^G < M. There the planar algebra contains exactly the same information as G itself. Thus in that special case your question becomes:

Under what circumstances is there exactly one outer action of a group G on a factor M up to equivalence?

For the hyperfinite II_1 factor a finite group can only outerly act in one way (this is Jones's very early work), and indeed an amenable group can only outerly act on the hyperfinite in one way (this is Ocneanu's very early work). However, Jones showed that if you have a non-amenable group then it always outerly acts on the hyperfinite II_1 factor in more than one ways. Once you move away from the hyperfinite life becomes substantially more complicated. In general it's quite hard to figure out all ways that a group can act on an arbitrary factor. For example, by work of Ioana, Peterson, and Popa there are II_1 factors which have no outer automorphisms at all!

share|improve this answer
    
BTW: I'm really looking forward to your answer on planar algebra questions not relying on the relation to subfactors. Thanks again for helping collect the state of the art on this invertibility question! –  Jon Bannon Jan 23 '11 at 14:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.