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

It is a well known result that any subfactor of the hyperfinite $II_{1}$ factor is hyperfinite. I wonder if there is any finite index version of this for free group factors. In particular is it true that if $N \subset L(F_{\infty})$ is a finite index subfactor then $N \cong L(F_{\infty})?$

share|improve this question
As of jstor.org/discover/10.2307/… it was still open as to what the finite index subfactors of free group factors are. In that paper it was proved that all such subfactors are prime factors. Since then we have solidity, but I'm not sure if this will help. –  Jon Bannon Jun 5 '12 at 22:31
Hi Mike, I don't think this is known. In fact, I don't even think it is known if any finite index subfactor is an interpolated free group factor. The problem is that there isn't alot of ways to prove something is a free group factor. One possibility, which might not be too far out of reach, is to show that if a factor has lots of "free malleable deformations" in the sense of Popa (which is something like a one parameter group of automorphisms that takes a large subalgebra to a free copy of itself) then it is isomorphic to a free product. This still seems somewhat out of reach though. –  Owen Sizemore Jun 6 '12 at 2:15
Thanks for the comments! I have never heard any results one way or another on this question so I figured that it might not be known. –  Mike Hartglass Jun 7 '12 at 4:51
Hi Mike, just a few comments: (a) your question is, as far as I know, open even for index $2$, i.e., in the case that $N=L(F_\infty)^G$ is the fixed point algebra of an outer action of the group $G$ with $2$ elements; (b) If you construct a tunnel $P\subset N\subset L(F_\infty)$, then $P\cong L(F_\infty)$ since it is stably isomorphic to it. (c) One could even ask if what you propose is true for any non-hyperfinite subfactor $N$ without the assumption of finite index. Note that (c) is true for equivalence relations: an equivalence subrelation of a treeable one is treeable. –  Dima Shlyakhtenko Jun 10 '12 at 4:48
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer


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

Browse other questions tagged or ask your own question.