# If the flip automorphism of a finite factor can be connected to the identity is it approximately inner?

This intriguing question is due to Sorin Popa, so I'm marking it community wiki. I'd like to know the status of the question, and if it is still open, perhaps gather some new ideas for attacking it.

An automorphism of a $II_{1}$ factor is called approximately inner if it is the pointwise strong operator limit of inner automorphisms.

Let $M$ be a type $II_{1}$ factor, and let $M\overline{\otimes}M$ denote its tensor square.

Question: If the flip automorphism $\phi(x \otimes y)=y \otimes x$ of $M\overline{\otimes}M$ is path connected to the identity automorphism (in $Aut(M\overline{\otimes}M)$ with the pointwise strong operator topology), does it follow that the flip is approximately inner?

• Any links to what's already known? Does this work for the hyperfinite II_1 (perhaps using Property P or similar for its tensor square)? – Yemon Choi Feb 2 '11 at 19:47
• @Yemon: I wish I had such a link. This is sort of what I'm trolling for. – Jon Bannon Feb 2 '11 at 20:59
• Starting with the hyperfinite II_1 begs the question, as M being the hyperfinite II_1 factor is equivalent to the flip being approximately inner. – Jon Bannon Feb 2 '11 at 21:08
• So, when you mean "pointwise", you mean "evaluating the automorphisms on elements of the factor"? That's how I understand it, but in part of the literature (cfr. work of Haagerup and Stormer) "pointwise" means something else (i.e. evaluation on states). In the case of the first definition, I think that any automorphism in a II_1 factor is a pointwise-norm limit of inner automorphisms. – Martin Argerami Feb 4 '11 at 22:31