# realizing fusion categories as subfactors of the hyperfinite

Let R be the hyperfinite II_1 or the hyperfinite III_1 factor (pick which ever one you prefer), and let Bim(R) denote the tensor category of R-R-bimodules.

This question is inspired by the recent article The classification of subfactors of index at most five by Jones, Morrison, and Snyder. More precisely, I am interested in Thm 1.1 and Thm 2.10 of the above paper (both of them are older results).

Given the above, I expect the following to be true:

(1) Any unitary fusion category can be embedded in Bim(R).
(2) Any two embeddings are conjugated by an automorphism of R.

however, I am not sure if things have ever been formulated in this way.

My questions are:
• What is the closest result to (1) and (2) available in the literature?
• is it easy to adapt/use some existing proofs to get (1) and (2), and how?

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This might be relevant: In the paper at arxiv.org/abs/1112.4088v2 Sébastian Falguières and Sven Raum show that every finite tensor C*-tensor category is the bimodule category of a $II_1$-factor. – Ulrich Pennig Sep 15 '13 at 17:34
Ulrich: your reference is interesting but does not do in the direction that I want. I want to know that Bim(R) is very big (R hyperfinite). You're telling me that there exist factors M such that Bim(M) is very small. – André Henriques Sep 15 '13 at 18:54
The best persons to answer your question are surely the authors of the article, in the meantime, here are some interesting things : slides p 15, and paper p 7. – Sebastien Palcoux Sep 15 '13 at 20:32

In the fusion case, if the universal grading group is trivial, I think you can just look at the algebra object $\oplus_x x \otimes x^*$ in $\mathcal{C}$ and use the usual reconstruction and uniqueness for subfactors. It's possible I'm missing a technicality, but I think this should work (in particular, Popa doesn't assume irreducibility). If the grading group is non-trivial there still may be some trick that reduces it to the known cases, but it's not clear to me.
Ah, sorry, that trick doesn't work if $\mathcal{C}$ has a nontrivial grading group, since then the algebra does not tensor generate. – Noah Snyder Sep 24 '13 at 4:51