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?

  • $\begingroup$ 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. $\endgroup$ – Ulrich Pennig Sep 15 '13 at 17:34
  • $\begingroup$ 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. $\endgroup$ – André Henriques Sep 15 '13 at 18:54
  • $\begingroup$ 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. $\endgroup$ – Sebastien Palcoux Sep 15 '13 at 20:32

1, and indeed its generalization to the amenable case, is in "Amenable tensor categories and their realizations as AFD bimodules" by Hayashi and Yamagami, see Section 7. I don't think 2 has appeared in the literature, though I'd expect that it's true.

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.

  • $\begingroup$ Ah, sorry, that trick doesn't work if $\mathcal{C}$ has a nontrivial grading group, since then the algebra does not tensor generate. $\endgroup$ – Noah Snyder Sep 24 '13 at 4:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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