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Let $(N \subset M)$ be a subfactor and $N \subset M \subset M_1$ the basic construction.

Question: Is $(M \subset M_1) \simeq (M' \subset N')$? Else in which generic case it's true?
What's the proof?

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  • $\begingroup$ I assume you want something beyond the obvious anti-isomorphism given by $J_M$? $\endgroup$ – Martin Argerami Sep 1 '14 at 19:01
  • $\begingroup$ @MartinArgerami: I'm looking for the most general case possible giving an isomorphism of subfactors (in fact, planar algebra equivalence would be enough for me). Nevertheless, a large generic case would be nice. $\endgroup$ – Sebastien Palcoux Sep 1 '14 at 19:14
  • $\begingroup$ @MartinArgerami: A von Neumann algebra is anti-isomorphic to its opposite, but not isomorphic to it in general, isn't it? Do you have an example of factor not isomorphic to its opposite? Anyway "being isomorphic to its opposite" for the factors seems to be the good assumption for the isomorphism of subfactors, but it's not clear to me if it's necessary for the planar algebra equivalence. $\endgroup$ – Sebastien Palcoux Sep 1 '14 at 19:34
  • $\begingroup$ @MartinArgerami: I've found the following paper of Alain Connes: A Factor not Anti-Isomorphic to itself. The example of Alain is a type ${\rm III}$ factor. I don't know yet if there is an example of type ${\rm II}_1$. $\endgroup$ – Sebastien Palcoux Sep 1 '14 at 20:14
  • $\begingroup$ There is. Some information here and its references: ams.org/mathscinet/search/… I still don't think that answers your question, though. $\endgroup$ – Martin Argerami Sep 1 '14 at 20:16

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