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This post is a sequel of: Are all the R-R-bimodules completely reducible?

Question: For which (as general as possible) class of subfactors $(N \subset M)$, the bimodule $_NM_M$ is known completely reducible? Is there counter-examples or open cases?

Following the comments of Marcel it seems that $_NM_M$ is completely reducible iff $N' \cap M$ is a type ${\rm I}$ von Neumann algebra. How to prove that in details?

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  • $\begingroup$ Finite dimensional relative commutant should be sufficient. $\endgroup$
    – Marcel Bischoff
    Commented May 9, 2015 at 2:57
  • $\begingroup$ @MarcelBischoff: where is the proof? and for the general case, do you know (or expect) a counter-example? $\endgroup$
    – Sebastien Palcoux
    Commented May 9, 2015 at 3:03
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    $\begingroup$ The $\mathrm{Hom}(_NM_M,_NM_M)$ can be identified with the relative commutant. If you take a standard half-sided modular inclusion the relative commutant is a type III factor. $\endgroup$
    – Marcel Bischoff
    Commented May 9, 2015 at 3:09
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    $\begingroup$ Corollary 2.9 in journals.math.ac.vn/acta/pdf/0803209.pdf $\endgroup$
    – Marcel Bischoff
    Commented May 9, 2015 at 3:17
  • $\begingroup$ @MarcelBischoff; Nice! So if I understand well, $_NM_M$ is completely reducible iff $N' \cap M$ is a type I von Neumann algebra, right? $\endgroup$
    – Sebastien Palcoux
    Commented May 9, 2015 at 3:37

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