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I have a question on subfactors of $B(H)$ (the von Neumann algebra of bounded operators on a complex Hilbert space).

Say I have a subfactor $M$ of $B(H)$. Is it true that another subfactor $N \subset B(H)$ which commutes with $M$ and is such that the vN algebra $(M\cup N)''$ generated by $M$ and $N$ is all of $B(H)$, is automatically the commutant of $M$?

I think that it automatically follows that $M \cap N=\mathbb C \,1\!\!1$ because the intersection is contained in the center of, say, $M$.

EDIT: Another way to put this question would be as follows. Consider pairs of commuting subfactors $M,N \subset B(H)$. A pair is trivial if $M=N'$ and it is generating if $(M\cup N)''=B(H)$. The question is now: Is there a non-trivial generating pair of subfactors $M,N \subset B(H)$?

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  • $\begingroup$ Did you mean to ask whether $M \cup N$ generates $B(H)$ as a von Neumann algebra? Because I can't see how the sum could ever be dense. $\endgroup$
    – Nik Weaver
    Commented Jul 3, 2022 at 1:27
  • $\begingroup$ @NikWeaver Thanks for pointing that out, I meant the algebra generated by $M$ and $N$. I edited the question. An example would be any subfactor $M$ and its commutant $N=M'$ and my question is whether this is the only example $\endgroup$
    – Lau
    Commented Jul 3, 2022 at 7:59

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Let $N \subset M'$ be an irreducible inclusion of subfactors, meaning that $N \neq M'$ but the relative commutant is trivial, i.e., $N' \cap M' = \mathbb{C}\cdot I$. There are lots of examples of such things (Google "example of irreducible subfactor"). Then the commutant of the von Neumann algebra generated by $M \cup N$ is contained in both $M'$ (since it contains $M$) and $N'$ (since it contains $N$), so it must be trivial. Therefore this von Neumann algebra must be $B(H)$.

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  • $\begingroup$ Thank you very much for the quick answer! I will look into the examples. If I understand correctly then $N,M$ are a non-trivial generating pair of commuting subfactors (defined as in the question), if and only if $N$ is an irreducible subfactor of $M'$. Do you know if any conditions are known which imply that the subfactor $M'$ has no (non-trivial) irreducible subfactors? My guess would be that for type I it could work. $\endgroup$
    – Lau
    Commented Jul 3, 2022 at 17:11
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    $\begingroup$ You are welcome! I agree with everything you say here: trivial relative commutant is necessary and sufficient, and $B(H)$ indeed has no irreducible subfactors. Those may be the only examples, though. $\endgroup$
    – Nik Weaver
    Commented Jul 3, 2022 at 17:31

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