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 $M,N \subset B(H)$?



Commuting and generating subfactors of $ B(H)$
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 $B(H)$.


