the type of the intermediate von Neumann subalgebra

Question

Let $M$ be a type III$_1$ factor and $N$ be a type III$_1$ subfactor of $M$.

If we know that $P$ contains $N$ amd also is a subalgebra of $M$. Can we determine the type of $P$. Is it of type III?

Answer 1

No. $M$ is of type III, so $M = M \otimes B(l^2)$. Then $N$ could be embedded in the $B(l^2)$ tensor component, and $P$ could be intermediate between $N$ and $B(l^2)$, which means it could be type I$_\infty$ ($P$ = $B(l^2)$), type II$_\infty$ ($B(l^2) = B(l^2) \otimes B(l^2)$, then $N$ could be embedded into the first tensor component and $P$ could be $B(l^2) \otimes Q$ for a II$_1$ factor $Q \subset B(l^2)$), or type III$_\lambda$ for any $\lambda$ (again, $B(l^2) = B(l^2) \otimes B(l^2)$, embed $N$ into the first tensor component and $P$ could be $B(l^2) \otimes Q \simeq Q$ for a III$_\lambda$ factor $Q \subset B(l^2)$). Basically, $P$ could be any properly infinite factor.