types of von Neumann subalgebras

Question

Let $N$ be a von Neumann subalgebra of the von Neumann algebra $M$. Is the type of $N$ at most the type of $M$?

Answer 1

If you refer to the main types I, II, III, then all inclusions are possible.

• $B(H)$ is type I and it contains all types.

• Any type II$_\infty$ algebra is of the form $M\otimes B(H)$, and so by embedding $I\otimes N$ you can get the subalgebra to be of any type.

• Any type III algebra $M$ satisfies $M\simeq M\otimes B(H)$, so again any type can be embedded as $I\otimes N$.