Irreducible subfactor inclusion and von Neumann type

Question

Let $M$ and $N$ be factors and $N\subset M$ be an irreducible subfactor inclusion, i.e., $N$ has trivial relative commutant in $M$.

Does it follow that $M$ and $N$ have the same type?

Answer 1

Yes.

• if $M$ is type I then $M=B(H)$ and no subfactor can be irreducible.

• if $M$ is type II$_1$ it cannot have neither II$_\infty$ nor III subfactors (as these are infinite). And a type I subfactor cannot be irreducible: if $\{e_{kj}\}_{k,j=1}^n$ are matrix units in for $N$, then $\sum_ke_{k1}xe_{1k}\in N'\cap M$ for all $x\in M$.

• if $M$ is type II$_\infty$ then $N$ cannot be type III. If $N$ is type I then $N'\cap M$ is non-trivial as in the previous case. And if $N$ is II$_1$ then we can embed $N\otimes B(H)$ in $M$, and so $N'\cap M\supset 1\otimes B(H)$.

• if $M$ is type III, then $N$ cannot be type I nor type II$_1$ for the same reasons as in the previous case. And it cannot be II$_\infty$ either, because as you noted the inclusion $M'\subset N'$ would be an irreducible inclusion of a type III inside a type II.

Answer 2

Maybe the discrete decomposition of a type III$_{\lambda}$ factor $M$, $0<\lambda<1$, provides an example of an infinite-index, irreducible inclusion of II$_{\infty}$ in III$_{\lambda}$, i.e., we have:

$M = M^{\phi}\rtimes_{\alpha}\mathbb{Z}$,

where $\phi$ is a generalized trace on $M$, $M^{\phi}$ its centralizer (a II$_{\infty}$ factor), and $\alpha$ is the automorphism scaling the trace by $\lambda$.

Since $\alpha$ is outer, the inclusion of $M^{\phi}$ in $M$ should be irreducible by the relative commutant theorem(?).