Irreducible subfactor inclusion and von Neumann type

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?

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(?).

– Community Bot
Mar 3, 2023 at 10:00

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.

• Thanks Martin! Do you know of a good reference for learning about this? The material that I know focuses on subfactor inclusions of type II.
– Lau
Feb 17, 2023 at 10:11
• Many years ago I would have had a better answer, I guess, but I've never been a specialist and I've forgotten lots of stuff. From the Takesaki/Connes/Haagerup days in the 70s there have been many papers about type III factors, though not so many specifically dealing with inclusions. Longo's papers from the 90s could be a place to look, but there are others, too. Feb 17, 2023 at 12:03
• Why finite von Neumann algebras cannot contain infinite von Neumann algebras? Feb 17, 2023 at 16:10
• If we cannot have a irreducible type III factor inside type II factor inclusion that we also cannot have one the other way round, right? Just because $N \subset M$ irreducible implies $M' \subset N'$ irreducible.
– Lau
Feb 20, 2023 at 13:39
• I'll try to find a good argument (I don't have one right now), but here is my intuition: $M=R\otimes B(H)$, with $R$ a II$_1$-factor. If $N$ is type III, then it has to be contained in the $B(H)$ "side", and so it cannot be irreducible. Feb 20, 2023 at 14:32