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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?

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2 Answers 2

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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(?).

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Mar 3, 2023 at 10:00
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Lau
    Feb 17, 2023 at 10:11
  • $\begingroup$ 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. $\endgroup$ Feb 17, 2023 at 12:03
  • $\begingroup$ Why finite von Neumann algebras cannot contain infinite von Neumann algebras? $\endgroup$ Feb 17, 2023 at 16:10
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    $\begingroup$ 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. $\endgroup$
    – Lau
    Feb 20, 2023 at 13:39
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    $\begingroup$ 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. $\endgroup$ Feb 20, 2023 at 14:32

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