Let $M$ be a type III$_1$ factor and $N$ be the type II$_1$ subalgebra of $M$.

What is the type of $N'\cap M$? Can it be any type?


1 Answer 1


Such a relative commutant $N' \cap M$ can even be any factor. Choose any type III$_1$ factor $P$ with a type II$_1$ subfactor $N \subset P$ with trivial relative commutant $N' \cap P = \mathbb{C} 1$. When $Q$ is an arbitrary factor, we get that $P \overline{\otimes} Q$ is a type III$_1$ factor that contains the type II$_1$ subfactor $N \otimes 1$ with relative commutant $1 \otimes Q$.

  • 1
    $\begingroup$ What is a good example to keep in mind of a type III_1 factor ๐‘ƒ with a type II_1 subfactor ๐‘โŠ‚๐‘ƒ with trivial relative commutant? $\endgroup$ Feb 21 at 9:42
  • 7
    $\begingroup$ When $P_1$ is any type III$_1$ factor, one can realize $P_1$ as the crossed product of a II$_\infty$ factor $N_1$ and a trace scaling action of $\mathbb{R}$. Then $N_1$ is a type II$_\infty$ subfactor of $P_1$ with trivial relative commutant. Now take a projection $p \in N_1$ that is finite in $N_1$. Then $N = p N_1 p$ is a type II$_1$ subfactor of $P = p P_1 p$ with trivial relative commutant. The conclusion is that any type III$_1$ factor has a type II$_1$ subfactor with trivial relative commutant. $\endgroup$ Feb 21 at 10:09

Your Answer

By clicking โ€œPost Your Answerโ€, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.