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Given two von-Neumann algebra factors $\mathcal M,\mathcal N$, is $\mathcal M\cap\mathcal N$ a factor?

And how about the intersection of infinitely many factors?

Notes:

  • I know that the intersection is a von-Neumann algebra. (This is immediate from the definition of a von-Neumann algebra as a SOT-closed algebra with adjoints and 1.)
  • I know that this does not hold for Type I factors. (See comments here.)
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    $\begingroup$ If I'm not mistaken, given any tracial vN algebra $A$, one could form free products $1*A*L\mathbb{F}_2=\mathcal{N}$ and $L\mathbb{F}_2*A*1=\mathcal{M}$ which provide $\mathcal{M}\cap\mathcal{N}\simeq A$, whereas $\mathcal{M}$ and $\mathcal{N}$ are II$_1$ factors by virtue of being the free product of a II$_1$ factor with something. $\endgroup$
    – Ben Johnsrude
    Commented May 1, 2023 at 19:38

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The answer is no. There are subfactors $N\subset M$ with finite Jones index $[M:N]$ with $N^{\prime}\cap M=\mathbb{C}\oplus \mathbb{C}$. For example, consider a type $II_1$ factor $P$ and let $\alpha$ be an outer automorphism of $P$. Put $M= P\otimes M_2(C)$ and $N$ be the algebra of all diagonal matrices $(x,\alpha(x)),$ for $x\in P$.

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