# Intersection of von-Neumann algebra factors

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.)
• 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. May 1, 2023 at 19:38

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 $$x\in P$$.