The form of the intersection of two von neumann algebras

Question

Let $A$ and $B$ be Type I factor von Neumann algebras of operators on a separable Hilbert space $\mathcal{H}$. Let $B'$ denote the commutant of $B$. Can $A \cap B'$ always be written as a discrete (but possibly infinite) direct sum of type I factors; i.e., as \begin{equation} A\cap B' = \bigoplus_i \mathcal{B}(\mathcal{H}_L^i) \otimes I_{\mathcal{H}_R^i}? \end{equation}

Answer 1

The commutant is playing no role here, for the commutant of a type I factor is again a type I factor.

So the question can be phrased as whether $A\cap B$ is of the form you want. It turns out that the answer is a resounding "no". In fact, $A\cap B$ doesn't even have to be type I.

Let $A_0=B_0=M_3(\mathbb C)$. By Proposition 3.5 of this paper by Ken Dykema, the free product $A_0*B_0$ is isomorphic to $L(\mathbf F_s)$, an interpolated free group factor; that is, a II$_1$ factor.

The construction of the free product gives us a Hilbert space $H$ such that $A_0,B_0\subset B(H)$ and $A_0*B_0$ is the von Neumann algebra generated by $A_0\cup B_0$. Now put $A=A_0'$, $B=B_0'$. Then $A,B$ are type I factors (because they are commutants of type I factors) and $$ A\cap B=A_0'\cap B_0'=(A_0\cup B_0)'=W^*(A_0\cup B_0)'. $$ The commutant of a type II factor is type II, so $A\cap B$ is type II.