Consider the von Neumann subalgebra of $M\otimes M$ by $ B= \mathrm{vN} \{T\otimes T: T\in M\}$. What is the commutant of B?



Commutant of subalgebra of tensor product
We need $M \subseteq B(H)$ in order for the commutant to make sense. So $B \subseteq B(H\otimes H)$. The commutant of $B$ is the von Neumann algebra $C$ generated by $M' \otimes M'$ and the flip unitary $u$ acting on $H \otimes H$. It's clear that $B$ is contained in $C'$; conversely, if $x \in C'$ then $\phi(x) = x$ where $\phi$ is the flip automorphism of $B(H\otimes H)$ (= conjugation by $u$), which implies that $x$ lies in the symmetric part of $B(H)\otimes B(H)$ (see the argument [here](https://mathoverflow.net/questions/325163/on-diagonal-part-of-tensor-product-of-c-algebras/325182#325182), replacing norm limits by bounded weak\* limits), and also, commuting with $M'\otimes M' \subseteq C$ implies that $x \in M\otimes M$. Thus $x$ lies in $B =$ the symmetric part of $M\otimes M$. This shows that $B = C'$.


