# Commutant of subalgebra of tensor product

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?

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, 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 $$B = C'$$.
• Assume $M$ is in standard form sitting inside $B(L^2(M,\tau))$, sorry for not mentioning – user136400 Apr 15 at 13:37
• Does there exist canonical trace on $C$ if M is tracial? – user136400 Apr 15 at 14:16
• I think so, if $H = L^2(M)$ then $H\otimes H = L^2(M\otimes M)$ and there is a canonical trace on $(M\otimes M)' = M' \otimes M'$. I guess every element of $C$ has the form $x + uy$ where $x,y \in M'\otimes M'$, and then you just define $\tau(x + uy) = \tau(x)$. – Nik Weaver Apr 15 at 14:41
• You sure about that? Note that every positive element has $y=0$. – Nik Weaver Apr 16 at 11:36