Commutant of tensor product of commutants contains the tensor product of von Neumann algebra

Question

Let $A_1$ and $A_2$ are two von Neumann algebras acting on the Hilbert spaces $H_1$ and $H_2$ respectively. Let $A_k'$ denotes the commutant of $A_k$ in $B(H_K)$, for $k=1,2$. I want to show that $A_1 \bar \otimes A_2 \subset \left(A_1' \bar \otimes A_2'\right)'$, where $A_1 \bar \otimes A_2$ is defind to be $$A_1 \bar \otimes A_2:=\left\{a_1\otimes a_2~|~a_i\in A_i, i=1,2\right\}''.$$ Also note that $A''$ denotes the double commutant of $A$, that is, $A''=(A')'$. Please help me to solve. Thank you.

Answer 1

I will change the notation to $M=A_1$ and $N=A_2$. Note that, for any subsets of the bounded operators, say $S,T$, if $S\subset T$, then $T'\subset S'$.

It suffices to show that $M'\bar{\otimes}N'\subset\{a\otimes b: a\in M,b\in N\}'$, since taking commutants reverses the inclusion. Since by definition $M'\bar{\otimes}N'=\{x\otimes y: x\in M', y\in N'\}''$, it further suffices to show that $\{a\otimes b: a\in M,b\in N\}\subset\{x\otimes y: x\in M', y\in N'\}'$, again the reason being that taking commutants reverses the inclusion.

But this is obvious: let $a\in M,b\in N$ and $x\in M',y\in N'$. We have $ax=xa$ and $by=yb$, so $$(a\otimes b)(x\otimes y)=ax\otimes by=xa\otimes yb=(x\otimes y)(a\otimes b)$$ as we wanted.