# Commutator of bounded operators and tensor product

Let $E$ be an infinite-dimensional complex Hilbert space, $E\otimes E$ be the Hilbert space tensor product and $$\mathcal{L}(E)^+=\left\{A\in \mathcal{L}(E);\,\langle Ax,x\rangle\geq 0,\;\forall\;x\in E\;\right\}.$$

Let $A,B,C,D\in \mathcal{L}(E)$ and $S_1,S_2\in \mathcal{L}(E)^+$ be non zero operators such that $$(S_1\otimes S_2)[A\otimes C,B\otimes D]=0.$$ I want to find sufficient conditions under which $S_1[A,B]=S_2[C,D]=0$ i.e. $S_1AB=S_1BA$ and $S_2CD=S_2DC$.

My attempt: Since $(S_1\otimes S_2)[A\otimes C,B\otimes D]=0$, then $(S_1\otimes S_2)(A\otimes C)(B\otimes D)=(S_1\otimes S_2)(B\otimes D)(A\otimes C)$. Hence $$S_1AB\otimes S_2CD=S_1BA\otimes S_2DC.$$ By using the following result:

Lemma: Let $A_1, A_2,B_1, B_2\in \mathcal{L}(E)$ be non-zero operators. The following conditions are equivalent:

• $A_1\otimes B_1=A_2\otimes B_2$.

• There exists $z\in \mathbb{C}^*$ such that $A_1 =zA_2$ and $B_1= z^{-1}B_2$.

We deduce the existence of $z\in \mathbb{C}^*$ such that $S_1AB=zS_1BA$ and $S_2CD=z^{-1}S_2DC$. When we get $$z=1?$$

A tensor product cannot distinguish that. You always have, for any $A,B$ and for any $z\in\mathbb C$, $$zA\otimes B=A\otimes zB.$$ You can force $z=1$ with very strong conditions, like for example $A_1,A_2\geq0$ and $\|A_1\|=\|A_2\|$.
• The lemma is right. But you will never get it to force $z=1$. – Martin Argerami Mar 23 '18 at 5:01
• Impossible, no. If for instance you require $A_1,A_2\geq0$ and $\|A_1\|=\|A_2\|$, then your $z$ will be $1$. – Martin Argerami Mar 23 '18 at 13:09