# 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