# $[A\otimes C,B\otimes D]=0$ but neither $[A,B]= 0$ nor $[C,D]=0$

Let $E$ be a complex Hilbert space, $E\otimes E$ be the Hilbert space tensor product.

I want to find $A,B,C,D\in \mathcal{L}(E)$ such that

$[A\otimes C,B\otimes D]=0$ but neither $[A,B]= 0$ nor $[C,D]=0$.

Note that $[A\otimes C,B\otimes D]=0$ implies that $$AB\otimes CD=BA\otimes DC.$$ 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 $B_1= z^{-1}B_2$.

By this lemma, we deduce the existence of $z\in \mathbb{C}^*$ such that $AB=zBA$ and $CD=z^{-1}DC$.

So you need operators $A,B$ with $AB\ne BA$, but $AB=zBA$ for some $z\ne1$. One way to get this is to get a unitary with powers of a matrix unit in the diagonal, like $$B=\begin{bmatrix} -1&0\\0&1\end{bmatrix},$$ and $A$ the unitary that permutes the diagonal, namely $$A=\begin{bmatrix} 0&1\\1&0\end{bmatrix}.$$ Then $$AB=\begin{bmatrix} 0&1\\ -1&0\end{bmatrix},$$ while $$BA=\begin{bmatrix} 0&-1\\ 1&0\end{bmatrix} .$$ That is, $AB=-BA$. For these two matrices, $$[A\otimes A,B\otimes B]=AB\otimes AB-BA\otimes BA=AB\otimes AB-AB\otimes AB=0,$$ but $$[A,B]=AB-BA=2AB\ne0.$$