# $[A\otimes C,B\otimes D]=0$ 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 $$[A,B]=AB-BA=2AB\ne0.$$