Consider two Hilbert spaces $H_1$ and $H_2$, and $A$, $B$ unbounded operators on $H_1$, $H_2$ respectively. $(A \otimes I)$ is classically defined as the closure of the operator defined on the set of linear combinations of elements of the form $f \otimes g$ where $f\in D(A)$, $g \in H_2$. The same goes on for $I \otimes B$.
Does one have $(A \otimes I)(I \otimes B) = (I\otimes B)(A\otimes I)$ on the natural domain of the commutator ? One would expect that both are included in $A \otimes B$.
I have a hard time proving it. On the set of linear combinations of elements of the form $f\otimes g$ with $f \in D(A)$, $g\in D(B)$, one easily gets $(A \otimes I)(I \otimes B) = (I\otimes B)(A\otimes I) = A \otimes B$. But extending this to the closures seems non trivial.
However, one could expect this to be true. It is true for bounded operators and it is even true for self-adjoint operators, since the spectral families commute. I thought about using polar decomposition but I didn't get anywhere.
