The expression of the FRW metric in Cosmology in usually written as:
$$ds^2=-dt^2+a^2(t)d\vec{x}^2$$
where $c=1$. However, $dt^2$ is a shortening of $dt\otimes dt$, that is, of the tensor product of $dt$ with itself. Since the cross terms are zero, we do not have terms of the form $dtd\vec{x}\equiv dt\otimes d\vec{x}$, but it could be the case after performing a coordinate change.
My question is, is this tensor product symmetric? In the case of having nonzero cross terms of the form $dtd\vec{x}$, would it be true that $dtd\vec{x}=d\vec{x}dt$?
