I've just learned a bit about the tensor product and I couldn't find a real answer to this. I've read something about, that in some cases it could be or not. Let's consider next example:
In the vector space $\mathbb{R}^n\otimes_\mathbb{R}\mathbb{R}^n$ with standard basis $\mathbb{B}=(e_1,…,e_n)$ of $\mathbb{R}^n$, can we say that
$e_1\otimes e_2=e_2\otimes e_1$?
If yes can we say that $\otimes$ is commutative in a vector space $V\otimes V$ generated by the tensor product of a vector space $V$ with itself?
If not, when can it be considered?
Best Answer
No, it is not commutative. It would imply that all bilinear maps are symmetric.
For any vector space $V$ over a field $K$, we only have an isomorphism \begin{align}V\otimes _KV&\longrightarrow V\otimes_KV, \\v_1\otimes v_2&\longmapsto v_2\otimes v_1. \end{align}
Furthermore, the quotient of $V\otimes_K V$ by the subspace generated by all tensors $v_1\otimes v_2 - v_2\otimes v_1$ is called the symmetric product of $V$ by itself.