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?

  • 6
    $\begingroup$ Definitely not. In fact, those are both basis elements of $V\otimes V$. You need to symmetrize by taking $e_1\otimes e_2+e_2\otimes e_1$. $\endgroup$ – Ted Shifrin May 7 at 17:12
  • 9
    $\begingroup$ It is worth noting that the tensor product is symmetric in the sense that $V\otimes W$ and $W\otimes V$ are isomorphic. $\endgroup$ – Inactive - avoiding CoC May 7 at 17:31
  • 1
    $\begingroup$ I was once told that $\otimes$ symbolizes a stop sign which says “Stop! Does not commute!” $\endgroup$ – Jendrik Stelzner May 7 at 22:51

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.


Certainly not. We can dualize an element of $\Bbb R^n \otimes \Bbb R^n$ and then view it as a map $(\Bbb R^n)^* \times (\Bbb R^n)^* \to \Bbb R$. Then, if $(\epsilon^a)$ denotes the basis of $(\Bbb R^n)^*$ dual to $(e_a)$, concretely we have $$(e_1 \otimes e_2)(\epsilon^1, \epsilon^2) = e_1(\epsilon^1) e_2(\epsilon^2) = (1) (1) = 1$$ but $$(e_2 \otimes e_1)(\epsilon^1, \epsilon^2) = e_2(\epsilon^1) e_1(\epsilon^2) = (0) (0) = 0 .$$ Therefore $$e_1 \otimes e_2 \neq e_2 \otimes e_1 .$$

  • 1
    $\begingroup$ You introductory sentence talks about $(\epsilon^a)$, while in the main formula no $\epsilon$ occurs at all, it is using $E_a$ instead. $\endgroup$ – Paŭlo Ebermann May 7 at 22:29
  • $\begingroup$ Thanks, Paŭlo, I've fixed the error. $\endgroup$ – Travis Willse May 7 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.