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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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ I was once told that $\otimes$ symbolizes a stop sign which says “Stop! Does not commute!” $\endgroup$ – Jendrik Stelzner May 7 at 22:51
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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.

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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 $$e_1 \otimes e_2 \neq e_2 \otimes e_1 .$$

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    $\begingroup$ 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

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