# Is the tensor product (of vector spaces) commutative?

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?

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

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 $$e_1 \otimes e_2 \neq e_2 \otimes e_1 .$$
• You introductory sentence talks about $(\epsilon^a)$, while in the main formula no $\epsilon$ occurs at all, it is using $E_a$ instead. – Paŭlo Ebermann May 7 at 22:29