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I find tensor products a bit difficult to grasp. In particular, I would like to be sure to understand the concept of universal property. Here is a tensor product description of a (0,2)-tensor, the dot product between two vectors belonging to the same K-vector space (with $K = \mathbb{R}$). I would like to know if there are any errors in my reasonning.

According to Wikipedia: "The tensor product of V with itself can be defined (up to isomorphism) by any pair (L, $\varphi$), with L a vector space on K and $\varphi:V\times V\to L$ a bilinear map such that for any K-vector space Z and any bilinear map $h:V\times V\to Z$, there exists a unique linear map $\tilde{h}:L\to Z$ verifying $h=\tilde{h}\circ\varphi$."

The image of $(a^1 e_1+a^2 e_2,b^1 e_1+b^2 e_2)$ by $\varphi$ is:

\begin{equation} a^1b^1 e_1\otimes e_1 + a^1b^2 e_1\otimes e_2 + a^2b^1 e_2\otimes e_1 + a^2b^2 e_2\otimes e_2 \end{equation}

In the specific case of the inner product, the function $\tilde{h}$ linearly associates the following elements of L to elements of K: \begin{align} e_1\otimes e_1\to1 \\ e_1\otimes e_2\to0 \\ e_2\otimes e_1\to0 \\ e_2\otimes e_2\to1 \end{align}

Using the bilinearity of $\varphi$ and the linearity of $\tilde{h}$, the value of this dot product therefore is $a^kb^k$.

Is this a correct way to describe what is going on in this dot product case ?

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I imagine by $W$ you mean $V.$ If you're using $a^kb^k$ as shorthand for $a^1b^1 + a^1b^2$ then nothing you've said is incorrect. Uniqueness is already guarenteed, but it isn't so hard to verify in this example. –  Tim Duff Jul 11 '12 at 20:58
    
@TimDuff: if we're fishing for typos, it should be $a^1b^1+a^2b^2$. ;) Otherwise, yeah, it seems correct. Was there anything in particular you were uneasy about? –  tomasz Jul 11 '12 at 21:32
    
Thanks, I corrected the typo. All the worked out examples I could find on the Internet where more abstract that what, as an engineer, I am used to for tensor use. So I wanted to find the simplest case of application for me, a kind of mathematical equivalent of an "hello world" :). Thank you for your comments ! –  vkubicki Jul 12 '12 at 6:52
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