# Isomorphism between tensor product and set of all tensors

I'm new to tensor and quite confused.

First of all, could anyone provide any friendly reference about tensors which explains the idea behind those boring-looking definition?

Then is my problem.

Let $V$ be a $R$-module and we have homomorphism $$h:\left(\bigotimes^r V\right)\otimes\left(\bigotimes^s V^*\right)\rightarrow T^r_s(V;R)$$

by mapping $u_1\otimes u_2\cdots\otimes u_r\otimes\beta^1\otimes\beta^2\cdots\otimes\beta^s$ to a tensor:

$$(\alpha^1,\cdots,\alpha^r,v_1,\cdots,v_s)\mapsto\alpha^1(u_1)\cdots\alpha^r(u_r)\beta^1(v_1)\cdots\beta^s(v_s)$$

where $$\bigotimes^r V=\underbrace{V\otimes V\cdots\otimes V}_r$$ and $T^r_s(V;R)$ is the set of all tensors from $V$ to $R$ with the form $$\tau:\underbrace{V^*\times V^*\cdots\times V^*}_{r}\times\underbrace{V\times V\cdots\times V}_{s}\rightarrow R$$

It is said that when $V$ is a finite-dimensional vector space or space of sections of some vector bundle over $M$ with finite-dimensional fibers, $h$ is an isomorphism. But in general, it maybe not.

I wonder why it is an isomorphism under those two condition?

• Are you asking why $h$ is an isomorphism in the finite dimensional case, or are you asking why $h$ may not be in the infinite dimensional case? For the latter, imagine an infinite dimensional Hilbert space, the inner product is an element of $(V\otimes V)^*$, but it is not in $V^*\otimes V^*$ (It cannot be written as a finite linear combination of pure tensors.) – Willie Wong Oct 30 '12 at 15:27
• I'm asking the former one, why it is isomorphism under such mentioned two circumstance?... – hxhxhx88 Oct 31 '12 at 4:39
• In the definition of $\tau$, did you mean to write the cartesian product $\times$ or did you intent the tensor product $\otimes$? – Willie Wong Oct 31 '12 at 8:41