Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.