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In S.S.Chern's Lectures on Differential Geometry, I don't understand the following text in Chapter 2, which introduces the tensor product:

The tensor product $V^*\otimes W^*$ of the vector spaces $V^*$ and $W^*$ refers to the vector space generated by all elements of the form $v^*\otimes w^*$, $v^*\in V^*$, $w^*\in W^*$. It is a subspace of ${\mathcal L}(V,W;{\mathbb F})$. We need to point out that any element in $V^*\otimes W^*$ is a finite linear combination of elements of the form $v^*\otimes w^*$, but generally cannot be written as a single term $v^*\otimes w^*$ (the reader should construct examples).

Here are my questions:

  • What does the first sentence mean? Does it mean $$V^*\otimes W^*:=span\{v^*\otimes w^*|v^*\in V^*, w^*\in W^*\} $$ or $$V^*\otimes W^*:=\{v^*\otimes w^*|v^*\in V^*, w^*\in W^*\} ?$$
  • In the context, it is only defined that $$v^*\otimes w^*(v,w)=v^*(v)\cdot w^*(w).$$ What's the "finite linear combination of elements of the form $v^* \otimes w^* $" supposed to be defined? And what's the example "the reader needs to construct"?
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By the way: when $V$ and $W$ are finite-dimensional, the span of these elements is all of $L(V, W; F)$. –  Dylan Moreland Jul 12 '11 at 2:36
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up vote 11 down vote accepted

The exposition you're quoting may be somewhat confusing if you're used to how tensor products are usually defined in a general setting (see e.g. the Wikipedia article). Usually, the tensor product is defined as a new vector space, either through a universal property or by explicit construction using equivalence classes. In your case however, it's being regarded as a subspace of an existing vector space, the space of all bilinear functions from $V\times W$ to $\mathbb F$. That allows the book to speak of linear combinations without introducing these as formal expressions, since it's already known how to form linear combinations of bilinear functions from $V\times W$ to $\mathbb F$.

To answer your questions specifically: Yes, the vector space generated by a set of elements is the span of those elements. There's a subtle difference in that the formulation "the vector space generated by $S$" can also be used to refer to the free vector space over $S$, whereas the formulation using the span can't be thus used and only serves to identify a subspace of a vector space already otherwise defined.

Concerning the examples to be constructed, consider linearly independent functions $v^*_1,v^*_2\in V^*$ and $w^*_1,w^*_2\in W^*$, and form $v^*_1\otimes w^*_1+v^*_2\otimes w^*_2$; I think you'll find that you can't express this in the form $v^*\otimes w^*$ for any $v^*\in V^*$ and $w^*\in W^*$.

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It's unfortunate that we (collectively)have inherited some bit of a culture in which "formal expression" can mean "marks on a page" or "meaningless symbols", without protest. It is true that the set-theoretic "construction" of many things is unsatisfactory, but one could, and I'd strongly recommend it, refer to the property one requires/needs, so that the object-in-question is unique-up-to-unique-isomorphism, at least. Existence/construction truly is a different matter. E.g., "polynomials" are not "expressions", but elements of a universal object in a category of commutative rings... –  paul garrett Jul 11 '11 at 23:06
    
To be honest, I'm not sure which part of the answer, if any, you're criticizing. I wrote first that the tensor product can be defined "either through a universal property or by explicit construction using equivalence classes". Then I specifically addressed the OP's concern how the finite linear combinations are defined. Does your complaint concern the fact that I wrote only "introducing these as formal expressions" and not "either introducing these as formal expressions or postulating their existence in a universal object"? –  joriki Jul 11 '11 at 23:16
    
I'm sorry, I should have been more careful not to imply (negativist) criticism of your answer. I mean to criticize something widespread in the mathematical environment, that amounts to laziness or ignorance, and/but should surely lead to confusion on the part of any serious, thinking reader. So, in fact, it is necessary, and, therefore, a good thing, to address a "questioner"'s confusion derived from the failings of the environment. I did not mean to be negative to you about addressing this! Sorry to have left such an impression! And/but we (collectively) might change the feeble/lazy texts... –  paul garrett Jul 12 '11 at 0:39
    
@paul: can't we define tensor algebra as the largest model of associative algebra axioms that contains the given vector space? –  Alexei Averchenko Aug 2 '11 at 6:05
    
@Alexei A... There are indeed various possibilities. Without in any way faulting joriki's response to the question as posed, for a definition/characterization I was thinking of a "tensor" algebra of a v.s. V as being as assoc alg A with a v.s. map V->A such that for any v.s. map V->B to an assoc alg B, there is a unique assoc alg map A->B making the obvious triangle commute. No coordinates, no interpretation of symbols, etc. But, as in the question as posed, usually one must work through a transitional viewpoint, as well. –  paul garrett Aug 2 '11 at 12:38
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