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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^*:=\operatorname{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 $L(V, W; F)$.2011-07-12

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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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    @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.2011-08-02