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In Gowers's article "How to lose your fear of tensor products", he uses two ways to construct the tensor product of two vector spaces $V$ and $W$. The following are the two ways I understand:

  1. $V\otimes W:=\operatorname{span}\{[v,w]\mid v\in V,w\in W\}$ where $[v,w]:{\mathcal L}(V\times W;{\mathbb R})\to {\mathbb R}$ such that $[v,w](f)\mapsto f(v,w)$
  2. $V\otimes W:=Z/E$ where $Z:=\operatorname{span}\{[[v,w]]\mid v\in V,w\in W\}$ and $E$ is the subspace of $Z$ generated by all vectors of one of the following four forms: $\begin{align} & [[v,w+w']]-[[v,w]]-[[v,w']]\\ & [[v+v',w]]-[[v,w]]+[[v',w]] \\ & [[av,w]]-a[[v,w]] \\ & [[v,aw]]-a[[v,w]] \end{align}$

Here are my questions:

  • Are the definitions I wrote above correct?
  • They look so different. How are they essentially the same?
  • The set $\operatorname{span}\{[v,w]\mid v\in V,w\in W\}$ in (1) and $Z$ in (2) seem to be the "same". Do we have $Z\cong Z/E$ here?
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    http://hitoshi.berkeley.edu/221a/tensorproduct.pdf2015-05-21

4 Answers 4

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The first definition comes from the philosophy that students are bad at understanding abstract definitions and would prefer to see the tensor product defined as a space of functions of some kind. This is the reason that some books define the tensor product of $V$ and $W$ to simply be the space of bilinear functions $V \times W \to k$ ($k$ the underlying field), but this defines what in standard terminology is called the dual $(V \otimes W)^{\ast}$ of the tensor product.

For finite-dimensional vector spaces, $(V^{\ast})^{\ast}$ is canonically isomorphic to $V$, and that is the property that Gowers is taking advantage of in the first definition, which is basically a definition of $((V \otimes W)^{\ast})^{\ast} \cong V \otimes W$. The second definition is essentially the standard definition.

To answer your last question, no, we do not. $Z$ is infinite-dimensional whenever the underlying field is infinite. It is really, really huge, in fact pointlessly huge; the relations are there for a reason.

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    @Qiaochu: Yes, that _might_ be the philosophy behind the first definition... Or, it could be that that's often the way that many differential geometers think about the tensor product.2011-07-13
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I'm going to go way out on a limb and instead of answering the questions actually posed, I'll propose a way to think about..... um OK, here it is: what's the difference between an ordered pair of vectors and a tensor product of two vectors? It is this: If you multiply one of the two vectors by $c$ and the other by $1/c$, then you've got a different ordered pair of vectors, but you've got the same tensor product. (Hence the tensor product is a quotient of the Cartesian product.)

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    .....and then you want to say that the tensor product of two $v$ec$t$or spaces is the set of all _sums_ of tensor products of two vectors. And then show that for finite-dimensional spaces, you only need sums of boundedly many terms.2011-07-13
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As for your first question: yes.

As for your second question: for instance, elements of the basis of the "first" $V\otimes W$, make the expression

[v, w+w'] - [v,w]- [v,w']

to be equal to zero. Indeed, by definition, this guy evaluated on any bilinear map $f: V \times W \longrightarrow \mathbb{R}$ is

\begin{align} ([v, w+w'] - [v,w]- [v,w']) (f) &= [v, w+w'] (f) - [v,w] (f)- [v,w'] (f) \\ &= f(v,w+w') - f(v,w) - f(v,w') \\ &= 0 \end{align}

by the bilinearity of $f$.

In the same way, you can verify that elements of the basis of the "first" $V\otimes W$ make all the expression you're quotienting out in the second $V\otimes W$ to be zero. Hence, it is true that

[v, w+w'] - [v,w]- [v,w'] = 0

as well as

[[v, w+w']] - [[v,w]]- [[v,w']] = 0

in $Z/E$.

As for your third question: no.

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    Thank you. You're very kind.2011-07-13
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For each pair $(V,W)$ of vector spaces (over a fixed ground field), let $T(V,W)$ be their tensor product, and $F(V,W)$ the vector space of bilinear forms on $V^*\times W^*$. One checks that

(a) there is a unique linear map $e(V,W)$ from $T(V,W)$ to $F(V,W)$ satisfying $ e(V,W)(v\otimes w)(b)=b(v,w) $ for all $v\in V,w\in W$,

(b) $e(V,W)$ is injective,

(c) $T$ and $F$ are functors,

(d) $e$ is a natural transformation from $T$ to $F$,

(e) $T,F$ and $e$ are compatible (in an obvious sense) with finite direct sums.

Claim 1: $e(V,W)$ is surjective $\iff$ the cardinal number $\dim(V)\dim(W)$ is finite.

In view of (b), implication "$\Leftarrow$" follows by dimension counting. It suffices thus to prove the non-surjectivity when $V$ is infinite dimensional and $W$ nonzero. Writing $W$ as $W_1\oplus W_2$ with $\dim W_1=1$ and using (b), we are reduced to

Claim 2: if $V$ is infinite dimensional, then the canonical embedding $V\to V^{**}$ is not surjective.

To prove this, we'll use an embedding of $V$ in $V^*$, and an embedding of $V^*$ in $ V^{**}$. None of these two embeddings will be canonical, but their composition will.

Choose a basis $B$ of $V$, and identify $V$ to the space $K^{(B)}$ of finitely supported $K$-valued functions on $B$. Then $V^*$ can be identified to the space $K^B$ of all $K$-valued functions on $B$. Similarly, we can identify $V^*$ to $K^{(B\coprod C)}$, where $C$ is a set and $\coprod$ means "disjoint union". As $B$ is infinite, $C$ is nonempty. Using the same trick once more, we can identify $V^{**}$ to $K^{(B\coprod C\coprod D)}$, where $D$ is a nonempty set. Then the natural embedding of $K^{(B)}$ in $K^{(B\coprod C\coprod D)}$, which is clearly not surjective, corresponds to the natural embedding of $V$ in $V^{**}$. This completes the proof.

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    Pierre: Thank you for this thorough answer. @Community: I am sorry for the duplicate question.2011-10-10