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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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Yes, those definitions are correct. And no, $Z$ is not isomorphic to $Z/E$. In fact, $Z$ is infinite dimensional (even for finite dimensional $V$ and $W$!). Notice that the quotient map $Z \to Z/E$ does not give you an isomorphism: for example, $[[u,v+w]] - [[u,v]] - [[u,w]]$ is a non-zero vector in $Z$ whose image in $Z/E$ is $0$. – Omar Antolín-Camarena Jul 13 '11 at 3:59
That's an interesting way to lose fear... – Alexei Averchenko Jul 13 '11 at 4:01
Isn't it easier to just define tensor algebra as the largest model (in the sense of a universal property) of the unital associative algebra oer $k$ that contains $V$ as a subspace? Tensor product of vector spaces becomes easier to understand then. – Alexei Averchenko Jul 13 '11 at 4:11
Could you define ${\mathcal L}$ and $[v,w]$? – Garrett Feb 23 '14 at 3:07
up vote 18 down vote accepted

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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Hmm, so one may define, with the philosophy of the first definition, $V\otimes W:={\mathcal L}(V^*,W^*;{\mathbb R})$, right? And here is my confusion for the last question: since the two definitions are the same for the tensor product, why the size of $Z$ and $Z/E$ cannot be the same? And since they are not the same, why are both of the definitions correct? – Jack Jul 13 '11 at 4:04
@Jack: yes to the first question, if the notation $L$ means what I think it means. To the second question, I don't think you understand the definition of $Z$. It is literally the free vector space on the Cartesian product $V \times W$. It is humongous. $Z/E$, on the other hand, has dimension $(\dim V)( \dim W)$. – Qiaochu Yuan Jul 13 '11 at 4:06
OK. So as I understand what I put in the post, the first definition is $V\otimes W=Z$ and the second one is $V\otimes W=Z/E$. Now that these two sets are of different sizes, why are both of them correct for defining tensor product? – Jack Jul 13 '11 at 4:13
@Jack. No, the first definition is NOT $V\otimes W = Z$. In the first definition you're adding a restriction: you are saying that $[v,w]$ is a linear map from the set of bilinear maps $V\times W \longrightarrow \mathbb{R}$ to $\mathbb{R}$ such that $[v,w] (f) = f(v,w)$. Whereas there is no restriction at all for those $[[v,w]]$: they are just pairs of vectors. – a.r. Jul 13 '11 at 4:19
@Agu: +1. Ah, I see. So the restriction of $[v,w]$ in the first definition that it is a linear map on ${\mathcal L}(V,W;{\mathbb R})$ is kind of the same as $E$ in the second definition, correct? – Jack Jul 13 '11 at 4:24

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 vector 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. – Michael Hardy Jul 13 '11 at 6:46

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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+1. Your answer is nice for the second question. And this resolve my confusion which I said in the comment after Qiaochu's answer. – Jack Jul 13 '11 at 4:48
Thank you. You're very kind. – a.r. Jul 13 '11 at 5:18

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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It is much better if you provide a link to the earlier answer, – Mariano Suárez-Alvarez Oct 10 '11 at 6:35
Dear @Mariano: The other question has been closed. My (perhaps incorrect) understanding is that closed questions get eventually deleted. The users who closed the other question wrote: "its answers may be merged with another identical question". I took this as an invitation to do what I did. I hesitated to delete the other answer. Please tell me what I should do. Thank you in advance. – Pierre-Yves Gaillard Oct 10 '11 at 6:43
Dear @Mariano: I'd be most grateful if you could go there to see what I did, and tell me if it's ok. – Pierre-Yves Gaillard Oct 10 '11 at 7:41
that's perfectly fine :) The problen with having two equal answers is that one might later change or be upated without the other knowing about it (moreover, the connected between the two questions is useful for people readong both of them) Thanks a bunch. – Mariano Suárez-Alvarez Oct 10 '11 at 16:08
Pierre: Thank you for this thorough answer. @Community: I am sorry for the duplicate question. – Giuseppe Negro Oct 10 '11 at 22:51

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