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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    $\begingroup$ 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 $0$. $\endgroup$ Jul 13, 2011 at 3:59
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    $\begingroup$ That's an interesting way to lose fear... $\endgroup$ Jul 13, 2011 at 4:01
  • $\begingroup$ 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. $\endgroup$ Jul 13, 2011 at 4:11
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    $\begingroup$ Could you define ${\mathcal L}$ and $[v,w]$? $\endgroup$
    – Garrett
    Feb 23, 2014 at 3:07
  • 2
    $\begingroup$ hitoshi.berkeley.edu/221a/tensorproduct.pdf $\endgroup$ May 21, 2015 at 13:30

6 Answers 6


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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    $\begingroup$ @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 $(\dim V)( \dim W)$. $\endgroup$ Jul 13, 2011 at 4:06
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    $\begingroup$ @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. $\endgroup$ Jul 13, 2011 at 4:19
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    $\begingroup$ @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? $\endgroup$
    – user9464
    Jul 13, 2011 at 4:24
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    $\begingroup$ @Jack. I think that, if you look into my answer below, you'll see what "produces" relations $E$ is the formula by which $[v,w]$ acts on bilinear maps. Namely, $[v,w] (f) = f(v,w)$. $\endgroup$ Jul 13, 2011 at 4:43
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    $\begingroup$ @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. $\endgroup$ Jul 13, 2011 at 6:37

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.

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    $\begingroup$ .....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. $\endgroup$ Jul 13, 2011 at 6:46
  • $\begingroup$ "Hence the tensor product is a quotient of the Cartesian product." Could you explain this a bit? The dimension of $V\times W$ is the sum of their dimensions and $V\tensor W$ is the product of their dimensions. Product could be more than sum sometimes. Then how can we raise the dimension by quotienting? Thanks. $\endgroup$
    – Boka Peer
    Sep 10, 2021 at 18:54
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    $\begingroup$ @BokaPeer : It appears that what I should have said is that the set of all tensor products of two vectors is a quotient of the Cartesian product, in that many different members of the Cartesian product of two spaces correspond to just one tensor product of two vectors. $\endgroup$ Sep 10, 2021 at 22:35

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.


Edit. This answer had been moved here from this question, question which had been closed and then reopen. After the reopening I rolled back the original answer with the help of Will Orrick.

  • $\begingroup$ It is much better if you provide a link to the earlier answer, math.stackexchange.com/questions/71360 $\endgroup$ Oct 10, 2011 at 6:35
  • $\begingroup$ 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. $\endgroup$ Oct 10, 2011 at 6:43
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    $\begingroup$ I know this answer was meant for @GiuseppeNegro's question, but as it's ended up being moved here instead, would you add something addressing the relationship between Gowers' two definitions? As it stands, your answer might be misread as saying that Gowers' Definition 1 is only the same as his Definition 2 in the case $V$ and $W$ finite-dimensional, which, I think, is not true. I don't think it's being pedantic to point out that Giuseppe Negro's Def 2 is not the same as Gowers' Def 1. $\endgroup$ Apr 14, 2020 at 13:39
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    $\begingroup$ Thanks! That helps. Ideally the other question would be reopened and this answer restored to its proper place. We'll see whether that happens. $\endgroup$ Apr 15, 2020 at 12:23
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    $\begingroup$ I want to let you know that the other question has been reopened so, if you wish, it should be safe to roll back to the answer that was originally there. $\endgroup$ Apr 20, 2020 at 3:35

The definitions aren't actually that different. In both, the elements of $V\otimes W$ are equivalence classes of linear combinations of objects labeled by pairs of vectors (one from $V$ and one from $W$). The definitions seem different because in the second one the equivalence classes are seen directly in the notation $Z/E$ and explicitly defined in the process of forming the quotient, whereas in the first one you have to think a bit to realize that equivalence classes are even there.

The equivalence relation in the first definition is, in fact, simple equality: each equivalence class contains those linear combinations of the $[v,w]$ that represent the same function from $\mathcal{L}(V,W;\mathbf{R})$ to $\mathbf{R}$. Remember that an expression like $a_1[v_1,w_1]+\ldots+a_n[v_n,w_n]$ is not the unique name of a function: due to bilinearity the same function will be named by many other expressions of that form.

If you have the impression that these equivalence classes are large sets, I've misled you. In fact these equivalence classes are sets of size $1$ since all the different linear combinations they contain are equal. They represent one and the same function, and it is functions that comprise the set, not formal expressions representing functions. Here we are simply following the usual convention that when we define a set, as, for example, you did when you wrote $\operatorname{span}\{[v,w]\mid v\in V,\ w\in W\}$ as a shorthand for the set of linear combinations of the functions $[v,w]$, equal expressions represent a single element. In general, when the equivalence relation is equality, the equivalence classes will be of size $1$ because of what it means to be an element of a set.

This is the point that necessitates introducing $[[v,w]]$ in the first place: we need a mathematical way of talking about an unevaluated linear combination, that is, of talking about the form of a linear combination (the terms and coefficients it contains) rather than the function from $\mathcal{L}(V,W;\mathbf{R})$ to $\mathbf{R}$ that it represents. We can't do this with $Z'=\operatorname{span}\{[v,w]\mid v\in V,\ w\in W\}$ since its elements are just functions and the usual ways of understanding and manipulating sets don't distinguish between the same element, described differently. We can do it in $Z=\operatorname{span}\{[[v,w]]\mid v\in V,\ w\in W\}$, however, since the elements are the formal linear combinations of pairs $[[v,w]]$ and therefore all distinct.

This answers your third question too: $Z/E$, where $E$ is the subspace of $Z$ that you defined above, is isomorphic to $Z'$. In the case of $Z/E$, vast numbers of elements of the enormous, infinite-dimensional set $Z$ are declared equivalent in forming the quotient by $E$. In the case of $Z'$, vast numbers of expressions appearing in the definition of $Z'$ represent the same function and hence contribute just one element to $Z'$. And there is an obvious correspondence between equivalence classes in $Z/E$ and functions in $Z'$: $[[v,w]]+E\mapsto [v,w]$.

Gowers's first defintion.

$\def\bbf{\mathbf{R}}$Given any vector spaces $V,W,X$ over the field $\bbf$, the notation $L(V;X)$ is commonly used as the set of all linear functions from $V$ to $X$ and $L(V,W;X)$ denotes the set of all bilinear functions from the product space $V\times W$ to $X$.

In Gowers's article, $B$ denotes "the set of all bilinear maps from $V\times W$ to $\bbf$". By the above convention, $B=L(V, W;\bbf)$. His first definition can be interpreted as follows:

Given any $(v,w)\in V\times W$, define $[v,w]:B\to\bbf$ as a function with $ [v,w](f):=f(v,w)\;. $ $\def\span{\operatorname{span}}$Define $$V\otimes W=\span\{[v,w]\mid (v,w)\in V\times W\}$$ where the "span" takes place in the vector space of all real-valued functions on $B$. (See the last sentence of the first paragraph in the section "Back to the main discussion" of his arcticle.)

According to the definition above, we have $V\otimes W=B^*=(L(V,W;\bbf))^*$ where $B^*$ is the dual of $B$.

Gowers's second definition.

In order to talk about the notion of "span", one needs a vector space in the first place. Your interpretation of the set $Z$ is not quite right. He writes in the article that

We can define a rather large vector space $Z$ by taking formal linear combinations of these symbols. By that I mean that Z consists of all expressions of the form $$a_1[[v_1,w_1]]+ a_2[[v_2,w_2]]+...+ a_n[[v_n,w_n]]$$ with obvious definitions for addition and scalar multiplication.

Note that the space $Z$ is not isomorphic to $V\times W$. For instance, $(v,w)$ and $(2v,2w)$ are linearly dependent in $(v,w)$ but $[[v,w]]$ and $[[2v,2w]]$ are linearly independent in $Z$. One can think $Z$ as a vector space that has the Cartesian product $V\times W$ as a basis. So every element in $V\times W$, including $(0,0)$, identified as $[[0,0]]$, is a basis element of $Z$. This is why Gowers says it "rather large". It is in this huge space that the quotient is taken. This also gives a NO to your third question.

The two definitions are equivalent in the sense that one can establish an isomorphism between the spaces in the two definitions. As Gowers writes in his article:

The usual notation for the tensor product of two vector spaces V and W is V followed by a multiplication symbol with a circle round it followed by W. Since this is html, I shall write V@W instead, and a typical element of V@W will be a linear combination of elements written v@w. You can regard v@w as an alternative notation for [v,w], or for [[v,w]]+E - it doesn't matter which as the above discussion shows that the space spanned by [v,w] is isomorphic to the space spanned by [[v,w]]+E, via the (well-defined) linear map that takes [v,w] to [[v,w]]+E and extends linearly.


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