## Tensor product and category theory

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Hello. I'm trying to understand the definition of tensor product of two vector spaces. So far, I've read the one using free vector spaces and a quotient space (this one http://en.wikipedia.org/wiki/Tensor_product#Tensor_product_of_vector_spaces) , and I think I understand it well. However, I want to understand the other definitions I can find, and it seems as a very common way to define it is through the universal property (some category theory included, I suspect). Does anyone here know of a good treatment of this? I have no knowledge of category theory though, but would love to read some about it. I'm a second-year undergrad, so not too much of a high level would be nice.

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The definition of $A \otimes B$ as a universal object in a category is treated in most graduate-level algebra books, e.g. Section IV.5 of Hungerford's Algebra. Here's a link: books.google.co.uk/… – Chris Phan Jul 14 2010 at 12:30

## 6 Answers

Just some definitions, in case you're unfamiliar with them: Let $\hat{V}$ denote the vector space of linear functions from a vector space $V$ to the scalar field. Remember, a multilinear map is one of the form $V \times V \times \cdots \times V \to W$ (with $n$ copies of $V$), where $W$ is another vector space, such that if we fix $n-1$ of the arguments, the function becomes a linear function from $V$ to $W$ in the argument not fixed. A multilinear form is one in which $W=K$, the scalar field (you can replace $K$ with $\mathbb{R}$ or $\mathbb{C}$ if you like). For example, the inner product on $\mathbb{R}^n$ is a bilinear form on $\mathbb{R}^n$, since if we fix one argument, it becomes linear in the other. If we view an $n \times n$ matrix as a conglomeration of $n$ columns, then the determinant is an $n$-form.

Then $\hat{V} \otimes \hat{V}$ corresponds to the set of bilinear forms, and in general, a tensor product of multiple copies of $\hat{V}$ corresponds to the set of $n$-linear forms (i.e. multilinear forms with $n$ arguments). That, there is a concrete description of tensor products of the dual space with itself, and many books which do not wish to develop the notion of tensor product will use this in place of tensor products. That is, all they must do is define a certain kind of map, and then the tensor product is just the set of maps of that kind. Then how do we explain the tensor product $V \otimes V$ (or more generally $V \otimes U$, where $U$ is another vector space)? We could note that $V$ is canonically isomorphic to its double dual, i.e. the dual space of $\hat{V}$, and then view $V \otimes V$ as the set of bilinear forms on $\hat{V}$. But there is a nicer way, and this uses the universal property.

A bilinear map $V \times V \to W$ corresponds to a linear map $V \otimes V \to W$. If $f(-,-)$ denotes the bilinear map, and $x,y \in V$, then our linear map sends $x \otimes y$ to $f(x,y)$. You could try to think of the tensor product as pairs of vectors, but the tensor product contains elements which are not $x \otimes y$ for some $x,y \in V$. We do have that $x_1 \otimes y_1 + x_2 \otimes y_2$ maps to $f(x_1,y_1)+f(x_2,y_2)$. In more generality, if $W$ and $U$ are two other vector spaces, linear maps $U \otimes V \to W$ correspond to bilinear maps $U \times V \to W$. Then what is an element of $U \otimes V$? It is a thing you stick into a bilinear map. This is the key idea which helped me understand tensor products. I repeat, an element of a tensor product is simply a thing you stick into a bilinear map. In general, elements of some universal construction defined by maps going out of a certain object have some description as "things you stick into some kind of map (or a collection of multiple maps)."

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Does (dual of V) $\otimes$ (dual of V) give all the bilinear forms on $V$ when $V$ is infinite-dimensional? Certainly if one starts wanting to put norms on such spaces (when using continuous duals), care is needed. – Yemon Choi Jul 16 2010 at 8:39
Right, I'm thinking about finite-dimensional spaces. In the infinite dimensional case, it still gives some of them. – David Corwin Jul 17 2010 at 21:30

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You might like Brian Conrad's handouts for a sophomore differential geometry course. Especially relevant are Construction of tensor products and the two handouts after that one. They have some nice examples and a heavy emphasis on the universal property.

(I don't think this warranted more than a comment, but I can't post those yet.)

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You should check Keith Conrad's handouts on the tensor product too. They are really good. math.uconn.edu/~kconrad/blurbs – Gonçalo Marques Jul 14 2010 at 13:17
The only reason I would say that Kieth Conrad's exposition of tensor products might not be appropriate is because the reader seems to be interested in vector spaces and, as a second year undergraduate, may not know anything about modules yet. – Keenan Kidwell Jul 14 2010 at 15:50

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Gonçalo, there are quite a few notes posted on that page, so it may be better to give links directly to the files: math.uconn.edu/~kconrad/blurbs/linmultialg/… and math.uconn.edu/~kconrad/blurbs/linmultialg/…. Concerning Keenan's comment, I suppose if the OP replaces "ring" with "field", "module" with "vector space", and ignores all examples that don't make sense with vector spaces (e.g., Q/Z), then my notes may make sense. But that could be asking too much. If Dedalus tries that experiment I'd be curious to know if it works. – KConrad Jul 14 2010 at 23:32
I'll try your notes Keith. I do however know about modules, so I think it'll be alright. – Dedalus Jul 16 2010 at 10:42

My view of the pedagogy, based on teaching this to second year undergraduates at Cambridge.

The tensor product of vector spaces is defined by generators and relations. Also generators and relations, as a way of defining anything, is a method depending on a universal property (to make much sense).

If you take these two parts one at a time, you have a chance of understanding what is happening. The generators and relations are just bilinearity spelled out. The remark about generators and relations as a mode of defining anything can be learned anywhere you like (e.g. group theory): the reason that there is a universal property is just "stuff", "abstract nonsense", "mathematical maturity" even.

I believe, quite strongly, that the eliding of the punctuation between the two sentences is a negative in teaching this material. (I really do not care if this spoils Mac Lane's or anyone else's view of category theory and its role: "universal property" is only a stepping stone there, not the ultimate goal.)

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Pedagogically, do you find your two sentences should go in the order you wrote them? For myself, I understood it the other way around. My mental story went "Wouldn't it be nice if we could somehow turn bilinear maps into linear maps, so we could reuse all our theorems? (ie, the desired universal property) Now let's go build a vector space to make this happen (ie, the generators and relations)." But I haven't ever taught this, so I'm curious which way around you found better. – Neel Krishnaswami Jul 14 2010 at 16:57
When I learned it all, the universal property came first: i.e. I learned the orthodox post-Bourbaki version of tensor products. But I only really felt I had understood it properly after I had taught it. I was particularly struck by a student's remark that generators and relations was "more sensible" than what they were taught in lectures (I wasn't lecturing, but picking up the pieces.) Now, you do need both sides. The real test came when I tried using tensor products of fields in Galois theory lectures ... another story. – Charles Matthews Jul 14 2010 at 17:56
I think this is definitely a “different strokes for different folks” issue. I was first taught it just in terms of the generators and relations (as a second-year undergraduate at Cambridge!) and while I was able to use it then, I didn't begin to grok it until someone pointed out the universal property to me, transforming it from something fiddly and ad hoc into something natural and tractable. I appreciate that some people may find it clearer to learn with generators and relations emphasised, but not all of us did! – Peter LeFanu Lumsdaine Jul 14 2010 at 18:26

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I didn't understand what the universal property was for, and so didn't appreciate the construction, until the monoidal closure of vector spaces was pointed out to me. That's when it all fell into place -- I could now see why we wanted this particular universal property (so we could curry and uncurry linear maps a la functional programming). Then each piece of the construction with generators and relations made sense as the minimal construction to meet this requirement. – Neel Krishnaswami Jul 14 2010 at 18:42

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@Peter: Yes, different types of students are going to be worried by the different questions (a) how do you manipulate this gadget, and (b) how do you know it isn't 0, anyway? Because you need to able to answer both to do any serious work, it is no good telling just half the story. But my point is that the "economy" of saying you can teach both at once is a false one. – Charles Matthews Jul 14 2010 at 19:48

I'm pretty much in your spot. I think part of the way there is learning to think with universal properties. I recently found a really good book (Algebra: Chapter 0, link below) on 'basic' algebra using category theory to unify things. All the basic stuff like products, disjoint union, surjections and injections are treated rigorously and in great generality through their universal properties. If you already know your group and set theory reading through the first few chapters can be done quickly, and should get you in the right mode of thought. I'm doing this myself right now, and so far I recommend you do the same.

http://www.amazon.com/Algebra-Chapter-Graduate-Studies-Mathematics/dp/0821847813/ref=sr_1_1?ie=UTF8&s=books&qid=1279112196&sr=8-1

EDIT: A nice application of the tensor product can be found in the first few pages of 'Differential Forms in Topology', that is, if $\Omega^*$ is the algebra generated by the formal symbols $dx_j,j=1,\dots,n$ under the relations $dx^2=0$ and $dx_idx_j=-dx_jdx_i$, then `$\Omega^*(U)=C^\infty(U)\otimes\Omega^*$` is the algebra of differential forms on the open set $U$ (under the wedge product). I'm not sure if that's how it's primarily used.

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Fixed the LaTeX for you – Yemon Choi Jul 15 2010 at 7:55
Thank you Choi! – Eivind Dahl Jul 15 2010 at 7:58

A fully categorical approach that emphasizes the universal properties of the tensor product ,as well as a great deal of multilinear algebra, can be found in T.S.Blyth's Module Theory:An Approach to Linear Algebra. There's also a discussion in Steven Roman's Advanced Linear Algebra,but the presentation in Blyth's book isn't as dry and formal.

By the way,if anyone has a serious interest in algebra,Blyth's books are some of the great unsung textbooks in the subject. They really should be better known and used in the U.S. then they are.

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Thank you all. Your documents has been most helpful. I also saw some papers on the tensor product of modules, especially: http://www.math.ucsb.edu/~mckernan/Teaching/05-06/Winter/220B/l_7.pdf was helpful, and http://www.dpmms.cam.ac.uk/~wtg10/tensors3.html gave some good info too.

Now I'm considering TeXing a file where I try to motivate why one defines the tensor product in the first place. I think that might help me learn the definition even more. I really like the definition in some strange way, even though I find it kind of hard. I want to learn.

So once again, Thank you.

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Everyone finds the definition hard the first time.  A reason why the tensor product is defined is to base extend a module over one ring to become a module over a second ring, subsuming at the same time two classical operations on polynomials: polynomials in Z[x] can be viewed in Q[x] and can be reduced mod p to be in (Z/p)[x]. Each of these is useful for different irreducibility tests, for instance, and the passage Z[x] --> Q[x] and Z[x] --> (Z/p)[x] are both examples of tensor products. See Section 6 of the first link I made in a comment to Dylan's answer.  – KConrad Jul 16 2010 at 14:49
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