Tensor product and category theory - MathOverflow

Tensor product and category theory

2010-07-14T12:17:46Z

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.

Answer by Dylan Moreland for Tensor product and category theory

2010-07-14T12:35:21Z

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.)

Answer by Eivind Dahl for Tensor product and category theory

2010-07-14T13:03:12Z

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.

http://books.google.com/books?id=S6Ve0KXyDj8C&printsec=frontcover&dq=bott&hl=en&ei=ALs-TJLFF4SI4QaRuvCFCw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCcQ6AEwAA#v=onepage&q&f=false

Answer by Charles Matthews for Tensor product and category theory

2010-07-14T13:51:22Z

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.)

Answer by Andrew L for Tensor product and category theory

2010-07-14T19:20:08Z

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.

Answer by David Corwin for Tensor product and category theory

2010-07-15T10:17:37Z

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)."

Answer by Dedalus for Tensor product and category theory

2010-07-16T08:21:18Z

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.