Vector bundles, Grassmannians, and characteristic classes

I’ve been reading about spectra and stable homotopy theory lately, but don’t feel ready to start talking about them here. Instead, I shall say a few words on characteristic classes. The present post will be quite general and preparatory — the more difficult matter is to actually construct such characteristic classes. Our goal is to see that characteristic classes essentially boil down to computing the cohomology of the infinite Grassmannian.

A lot of problems in mathematics involve the existence of sections to vector bundles. For instance, there is the old question of when the sphere is parallelizable. A quick Euler characteristic argument shows that even-dimensional spheres can’t be—then there would be an everywhere nonzero vector field, whose infinitesimal flows would be homotopic to the identity (and consequently having nonzero Lefschetz number by the even-dimensionality) while having no fixed points. In fact, much more is known. Using the group or group-like structures on ${S^1, S^3, S^7}$ (coming from the complex numbers, quarternions, and octonions), it is easy to see that these manifolds are parallelizable. But in fact no other sphere is.

A characteristic class is a means of assigning some invariant to a vector bundle. Ideally, it should be trivial on trivial bundles, so the characteristic class can be thought of as an “obstruction” to finding large numbers of linearly independent sections.

More formally, let ${p: E \rightarrow B}$ be a vector bundle. A characteristic class assigns to this bundle (of some fixed dimension, say ${n}$ ) an element of the cohomology ring ${H^*(B)}$ (with coefficients in some ring). To be interesting, the characteristic class has to be natural. That is, if ${f: B' \rightarrow B}$ is a map, then the characteristic class of the pull-back bundle ${f^*E \rightarrow B'}$ should be the pull-back of the characteristic class of ${E \rightarrow B}$ .

This contrafunctoriality is a good reason to prefer cohomology to homology here: one can’t push vector bundles forward, but one can pull them back by a map of spaces along the base. It is with this that we can talk about naturality.

So what does a characteristic class really mean? I claim that a characteristic class is really the same thing as an element of the cohomology ring of the Grassmannian ${\mathrm{Gr}_n(\mathbb{R}^{\infty})}$ of ${n}$ -planes in ${\mathbb{R}^{\infty}}$ .

The reason for this is fundamentally categorical, and it boils down to the fact that there is a universal ${n}$ -dimensional bundle ${U \rightarrow B_U}$ for any integer ${n}$ . Any vector bundle of dimension ${n}$ on a paracompact base space ${B}$ can be obtained by pulling back this bundle in some way, from some map ${B \rightarrow B_U}$ . Further, this map ${B \rightarrow B_U}$ is unique up to homotopy. (It is a basic fact that a pull-back of a vector bundle ${E \rightarrow B}$ by a map ${B' \rightarrow B}$ depends only on the homotopy class of the map ${B' \rightarrow B}$ .)

Another way of stating this is that the contravariant functor ${F_n}$ that assigns to each ${B}$ (a reasonable space, say paracompact) the set of isomorphism classes of ${n}$ -dimensional bundles is representable on the homotopy category.

Let’s state this more precisely. First, what do we mean for ${F_n}$ to be a functor? Well, first it has to assign to every ${B}$ some set ${F_n(B)}$ —OK, we have that. But, it also has to assign to every homotopy class of maps ${B' \rightarrow B}$ a map ${F_n(B) \rightarrow F_n(B')}$ . Fortunately, we have a way of doing that: the pull-back. Given a bundle over ${B}$ , we can pull it back to ${B'}$ from the map ${B' \rightarrow B}$ . As I said, this depends only on the homotopy class of ${B'\rightarrow B}$ , so this is not surprising.

Now in fact we have proved a pretty general result about representable functors on the pointed homotopy category of CW complexes. Indeed, Brown representability can be used to tell us that ${F_n}$ is representable on that category. But we are not working with pointed spaces, and we don’t want to restrict only to CW complexes. So this will be a separate result.

What is this universal bundle going to look like? There is a very clean picture of it. Namely, it is going to be the tautological ${n}$ -plane bundle over ${\mathrm{Gr}_n(\mathbb{R}^{\infty})}$ . That is, the fiber over a point ${x}$ in the Grassmannian is just the collection of vectors in ${\mathbb{R}^{\infty}}$ that lie in the plane corresponding to that point. One can check that this is indeed a vector bundle, and in Milnor-Stasheff’s “Characteristic classes” it is proved that it is universal.

The key idea of the proof is that, on a compact space, any vector bundle ${p: E \rightarrow B}$ can be imbedded inside a trivial bundle ${B \times \mathbb{R}^{N}}$ for some ${N}$ large. As a result, over each ${b \in B}$ , the fiber ${p^{-1}(b)}$ is identified with a subspace of ${\mathbb{R}^N}$ whose dimension is ${n}$ . In this way, we can get a map from ${B}$ into the Grassmannian of ${n}$ -planes in ${\mathbb{R}^N}$ by sending each ${b \in B }$ to the corresponding ${n}$ -plane. The reason we had to use the infinite Grassmannian in general is that ${N}$ could be very large.

Alright. So we know that there is a universal ${n}$ -bundle ${U}$ over ${\mathrm{Gr}_n(\mathbb{R}^{\infty})}$ for each ${n}$ . Let’s say we have a characteristic class ${c}$ that sends each ${n}$ -bundle ${p: E \rightarrow B}$ to an element ${c(E) \in H^*(B)}$ . Then, by definition, there is ${c(U) \in H^*(\mathrm{Gr}_n(\mathbb{R}^{\infty}))}$ . This is the universal characteristic class. Since any bundle is a pull-back of ${U}$ , any characteristic class is a pull-back of this. Thus ${c}$ is determined by ${c(U)}$ .

Conversely, if we prescribe ${c(U)}$ , then we can define ${c(E)}$ by writing ${E}$ as a pull-back of some ${B \rightarrow \mathrm{Gr}_n(\mathbb{R}^{\infty})}$ (unique up to homotopy) and then just setting ${c(E) }$ to be the pull-back of ${c(U)}$ .

What we have really done here is, of course, the Yoneda lemma, and nothing more. Namely, we have the functor ${F_n}$ on the homotopy category which is representable, and we have the functor ${B \rightarrow H^*(B)}$ on the homotopy category. A characteristic class is just a natural transformation

$\displaystyle F_n \rightarrow H^*$

and since ${F_n}$ is representable by ${\mathrm{Gr}_n(\mathbb{R}^{\infty})}$ , Yoneda’s lemma states that these characteristic classes are in bijection with

$\displaystyle H^*(\mathrm{Gr}_n(\mathbb{R}^{\infty})).$

Thom’s construction of the Stiefel-Whitney classes « Climbing Mount Bourbaki Says:

October 18, 2011 at 9:44 pm

[…] be a real vector bundle. There are numerous constructions for the characteristic classes of . Recall that these are elements in the cohomology ring (for some ring) that measure, in some sense, the […]

Qiaochu Yuan Says:

February 7, 2012 at 1:50 am

This discussion reveals something that I didn’t really understand about the word “pullback” until I started thinking about vector bundles. Namely, “pullback” can refer both to precomposition and to fiber products, and it’s not clear what the connection between these two concepts is until you start to think about vector bundles, since the pullback of vector bundles can be defined either by precomposition (using homotopy classes of maps into Grassmannians) or by the fiber product (using the ordinary definition).

Akhil Mathew Says:

February 7, 2012 at 8:50 pm
That’s true. It also applies to other situations when you use some geometric objects to build a functor (e.g. in algebraic geometry the functor sending a scheme $X$ to the closed subschemes of $X$ flat over $X$ ), such that the functoriality comes from the fiber product, and use other means to prove that the functor is representable (so we can also think of pre-composition as defining it).

Climbing Mount Bourbaki