Suppose is a topological space, and is a vector bundle over . Here is a “reasonable” space, for instance a manifold or a CW complex. There are many familiar examples of how such bundles arise “in nature,” for instance the tangent bundle to a manifold. It is of interest to understand how the apparatus of algebraic topology applies to the bundle.

Now, clearly there is a deformation retraction of given by that deformation retracts to the image of the zero section. In particular, is homotopy equivalent to . So by itself might not all be that interesting.

Nonetheless, let us suppose that is a *normed* bundle, e.g. the tangent bundle to a manifold with a Riemannian metric. In this case, we can consider the set of elements of norm one; it is no longer a vector bundle over , but it is a fiber bundle (with fibers various spheres). This is homotopically very different from in general.

For a general bundle , we can still consider the subset consisting of nonzero elements. Then there is a map whose fibers are of the form . It is easy to see that if is a normed bundle, then deformation retracts onto the associated sphere bundle. So we might as well see what is like. By using the long exact sequence, we might try studying the pair .

Today, I want to talk about the cohomology of the pair . Namely, the main result is:

Theorem 1 (Thom)For all , there is an isomorphism (to be described)

if has dimension .

(**To ease notation, all cohomology will henceforth be with -coefficients.**)

To prove this fact, we are going to use the fact that vector bundles are, by definition, *locally trivial.* In particular, we are going to cover with open sets on which the result is true, then use a Mayer-Vietoris argument to deduce it for the whole space.

So that’s the plan. There are still several details remaining. For one thing, we will need to have some kind of description of what the Thom isomorphism actually looks like.

There is an explicit form that this isomorphism is going to take. Namely, it will involve pulling back to and then cupping with a “fundamental” class The following will describe the fundamental class:

Theorem 2 (Thom II)There is a unique that restricts to the nonzero element of

So is just because it is the cohomology of a sphere. The point is that we can find something defined globally on the manifold which becomes nonzero on every fiber. The Thom isomorphism will be

**1. The case of a trivial bundle **

The first thing to do will be to figure out what is going on for a trivial bundle. In this case, we have and , so our spaces are pretty concrete.

There is a unique generator in which becomes, by pulling back, an element which restricts to a generator on each fiber. The claim is that this is the fundamental class.

Now let us recall the Künneth theorem.

Theorem 3Let and be pairs of topological spaces. Suppose that is finitely generated. Then there is an isomorphism of graded algebras

that sends (for the two projections).

The Künneth theorem as stated is valid with coefficients in a field (e.g. ). While I won’t recall the proof, it relies on basically the Eilenberg-Zilber equivalence (for homology) together with some dualization.

It is easy to see that it applies to and . The cohomology of the latter is just in dimension and zero elsewhere. In particular, we see that the cohomology is given by where the isomorphism is given by cupping with the pull-back of a generator in . This generator is just what we called above.

If we apply this to , we see that the one element in that can restrict to a generator on each fiber is cupped with something in that is nonzero when restricted to each point. In particular, it is just cupped with , or . So there is a *unique* fundamental class.

Moreover, the Thom isomorphism is also clear. Indeed, the Künneth theorem just states that the map

obtained by pulling back and cupping with is an isomorphism.

**2. The Mayer-Vietoris argument **

Let be a bundle, as before. By definition, there is an open cover of such that is trivial. By the previous section, we know the Thom theorems for each . The problem now is to glue the results to the whole space.

We shall say that Thom and Thom II are **true for ** (an open set) if the conclusions of both the Thom theorems apply to the bundle . So there is an open cover of by sets on which Thom and Thom II are true.

We are now going to use a Mayer-Vietoris argument to handle to the case where can be covered by *finitely * many open sets on which the bundle is trivial.

Namely, let us argue:

Lemma 4If Thom and Thom II are true for , as well as for , then they are true for .

This is a fairly standard argument. Namely, let us first show Thom II. That is, let us show that there is a unique fundamental class on . For this we need notation: if is open, we write for the pull-back of to (i.e. the preimage of ). Similarly denotes the nonzero elements in .

Now by Mayer-Vietoris, there is an exact sequence

But the first term is zero! Indeed, this is an immediate consequence of the Thom isomorphism for —we have that . This implies at once that if the fundamental class exists, it is unique. Indeed, by definition, it’s going to have to restrict to the fundamental classes on and (it’s going to restrict to something nonzero on all the fibers), but the exact sequence shows that only one class can do that.

The same argument also shows that the fundamental class over *does* exist. Indeed, the two fundamental classes over must both restrict to the fundamental class of , because that’s unique by hypothesis.

So the fundamental class does exist. OK. That’s good. Now we have to deduce the Thom isomorphism. That is, for each , we want to know that the map

obtained by cupping with is an isomorphism. However, for this we can draw a commutative and exact diagram (click on it to make it bigger):

The reason for commutativity is that cup products commute (or sign-commute, I forget which) with boundary morphisms. This is a general fact which can be checked directly on the level of chain complexes.

Anyway, we know by hypothesis that every downward map is an isomorphism except possibly the middle one. So the middle one is by the five lemma. And this, itself, establishes the lemma.

So now the goal is to see that Thom and Thom II are true for spaces which are finite unions where the bundle is trivial on each . We can do this by induction on ; we know it for , and suppose it true for . Then Thom and Thom II are true for and by the inductive hypothesis. They are also true for by the same logic. So the above lemma shows that they are true on the union.

**3. Finishing the proof **

OK. This handles the case of a compact base space. What about in general? Well, here I’m just going to invoke a fact that cohomology sends infinite direct limits to inverse limits since we are working with field coefficients. In particular, we find that there is a unique fundamental class that restricts to the fundamental classes on each of the open sets on which the bundle is trivial. Since the inverse limit of isomorphisms is an isomorphism, cupping with it gives an isomorphism as in Thom’s theorem.

This is a familiar trick in algebraic topology: one proves a result for basic open sets of a certain form, uses a Mayer-Vietoris argument to deduce it for finite unions, and then appeals to some kind of direct limit argument. This sort of argument, for instance, can be used to prove the isomorphism between de Rham and singular cohomology (with -coefficients) for a smooth manifold; see Bredon’s book *Topology and Geometry.*

**4. The Thom space **

The way the present theorem has been stated is not quite how the Wikipedia article on the Thom result states it. In the Wikipedia article, one starts with a vector bundle , and constructs the **Thom space** ; this is obtained by taking the one-point compactifications of the fibers and gluing them together.

The claim is that for a paracompact base and an -dimensional , we have an isomorphism

given by cupping with a class which restricts to a generator on each fiber.

We could actually prove this by *exactly* the same argument that we used to prove the other version of the Thom theorem. Alternatively, we can deduce it from the theorem as stated above. Namely, if we put a metric on , then becomes homotopy equivalent to the fiber bundle of vectors of norm . So the cohomology of is what was claimed; this is the same as the reduced cohomology of by standard arguments.

October 18, 2011 at 9:28 pm

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