I have been planning to say a few words about local cohomology, and I stand by that still–but I got sidestracked.

I would now like to explain an interesting phenomenon that we talked about today in topology class. In algebraic topology, one often considers *fiber bundles* . The idea of a fiber bundle is to generalize the notion of a product space . A fiber bundle is, loosely, something that locally looks like the projection from a product . The reason that we only require it to look *locally* like a product is, for instance, that we want the *tangent bundle* to be considered a fiber bundle. However, as is well known, the tangent bundle to a manifold may admit no nonzero global sections—which means that it cannot be trivial. Nonetheless, given a fiber bundle, one may ask in some sense how strongly it fails to be nontrivial.

A detailed study of such questions belongs to the theory of characteristic classes, which I am not currently ready to talk about. Nonetheless, let us start with a very simple question. Let be a fiber bundle over a compact space . There is then, by definition, an open covering of such that the restricted fiber bundles is trivial, isomorphic to for the fiber. Clearly we can take this cover finite. One might naturally ask how small we can take this cover.

We shall describe the answer in a specific important case. One of the standard examples of a fiber bundle is the ** Hopf bundle**

The description is straightforward. An element in can be viewed as an -tuple of complex numbers. This spans a line through the origin and maps to a point in . The fiber corresponds to all the points on a circle that span the same fixed complex line, which is traced out by a circle. In fact, this is more than a simple fiber bundle.

Since is a group, this is a principal -bundle. This has the following consequence. If the bundle admits a section, then it is trivial. This is a general fact about principal bundles, and it is also true locally (i.e. on open sets). Let us now check that this is in fact locally trivial, and in fact find sections locally. Consider the open sets of points of represented by homogeneous coordinates with .

Given an element , we map this to . If we divide this by its norm, we get an element of which maps down to it. So we have sections over the . So the bundle is trivial over each , by principalness.

Of course, there is an easier way to see that the bundle is trivial over : namely, is contractible, and any fiber bundle over a contractible space is trivial. But in any case, we see the claim. In total, we find that there is a covering of the base space by open sets on which the Hopf bundle is trivial. Is this the best possible?

Theorem 1Any trivializing cover of the base of the Hopf bundle is of size at least .

The proof is not difficult, but it is ingenious. The first thing to recall is that there is a homological theory of fiber bundles. When one has a product space, one can use the Kunneth formula to calculate the cohomology of a product from that of its components. Under reasonable assumptions, the Kunneth theorem extends to fiber bundles as well; it is called the Leray-Hirsch theorem. This requires cohomology classes on the total space that generate the cohomology of the fiber at each point. But even that fails here because of is trivial while is nontrivial. The more powerful general theorem, called the Serre spectral sequence, is still valid here, but we shall not use anything so fancy. So we shall try something else.

Namely, we shall use even more elementary methods. Let be an open cover of over which the Hopf bundle is trivial. Suppose . Note that we have a very good understanding of the singular cohomology rings of all these three spaces.

The most interesting of , which can be shown, either by elementary means or even by appealing to the Serre spectrnal sequence (I’ll have to explain this someday), to be where is the generating class in degree two. Intuitively, one should think of this as a complex plane.

Consider the element . It pulls back to zero in as has no cohomology in degree two. This is actually unexpected. If one has a product map , the map is actually injective (if have, say, finitely generated cohomology) by the Kunneth formula.

In particular, this means that is injective. So the fact that goes to zero in means that it goes to zero in for each . This in turn implies that goes to zero in . By the long exact sequence in cohomology, this means that comes from an element in .

Now one of the properties of the cup product is that if is a space and are open sets, the cup product of classes is actually in . The reason is essentially that a homology cycle lying in can be split into pieces in and by the excision property.

Now this means that for each . In particular, the cup-product of all of them is in . But if , we know that , as it is the generator of . This is the contradiction. And that completes the proof.

The essential idea of the above proof is to use the algebra structure on the singular cohomology, and in particular the fact that the cohomology ring of projective space is highly nontrivial.

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