Suppose ${B}$ is a topological space, and ${p: E \rightarrow B}$ is a vector bundle over ${B}$. Here ${B}$ 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 ${E}$ given by ${(t,v) \rightarrow tv}$ that deformation retracts ${E}$ to the image of the zero section. In particular, ${E}$ is homotopy equivalent to ${B}$. So ${E}$ by itself might not all be that interesting.

Nonetheless, let us suppose that ${E}$ 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 ${B}$, but it is a fiber bundle (with fibers various spheres). This is homotopically very different from ${B}$ in general.

For a general bundle ${E}$, we can still consider the subset ${\dot{E}}$ consisting of nonzero elements. Then there is a map ${p: \dot{E} \rightarrow B}$ whose fibers are of the form ${\mathbb{R}^n - \left\{0\right\}}$. It is easy to see that if ${E}$ is a normed bundle, then ${\dot{E}}$ deformation retracts onto the associated sphere bundle. So we might as well see what ${\dot{E}}$ is like. By using the long exact sequence, we might try studying the pair ${(E, \dot{E})}$.

Today, I want to talk about the cohomology of the pair ${(E, \dot{E})}$. Namely, the main result is:

Theorem 1 (Thom) For all ${k}$, there is an isomorphism (to be described)$\displaystyle H^k(B; \mathbb{Z}/2) \simeq H^{k+n}(E, \dot{E}; \mathbb{Z}/2)$

if ${E}$ has dimension ${n}$.

(To ease notation, all cohomology will henceforth be with ${\mathbb{Z}/2}$-coefficients.) (more…)