Let $p: E \rightarrow B $ be an an oriented real vector bundle or rank $n$. Then there exists a unique class $u \in H^n(E, E - B; \mathbb{Z})$, where $B$ is embedded into $E$ as the zero section, such that for any fiber $F$ the restriction of $u$ to $F$ is the class induced by the orientation of $F$. $H^k(E) \cong H^{k+n}(E,E - B)$ and this isomorphism is given explicitly by $\, x \mapsto x \cup u$. I would like to get a feel for this isomorphism. For the record: I am only really interested in smooth vector bundles over manifolds.

If $k+n$ is greater than the dimension of the base space then I think this is easily interpreted (though please correct me if I say something wrong or misleading). Writing out the the long exact sequence associated with the relative cohomology, we see that $H^{k+n}(E)$ is zero as $E$ deformation retracts to $B$. We see that in this case $H^{k+n}(E;E - B) \cong H^{k+n-1}(E - B)$. The isomorphism $H^k(E) \cong H^{k+n}(E,E - B)$ then tells us about the cohomology of $H^{k+n-1}(E - B)$, which I can "feel" is the twisting of the vector bundle.

For the case $k+n$ less than the dimension of the manifold, it is less clear for me how to think of relative cohomology.

I know that Bott and Tu have some interpretation like this and a lot of people swear by that book, but to be honest that book just confused me. Maybe I should try again... Any other ideas?


1 Answer 1

  1. Sometimes these things make more intuitive sense by passing to the representing spectra, but I do not know if you are interested in that degree of generality. If you want a more "low-tech" approach, I can recommend Milnor-Stasheff's Characteristic Classes.
  2. By excision, that relative cohomology is the same as the disk bundle relative its boundary. In that case, we can think about it, with coefficients in $\mathbb{R}$, as having to do with forms on the disk bundle that vanish on the boundary. We are looking for a form which integrates to $1$ on each fiber. There is always a choice of such a thing when the bundle is trivial, but when we glue, we might worry about a $1$ becoming a $-1$. An orientation says we can make choices that do not cancel out. Once we have such a thing, cupping with it is an isomorphism with inverse given by "integrating it out."

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