When is the Thom class the Poincare dual of the zero section?

Question

As the title suggests, when is the Thom class the Poincare dual of the zero section? For starters, it's true for the normal bundle of an immersion...

Answer 1

I don't know exactly what you are asking (Thom class of what? Poincare duality in what manifold?), but here is an answer to some question in this family:

If $E\to M$ is a smooth oriented vector bundle of rank $r$ on a compact smooth manifold of dimension $m$, then the Thom class of the bundle can be regarded as an element of $H^r(D(E),S(E))$ where $D(E)$ is the disk bundle and $S(E)=\partial D(E)$ is the sphere bundle. This corresponds by the Poincare duality isomorphism to the element $H_{m}(D(E))$ given by the submanifold $M\subset D(E)$ (the zero section or any other section).

This can be generalized to the case where $M$ is noncompact, or has nonempty boundary. For example, in the former case the duality isomorphism goes from $H^r(D(E),S(E))$ to the locally finite homology group $H_{m}^{lf}(D(E))$, in which there is a class given by the noncompact manifold $M$.

It can also be generalized to unoriented situations, but here the homology and cohomology groups involve nontrivial coefficient systems in general.

Answer 2

For De Rham cohomology, this is always true. It is theorem 6.24 in Bott and Tu.