Intuition behind the Thom Isomorphism.

Question

For a talk about Euler and Chern classes I need the Thom isomorphism with $\mathbb{Z}$ coefficients at some point (when proving the Gysin exact sequence). Recall that for an oriented real $n$-plane bundle $\xi$ with projection $\pi: E \to B$ the Thom isomorphism is given by: $$\Phi: H^k\left(B\right)\to H^{n+k}\left(E,E_0\right), \; x \mapsto\pi^{*}\left(x\right) \cup u$$

where $u$ denotes the fundamental class of $H^n \left(E,E_0\right)$. I don't have enough time to prove or even sketch the proof of this fact but I want to explain why this makes sense in 2 or 3 sentences. I'll prove the fact that $\pi$ induces an isomorphism in cohomology anyway, so I'd like to understand the cup product part of this isomorphism. So far I consulted Milnor's Characteristic classes but without success. I'd be glad if someone could give me a reference or explain this to me.

Answer 1

$u$ is the Thom class, it is Kronecker dual to those classes "orthogonal" to $E_0$. That is, think of those simplices $\nu$ which intersect $E_0$ in one point, then $u(\nu)=1$. Now $E_0$ is homeomorphic to $B$, so if we have a simplex $\tau$ in $B$ we can think of it is being in $E_0$. Now take $\tau\times \nu$, then the image of the $f$ where $f$ is dual to $\tau$ (that is $f(\tau)=1$) under the Thom isomorphism will be the function dual to $\tau\times \nu$, so $\Phi(f)( \tau\times \nu)=1$. Cup product comes from the external product operation $\times$, thus that is how it enters into the formula.

In Milnor, he proves Thom, via a lemma which is basically "Thom for $\mathbb{R}^n$", this is the local picture for the Thom map so I would go back and study that lemma. One can also try to analyse the theorem in terms of Poincare duality, in which case you are taking an intersection with $E_0$ and then dualising.

Answer 2

There is a very nice geometric explanation for the cup product part by the de Rham theory.

Let $B$ be an oriented $n$-manifold($\mathbb{Z}$-coefficients). First recall that in de Rham cohomology, the cup product corresponds to the wedge product of differential forms. For oriented vector bundles there is another cohomology called compact vertical cohomology. Let $p: E \to B$ be an oriented vector bundle. Then cochain groups for this cohomology are given by $$ \Omega_{cv}^n(E)=\{ \omega \in \Omega^n (E) \ | \ p^{-1}(K) \cap \text{Supp}(\omega) \text{ is compact, } \forall \text{ compact } K \subset B\} $$ In particular, $\text{Supp}(\omega|_{p^{-1}(x)})$ is compact. Choose a trivialization for $E$ with coordinates $(t_1,\ldots,t_n)$ in the fibers and define the integration along the fiber as $$ p_{\ast} : \Omega_{cv}^{\ast} (E) \to \Omega^{\ast -n}(B) \\ (p^{\ast}\phi) f(x,t_1,\ldots,t_n)dt_1 \ldots dt_n \mapsto \phi \int_{\mathbb{R}^n} f(x,t_1,\ldots,t_n)dt_1 \ldots dt_n \ \text{ where } \phi \in \Omega^{\ast-n}(B)$$ and as $0$ for different types of forms. This gives an isomorphism $\Phi^{-1}$ in the question. Then Thom class $u$ is given by $u = \Phi(1)$ where $1 \in H^0(B)$(can think of bump form in each fiber) and $\Phi$ is exactly given by $\eta \mapsto p^{\ast}(\eta) \wedge u$.

More Geometry: Let $M \subset B$ be closed oriented $m$-dim submanifold. We have a natural $(n-m)$-vector bundle which is the normal bundle $\nu(M)$. Then the Thom class of the normal bundle is the Poincare dual of $S$ and Thom isomorphism relates cohomology of $M$ with the relative cohomology of the tubular neighborhood of $M$ in $B$.

Now as Rene Schipperus pointed out in his answer one can see that intersection of oriented submanifolds corresponds to wedge product.

Answer 3

The Thom isomorphism is the intersection with the zero section. To see this, we need two observations:

1. First, recall the homological version of the Thom isomorphism; if $u$ is the Thom class of $E$, then \begin{align*} \eta \mapsto p_*(\eta\cap u), \quad H_{\bullet+n}(E, E_0) \to H_{\bullet}(E) \to H_\bullet(M) \end{align*} is an isomorphism. I think this is stated in Milnor-Stasheff.

2. The Thom class is Poincaré dual to the the zero section in $E$: Let $E\to M$ is a smooth orientable rank $r$ vector bundle over a compact orientable manifold with boundary $\partial M$ and let $DE$ and $SE$ be the unit ball resp. sphere bundles of $E$. Let $u=u(E)\in H^r(DE, SE)$ be the Thom class of $E$, $[M]\in H_n(DE)$ be the class of the zero section (or any other section) and $[DE]\in H_{r+n}(DE, SE)$ the fundamental class, then \begin{align*} [DE] \cap u = [M]\in H_n(DE) \cong H_n(M) \end{align*} To prove this, verify the characterizing local property of the fundamental class $[M]$. By excision you can reduce it to a local statement in a trivialization, but there DE is a cartesian product, so it isn't hard to see.

Combining these observations, the cap product with $u$ should be the same as taking the intersection product with $[M]$. This doesn't really tell you why it should be an isomorphism, but maybe it tells you why it is interesting. It explains why the intersection product can be constructed using the Thom isomorphism: The intersection product can be seen as an intersection with the diagonal $\Delta\subset M\times M$, and now you just have fiddle with your spaces so that the diagonal is translated to the zero section of some bundle.