Let $E$ be a smooth oriented vector bundle over a smooth manifold $M$ and let $E^0$ be the complement of the zero section in $E$. I would like a reasonably explicit isomorphism between the relative De Rham cohomology group $H^p(E,E^0)$ and $H^p(M)$. Perhaps this is some version of the Thom isomorphism?

In case anyone needs a refresher, $H^*(E,E^0)$ is the cohomology of the complex whose chain groups are $\Omega^p(E,E^0) := \Omega^p(E) \oplus \Omega^{p-1}(E^0)$ and whose differential is given by $d(\omega_1, \omega_2) = (d \omega_1, i^* \omega_1 - d \omega_2)$.

EDIT: I actually want to prove that $H_{cv}^P(E,E^0) \cong H^p(M)$, where $H_{cv}^p(E,E_0)$ is the cohomology of the complex described above where all forms have compact support in the vertical direction.

  • $\begingroup$ I'm a bit confused. If $E \to M$ is a vector bundle, then $M$ is a retract of $E$ and they have the same cohomology. It seems to me that the relative cohomology $H^\ast(E, E^o)$ should be isomorphic to the (reduced?) cohomology of the Thom space of $E$, which is certainly not isomorphic to $H^\ast(M)$ in general. $\endgroup$ – Jonathan Apr 27 '12 at 0:14
  • $\begingroup$ Sorry, I actually want the relative cohomology groups with compact support in the vertical direction. I think that fixes it. $\endgroup$ – Paul Siegel Apr 27 '12 at 5:17
  • 1
    $\begingroup$ are you sure you do not expect an index shift in your cohomology? I suppose that your $H_{cv}^p(E,E^0)$ is isomorphic to singular cohomology with compact support (at least if $M$ is compact) $H^p_{c}(E,E^0) \cong \tilde{H}^p(M(E))$ but by the Thom isomorphism this is the shifted cohomology $H^{p-rk(E)}(M)$. $\endgroup$ – mland Apr 27 '12 at 9:10

I don't think that what you're asking is true exactly as stated, but what you're getting at is the Thom isomorphism. The proof of the isomorphism really only uses basic facts about vector bundles and axioms of homology, so pick any proof that you like and you should be able to translate it to relative de Rham cohomology. I won't prove it, but I can tell you explicitly what the maps are.

Given a vector bundle $p:E \to M$ of rank $r$, pick some metric on $E$ and use it to define the disc bundle $D \to M$ and unit sphere bundle $S \to M$. An orientation of $E$ gives us a class $\alpha \in H^r(D, S)$ that satisfies the property that for all $x \in M$, the restriction of $\alpha$ to the fiber above $x$ yields a generator of $H^r(D_x, S_x)$. Then we can define a map $H^k(M) \to H^{k+r}(D, S)$ via $\omega \mapsto p^\ast(\omega) \cup \alpha$. The Thom isomorphism theorem tells us that this map is an isomorphism of graded rings. If we use the de Rham model, then elements of $H^\ast(D, S)$ can be represented by differential forms, and the inverse map $H^{k+r}(D,S) \to H^k(M)$ is given by fiberwise integration.

Note that by homotopy invariance, $H^\ast(D, S) \cong H^\ast(E, E^o)$, but I prefer to use $(D,S)$ since it makes the fibers compact.

The best reference I can think of for the de Rham approach is Bott and Tu, Differential Forms in Algebraic Topology.

  • 1
    $\begingroup$ This is indeed the sort of thing I'm looking for (sorry I got the degrees wrong), but I'm still confused. According to Bott and Tu the Thom class is in the cohomology of $E$ with compact support in the vertical direction rather than in $H^*(D,S)$. Similarly, the target of the Thom isomorphism is $H_{cv}^*(E)$. So implicit in your answer is the assertion that $H^*(D,S) \cong H_{cv}^*(E)$, and it's not clear to me why this is true. Can you elaborate on this point? $\endgroup$ – Paul Siegel May 1 '12 at 21:47
  • 1
    $\begingroup$ There are a few versions of the Thom isomorphism, depending on whether you're working smoothly or otherwise. When you're dealing with differential forms, you can define relative cohomology by requiring that your forms vanish on the submanifold; one-point compactification is simulated by requiring your forms to be supported away from the "ends". This should follow from the de Rham theorem. $\endgroup$ – Aaron Mazel-Gee May 2 '12 at 1:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.