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Suppose that $S$ is an oriented smooth $s$-manifold, and $\pi :E\to S$ is an oriented real vector bundle over $S$. The Thom isomorphism asserts that the integration along the fiber defines isomorphisms$$H_{cv}^*(E)\cong H^{*-n}_{dR}(S).$$The Thom class $\Phi_E$ is defined to be the element $\Phi_E\in H^n_{cv}(E)$ which corresponds to $1\in H_{dR}^0(S)$ in the above isomorphism. In several sources I have read, such as Bott-Tu (around p.65) and this pdf, it is stated that the Poincare dual of $S$ in $E$ is equal to $\Phi_E$, that is, if we equip $E$ an orientation by the canonical isomorphism $T_xE_x\oplus T_xS\cong T_x E$, then we have$$\int_E \omega\wedge \Phi_E=\int _S\omega$$for all $\omega\in H^s_{c}(E)$. However, I have some trouble understanding the proof of this claim, and I need someone's help.

The proof goes as follows: if $i:S\to E$ denotes the zero section, then $i\circ\pi$ is homotopy equivalent to the identity on $E$, so $\omega -\pi^*i^*\omega =d\tau$ for some $\tau\in\Omega^{s-1}(E)$. Thus we have $$ \int_E\omega\wedge \Phi_E=\int_E\pi^*i^*\omega\wedge\Phi_E+\int_E d\tau\wedge \Phi_E. $$ The first term on the RHS of the above equation can be computed, using the projection formula, and it equals $\int_S\omega$. The problem is the second term, which involves the integral of $d\tau\wedge \Phi_E=d(\tau\wedge \Phi_E)$. If we can show that this integral vanishes, then we are done. But we cannot blindly apply the Stokes's theorem, for $\tau\wedge \Phi_E$ may not have compact support.

  • In Bott-Tu, it is simply stated that this integral equals zero by the Stokes's theorem. But for the reasons stated above, I think we need to ensure compactness of the support of $\tau\wedge \Phi _E$
  • In the pdf I cited above, it is stated that $i\circ\pi$ is properly homotopic to the identity map of E via the homotopy $H:E\times [0,1]\to E, \,(v,t)\mapsto (1-t)v$, and hence $i$ and $\pi$ induce isomorphism in the compact cohomology and thus we can actually assume $\tau $ to have compact support. However, if I understand it correctly, $H$ is not proper because, for example, if $x\in S$ is any point in $S$ and $0_x$ denotes the zero vector in $E_x$, then $H^{-1}(0_x)\cap(E\times\{1\})=E_x\times\{1\}$ is not compact, unless $E$ has rank $0$.
  • Actually, since $\Phi_E\in\Omega_{cv}^n(E)$, it is enough to ensure the compactness of $\pi(\operatorname{supp}\tau)$. But I have trouble proving even this.

I must be missing something. Can anyone help me? Thanks in advance.

Ken
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1 Answers1

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I had to think about this a while to untangle the issue.

First, Bott/Tu have already restricted to a compact disk bundle, so the homotopy is proper. But, second, I believe we don't even need that. If you look at the construction of the chain homotopy (around p. 34), the form $\tau$ is given in our case by $\tau = KH^*\omega$, where $K$ is integration over the fiber of the homotopy $H$. Since $\omega$ has compact support in the base manifold, so will $\tau$.

Ted Shifrin
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  • Thank you! Of course, I should have looked at the chain homotopy... By the way, I believe that the chain homotopy $K$ you are talking about corresponds to $s_1^*K$ in the notation of Bott-Tu, pp.34-35, correct? – Ken Jan 04 '20 at 01:21
  • I usually write $I$ (for integration over the fiber). Yeah, it looks like you're right, except I'm not sure what $s_1$ is here. – Ted Shifrin Jan 04 '20 at 01:24
  • $s_1$ denotes the section $x\mapsto (x,1)$. This symbol was used in p.35 of Bott-Tu. – Ken Jan 04 '20 at 01:56
  • Oh, well, for me the map is $\Omega^{k+1}(\Bbb R^n\times [0,1]) \to \Omega^k{\Bbb R^n}$. – Ted Shifrin Jan 04 '20 at 02:00
  • which is given by $f(x,t)dx^I\mapsto 0$ and $f(x,t)dx^I\wedge dt\mapsto (\int_0^1f(x,t)dt )dx^I$ ? – Ken Jan 04 '20 at 02:04
  • Yup. This standardly appears in most texts. :) – Ted Shifrin Jan 04 '20 at 02:10
  • Thanks! I cannot tell you how grateful I am! – Ken Jan 04 '20 at 02:14
  • You're most welcome, Ken. Having been through Bott/Tu several times with graduate students, I was stunned that I had not considered your question before. All power to you! – Ted Shifrin Jan 04 '20 at 02:16