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Bott and Tu, Proposition 6.15:

Let $\pi: E\rightarrow M$ be an oriented rank $n$ vector bundle, $\tau$ a form on $M$ with compact support and $\omega$ a form with compact support along fiber with $\omega \in \Omega^q_{cv}(E)$ and $\tau \in \Omega_c^{m+n-q}(M)$. Then, with local product orientation on $E$

$$\int_{E} (\pi^* \tau) \wedge \omega=\int_{M}\tau\wedge \pi_{*}\omega$$

I do not understand why the integrant has compact support on $E$. I understand that $\pi^* \tau$ is zero outside a closed set and so product by $\omega$ in each fiber has compact support but this does not imply that it has compact support over all of $E$. Has Bott&Tu made a mistake here??

Edit: As the answer below shows, it is an error. And also see this:

Tubular neighborhood: compact support for the pullback of a form with compact supoprt

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  • $\begingroup$ I think the book defines $\Omega_cv (E)$ as forms on $E$ which have compact support on $E|_K$ where $K$ is compact in $M$. If you go by that definition, $\omega|_{supp(\tau)}$ is compactly supported. $\endgroup$ – Soham Nov 7 '18 at 4:50
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This does indeed appear to be an error. For instance, consider the case where $m=n=1$ and $q=1$, $M=\mathbb{R}$, $E=M\times\mathbb{R}$ is the trivial bundle, and $\tau$ is nonzero on all of $[0,1]$. We could then have $\omega$ be a vertical $1$-form on $E$ which for each positive integer $n$ has a little bump on the set $[1/(n+1),1/n]\times[n,n+1]$, and is $0$ outside these sets. Then $\omega$ has compact support on each fiber, but $\pi^*\tau\wedge\omega$ does not have compact support, and may not even be integrable if $\omega$ gets large enough on its bumps.

I believe the fix is to change the definition of "compact support along the fibers". You need to require not just that the intersection of the support of $\omega$ with each fiber is compact, but that the map from the support of $\omega$ to $M$ is a proper map. That is, for any compact set $K\subseteq M$, the support of $\omega$ on $\pi^{-1}(K)$ is compact. This certainly would solve the issue you have observed, since you can just take $K$ to be the support of $\tau$.

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  • $\begingroup$ please see this: math.stackexchange.com/questions/2026883/… $\endgroup$ – alireza Nov 23 '16 at 6:58
  • $\begingroup$ would you see this please? I think there is another mistake. math.stackexchange.com/questions/2063380/… $\endgroup$ – alireza Dec 18 '16 at 13:35
  • $\begingroup$ What can happen is that $\pi_*\omega$ is not integrable. In fact in your example $\pi_*\omega$ isn't even continuous at $0$. A less restrictive condition than what you propose, is to just add the hypothesis that $\pi_*\omega$ be integrable. $\endgroup$ – Charlie Frohman Aug 15 '19 at 12:07

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