# Projection formula, Bott and Tu

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

• 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. – Soham Nov 7 '18 at 4:50

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 $$\tau$$.
• 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. – Charlie Frohman Aug 15 '19 at 12:07