The following statements are from the book heat kernel and dirac operator chapter 1.

" Let $\pi : M \rightarrow B $ be a fiber bundle with n-dimensional fiber, such that both M and B are oriented. If $\alpha \in {A}_c^k(M)$ is a compactly-supported differential form on M, its integral over the fibers of $ M \rightarrow B $ is the differential form $\int\limits_{M/B} \alpha \in A^{k-n}(B)$ such that

\begin{equation} \int\limits_B (\int\limits_{M/B} \alpha) \wedge \beta = \int\limits_M \alpha \wedge \pi^* \beta \quad ... (1.15) \end{equation} for all differential forms $\beta$ on the base B. We sometimes write $ \pi_* \alpha $ instead of $ \int\limits_{M/B} $. It follows easily from the (1.15) that

\begin{equation} \pi_*(\alpha \wedge \pi^* \beta ) = \pi_* \alpha \wedge \beta \quad ...(1.16) \end{equation} for all $\alpha \in A_c(M) $ and $\beta \in A(\beta).$ "

My questions are the following:

  1. what is the intuition behind defining the integral of $\alpha$ over the fibers by the equation (1.15), and why the notion of integration along fiber is important ?

  2. how to prove that equation 1.15 implies equation 1.16 ?

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    $\begingroup$ It's just a generalization of the Fubini theorem for integrating over the product $B\times F$. Indeed, locally any fiber bundle is a product and you reduce to this case with the usual partition of unity argument. $\endgroup$ – Ted Shifrin Jan 26 at 18:12
  • $\begingroup$ Thank you so much for your comment! This answer my first question, and could you please help with the second question as well, i don't see how the first equation implies the second ? $\endgroup$ – asma Jan 26 at 21:05
  • $\begingroup$ Write it down in $(x,y)$-coordinates ($x$ on base, $y$ on fiber). You're integrating out all the $dy$. But $\pi^*\beta$ has only $dx$ in it. $\endgroup$ – Ted Shifrin Jan 27 at 0:28
  • $\begingroup$ @Ted Shifrin, so we can prove it without the use of equation 1.15, is that what you meant ? Please, could you give me some more details because my background in differential geometry is poor, and could you please recommend some good references to understand the notion of integration along fiber ? $\endgroup$ – asma Jan 27 at 9:48

For your first question about why integration along the fiber is important, note that it gives a cochain map of degree $-n$, $$\pi_*: A^k(M) \to A^{k-n}(B).$$ If $M$ was an oriented rank $n$ vector bundle over $B$ then it can be shown that this map induces isomorphism at the level of cohomology, $H^k(M) \to H^{k-n}(B)$. This is called the Thom isomorphism. You can find a proof in Bott & Tu's classic.

Now for your second part, notice that (1.15) is the defining property for $\pi_*\alpha$. That is if there were any other form $\omega$, which satisfied (1.15): $\int_B \omega\wedge\beta = \int_M \alpha\wedge\pi^*\beta$ for all $\beta$ (of appropriate degree) on the base, then we would have $\omega = \pi_*\alpha$. So, (1.15) uniquely characterizes $\pi_*\alpha$.

T show that $\pi_*(\alpha\wedge\pi^*\beta) = \pi_*\alpha\wedge\beta$, it suffices to show that $\pi_*\alpha \wedge \beta$ indeed satisfies the defining property (1.15) for $\alpha\wedge\pi^*\beta$. That is we need to show that, for any $\phi \in A^*(B)$.

$$ \int_B (\pi_*\alpha\wedge\beta)\wedge\phi= \int_M (\alpha\wedge\pi^*\beta)\wedge\pi^*\phi $$

but this is obvious because of the defining property of $\pi_*\alpha$ we have, $$ \int_B (\pi_*\alpha\wedge\beta)\wedge\phi = \int_B \pi_*\alpha\wedge(\beta\wedge\phi) = \int_M \alpha \wedge \pi^*(\beta\wedge\phi) = \int_M (\alpha \wedge \pi^*\beta)\wedge\pi^*\phi. $$

Lastly, regarding your comment to @Ted about references, I would suggest Bott & Tu's book. They define integration along the fiber more explicitly using local charts (similar to what Ted was suggesting you to do). Then both (1.15) and (1.16) follow easily from that definition. FWIW, these formulas together are called the Projection Formulas (see Proposition 6.15 in Bott & Tu).

  • $\begingroup$ I don't think the Thom isomorphism holds in this situation, since we don't even know what the fiber is. $\endgroup$ – Jason DeVito Jan 28 at 14:13
  • $\begingroup$ @JasonDeVito Yes, Thom isomorphism doesn't hold in this situation. I misread 'n-dimensional fiber' as 'n-dimensional vector space as fiber'. I will correct my answer. Thanks. $\endgroup$ – feynhat Jan 28 at 16:20
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    $\begingroup$ @asma I should add that I haven't seen the book you mentioned in your OP so I don't know what is the motivation behind introducing integration along the fiber at this level of generality (fiber bundles). However, in the special case where your fiber bundles are oriented vector bundles, the importance of the "integration along the fiber"-map is that it gives you the Thom isomorphism. $\endgroup$ – feynhat Jan 28 at 16:31
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    $\begingroup$ @asma If there were such an $\omega$ then we would have $\int_B (\omega - \pi_*\alpha)\wedge \beta = 0$ for all $\beta$, but $\int_B (- \wedge -)$ is a non-degenerate bilinear form (by Poincaré duality), so $\omega - \pi_* \alpha = 0$. $\endgroup$ – feynhat Feb 23 at 16:06
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    $\begingroup$ Well this only shows that they are cohomologous but your definition $\pi_*\alpha$ itself only makes sense upto cohomology. $\endgroup$ – feynhat Feb 23 at 16:10

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