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I want to use a version of Stokes theorem for integration along the fiber and I need some help in proving a general statement.

Let $F$ be a $k$-manifold with boundary and let $E \to M$ be a smooth fiber bundle with fiber $F$. Let $\omega$ be a differential form on $E$, let $\int_F: \Omega^*(E) \to \Omega^{*-k}(E)$ denote integration along the fiber and $i: \partial F \to F$ the inclusion map. Can I have some help in formalizing and proving a statement of the following form:

$d \int_F \omega=\int_F d\omega \pm \int_{\partial F}i^*\omega$

Thanks!

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up vote 3 down vote accepted

Choose some open $U\subset M$ such that $\pi^{-1}(U) \cong U\times F$, where $\pi: E\to M$ is the projection.

Let us assume first, that $\omega$ is compactly supported in some chart, so that we can work in local coordinates. By shrinking $U$ we can assume that this chart is of the form $\phi: U\times V \to \mathbb R^m\times \mathbb R^k$, with $ U\subset M$ and $V\subset F$. In this case we can write $\omega$ as

$$\omega = \sum_{I} f_I(x,y) \, dx^I \wedge dy^1 \wedge \ldots \wedge dy^k$$

where summation is over all multiindices $I = (i_1, \ldots, i_s)$. (Here $x^1, \ldots, x^m$ are the coordinates on $U$, $y^1, \ldots, y^k$ those on $V$.)

Integration along the fiber is now simply given by

$$\int_F \omega = \sum_{I} \left(\int_{\mathbb X} f(x,y) \; dy^1 \dots dy^k\right) \, dx^I$$

where $\mathbb X = \mathbb R^k$ or $\mathbb R^{k-1}\times \mathbb R_{\ge 0}$, depending on $V$ (the first case corresponding to $V \cap \partial F = \varnothing$, the second to $V \cap \partial F \ne \varnothing$). By an argument analogous to that in the proof of Stokes' theorem, we find $d\int_F \omega = \int_F d\omega + (-1)^{q-k}\int_{\partial F} \omega$ in this special case. (Where $\omega$ is supposed to be a $q$-form.)

In general, we can cover $E$ by charts of the form $\phi: U\times V \to \mathbb R^m\times \mathbb R^k$ as above and take a partition of unity $(\psi_i)_{i\in I}$ subordinate to this cover to conclude

$$ \begin{align} d\int_F \omega &= \sum_{i\in I}\; d\left( \int_F \psi_i \omega \right) \\ &= \sum_{i\in I}\; \int_F d \psi_i\omega + (-1)^{q-k} \sum_{i\in I}\; \int_{\partial F} \psi_i \omega \\ &= \int_F d\left( \sum_{i\in I} \psi_i\omega \right)+ (-1)^{q-k} \sum_{i\in I}\; \int_{\partial F} \psi_i \omega \\ &= \int_F d\omega + (-1)^{q-k} \int_{\partial F} \omega \end{align} $$

for arbitrary $\omega$.


Edit: The $(-1)^{q-k}$ comes from the fact that we need $\int_{\partial F}$ only for the case that the chart is a boundary chart and $$\omega = f(x,y) \, dx^I \wedge dy^1\wedge \ldots \wedge dy^{k-1}$$

Otherwise this integral vanishes. Then

$$\begin{align} \int_{F} d\omega &= (-1)^{q}\left( \int_{\mathbb R^{k-1}\times \mathbb R_{\ge 0}} \frac{\partial f}{\partial y_k}(x,y) \, dy^1 \dots dy^k\right)\, dx^I \\ &\quad + \sum_{i = 1}^n\, \int_{F} \frac{\partial f}{\partial x_i}(x,y) \, dx^i \wedge dx^I \wedge dy^1\wedge \ldots \wedge dy^{k-1} \\ &= (-1)^{q+1}\left(\int_{\mathbb R^{k-1}} f(x,y^1, \dots,y^{k-1}, 0) \, dy^1 \dots dy^{k-1}\right)\, dx^I \\ \end{align}$$

whereas $\int_{\partial F} \omega = \left[(-1)^{k}\int_{\mathbb R^{k-1}} f(x,y^1, \dots,y^{k-1}, 0) \, dy^1 \dots dy^{k-1}\right] \, dx^I = (-1)^{q-k+1} \int_F d\omega$. So

$$d\int_F \omega = 0 = \int_{\partial F} \omega + (-1)^{q-k}\int_F d\omega$$

Note that in particular, for $\omega = f(x,y) \, dy^{1} \wedge \ldots \wedge dy^{{k-1}}$, we get $\int_{\partial F}\omega = \int_F d\omega$, which is just Stokes' theorem.

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I guess, you meant $d \int_F \omega= d(\sum_{i\in I} \int_F \phi_i \omega)=...$ –  Ehsan M. Kermani Dec 30 '11 at 21:50
    
and where does $\pm$ come from as OP has written? –  Ehsan M. Kermani Dec 30 '11 at 21:52
    
@ehsanmo: Yes, you're right. Thanks. I thought about the $\pm$ issue briefly (and not properly it seems...). I will fix it! Sorry. –  Sam Dec 30 '11 at 22:43
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