General Stokes' Theorem
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Theorem
Let $\omega$ be a smooth $\paren {n - 1}$-form with compact support on a smooth $n$-dimensional oriented manifold $X$.
Let the boundary of $X$ be $\partial X$.
Then:
- $\ds \int_{\partial X} \omega = \int_X \rd \omega$
where $\d \omega$ is the exterior derivative of $\omega$.
Proof
A Special Case
First we suppose that there is a chart:
- $x = \tuple {x_1, \ldots, x_n}: V \subseteq X \to \R^n$ such that $\map \supp \omega \subseteq V$
Here, $\map \supp \omega = \overline {\set {p \in M : \map \omega p \ne 0} }$.
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We may suppose that $V$ is relatively compact.
Thus, by composing $x$ with a translation, we may suppose that:
- $\ds \map x V \subseteq \mathbb H^n = \set {\tuple {x_1, \ldots, x_n} \in \R^n : x_1 < 0}$
We have, in the coordinates $x$:
- $\ds \omega = \sum_{i \mathop = 1}^n f_i \rd x_1 \wedge \cdots \wedge \hat {\d x}_i \wedge \cdots \wedge \d x_n$
The forms $\hat {\d x}_i := \d x_1 \wedge \cdots \wedge \hat {\d x}_i \wedge \cdots \wedge \d x_n$ vanish on the tangent space to $\mathbb H^n$ for $i > 1$, so we have:
- $(1):\quad \ds \int_{\partial \mathbb H^n} \omega = \int_{\partial \mathbb H^n} f_1 \hat {\d x}_1$
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Moreover:
| \(\ds \d \omega\) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \rd f_i \wedge \hat {\d x}_i\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \frac {\partial f} {\partial x_i} \d x_i \wedge \hat {\d x}_i\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\sum_{i \mathop = 1}^n \frac {\partial f} {\partial x_i} } \rd x_1 \wedge \cdots \wedge \d x_n\) |
so that:
- $\ds \int_{\mathbb H^n} \rd \omega = \sum_{i \mathop = 1}^n \int_{\mathbb H^n} \frac {\partial f_i} {\partial x_i} \rd x_1 \wedge \cdots \wedge \d x_n$
If $i > 1$:
| \(\ds \int_{\mathbb H^n} \frac {\partial f_i} {\partial x_i} \rd x_1 \wedge \cdots \wedge \d x_n\) | \(=\) | \(\ds \int \cdots \int \paren {\int_{-\infty}^\infty \frac {\partial f_i} {\partial x_i} } \hat {\d x}_i\) | Fubini's Theorem | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
For $i = 1$:
| \(\ds \int_{\mathbb H^n} \frac {\partial f_1} {\partial x_1} \d x_1 \wedge \cdots \wedge \d x_n\) | \(=\) | \(\ds \int \cdots \int \paren {\int_{-\infty}^0 \frac {\partial f_i} {\partial x_i} } \hat {\d x}_i\) | Fubini's Theorem | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_{\partial \mathbb H^n} f_1 \hat {\d x}_1\) |
So:
- $\ds \int_{\mathbb H^n} \rd \omega = \int_{\partial \mathbb H^n} f_1 \, \hat {\d x}_1$
Together with $(1)$, this establishes the result.
$\Box$
General Case
Choose a finite family of relatively compact charts $V_1, \ldots, V_k$ on $X$ such that
- $\ds \supp \omega \subseteq \bigcup_{i \mathop = 1}^k V_i$
Choose a partition of unity:
- $\chi_1, \ldots, \chi_k$
with $\chi_1 + \cdots + \chi_k = 1$ subordinate to the cover $\set {V_1, \ldots, V_k}$.
Put $\omega_i = \chi_i \omega$.
Then we have:
| \(\ds \omega\) | \(=\) | \(\ds \paren {\chi_1 + \cdots + \chi_k} \omega\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \omega_1 + \cdots + \omega_k\) |
Moreover, $\supp \omega_i \subset V_i$ by definition.
Therefore by the special case above, Stokes' theorem holds for each $\omega_i$, so we have:
- $\ds \int_X \rd \omega = \sum^k_{i \mathop = 1} \int_x \rd \omega_i = \sum^k_{i \mathop = 1} \int_{\partial X} \omega_i = \int_{\partial X} \omega$
$\blacksquare$
Also see
Source of Name
This entry was named for George Gabriel Stokes.