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

Special Case

Let there exist a chart:

$x = \tuple {x_1, \ldots, x_n}: V \subseteq X \to \R^n$

such that:

$\map \supp \omega \subseteq V$

where:

$\map \supp \omega = \overline {\set {p \in M : \map \omega p \ne 0} }$


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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 $i > 1$

Hence 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:

$\ds \int_{\mathbb H^n} \rd \omega = \sum_{i \mathop = 1}^n \int_{\mathbb H^n} \frac {\partial f_i} {\partial x_i} \rd x_1 \wedge \cdots \wedge \d x_n$


Let $i > 1$.

Then:

\(\ds \int_{\mathbb H^n} \frac {\partial f_i} {\partial x_i} \rd x_1 \wedge \cdots \wedge \d x_n\) \(=\) \(\ds \int \cdots \int \paren {\int_{-\infty}^\infty \frac {\partial f_i} {\partial x_i} } \hat {\d x}_i\) Fubini's Theorem
\(\ds \) \(=\) \(\ds 0\)


Let $i = 1$.

Then:

\(\ds \int_{\mathbb H^n} \frac {\partial f_1} {\partial x_1} \d x_1 \wedge \cdots \wedge \d x_n\) \(=\) \(\ds \int \cdots \int \paren {\int_{-\infty}^0 \frac {\partial f_i} {\partial x_i} } \hat {\d x}_i\) 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

Let $\omega$ be a smooth $\paren {n - 1}$-form with compact support on a smooth $n$-dimensional oriented manifold $X$.

Choose a finite family of relatively compact charts $V_1, \ldots, V_k$ on $X$ such that:

$\ds \map \supp \omega \subseteq \bigcup_{i \mathop = 1}^k V_i$

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, $\map \supp {\omega_i} \subset V_i$ by definition.


This needs considerable tedious hard slog to complete it.
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Let the boundary of $X$ be $\partial X$.

Therefore, by the special case above, Stokes' theorem holds for each $\omega_i$.

Hence 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.


Sources

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