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:

$\displaystyle \int_{\partial X} \omega = \int_X \rd \omega$

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 $\operatorname{supp} \omega \subseteq V$.



We may suppose that $V$ is relatively compact.

Thus, by composing $x$ with a translation, we may suppose that:

$\displaystyle \map x V \subseteq \mathbb H^n = \set {\tuple {x_1, \ldots, x_n} \in \R^n : x_1 < 0}$

We have, in the coordinates $x$:

$\displaystyle \omega = \sum_{i \mathop = 1}^n f_i \rd x_1 \wedge \cdots \wedge \hat {\d x}_i \wedge \cdots \wedge \d x_n$

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:

$(1):\quad \displaystyle \int_{\partial \mathbb H^n} \omega = \int_{\partial \mathbb H^n} f_1 \hat {\d x}_1$




Moreover:

\(\displaystyle \d \omega\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \rd f_i \wedge \hat {\d x}_i\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \frac {\partial f} {\partial x_i} \d x_i \wedge \hat {\d x}_i\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\sum_{i \mathop = 1}^n \frac {\partial f} {\partial x_i} } \rd x_1 \wedge \cdots \wedge \d x_n\)

so that:

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


If $i > 1$:

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


For $i = 1$:

\(\displaystyle \int_{\mathbb H^n} \frac {\partial f_1} {\partial x_1} \d x_1 \wedge \cdots \wedge \d x_n\) \(=\) \(\displaystyle \int \cdots \int \paren {\int_{-\infty}^0 \frac {\partial f_i} {\partial x_i} } \hat {\d x}_i\) Fubini's Theorem
\(\displaystyle \) \(=\) \(\displaystyle \int_{\partial \mathbb H^n} f_1 \hat {\d x}_1\)

So:

$\displaystyle \int_{\mathbb H^n} \rd \omega = \int_{\partial \mathbb H^n} f_1 \, \hat {\d x}_1$

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

$\displaystyle \operatorname {supp} \omega \subseteq \bigcup_{i \mathop = 1}^k V_i$

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:

\(\displaystyle \omega\) \(=\) \(\displaystyle \paren {\chi_1 + \cdots + \chi_k} \omega\)
\(\displaystyle \) \(=\) \(\displaystyle \omega_1 + \cdots + \omega_k\)

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

Therefore by the special case above, Stokes' theorem holds for each $\omega_i$, so we have:

$\displaystyle \int_X \rd \omega = \sum^k_{i \mathop = 1} \int_x \rd \omega_i = \sum^k_{i \mathop = 1} \int_{\partial X} \omega_i = \int_{\partial X} \omega$

$\blacksquare$

Also see


Source of Name

This entry was named for George Gabriel Stokes.