General Stokes' Theorem

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Theorem

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

Let the boundary of $X$ be $\partial X$.


Then:

$\displaystyle \int_{\partial X} \omega = \int_X \mathrm d \omega$

where $\mathrm d \omega$ is the exterior derivative of $\omega$.


Proof

A Special Case

First we suppose that there is a chart $x = \left({x_1, \ldots, x_n}\right) : V \subseteq X \to \R^n$ 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:

$\displaystyle x \left({V}\right) \subseteq \mathbb H^n = \left\{ {\left({x_1, \ldots, x_n}\right) \in \R^n : x_1 < 0}\right\}$

We have, in the coordinates $x$:

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

The forms $\hat{\mathrm d x}_i := \mathrm d x_1 \wedge \cdots \wedge \hat{\mathrm d x}_i \wedge \cdots \wedge \mathrm d x_n$ vanish on the tangent space to $\mathbb H^n$ for $i > 1$, so we have:

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




Moreover:

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

so that:

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


If $i > 1$:

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


For $i = 1$:

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

So:

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

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 $\left\{ {V_1, \ldots, V_k}\right\}$.

Put $\omega_i = \chi_i \omega$.

Then we have:

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

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:

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

$\blacksquare$

Also see


Source of Name

This entry was named for George Stokes.