General Stokes' Theorem

Theorem

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

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

Then:

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

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

Proof

A Special Case

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

We may suppose that $V$ is relatively compact.

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

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

We have, in the coordinates $x$,

$\displaystyle \omega = \sum_{i = 1}^n f_i dx_1\wedge \cdots \wedge \hat{dx}_i \wedge \cdots \wedge dx_n$

The forms $\hat{dx}_i := dx_1\wedge \cdots \wedge \hat{dx}_i \wedge \cdots \wedge dx_n$ vanish on the tangent space to $\mathbb H^n$ for $i > 1$, so we have

$(1)\qquad\displaystyle \int_{\partial \mathbb H^n} \omega = \int_{\partial \mathbb H^n} f_1\hat{dx}_1$

Moreover,

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle d\omega\)	\(=\)	\(\displaystyle \)	\(\displaystyle \sum_{i = 1}^n df_i \wedge \hat{dx}_i\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \sum_{i = 1}^n \frac{\partial f}{\partial x_i} dx_i \wedge \hat{dx}_i\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \left( \sum_{i = 1}^n \frac{\partial f}{\partial x_i}\right) dx_1 \wedge \cdots \wedge dx_n\)	\(\displaystyle \)	\(\displaystyle \)

so that

$\displaystyle \int_{\mathbb H^n}d\omega = \sum_{i = 1}^n \int_{\mathbb H^n}\frac{\partial f_i}{\partial x_i}dx_1 \wedge\cdots dx_n$

If $i > 1$,

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int_{\mathbb H^n}\frac{\partial f_i}{\partial x_i}dx_1 \wedge\cdots dx_n\)	\(=\)	\(\displaystyle \)	\(\displaystyle \int\cdots \int \left( \int_{-\infty}^\infty \frac{\partial f_i}{\partial x_i} \right) \hat{dx}_i\)	\(\displaystyle \)	\(\displaystyle \)	By Fubini's theorem
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)

For $i = 1$,

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int_{\mathbb H^n}\frac{\partial f_1}{\partial x_1}dx_1 \wedge\cdots dx_n\)	\(=\)	\(\displaystyle \)	\(\displaystyle \int\cdots \int \left( \int_{-\infty}^0 \frac{\partial f_i}{\partial x_i} \right) \hat{dx}_i\)	\(\displaystyle \)	\(\displaystyle \)	By Fubini's theorem
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \int_{\partial \mathbb H^n} f_1 \hat{dx}_1\)	\(\displaystyle \)	\(\displaystyle \)

So

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

Together with $(1)$, this establishes the result.

$\Box$

General Case

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

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

Choose a partition of unity $\chi_1,\dots,\chi_k$ with $\chi_1+\dots+\chi_k=1$, subordinate to the cover $\left\{ V_1,\ldots,V_k \right\}$.

Put $\omega_i = \chi_i \omega$.

Then we have

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

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

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

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

$\blacksquare$

Also see

• Implications of Stokes' Theorem

Source of Name

This entry was named for George Stokes.