General Stokes' Theorem

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

We may suppose that $V$ is relatively compact.

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

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

We have, in the coordinates $x$,

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

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:

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

Moreover:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle d \omega\) \(=\) \(\displaystyle \) \(\displaystyle \sum_{i \mathop = 1}^n df_i \wedge \hat{d x}_i\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \sum_{i \mathop = 1}^n \frac{\partial f}{\partial x_i} dx_i \wedge \hat{d x}_i\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \left({\sum_{i \mathop = 1}^n \frac {\partial f} {\partial x_i} }\right) d x_1 \wedge \cdots \wedge dx_n\) \(\displaystyle \) \(\displaystyle \)                    

so that:

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


If $i > 1$:

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


For $i = 1$:

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

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

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

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

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

$\blacksquare$

Also see


Source of Name

This entry was named for George Stokes.