ProofOfGeneralStokesTheorem 2003-06-16 10:56:41 2003-06-16 13:17:40 Proof general Stokes theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here We divide the proof in several steps. \emph{Step One.} Suppose $M=(0,1]\times (0,1)^{n-1}$ and \[ \omega(x_1,\ldots,x_n)=f(x_1,\ldots,x_n)\, d x_1\wedge\cdots\wedge \widehat{d x_j}\wedge \cdots \wedge d x_n \] (i.e. the term $dx_j$ is missing). Hence we have \begin{eqnarray*} d\omega(x_1,\ldots,x_n) &=& \left(\frac{\partial f}{\partial x_1} d x_1+\cdots +\frac{\partial f}{\partial x_n}d x_n\right)\wedge d x_1\wedge \cdots \wedge \widehat{d x_j}\wedge \cdots \wedge d x_n \\ &=& (-1)^{j-1} \frac{\partial f}{\partial x_j} d x_1\wedge \cdots \wedge d x_n \end{eqnarray*} and from the definition of integral on a manifold we get \[ \int_M d\omega = \int_0^1\cdots \int_0^1 (-1)^{j-1} \frac{\partial f}{\partial x_j} d x_1 \cdots d x_n. \] From the fundamental theorem of Calculus we get \[ \int_M d\omega =(-1)^{j-1} \int_0^1\cdots\widehat{\int_0^1}\cdots\int_0^1 f(x_1,\ldots,1,\ldots,x_n)-f(x_1,\ldots,0,\ldots,x_n) d x_1\cdots \widehat{ d x_j}\cdots d x_n. \] Since $\omega$ and hence $f$ have compact support in $M$ we obtain \[ \int_M d\omega=\left\{ \begin{array}{lcl} \int_0^1\cdots\int_0^1 f(1,x_2,\ldots,x_n) d x_2 \cdots \d x_n & \text{if} & j=1 \\\\ 0 &\text{if} & j>1 . \end{array}\right. \] On the other hand we notice that $\int_{\partial M}\omega$ is to be understood as $\int_{\partial M}i^* \omega$ where $i:\partial M\to M$ is the inclusion map. Hence it is trivial to verify that when $j\neq 1$ then $i^*\omega=0$ while if $j=1$ it holds \[ i^*\omega(x) = f(1,x_2,\ldots,x_n) d x_2\wedge\ldots\wedge d x_n \] and hence, as wanted \[ \int_{\partial M}i^*\omega = \int_0^1\cdots\int_0^1 f(x) d x_2 \cdots d x_n. \] \emph{Step Two.} Suppose now that $M=[0,1)\times (0,1)^{n-1}$ and let $\omega$ be any differential form. We can always write \[ \omega(x)=\sum_j f_j(x) d x_1 \wedge\cdots \wedge \widehat{ d x_j} \wedge \cdots \wedge d x_n \] and by the additivity of the integral we can reduce ourself to the previous case. \emph{Step Three.} When $M=(0,1)^n$ we could follow the proof as in the first case and end up with $\int_M d \omega = 0$ while, in fact, $\partial M=\emptyset$. \emph{Step Four.} Consider now the general case. First of all we consider an oriented atlas $(U_i,\phi_i)$ such that either $U_i$ is the cube $(0,1]\times (0,1)^{n-1}$ or $U_i=(0,1)^n$. This is always possible. In fact given any open set $U$ in $[0,+\infty)\times \mathbb R^{n-1}$ and a point $x\in U$ up to translations and rescaling it is possible to find a ``cubic'' neighbourhood of $x$ contained in $U$. Then consider a partition of unity $\alpha_i$ for this atlas. From the properties of the integral on manifolds we have \begin{eqnarray*} \int_{M} d \omega &=& \sum_i \int_{U_i} \alpha_i \phi^* d \omega = \sum_i \int_{U_i} \alpha_i d (\phi^* \omega) \\ &=& \sum_i \int_{U_i} d (\alpha_i\cdot \phi^*\omega) - \sum_i \int_{U_i} (d\alpha_i)\wedge(\phi^*\omega). \end{eqnarray*} The second integral in the last equality is zero since $\sum_i d\alpha_i= d \sum_i \alpha_i = 0$, while applying the previous steps to the first integral we have \[ \int_{M} d\omega = \sum_i \int_{\partial U_i}\alpha_i\cdot \phi^*\omega. \] On the other hand, being $(\partial U_i,\phi_{|\partial U_i})$ an oriented atlas for $\partial M$ and being ${\alpha_i}_{|\partial U_i}$ a partition of unity, we have \[ \int_{\partial M} \omega = \sum_i \int_{\partial U_i} \alpha_i \phi^* \omega \] and the theorem is proved.