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Homeproof of general Stokes theorem

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# proof of general Stokes theorem

We divide the proof in several steps.

*Step One.*

Suppose $M=(0,1]\times(0,1)^{{n-1}}$ and

$\omega(x_{1},\ldots,x_{n})=f(x_{1},\ldots,x_{n})\,dx_{1}\wedge\cdots\wedge% \widehat{dx_{j}}\wedge\cdots\wedge dx_{n}$ |

(i.e. the term $dx_{j}$ is missing). Hence we have

$\displaystyle d\omega(x_{1},\ldots,x_{n})$ | $\displaystyle=$ | $\displaystyle\left(\frac{\partial f}{\partial x_{1}}dx_{1}+\cdots+\frac{% \partial f}{\partial x_{n}}dx_{n}\right)\wedge dx_{1}\wedge\cdots\wedge% \widehat{dx_{j}}\wedge\cdots\wedge dx_{n}$ | ||

$\displaystyle=$ | $\displaystyle(-1)^{{j-1}}\frac{\partial f}{\partial x_{j}}dx_{1}\wedge\cdots% \wedge dx_{n}$ |

and from the definition of integral on a manifold we get

$\int_{M}d\omega=\int_{0}^{1}\cdots\int_{0}^{1}(-1)^{{j-1}}\frac{\partial f}{% \partial x_{j}}dx_{1}\cdots dx_{n}.$ |

From the fundamental theorem of Calculus we get

$\int_{M}d\omega=(-1)^{{j-1}}\int_{0}^{1}\cdots\widehat{\int_{0}^{1}}\cdots\int% _{0}^{1}f(x_{1},\ldots,1,\ldots,x_{n})-f(x_{1},\ldots,0,\ldots,x_{n})dx_{1}% \cdots\widehat{dx_{j}}\cdots dx_{n}.$ |

Since $\omega$ and hence $f$ have compact support in $M$ we obtain

$\int_{M}d\omega=\left\{\begin{array}[]{lcl}\int_{0}^{1}\cdots\int_{0}^{1}f(1,x% _{2},\ldots,x_{n})dx_{2}\cdots dx_{n}&\text{if}&j=1\\ \\ 0&\text{if}&j>1.\end{array}\right.$ |

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

$i^{*}\omega(x)=f(1,x_{2},\ldots,x_{n})dx_{2}\wedge\ldots\wedge dx_{n}$ |

and hence, as wanted

$\int_{{\partial M}}i^{*}\omega=\int_{0}^{1}\cdots\int_{0}^{1}f(1,x_{2},\ldots,% x_{n})dx_{2}\cdots dx_{n}.$ |

*Step Two.*

Suppose now that $M=(0,1]\times(0,1)^{{n-1}}$ and let $\omega$ be any differential form. We can always write

$\omega(x)=\sum_{j}f_{j}(x)dx_{1}\wedge\cdots\wedge\widehat{dx_{j}}\wedge\cdots% \wedge dx_{n}$ |

and by the additivity of the integral we can reduce ourself to the previous case.

*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, $\partial M=\emptyset$.

*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

$\displaystyle\int_{{M}}d\omega$ | $\displaystyle=$ | $\displaystyle\sum_{i}\int_{{U_{i}}}\alpha_{i}\phi^{*}d\omega=\sum_{i}\int_{{U_% {i}}}\alpha_{i}d(\phi^{*}\omega)$ | ||

$\displaystyle=$ | $\displaystyle\sum_{i}\int_{{U_{i}}}d(\alpha_{i}\cdot\phi^{*}\omega)-\sum_{i}% \int_{{U_{i}}}(d\alpha_{i})\wedge(\phi^{*}\omega).$ |

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.

## Mathematics Subject Classification

58C35*no label found*

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