proof of general Stokes theorem

We divide the proof in several steps.

Step One.

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

$$\omega (x_1,\mathrm{\dots },x_n)=f(x_1,\mathrm{\dots },x_n)d{x}_1\wedge \mathrm{\cdots }\wedge \widehat{d{x}_j}\wedge \mathrm{\cdots }\wedge d{x}_n$$

(i.e. the term $d{x}_j$ is missing).Hence we have

$d\omega (x_1,\mathrm{\dots },x_n)$$=$$\left({\displaystyle \frac{\partial f}{\partial x_1}}d{x}_1+\mathrm{\cdots }+{\displaystyle \frac{\partial f}{\partial x_n}}d{x}_n\right)\wedge d{x}_1\wedge \mathrm{\cdots }\wedge \widehat{d{x}_j}\wedge \mathrm{\cdots }\wedge d{x}_n$

$=$${(-1)}^{j-1}{\displaystyle \frac{\partial f}{\partial x_j}}d{x}_1\wedge \mathrm{\cdots }\wedge d{x}_n$

and from the definition of integral on a manifold we get

$${\int }_M𝑑\omega ={\int }_0^1\mathrm{\cdots }{\int }_0^1{(-1)}^{j-1}\frac{\partial f}{\partial x_j}𝑑x_1\mathrm{\cdots }𝑑x_n.$$

From the fundamental theorem of Calculus we get

$${\int }_M𝑑\omega ={(-1)}^{j-1}{\int }_0^1\mathrm{\cdots }\widehat{{\int }_0^1}\mathrm{\cdots }{\int }_0^{1}f(x_1,\mathrm{\dots },1,\mathrm{\dots },x_n)-f(x_1,\mathrm{\dots },0,\mathrm{\dots },x_n)d{x}_1\mathrm{\cdots }\widehat{d{x}_j}\mathrm{\cdots }d{x}_n.$$

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

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 $j\ne 1$ then $i^{*}\omega =0$while if $j=1$ it holds

$$i^{*}\omega (x)=f(1,x_2,\mathrm{\dots },x_n)d{x}_2\wedge \mathrm{\dots }\wedge d{x}_n$$

and hence, as wanted

$${\int }_{\partial M}{i}^{*}\omega ={\int }_0^1\mathrm{\cdots }{\int }_0^{1}f(1,x_2,\mathrm{\dots },x_n)𝑑x_2\mathrm{\cdots }𝑑x_n.$$

Step Two.

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

$$\omega (x)=\sum _j{f}_j(x)d{x}_1\wedge \mathrm{\cdots }\wedge \widehat{d{x}_j}\wedge \mathrm{\cdots }\wedge d{x}_n$$

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 caseand end up with${\int }_M𝑑\omega =0$while, in fact, $\partial M=\mathrm{\varnothing }$.

Step Four.

Consider now the general case.

First of all we consider an oriented atlas $(U_i,{\varphi }_i)$ 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 $[0,+\mathrm{\infty })\times {ℝ}^{n-1}$ and a point $x\in U$ up to translations and rescaling it is possibleto 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 \sum _i}{\displaystyle {\int }_{U_i}}{\alpha }_i{\varphi }^{*}𝑑\omega ={\displaystyle \sum _i}{\displaystyle {\int }_{U_i}}{\alpha }_{i}d({\varphi }^{*}\omega )$

$=$${\displaystyle \sum _i}{\displaystyle {\int }_{U_i}}d({\alpha }_i\cdot {\varphi }^{*}\omega )-{\displaystyle \sum _i}{\displaystyle {\int }_{U_i}}(d{\alpha }_i)\wedge ({\varphi }^{*}\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

$${\int }_M𝑑\omega =\sum _i{\int }_{\partial U_i}{\alpha }_i\cdot {\varphi }^{*}\omega .$$

On the other hand, being$(\partial U_i,{\varphi }_{|\partial U_i})$ an oriented atlas for $\partial M$ and being $\alpha _i{}_{|\partial U_i}$ a partition of unity, we have

$${\int }_{\partial M}\omega =\sum _i{\int }_{\partial U_i}{\alpha }_i{\varphi }^{*}\omega$$

and the theorem is proved.