Stokes theorem.

Let $M$ be a manifold of dimension $n$ with boundary $\partial M$ and let $\omega$ be an $n-1$ form on $M$ with compact support. We want to prove

Theorem 6.1 (Stoke's theorem)   Let $M$ be an oriented manifold with boundary of dimension $n$ and let $\omega$ be a differential form of degree $n-1$ with compact support then

$$\int_M d\omega = \int_{\partial M} \omega.$$

Proof. We cover $M$ with a covering by co-ordinate charts $(U_\alpha, \psi_\alpha)$ and choose a partition of unity $\phi_\alpha$ subordinate to this cover. Notice that because $\sum_\alpha \phi_\alpha =1$ we have $\sum_\alpha d\phi_\alpha = 0$ and hence

$$\int_M d\omega = \sum_\alpha \int_M \phi_a d\omega$$

$$= \sum_\alpha \int_M d(\phi_a \omega)$$

$$= \sum_\alpha \int_{U_\alpha } d(\phi_a \omega)$$

$$= \sum_\alpha \int_{\psi_\alpha (U_\alpha )} (\psi^{-1})^* d(\phi_a \omega)$$

$$= \sum_\alpha \int_{\psi_\alpha (U_\alpha )} d((\psi^{-1})^* (\phi_a \omega) )$$

and

$$\int_{\partial M} \omega =\sum_\alpha \int_{\partial M} \phi_\alpha \omega$$

$$= \sum_\alpha \int_{\partial U_\alpha } \phi_\alpha \omega$$

$$= \sum_\alpha \int_{\partial \psi_\alpha (U_\alpha )} (\psi^{-1})^*(\phi_\alpha \omega).$$

So it suffices prove that

$$\int_{\psi_\alpha (U_\alpha )} d((\psi^{-1})^* (\phi_a \omega) ) ... ...lpha \int_{\partial \psi_\alpha (U_\alpha )} (\psi^{-1})^*(\phi_\alpha \omega)$$

or equivalently Stoke's theorem for differential forms with compact support on $\mathbb{R}^n_+$. Let us assume then that $\omega$ is a differential form on $U$, where $U$ is of the form $U = \mathbb{R}^n_+ \cap V$ for $V$ open in $\mathbb{R}^n$. As as $\omega$ is has compact support it has bounded support so there is an $R > 0$ such that if $\vert x^i\vert > R$ for all $i = 1, \dots, n$ then $\omega(x) = 0$. Write $x = (t, y)$ for $y \in \mathbb{R}^{n-1}$ and $\omega = f dt\wedge dy^1 \dots dy^{n-1}$ so we have

$$\int_U d\omega = \int_{U} \frac{\partial f}{\partial t} dt \wedge dy^{1} \dots d... ...} \frac{\partial f}{\partial x^{i}} dy^{i}\wedge dt \wedge dy^{1} \dots dy^{n}.$$

$$= \int_{U} \frac{\partial f}{\partial t} dt \wedge dy^{1} \dots d... ...nt_{U} \sum_{i=1}^{n-1} [f(t,y^{1}, \dots, y^{i-1}, R, y^{i+1}, \dots, y^{n-1})$$

$$\quad - \dots f(t,y^{1}, \dots, y^{i-1}, -R, y^{i+1}, \dots, y^{n-1})] dt \wedge dy^{1} \dots dy^{n}.$$

$$=\int_{U} \frac{\partial f}{\partial t} dt \wedge dy^{1} \dots dy^{n}$$

$$=\int_{\partial U}f(0, y^{1},\dots, y^{n-1}) dy^{1}\dots dy^{n}$$

$$= \int_{\partial U} \omega$$

$\qedsymbol$