Stokes' Theorem (Proof)

The proof of Stokes' theorem on manifolds reduces the global statement to the local version in $\mathbb{R}^n$ via partitions of unity, then verifies the local case by direct computation.

Statement

Theorem4.3Stokes' theorem

Let $M$ be a compact oriented $n$ -manifold with boundary, and let $\omega \in \Omega^{n-1}(M)$ . Then

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

Proof

Step 1: Reduction to local charts. Let $\{(U_\alpha, \varphi_\alpha)\}$ be a finite atlas (finite since $M$ is compact) and $\{\rho_\alpha\}$ a subordinate partition of unity. Write $\omega = \sum_\alpha \rho_\alpha \omega$ . Then

\int_M d\omega = \sum_\alpha \int_M d(\rho_\alpha \omega), \quad \int_{\partial M} \omega = \sum_\alpha \int_{\partial M} \rho_\alpha \omega.

It suffices to prove the theorem for each $\rho_\alpha \omega$ , which is compactly supported in $U_\alpha$ .

Step 2: Local case — interior chart. If $U_\alpha$ does not meet $\partial M$ , then $\varphi_\alpha(U_\alpha) \subset \mathbb{R}^n$ . Write $\rho_\alpha \omega = \sum_i f_i \, dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^n$ in local coordinates. Then

d(\rho_\alpha \omega) = \sum_i (-1)^{i-1} \frac{\partial f_i}{\partial x^i} dx^1 \wedge \cdots \wedge dx^n.

Each integral $\int_{\mathbb{R}^n} \frac{\partial f_i}{\partial x^i} dx^1 \cdots dx^n = 0$ by the fundamental theorem of calculus (since $f_i$ has compact support, it vanishes at $\pm \infty$ ). Also $\int_{\partial M} \rho_\alpha \omega = 0$ since $\operatorname{supp}(\rho_\alpha \omega) \cap \partial M = \emptyset$ .

Step 3: Local case — boundary chart. If $U_\alpha$ meets $\partial M$ , the chart maps to $\mathbb{H}^n = \{x^n \geq 0\}$ . Similarly, for $i < n$ :

\int_{\mathbb{H}^n} \frac{\partial f_i}{\partial x^i} dx^1 \cdots dx^n = 0

by integrating in $x^i$ first (compact support gives vanishing boundary terms). For $i = n$ :

\int_{\mathbb{H}^n} \frac{\partial f_n}{\partial x^n} dx^1 \cdots dx^n = \int_{\mathbb{R}^{n-1}} \left[ f_n \Big|_{x^n=0}^{x^n=\infty} \right] dx^1 \cdots dx^{n-1} = -\int_{\mathbb{R}^{n-1}} f_n(x', 0) dx^1 \cdots dx^{n-1}.

The boundary integral is $\int_{\partial \mathbb{H}^n} \rho_\alpha \omega = (-1)^{n-1} \int_{\mathbb{R}^{n-1}} f_n(x', 0) dx^1 \cdots dx^{n-1}$ (accounting for the induced orientation). With the sign $(-1)^{n-1}$ from $d\omega$ , the terms match.

Step 4: Summing. Summing over all $\alpha$ : $\int_M d\omega = \int_{\partial M} \omega$ . $\blacksquare$

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Remarks

RemarkRole of orientation

The proof crucially uses the orientation: the induced boundary orientation determines the sign convention. If $M$ is not orientable, the theorem fails for forms but holds for densities (with the appropriate formulation).

RemarkNon-compact manifolds

For non-compact manifolds, Stokes' theorem still holds provided $\omega$ has compact support. More generally, it holds if $\omega$ and $d\omega$ are integrable with appropriate decay conditions at infinity.