Proof of Well-Definedness of Integration

The integral of a differential form on an oriented manifold must be shown to be independent of the choice of atlas and partition of unity. This fundamental result ensures that integration is a geometric invariant.

Statement

Theorem4.1Well-definedness of the integral

Let $M$ be an oriented smooth $n$ -manifold, $\omega$ a compactly supported $n$ -form on $M$ , $\{(U_\alpha, \varphi_\alpha)\}$ and $\{(V_\beta, \psi_\beta)\}$ two oriented atlases, and $\{\rho_\alpha\}$ , $\{\sigma_\beta\}$ subordinate partitions of unity. Then

\sum_\alpha \int_{\varphi_\alpha(U_\alpha)} (\varphi_\alpha^{-1})^*(\rho_\alpha \omega) = \sum_\beta \int_{\psi_\beta(V_\beta)} (\psi_\beta^{-1})^*(\sigma_\beta \omega).

Proof

Step 1: Double partition. Since $\sum_\alpha \rho_\alpha = 1$ and $\sum_\beta \sigma_\beta = 1$ , we have

\rho_\alpha \omega = \sum_\beta \rho_\alpha \sigma_\beta \omega, \quad \sigma_\beta \omega = \sum_\alpha \rho_\alpha \sigma_\beta \omega.

Thus both sums equal $\sum_{\alpha, \beta} \int_M \rho_\alpha \sigma_\beta \omega$ (with appropriate chart interpretation). It suffices to show that for a single term $\eta = \rho_\alpha \sigma_\beta \omega$ (compactly supported in $U_\alpha \cap V_\beta$ ), the two chart integrals agree.

Step 2: Change of variables. Let $\eta$ be an $n$ -form compactly supported in the overlap $U_\alpha \cap V_\beta$ . In the chart $\varphi_\alpha$ :

\int_{\varphi_\alpha(U_\alpha \cap V_\beta)} (\varphi_\alpha^{-1})^*\eta.

In the chart $\psi_\beta$ :

\int_{\psi_\beta(U_\alpha \cap V_\beta)} (\psi_\beta^{-1})^*\eta.

The transition map $\tau = \psi_\beta \circ \varphi_\alpha^{-1}$ relates these. Write $(\varphi_\alpha^{-1})^*\eta = f(x) dx^1 \wedge \cdots \wedge dx^n$ in coordinates $x$ . Under the change of variables $y = \tau(x)$ :

(\psi_\beta^{-1})^*\eta = f(\tau^{-1}(y)) \det(D\tau^{-1}(y)) \, dy^1 \wedge \cdots \wedge dy^n.

Step 3: Orientation consistency. Since both atlases are oriented-compatible, the transition map $\tau$ is orientation-preserving, meaning $\det(D\tau) > 0$ everywhere. By the standard change of variables formula for Lebesgue integrals:

\int f(x) dx^1 \cdots dx^n = \int f(\tau^{-1}(y)) |\det(D\tau^{-1}(y))| dy^1 \cdots dy^n.

Since $\det(D\tau^{-1}) > 0$ , we have $|\det(D\tau^{-1})| = \det(D\tau^{-1})$ , so the two expressions agree. $\blacksquare$

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Consequences

RemarkRole of orientation

If the transition maps could reverse orientation (negative Jacobian determinant), the absolute value in the change of variables formula would differ from the signed determinant, and the integral would depend on the atlas. This is precisely why orientability is necessary for integrating forms. On non-orientable manifolds, one integrates densities instead, which transform by $|\det J|$ .

ExampleFailure of well-definedness without orientation

Consider the Mobius band $M$ with a "top form" $\omega$ defined on charts. Following $\omega$ around the band, the form returns with opposite sign. Any global "volume form" would need to vanish somewhere, so $M$ has no consistent orientation. Attempting to define $\int_M \omega$ via charts gives contradictory signs in overlapping regions.

RemarkExtension to manifolds with boundary

The same proof shows well-definedness of integration on oriented manifolds with boundary, using half-space charts $\mathbb{H}^n$ near the boundary. The transition maps between half-space charts preserve orientation and respect the boundary, ensuring the integral remains well-defined.