CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-30

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Metric independence of total Gaussian curvature

Statement

Assume the axiom of choice. Let $M$ be a closed oriented smooth surface and let $g_{0}, g_{1}$ be two smooth Riemannian metrics on $M$ . Then $\int_{M} K_{g_{0}} d A_{g_{0}} = \int_{M} K_{g_{1}} d A_{g_{1}} = 2 π χ (M),$ where $K_{g_{i}}$ is the Gaussian curvature of $g_{i}$ , $d A_{g_{i}}$ the area form of the orientation, and $χ (M)$ the Euler characteristic of Topological well-definedness of the surface Euler characteristic. For $M = \emptyset$ the identity reads $0 = 0$ . If $\partial M \neq \emptyset$ the metric-independent quantity is the full Gauss-Bonnet sum including the geodesic-curvature boundary term, so the closedness hypothesis is not decorative and the boundary case is not asserted here.

Facts & Assumptions

Given: A closed oriented smooth surface $M$ (possibly empty, possibly disconnected) and two smooth Riemannian metrics $g_{0}, g_{1}$ on $M$ .

[A1]

full AC is assumed; it is inherited exactly through the two invocations of the global Gauss-Bonnet theorem and is used nowhere else (The Axiom of Choice).

[F1]

For every closed oriented Riemannian surface $(M, g)$ , $\int_{M} K_{g} d A_{g} = 2 π χ (M)$ (Global Gauss-Bonnet for closed oriented surfaces).

[F2]

Any two finite face-to-face piecewise $C^{2}$ curvilinear triangulations of a compact smooth surface have the same $V - E + F$ ; geodesic triangulations with respect to possibly different smooth Riemannian metrics in particular agree, and the common value $χ (M)$ is independent of any Riemannian metric used to compute it (Topological well-definedness of the surface Euler characteristic).

Proof

technique · apply the global identity to each metric and compare the common right-hand side

1.1F1F2given

Applying [F1] to $(M, g_{0})$ gives $\int_{M} K_{g_{0}} d A_{g_{0}} = 2 π χ (M)$ , and applying it to $(M, g_{1})$ gives $\int_{M} K_{g_{1}} d A_{g_{1}} = 2 π χ (M)$ ; the symbol $χ (M)$ denotes in both cases the common value of [F2], which is independent of the metric.

2.1step 1.1algebra

Equating the two expressions of step 1.1 gives $\int_{M} K_{g_{0}} d A_{g_{0}} = \int_{M} K_{g_{1}} d A_{g_{1}} = 2 π χ (M)$ . When $M = \emptyset$ , [F2] gives $χ (\emptyset) = 0$ from the empty triangulation and both integrals are $0$ .

3.1A1step 2.1∎

No new choice is made: full AC entered only through the two applications of the global theorem, once for each metric.

Source locator

Lee, Riemannian Manifolds: An Introduction to Curvature, Chapter 9, Theorem 9.7 (printed pp. 167-172), and Datar, Lectures on Riemannian Geometry, Lecture 2, Theorem 2.2.4 (printed pp. 14-15), prove that the total curvature of a closed oriented Riemannian surface equals $2 π χ (M)$ ; since the right-hand side is the metric-independent invariant of Topological well-definedness of the surface Euler characteristic, the total curvature is the same for $g_{0}$ and $g_{1}$ . The two separate applications are kept explicit rather than treating the identity as automatic in the metric.

Depends on

Used by

A deformed sphere has the same total curvature Example

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John M. Lee, Riemannian Manifolds: An Introduction to Curvature (1997) (standard reference, not scraped)
Ved Datar, Lectures on Riemannian Geometry (standard reference, not scraped)