TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

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Gauss-Bonnet for a geodesic triangle

Statement

Assume the axiom of choice. Let $(M, g, J)$ be an oriented Riemannian surface and let $T \subseteq M$ be a compact regular oriented disk region with exactly three vertices whose boundary is the cyclic concatenation of three regular $C^{2}$ embedded geodesic segments of the interior metric; at each vertex the incoming and outgoing one-sided unit tangents are not antipodal. Suppose that a neighbourhood of $T$ carries a smooth positive orthonormal frame and that $T$ is positively oriented as a disk region. If $α, β, γ \in (0, 2 π)$ are the interior sector angles at the three vertices, then

$\int_{T} K d A = α + β + γ - π .$

The boundary orientation is the outward-normal-first orientation of Regular oriented surface regions with corners, and no claim is made about the existence of such a triangle on a general surface.

Facts & Assumptions

Given: Full AC through the local disk Gauss–Bonnet supplier (The Axiom of Choice); An oriented Riemannian surface, a positively oriented compact geodesic triangular disk region with ordinary corners, contained in a frameable neighbourhood.

[F1]

For a positively oriented compact regular disk region with ordinary corners carrying a smooth positive orthonormal frame on a neighbourhood, $\int_{D} K d A + \int_{\partial D} k_{g} d s + \sum_{j} α_{j} = 2 π$ (Local Gauss-Bonnet for a frameable disk region).

[F2]

An affinely parametrized geodesic segment has $\nabla_{T} T = 0$ along its interior (Geodesic of an affine connection).

[F3]

The signed geodesic curvature is the scalar with covariant acceleration $A_{γ} = k_{g} J T$ , so $k_{g} = 0$ wherever $A_{γ} = 0$ (Signed geodesic curvature).

[F4]

At a positively oriented boundary corner with interior sector angle $β \in (0, 2 π)$ , the signed exterior angle is $α = π - β$ (Signed exterior angle at an ordinary corner).

Proof

technique · insert the vanishing geodesic curvatures and the exterior angle identities into the frameable disk formula

1.1F2F3given

Each of the three sides of $T$ admits a regular $C^{2}$ geodesic parametrization whose interior is affinely parametrized, so its covariant acceleration vanishes identically on the side by [F2]; therefore its signed geodesic curvature vanishes identically on that side by [F3].

1.2F1given

The hypotheses make $T$ a compact regular oriented disk region with ordinary corners in a frameable neighbourhood, so [F1] applies and gives $\int_{T} K d A + \int_{\partial T} k_{g} d s + α_{1} + α_{2} + α_{3} = 2 π$ , where the $α_{j}$ are the signed exterior angles at the three vertices.

1.3F4given

Each vertex has interior sector angle in $(0, 2 π)$ by the regular-region hypothesis, so [F4] identifies its exterior angle with $π$ minus the interior angle: writing the interior angles as $α, β, γ$ , the three exterior angles are $π - α$ , $π - β$ , $π - γ$ .

2.1F1step 1.1step 1.3algebra∎

The boundary integral in step 1.2 vanishes by step 1.1, so $2 π = \int_{T} K d A + (π - α) + (π - β) + (π - γ)$ and hence $\int_{T} K d A = α + β + γ - π$ .

Source locator

Lee, Riemannian Manifolds: An Introduction to Curvature, Chapter 9, Theorem 9.3, printed pp. 165-167, gives the local formula for a curved polygon, of which the geodesic triangle with vanishing boundary curvature is the special case computed here; Datar, Lectures on Riemannian Geometry, Lecture 2, Theorem 2.0.1, printed pp. 10-13, states the same computation. The identification of the corner jumps with $π$ minus the interior angles is the library definition Signed exterior angle at an ordinary corner, and the vanishing of $k_{g}$ along geodesics is the library definition Signed geodesic curvature together with Geodesic of an affine connection.

Depends on

Used by

Angle-sum comparison for geodesic triangles Corollary
Area defect of a hyperbolic geodesic triangle Example
Area excess of a spherical geodesic triangle Example
Geodesic triangles need not have Euclidean angle sum False statement

Dependency tree · two levels

26 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)