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CorollaryStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30
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Angle-sum comparison for geodesic triangles

Statement

Assume the axiom of choice. Let (M,g,J) be an oriented Riemannian surface and let T⊆M be a positively oriented geodesic triangular disk of the kind considered in Gauss-Bonnet for a geodesic triangle, with interior angles α,β,γ∈(0,2π). Then

α+β+γ>π,α+β+γ=π,α+β+γ<π

according as the total Gaussian curvature ∫TK dA is positive, zero, or negative. The comparison concerns the curvature integral and not merely the sign of K at a single point.

Facts & Assumptions

Given: Full AC through the local disk Gauss–Bonnet supplier (The Axiom of Choice); A positively oriented geodesic triangular disk in a frameable neighbourhood with interior angles.

[F1]

For such a triangle ∫TK dA=α+β+γ−π (Gauss-Bonnet for a geodesic triangle).

Proof

technique · rearrange the exact triangle identity and compare signs
1.1F1algebra

Subtracting π from both sides of [F1] gives α+β+γ−π=∫TK dA.

2.1step 1.1algebra∎

By step 1.1, the real number α+β+γ−π is positive, zero or negative exactly when the real number ∫TK dA is positive, zero or negative; adding π to each comparison gives the three stated alternatives.

Source locator

Lee, Riemannian Manifolds: An Introduction to Curvature, Chapter 9, printed pp. 165-172, notes the sign interpretation of the local formula: the curvature integral measures the excess of the angle sum over π. Datar, Lectures on Riemannian Geometry, Lecture 2, printed pp. 10-13, records the same consequence. The rearrangement is immediate from the library theorem Gauss-Bonnet for a geodesic triangle.

Depends on

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Sources