Local Gauss–Bonnet from a moving frame

Try

For a north cap of the unit sphere bounded at colatitude θ, the area is 2π(1−cos θ). Since K=1, this is its integrated Gaussian curvature. The latitude boundary contributes ∫k_g ds=2πcos θ. Their sum is 2π.

This exact calculation suggests a general disk formula, but the proof requires a moving frame and Stokes’ theorem. We will prove it for a simple piecewise smooth disk region contained in one oriented chart, with corner angles included.

Predict

Predict what happens for a geodesic triangle. Its smooth sides have k_g=0, leaving only the exterior angles and interior curvature. The resulting identity will supply the angle-excess formula previously used as a theorem dependency in the axiomatic geometry course.

Predict why a smooth frame extending over the disk matters. A frame with a singularity inside can contribute additional winding, so its boundary angle cannot be treated as a globally trivial reference.

Experiment

Use a positively oriented orthonormal frame e₁,e₂ and define its connection form by ∇e₁=ωe₂.

Quantity	Formula with the stated convention
Curvature two-form	dω=−K dA
Tangent angle T=cos α e₁+sin α e₂	k_g=α′+ω(T)
Disk boundary turning, including corners	2π

Check the sphere frame away from the poles: ω=cos θ dφ gives dω=−sin θ dθ∧dφ, agreeing with K=1. That frame does not extend over the north pole, so it cannot be used there without accounting for its frame singularity.

Understand

First derive the structure equation. Expanding R(X,Y)e₁ with ∇e₁=ωe₂ and ∇e₂=−ωe₁, the coefficient of e₂ is Xω(Y)−Yω(X)−ω([X,Y])=dω(X,Y). The e₁ terms cancel. On the oriented orthonormal pair e₁,e₂, our curvature convention gives ⟨R(e₁,e₂)e₁,e₂⟩=−K. Therefore dω=−K dA.

Along a unit-speed boundary write T=cos α e₁+sin α e₂. Covariant differentiation gives coefficient α′+ω(T) in direction N×T, so k_g=α′+ω(T).

The total change of α along the smooth arcs plus its signed corner jumps is 2π. This uses the planar turning theorem and the fact that the smooth oriented frame extends across the disk: its boundary winding relative to a chart frame is zero. Integrating the previous identity gives ∫∂D k_g ds+Σ exterior turns=2π+∮∂Dω. Stokes’ theorem and dω=−K dA now yield ∫D KdA+∫∂D k_g ds+Σ exterior turns=2π. This completes the local proof with the turning theorem and Stokes identified as inputs.

Apply

For a geodesic triangle with interior angles A,B,C, the exterior turns sum to 3π−(A+B+C). Hence ∫K dA=A+B+C−π. Constant curvature then converts angle excess or defect to area. No finite triangle plot was used to establish the identity.

Repair “The right side is 2π for every region.” An annulus has two boundary components; their oriented turning contributions cancel in the flat case. The general formula uses Euler characteristic, which the next lesson derives by summing disk pieces.

For optional depth, the frame proof is local but coordinate-independent. It does not require principal directions, which may fail to vary smoothly at umbilics.

Continue

State the region topology, orientation, smoothness and corner conventions before applying the formula. Continue to global Gauss–Bonnet to turn local disk identities into a theorem for compact surfaces.