Solution

For a geodesic triangle with interior angles

α, β, γ

\int_{T} K d A = α + β + γ - π,

because its geodesic sides have zero geodesic curvature.

Triangulate a closed oriented surface into

F

geodesic triangles, with

E

edges and

V

vertices. Summing the local formula, the angles around each vertex total

2 π

, so

\int_{S} K d A = 2 πV - π F .

Since every triangular face has three edges and every edge belongs to two faces,

3 F = 2 E

. Hence

\int_{S} K d A = 2 π (V - E + F) = 2 π χ (S),

which is the global Gauss-Bonnet theorem.

For the sphere

S_{r}

, the unit normal is

N (p) = p / r

. Its shape operator is, up to the conventional sign,

(1/ r) I

on each tangent plane. Both principal curvatures therefore have magnitude

1/ r

K = \frac{1}{r ^{2}} .

An octant has one eighth of the sphere's area:

area (T) = \frac{1}{8} (4 π r^{2}) = \frac{π r ^{2}}{2} .

Thus

\int_{T} K d A = π /2

. Its three great-circle sides meet at three right angles, so

α + β + γ - π = 3 \frac{π}{2} - π = \frac{π}{2} .

The two sides of the local Gauss-Bonnet formula agree directly.

Solved by gpt-5.6-sol high.

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