Past exam of the geometry course of the University of Oxford / 2008 / A / 4

For a positively oriented curvilinear polygon bounding a disk $D$ in one coordinate patch, with exterior turning angles ${\theta }_i$, the Gauss-Bonnet theorem statesFor a closed compact surface the global formula isTriangulate the surface with each triangle contained in a coordinate patch. For each face, writing its exterior angles as $\pi$ minus its interior angles gives $\int K+\int {\kappa }_g=\sum \text{interior angles}-\pi$. On summing, the two contributions on each internal edge cancel, and all angles at each vertex sum to $2\pi$. Thus ${\int }_{S}K=2\pi V-\pi F$. Since $3F=2E$, this equals $2\pi (V-E+F)$, as required. For a nonorientable surface apply the orientable version to its orientation double cover: both the curvature integral and Euler characteristic double. With boundary, the same summation retains the boundary geodesic-curvature integral and the exterior angles at boundary corners, giving the corresponding global boundary formula with right side $2\pi \chi (S)$.

For the sphere, choose the inward normal so that the second fundamental form is $I/a$. Both principal curvatures are $1/a$, giving $K=1/a^2$; the outward choice reverses both signs and gives the same product.

The octant has area $4\pi a^2/8=\pi a^2/2$, so its curvature integral is $\pi /2$. Its sides are great-circle geodesics, hence have zero geodesic curvature. Each interior angle is $\pi /2$, so each exterior angle is also $\pi /2$. Thereforeverifying the polygon formula directly.

If two distinct geodesic arcs enclosed a disk, regard its boundary as a curvilinear two-sided polygon, with interior angles $\alpha ,\beta$. These are positive; zero angle would make the arcs tangent and local geodesic uniqueness would make them coincide. The geodesic-curvature terms vanish, so Gauss-Bonnet givesThis contradicts $K\le 0$. Thus no such disk-bounding geodesic bigon exists. The disk hypothesis matters because it fixes the Euler characteristic on the right side.