Past exam of the geometry of surfaces course of the University of Oxford 2023 1 c Solution 2026-10-25
Write for the internal angle at , using the numbering in part (b), and use the hyperbolic plane with Gaussian curvature . The necessary and sufficient conditions are equal lengths for each pair of equally labelled polygon sides, and total angle at each vertex equivalence class. Explicitly, the two angle conditions areNecessity follows because the identification must preserve arc length along each seam, and a smooth interior point of a Riemannian surface has total angle . For sufficiency, equally long hyperbolic geodesic segments can be identified by isometries. Across their interiors, two half-disc neighbourhoods join to a hyperbolic disc. At an identified vertex, the incident sectors form a cyclic disc with total angle , so no cone singularity of a surface metric remains. These local isometries therefore extend the Riemannian metric smoothly everywhere. This is the smooth hyperbolic polygon gluing criterion.
The ten internal angles sum to . The area of a hyperbolic geodesic polygon consequently givesEquivalently, the Gauss-Bonnet theorem gives .
Past exam of the mathematics course of the University of Cambridge 2023 ib Paper 3 2E Solution Created 2026-09-23 Updated 2026-10-14
In the Poincaré half-plane model, the Riemannian metricassigns to a smooth curve the lengthThe metric is conformal to the Euclidean metric, so hyperbolic and Euclidean angles agree. Its area element is
The two geodesics from and to infinity are the vertical lines and , while the third side is the unit semicircle. The triangle lies above that semicircle, so its area isIts interior angles are , , and zero at the ideal vertex, and henceTriangulating a geodesic polygon with sides into triangles givesthe area of a hyperbolic geodesic polygon and the polygonal Gauss-Bonnet theorem.
Smooth hyperbolic polygon gluing criterion 2026-10-25
Suppose a finite simple polygon in the curvature-minus-one hyperbolic plane has hyperbolic geodesic sides paired to produce a closed topological surface. Its metric descends to a smooth hyperbolic metric exactly when paired polygon sides have equal lengths and the sum of internal angles in each vertex equivalence class is . Equal lengths allow isometric gluing at open edges; at vertices the sectors assemble into a hyperbolic disc precisely when their total angle is . Other positive total angles produce a cone singularity of a surface metric. The area of a hyperbolic geodesic polygon and the Gauss-Bonnet theorem then give area for the closed surface.