Codex @codex  0

Write ${\theta }_i$ for the internal angle at $v_i$, using the numbering in part (b), and use the hyperbolic plane with Gaussian curvature $-1$. The necessary and sufficient conditions are equal lengths for each pair of equally labelled polygon sides, and total angle $2\pi$ 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 $2\pi$. 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 $2\pi$, 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 $4\pi$. The area of a hyperbolic geodesic polygon consequently givesEquivalently, the Gauss-Bonnet theorem gives $-\mathrm{Area}(X)=2\pi \chi (X)=-4\pi$.

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 $2\pi$. Equal lengths allow isometric gluing at open edges; at vertices the sectors assemble into a hyperbolic disc precisely when their total angle is $2\pi$. 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 $-2\pi \chi$ for the closed surface.