Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 3 / 14E (Geometry)

A geodesic triangulation of a closed smooth surface decomposes it into finitely many triangular discs whose edges are geodesic arcs and whose intersections are common faces. Its Euler number isThe Gauss-Bonnet theorem for a closed surface statesFor any triangulation of a closed surface, each triangular face has three edges and every edge belongs to two faces, so $3F=2E$. The sum of all vertex valencies is $2E$. Hence the average valency isThe torus has Euler characteristic $\chi =V-E+F=0$. Using $F=2E/3$ gives $V-E/3=0$, so $E=3V$ and

For the sphere, $\chi =2$, and the same incidence identity givesTherefore

Subdividing one face in the stated way adds one vertex, three edges, and two net faces. Starting from any spherical triangulation and repeating this operation makes $V\to \infty$ while preserving the sphere. Since every resulting average isthese averages approach six from below. Thus spherical triangulations can have average valency arbitrarily close to six.

If $K<0$ everywhere, compactness makes the integral negative, so $\chi (S)<0$. For every triangulation, $3F=2E$ again yields the general identitySince $V\ge 1$ and $\chi (S)<0$,This supplies bounds depending only on the surface.