Past exam of the geometry course of the University of Oxford / 2011 / 1 / c

Count incidences of triangular faces with edges. Every triangle contributes three and every edge of a closed surface occurs in two triangles, hence $3F=2E$. Substituting $F=2E/3$ into the Euler characteristic gives

The simplicial intersection conditions prohibit loops and multiple edges between the same pair of vertices, so the one-skeleton is a simple graph andUsing the preceding identity gives $V^2-7V+6\chi \ge 0$. A closed surface has $\chi \le 2$ and a simplicial triangulation has $V\ge 4$. The smaller root is at most three, so this inequality forces $V$ to lie above the larger root:For completeness, fewer than four vertices could give at most one triangle under the intersection rules, which cannot form a closed surface.

Here the surface is the real projective plane, with Euler characteristic one. The bound gives $V\ge 6$. The following ten triangles on vertices $1,\dots ,6$ attain it:Take these as the maximal simplices of an abstract simplicial complex, identifying edges with the same vertex labels. Every one of the 15 pairs of vertices occurs as an edge in exactly two triangles. The links of the six vertices are single five-cycles; listed cyclically they areThus every point has a disk neighborhood, and the realization is a connected closed surface. Its Euler characteristic is $6-15+10=1$, so the classification of compact surfaces identifies it with the real projective plane. This is an explicit triangulation, proving $\boxed{V_{\min }=6}$.