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

For a finite CW complex with $V$ vertices, $E$ edges and $F$ faces,This Euler characteristic is unchanged by subdividing an edge, which adds one vertex and one edge, or subdividing a face by an edge, which adds one edge and one face. Common refinements therefore give the same value for different subdivisions of the same surface.

Form the dual graph of the finite triangulation: its vertices are the triangles, and two vertices are joined when their triangles share an edge. Connectedness of the surface makes this graph connected. Choose a spanning tree and begin with one disjoint triangle. Attach the remaining triangles along the tree edges, one at a time, always joining a new triangle along one boundary edge. At every step the result is a topological disk, so the final union is a polygonal disk.

The edges not used for these attachments still occur twice on the boundary, once for each incident triangle. Identify each pair as in the original triangulation. The result recovers the original surface. This proves a compact connected closed surface is a polygon with its boundary edges identified in pairs. Closedness is the convention here; a boundary would leave some edges unpaired.

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 $\boxed{V_{\min }=6}$.