Classifying Surface Given a Triangulation.

Question

So I have a compact surface $M$ without boundary with a triangulation such that $q$ edges meet at each vertex. I want to give possibilities for classifying $M$ for $3 \leq q \leq 6,$ and specify the number of vertices if possible.

My first attempt is to look at the Euler characteristic, $\chi(M) = V - E + F$. Because each vertex is part of $q$ edges and each edge has two vertices in it, we have $E = \frac{q}{2}V$. Similarly, because each face has three edges and each edge is part of two faces, we have $F = \frac{3}{2}E = \frac{3q}{4}V$. Therefore, the Euler characteristic is $\chi(M) = \frac{4 + q}{4} V.$ Because $\chi(M) \leq 2$, this only makes sense for $q = 4$ and $V = 1$. Have I made some very silly mistake somewhere? I have done the computation multiple times to check my work but can't seem to find anything.

Answer 1

The simplest triangulation to check for these is the tetrahedron, which has $q=3$, and $V=4$, $E=6$, $F=4$. It's then clear that while $ qV=2E $, the other relationship you posit cannot be correct: if there are $3$ edges for every face, and this has counted each edge twice, we should have $ 2E = 3F $, rather than vice versa as you have. We then have $$ \chi(M) = V - \frac{q}{2}V + \frac{2}{3} \frac{q}{2}V = \left( 1-\frac{q}{6} \right) V. $$ Checking with the tetrahedron gives $(1-3/6)4 = 2$, so it's believable, at least!