Minimum Number of Triangles for Triangulation of Closed Surfaces

Question

I am studying Topology, and recently I learned that every closed surface (2 dimensional manifold) can be triangulated. I understood triangulation as making polygons (with triangles) homeomorphic to the surface, and I started wondering the minimum case.

In the case of sphere, it is easy to know that at least four triangles are needed to triangulate, which will form a tetrahedron. There is a unique form of polygon with the minimum number of triangles (up to the structure of polygon? I don’t know how to explain).

In the case of torus, there is an easy way to triangulate with 18 triangles, and I reduced the number to 14. However I am not sure that it is the minimum.

In the case of projective plane or Klein bottle, I have no idea.

In short, I want to know the minimum numbers of triangles for triangulations of well-known sufaces (torus, projective plane, and Klein bottle), or the way to calculate them. Also I want to know the triangulation with the minimum number of triangles is unique for each surface.

Answer 1

7 vertices, 14 triangles is minimal for a torus.

Massey's intro to algebraic topology has a nice exercise in which you prove that any triangulation of a compact surface with $v$ vertices, $e$ edges, and $f$ faces, must satisfy:

1. $3 f = 2 e$

2. $e = 3 (v - \chi)$

3. $v\ge \frac{1}{2} \left(\sqrt{49-24 \chi }+7\right)$

where $\chi = v - e + f$ is the Euler characteristic. For the klein bottle $K$ we have $\chi = 0$, so this last tells us that $$ v \ge \frac12 (14) = 7 $$ with $e = 21$ and $3f = 2e = 42$ so $f = 14$. I leave it to you to attempt to find a 14-triangle triangulation of $K$. Note that Massey doesn't say that there is one, only that any triangulation has to have at LEAST 14 triangles.

For the projective plane, the Euler characteristic is $1$, so Massey's equations read \begin{align} 3f &= 2e \\ e &= 3(v-1) \\ v &\ge \frac12 \left(\sqrt{49 -24} + 7 \right) \\ &= \frac12 \left(5 + 7 \right) = 6 \end{align} Assuming the minimal case is possible (i.e., $v = 6$) we get that $e= 15$ and $f = 10$.

Once again I leave it to you to enjoy trying to find a 6-vertex triangulation of the projective plane (or decide, perhaps, that 7 is the smallest possible -- I'm not going to take away the fun of discovery).

The first two formulas for the exercise in Massey comes from nothing more that the definition of the Euler characteristic, the observation that every edge lies in 2 faces and every face has three edges, and some algebraic fiddling. I can't immediately reconstruct the source of the 3rd formula...but then again, I did that problemabout half a century ago, so I forgive myself this lapse.

Late-breaking edit

I worked out that third claim!

In a triangulation, there cannot be two edges joining the same pair of vertices. The number of vertex pairs is ${v \choose 2} = \frac{v(v-1)}{2}$, so we have $$ \frac{v(v-1)}{2} \ge e. $$ Now let's algebraically transform that: \begin{align} \frac{v(v-1)}{2} &\ge e. \\ v^2 - v -2e &\ge 0 \\ v^2 - v + 4e - 6e &\ge 0 \\ v^2 - v + 6f - 6e &\ge 0 & \text{ because $6f = 4e$ using fact 1 }\\ v^2 - 7v + 6f - 6e + 6v &\ge 0 & \text{ add and subtract $6v$ }\\ v^2 - 7v + 6\chi &\ge 0 \\ 4v^2 - 28v + 24\chi &\ge 0 & \text{ multiply both sides by 4}\\ 4v^2 - 28v &\ge -24\chi \\ 4v^2 - 28v + 49 &\ge 49-24\chi & \text{ add 49 to both sides}\\ (2v-7)^2 &\ge 49-24\chi \\ 2v-7 &\ge \sqrt{49-24\chi} \\ 2v &\ge 7 + \sqrt{49-24\chi} \\ v &\ge \frac12 \left( 7 + \sqrt{49-24\chi} \right) \end{align}

Of course, some of the steps here need sign-checking: you have to make sure the thing on the right is positive before taking a square root, for instance. But that's the gist of it. Also (as must be obvoius) I worked this out by starting from Massey's claim and simplifying it until I got to $v(v-1)/2 \ge e$, and then realized why that had to be true, and then wrote down everything in reverse order to make a proof.

Answer 2

Edit: I misread the accepted answer, and did not realise it had already attained the same result as below. However, as my method was entirely different, I leave it up.

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For the projective plane, we know that $3F=2E$, by counting edge-face incidences, so $F$ must be even.

If $F=2$ then $E=3F/2=3$ and $V=3+1-2=2$, so we would be embedding the projective plane in a line which is impossible.

If $F=4$ then $E=3F/2=6$ and $V=6+1-4=3$, so we would be embedding the projective plane in a triangle which is impossible (a triangular surface has no orientation reversing paths).

If $F=6$, then $E=3F/2=9$ and $V=E+1-F=4$. But then you would be embedding a projective plane in the surface of a tetrahedron, which is impossible (a tetrahedron's surface has no orientation reversing paths).

This leaves $F=8, E=12, V= 5$ as the next lowest possibility (as $F$ even). Suppose we had such a triangulation. Pick a vertex. Its neighbourhood must be a polygon, with distinct vertices around the boundary (if a vertex is repeated, then we would have at least four faces meeting at an edge, so we would not have a surface at all).

Thus removing one vertex (and all its edges and faces), removes a polygon from the surface, which leaves a Mobius strip which cannot embed on the surface of a tetrahedron. So after removing a vertex we must have more than $4$ vertices left. That is $V>5$, so $F=8$ is impossible.

Thus the minimal number of faces is $10$, as we attain $F=10, E=15, V=6$ by taking the projective icosahedron.