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I'm reading Thurston's article "Shapes of polyhedra and triangulations of the sphere." In the introduction he claims the following:

"${}^{(1)}$There are procedures to refine and modify any triangulation of a surface until every vertex has either 5, 6 or 7 triangles around it, or with more effort, ${}^{(2)}$so that there are only 5 or 6 triangles if the surface has positive Euler characteristic, only 6 triangles if the surface has zero Euler characteristic, or only 6 or 7 triangles if the surface has negative Euler characteristic..."

I divide the quote into two claims. I proved the first claim, but I'm stuck with the second one. There must be a global argument I'm not seeing. Does anyone know how to do it?

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  • $\begingroup$ Since Thurston does not define "refine" or "modify" we have to make up our own operations. What do you suggest? Even better - what considerations should guide our choices? $\endgroup$
    – Sam Nead
    Commented Mar 2, 2022 at 0:11

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Suppose that $S$ is a closed, connected surface with negative Euler characteristic. Suppose that $T$ is a triangulation of $S$.

Define "refine" to mean "replace each triangle by four triangles" (so that edge midpoints become vertices of valence six). This does not improve any of the vertices of "concentrated positive or negative curvature", but it does isolate them.

Suppose that there are no vertices of degree five or lower. Refine as above to isolate all vertices of degree greater than six. Suppose that the vertex $v$ has degree eight or higher. We "split" at $v$ - we choose two edges $e$ and $e'$ at $v$, cut $T$ along $e \cup e'$ and insert a pair of triangles. This increases the number of triangles by two, the number of vertices by one, and the number of edges by three. The new vertices $u$ and $u'$ have degree less than that of $v$. Also, two old vertices of degree six will now have degree seven. Repeat.

As a bit of motivation: the original triangulation gives a metric on $S$. Refinement has the effect of scaling the metric up, which "brings the curvature down". Cutting and inserting triangles disperses concentrated negative curvature (at a vertex) into adjacent (almost) flat regions.

In general, if there are vertices of degree less than six, we refine and then pair them with vertices of higher degree, and (carefully!) cancel curvature.

In the case of Euler characteristic zero, after pairing in this way, only vertices of degree six remain.

In the case of positive Euler characteristic, we have some vertices of degree five (either six or twelve) left at the end.

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    $\begingroup$ Never thought about it that way! Thank you, I feel I can complete the argument following your answer $\endgroup$
    – Joaquin Lema
    Commented Mar 2, 2022 at 1:33
  • $\begingroup$ In the end you must use a "combinatorial Gauss-Bonnet theorem" which relates the number of edges coming from a "non-flat vertex" (ie. a vertex of degree $\neq 6$) and the Euler characteristic. That is easy to prove by hand but, is that result also implied by the usual Gauss-Bonnet theorem? I think that if you modify a metric so that the curvature stays close to zero everywhere except near finitely many points you get the result, but I'm not sure. $\endgroup$
    – Joaquin Lema
    Commented Mar 2, 2022 at 17:52
  • $\begingroup$ If you declare each triangle to be an equilateral Euclidean triangle, then this induces a flat cone metric on the sphere. McMullen proved a Gauss-Bonnet theorem for cone manifolds which would apply in this situation. mathscinet.ams.org/mathscinet-getitem?mr=3619915 $\endgroup$
    – Ethan Dlugie
    Commented Oct 13, 2022 at 23:53

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