I have two questions from Thurston's paper [1].

In the paper [1], Thurston talks about classifying certain classes of triangulations of the sphere. Here a triangulation of a sphere a Topological sphere that is formed by glueing together equilateral triangles. Two triangulations are isomorphic if there is a metric space isometry between the two spheres that are also orientation preserving, maps vertices to vertices and edges to edges. Thurston studies about a special class of triangulations, $P(n; k_1, \cdots, k_r)$ where there are $r$ vertices (called cone points) with $6 - k_i$ number of triangles around them and the rest with exactly $6$ triangles.

Given $T \in P(n; k_1, \cdots, k_r)$, we'll define the developing map $D_T \colon \widetilde{T} \rightarrow \mathbb{C}$. Where $\widetilde{T}$ is the universal cover of $T - \{x_1, \cdots, x_r\}$ where $x_i$ is a singular point. The developing map maps one triangle to the triangle $(0, 1, \omega)$ in $\mathbb{C}$ where $\omega$ is the 6th root of unity; the rest of the map is an analytic continuation.

Here's the part that is hard to understand (pp. 8/ pp. 518):

"A particularly interesting phenomenon happens when the number of triangles around a singular vertex is $1$, $2$ or $3$. Consider the component $N_v$ of the inverse image in $\widetilde{T}$ of a small neighborhood of any such vertex $v$. It develops into the vicinity of some vertex $w$ in Eis (Eis is the set $a + b\omega$ for $a, b \in \mathbb{Z}$ and $\omega$ the 6th root of unity) In these cases, the number of triangles around $v$ is a divisor of $6$, so the developing map repeats itself when it first wraps around the vertex $w$, along a path in $\widetilde{T}$ which maps to a curve in $T$ wrapping respectively $6, 3$ or $2$ times around $v$. Therefore the developing map is defined from a smaller covering of $T$ minus its singular vertices which can be obtained as a certain quotient space $S(T)$ of $\widetilde{T}$. In $S(T)$, each component of the preimage of a small neighborhood of $v$ only intersects in six triangles. In fact $S(T)$ is isomorphic to $\mathbb{C}$. Therefore $T$ is a quotient space of a discrete group $\Gamma(T)$ acting on $\mathbb{C}$ such that only elements of Eis are fixed points of elements of $\Gamma(T)$

For $P(n; 4, 4, 4)$ or $P(n; 3, 4, 5)$, the group $\Gamma(T)$ is a triangle group"


  1. How does Thurston go from "there exists a smaller cover" to "$T$ is a quotient of a discrete group"? Thurston is claiming that the space $P(n; k_1, \cdots, k_r)$ can be thought of as a quotient of $\mathbb{C}$.

  2. For the specific examples with $3$ cone points, why is the group a triangle group?

[1] William P. Thurston, Shapes of polyhedra and triangulations of the sphere. https://arxiv.org/abs/math/9801088


1 Answer 1


Pass to the universal cover that branches 6, 3 or 2 times around the singular points --- you get a plane with an isometric action of the group of deck transformations.

  • $\begingroup$ Why does deck transformations correspond to different triangulations and vice versa? $\endgroup$
    – hrkrshnn
    Aug 22, 2019 at 11:50
  • $\begingroup$ @jdoicj they do not, or I do not understand your question. $\endgroup$ Aug 23, 2019 at 15:03
  • $\begingroup$ My comment doesn't make sense. I think I get it now: If $\Gamma$ is the group of deck transformations, then every element of $\mathbb{C}/\Gamma$ corresponds to a unique triangulation. This correspondence is also surjective. Thus $\mathbb C/\Gamma$ is the space of triangulations. The triangle group part follows from the fact that the fundamental group of a thrice punctured sphere is a triangle group. $\endgroup$
    – hrkrshnn
    Aug 24, 2019 at 14:47

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