## Lecture 8: Friday, September 6, 2013

Over the next couple of lectures, we will generalize cyclic groups $Z_n$ and dihedral groups $D_n$. Given positive integers $m$, $n$, and $k$, formally define the abstract group
$D(m,n,k) = \left \langle \gamma_0, \, \gamma_1, \, \gamma_\infty \ \biggl| \ {\gamma_0}^m = {\gamma_1}^n = {\gamma_\infty}^k = \gamma_0 \, \gamma_1 \, \gamma_\infty = 1 \right \rangle.$
Such a group is called a Triangle Group, although it is also known as a von Dyck Group after the German mathematician Walther Franz Anton von Dyck (1856 — 1934). Today, we focus on why these are called “triangle groups.”

### Basic Properties

It is easy to see that $D(m, n, k) = D(k, m, n) = D(k, n, m)$, that is, the ordering of $m$, $n$, and $k$ does not matter. Indeed, using the substitutions $\gamma_0 = s$, $\gamma_1 = r$, and $\gamma_{\infty} = ( s \, r)^{-1}$, we have the following diagram:

\begin{matrix} {\begin{aligned} u & = (s \, r)^{-1} \\ v & = s \\ u \, v & = r^{-1} \end{aligned}} & {\begin{aligned} u^k & = 1 \\ v^m & = 1 \\ (u \, v)^n & = 1 \end{aligned}} & D(k,m,n) = \left \langle u, v \ \biggl| \ u^k = v^m = (u \, v)^n = 1 \right \rangle \\ & & \uparrow \\ {\begin{aligned} s & = \gamma_0 \\ r & = \gamma_1 \\ s \, r & = {\gamma_\infty}^{-1} \end{aligned}} & {\begin{aligned} s^m & = 1 \\ r^n & = 1 \\ (s \, r)^k & = 1 \end{aligned}} & D(m,n,k) = \left \langle r, s \ \biggl| \ s^m = r^n = (s \, r)^k =1 \right \rangle \\ & & \downarrow \\ {\begin{aligned} z & = s \, r \\ w & = r^{-1} \\ z \, w & = s \end{aligned}} & {\begin{aligned} z^k & = 1 \\ w^n & = 1 \\ (z \, w)^m & = 1 \end{aligned}} & D(k,n,m) = \left \langle z, w \ \biggl| \ z^k = w^n = (z \, w)^m =1 \right \rangle \end{matrix}

### Triangles in the Affine Real Plane

We give a geometric interpretation of this group using matrices. Given positive integers $m$, $n$, and $k$ which satisfy the identity $\dfrac {1}{m} + \dfrac {1}{n} + \dfrac {1}{k} = 1$, define the quantities
\begin{aligned} x_P & = 0 & & & y_P & = 0 & & & & & A & = \dfrac {\pi}{m} \\[10pt] x_Q & = \cos A \, \sin B + \sin A \, \cos B & & & y_Q & = 0 & & & \text{in terms of} & & B & = \dfrac {\pi}{n} \\[10pt] x_R & = \cos A \, \sin B & & & y_R & = \sin A \, \sin B & & & & & C & = \dfrac {\pi}{k}. \end{aligned}

Denoting $\mathbb A^2(\mathbb R)$ as the affine real plane, consider the affine points $P = (x_P, y_P)$, $Q = (x_Q,y_Q)$ and $R = (x_R, y_R)$ as well as the transformations $\gamma_0, \, \gamma_1, \, \gamma_\infty: \mathbb A^2(\mathbb R) \to \mathbb A^2(\mathbb R)$ defined by

\begin{aligned} \gamma_0: \qquad \left[ \begin{matrix} x \\[5pt] y \end{matrix} \right] & \qquad \mapsto \qquad \left [ \begin{matrix} \cos \, 2 A & -\sin \, 2 A \\[5pt] \sin \, 2 A & \cos \, 2 A \end{matrix} \right ] \left[ \begin{matrix} x - x_P \\[5pt] y - y_P \end{matrix} \right] + \left[ \begin{matrix} x_P \\[5pt] y_P \end{matrix} \right] \\[10pt] \gamma_1: \qquad \left[ \begin{matrix} x \\[5pt] y \end{matrix} \right] & \qquad \mapsto \qquad \left [ \begin{matrix} \cos \, 2 B & -\sin \, 2 B \\[5pt] \sin \, 2 B & \cos \, 2 B \end{matrix} \right ] \left[ \begin{matrix} x - x_Q \\[5pt] y - y_Q \end{matrix} \right] + \left[ \begin{matrix} x_Q \\[5pt] y_Q \end{matrix} \right] \\[10pt] \gamma_\infty: \qquad \left[ \begin{matrix} x \\[5pt] y \end{matrix} \right] & \qquad \mapsto \qquad \left [ \begin{matrix} \cos \, 2 C & -\sin \, 2 C \\[5pt] \sin \, 2 C & \cos \, 2 C \end{matrix} \right ] \left[ \begin{matrix} x - x_R \\[5pt] y - y_R \end{matrix} \right] + \left[ \begin{matrix} x_R \\[5pt] y_R \end{matrix} \right] \end{aligned}

Proposition.
Continue notation as above.

• The angles $A$, $B$, and $C$ sum to $180^\circ$, that is, $A + B + C = \pi$. Moreover, $\cos^2 A + \cos^2 B + \cos^2 C + 2 \, \cos A \, \cos B \, \cos C = 1$.

• $V = \{ P, \, Q, \, R \} \subseteq \mathbb A^2(\mathbb R)$ is a triangle with angles $A$, $B$, and $C$ and area $\dfrac {1}{2} \, \sin A \, \sin B \, \sin C$.

• The transformation $\gamma_0$ is a rotation around $P$ by $(2 \pi/m)$ radians counterclockwise, the transformation $\gamma_1$ is a rotation around $Q$ by $(2 \pi/n)$ radians counterclockwise, and the transformation $\gamma_\infty$ is a rotation around $R$ by $(2 \pi/k)$ radians counterclockwise.

• The compositions ${\gamma_0}^m = {\gamma_1}^n = {\gamma_\infty}^k = \gamma_0 \, \gamma_1 \, \gamma_\infty = 1$ are the identity transformation, while the composition

$\gamma_1 \circ \gamma_0: \qquad \left[ \begin{matrix} x \\[5pt] y \end{matrix} \right] \qquad \mapsto \qquad \left [ \begin{matrix} \cos \, 2 C & \sin \, 2 C \\[5pt] -\sin \, 2 C & \cos \, 2 C \end{matrix} \right ] \left[ \begin{matrix} x - x_R \\[5pt] y - y_R \end{matrix} \right] + \left[ \begin{matrix} x_R \\[5pt] -y_R \end{matrix} \right]$
has infinite order. In particular, $D(m,n,k)$ is a finitely generated group which is not abelian and has infinitely many elements.

We leave the proof as an exercise for the reader.

There are only finitely many positive integers $m$, $n$, and $k$ satisfying the conditions above, and up to symmetry they fit into the following table:

 Group $D(m, n, k)$ $m$ $n$ $k$ Triangle $V = \{ P, Q, R \}$ Symmetries of the kisrhombille 2 3 6 $30^\circ - 60^\circ - 90^\circ$ Symmetries of the kisquadrille 2 4 4 $45^\circ - 45^\circ - 90^\circ$ Symmetries of the deltille 3 3 3 $60^\circ - 60^\circ - 60^\circ$

These are examples of wallpaper groups. The images of the triangle $V$ by transformations of the elements in $D(m,n,k)$ form a tiling of the plane $\mathbb A^2(\mathbb R)$. This explains why $D(m,n,k)$ is called a “triangle group”.

### Triangles on the Unit Sphere

Triangle groups $D(m,n,k)$ are much more interesting when the integers $m$, $n$, and $k$ are a bit larger. For example,

\begin{aligned} Z_n & = \left \langle r \, \biggl| \, r^n = 1 \right \rangle & \simeq & \left \langle \gamma_0, \, \gamma_1, \, \gamma_\infty \, \biggl| \, {\gamma_0}^1 = {\gamma_1}^n = {\gamma_\infty}^n = \gamma_0 \, \gamma_1 \, \gamma_\infty = 1 \right \rangle = D(1,n,n); \\[10pt] D_n & = \left \langle r, \, s \, \biggl| \, s^2 = r^n = (s \, r)^2 = 1 \right \rangle & \simeq & \left \langle \gamma_0, \, \gamma_1, \, \gamma_\infty \, \biggl| \, {\gamma_0}^2 = {\gamma_1}^n = {\gamma_\infty}^2 = \gamma_0 \, \gamma_1 \, \gamma_\infty = 1 \right \rangle = D(2,n,2). \end{aligned}

We will use this to study triangles on the unit sphere $S^2(\mathbb R) = \left \{ (u,v,w) \in \mathbb A^2(\mathbb R) \, \biggl| \, u^2 + v^2 + w^2 = 1 \right \}$. Given positive integers $m$, $n$, and $k$ which satisfy the inequalities $\dfrac {1}{m} + \dfrac {1}{n} + \dfrac {1}{k} > 1$ and $m, \, n, \, k \geq 2$, define the quantities
\begin{aligned} x_P & = 0 & & & x_Q & = \dfrac {\sqrt{\delta}}{\sin A \, \sin B} & & & x_R & = \cos A \, \dfrac {\sqrt{\delta}}{\sin C \, \sin A} \\[5pt] y_P & = 0 & & & y_Q & = 0 & & & y_R & = \sin A \, \dfrac {\sqrt{\delta}}{\sin C \, \sin A} \\[5pt] z_P & = 1 & & & z_Q & = \dfrac {\cos C + \cos A \, \cos B}{\sin A \, \sin B} & & & z_R & = \dfrac {\cos B + \cos C \, \cos A}{\sin C \, \sin A} \end{aligned} \quad \text{where} \quad \begin{aligned} A & = \dfrac {\pi}{m} \\[5pt] B & = \dfrac {\pi}{n} \\[5pt] C & = \dfrac {\pi}{k} \end{aligned}

and $\delta = 1 - \bigl( \cos^2 A + \cos^2 B + \cos^2 C + 2 \, \cos A \, \cos B \, \cos C \bigr) > 0$. Denoting $S^2(\mathbb R)$ as the unit sphere, consider the affine points

\begin{aligned} P & = \left[ \begin{matrix} x_P \\[5pt] y_P \\[5pt] z_P \end{matrix} \right] = \gamma_P \left[ \begin{matrix} 0 \\[5pt] 0 \\[5pt] 1 \end{matrix} \right] \\[5pt] Q & = \left[ \begin{matrix} x_Q \\[5pt] y_Q \\[5pt] z_Q \end{matrix} \right] = \gamma_Q \left[ \begin{matrix} 0 \\[5pt] 0 \\[5pt] 1 \end{matrix} \right] \\[5pt] R & = \left[ \begin{matrix} x_R \\[5pt] y_R \\[5pt] z_R \end{matrix} \right] = \gamma_R \left[ \begin{matrix} 0 \\[5pt] 0 \\[5pt] 1 \end{matrix} \right] \end{aligned} \quad \text{where} \quad \begin{aligned} \gamma_P & = \left[ \begin{matrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{matrix} \right] \\[5pt] \gamma_Q & = \left[ \begin{matrix} z_Q & 0 & x_Q \\[5pt] 0 & 1 & 0 \\[5pt] -x_Q & 0 & z_Q \end{matrix} \right] \\[5pt] \gamma_R & = \left[ \begin{matrix} x_R \, z_R/\sqrt{x_R^2 + y_R^2} & - y_R/\sqrt{x_R^2 + y_R^2} & x_R \\[5pt] y_R \, z_R/\sqrt{x_R^2 + y_R^2} & x_R/\sqrt{x_R^2 + y_R^2} & y_R \\[5pt] -\sqrt{x_R^2 + y_R^2} & 0 & z_R \end{matrix} \right] \end{aligned}
as well as the rotations $\gamma_0, \, \gamma_1, \, \gamma_\infty: S^2(\mathbb R) \to S^2(\mathbb R)$ defined by

\begin{aligned} \gamma_0: & \qquad \left[ \begin{matrix} x \\[5pt] y \\[5pt] z \end{matrix} \right] \qquad & \mapsto \qquad \left( \gamma_P \left[ \begin{matrix} \cos 2 A & -\sin 2 A & 0 \\[5pt] \sin 2 A & \cos 2 A & 0 \\[5pt] 0 & 0 & 1 \end{matrix} \right] {\gamma_P}^{-1} \right) \left[ \begin{matrix} x \\[5pt] y \\[5pt] z \end{matrix} \right] \\[5pt] \gamma_1: & \qquad \left[ \begin{matrix} x \\[5pt] y \\[5pt] z \end{matrix} \right] \qquad & \mapsto \qquad\left( \gamma_Q \left[ \begin{matrix} \cos 2 B & -\sin 2 B & 0 \\[5pt] \sin 2 B & \cos 2 B & 0 \\[5pt] 0 & 0 & 1 \end{matrix} \right] {\gamma_Q}^{-1} \right) \left[ \begin{matrix} x \\[5pt] y \\[5pt] z \end{matrix} \right]\\[5pt] \gamma_\infty: & \qquad \left[ \begin{matrix} x \\[5pt] y \\[5pt] z \end{matrix} \right] \qquad & \mapsto \qquad \left( \gamma_R \left[ \begin{matrix} \cos 2 C & -\sin 2 C & 0 \\[5pt] \sin 2 C & \cos 2 C & 0 \\[5pt] 0 & 0 & 1 \end{matrix} \right] {\gamma_R}^{-1} \right) \left[ \begin{matrix} x \\[5pt] y \\[5pt] z \end{matrix} \right] \end{aligned}

Proposition.
Continue notation as above.

• The angles $A$, $B$, and $C$ sum to more than $180^\circ$, that is, $A + B + C > \pi$.

• $V = \{ P, \, Q, \, R \} \subseteq S^2(\mathbb R)$ is a spherical triangle with angles $A$, $B$, and $C$.

• The transformation $\gamma_0$ is a rotation around $P$ by $(2 \pi/m)$ radians counterclockwise, the transformation $\gamma_1$ is a rotation around $Q$ by $(2 \pi/n)$ radians counterclockwise, and the transformation $\gamma_\infty$ is a rotation around $R$ by $(2 \pi/k)$ radians counterclockwise.

• The compositions ${\gamma_0}^m = {\gamma_1}^n = {\gamma_\infty}^k = \gamma_0 \, \gamma_1 \, \gamma_\infty = 1$ are the identity transformation.

Sketch of Proof: In order to verify that the triangle $V = \{ P, \, Q, \, R \}$ makes angles $A$, $B$, and $C$ on the sphere, we first compute the angles the vectors make with each other with respect to the origin $(0,0,0)$. Denoting $a$, $b$, and $c$ as the angles between $Q$ and $R$, $R$ and $P$, and $P$ and $Q$, respectively, we use inner products and cross products to verify that
\begin{aligned} \cos a & = \dfrac {Q \cdot R}{\| Q \| \, \| R \|} & = \dfrac {\cos A + \cos B \, \cos C}{\sin B \, \sin C} \\[5pt] \cos b & = \dfrac {R \cdot P}{\| R \| \, \| P \|} & =\dfrac {\cos B + \cos C \, \cos A}{\sin C \, \sin A} \\[5pt] \cos c & = \dfrac {P \cdot Q}{\| P \| \, \| Q \|} & = \dfrac {\cos C + \cos A \, \cos B}{\sin A \, \sin B} \end{aligned} \qquad \qquad \begin{aligned} \sin a & = \dfrac {\| Q \times R \|}{\| Q \| \, \| R \|} & = \dfrac {\sqrt{\delta}}{\sin B \, \sin C} \\[5pt] \sin b & = \dfrac {\| R \times P \|}{\| R \| \, \| P \|} & = \dfrac {\sqrt{\delta}}{\sin C \, \sin A} \\[5pt] \sin c & = \dfrac {\| P \times Q \|}{\| P \| \, \| Q \|} & = \dfrac {\sqrt{\delta}}{\sin A \, \sin B} \end{aligned}
Using the Spherical Law of Cosines, we see that
$\cos A = \dfrac {\cos a - \cos b \, \cos c}{\sin b \, \sin c}$, $\cos B = \dfrac {\cos b - \cos c \, \cos a}{\sin c \, \sin a}$, and $\cos C = \dfrac {\cos c - \cos a \, \cos b}{\sin a \, \sin b}$. That is, $V$ is indeed a triangle with angle $A$ at vertex $P$, angle $B$ at vertex $Q$, and angle $C$ at vertex $R$. $\square$

The triangles $V = \{ P, \, Q, \, R \}$ such that their angles $A + B + C > \pi$ while $m = \pi/A$, $n = \pi/B$, and $k = \pi/C$ are integers are called Möbius triangles, named in honor of the German mathematician August Ferdinand Möbius (1790 — 1868).

In the next lecture, we’ll focus more of the geometry in this case. And we’ll show some pretty pictures!