Lecture 9: Monday, September 9, 2013

In the previous lecture, we discussed how to draw triangles in the plane and on the unit sphere such that the triangles tile these surfaces. In this lecture, we discuss triangle groups in more detail by focusing on discrete symmetries of the unit sphere.

Cyclic Groups Revisited

In the previous lecture, we focused on positive integers $m$, $n$, and $k$ such that $\dfrac 1m + \dfrac 1n + \dfrac 1k > 1$ and $m, \, n, \, k \geq 2$. We showed that there always exists a spherical triangle $V = \{ P, \, Q, \, R \} \subseteq S^2(\mathbb R)$ with angles $A = \dfrac {\pi}{m}$, $B = \dfrac {\pi}{n}$, and $C = \dfrac {\pi}{k}$. Unfortunately, these formulas from become degenerate when at least one of $m$, $n$, or $k$ is 1. Since $D(m,1,k) = D(N,1,N)$ for $N$ being the greatest common divisor of $m$ and $k$, we may consider $(m,n,k) = (N,1,N)$. In this case, we have $\sin B = \delta = 0$, but we can interpret the isosceles triangle $V = \{ P, \, Q, \, R \}$ as having angles $A = \dfrac {\pi}{N}$, $B = \pi$, and $C = \dfrac {\pi}{N}$. Such a region is sometimes called a lune. The cyclic group $Z_N = D(N, 1, N)$ is generated by $\gamma_0$, $\gamma_1$, and $\gamma_\infty$ satisfying ${\gamma_0}^N = {\gamma_1}^1 = {\gamma_\infty}^N = \gamma_0 \, \gamma_1 \, \gamma_\infty = 1$:

\begin{aligned} P & = \left[ \begin{matrix} 0 \\[10pt] 0 \\[10pt] 1 \end{matrix} \right] & \qquad \qquad & & \gamma_0: \left[ \begin{matrix} x \\[10pt] y \\[10pt] z \end{matrix} \right] & \mapsto \left[ \begin{matrix} \cos \dfrac {2 \pi}{N} & -\sin \dfrac {2 \pi}{N} & 0 \\[10pt] \sin \dfrac {2 \pi}{N} & \cos \dfrac {2 \pi}{N} & 0 \\[10pt] 0 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\[10pt] y \\[10pt] z \end{matrix} \right] \\[10pt] Q & = \left[ \begin{matrix} 1 \\[10pt] 0 \\[10pt] 0 \end{matrix} \right] & & & \gamma_1: \left[ \begin{matrix} x \\[10pt] y \\[10pt] z \end{matrix} \right] & \mapsto \left[ \begin{matrix} 1 & 0 & 0 \\[10pt] 0 & 1 & 0 \\[10pt] 0 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\[10pt] y \\[10pt] z \end{matrix} \right] \\[10pt] R & = \left[ \begin{matrix} 0 \\[10pt] 0 \\[10pt] -1 \end{matrix} \right] & & & \gamma_\infty: \left[ \begin{matrix} x \\[10pt] y \\[10pt] z \end{matrix} \right] & \mapsto \left[ \begin{matrix} \cos \dfrac {2 \pi}{N} & \sin \dfrac {2 \pi}{N} & 0 \\[10pt] -\sin \dfrac {2 \pi}{N} & \cos \dfrac {2 \pi}{N} & 0 \\[10pt] 0 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\[10pt] y \\[10pt] z \end{matrix} \right] \end{aligned}

We can give a different color to each copy $\gamma(V) = \bigl \{ \gamma(P), \, \gamma(Q), \, \gamma(R) \bigr \}$ of the triangle coming from $\gamma \in D(N,1,N)$. While in the plane $\mathbb A^2(\mathbb R)$ the group $D_N$ is associated with the rotations of the regular $N$-gon, the corresponding figure in $\mathbb A^3(\mathbb R)$ is called a hosohedron.

Dihedral Groups Revisited

As a non-degenerate case, consider $(m,n,k) = (N,2,2)$. Then we have $\delta = \sin^2 (2 \pi / N) > 0$ and the isosceles triangle $V = \{ P, \, Q, \, R \}$ with angles $A = \pi / N$, $B = \pi/2$, and $C = \pi/2$. The dihedral group $D_N = D(N, 2, 2)$ is generated by $\gamma_0$, $\gamma_1$, and $\gamma_\infty$ satisfying ${\gamma_0}^N = {\gamma_1}^2 = {\gamma_\infty}^2 = \gamma_0 \, \gamma_1 \, \gamma_\infty = 1$:

\begin{aligned} P & = \left[ \begin{matrix} 0 \\[10pt] 0 \\[10pt] 1 \end{matrix} \right] & \qquad \qquad & & \gamma_0: \left[ \begin{matrix} x \\[10pt] y \\[10pt] z \end{matrix} \right] & \mapsto \left[ \begin{matrix} \cos \dfrac {2 \pi}{N} & -\sin \dfrac {2 \pi}{N} & 0 \\[10pt] \sin \dfrac {2 \pi}{N} & \cos \dfrac {2 \pi}{N} & 0 \\[10pt] 0 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\[10pt] y \\[10pt] z \end{matrix} \right] \\[10pt] Q & = \left[ \begin{matrix} 1 \\[10pt] 0 \\[10pt] 0 \end{matrix} \right] & & & \gamma_1: \left[ \begin{matrix} x \\[10pt] y \\[10pt] z \end{matrix} \right] & \mapsto \left[ \begin{matrix} +1 & 0 & 0 \\[10pt] 0 & -1 & 0 \\[10pt] 0 & 0 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\[10pt] y \\[10pt] z \end{matrix} \right] \\[10pt] R & = \left[ \begin{matrix} \cos \dfrac {2 \pi}{N} \\[10pt] \sin \dfrac {2 \pi}{N} \\[10pt] 0 \end{matrix} \right] & & & \gamma_\infty: \left[ \begin{matrix} x \\[10pt] y \\[10pt] z \end{matrix} \right] & \mapsto \left[ \begin{matrix} \cos \dfrac {2 \pi}{N} & \sin \dfrac {2 \pi}{N} & 0 \\[10pt] \sin \dfrac {2 \pi}{N} & -\cos \dfrac {2 \pi}{N} & 0 \\[10pt] 0 & 0 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\[10pt] y \\[10pt] z \end{matrix} \right] \end{aligned}

You may wish to compare these formulas for $r = \gamma_0$ and $s = \gamma_1$ and those in Lecture 6. We can give a different color to each copy $\gamma(V) = \bigl \{ \gamma(P), \, \gamma(Q), \, \gamma(R) \bigr \}$ of the triangle coming from $\gamma \in D(N,2,2)$. While in the plane $\mathbb A^2(\mathbb R)$ the group $D_N$ is associated with the symmetries of the regular $N$-gon, the corresponding figure in $\mathbb A^3(\mathbb R)$ is called a bipyramid; such tilings of the sphere can be found below.

Platonic Solids

A Platonic solid is a regular, convex polyhedron. They are named after Plato (424 BC – 348 BC). Aside from the regular polygons, there are five such solids. The complete list of these regular solids can be found below.

 Platonic Solid No. of Vertices No. of Edges No. of Faces Edges at each Vertex Tetrahedron 4 vertices 6 edges 4 triangular 3 edges Cube 8 vertices 12 edges 6 square 3 edges Octahedron 6 vertices 12 edges 8 triangular 4 edges Dodecahedron 20 vertices 30 edges 12 pentagonal 3 edges Icosahedron 12 vertices 30 edges 20 triangular 5 edges

We have the following result.

Proposition.

• There are five Platonic Solids: The tetrahedron, cube, octahedron, dodecahedron, and icosahedron. In particular, the cube and octahedron are duals of each other, while the dodecahedron and icosahedron are duals of each other.

• The Platonic Solids have the following groups as sets of symmetries:

$G = \begin{cases} A_4 & \text{for the tetrahedron,} \\[5pt] S_4 & \text{for the cube,} \\[5pt] S_4 & \text{for the octahedron,} \\[5pt] A_5 & \text{for the dodecahedron,} \\[5pt] A_5 & \text{for the icosahedron.} \end{cases}$

• Each of the symmetry groups associated with the Platonic Solids is a triangle group. Explicitly,

\begin{aligned} D(2, 3, N) & = \left \langle \gamma_0, \, \gamma_1, \, \gamma_\infty \ \biggl| \ {\gamma_0}^2 = {\gamma_1}^3 = {\gamma_\infty}^N = \gamma_0 \, \gamma_1 \, \gamma_\infty = 1 \right \rangle \\[5pt] & \simeq \begin{cases} A_4 & \text{for} \ N = 3; \\[5pt] S_4 & \text{for} \ N = 4; \\[5pt] A_5 & \text{for} \ N = 5. \end{cases} \end{aligned}

• If $m$, $n$, and $k$ are positive integers such that $\dfrac {1}{m} + \dfrac {1}{n} + \dfrac {1}{k} > 1$, then $D(m,n,k) \simeq Z_N, \, D_N, \, A_4, \, S_4, \, A_5$.

The symmetric group $S_4$ is called the octahedral group while the alternating group $A_5$ is called the icosahedral group. In general, the groups $D(m,n,k)$ with $(1/m) + (1/n) + (1/k) > 1$ are examples of polyhedral groups.

Sketch of Proof: We sketch the ideas for the first two statements.

We consider only the tetrahedron, the octahedron, and the icosahedron since the rotation groups are the same for their duals. The tetrahedron has four vertices, so its rotation group is a subgroup of $S_4$; since any automorphism must be orientation preserving it is actually contained in $A_4$. But each of the four vertices has $N = 3$ edges attached, so there are at least $4 \cdot 3 = 12$ automorphisms of the tetrahedron. Since $|A_4| = 12$, this must be all of them. The octahedron has four pairs of opposing triangular faces, so its rotation group is a subgroup of $S_4$. Each of the six vertices has $N = 4$ edges attached, so there are at least $6 \cdot 4 = 24$ automorphisms of the octahedron. Since $|S_4| = 24$, this must be all of them. Finally, one can find an arrangement of the 20 triangular faces of the icosahedron so that their midpoints form five tetrahedra. The automorphisms of the icosahedron must be orientation preserving of this arrangement of the collections of midpoints, so the automorphisms form a subgroup of $A_5$. Each of the twelve vertices has $N = 5$ edges attached, so there are at least $12 \cdot 5 = 60$ automorphisms of the icosahedron. Since $|A_5| = 60$, again this must be all of them.

Each vertex $P \in V$ of the tetrahedron, octahedron, and icosahedron has $N$ edges attached for $N = 3, \, 4, \, 5$. We inscribe the vertices $V = \{ P_\infty, P_0, \, P_1, \, \dots \} \hookrightarrow S^2(\mathbb R)$ in the unit sphere by making the following $N+1$ choices:

\begin{aligned} P_\infty & = \bigl( 0, \ 0, \ 1 \bigr) \\[5pt] P_k & = \left( \sin \theta_N \, \cos \frac {2 \pi k}N, \ \sin \theta_N \, \sin \frac {2 \pi k}N, \ \cos \theta_N \right), \qquad k \in \mathbb Z; \end{aligned}
where the angle $\theta_N$ satisfies

$\cos \theta_N = \frac {\cos \dfrac {2 \pi}N}{1 - \cos \dfrac {2 \pi}N} = \begin{cases} -\dfrac 13 & \text{for} \ N = 3, \\[10pt] 0 & \text{for} \ N = 4, \\[10pt] \dfrac 1{\sqrt{5}} & \text{for} \ N = 5; \end{cases} \qquad \qquad \sin \theta_N = \begin{cases} \dfrac {2 \, \sqrt{2}}3 & \text{for} \ N = 3, \\[10pt] 1 & \text{for} \ N = 4, \\[10pt] \dfrac 2{\sqrt{5}} & \text{for} \ N = 5. \end{cases}$
Form the triangle $\{ P, \, Q, \, R \} \subseteq S^2(\mathbb R)$ via the three vertices

\begin{aligned} P & = \dfrac {P_\infty}{\| P_\infty \|} & = & \bigl(0, \ 0, \ 1 \bigr) \\[10pt] Q & = \dfrac {P_\infty + P_0}{\|P_\infty + P_0 \|} & = & \left( \dfrac {\sqrt{\delta}}{\sin A \, \sin B}, \ 0, \ \dfrac {\cos C + \cos A \, \cos B}{\sin A \, \sin B} \right) \\[10pt] R & = \dfrac {P_\infty + P_0 + P_1}{\|P_\infty + P_0 + P_1 \|} & = & \left( \cos A \, \dfrac {\sqrt{\delta}}{\sin C \, \sin A}, \ \sin A \, \dfrac {\sqrt{\delta}}{\sin C \, \sin A}, \ \dfrac {\cos B + \cos C \, \cos A}{\sin C \, \sin A} \right) \end{aligned}
in terms of the angles $A = \pi/N$, $B = \pi/2$ and $C = \pi/3$ and the positive real number

$\delta = 1 - \bigl( \cos^2 A + \cos^2 B + \cos^2 C + 2 \, \cos A \, \cos B \, \cos C \bigr) = \begin{cases} \dfrac 12 & \text{for} \ N = 3, \\[10pt] \dfrac 14 & \text{for} \ N = 4, \\[10pt] \left( \dfrac {1 - \sqrt{5}}{4} \right)^2 & \text{for} \ N = 5. \end{cases}$

Note that the affine point $P$ is the north pole, the affine point $Q$ is the midpoint of the edge between $P_\infty$ and $P_0$, while the point $R$ is the midpoint of the face formed by $P_\infty$, $P_0$, and $P_1$. Hence the triangle group $D(N, 2, 3)$ permutes around copies of this triangle on the sphere $S^2(\mathbb R)$. Since the automorphism group $G$ of either the tetrahedron, the octahedron, and the icosahedron must permute the triangle $\{ P, \, Q, \, R \}$ as well, we see that $G \simeq D(N, 2, 3)$. To be explicit, we have the $3 \times 3$ matrices

\begin{aligned} \gamma_0 & = \left[ \begin{matrix} \cos \dfrac {2\pi}{N} & - \sin \dfrac {2\pi}{N} & 0 \\[10pt] \sin \dfrac {2\pi}{N} & \cos \dfrac {2\pi}{N} & 0 \\[10pt] 0 & 0 & 1 \end{matrix} \right] \\[5pt] \gamma_1 & = \left[ \begin{matrix} \dfrac {1-m^2}{1+m^2} & 0 & \dfrac {2 \, m}{1+m^2} \\[10pt] 0 & -1 & 0 \\[10pt] \dfrac {2 \, m}{1+m^2} & 0 & -\dfrac {1-m^2}{1+m^2} \end{matrix} \right] \\[5pt] \gamma_\infty & = \left[ \begin{matrix} \dfrac {1-m^2}{1+m^2} \, \cos \dfrac {2 \pi}{N} & \dfrac {1-m^2}{1+m^2} \, \sin \dfrac {2 \pi}{N} & \dfrac {2 \, m}{1 + m^2} \\[10pt] \sin \dfrac {2 \pi}{N} & -\cos \dfrac {2 \pi}{N} & 0 \\[10pt] \dfrac {2 \, m}{1+m^2} \, \cos \dfrac {2 \pi}{N} & \dfrac {2 \, m}{1+m^2} \, \sin \dfrac {2 \pi}{N} & -\dfrac {1 - m^2}{1 + m^2} \end{matrix} \right] \end{aligned} \ \text{in terms of} \ m = \dfrac {1}{2 \sqrt{\delta}}

where ${\gamma_0}^N = {\gamma_1}^2 = {\gamma_\infty}^3 = \gamma_0 \, \gamma_1 \, \gamma_\infty = 1$.

For the final statement, the integers $m$, $n$, and $k$, such that $(1/m) + (1/n) + (1/k) > 1$ fit into the table below.

 Group $D(m, n, k)$ $m$ $n$ $k$ Triangle $V = \{ P, Q, R \}$ $Z_n$ 1 $n$ $n$ $360^\circ/n - 360^\circ/n - 180^\circ$ $D_n$ 2 $n$ 2 $360^\circ/n - 90^\circ - 90^\circ$ $A_4$ 2 3 3 $60^\circ - 60^\circ - 90^\circ$ $S_4$ 2 3 4 $45^\circ - 60^\circ - 90^\circ$ $A_5$ 2 3 5 $36^\circ - 60^\circ - 90^\circ$

This completes the proof. $\square$.

Consider again the triangle $\{ P, \, Q, \, R \} \subseteq S^2(\mathbb R)$ where the affine point $P$ is the north pole, the affine point $Q$ is the midpoint of the edge between $P_\infty$ and $P_0$, while the point $R$ is the midpoint of the face formed by $P_\infty$, $P_0$, and $P_1$. We can give a different color to each copy $\bigl \{ \gamma(P), \, \gamma(Q), \, \gamma(R) \bigr \}$ of the triangle coming from $\gamma \in D(N,2,3)$. Such tilings of the sphere can be found below.