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 SolidNo. of VerticesNo. of EdgesNo. of FacesEdges at each Vertex
Tetrahedron4 vertices6 edges4 triangular3 edges
Cube8 vertices12 edges6 square3 edges
Octahedron6 vertices12 edges8 triangular4 edges
Dodecahedron20 vertices30 edges12 pentagonal3 edges
Icosahedron12 vertices30 edges20 triangular5 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)mnkTriangle V = \{ P, Q, R \}
Z_n1nn360^\circ/n - 360^\circ/n - 180^\circ
D_n2n2360^\circ/n - 90^\circ - 90^\circ
A_423360^\circ - 60^\circ - 90^\circ
S_423445^\circ - 60^\circ - 90^\circ
A_523536^\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.

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About Edray Herber Goins, Ph.D.

Edray Herber Goins grew up in South Los Angeles, California. A product of the Los Angeles Unified (LAUSD) public school system, Dr. Goins attended the California Institute of Technology, where he majored in mathematics and physics, and earned his doctorate in mathematics from Stanford University. Dr. Goins is currently an Associate Professor of Mathematics at Purdue University in West Lafayette, Indiana. He works in the field of number theory, as it pertains to the intersection of representation theory and algebraic geometry.
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2 Responses to Lecture 9: Monday, September 9, 2013

  1. Pingback: Lecture 11: Friday, September 13, 2013 | Lectures on Dessins d'Enfants

  2. Pingback: MA 59800 Course Syllabus | Lectures on Dessins d'Enfants

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