Finite Subgroups of the Rotation Group 183

P e r m u t a t i o n r e p r e s e n t a t i o n s < p :G - + P e r m ( S ) a r e r a r e l y s u r j e c t i v e b e c a u s e t h e o r d e r o f P e r m ( S ) t e n d s t o b e v e r y l a r g e . B u t o n e c a s e i s g i v e n i n t h e n e x t e x a m p l e .

Example 6.11.5 T h e g r o u p G ^L2(lB'2) o f i n v e r t i b l e m a t r i c e s w i t h m o d 2 c o e f f i c i e n t s i s i s o m o r p h i c t o t h e s y m m e t r i c g r o u p S3.

W e d e n o t e t h e f i e l d lF2 b y F a n d t h e g r o u p G ^L2^(lF2^{) b y} G. T h e s p a c e F 2 o f c o l u m n v e c t o r s c o n s i s t s o f f o u r v e c t o r s :

' 0' V ' 0' ' 1 '

_0_ , = 0 , <?2 = _1_ , e\ + e2 = _1_

The group G operates on the set of three nonzero vectors S = {ei, e2, ei + ^2), and this gives us a permutation representation <p: G -+ S3. The identity is the only matrix that fixes both ei and e2, so the operation of G on S is faithful, and is injective. The columns of an invertible matrix must be an ordered pair of distinct elements of S. There are six such pairs, so |G | = 6. Since S3 also has order six cp is an isomorphism. □

6 . 1 2 FIN IT E S U B G R O U P S O F T H E R O T A T I O N G R O U P

In this section, we apply the Counting Formula to classify the finite subgroups of SO 3, the group of rotations of ]R3. As happens with finite groups of isometries of the plane, all of them are symmetry groups of familiar figures.

Theorem 6.U.1 A finite subgroup of S O 3 is one of the following groups:

Ck- the cyclic groupof rotations by multiples of 2n / k about a line, with karbitrary;

Dfc: the dihedral group of symmetries of a regular k-gon, with karbitrary;

T: the tetrahedral groupof 12 rotational symmetries of a tetrahedron;

O: the octahedral group of 24 rotational symmetries of a cube or an octahedron;

I: the icosahedral group of 60 rotational symmetries of a dodecahedron or an icosahedron.

N ote:The dihedral groups are usually presented as groups of symmetry of a regular polygon in the plane, where reflections reverse orientation. However, a reflection of a plane can be achieved by a rotation through the angle n in three-dimensional space, and in this way the symmetries of a regular polygon can be realized as rotations of K3. The dihedral group can be generated by a rotation x with angle 2n / n about the ei-axis and a rotation y with

angle rr about the e 2-axis. With c = cos 2rr/n and s = s in 2 rr/n , the matrices that represent these rotations are

'1 ' "-1 '

(6.12.2) x = c -s , and >’ = 1

s c -1

Let G be a finite subgroup of S O 3, of order N > 1 . W e’ll call a pole of an element g *-1 of G a pole of the group. Any rotation of ]R3 except the identity has two poles - the intersections of the axis of rotation with the unit sphere §2. So a pole of G is a point on the 2-sphere that is fixed by a group element g different from 1.

Example 6.U.3 The group T of rotational symmetries of a tetrahedron A has order 12. Its poles are the points of §2 that lie above the centers of the faces, the vertices, and the centers of the edges. Since A has four faces, four vertices, and six edges, there are 14 poles.

\poles\ = 14 = 1faces] + ^| vertices ^| + \edges\

Each of the 11 elements g *-1 of T has two spins - two pairs (g, p ), where p is a pole of g.

So there are 22 spins altogether. The stabilizer of a face has order 3. Its two elements 1 share a pole above the center of a face. Similarly, there are two elements with a pole above a vertex, and one element with a pole above the center of an edge.

|spins| = 22 = 2\faces\ + 2 ^| vertices ^| + \e d g e s |

□

Let P denote the set of all poles of a finite subgroup G. We will get information about the group by counting these poles. As the example shows, the count can be confusing.

Lemma 6.12.4 The set P of poles of G is a union of G-orbits. So G operates on P.

Proof Let p be a pole, say the pole of an element g 1 in G, let h be another element of G, and let q = h p . We have to show that q is a pole, meaning that q is fixed by some element g' of G other than the identity. The required element is h g h -1^ This element is not equal to

1 because g*- 1, and h g h _1q = h g p = h p = q. □

The stabilizer G p of a pole p is the group of all of the rotations about p that are in G.

It is a cyclic group, generated by the rotation of smallest positive angle 0. W e’ll denote its order by rp. Then 0 = 2 rr/rp .

Section 6.12 Finite Subgroups of the Rotation Group 185 Since p is a pole, the stabilizer G p contains an element besides 1, so rp > 1. The set of elements of G with pole p is the stabilizer G p, with the identity element omitted. So there are rp — 1 group elements that have p as pole. Every group element g except one has two poles. Since |G | = N, there are 2N - 2 spins. This gives us the relation

(6.12.5) I ( r p — 1) = 2(N - 1 ) .

peP

We collect terms to simplify the left side of this equation: Let n p denote the order of the orbit Op of p. By the Counting Formula (6.9.2),

(6.12.6) r pn p = N.

Iftw o poles p and p ' are in the same orbit, their orbits are equal, so n p = n pi, and therefore rp = r p . We label the various orbits arbitrarily, say as O \, O2, . . . Ok, and we let n j = n p and ri = rp for p in Oj, so that n ,r; = N. Since the orbit Oj contains n elements, there are H terms equal to rj - 1 on the left side of (6.12.5). We collect those terms together. This gives us the equation

k

I : n , ( n - 1) = 2N - 2 . i=-i

We divide both sides by N to get a famous formula:

(6.12.7) l ( l - ; r ) = 2 - ! •

This may not look like a promising tool, but in fact it tells us a great deal. The right side is between 1 and 2, while each term on the left is at least It follows that there can be at most three orbits.

The rest of the classification is made by listing the possibilities:

One orbit: 1 - * = 2 — ^ . This is impossible, because 1 - * < 1, while 2 — ^ 2: 1.

Two orbits-. (1 - * ) + (1 — = 2 - -|, t hat is, * + ± = ^ .

Because r; divides N, this equation holds only when r i = r2 = N, and then n i = n2 = 1.

There are two poles p i and P 2, both fixed by every element of the group. So G is the cyclic group C,v of rotations whose axis of rotation is the line l through p \ and P2.

Three orbits (1 - 1.) + (1 - 1.) + (1 — 1.) = 2 - 1. . . This is the most interesting case. Since ^ is positive, the formula implies that

1 1 1

(6.12.8) — + — + — > 1.

ri r2 r

We arrange the r (- in increasing order. Then ri = 2: If all rj were at least 3, the left side would be 1.

Case 1 \r \ = r2 = 2. The third order r 3 = k can be arbitrary, and N = 2k:

' = 2, 2, k; ni = k, k, 2; N = 2k.

There is one pair of poles {p, p'} making the orbit O 3. Half of the elements of G fix p, and the other half interchange p and p '. So the elements of G are rotations about the line .e through p and p ', or else they are rotations by rr about a line perpendicular to .e. The group G is the group of rotations fixing a regular k-gon A, the dihedral group Dfc. The polygon A lies in the plane perpendicular to .e, and the vertices and the centers of faces of A correspond to the remaining poles. The bilateral symmetries of A in K2 have become rotations through the angle rr in K3.

Case 2 r\ = 2 and 2 < r 2 n . The equation 1/2 + 1/4 + 1/4 = 1 rules out the possibility that r² :: 4. Therefore r² = 3. Then the equation 1/2 + 1/3 + 1/6 = 1 rules out 0 :: 6. Only three possibilities remain:

(6.12.9)

(i) r,- = 2, 3, 3; m = 6, 4, 4; N = 12.

The poles in the orbit O 3 are the vertices of a regular tetrahedron, and G is the tetrahedral group T of its 12 rotational symmetries.

(ii) ri = 2, 3, 4; ni = 12, 8, 6; N = 24.

The poles in the orbit O³ are the vertices of a regular octahedron, and G is the octahedral group O of its 24 rotational symmetries.

(iii) ri = 2, 3, 5; ni = 30, 20, 12; N = 60.

The poles in the orbit O3 are the vertices of a regular icosahedron, and G is the icosahedral group I of its 60 rotational symmetries.

In each case, the integers ni are the numbers of edges, faces, and vertices, respectively.

Intuitively, the poles in an orbit should be the vertices of a regular polyhedron because they must be evenly spaced on the sphere. However, this isn’t quite correct, because the centers of the edges of a cube, for example, form an orbit, but they do not span a regular polyhedron. The figure they span is called a truncated polyhedron.

We’ll verify the assertion of (iii). Let V be the orbit 03 of order twelve. We want to show that the poles in this orbit are the vertices of a regular icosahedron. Let p be one of the poles in V. Thinking of p as the north pole of the unit sphere gives us an equator and a south pole. Let H be the stabilizer of p. Since r 3 = 5, this is a cyclic group, generated by a rotation x about p with angle 2 rr/5. When we decompose V into H-orbits, we must get at least two H-orbits of order 1. These are the north and south poles. The ten other poles making up V form two H-orbits of order 5. We write them as {#q, . . . , q 4} and {q'0, . . . , q^}, where q (- = x'qQ and qi = x lq'Q. By symmetry between the north and south poles, one of these H-orbits is in the northern hemisphere and one is in the southern hemisphere, or else both are on the equator. Let’s say that the orbit {qd is in the northern hemisphere or on the equator.

Section 6.12 Finite Subgroups of the Rotation Group 187 Let |x, y| denote the spherical distance between points x and y on the unit sphere. We note that d = |p , qi \ is independent of i = 0, . . . , 4, because there is an element of H that carries qo q,, while fixing p. Similarly, d ' = Ip, qi l is independent of i. So as p ' ranges over the orbit V the distance Ip, p 'l takes on only four values 0, d, d ' and n. The values d and d' are taken on five times each, and 0 and n are taken on once. Since G operates transitively on V, we will obtain the same four values when p is replaced by any other pole in V.

We note that d :: n / 2 while d ' :: n /2 . Because there are five poles in the orbit {q,}, the spherical distance |q,-, q !+il is less than n /2 , so it is equal to d, and d < n /2 . Therefore that orbit isn’t on the equator. The three poles p, q,, q,+i form an equilateral triangle. There are five congruent equilateral triangles meeting at p , and therefore five congruent triangles meet at each pole. They form the faces of an icosahedron.

Note: There are just five regular polyhedra. This can be proved by counting the number of ways that one can begin to build one by bringing congruent regular polygons together at a vertex. One can assemble three, four, or five equilateral triangles, three squares, or three regular pentagons. (Six triangles, four squares, or three hexagons glue together into flat surfaces.) So there are just five possibilities. But this analysis omits the interesting question of existence. Does an icosahedron exist? Of course, we can build one out of cardboard. But when we do, the triangles never fit together precisely, and we take it on faith that this is due to our imprecision. If we drew the analogous conclusion about the circle of fifths in music, we’d be wrong: the circle of fifths almost closes up, but not quite. The best way to be sure that the icosahedron exists may be to write down the coordinates of its vertices and check

the distances. This is Exercise 12.7. □

Our discussion of the isometries of the plane has analogues for the group of isometries of three-space. One can define the notion of a crystallographic group, a discrete subgroup whose translation group is a three-dimensional lattice. The crystallographic groups are anal­

ogous to two-dimensional lattice groups, and crystals form examples of three-dimensional configurations having such groups as symmetry. It can be shown that there are 230 types of crystallographic groups, analogous to the 17 lattice groups (6.6.2). This is too long a list to be useful,.so crystals have been classified more crudely into seven crystal systems. For more about this, and for a discussion of the 32 crystallographic point groups, look in a book on crystallography, such as [Schwarzenbach].

Un bon heritage vaut mieux que Ie plus joli probleme de geometrie, parce qu'il tient lieu de methode generale, et sert a resoudre bien des problemes.

—Gottfried Wilhelm Leibnitz2

21 learned this q u ote from V .l. A rnold. I’HOpital had written to L eibniz, apologizing for a long silence, and saying that h e had b een in the country taking care o f an inheritance. In his reply, L eibniz told him not to worry, and continued with the sen ten ce quoted.

E X E R C IS E S

Section 1 Symmetry o f Plane Figures

1.1. Determine all symmetries of Figures 6.1.4, 6.1.6, and 6.1.7.

Section 3 Isometries o f the Plane 3.1. Verify the rules (6.3.3).

3.2. Let m be an orientation-reversing isometry. Prove algebraically that m 2 is a translation.

3.3. Prove that a linear operator on ]R2 is a reflection if and only if its eigenvalues are 1 and -1 , and the eigenvectors with these eigenvalues are orthogonal.

3.4. Prove that a conjugate of a glide reflection in M is a glide reflection, and that the glide vectors have the same length.

3.5. Write formulas for the isometries (6.3.1) in terms of a complex variable z = x + iy.

3.6. (a) Let s be the rotation of the plane with angle rr/2 about the point (1, 1)'. Write the formula for s as a product taPe.

(b) Let s denote reflection of the plane about the vertical axis x = 1. Find an isometry g such that grg-1 = s, and write s in the form taper.

Section 4 Finite Groups of Orthogonal Operators on tbe Plane

4.1. Write the product x2yx~1 y~!x3y3 in the form x' yj in the dihedral group D n.

4.2. (a) List all subgroups of the dihedral group D⁴, and decide which ones are normal.

(b) List the proper normal subgroups N of the dihedral group D 15, and identify the quotient groups D 15 / N.

(c) List the subgroups of Dg that do not contain x3.

4.3. (a) Compute the left cosets of the subgroup H = {l, xS} in the dihedral group Dio.

(b) Prove that H is normal and that D io /H is isomorphic to D5.

( c ) Is Dio isomorphic to D 5 x H?

Section 5 Discrete Groups of Isometries

5.1. Let i i and £2 be lines through the origin in JR. 2 that intersect in an angle rr/n , and let r< be the reflection about £,-. Prove that ri and r2 generate a dihedral group Dn.

5.2. What is the crystallographic restriction for a discrete group of isometries whose translation group L has the form Za with a -=1= O?

5.3. How many sublattices of index 3 are contained in a lattice L in ]R2?

5.4. Let (a, b) be a lattice basis of a lattice L in ]R2. Prove that every other lattice basis has the form (a', b') = (a, b )P , where P is a 2 x 2 integer matrix with determinant ± 1.

5.5. Prove that the group of symmetries of the frieze pattern is isomorphic to the direct product C2 X Coo of a cyclic group of order 2 and an infinite cyclic group.

5.6. Let G be the group of symmetries of the frieze pattern L, H L, H L, H L, r 1 . Determine the point group G of G, and the index in G of its subgroup of translations.