Twisted Honeycombs 0821816535, 9780821816530

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Conference Board of the Mathematical Sciences

REGIONAL CONFERENCE SERIES IN MATHEMATICS supported by the National Science Foundation

Number 4

TWISTED HONEYCOMBS by H. S. M. Coxeter

Published for the Conference Board of the Mathematical Sciences by the American Mathematical Society Providence, Rhode Island

Other Monographs in this Series No. l.

Irving Kaplansky:

Algebraic and analytic aspects of operator algebras.

2.

Gilbert Baumslag:

Lecture notes on nilpotent groups.

3,

Lawrence Markus:

Lectures in differentiable dynamics.

5.

George W. Whitehead:

6.

Walter Rudin:

Recent advances in homotopy theory.

Lectures on the edge-of-the-wedge theorem.

7. Yozo Matsushima:

Holomorphic vector fields of compact Kcihler manifolds.

International Standard Book Number 0-8218-1653-5 Library of Congress Catalog Card Number 75-145638 AMS 1970 Subject Classifications: 05B30, 05B45, 05C25, 16A40, 20B25, 20F05, 20Hl5, 50A10, 50B30, 50Cl5, 55A15, 57 A05, 57AlO. Key Words and Phrases: dodecahedron spaces, finite groups of quaternions, finite groups of rotations, fundamental group of a manifold, fundamental region, glide, hyperbolic geometry, linear fractional group, octahedron space, orthoscheme, Petrie polygon, Platonic solids and their symmetry groups, presentation of a group, pure quaternion, reflection groups, regular honeycomb, regular polytope, right and left screws, Schla."fli symbol, simplex, spherical geometry, torus, truncated cube space, vertex figure.

Copyright © 1970 by the American Mathematical Society Printed in the United States of America All rights reserved except those granted to the United States Government May not be reproduced in any form without permission of the publishers.

CONTENTS 1. Introduction . . . . . . . . . . • . . . • . . • . • . . . • • . . • • . . . . • . . . . . . • . • • • • • . . • • . . . . . . •

1

2. 'Regular tess~llations .•....•.•..•................•......••...•.•...... ,

2

3. Finite groups of rotations . . . . . . . . . . . . . . . • . . . . . . . . . . • • • . . . . • • . . • . . . . . . . .

3

4. Groups generated by two or three reflections •.•.....••.•.•..••.•.• , • . . • • . .

5

5. The Petrie polygon of a tessellation • . . . . . . . • . . . • • . . . . • . • • • • . • . . • . . . . . . • .

8

6. The regular tessellations {p, q}r •• ~ ••••••••••••••••••••••••••••••••••••• 11 7. Three dimensional honeycombs . • . . . . • • . . • . . • . • • . • • • . . • . • . • • . . • . • • . . • . . . 14 0

8, The simplex, cube, cross polytope and 24-cell .......••••.•••••...•..•••.• 17 9, The 120-cell and the 600-cell •••.•..••••.••.•.•••.••••..••••...••.••..•• 19 10. Groups generated by four reflections ••••..••.•••••..•••.••••.•.•••••••••• 24 11, The Petrie polygon of a honeycomb .•..•••.••..•.••••..•••••••.••••.••••. 25 12, The regular honeycombs {p, q, r}t ••••••••••••••••••••••••••••••••••••••• 28 13, Finite groups of quaternions •..•••...• , ..••••.••.• ,,, .•.••••••..••.•• ,, ••. 36 14, Quaternions and Petrie polygons ..•.•.••••.•••.• , .••••.••••••••..•••.••• 42 Table L Regular tessellations {p, q }r and their groups , • , , •••• , •••••••••• , 44 Table 2, Reflexible honeycombs and their groups ••.••••••••..••..••••••.• 45 Table 3. Twisted honeycombs and their groups .•.•..••••••••••..••.•••••• 46 References ••.••.•••.•••••.•••••••••..•.••••..••••.•.••••••••••••••.•. 47

iii

PREFACE Following D. M. Y. Sommerville [7, p~ 166] we define a honeycomb to be a symmet· rical subdivision of a three-dimensional manifold into a number of polyhedral cells, all alike, each rotation that is a symmetry operation of a cell being also a symmetry opera· tion of the whole configuration. The honeycomb is said to be twisted if it is not sym· metrical by reflection; thus a twisted honeycomb, like a screw, occurs in right-handed and left-handed varieties. The subject originated in 1933, when Weber and Seifert [IO] considered two one-celled honeycombs, each consisting of a regular dodecahedron with opposite faces identified. The work involves a combination of geometry and group theory, assisted by quaternions. I wish to thank the National Science Foundation and the University of Maine for sponsoring this series of lectures, and to thank Norman W. Johnson and Grattan P. Murphy for taking notes. I gratefully acknowledge also the help of John Leech and Abraham Sinkov, who computed the orders of certain groups presented in the form

L 2 = M2 = N 2 = (LM)P = (MNY

=

(LN/

=

(LMN)q

=

1,

and found representations for these groups.

H. S. M. Coxeter

February 1970

iv

TWISTED HONEYCOMBS

1. Introduction We begin by considering regular tessellations. One example is the tessellation of the Euclidean plane E 2 by squares whose vertices are the "lattice points" (x, y), x and y running independently over the ,set of all integers. This tessellation is de-

noted by the Schlafli symbol {4, 4} because it consists of squares, four at each vertex. More generally, {p, q} denotes a tessellation (of the Euclidean plane or hyperbolic plane or sphere) by p-gons {p}, q of which surround each vertex. The p-gons are called "faces". The centers of the faces are the vertices of the dual (or "reciprocal") tessellation {q, p}, whose edges cross those of {p, q}. A typical face of {4, 4} is the square

(0, 1),

(1, 1), ·

(0, 0),

(1, 0),

whose interior consists of the points (x, y) with O < x

< 1,

,0

< y < I.

By reducing

the coordinates of all points (x, y) to their fractional parts (or residues modulo 1), we represent all the infinitely many faces of {4, 4} by this single face with its pairs of opposite sides-abstractly identified. This is the familiar representation of a torus, cut along two circles that have just one common point, unfolded, and suitably distorted so as to lie flat. In this manner the infinite regular tessellation of squares in the Euclidean plane is reduced to a finite regular tessellation having one vertex, two edges, and one face, on the torus. The symmetry group S of the infinite tessellation has a normal subgroup T, generated' by the translations X and Y that increase x and y (separately) by I. The symmetry group of the finite tessellation on the toms is the quotient group

S/ T. Of

course, this is simply the symmetry group of the square itself. The group T has the presentation

XY = YX or, briefly, X t Y; it is the fundamental group of the torus. Analogously in, Euclidean space E 3, the lattice points (x, y, z), where x, y, z are integers, are the vertices of a regular honeycomb of cubes. A typical cube is shown in Figure I. Translations X, Y, Z, along the three edges .at the vertex (0, 0, 0), generate a group having the presentation

Y ~ Z, Z ;:! X, X t Y.

1

H. S. M. COXETER

2

(o,o,

(0,1,1) f'------------c..- 77, =

or

77,

< 77

q-l

according as the tessellation is on a

sphere S 2, the Euclidean plane E 2 , or the hyperbolic plane H 2 • Thus the criteria to distinguish these. three kinds of geometry are respectively 1

1

(p - 2) (q - 2)

< 4,

1

1

1

11

1

1

-+-> -2' -+-=- +-

4.

In particular, Euclid's famous problem of enumerating the five Platonic solids is re-

duced to the trivial problem of finding pairs of positive integers whose products are less than 4. One effect of considering spherical tessellations instead of solid polyhedra is to allow (p -:2) (q - 2) = O,

that is, to consider not only pairs of positive integers but pairs of nonnegative integers. The tessellation {p, 2 l, whose p vertices are evenly spaced along the "equator", is called the p-gonal dihedron. The dual tessellation {2, pl, whose p edges are evenly spaced "meridians" (great semicircles joining its two antipodal vertices), is called the p-gonal hos ohedron. As extreme cases we may even admit the "monogonal dihedron" {1, 2 l which has two faces (the northern and southern hemispheres) but only one vertex, and the "monohedron" {2, ll which has one face (covering the whole sphere) and one semicircular edge. However, for tessellations of the Euclidean plane, or of the hyperbolic plane, it is natural to insist that both p and q should be greater than 2. 3. Finite groups of rotations The spherical tessellations {p,

ql, where p and q are integers, play a role in

Klein's enumeration of the finite groups of rotations [4, pp. 270-276]: the cyclic group C

p

which cyclically permutes the vertices of a regular p-gon; the dihedral

group Dp (of order 2p) which combines Cp with p half-turns interchanging the two faces of the dihedron {p, 2l or the two vertices of the hosohedron {2, pl; the tetra-

hedral group A

4

(of order 12) which is the alternating group of degree 4, evenly per-

muting the 4 faces (or vertices) of the tetrahedron {3, 3l; the octahedral group S

4

( of order 24) which is the symmetric group permuting the 4 pairs of opposite faces of the octahedron {3, 4} (and thus also permuting the 4 diagonals that join pairs of opposite vertices of the cube {4, 3D; and the icosahedral group A

5

(of order 60) which

is the alternating group of degree 5, evenly permuting the 5 cubes that can be inscribed

H.S.M. COXETER

4

in the dodecahedron {5, 3} [6, p. 19, footnote]. This last remark can be elucidated as follows.

e

e

e

Figure 3: {3, 5}

Figure 2: {5, 3}

a

cL

b _____..,__ _ _ _ _-< C

~

, ~ - - - - - --- -

b

d1....r-------~ d.

Figure 4: {4, 3}

Each edge of the dodecahedron (Figure 2) visibly determines a square whose vertices are the four further ends of the other two edges through each end of the given edge. These four vertices of the dodecahedron, along with their opposites, form an

5

TWISTED HONEYCOMBS

inscribed cube. The same cube arises equally well from five other edges of the dodecahedron, namely, all the edges that are either parallel or perpendicular to the given edge. Accordingly, the 30 edges of the dodecahedron {5, 3 }, or of the reciprocal icosahedron {3, 5} (Figure 3), can be marked in sets of 6 with the 5 letters a, b, c, d, e in such a way that any two edges marked alike are either parallel (that is, opposite) or perpendicular. It thus becomes clear that the dodecahedron has just 5 inscribed cubes. (Each cube has 6 faces, exactly accounting for all the 30 edges of the dodecahedron [5, p. 32].) The even permutations (abcde), (adb), (be) (de)

are easily recognized as rotations that are symmetry operations of both {5, 3} and {3, 5 }. Any two of these three permutations will suffice to generate the alternating group A , which is thus seen to be isomorphic to the icosahedral group of rotations. 5

Figure 4 shows the "e" cube by itself, with its vertices systematically marked

a, b, c, d. The two opposite vertices marked a coincide with those vertices of the dodecahedron which belong to edges marked with the three remaining letters: b, c, d. The even permutations of a, b, c, d represent rotations that belong to the dodecahedron as well as the cube, exhibiting A

4

.

as a subgroup of A

•

5

Preserving the distinc-

tion between black and white vertices in the figures, these rotations are symmetries of either of the two tetrahedra that can be inscribed in the cube. On the other hand, the odd permutations of a, b, c, d represent rotations that belong only to the cube, not to the dodecahedron. They interchange the two inscribed tetrahedra. 4 .. Groups generated by two or three reflections

In §2 we defined, for each regular tessellation {p,

ql,

a characteristic triangle

(p q 2) whose angles are 7T

7T

7T

p'q'2· The angle rr/p occurs at the center I of a face ABC, the angJe rr/ q at the vertex C, and the right angle at the midpoint M of the edge BC. (See Figure 5 for the case when p = 4 and q = 3.) The complete symmetry group of {p, q}, say [p, q], is generiated by reflections Rv in the three sides of this triangle, as follows. R

1

,

reversing

the edge BC, is the reflection in IM; R 2, interchanging BC with another edge of the same face, is the reflection in /C; and R, ,reflecting the face ABC into an adjacent .

3

face, is the reflection in MC ( or in BC). Thus R

1

and R

2

satisfy the relations

Ri =(RI R)P = I and generate the dihedral group Dp or [p ], which is the symmetry group of the face {p }, while R

2

and R , ·Satisfying (R R ) q = I, generate the symmetry group [q] of 3

2

3

the vertex figure {q}. The combined relations

H. S.M. COXETER

6

4.1 [3, p. 80] suffice for a presentation of the whole group [p, q]. The rotations A

=

R 1 R 2 and B = R 2 R 3 (which appear in Figure 5 as rotations

through 2 77/p and 277/ q about the points / and C) generate the "positive" subgroup [p, q] +, of index 2, which has the presentation

AP= Bq =(AB) 2 = 1. Our discussion in §3 shows that the finite groups of rotations are the cyclic group of order p with the presentation AP = 1, and

[p, 2] + ~ D p , , of order 2p, [3,, 3] + ~ A [4, 3] + ~ S

4

4

, · of

order 12,

of order 24,

,

[5, 3] + ~ A , of order 60. 5

The isomorphism of [p, 2] + and [p] can be explained by the observation that the former is generated by half-turns about two intersecting lines ( diameters of the sphere), , and the effect of these half-turns on the plane spanned by the lines is the same as that of ,reflections in these lines, or of reflections in planes through the lines perpendicular to their plane. The characteristic triangle ICM (Figure 5) is a fundamental region for the group [p, q] generated by reflections in its sides Ml, IC, CM. In other words, there is one replica of the triangle for each element of the group. These replicas, bounded by the

lines of symmetry of the regular tessellation, fill and cover the plane (that is, the sphere, or the Euclidean or hyperbolic plane). After shading alternate triangles, so that each reflection interchanges shaded and unshaded triangles, we can assert that [p, q] + is the subgroup that preserves the shading. Each face of {p, q} is divided by its lines of symmetry into 2p triangles (p q 2), p shaded and p unshaded; each edge is surrounded by four of them, 2 shaded and 2 unshaded; and each face of the dual tessellation {q, p} is made up of 2 q of them. In the spherical case it follows that, if the polyhedron {p, q} has F faces, .£ edges and V vertices,

pF

4.2

=

2E

=

qV.

Combining this result with Euler's formula F - E + V = 2, we get

4.3

1

1

1

1

------+-. E P 2 q

The same expression can alternatively be obtained from the fact that the area of the whole spherical surface must be 4£ times the area of one spherical triangle (p q 2).

TWISTED HONEYCOMBS

Figure 5 : [4, 3] and (4 3 2)

7

H.S.M. COXETER

8

5. The Petrie polygon of a tessellation Among the vertices and edges of any regular tessellation {p,

qJ,

we can pick out

(in many equivalent ways) a Petrie polygon. This is a zigzag in which every .two con· secutive edges belong to a face but no three consecutive edges belong to the same face. For instance, in Figure 5, where ABC is one face and BCD (only partially visible) is an adjacent face, AB CD·,, is a Petrie polygon. The isometry

ABC-- BCD, which takes us one step along the Petrie polygon, may be expressed as the product of three reflections, namely

RI R2R3. Being conjugate to R R R 2

3

, 1

this is a glide reflection (which crystallographers prefer

to call simply a "glide"): the product of the reflection in the line lB and the halfturn about M. By regarding the half-turn as the product of two reflections, we obtain another expression for the glide reflection as the product of a translation along a line

l and the reflection in this same line [ 4, p. 43]. This line l, which passes through the midpoints of all the edges of the Petrie polygon ABCD ·, ·, is conveniently called an equator. In the spherical case, l is, of course, a great circle, the "translation" along it is a rotation about its pole, and the glide reflection R R R 1

2

3

is a rotatory reflection

[ 4, p. 99]. Let h denote the period of this rotatory reflection, so that (if p q

> 2 and

> 2) the Petrie polygon is a skew h-gon. Since the zigzag has equally many zigs

and zags, h is an even integer. We shall find three different expressions for it as a function of p and q, one trigonometric and two algebraic. Looking once more at Figure 5, we observe that the equator (which passes through the midpoints of the edges of the Petrie polygon} crosses h shaded and h unshaded triangles, each congruent to the characteristic triangle ICM (Figure 6). Thus the altitude from M ·of this right-angled spherical triangle is of length Since the angles at / and C are 77/p and spherical trigonometry that, if MC = ¢,, cos

5.1

,./..

'P

•

7T

sin -

q

= cos

sin

2

77/q,

7T

• ,./.. • 7T Sln 'P sin -

- ,

p

=

q

!!.. •= cos E.. + sin 2

q

2 77

p

h

and, finally,

5.2

cos 2 !!_ = cos 2

h

7T

p

77/

h.

it follows from classical formulae of

+ cos 2

77

q

• 7T Sl.n - . '

h

TWISTED HONEYCOMBS

9

C

Figure 6: (p q 2)

One algebraic expression for h can be obtained by counting the equators and their points of intersection. Every edge belongs to two faces, and so also to two Petrie polygons. Since there are E edges, where E is given by 4.3, there are 2E/h Petrie polygons and 2E/ h equators. Since the

[2E:l 2/! (3: ~ 1]

2

points of intersection of pairs of equators are just the E mid-edge points like M, we have

~E [

2: -IJ

= E,

whence

E = !:_

5.3

2

[!:..2 + I J'

and thus we see why the number of edges of a Platonic solid is always the product of two consecutive integers. The equation 5.3 can be solved for h, but a more agreeable expression, free from square roots, may be obtained by counting the lines of symmetry (or circles of symmetry, or planes of symmetry, which are mirrors reflecting the whole pattern into itself). We count these circles of symmetry by counting their pairs of antipodal points of intersection with a single equator. The equator crosses h hypotenuses like [B ( each being an arc of a circle of symmetry) and passes through h mid-edge points like M (each a crossing point of two perpendicular circles of symmetry). Thus there

H. S.M. COXETER

10

are altogether 3h/2 circles of symmetry. Pairs of these 3h/2 great circles intersect in

~h (3; _ J 1

points, not all distinct. These intersections consist of ( like I,

[

iJ

~

J

at each of the F points

at each of the V points like C, and one at each of the E points like

M. Using 4.2 to express F and V in terms of p, q and E, we deduce

1; [3; -1 J (n 2:-+ (fl3f =

+ E

(p - 1) E + (q - 1) E + E

(p + q - 1) E. Combining this with 5.3, we obtain 3 (3h - 2) 24 p+q-1=------=9----

h+2

h+2'

whence

h - +1 2

5.4

12

=-----10-p-q

and, using 5.3 again, 5.5 An interesting by-product of this last formula is an expression [3, p. 74] for the perimeter

¢ +

x + ijJ =MC+ CI+ IM

of the characteristic triangle (p q 2). Since the sides of all the 4£ congruent spherical triangles together cover twice the circumferences of the 3h/2 circles of symmetry, we have

4E (¢ + x + ijJ) = 3h · 277 = 6h17. Therefore

5.6

¢ +X+ Since R R R 1

2

3

i/I

3h17 IO - p - q = - - = ---:----77.

2E

4

is a rotatory reflection ( the product of the reflection in the equa-

tor and a rotation that slides the equator along itself), the odd powers of R R R 1

2

3

are

again rotatory reflections whereas the even powers are pure rotations. In particular, since

(R 1Ri 3)h = 1, the isometry (R 1R 2R

)h 12 , which takes us half-way round the 3

equator, is a half-turn if h/2 is even (that is, if {p, q} is the tetrahedron {3, 3 });

11

TWISTED HONEYCOMBS

but if h/2 is odd (as it is for the remaining Platonic solids), (R R R )h 12 is the 1

2

3

central inversion Z, which transforms each point on the sphere into the antipodal point. (In terms of Cartesian coordinates, this means that Z transforms (x, y, z) into (- x, - y, - z).) By abstractly identifying each pair of antipodal points of the sphere, we derive the elliptic plane (or projective plane): a nonorientable surface whose fundamental group is the group of order 2 generated by Z. In this manner, the centrally symmetric regular tessellations

{4, 3}, {3, 4}, {5, 3}, {3, 5} of the sphere yield the regular tessellations

of the elliptic plane. In each case the subscript is h/2, ,indicating that we are ident· ifying pairs of points separated by this number of steps along every Petrie polygon. The symmetry group of the spherical tessellation {p, q}, namely

5.7 has a normal subgroup of order 2 generated by the central inversion Z = (R

RR

1 2 3

)hl 2•

The quotient group is the symmetry group of the elliptic tessellation {p, q} hi r 6. The regular tessellations {p, q},

Analogously, whenever the symmetry group of a Euclidean or hyperbolic tessellation {p, q} has a normal subgroup generate_d by (R R R 1

2

)r 3

and its conjugates, we can

identify pairs of points separated by r steps along every Petrie polygon to obtain a regular tessellation {p, q}, of a multiply-connected surface. If the quotient group 6.1 has finite order g, the tessellation has g/2p faces. In fact, the reflections R

1

and

R 2 generate a dihedral subgroup or order 2p leaving one face invariant, and there is one face for each coset of this subgroup. Similarly, there are g/2q vertices and g/4 edges, in: agreement with 4.2. (Figure 7 shows {4, 5} , with g = 160, as an unfolded 5

net. Points numbered alike are to be identified. A typical Petrie polygon is the pentagon 56789.) Expressing the same group in terms of new generators

so that R 1 = BC, R 2

6.2

= BCA, R 3 = CA,

we obtain the more symmetrical presentation

AP= Bq = C' = (AB) 2 = (BC) 2 = (CA) 2 = (ABC) 2 = 1,

H.S.M. COXETER

12

for which a convenient symbol [1, p. 104] is GP,q,r. Whenever 9

Figure 7: {4, 5} 5 this group is finite, it is the symmetry group of each of the six tessellations

{p, q},, {q, p},, {q, r}P, {r, q}P, {r, p}q, {p, r}q, which are all distinct if the numbers p, q, r are all distinct. The known cases are listed in Table 1 where, for simplicity, just one tessellation is shown in each case, namely the one for which p ~ q ~ r. The relationship of {p, ql, and {r, qlP is interesting: their vertices and edges form the same topological graph, involving circuits of edges which are p-gons and r-gons; these circuits, being the faces and Petrie polygons of {p, ql , ,are also the Petrie polygons and faces of {r, q} • P

r

Since G 3 • 7 • 1 2 and C 3 • 7 ' 14 are isomorphic, it is remarkable that C 3, 7 • 13 is the smaller group P SL (2, 13) or, in Dickson's notation, LF (2, 13). (This was pointed out by John Leech.) Since it is known that G3,q,r is infinite whenever

4rr

6.3

4rr

.1

cos - - +cos--> q r - 2

with q and r even [2, p. 54], it was natural to conjecture that the same criterion would hold when q or r is odd. This notion was supported by whereas C 3 ' 7 ' 16 is finite, C 3 ' 7 ' 18 is infinite. Since cos

4rr 4rr 7 .... + cos IT=

0.515 ... ,

c.

C. Sim's proof that,

TWISTED HONEYCOMBS

13

the intermediate group G 3 • 7 • 17 also should be infinite. In 1969, M. J. T. Guy,

J.

Mc

Kay and J. H. Conway presented this question to the Atlas computer at Chilton, England. After enumerating more than fifteen thousand apparently distinct cosets of the subgroup D

17

generated by AB and C, the procedure collapsed, indicating that

G 3 • 7 ' 17 is of order 1. In 1970, Mrs. Jane Watson of Canberra, Australia, obtained the same result, reducing the number of cosets to 4755 (and the machine time to forty seconds). Apart from this sad case of total collapse, the only finite group GP,q,r discovered since 1962 is G 3 • 9 • lO (compare [1, p. 143, third footnote]). In fact, the relations 6.2 are easily seen to imply

If, as in the present case, r is even, GP,q,r has a subgroup of index 2 generated by

A and B, with the presentation 6.4 It was observed by Abraham Sinkov in 1967 that the group

A 3 = B 9 = (AB) 2 = (A 2 B 2 ) 5 = 1 is the direct product P SL (2, 19) x C , with generators 3

A= w !·

8 - 8

2 ],

-

B-w

9

2

L: -:1

(modl9),

where w 3 = 1 and the matrices are "homogeneous" so that we do not distinguish between

[

a C

l

b and [- a - b dj - C - d

l·

It follows that PSL (2, 19) itself is generated by

L =AB=

[O 1

-11 0

,

M=BA=

9] , N = AB

r-7

L

5

7

-1 A=

[6 7

- SJ.

-6

(mod 19),

with the presentation ·

6.5

L2 = M2 = N 2 = (LM) 5 = (MN) 5 = (NL) 5 = (LMN) 3 = 1,.

while the factor C is generated by 3

Adjoining to 6.4 a suitable C (such that BCA, of period 2, will transform A into A - l and B into B- 1), ·We obtain for G 3 • 9 • 10 the generators

[-88 -92].,

A=w.

B=w 2 [

2 - 9],

-8

8

C=

i

f5 [1

(mod 19),

H. S. M. COXETER

14 where w and

L ( commuting

with the matrices) generate D or S 3

3

( the

dihedral group

of order 6) in the form w 3 = L 2 = (wd 2 = 1.

Thus c3,9, 10 ~

6.6

PSL(2, 19) x D3°

The chart of groups G3,x,y plotted as points (x, y) [2, p. 54] requires correction in at least six places: at the points (7, 13) and (13, 7), the number 2184 should be 1092; at (9, 10) and (10, 9) the cross should be an o, with the number 20520; at (18, 7) there should be a big black dot (to indicate infinite order); and there should be a cross at (17, 7)•

7. Three-dimensional honeycombs As we saw in §2, each regular tessellation {p, q} has a vertex figure {q} whose vertices are the midpoints of the q edges that meet at one vertex. Analogously, the line-segment whose ends are the midpoints of two adjacent edges (or sides) of a regular polygon is the "vertex figure" of the polygon, and the midpoint of a line-segment is the "vertex figure" of the line-segment. Thus the vertex figure of a polygon is the line-segment whose end-points are the vertex figures of the two edges at one vertex, the vertex figure of a polyhedron or tessellation {p, q} is the polygon {q} whose edges are the vertex figures of the q faces at one vertex, and the analogous "vertex figure" of the honeycomb of cubes (§1) is the polyhedron whose faces are the vertex figures of the eight cubic cells that come together at one vertex. By taking the one vertex to be the origin (0, 0, 0), we see that this polyhedron is the octahedron {3, 4} whose vertices are the midpoints of the six edges that join the origin to the neighboring lattice points

( ± 1, 0, 0),

(0, ± 1, 0),

(0, 0, ± 1).

More generally, in Euclidean, spherical or hyperbolic 3-space [7, pp. 166, 169], a regular honeycomb {p, q, r} is a collection of nonoverlapping Platonic solids {p, q} such that every face {p} belongs to just two of them, and every edge to r of them. In other words, the Schlafli symbol {p, q, r} indicates that each solid cell (or "facet") is a {p, q} and that r such cells come together at each edge. By considering the vertex figures {q} at one vertex of all the cells that share this vertex, we obtain a polyhedron whose faces are {q }'s, of which r come together at each vertex; thus the vertex figure of the honeycomb is {q, r}. The symbol {p, q, r} is obtained by "telescoping" the symbols {p, q} and {q, r} which represent the cell and vertex figure, respectively. In particular, the familiar honeycomb of cubes is {4, 3, 4}.

From the {4, 3, 4} whose vertices have integers for their Cartesian coordinates,

15

TWISTED HONEYCOMBS

we can derive another {4, 3, 4} by applying the translation that adds 1/2 to each of the three coordinates. Each vertex of either honeycomb is the center of a cell of the other, and each edge of either honeycomb joins the centers of two adjacent cells of the other, and thus crosses the common face of these two cells. Accordingly we say that these two honeycombs are duals (of each other). More generally, every regular honeycomb {p, q, r} has a dual {r, q, p}: the vertex figure of each is the dual of the cell of the other.

B I)•

P, '

E

Figure 9

Figure 8

We saw, in §2, ·that each regular tessellation {p, q} has a "characteristic .triangle" whose angles are 1r/p, 1r/ q, 1r/2. Analogously, each regular honeycomb {p, q, r} has a characteristic tetrahedron P P P P O

1

2

3

(see Figure 8), where P

O

is a vertex,

P 1 is the midpoint of an edge containing this vertex, P 2 is the center of a face containing this edge, and P is the center of a cell containing this face. Such a tetra~ 3

hedron, having four right-angled triangles for its faces, is called an orthoscheme [ 4, p. 155] .. Figure 9 shows its net, obtained by cutting along the edges P P , P P , O

1

1

2

P2 P3 and foldingthe faces back to form a plane figure. Clearly P0 P1 , being half an edge of the honeycomb, is perpendicular to the plane P P P ; and P P , perpendicu1 2 3 3 2 lar to the plane PO P P2 , is half an edge of the dual honeycomb {r, q, p }.. Thus the 1 dual honeycomb has the same characteristic tetrahedron with its vertices named in the reverse order. The four face-planes of the tetrahedron are easily seen to be planes of symmetry of the honeycomb. To agree with existing literature, we let R

V

denote the reflection

H.S.M.COXETER

16

in the face opposite to the vertex P

R 1 , the reflection in P1 P2 P3 , reverses the edge CD (Figure 8); R , ,the reflection in PO P2 P3 , interchanges CD 2 with another edge of the same face BCD; R , the reflection in PO P1 P3 , interchanges 3 this face BCD with another face (namely CDE') of the same cell ABCD; and R 4 , interchanges this cell with its neighbor on the other side t he reflection in P P P 11

0

1

_

1. Thus

2'

of the plane BCD. On a sphere with center P , ,the three mirrors through this point cut out a spheri3

cal triangle which is the characteristic triangle (Figure 6) for the cell {p, q }. Thus the dihedral angles at the edges P2 P , POP , P1 P are 3 3 3 angles are

Similarly, on a sphere with center P may be denoted by

0

¢,', x', 'tj/ because

,

1T/ p, 1T/ q, TT/2,

and the face-

we obtain a spherical triangle whose sides it is the characteristic triangle for the vertex

figure {q, r}. Thus the dehedral angles at the edges P0 P , P0 P1, P0 P2 are 3

TT/q, TT/r, TT/2,

and the face angles are as marked in Figure 9. What makes {4, 3, 4} a Euclidean honeycomb, rather than a hyperbolic or spherical honeycomb, is the fact that the dihedral angle of the cube {4, 3} is exactly a right angle, allowing 4 cubes to come together at a common edge. (This 4 is .the third digit in the Schlafli symbol {4, 3, 4}.} If, instead of being Euclidean (£ 3), the space filled by a honeycomb is hyperbolic (li3) or spherical (S 3), ·the dihedral angle of the

hexahedron {4, 3}

(Should we still call it a cube?) is acute or obtuse, respectively.

An infinitesimal hexahedron is indistinguishable from a cube, but as the edge-length increases the dihedral angle departs more and more from the right angle. In the hyperbolic case the dihedral angle decreases steadily from

TT/2

to

TT/3.

When it reaches the value 2TT/5, just five hexahedra will fit together round a common edge, and we have the hyperbolic honeycomb {4, 3, 5} in which every vertex is surrounded by twenty hexahedra, arranged like the faces of an icosahedron {3, 5} (the vertex figure) ..

In the spherical case the dihedral angle of {4, 3} increases steadily from TT/2 to

TT.

When it reaches the value 2TT/3, ,just 3 hexahedra will fit together round a com-

mon edge, and we have the spherical honeycomb {4, 3, 3} iri which every vertex is surrounded by four hexahedra arranged like the faces of a tetrahedroQ {3, 3} .. When the spherical space S 3 is regarded as a 3-sphere (or "hypersphere")-in Euclidean 4-space £4, ,the great-circle arcs that are the edges of this finite honeycomb' can be replaced by their chords to form the edges of the corresponding 4-dimensional polytope, for which the Schlafli symbol {4, 3, 3} is still appropriate. The hexahedraf cells now become cubes, iri fact 8 cubes, as we see in Figure IO. We recognize this polytope as the 4-dimensional cube (or "hypercube", or "tessaract" or "S·cell").

17

TWISTED HONEYCOMBS

8. The simplex, cube, cross polytope and 24·cell Let a finite honeycomb ot polytope of type {p, q, r} ha.ve N

O

vertices, N

1

edges,

N 2 faces, and N 3 cells. Let the cell {p, q} have V vertices, E edges and F faces, as in 4.2 or 4.3; and let the analogous numbers for the vertex figure be V', E', F'. Since the FN 3 faces of the N cells are just all the N faces of {p, q, r}, each 2 3 counted twice, we have

Similarly

EN 3 = rN 1 = pN 2 = E' N 0 [4, p. 398]. Hence

11111

8.1

1

1

1

N :N :N :N = - - - + - · : - : - · : - · - - + o 1 2 3 q 2 TT pp 2 q ={4-(q-2) (r- 2)}p: 2pq: 2qr: {4 -(p - 2) (q- 2)}r,

and we easily verify that

N 0 - N 1 + N 2 - N 3 = o. This is the analogue of Euler's formula V - E + F = 2.

The simplest regular polytope in 4 dimensions is the simplex (or "5-cell") {3, 3, 3 }, whose 5 vertices are all equidistant from one another. Since every 2 of the

NI

j

6/

C

.D

0

,I

E

F

I

10.1

p

q

7

q

2

2

Thus the possible regular honeycombs are

{3, 3, 3},

{4, 3, 3},

{3, 4, 3},

{5, 3, 3},

{4, 3, 4 }, {5, 3, 4},

{3, 5, 3},

{5, 3, 5}

and their duals. We have seen that the .first four are finite (spherical) and the next one is Euclidean. It remains to be proved (more rigorously than in

§s)

that the last

three are hyperbolic. One approach uses the complete symmetry group [p, q, r], ,which is generated by

R

1

,

R

2

,

R

3

,

R

4

(§7) with the presentation

R~= (RlR2)P = (R2R3)q = (R3R4)' 10.2

= (R i R 3) 2 = (RI R 4) 2 = (R 2R 4) 2 = 1 [3, p. 188]. Since [p, q] is a subgroup of index N

3

in [p, q, r ],- we see from §§8 and

9 that

[3, 3, 3], · [4, 3, 3], [3, 4, 3], [5, 3, 3] are finite groups, their orders being respectively 120, 384, 1152, 14400. We see also that [4, 3, 4], containing translations, must be infinite. It is less easy to prove that

[3, 5, 3], · [5, 3, 4], [5, 3, 5] are infinite. A more agreeable approach involves a geometric examination of the characteristic tetrahedron (Figure 9). The space in which this orthoscheme lies is clearly S 3, E 3 or H 3 according as the four faces are spherical triangles, Euclidean triangles or hyperbolic triangles. Hence, by adding together all the twelve face-angles and subtracting 477, we obtain a positive, zero, or negative number according to the nature of the space. Applying 5.6, we see that this number is ,.t. 'P

+X+

,/,

V'

[ _!_?_

+

r1-.'

'f-'

t

+X +

,,,u V'

77

77

-1-=-i. + ]-9-= r=·_!:_ + ~

[ 12-p-2q-r+± p

77

+ -p + -r + 4 -2 - 477

+i·j' !!... r 4

+

~

- 2 J77

25

TWISTED HONEYCOMBS

Therefore {p, q, r} is spherical, Euclidean, or hyperbolic according as p + 2q + r - 4/ p - 4/ r is less than 12, equal to 12, or greater than 12• The actual values are as follows: for {3, 3, 3}, 9 ~ ; for {4, 3, 3}, 10; ; for {3, 4, 31, 11

+;

for {5, 3, 3}, 11~;

for {4, 3, 4}, 12; for {5, 3, 4}, 13 ~·; for {3, 5, 3}, 13

½-;

for {5, 3, 5}, 14; .

11. The Petrie polygon of a honeycomb Among the vertices and edges of a regular honeycomb {p, q,

d,

we can pick out

a new kind of Petrie polygon in which every three consecutive edges belong to the Petrie polygon of a cell but no four consecutive edges belong to the same cell. For instance, in Figure 8, where AB CDE' · • · :is a Petrie polygon of one cell and BCD.E .. .. is a Petrie polygon of an adjacent cell, ABCDE ·., is a Petrie polygon of the whole honeycomb. By taking the characteristic tetrahedron in such a position that R R R 1

transforms ABCD into BCDE', while R

4

,

2

3

leaving the plane BCD invariant, trans-

forms BCDE' into BCDE, we see that the isometry ABCD--. BCDE, which takes us one step along the Petrie polygon, is R IR 2R R 3

4

•

Being conjugate to

which is the product of half-turns about two opposite edges of the characteristic tetrahedron, R IR R R 2

3 4

is the product of half-turns about two skew lines, that is, a twist

[9, p. 320]: the product of a translation along a line l (which measures the shortest distance between the two skew lines) and a rotation about the same line. Thus the Petrie :(X)lygon is a "helical" polygon: its edges are chords of a helix. This description is valid in hyperbolic space as well as in Euclidean space.

it is a rotation about the polar great circle, and the twist R R R R is a compound rotaI 2 3 4 tion [5, p. 36]: the product of two rotations whose axes are polar great circles (lying In spherical space,

.Z

is, of course, a great circle, the "translation" along

in completely orthogonal planes of the Euclidean 4-space). Let h denote the period of this compound rotation, so that the Petrie polygon is a skew h-gon. One way to compute h (possibly the simplest) is to seek the period of R R R R I

2 3 4

in the abstract group 10.1, or in its "positive" subgroup [p, q, r] +. This subgroup is generated by three rotations such as

A= RIR 2,

B = R 2 R 3,

C = R 3R 4

(which are rotations through 27T/ p, 2TT! q, 2TT! r about the edges P P , P P , P P of 2 3 0 3 0 1 the characteristic tetrahedron shown in Figure 8) or

26

Ho S. M. COXETER

(namely, half-turns about the remaining edges P0 P 2, P 1 P2 , P 1 P

3

dral angles are right angles). In terms of A, B, C, [p, q, r]

+

,

at which the dihe-

has the presentation

II.I which was discovered by

J.

A. Todd [8, p. 217] ;. h is the period of AC or CA. In

terms of L, M, N,

L 2 = M2 = N 2 = (LM)P = (LMN)q = (MN)r = I,

11.2

and h is the period of LN or NL. Note that

11.3

A= ML, B = LMN, C = NM; L = BC, M = ABC, N = AB. By enumerating cosets of the subgroup [q, r]

+

generated by L and MN (or by B

and C), we obtain a representation of [p, q, r] + by permutations of the vertices of ·{p, q, rl. In this representation, NL (or CA) appears as a product of cycles of h vertices, one of these cycles being a Petrie polygon. The result is that h has the value 5 for {3, 3, 31,

8 for 14, 3, 31 and {3, 3, 41,

12 for{3, 4, 31, 30 for{5, 3, 3land {3, 3, 51. In the notation of Figures 10, 11 and 12, one Petrie polygon for 14, 3, 31 is

ABCDEFGII. In the notation of §9, one for {3, 3, 51 is Eo,o Eo ,4 DIE 2, 2 E 2,6 D3E 4,4E 4,8 D 5E6,6 E 6, 1 o D 7E s, s Es, 12 D 9E 1 o, 10 E 1 o, 14 D 11E 12, 12 E 12, 16 D 1 3E 1 4, 1 4 E 1 4, 1 s

D 1 5E

16 , 16 E 16 , o D 1 7 E 1 s , 1 s E 1 8 , 2D 1 9 •

In Figure 8, the twist ABCD---+ BCDE is right-handed, yielding a right-handed Petrie polygon whose sides are chords of a right-handed helix. The same honeycomb has, of course, left-handed Petrie polygons which are reflected images of the righthanded ones. For instance, the reflection R 4 ( whose mirror is the plane BCD) transforms the ri'ght-handed twist R 1 R 2 R 3 R 4 into a left-handed twist R 4 R 1 R 2 R

3

.

Similarly

R 4 transforms the right-handed twist LN

=

R 4 R 2 R3R 1

=

R 2 R 1 R 4 R 3 = (R 3 R 4 R 1 R)- 1

into the left-handed twist R 2 R R 4 R 1 = LMNM. This transformation by R 4 , which is an 3

inner automorphism of [p, q,

rl,

is an outer automorphism of [p, q, r l :, replacing L,

M, N by L, M, MNM. Thus, when

11.4

4

4

p + 2q + r - - - p r

< 12

§ 10), so that [p, q, rt is finite, the relations 11.2 imply not only (LN) h = I but also (LMNM) h = I. The twist LN (and similarly LMNM) is now a compound rotation

(see

27

TWISTED HONEYCOMBS

through two angles: 2rr /h and

a

certain multiple of 2rr/h. Hence, when h is even,

(LN) hi 2 is the central inversion Z, which transforms each vertex of {p, q, r} into the opposite vertex. (In terms of Cartesian coordinates, this Z transforms (xl' x , x , x 2

3

) 4

into (- x 1, - x 2, - x 3, - x /) By abstractly identifying each pair of antipodal points of the 3-sphere

we derive the three-dimensional elliptic space: an orientable manifold whose funda mental group is the group of order 2 generated by Z. In this manner, the centrally symmetric regular honeycombs {4, 3, 3}, {3, 3, 4}, {3, 4, 3}, {5, 3, 3}, {3, 3, 5} on the 3-sphere yield the regular honeycombs {4, 3, 4}4, {3, 3, 4}4, {3, 4, 3}6, {5, 3, 3} 15' in the elliptic space. In each case the subscript is

h/2,

{3, 3, 5}15

indicating that we are iden-

tifying pairs of points separated by this number of steps along every right-handed Petrie polygon. The group [p, q, r]+, given by 11.2, has a normal subgroup of order 2 generated by Z = (LN) hi 2• The quotient group is the group of direct (sense-preserving) symmetry operations of the elliptic honeycomb {p, q, r} hi 2 which has half as many vertices, edges, faces, and cells as {p, q, r} itself. (See Table 2 on page 45.) Since Z can also be expressed as (LMNM) hi 2, the relations

L 2 = M2 = N 2 = (LM)P = (LMN)q = (MNV = (LN) hl 2 = 1 must imply (LMNM)h 12

A

= 1.

The elliptic honeycomb, like the spherical one, is

l

B

Figure 18: {4, 3, 3} 4

H.S.M.COXETER

28

"teflexible ", having planes of symmetry which reflect right-handed Petrie polygons into left-handed Petrie polygons. By dissecting a rhombic dodecahedron [4, p. 407] along the planes joining its cen• ter C to a suitable set of twelve edges (as in Figure 18), we obtain four oblate rhom· bohedra which may be regarded as the cells of an unfolded {4, 3, 3} 4 • The eight ver- , tices of this elliptic honeycomb are represented by the center C and the pairs of op· posite vertices of the rhombic dodecahedron. The two points A are joined by the right-handed Petrie polygon ABCDA and by the left-handed Petrie polygon ABCJA. 12. The regular honeycombs {p, q, r}t Analogously, whenever the direct symmetry group [p, q, r] + of a Euclidean or hyperbolic tessellation has a normal subgroup generated by (LN)t and its conjugates, we can identify pairs of points separated by t steps along every right-handed Petrie polygon to obtain a regular honeycomb {p, q, r }t which fills and covers a certain threedimensional manifold. If the quotient group 12.1

L 2 = M2 = N 2 = (LM)P = (MN)T = (LN)t = (LMN)q = 1

is finite, the honeycomb has a number of vertices equal to the index of the subgroup

[q, r] + generated by L and MN, and a number of cells equal to ,the index of the, subgroup [p, q ]+ generated by LM and N. The group 12.1 is conveniently denoted by ((p, r, t; q )) [1, p. 142], because it involves p, r, t symmetrically. Whenever these numbers (and q) are sufficiently small, it is the symmetry group of each of the six honeycombs {p, q, r }t, {r, q, p }t, {r, q, t }P, {t, q, rJP, {t, q, p }r, {p, q, t }r

[1, p. 145], which are all distinct if the numbers p, r, ·t are all distinct. The known cases are listed in Tables 2 and 3 (pages 45 and 46). Since the order of [4, 3, 3] is 2 4 4! = 384, that of [4, 3, 3]+ is 192, and that of ((4, 3, 4; 3)) or ((4, 4, 3; 3)) is 96, in agreement with our .observation that {4, 3, 3}

4

has 4 cells, each of which is a cube. The "Euclidean" honeycomb {4, 3, 4} , having 3

the same symmetry group, has likewise 4 cubic cells. In fact, we can construct it as follows. Color the cells of the infinite honeycomb {4, 3, 4} with 4 colors so that the 4 cells round each edge all have different colors. Then, by identifying all the cubes that are colored alike (without rotating them), we obtain the 4 cells of {4, 3, 4}3" In this case the twist LN is of such a nature that its third power is a translation (from one vertex of a cube to the opposite vertex). For any positive integer n, we can replace (LN) 3 by the magnified translation (LN) 3 n so as to obtain the tessellation {4, 3, 4} 3 n which has 4n 3 cells (and the same number of vertices). Thus the order of ((4, 4, 3n; 3)) is 96n 3 , and we have completed the explanation of Table 2~

29

TWISTED HONEYCOMBS

Turning now to Table 3, (page 46), we observe that, in terms of A, B, C (see 11.3), the presentation 12.1 for ((p, r, t; q)) becomes

AP= Bq

=

C'

=

(AB) 2 = (BC) 2 = (C A)t = (ABC) 2 = I.

It follows, by comparison with 6.2, that

((p, r, 2; q))

12.2

~

GP, q,r.

This isomorphism has a simple geometric interpretation: ((p, r, 2; q)) is the direct symmetry group of the honeycomb

which is a sort of "dihedron" based on the tessellation {p, q}, (Compare the description of {p, 2} in§ 2.) The reflections belonging to GP,q,r are reinterpreted as halfturns. It follows that, if the numbers p, q, r are small enough to satisfy 10.1, GP,q,r is also the symmetry group of a honeycomb {p, q, r}

q,

r. In this manner, the octahedral group

2

,

and of course we can permute p,

S 4 ~ G 3 • 3 • 4 yields {4, 3, 3} 2 (Figure 19),

{3, 3, 4} 2 (its dual), and {3, 4, 3} 2 (Figure 20).

b

a.

~d C

a.

b·

I I

/

)>_ __ - C

/ /b

d a

Figure 19: {4, 3, 3}

2

Figure 20: {3, 4, 3} 2

H.S.M.COXETER

30

Since the honeycombs {4, 3, 3} 2 and {3, 4, 3} 2 each have one cell and four edges, it is natural to mark these edges with letters a, b,. c, d (each occurring three times, because r = 3). In this representation, the generating half-turns are: for {4, 3, 3} 2,

L = (ab) (cd), M = (be), N.= (cd), implying LM

=

(acdb), MN= (bdc), LN = (ab), LMN

=

(adb), LMNM

=

(adcb),

and for {3, 4, 3} 2, L = (ab), M = (be), N = (cd), implying

LM = (acb), MN= (bdc), LN = (ab) (cd), L MN= (adcb), LMNM = (adb). Notice that LM cyclically permutes the edges of one face (clockwise), LMN cyclically permutes the edges at one vertex (counterclocl:wise), MN cyclically permutes the faces that surround one edge, and N appears as the half-turn that reverses this edge. In both cases, the period t' of LMNM is different from the period t of LN. Ac-

cordingly, these are twisted honeycombs. Opposite faces of the single cell have to be identified after the application of a (left=handed) twist. All the· honeycombs in Table 3 are of this nature: the relations 12.1 imply the period t' for LMNM. On the other hand, the reflexible honeycombs (Table 2) have t' = t. When t'

> t, ((p, r, t; q )) is a factor group in a larger group having t' in place

of t. In this manner, the one celled honeycomb {4, 3, 3} 2 yields the four celled 0

0

honeycomb {4, 3, 3} 4 which (as we have seen) is not twisted but reflexible. On the other hand,

l 3,

4, 3} 2 yields the three-celled honeycomb {3, 4, 3} 3 (Figure 21) which

is again twisted, having t' = 6. A representation by permutations of the twelve edges can be obtained by enumerating cosets of the dihedral subgroup generated by M and

N, which leaves one edge invariant. In this manner we find

L = (ac) (ef) (gl) (hk) (ij), N = (bg) (cd) (e j) (fh) (ik), MN= (bdj) (cge) (fhl), LMN = (agfc) (bdji) (ehkl),

M = (be) (de) (gj) (hl) (ik),

LM = (abc) (def) (ghi) (jkl), LN = (adc) (bgl) (ehi) (fjk ), LMNM = (ajkhic) (dgfbel ).

One right handed Petrie polygon appears as cad in the first octahedron, dca in the second, and adc in the third. 0

When the numbers p, q, r satisfy 11.4, so that the relations 11.2 imply (LN)h= 1, a necessary condition for the existence of {p, q,. r} t is that t divides h. Since there . is no group G 3• 3 • 5, 12.2 shows that {5, 3,31 2 and {3, 3,51

2

cannot exist. It may

be seen without much trouble that the relations described by the symbol ((4, 3, 3; 4)) imply the same relations with the first 4 replaced by 2 (to make G 3 • 3 , 4). Similarly, ((5, 3, 6; 3)) reduces to ((5, 3, 3; 3)), and ((5, 3, 10; 3)) to ((5, 3, 5; 3)) or

((5, 5, 3; 3)).

1 ' . W o 1 S N虹 ' l￡Du

coM 砕

3 1

玉 " 一 vふ tt 31n3●i

H.S.M.COXETER

32

Since the 5 vertices of the regular simplex {3, 3, 3} form (in any rder) a Petrie polygon, ((3, 3, 5; 3)) is the icosahedral group A 5 • Since ((5, 3, 3; 3)) is the same group, with generators

L = (ab) (ed), M = (be) (de), N = (ab) (de), we can associate the 5 vertices of the simplex, its 10 edges and 6 of its 12 Petrie polygons with the 5 vertices, 10 edges and 6 faces of a one-celled honeycomb {5, 3, 3} 3" (See Figure 22, also [5, p. 84], Notice that we are now lettering the vertices, not the edges.) Since

LMNM = (abeed), t' == 5. Surprisingly, {5, 3, 3} 5 is not a more complicated honeycomb but merely a re-

flected image of {5, 3, 3} 3; in fact, ((5, 3, 5; 3)) is the icosahedral group again, with generators

L = (ab) (ed), M = (be) (de), N = (ae) (de). Since LMNM = (eed), { 5, 3, 3} 5 has t' = 3.

b

a.

6

e

Figure 2 2 : {5 , 3, 3 }3

Interchanging L and M, so that now

L = (be) (de), M = (ab) (ed}, N = (ae) (de), we obtain the same icosahedral group in the form ((5, 5, 3; 3)), and a one=celled honeycomb {5, 3, 5}

3

(arising from the infinite hyperbolic honeycomb {5, 3, 5}). Since

LMNM = (be) (cd), we now have t' = 2. In fact, the reflected image of {5, 3, 5} {5, 3, 5} 2 (Figure 23, see also [IO, p, 242], whose group

G3 • 5• 5 ~ ((5, 5, 2; 3)) is generated hy

3

is

TWISTED HONEYCOMBS

L

=

33

(ab) (cd), M = (ac) (be), N = (ad) (be),

so that LMNM = (abe) and t' = 3. Since {S, 3, 5} 2 has 6 edges, there is some advantage in replacing the above representation of degree 5 by one of degree 6. Such a representation (see the lettering

f

e

Figure 23: {S, 3, 5} 2

in Figure 23) can be found by enumerating cosets of the subgroup D and N. Calling this subgroup a, and its cosets b, c, d, e,

f,

5

generated by M

we find

L = (ab) (cd), M = (be) (de), N = (cd) (ef), whence

LM = (acedb), MN= (bdfec), LN = (ab) (ef), LMN = (adb) (cfe), LMNM = (aeb) (cfd). The apparently possible group ((5, 5, 4; 3)) reduces to (( 5, 5, 2; 3)); that is, the relations, including (LN) 4

=

1, imply (LN) 2

=

I.

For the next entry in Table 3, we make use of

J.

M. Kingston's two presentations

for the alternating group of degree six [1, p. 144, footnote]:

A6

~ ((3, 3, 4; 5)) ~ ((3, 3, 5; 5)).

The case of ((3, 3, 5; 5)) may perhaps deserve a full description, because it illustrates the phenomenon of partial collapse. Letting a denote the subgroup generated by L and MN, so that aL = a and aMN

= a,

we define cosets

aM = b, bL = c, cN = d, dM = e, dL = f, e L = g, fM == h, gN = i,

and arrange the work as follows:

hN = j, iM = k,

H, S, M. COXETER

34

LL

MM

NN

LMLMLM

aaa

aba

aba

aabccba

bcb

ccc

cdc

dfh

dfd

ded

eee

iikjjk

e g e

fhf

ff f

i i i

i k i

gi g

MNMNMN

j k j

j j j

h j h

a b a b a b a

ged

ccdeedc fhjjhff

LNLNLNLNLN

LMNLMNLMNLMNLMN

aabcdf fdcba

a a b a a b a a b a a b a a b a

e g

b

g e e g i i g

d

C C

f

h j k

g e d

b a b

C

At a first glance this enumeration of cosets seems to be complete; but then we notice that the bottom row of the (LN) 5 table begins with e and ends with g, indicat-

e (LN) 5 = g. Thus, although we defined the coset g as eL, it is really the same as the coset e that we defined J?efore, that is; g = e. This one identification entails others: i = e, k = d, j = f = h. However, the collapse is not total: we still have ing that

the 6 genuinely distinct cosets a, b, c, d, e,

f,

in terms of which

L = (be) (df), M = (ab) (de), N = (ab) (cd). Since the generators

L = (be)

(df) and

MN = (cde)

of the subgroup a are even permutations of the 5 letters b, c, d, e,

f

and satisfy

L 2 = (MN)3 = (L · MN) 5 = 1, this subgroup is the simple group A

5

(the icosahedral group of order 60). Since there

are 6 cosets (including a itself), the order of the whole group is 6 x 60

=·

360. Since

L, M, N are even permutations of the 6 letters, this group ((3, 3, 5; 5)) is precisely the alternating group A

6

•

Thus { 3, 5, 3} 5 is a honeycomb whose cells consist of 6 icosahedra. Since

LMNM = (abec)

(df), t'

=

4. Like {5, 3, 3} 5 and {5, 3, 3} 3, {3, 5, 3 J 5 ( with t'

and {3, 5, 3} 4 ( with t' = 5) are reflected images of each other. Replacing

=

4)

N by MNM,

we obtain for ((3, 3, 4; 5)) the representation

L = (be) (df), M = (ab) (de), N

=

(ab) (ce).

As we saw in 6.5, ((5, 5, 5; 3)) is another simple group: the linear fractional group PSL(2, 19) of order 3420, generated by the homogeneous matrices

35

TWISTED HONEYCOMBS

-[o -1.J'

L-

I

0

9] M=lr- 7 5 7'

[6 -8]

(mod 19).

N= 7 -6

Leech checked its order by finding 57 cosets for the subgroup of order 60 generated by LM and N. Thus 15, 3, 5 l5 has 57 dodecahedral cells. Working out the period of the matrix

LMNM, we find t' = 9. Thus 15, 3, 5 l5 has right-handed Petrie polygons with 5 sides and left-handed Petrie polygons with 9 sides. However, there is no reason to expect ((5, 5, 9; 3)) to be a finite group.

In ((5, 4, 6; 3)), there is again an icosahedral subgroup generated by LM and N, but this time Leech found 114 cosets: just twice as many as before! Sinkov supplied the representation

01

7

-1 J

-2

J,

] (mod 19).

Since both L and M have determinant -1, it follows that this group is PGL (2, 19), of order 6840, and {5, 3, 4 l6 has 114 dodecahedral cells. The representation yields

l- 95 93j' (LMNM)9 = [ 0 8] (lMNM)IB = [I 01]; -7 0 ' . 0

LMNM =

., [-9-3 -2J - 9

(LMNN) 6 = r

thus t'= 18. (It would be interesting to see a direct proof that ((5, 4, 6; 3)) contains ((5, 5, 5; 3)) as a subgroup of index 2.) Similarly, Sinkov discovered that, in ((5, 4, 8; 3)), the icosahedral subgroup generated by LM and N has 496 cosets. Thus the order is 60

X

496 = 30

X

31

X

32,

which is twice that of P SL (2, 31 ). He found the representation

L = t [- 2 - 4 9 2

l ,

-ll

0 '

N=

113 -5

3

-13

] (mod 31)

where t 2 = I. This shows that ((5, 4, 8; 3)) is the direct product P SL (2, 31) x C 2 •

Since the factor C 2 is generated by t = (LMNM) 15 , we have t' = 30. We cannot expect to .find any further regular honeycombs, because ((5, 5, 6; 3)) shows no sign of being finite, and Sinkov has found that ((5, 4, 7; 3)) collapses totally.

36

H.S.M.COXETER 13. Finite groups of quaternions The one-celled honeycombs

{4, 3, 3J.2, {3, 4, 3J2, {5, 3, 3}3 were first investigated (by Seifert and Threlfall) as manifolds whose fundamental groups are

( 2, 2, 2 ),,

( 3, 3, 2) · ( 5, 3, 2)

in the following notation: ( p, q, r) means the group defined by

AP = Bq = C =ABC=

z,

Z 2 = 1.

As these groups arise most naturally as finite multiplicative groups of quaternions, it seems desirable to give a brief account of this theory. The corpus (or "division ring", or "skew field") of quaternions is derived from the field of real numbers by adjoining three elements i, j, k which satisfy Hamilton's famous relations i 2 =/=k 2 =ijk=-l.

13.1 A quaternion a = a 0

+ a 1 i + a 2 j + ai has a scalar part Sa = a 0 , a vector part

Va= a 1 i + a 2 j + a 3 k, a conjugate ~=Sa - Va and a norm rv

rv

2

. 2

2

2

Na= aa = aa = a 0 + a 1 + a 2 + a 3 • Since a= Sa+ Va, Sa=½ (a+~). If Sa= 0:, we call a a pure quaternion; thus, for any a, Va is pure.

A unit quaterion has Na= 1; in this case we can express a as

P

cos a+ P sin a, where P is a unit pure quaternion so t9-at = - P and P 2 = - 1. Analogy with Euler's expression for eai suggests the notation e ap for the unit quaternion a. More generally, any quaternion a (except zero) has its own "unit'·' Ua =

a/.[N';,

and we can writ~

a = ~ eaP,

a= arc cos SUa,,

P = UVa.

Let two unit vectors x and y (in E 3), variable and fixed, respectively, be represented by unit pure quaternions X

=

X li

+ X 2 j + X 3k, y = y l i + y 2 j + Y 3 k.

Then the plane y 1x 1 + y 2 x 2 + y 3 x 3 = 0, perpendicular to. y, may be expressed as s (xy) = o, where S(xy) = ½(xy + yx) = - x•y. The image of any vector x by reflection in this plane (see Figure 24) is

x- 2.x· yy. Remembering that y 2 = -1, we see that the corresponding quaternion is

37

TWISTED HONEYCOMBS

x + (xy + yx )y = x + xy 2 + yxy = yxy. Thus the reflection is represented by the transformation

13.2

X

-yxy.

Consider two planes with respective normal vectors y and z. If the angle between them is a and they intersect in the line OP with direction cosines (, 1/, (, we have

P = (i + 71j + (k = UV(yz). . The rotation through 2a about Op ta k es x mto zyxyz = e -aP xe aP , where

y z = e aP = cos a + P s·in a. (For a half-turn, a= rr/2, and x -

- PxP.) Note that a=

11

gives x - ( - l)x(-1)=~.

Therefore, the continuous group of rotations about lines OP through a fixed point 0 is two-to-one homomorphic with the group of unit quaternions e aP, the kernel being the group of order 2 generated by - I.

,S ( X: y) =

0

Figure ·24 Considering finite subgroups of these continuous groups, we see that each finite group of rotations is two-to-one homomorphic with a finite group of quaternions including - I. Bearing in mind that the only finite groups of rotations not including a half-turn are cyclic groups of odd order, we find that the only finite groups of quaternions not including - 1 are the cyclic groups of odd order generated by such quater4

nions as e 1Ti/p (p odd). We thus obtain the following correspondence between finite groups of rotations (§ 4) and finite groups of quaternions:

H.S.M.COXETER

38 Rotation group

Order

Group of quaternions

Order

p (p odd)

Cyclic

Cp

Cyclic

C 2p

2p

Cp

p

Dihedral

[p, 2] +

2p

Dicyclic

( p, 2, 2)

4p

Tetrahedral

[3, 3] +

12

Binary tetrahedral

( 3, 3, 2 )

24

Octahedral

[4, 3] +

24

Binary octahedral

( 4, 3, 2 )

48

le os ahedral

[5, 3] +

60

Binary icosehedral ( 5, 3, 2 )

120

Cyclic

.

J

K.

t

Figure 25 Let the characteristic triangle ICM (Figure 6) be pushed into one corner of the octant ijk (Figure 25) so that its vertices are represented by unit pure quaternions

P,

Q, k.

By 5.1 and the analogous formulae ,/, • 7T COS 'fJ Sln -

p

= COS

7T - ,

q

•

sin

,/, 'fJ

•

7T

•

7T

sin - = sin - , p h

we find

P=k and

,/, + i. Sln . ,/, 'fJ =

COS 'fJ

l'"

k

COS -7T

q

+ l•

• -7T Sln

h

J/ .

Sln -7T

p

Q = k cos ¢ + j sin ¢ = [ k cos !!_ + j sin .!!_ J/ sin !!... , p

h

q

39

TWISTED HONEYCOMBS

where h is given by 5.2. Let A, B, C be the quaternions that represent rotations through angles 211/p, 211/ q,

77

Q, k

about the vertices P,

of this triangle (p q 2).

Then ( p, q, 2) is an appropriate symbol for the group of quaternions generated by

A=

13 · 3

e 77 P /p

= cos !!_ + P sin!!_= cos !!_ + k cos !!_ + i sin!!_ p p p q h '

B = e 11Q I q = cos -11 + Q sin . -11 = cos -11 + k cos -11 + J. sin-:, . 11 q q q p h C = e11k!2 = k.

It follows that

BC = k cos

!!_ - cos !!_

p

q

+ i sin!!_= - A h

I

and

AP=Bq=C 2 =ABC=-1.

13.31

When p = q = 2, this is simply the "quaternion group'' ( 2, 2, 2 ) of order 8 , generated by i, j, k in the form 13.1. Since the relation C 2 = ABC (in 13.31) implies C = AB, we can reduce the num ber of generators of ( p, q, 2) from three to two and write

Av = B q = (AB) 2 = - I. It is an interesting fact, first noticed by William Threlfall, that the two relations

AP = Bq = (AB) 2 , or the three relations AP=Bq=C 2 =ABC,

13.4

suffice for a presentation of the group ( p, q, 2) whenever this group is finite, that is, whenever p

> 1,

q

>l

and p- 1 + q- 1

>

I/2. The only difficult step is to show

that the relations AP = Bq = (AB) 2 = Z imply Z 2 = I. When p = 5 and q = 3, this can be done as follows. Since

and. Z commutes with both A and B, we have

z =As=

BAsB-1 = B(B-1A-1B2)sB-1 = (A-1B)5 = (A3B-1A-1)s = (A2B-1)s

= {(B-1A-1B2)2 8 -1p = (B-1A-1BA -11 s = (A-lBA-1)5 = (BA- 2)5 == (A 2 B - I) - 5 =

z - l.

The other cases can be proved similarly (and more easily). Another interesting observation is that, for any odd prime p, the matrices

13.5

- I

A= [ - I

o]

-11'

B ==

[o -Ii, I I

C=

[ 0 I] - I

0

(mod p)

H.S.M.COXETER

40

generate the "special linear group" SL (2, p) of order p (p 2 - 1), and satisfy

n.s1

AP-B 3 -c 2 -ABc-[-~ _:].

Thus SL (2, p) is a factor group of ( p, 3, 2) . In particular, when p = 3 or 5,

SL (2, p)

13.6

~ ( p,

3, 2).

Any quaternion can be regarded as a point in Euclidean 4-space E 4 : x

= x 0 + x 1 i + x 2 j + x 3 k.

Take b to be a fixed unit quaternion: b = e/3 Q, Q = UVb. Since

N (xb) = NxNb = Nx, the transformation X -

is an isometry. By allowing

/3

xb

to increase continuously from zero, we see that it is a

direct isometry. We call this a right screw. Likewise x -

"' ax, where ""' a =

e -a p , is a

le ft screw. It is evident that left screws commute with right screws. The general direct isometry is the product of a left screw and a right screw: x -- ""' axb. (Compare "Clifford translations" in elliptic 3-s pace.) The special case x -

~xa is a rotation

through 2a about the plane through points 0, 1, P. In particular, x -

- PxP is a

half-turn about this plane, while x -- PxP is the half-turn about the completely orthogonal plane. Their product is x -- - x, the central inversion. More generally, x - - eaP xeaP is a rotation through 2a about the completely orthogonal plane. When vectors x and y are interpreted as quaternions x and y, their inner product satisfies x ·y = S(~y) = ½ (xy + y~ ). This enables us to prove that the general rota'li:xb, where Na= Nb = 1 and Sa= Sb = cos a. In fact, the

tion through 2a is x -

reflection in the hyperplane through O normal to a unit vector y transforms each vector

x into x - 2x · yy, which is represented by x -

2S (xy"' ) y

=

""' + yx"') y x - (xy

=

"' + yxy "' x - xyy

= -

""' yxy.

The product of two such reflections is x or, in terms of a=

y?

and b =

"'"' - z ( - ""'""' y xy )z = zy xy z

yz,

13.7

X -

~xb.

Since ~xb = e -aP xe aQ, this is a rotation through angle 2a about the plane OPQ. A compound rotation has the form x -

e-aP xef3Q, which is the commutative prod·

uct of the ·two rotations x -- e -(a +/3 )P /2 xe (a +/3)Q /2 and x -

e