1. Introduction
A convex d-polytope in Euclidean space is uniform if its symmetry group is transitive on its vertices, and if, furthermore, each facet of is uniform. To initiate this recursive condition in a geometrically pleasing way, we agree that a uniform polygon should be regular.
It is easy to see that all edges of
have the same length. However, for
,
may well have different kinds of facets. For example, the pentagonal antiprism
on the right in
Figure 1 is bounded by two regular pentagons
and ten equilateral triangles
. A
regular polytope , which by definition has a symmetry group transitive on flags, is certainly uniform. Consider the regular tetrahedron
, also in
Figure 1.
In ordinary space
, the uniform convex polyhedra include the five Platonic solids, the thirteen Archimedean solids, as well as
n-gonal prisms and antiprisms, for
. There is a little redundancy here: the 3-gonal antiprism and 4-gonal prism have more symmetry than first expected, being the regular octahedron
and cube
, respectively. For an excellent discussion of these polyhedra, their groups, as well as uniform tessellations of the plane, we refer to Coxeter’s paper [
1]. After a remarkable break starting with World War II, Coxeter explored uniform polytopes of higher dimension in two follow-up articles [
2,
3] appearing in the 1980s. An essential tool throughout is
Wythoff’s construction for uniform polytopes.
In Section 2.8 of [
2], we find a discussion of the
grand antiprism , discovered by J. H. Conway and M. Guy in 1965 [
4]. This remarkable object is the only uniform 4-polytope which cannot be constructed by Wythoff’s construction, even accepting Coxeter’s extension of the method to rotation groups. Coxeter also described the symmetry group
as
In fact, this is the wrong group of order 400, an error which has percolated into the literature.
In
Section 2, we use Wythoff’s construction to construct the 600-cell
, then find
inscribed in it. A correct description of the symmetry group
(as a semidirect product
) appears in Proposition 1. Actually, this was already derived in a slightly different way in [
5]. Nevertheless, that paper still seems to suggest
as the group.
Now, the finite subgroups of
have been variously classified, but it seems that the catalogue recently appearing in Chapter 4 of [
6] is complete and corrects small errors or oversights in earlier attempts, such as that in [
7]. In order to help the reader understand all this, we have reviewed in
Section 4 how unit quaternions are used to describe isometries in
and
. At the end of this long but necessary digression, we show in Example 3 (and Proposition 4) that
where here, we use the notation of Table 4.3 of [
6].
2. The 600-Cell and the Grand Antiprism
A useful way to understand the grand antiprism
is to see it inscribed in the 600-cell
, so we begin by describing the latter regular 4-polytope. The symmetry group
is the (linear) Coxeter group
, with generating reflections
corresponding to the nodes of the diagram:
The ring decorating the first node is an instruction to perform
Wythoff’s construction. In this instance, we choose a non-zero
base vertex fixed by
. The regular polytope
is then the convex hull of the
-orbit of
.
If, as in Section 5A of [
8], we identify an involutory isometry like
with its fixed space, or
mirror, we see that
spans the
Wythoff space
corresponding to the unringed nodes in diagram (
1).
A linear Coxeter group like
has special properties which serve to make the construction recursive. In particular, the subgroup of
, which fixes
W pointwise, is generated by the reflections indicated in (
2). Thus, the number of vertices in
is the index of the subgroup
. Furthermore, this subgroup is itself the Coxeter group
corresponding to the diagram obtained by deleting the first node:
We conclude that there are 14,400/120 = 120 vertices. The diagram in (
3) arises by transferring the ring in (
1) to the second node. This means that the vertex-figure at each vertex of
is a regular icosahedron
. The orthogonal projection behind
Figure 2 maps
to the centre of this icosahedron. The red edges
and
serve as a reminder that
lies outside the hyperplane supporting the vertex-figure. We shall soon see that
is really part of a planar decagon.
One can read much more from diagram (
1). For instance, just by deleting the right-most node, we find that all facets of
are regular tetrahedra
, and that there are 600 = 14,400/24 of them.
We now draw on [
9,
10] to give a more explict description of both
and its group
(as a subgroup of
). Depending on our algebraic needs, it will be useful at times to regard a point
as either a pair
of complex numbers (so
) or as a single quaternion
. In this spirit, we find in Section 4.6 of [
10] a description of the 120 vertices of
as pairs of complex numbers. We need
and the related angle
, so that
and
, with
as the
Golden ratio.
Therefore, here are the 120 vertices of
in a slight modification of Coxeter’s notation. The parameters
are residues modulo 10:
Remark 1. We have indeed 120 points of norm 1 in . Since is centrally symmetric, the vertices occur in 60 antipodal pairs. A special property of is that each pair is normal to a hyperplane of symmetry for the polytope. These 60 reflections comprise the single conjugacy class of reflections in . Thus, (in 14,400 ways) we can extract from the vertices a simple system of roots for Chapter 1.3 of [11]. That is, we can find four vertices to serve as ‘outer’ unit normals for the mirrors of the generating reflections , (j = 0, 1, 2, 3). We chooseNote, for instance, thatA suitable base vertex (fixed by ) is then . The base edge joins to . Clearly, the angle between (vectors) is , and each edge of has the lengthThere is now enough algebraic detail in place for the reader to check, with effort, our subsequent calculations. (We often seek refuge in GAP [12]). Firstly, the central symmetry
factors as
in
[
9] (p. 226). Consider that the icosahedral vertex-figure at
has its own central symmetry
. Using
Figure 2 and our earlier calculations (Section 4.6 of [
10]), we see that
cyclically moves
one step along a planar convex decagon
A (contained in the 1-skeleton of
). We note that
Comparing (
4a), we see that the vertices
of
A lie in the
-plane, while the vertices
of an orthogonal convex decagon
B lie in the
-plane.
Remark 2. Since the icosahedron has six pairs of antipodal vertices, each vertex of lies on six planar decagons, and altogether there are 72 such decagons. Furthermore, one can select 12 vertex-disjoint decagons to exhaust the vertices of . The 12 circumcircles belong to a Hopf fibration of (Section 4.9 of [10]). Definition 1. The grand antiprism is the convex hull of the 100 vertices of which remain after deleting two orthogonal decagons.
Let us remove
A and
B, leaving the points
. Since
is inscribed in
, these 100 points survive as the vertices of their convex hull
. To survey the facets of
, we consult Section 4.6, Exercise 2 of [
10].
Each edge of decagon
A is surrounded in
by five tetrahedral facets; and a vertex such as
is common to ten further tetrahedra whose bases form a belt running in a zig-zag fashion around the middle of the icosahedral vertex-figure, as in
Figure 2. In this way,
A and
B each meet 150 tetrahedra. These 300 facets of
are lost when we construct
.
If we first remove
from
, its icosahedral vertex-figure (
Figure 2) becomes a facet of the new convex hull. If we next remove
adjacent to
, we further truncate this icosahedron back to the pentagonal antiprism
, whose lateral triangles are those in the belt just described. In this way, the facets of
include a ring
of 10 copies of
. One pentagonal face on an antiprism arising this way has vertices
while the other pentagon is
with
alternating
as we run round the ring. The 50 vertices of
are the points
, with
even, from (
4b). The symmetry
s in (
6) moves
one step along itself.
The complementary ring
derived from
B is disjointed from
and provides 10 more copies of
. Its 50 vertices are the points
, with
odd, found in (
4c).
The 100 triangular faces in each ring form a non-regular toroidal map of Schläfli type
(Figure 4.6B, 4.6C of [
10]). Each triangle on
is the base of a tetrahedral facet of
whose apex is on
. In this way,
inherits 100 tetrahedral facets, let us say of type
A. In complementary fashion,
acquires from
the 100 tetrahedral facets of type
B. The final 100 facets of
are tetrahedra of type
. Each has one edge on
with the opposite edge on
. Tetrahedra of type
have vertices
Altogether,
has 500 edges, 20 regular pentagons, and 700 equilateral triangles as faces of lower rank. Each vertex-figure is non-uniform and arises as the convex hull of the 10 points which remain when an edge is deleted from an icosahedron
.
It is still not quite clear that is uniform, so we take a closer look at its symmetry group . Notice that G is a subgroup of . It coincides with the (set-wise) stabilizer of the decagons .
Let be the subgroup that takes A into A (and thus B into B). First of all, K contains every reflection r in a hyperplane orthogonal to a pair of antipodal vertices of A. This r induces a reflection symmetry of A while fixing B pointwise; and the five reflections coming from A this way generate a dihedral group of order 10.
In addition, the central symmetry , so K contains , which also acts by reflection on A, though as a half-turn on B. (One can view as a half-turn about a vertex of A in the 3-space spanned by A and some vertex of B.)
Let us choose the new reflection r to have normal . Then, acts on A as the full dihedral symmetry group of order 20, though half its elements act as half-turns on B. Similarly, we have acting on decagon B.
Note that . These two dihedral groups commute with one another and intersect in a centre of order 2. Thus, K has order 200.
In [
2] (p. 590), Coxeter observed that
the ‘ionic’ subgroup of the Coxeter group
with diagram
Compare [
2] (p. 569) and [
13] (p. 239). The whimsical adjective ‘ionic’ comes from the fact that the reflections
have determinant
, so that words of even length like
give determinant
, thereby reducing the ‘negative charge’.
To verify (
8), first take
; but let
be the reflection acting on
A as
but
fixingB. Likewise, let
act as
on
B but fix
A (note that
do
not belong to
G). We obtain (
8) upon noting that
.
It is curious that with the involutory generators
,
K is isomorphic to the full-automorphism group of the regular map
(Section 8.5 of [
14]).
Any which take A to B must also take B to A, so . Thus, G has order 400. The crucial question is how G extends K.
In Section 2.8 of [
2], Coxeter describes a half-turn
t which is meant to perform the task. Certainly, various half-turns
t swap
A and
B. However, no such
t can lie in
G (or in
)! To verify this, we note that the supposed half-turn would have to map
to either
or
for some complex number
y of norm 1. But,
must map to some
, with
odd. We would need
, which is impossible for
odd. This fact is key to what follows.
If we do move sideways and adopt the half-turn
, with
, then we have an involution which (by conjugation) swaps
while fixing
. This is just what is needed to ‘double’ the group
and so arrive at
See [
13] (pp. 255ff) and [
2] (p. 590). The group on the left denotes the semidirect product
, which indeed is isomorphic to
, one of a family of groups defined by a special sort of presentation (p. 96 [
14]). In this case, in terms of the generators
, we have the following defining relations:
(Equation 2.39 of [
2]).
On the other hand, we can exhibit a symmetry
of period 4, which swaps
A and
B. Taking
in (
7), we see that
are vertices of a facet of type
for
. This regular tetrahedron is a facet of
, so it admits the Petrie symmetry
p which cyclically permutes the vertices as they appear in (
9). Thus,
p has order 4, and in fact also permutes the roots
in a 4-cycle. Moreover,
p swaps
A and
B, and
.
It is now finally clear that is vertex-transitive, so that really is uniform!
Note that the subgroup of K is the linear Coxeter group of order 100. Conjugation by p in G will transform its generators in a 4-cycle . Furthermore, lies in K but not in its subgroup . We thus have the below proposition.
Proposition 1. The grand antiprism is uniform. Its symmetry group is the semidirect productThis group is not isomorphic to (or to ). Remark 3. It is easy to check that has defining relations: The group was correctly described as such a semidirect product in Section 2 of [5]. The authors there used quaternion methods, which we turn to in Section 4. However, they seem to continue the mislabelling of as ‘the ionic diminished Coxeter group ’. Considering the toroidal maps on the surfaces of the rings , it is quite natural that the symmetry p is induced by an affine function of the vertex symbols: We conclude this section by describing the subgroups of , which preserve some substructures of .
The vertex is typical and is fixed in by the subgroup .
The point belongs to two facets of type B. One of these has the base triangle on and is fixed in by . Each tetrahedron of type A or B in has, in this way, a stabilizer generated by a single reflection.
However, a tetrahedron of type
has a stabilizer of order 4 generated by a Petrie symmetry, just as
p does for the tetrahedron with the vertices in (
9).
It is clear that
acts transitively and faithfully on the 20 pentagonal antiprisms. Thus, each such facet must inherit its full symmetry group of order 20 from
. For instance, the group of the pentagonal antiprism with vertices
is generated by the reflection
and the half-turn
about the centre of the edge
(the reflection
fixes the upper pentagon point-wise but maps the pentagonal antiprism itself to one of its neighbours in the ring
). We refer to
Section 3 for more on the symmetry group
for a pentagonal antiprism.
From the action of
on the 20 antiprismatic facets, we obtain the following faithful permutation representation:
Note that
simultaneously rotates each ring through a tenth of a turn.
4. Quaternions and Finite Isometry Groups in
In order to locate
, or
for that matter, in a catalogue of all finite isometry groups on
, that is, within the finite subgroups of
, we need some tools from the algebra of quaternions. We follow Chapter 4 of [
6] and Chapter 6 of [
10].
Recall first that the conjugate of
is
, for which we have
. The
norm or squared length of
is
which, crucially, is multiplicative. For a
unit quaternion , we have
.
The group
of
unit quaternions (also known as
) is a double cover of
6.43 [
10]. To see this, we first identify
with the space of
pure quaternions
(for which
). Note that
.
For each
, one can find a unit pure quaternion
and then a unique angle
(
), so that
Next, we observe that the mapping of pure quaternions given by
effects a rotation through angle
about the axis spanned by
. Noting that we compose such mappings from left to right, we have the below proposition:
Proposition 2. There is a 2:1 surjectionThe kernel of this epimorphism is . A finite multiplicative group of quaternions must be a subgroup of
. Using Proposition 2 and the known classification of finite rotation groups in
, we easily verify that the finite groups of quaternions are those described in
Table 1.
From Theorem 12 of [
6], we have the generators
,
,
and
.
Example 1. For the moment, let us view the vertices of as unit quaternions. Since the identity quaternion is not one of these, we do not quite have a multiplicative group. However, if we premultiply vertices by (essentially ), then we do obtain the binary icosahedral group . The notation is a reminder that this group is a double cover of the icosahedral group , of order 60. Consider also quaternions . Then, the two generators satisfy the defining relationsfor (Chapter 6.5 of [10]). Derived as they are from (Section 2), these alternate generators are slightly messier than from Table 1: Let us move on to
. The reflection in the hyperplane orthogonal to the unit quaternion
is described by the mapping
which we denote by
] (recall that
). It follows that any direct isometry on
can be described as
The notation is meant to suggest a pair of left and right unit quaternions, and so we need the direct product
Proposition 3. There is a 2:1 surjectionThe kernel of this epimorphism is . Any opposite symmetry is likewise described by
For instance, ordinary conjugation is given by either
or
. This effects a central symmetry in the real subspace of pure quaternions.
To put all this in one package, it is useful to extend
by an involution, which we label * and which acts on
by swapping entries:
Using the semidirect product
we now have a
epimorphism
still with kernel
.
These results provide the first step to determining all geometrically distinct finite subgroups
G of
. We must first find finite subgoups
H of
if we seek subgroups
of
. Clearly,
H is a subdirect product of some
, where the left and right groups
are, up to conjugacy, amongst the finite groups listed in
Table 1. In fact, we could assume, if it helps, that
are just as given in the table, and we can further assume
, though this need not be so for
H. To organize the many possibilities, we can use Goursat’s Theorems on subdirect products, as described, for instance, in [
17]. The upshot is that the
H’s are parametrized by triples
such that the normal subgroups
and
admit an isomorphism
Then,
The group
has the order
where
is the order of the common quotient, and
is the order of
.
The actual cases are bewildering and are outlined in Chapter 4, Tables 4.1 and 4.2 of [
6]. In those tables of groups
, typical entries look like
for the so-called ‘diploid’ or ‘haploid’ cases, respectively, for which the central symmetry
z does or does not lie in the group.
The convention in the tables is that denote subgroups of , not their quaternionic covers . Likewise, we used above rather than f, since in some cases (not of concern here), one has .
If we seek a finite subgroup
G of
with opposite isometries, then we work in
and adjoin to one of the subdirect products
some element
. Here, there are simplications, mainly because
is forced. If desired, we can even take
. Up to conjugacy in
, there can be various choices for
, though often
is usable. The finite subgroups of this kind in
are listed in Table 4.3 of [
6].
We shall look more closely only at a few of the “diploid, achiral groups”, which appear in Table 4.3 of [
6] as
though perhaps with some decorations to distinguish, for instance, choices for
. In such cases,
,
, and
, so the order is
We have reviewed all this machinery purely so that the reader can make sense of the following brief results. It can take a great deal of work to fit a well-known linear group into the scheme underlying the tables in [
6].
Example 2. From the very first entry in Table 4.3 of [6], we haveThe “” indicates that we have doubled the order of the rotation group by adjoining an opposite symmetry, in fact, . We have , so , the icosahedral group of order 60. The parameter , and , with θ trivial. The order of does indeed equal from (11). Now, compare this with our construction of in Section 2. There, we chose the basic roots in (5) for the generated reflections, . Thus, , so that the subgroup of direct isometries is generated by rotationsWe find that is the binary icosahedral group generated by quaternions in Example 1. But now, is the conjugate subgroup in . This has no effect on the conjugacy class of in . Example 3. The rotations in generate a copy of the icosahedral group . From Figure 2, we see that I contains the dihedral group , generated, for instance, by , a rotation through about , and our half-turn about the midpoint of edge . This edge belongs to the Petrie polygon preserved by in Figure 2. Lift the binary icosahedral group of Example 2, now generated by . This group of order 120 contains the dicyclic group of order 20, where But from the previous example, we now know that contains its own copy of the dicyclic group. Thus, contains the direct product of commuting dicyclic groups. This group of order 400 projects to the rotation groupof order 200 in (Table 4.2 of [6]). We can adjoin the opposite symmetry to finally see that(so, take in line 19 of Table 4.3 of [6]). We see that
appears as a subgroup of
in a quite natural way. Indeed, this is essentially the approach taken in [
5]. However, as mentioned earlier, the mislabelling of
is at least suggested there. To recap, we have the below proposition.
Proposition 4. The symmetry group of the grand antiprism is in the notation of Table 4.3 of [6]. Example 4. We will not include the details needed to correctly classify our unneeded ‘ionic diminished Coxeter group’:(line 21 of Table 4.3 of [6]). The adjoined opposite symmetry can again be taken to be . The bar in is merely a notational device signalling the fact that is special among dihedral groups in having automorphisms freely permuting the non-identity elements. See p. 50 and footnote 3 of [6] for more. Remark 6. Conway and Smith describe in Chapter 4.5 of [6] errors or omissions in previous catalogues of the isometry groups in . Perhaps the best known earlier enumeration of the groups is that of Du Val in Section 21–22 of [7]. Apparently, there are some redundancies to be found there.