The Symmetry Group of the Grand Antiprism

1. Introduction

A convex d-polytope $\mathbf{Q}$ in Euclidean space is uniform if its symmetry group $G\left(\mathbf{Q}\right)$ is transitive on its vertices, and if, furthermore, each facet of $\mathbf{Q}$ 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 $\mathbf{Q}$ have the same length. However, for $d⩾3$, $\mathbf{Q}$ may well have different kinds of facets. For example, the pentagonal antiprism ${\mathbf{P}}_5$ on the right in Figure 1 is bounded by two regular pentagons $\left\{5\right\}$ and ten equilateral triangles $\left\{3\right\}$. A regular polytope $\mathbf{Q}$, which by definition has a symmetry group transitive on flags, is certainly uniform. Consider the regular tetrahedron $\{3,3\}$, also in Figure 1.

In ordinary space ${\mathbb{E}}^3$, the uniform convex polyhedra include the five Platonic solids, the thirteen Archimedean solids, as well as n-gonal prisms and antiprisms, for $n⩾3$. There is a little redundancy here: the 3-gonal antiprism and 4-gonal prism have more symmetry than first expected, being the regular octahedron $\{3,4\}$ and cube $\{4,3\}$, respectively. For an excellent discussion of these polyhedra, their groups, as well as uniform tessellations of the plane, we refer to Coxeter’s paper [7]. After a remarkable break starting with World War II, Coxeter explored uniform polytopes of higher dimension in two follow-up articles [9,10] appearing in the 1980s. An essential tool throughout is Wythoff’s construction for uniform polytopes.

In [9], we find a discussion of the grand antiprism $\mathbf{A}$, 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 $G\left(\mathbf{A}\right)$ as

$$\left[[10,2^{+},10]\right]\simeq G^{4,4,10}.$$

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 $\mathbf{S}$, then find $\mathbf{A}$ inscribed in it. A correct description of the symmetry group $G\left(\mathbf{A}\right)$ (as a semidirect product $[5,2,5]⋊C_4$) appears in Proposition 1. Actually, this was already derived in a slightly different way in [17]. Nevertheless, that paper still seems to suggest $\left[[10,2^{+},10]\right]$ as the group.

Now the finite subgroups of $G{O}_4\left(\mathbb{R}\right)$ have been variously classified, but it seems that the catalogue recently appearing in [5] is complete and corrects small errors or oversights in earlier attempts, such as that in [14]. In order to help the reader understand all this, we have reviewed in Section 4 how unit quaternions are used to describe isometries in ${\mathbb{E}}^3$ and ${\mathbb{E}}^4$. At the end of this long but necessary digression, we show in Example 3 that

$$G\left(\mathbf{A}\right)\simeq \pm [D_{10}\times D_{10}]·2,$$

here using the notation of [5].

2. The 600-cell $\mathbf{S}=\{3,3,5\}$ and the Grand Antiprism $\mathbf{A}$

A useful way to understand the grand antiprism $\mathbf{A}$ is to see it inscribed in the 600-cell $\mathbf{S}=\{3,3,5\}$, so we begin by describing the latter regular 4-polytope. The symmetry group $G\left(\mathbf{S}\right)$ is the (linear) Coxeter group $H_4=[3,3,5]$, with generating reflections $r_0,r_1,r_2,r_3$ corresponding to the nodes of the diagram

$$◯•{\displaystyle \frac{3}{}}•{\displaystyle \frac{3}{}}•{\displaystyle \frac{5}{}}•$$

(1)

The ring decorating the first node is an instruction to perform Wythoff’s construction. In this instance, we choose a non-zero base vertex $\mathbf{v}$ fixed by $r_1,r_2,r_3$. The regular polytope $\mathbf{S}$ is then the convex hull of the $H_4$-orbit of $\mathbf{v}$.

If, as in [18], we identify an involutory isometry like $r_j$ with its fixed space, or mirror, we see that $\mathbf{v}$ spans the Wythoff space

$$W=r_1\cap r_2\cap r_3$$

(2)

corresponding to the unringed nodes in diagram (1).

A linear Coxeter group like $H_4$ has special properties which serve to make the construction recursive. In particular, the subgroup of $H_4$ which fixes W pointwise is generated by the reflections indicated in (2). Thus the number of vertices in $\mathbf{S}$ is the index of the subgroup $\langle r_1,r_2,r_3\rangle$. Furthermore, this subgroup is itself the Coxeter group $[3,5]$ corresponding to the diagram obtained by deleting the first node:

$$·◯•{\displaystyle \frac{3}{}}•{\displaystyle \frac{5}{}}•$$

(3)

We conclude that there are $14400/120=120$ 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 $\{3,3,5\}$ is a regular icosahedron $\{3,5\}$. The orthogonal projection behind Figure 2 maps $\mathbf{v}$ to the centre of this isosahedron. The red edges $\mathbf{v}\mathbf{u}$ and $\mathbf{v}\mathbf{w}$ serve as a reminder that $\mathbf{v}$ lies outside the hyperplane supporting the vertex-figure. We shall soon see that $...\mathbf{w}\mathbf{v}\mathbf{u}...$ is really part of a planar decagon.

One can read much more from the diagram (1). For instance, just by deleting the right-most node, we find that all facets of $\mathbf{S}$ are regular tetrahedra $\{3,3\}$, and that there are $600=14400/24$ of them.

We now draw on [11] and [8] to give a more explict description of both $\mathbf{S}$ and its group $H_4$ (as a subgroup of $G{O}_4\left(\mathbb{R}\right)$). Depending on our algebraic needs, it will be useful at times to regard a point $\mathbf{x}=(x_0,x_1,x_2,x_3)\in {\mathbb{E}}^4$ as either a pair $(u,v)$ of complex numbers (so $u=x_0+x_1ı,v=x_2+x_3ı$) or as a single quaternion $x_0+x_1\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{k}=u+v\mathbf{j}$. In this spirit, we find in [11] a description of the 120 vertices of $\mathbf{S}$ as pairs of complex numbers. We need $ϵ=exp(\pi /10)$ and the related angle $\lambda ={\displaystyle \frac{1}{2}}arctan2\doteq 31.{72}^{\circ }$, so that $cos\lambda ={\tau }^{{\displaystyle \frac{1}{2}}}{5}^{-{\displaystyle \frac{1}{4}}}$, $sin\lambda ={\tau }^{-{\displaystyle \frac{1}{2}}}{5}^{-{\displaystyle \frac{1}{4}}}$, with $\tau =(1+\sqrt{5})/2$ the Golden ratio.

Here then are the 120 vertices of $\mathbf{S}$ in a slight modification of Coxeter’s notation. The parameters $\mu ,\nu$ are residues modulo 10:

$$2\mu +1|\underline{ }=({ϵ}^{2\mu +1},0),\underline{ }|2\nu +1=(0,{ϵ}^{2\nu +1})$$

(4a)

$$2\mu |2\nu =({ϵ}^{2\mu }cos\lambda ,{ϵ}^{2\nu }sin\lambda )(\mu +\nu \mathrm{even})$$

(4b)

$$2\mu |2\nu =({ϵ}^{2\mu }sin\lambda ,{ϵ}^{2\nu }cos\lambda )(\mu +\nu \mathrm{odd})$$

(4c)

Remark 1. 

We have indeed 120 points of norm 1 in ${\mathbb{E}}^4$. Since $\mathbf{S}$ is centrally symmetric, the vertices occur in 60 antipodal pairs. A special property of $\mathbf{S}$ 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 $H_4$. Thus (in 14400 ways) we can extract from the vertices a simple system of roots for $H_4$ [15, Chapter 1.3]. That is, we can find four vertices to serve as `outer’ unit normals ${\mathbf{n}}_j$ for the mirrors of the generating reflections $r_j$, ($j=0,1,2,3$). We choose

$${\mathbf{n}}_0=7|\underline{ },{\mathbf{n}}_1=16|2,{\mathbf{n}}_2=\underline{ }|9,{\mathbf{n}}_3=\underline{ }|17.$$

(5)

Note, for instance, that

$$r_0:(u,v)\mapsto (-\overline{u}{ϵ}^{14},v).$$

A suitable base vertex (fixed by $r_1,r_2,r_3$) is then $\mathbf{v}=1|\underline{ }$. The base edge joins $\mathbf{v}$ to $\mathbf{u}=\left(\mathbf{v}\right)r_0=3|\underline{ }$. Clearly, the angle between (vectors) $\mathbf{v},\mathbf{u}$ is $2\pi /10$, and each edge of $\mathbf{S}$ has length

$$2sin{\displaystyle \frac{\pi }{10}}={\tau }^{-1}.$$

There is now enough algebraic detail in place for the reader to check, with effort, our subsequent calculations. (We often seek refuge in GAP [1].) □

First off, the central symmetry $\mathbf{x}\mapsto -\mathbf{x}$ factors as

$$z={\left(r_0{r}_1{r}_2{r}_3\right)}^{15}$$

in $H_4$[8]. The icosahedral vertex-figure at $\mathbf{v}$, say, has its own central symmetry ${\left(r_1{r}_2{r}_3\right)}^5$. Using Figure 2 and our earlier calculations, we see [11] that

$$s={\left(r_1{r}_2{r}_3\right)}^5{r}_0$$

(6)

cyclically moves $...\mathbf{w}\mathbf{v}\mathbf{u}...$ one step along a planar convex decagon A (contained in the 1-skeleton of $\mathbf{S}$). We note that

$$s:(u,v)\mapsto (u{ϵ}^2,-v).$$

Comparing (4a), we see that the vertices $2j+1|\underline{ }$ of A lie in the $x_0{x}_1$-plane, while the vertices $\underline{ }|2j+1$ of an orthogonal convex decagon B lie in the $x_2{x}_3$-plane.

Remark 2. 

Since the icosahedron has 6 pairs of antipodal vertices, each vertex of $\mathbf{S}$ lies on 6 planar decagons, and altogether there are 72 such decagons. Furthermore, one can select 12 vertex-disjoint decagons to exhaust the vertices of $\mathbf{S}$. The 12 circumcircles belong to a Hopf fibration of ${\mathbb{S}}^3$ [11, Section 4.9]. □

Definition 1. 

The grand antiprism $\mathbf{A}$ is the convex hull of the 100 vertices of $\mathbf{S}$ which remain after deleting two orthogonal decagons.

Let us remove A and B, leaving the points $2\mu |2\nu$. Since $\mathbf{S}$ is inscribed in ${\mathbb{S}}^3$, these 100 points survive as the vertices of their convex hull $\mathbf{A}$. To survey the facets of $\mathbf{A}$, we consult [11, Section 4.6, Exercise 2].

Each edge of the decagon A is surrounded in $\mathbf{S}$ by 5 tetrahedral facets; and a vertex such as $\mathbf{v}$ is common to 10 further tetrahedra whose bases form a belt running in 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 $\mathbf{S}$ are lost when we construct $\mathbf{A}$.

If we first remove $\mathbf{v}$ from $\mathbf{S}$, its icosahedral vertex-figure (Figure 2) becomes a facet of the new convex hull. If we next remove $\mathbf{u},\mathbf{w}$ adjacent to $\mathbf{v}$, we further truncate this icosahedron back to the pentagonal antiprism ${\mathbf{P}}_5$ whose lateral triangles are those in the belt just described. In this way, the facets of $\mathbf{A}$ include a ring ${\mathcal{R}}_A$ of 10 copies of ${\mathbf{P}}_5$. One pentagonal face on an antiprism arising this way has vertices

$$2\mu |0,2\mu |4,2\mu |8,2\mu |12,2\mu \left|16\right(\mu \mathrm{even}),$$

while the other pentagon is

$$2\mu +2\alpha |2,2\mu +2\alpha |6,2\mu +2\alpha |10,2\mu +2\alpha |14,2\mu +2\alpha |18,$$

with $\alpha$ alternating $\pm 1$ as we run round the ring. The 50 vertices of ${\mathcal{R}}_A$ are the points $2\mu |2\nu$ with $\mu +\nu$ even, from (4b). The symmetry s in (6) moves ${\mathcal{R}}_A$ one step along itself.

The complementary ring ${\mathcal{R}}_B$ derived from B is disjoint from ${\mathcal{R}}_A$ and provides 10 more copies of ${\mathbf{P}}_5$. Its 50 vertices are the points $2\mu |2\nu$, with $\mu +\nu$ odd, found in (4c).

The 100 triangular faces in each ring form a non-regular toroidal map of Schläfli type $\{3,6\}$ [11]. Each triangle on ${\mathcal{R}}_A$ is the base of a tetrahedral facet of $\mathbf{S}$ whose apex is on ${\mathcal{R}}_B$. In this way, $\mathbf{A}$ inherits 100 tetrahedral facets, let us say of type A. In complementary fashion, $\mathbf{A}$ acquires from $\mathbf{S}$ the 100 tetrahedral facets of type B. The final 100 facets of $\mathbf{A}$ are tetrahedra of type $AB$. Each has one edge on ${\mathcal{R}}_A$ with the opposite edge on ${\mathcal{R}}_B$. Tetrahedra of type $AB$ have vertices

$$2\mu |2\nu ,2\mu |2\nu +2,2\mu +2|2\nu ,2\mu +2|2\nu +2(\mathrm{any}\mu ,\nu ).$$

(7)

Altogether, $\mathbf{A}$ 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 $\{3,5\}$.

It is still not quite clear that $\mathbf{A}$ is uniform, so we take a close look at its symmetry group $G=G\left(\mathbf{A}\right)$. Notice that G is a subgroup of $H_4$. It coincides with the (set-wise) stabilizer of the decagons $\{A,B\}$.

Let $K⩽G$ 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 $z\in K$, so K contains $rz$, which also acts by reflection on A, though as a half-turn on B. (One can view $rz$ 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 $\mathbf{n}=15|\underline{ }$. Then $D\left(A\right)=\langle r_0,rz\rangle$ acts on A as the full dihedral symmetry group $D_{20}$ of order 20, though half its elements act as half-turns on B. Similarly, we have $D\left(B\right)=\langle r_{2}z,r_3\rangle \simeq D_{20}$ acting on decagon B.

Note that $z={\left(r_{0}rz\right)}^5={\left(r_{2}z{r}_3\right)}^5\in D\left(A\right)\cap D\left(B\right)$. These two dihedral groups commute with one another and intersect in a centre $\langle z\rangle$ of order 2. Thus K has order 200.

In [9, p. 590], Coxeter observed that

$$K\simeq [10,2^{+},10]=\langle a_0,a_1{a}_2,a_3\rangle ,$$

(8)

the `ionic’ subgroup of the Coxeter group $[10,2,10]=\langle a_0,a_1,a_2,a_3\rangle$ with diagram

$$•{\displaystyle \frac{10}{}}••{\displaystyle \frac{10}{}}•.$$

(Compare [9, p. 569] and [16, p. 239]. The whimsical adjective `ionic’ comes from the fact that the reflections $a_j$ have determinant $-1$, so that words of even length like $a_1{a}_2$ give determinant $+1$, thereby reducing the `negative charge’.)

To verify (8), first take $a_0=r_0,a_3=r_3$; but let $a_1$ be the reflection acting on A as $rz$ but fixingB. Likewise let $a_2$ act as $r_{2}z$ on B but fix A. (Note that $a_1,a_2$ do not belong to G.) We get (8) upon noting that $a_1{a}_2=r{r}_{2}z$.

It is curious that with the involutory generators $a_0,a_1{a}_2,a_3$, K is isomorphic to the full automorphism group of the regular map $\{10,10|2\}$ [12, Section 8.5].

Any $g,h\in G$ which take A to B must also take B to A, so $gh\in K$. Thus G has order 400. The crucial question is how G extends K.

In [9, Section 2.8], Coxeter describes a half-turn t which is meant to do the job. Certainly various half-turns t swap A and B. However, no such t can lie in G (or in $H_4$)! To verify this, we note that the supposed half-turn would have to map $(u,v)$ to either $(v{y}^{-1},uy)$ or $(\overline{v}y,\overline{u}y)$, for some complex number y of norm 1. But $0|0=(cos\lambda ,sin\lambda )$ must map to some $2\mu |2\nu =({ϵ}^{2\mu }sin\lambda ,{ϵ}^{2\nu }sin\lambda )$, with $\mu +\nu$ odd. We would need ${ϵ}^{2\mu \pm 2\nu }=1$, which is impossible for $\mu +\nu$ odd.

If we do move sideways and adopt the half-turn $t:(u,v)\mapsto (\overline{v}y,\overline{u}y)$, with $y={ϵ}^4$, then we have an involution which (by conjugation) swaps $a_0,a_3$ while fixing $a_1{a}_2$. This is just what is needed to `double’ the group $K\simeq [10,2^{+},10]$ and so arrive at

$$\left[[10,2^{+},10]\right]\simeq G^{4,4,10}.$$

(See [16, pp. 255ff] and [9, p. 590].) The group on the left denotes the semidirect product $[10,2^{+},10]⋊\langle t\rangle$, which indeed is isomorphic to $G^{4,4,10}$, one of a family of groups defined by a special sort of presentation [12, p. 96]. In this case, in terms of the generators $a=a_1{a}_2{a}_0,b=a_{0}t,c=t{a}_0{a}_1{a}_2$, we have defining relations

$$a^{10}=b^4=c^4={\left(ab\right)}^2={\left(bc\right)}^2={\left(ca\right)}^2={\left(abc\right)}^2=1$$

[9, Equation 2.39]. Note that $\langle t,a_2\rangle \simeq D_8$. Since $G\left(\mathbf{A}\right)$ has no such subgroup, we confirm once more that $G\left(\mathbf{A}\right)$ cannot be $\left[[10,2^{+},10]\right]$.

On the other hand, we can exhibit a symmetry $p\in H_4$ of period 4 which swaps A and B. Taking $\mu =0,\nu =1$ in (7), we see that

$$0|2,0|4,2|4,2|2$$

(9)

are vertices of a facet of type $AB$ for $\mathbf{A}$. This regular tetrahedron is a facet of $\mathbf{S}$, 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 ${\mathbf{n}}_0,{\mathbf{n}}_3,\mathbf{n},{\mathbf{n}}_2$ in a 4-cycle. Moreover, p swaps A and B, and $p\in G\left(\mathbf{A}\right)$.

It is now finally clear that $G\left(\mathbf{A}\right)$ is vertex-transitive, so that $\mathbf{A}$ really is uniform!

Note that the subgroup $\langle r_0,r,r_2,r_3\rangle$ of K is the linear Coxeter group $[5,2,5]$ of order 100. Conjugation by p in G will transform its generators in a 4-cycle $(r_0,r_3,r,r_2)$. Furthermore, $p^2$ lies in K but not in its subgroup $[5,2,5]$. We have

Proposition 1. 

The grand antiprism is uniform. Its symmetry group $G\left(\mathbf{A}\right)$ is the semidirect product

$$[5,2,5]⋊C_4.$$

Remark 3. 

It is easy to check that $G\left(\mathbf{A}\right)$ has defining relations

$$r^2=p^4={\left(p^{-2}r{p}^{2}r\right)}^5={\left(p^{-1}rpr\right)}^2=1.$$

The group $G\left(\mathbf{A}\right)$ was correctly described as such a semidirect product in [17, Section 2]. The authors there used quaternion methods, which we turn to in Section 4. However, they seem to continue the mislabelling of $G\left(\mathbf{A}\right)$ as `the ionic diminished Coxeter group $[10,2^{+},10]$’.

Considering the toroidal maps on the surfaces of the rings ${\mathcal{R}}_A,{\mathcal{R}}_B$, it is quite natural that the symmetry p is induced by an affine function of the vertex symbols:

$$p:2\mu |2\nu \mapsto (2\nu -2)|(4-2\mu )(mod20).$$

□

We conclude this section by describing the subgroups of $G\left(\mathbf{A}\right)$ which preserve some substructures of $\mathbf{A}$.

The vertex $2|2$ is typical and is fixed in $G\left(\mathbf{A}\right)$ by the subgroup $\langle r_0,r_3\rangle \simeq C_2\times C_2$.

The point $2|2$ belongs to 2 facets of type B. One of these has base triangle $0|2,2|4,4|2$ on ${\mathcal{R}}_B$ and is fixed in $G\left(\mathbf{A}\right)$ by $\langle r_0\rangle$. Each tetrahdron of type A or B in $\mathbf{A}$ has, in this way, a stabilizer generated by a single reflection.

However, a tetrahedron of type $AB$ 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 $G\left(\mathbf{A}\right)$ acts transitively and faithfully on the 20 pentagonal antiprisms. Thus each such facet must inherit its full symmetry group of order 20 from $G\left(\mathbf{A}\right)$. For instance, the group of the pentagonal antiprism with vertices

	$2|2$		$2|6$		$2|10$		$2|14$		$2|18$
$0|0$		$0|4$		$0|8$		$0|12$		$0|16$		$0|0$

is generated by the reflection $r_3$ and the half-turn $h=p^2$ about the centre of the edge $2\left|20\right|4$. (The reflection $r_0$ fixes the upper pentagon point-wise but maps the pentagonal antiprism itself to one of its neighbours in the ring ${\mathcal{R}}_A$.) We refer to Section 3 for more on the symmetry group $[5,2^{+}]$ for a pentagonal antiprism.

From the action of $G\left(\mathbf{A}\right)$ on the 20 antiprismatic facets we obtain this faithful permutation representation:

$$\begin{array}{ccc}\hfill r& \mapsto & (1,10)(2,9)(3,8)(4,7)(5,6)\hfill \\ \hfill p& \mapsto & (1,12)(2,11,10,13)(3,20,9,14)(4,19,8,15)(5,18,7,16)(6,17)\hfill \end{array}$$

Note that

$${\left(rp\right)}^2\mapsto (1,2,3,4,5,6,7,8,9,10)(11,12,13,14,15,16,17,18,19,20)$$

simultaneously rotates each ring through a tenth of a turn.

3. More on Wythoff’s Construction

In [6] Coxeter extended Wythoff’s construction to general Coxeter groups G of finite or affine type, with nodes of the diagram ringed in any way. (See also [7 Section 1.5], [9, Section 2.4] and [11, Section 2.4].) All this is aimed at enumerating uniform polytopes and tessellations in Euclidean space. Actually, Coxeter also employed a variant of the construction based on just the rotation subgroup $G^{+}$, indicated by replacing all nodes in the diagram by empty rings. The choice of base vertex $\mathbf{x}$ required to guarantee uniformity is now trickier and sometimes impossible, although the construction always works in ${\mathbb{E}}^3$. In particular, the n-gonal antiprism ${\mathbf{P}}_n$ is produced by the diagram

$$◯{\displaystyle \frac{n}{}}◯◯$$

(10)

The underlying Coxeter group $G=[n,2]$ has order $4n$, so $G^{+}$ has order $2n$ and indeed is isomorphic to the dihedral group $D_{2n}$ (of order $2n$). We have $G^{+}=\langle q,h\rangle$, where q is a rotation through $2\pi /n$ about some axis l, while h is a half-turn about an axis m meeting l at right angles. A base vertex $\mathbf{x}$ can now be chosen to produce a pair of regular n-gons separated by a belt of $2n$ equilateral triangles running in zig-zag fashion. The pentagonal antiprism ($n=5$) is illustrated in Figure 1. For $n>3$ the full symmetry group of the antiprism does have order $4n$ but is not quite the group G. Instead, $G\left({\mathbf{P}}_n\right)$ is the group $[2n,2^{+}]$, once more following the notation of [16, Section 11.4] and [9]. Beginning with the larger Coxeter group $[2n,2]=\langle u_0,u_1,u_2\rangle$, we need the ionic subgroup

$$[2n,2^{+}]=\langle u_0,u_1{u}_0{u}_1,u_1{u}_2\rangle ,$$

of order $4n$. In fact, as we saw in Section 2, we find that this group is generated by the reflection $u_0$ and the half-turn $u_1{u}_2$. When n is odd, this group actually is isomorphic to $[n,2]$, though geometrically different.

Remark 4. 

We noted earlier that Wythoff’s construction, including the extension to rotation groups, gives all uniform polyhedra in ${\mathbb{E}}^3$. In [4], Conway and Guy apparently used a local approach to construct all uniform 4-polytopes [9, p. 588]: they computed the dihedral angles for all uniform polyhedra, then tried to assemble these facets so that around each edge the dihedral angles sum to less than $2\pi$. They found that only one uniform polytope in ${\mathbb{E}}^4$, the grand antiprism, eludes the more general Wythoff’s construction. This is simply because $G\left(\mathbf{A}\right)$ is neither a Coxeter group nor a rotation subgroup.

The grand antiprism is discussed in [3], with some useful figures. However, I could not find there a description of the group, although Conway surely knew all about it. Nor can I find that he wrote about the group elsewhere, so it was Coxeter who initiated the discussion in [9].

There are also some very fine illustrations in the Wikipedia article Grand Antiprism. However, at the time of writing, that article also mislabels the group $G\left(\mathbf{A}\right)$. □

Remark 5. 

One can actually use the method to manufacture a vertex-transitive polytope $\mathbf{Q}$ by applying the `incorrect’ group $G=\left[[10,2^{+},10]\right]$ to the same base vertex $0|0$, as for $\mathbf{A}$. The orbit still has size 100, containing the vertices of ring ${\mathcal{R}}_A$ along with their images under the spurious half-turn t. We seem to obtain the points described in (4b) and (4c); however, we must now take $\mu +\nu$ even in both cases. The convex hull $\mathbf{Q}$ of this orbit still has the 100 vertices; but it cannot be uniform since it has edges of different lengths. For instance, there is an edge of length ${\tau }^{-1}$ in $\mathbf{Q}$ from $0|4$ to $0|0$, just as in $\mathbf{A}$. However, there is also a slightly longer edge of length $2^{{\displaystyle \frac{1}{2}}}{\tau }^{-{\displaystyle \frac{1}{2}}}{5}^{-{\displaystyle \frac{1}{4}}}$ from $0|4$ to the new vertex $\left(2\right|2)t$.

The geometric effect of the spurious half-turn t is to map each pentagon from ring ${\mathcal{R}}_A$ to a pentagon coplanar and concentric with a pentagon in ring ${\mathcal{R}}_B$, but turned a half-turn with respect to the latter. We see again why t cannot lie in $H_4$.

Disappointed, we conclude that there is no undiscovered uniform 4-polytope missed by Conway and Guy! □

4. Quaternions and Finite Isometry Groups in ${\mathbb{E}}^4$

In order to locate $G\left(\mathbf{A}\right)$, or $\left[[10,2^{+},10]\right]$ for that matter, in a catalogue of all finite isometry groups on ${\mathbb{E}}^4$, that is, within the finite subgroups of $G{O}_4\left(\mathbb{R}\right)$, we need some tools from the algebra of quaternions. We follow [5, Chapter 4] and [11, Chapter 6].

Recall first that the conjugate of $\mathbf{x}=x_0+x_1\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{k}\in {\mathbb{E}}^4$ is $\tilde{\mathbf{x}}=x_0-x_1\mathbf{i}-x_2\mathbf{j}-x_3\mathbf{k}$, for which we have $\widetilde{\mathbf{x}\mathbf{z}}=\tilde{\mathbf{z}}\tilde{\mathbf{x}}$. The norm or squared length of $\mathbf{x}$ is

$$N\left(\mathbf{x}\right)=\mathbf{x}\tilde{\mathbf{x}}=\tilde{\mathbf{x}}\mathbf{x},$$

which, crucially, is multiplicative. For a unit quaternion $\mathbf{x}$ we have ${\mathbf{x}}^{-1}=\tilde{\mathbf{x}}$.

The group $\mathbb{P}$ of unit quaternions (also known as ${\mathbf{Spin}}_3$) is a double cover of $S{O}_3\left(\mathbb{R}\right)$[11, 6.43]. To see this we first identify ${\mathbb{E}}^3$ with the space of pure quaternions $\mathbf{z}$ (for which $z_0=0$). Note that ${\mathbf{z}}^2=-N\left(\mathbf{z}\right)$.

For each $\mathbf{a}\in \mathbb{P}$, one can find a unit pure quaternion $\mathbf{u}$ and then a unique angle $\alpha$ ($0⩽\alpha ⩽\pi$), so that

$$\mathbf{a}=exp\left(\alpha \mathbf{u}\right):=cos\left(\alpha \right)+sin\left(\alpha \right)\mathbf{u}.$$

Next we observe that the mapping of pure quaternions given by

$$\left[\mathbf{a}\right]:\mathbf{z}\mapsto \tilde{\mathbf{a}}\mathbf{z}\mathbf{a},(\mathbf{z}\in {\mathbb{E}}^3)$$

effects a rotation through angle $2\alpha$ about the axis spanned by $\mathbf{u}$. Noting that we compose such mappings left to right, we have

Proposition 2. 

There is a $2:1$ surjection

$$\begin{array}{ccc}\hfill \mathbb{P}& \to & S{O}_3\left(\mathbb{R}\right)\hfill \\ \hfill \mathbf{a}& \mapsto & \left[\mathbf{a}\right]\hfill \end{array}$$

The kernel of this epimorphism is $\pm 1$.

A finite multiplicative group of quaternions must be a subgroup of $\mathbb{P}$. Using Proposition 2 and the known classification of finite rotation groups in ${\mathbb{E}}^3$, we easily verify that the finite groups of quaternions are those described in Table 1.

From [5, Theorem 12] we have the generators $\mathbf{a}=(-1+\mathbf{i}+\mathbf{j}+\mathbf{k})/2$, ${\mathbf{b}}_T=\mathbf{i}$, ${\mathbf{b}}_O=(\mathbf{j}+\mathbf{k})/\sqrt{2}$ and ${\mathbf{b}}_I=(\mathbf{i}+{\tau }^{-1}\mathbf{j}+\tau \mathbf{k})/2$.

Example 1. 

For the moment, let us view the vertices of $\mathbf{S}$ 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 ${\mathbf{v}}^{-1}$ (essentially ${ϵ}^{-1}$), then we do get the binary icosahedral group $2I=\langle 5,3,2\rangle$. The notation is a reminder that this group is a double cover of the icosahedral group $(5,3,2)={[5,3]}^{+}$, of order 60. Consider also quaternions $\mathbf{u}=3|\underline{ },\mathbf{z}=4|0$. Then the two generators $\mathbf{d}={\mathbf{v}}^{-1}\mathbf{u},\mathbf{e}={\mathbf{v}}^{-1}\mathbf{z}$ satisfy the defining relations

$${\mathbf{d}}^5={\mathbf{e}}^3={\left(\mathbf{d}\mathbf{e}\right)}^2$$

for $\langle 5,3,2\rangle$ [11, Chapter 6.5]. Derived as they are from (4), these alternate generators are a bit messier than $\mathbf{a},{\mathbf{b}}_I$ from Table 1:

$$\begin{array}{ccc}\hfill \mathbf{d}& =& {\displaystyle \frac{1}{2}}(\tau +5^{{\displaystyle \frac{1}{4}}}{\tau }^{-{\displaystyle \frac{1}{2}}}\mathbf{i}),\hfill \\ \hfill \mathbf{e}& =& {\displaystyle \frac{1}{2}}(1+5^{-{\displaystyle \frac{1}{4}}}{\tau }^{{\displaystyle \frac{3}{2}}}\mathbf{i}+\mathbf{j}-5^{-{\displaystyle \frac{1}{4}}}{\tau }^{-{\displaystyle \frac{3}{2}}}\mathbf{k}).\hfill \end{array}$$

□

Let us move on to ${\mathbb{E}}^4$. The reflection in the hyperplane orthogonal to the unit quaternion $\mathbf{a}$ is described by the mapping

$$\mathbf{z}\mapsto -\mathbf{a}\tilde{\mathbf{z}}\mathbf{a},$$

which we denote by $*[-\tilde{\mathbf{a}},\mathbf{a}$]. (Recall that $\tilde{\mathbf{a}}={\mathbf{a}}^{-1}$.) It follows that any direct isometry on ${\mathbb{E}}^4$ can be described as

$$[\mathbf{l},\mathbf{r}]:\mathbf{z}\mapsto \tilde{\mathbf{l}}\mathbf{z}\mathbf{r},\mathbf{z}\in {\mathbb{E}}^4.$$

The notation is meant to suggest a pair of left and right unit quaternions, and so we need the direct product

$$\Delta =\mathbb{P}\times \mathbb{P}.$$

Proposition 3. 

There is a $2:1$ surjection

$$\begin{array}{ccc}\hfill \Delta & \to & S{O}_4\left(\mathbb{R}\right)\hfill \\ \hfill (\mathbf{l},\mathbf{r})& \mapsto & [\mathbf{l},\mathbf{r}]\hfill \end{array}$$

The kernel of this epimorphism is $\pm (1,1)$.

Any opposite symmetry is likewise described by

$$*[\mathbf{l},\mathbf{r}]:\mathbf{z}\mapsto \tilde{\mathbf{l}}\tilde{\mathbf{z}}\mathbf{r},\mathbf{z}\in {\mathbb{E}}^4.$$

For instance, ordinary conjugation is given by either $*[1,1]$ or $*[-1,-1]$. (This effects a central symmetry in the real subspace of pure quaternions.)

To put all this in one package, it is useful to extend $\Delta$ by an involution, which we label * and which acts on $\Delta$ by swapping entries:

$$*(\mathbf{l},\mathbf{r})*={(\mathbf{l},\mathbf{r})}^{*}=(\mathbf{r},\mathbf{l}).$$

Using the semidirect product

$${\Delta }^{\prime }=\Delta ⋊C_2=(\mathbb{P}\times \mathbb{P})⋊\langle *\rangle ,$$

we now have a $2:1$ epimorphism

$$\begin{array}{ccc}\hfill {\Delta }^{\prime }& \to & G{O}_4\left(\mathbb{R}\right)\hfill \\ \hfill (\mathbf{l},\mathbf{r})& \mapsto & [\mathbf{l},\mathbf{r}],\hfill \\ \hfill (\mathbf{l},\mathbf{r})& \mapsto & *[\mathbf{l},\mathbf{r}],\hfill \end{array}$$

still with kernel $\pm (1,1)$.

These results provide the first step to determining all geometrically distinct finite subgroups G of $G{O}_4\left(\mathbb{R}\right)$. We must first find finite subgoups H of $\Delta$ if we seek subgroups $G^{+}$ of $S{O}_4\left(\mathbb{R}\right)$. Clearly, H is a subdirect product of some $L_{*}\times R_{*}$, where the left and right groups $L_{*},R_{*}$ are, up to conjugacy, amongst the finite groups listed in Table 1. In fact, we could assume, if it helps, that $L_{*},R_{*}$ are just as given in the Table; and we can further assume $(-1,-1)\in L_{*}\times R_{*}$, 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 [2]. The upshot is that the H’s are parametrized by triples $(K_{L_{*}},K_{R_{*}},\theta )$ such that the normal subgroups $K_{L_{*}}⊴L_{*}$ and $K_{R_{*}}⊴R_{*}$ admit an isomorphism

$$\theta :L_{*}/K_{L_{*}}\to R_{*}/K_{R_{*}}.$$

Then

$$H=\{(\mathbf{l},\mathbf{r})\in L_{*}\times R_{*}:\left(\mathbf{l}{K}_{L_{*}}\right)\theta =\mathbf{r}{K}_{R_{*}}\}.$$

The group $G^{+}$ has order

$${\displaystyle \frac{|L_{*}|·|R_{*}|}{e_{*}{f}_{*}}},$$

where $f_{*}=[L_{*}:K_{L_{*}}]=[R_{*}:K_{R_{*}}]$ is the order of the common quotient, and $e_{*}\in \{1,2\}$ is the order of $H\cap \langle (-1,-1)\rangle$.

The actual cases are bewildering and are outlined in [5]. In those Tables of groups $G^{+}$, typical entries look like

$$\pm {\displaystyle \frac{1}{f}}[A\times B]\mathrm{or}+{\displaystyle \frac{1}{f}}[A\times B],$$

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 $A,B$ denote subgroups of $S{O}_3\left(\mathbb{R}\right)$, not their quaternionic covers $L_{*},R_{*}$. Likewise, we used $f_{*}$ above rather than f, since in some cases (not of concern here), one has $f=f_{*}/2$.

If we seek a finite subgroup G of $G{O}_4\left(\mathbb{R}\right)$ with opposite isometries, then we work in ${\Delta }^{\prime }$ and adjoin to one of the subdirect products $H⩽L_{*}\times R_{*}$ some element $*(\mathbf{a},\mathbf{b})$. Here there are simplications, mainly because $L_{*}\simeq R_{*}$ is forced. If desired, we can even take $L_{*}=R_{*}$. Up to conjugacy in ${\Delta }^{\prime }$ there can be various choices for $*(\mathbf{a},\mathbf{b})$, though often $*(1,1)$ is usable. The finite subgroups of this kind in $G{O}_4\left(\mathbb{R}\right)$ are listed in Table 4.3 of [5].

We shall look more closely only at a few of the “diploid, achiral groups”, which appear in [5] as

$$G=\pm {\displaystyle \frac{1}{f}}[A\times A]·2,$$

though perhaps with some decorations to distinguish, for instance, choices for $*[a,b]$. In such cases, $L_{*}=2A$, $e_{*}=2$ and $f=f_{*}$, so the order is

$${\displaystyle \frac{{4\left|A\right|}^2}{f}}.$$

(11)

We have reviewed all this machinery just 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 [5].

Example 2. 

From the very first entry in [5] we have

$$H_4=[3,3,5]=\pm [I\times I]·2.$$

The “$·2$” indicates that we have doubled the order of the rotation group ${[3,3,5]}^{+}$ by adjoining an opposite symmetry, in fact, $*[1,1]$. We have $L_{*}=R_{*}=2I$, so $A=B=I$, the icosahedral group of order 60. The parameter $f=f_{*}=1$, and $K_{L_{*}}=K_{R_{*}}=2I$, with θ trivial. The order of $H_4$ does indeed equal $4·{60}^2=14400$ from (11).

Now compare this with our construction of $H_4$ in Section 2. There we chose the basic roots ${\mathbf{n}}_j$ in (5) for the generating reflections $r_j$. Thus $r_j=*[{\mathbf{n}}_j^{-1},{\mathbf{n}}_j]$, so that the subgroup of direct isometries $H_4^{+}$ is generated by rotations

$$s_j:=r_{j-1}{r}_j=[-{\mathbf{n}}_{j-1}{\mathbf{n}}_j^{-1},-{\mathbf{n}}_{j-1}^{-1}{\mathbf{n}}_j],(j=1,2,3).$$

We find that $R_{*}$ is the binary icosahedral group generated by quaternions $\mathbf{d},\mathbf{e}$ in Example 1. But now $L_{*}$ is the conjugate subgroup $\mathbf{v}{R}_{*}{\mathbf{v}}^{-1}$ in $\mathbb{P}$. This has no effect on the conjugacy class of $H_4$ in $G{O}_4\left(\mathbb{R}\right)$. □

Example 3. 

The rotations $s_2=r_1{r}_2,s_3=r_2{r}_3$ in $H_4$ generate a copy of the icosahedral group $I={[3,5]}^{+}$. From Figure 2, we see that I contains the dihedral group $D_{10}$, generated for instance by $s_3$, rotation through $2\pi /5$ about $\mathbf{u}$, and our half-turn $h=s_2{s}_3^{-2}{s}_2^{-1}{s}_3{s}_2^{-1}$ about the midpoint of edge $2\left|20\right|4$. (This edge belongs to the Petrie polygon preserved by $s_3$ in Figure 2.)

Lift to the binary icosahedral group $R_{*}=\langle {\mathbf{a}}_2,{\mathbf{a}}_3\rangle$ of Example 2, now generated by ${\mathbf{a}}_2=-{\mathbf{n}}_1^{-1}{\mathbf{n}}_2,{\mathbf{a}}_3=-{\mathbf{n}}_2^{-1}{\mathbf{n}}_3$. This group of order 120 contains the dicyclic group $\langle 5,2,2\rangle =\langle {\mathbf{a}}_3,\mathbf{b}\rangle$ of order 20, where

$$\mathbf{b}={\mathbf{a}}_2{\mathbf{a}}_3^{-2}{\mathbf{a}}_2^{-1}{\mathbf{a}}_3{\mathbf{a}}_2^{-1}.$$

But from the previous Example we now know that $L_{*}=\mathbf{v}{R}_{*}{\mathbf{v}}^{-1}$ contains its own copy $\langle \mathbf{v}{\mathbf{a}}_3{\mathbf{v}}^{-1},\mathbf{v}\mathbf{b}{\mathbf{v}}^{-1}\rangle$ of the dicyclic group. Thus $L_{*}\times R_{*}$ contains the direct product of commuting dicyclic groups. This group of order 400 projects to the rotation group

$$\pm [D_{10}\times D_{10}]$$

of order 200 in $S{O}_4\left(\mathbb{R}\right)$ [5]. We can adjoin the opposite symmetry $*[1,1]$ to finally see that

$$G\left(\mathbf{A}\right)\simeq \pm [D_{10}\times D_{10}]·2,$$

(so take $p=5$ in line 19 of [5]).

We see that $G\left(\mathbf{A}\right)$ appears as a subgroup of $H_4$ in a quite natural way. Indeed, this is essentially the approach taken in [17]. However, as mentioned earlier, the mislabelling of $G\left(\mathbf{A}\right)$ is at least suggested there. □

Example 4. 

We will not include the details needed to correctly classify our unneeded `ionic diminished Coxeter group’:

$$\left[[10,2^{+},10]\right]\simeq \pm {\displaystyle \frac{1}{4}}[D_{20}\times {\overline{D_{ }}}_{20}]·2$$

(line 21 of [5, Table 4.3]). The adjoined opposite symmetry can again be taken to be $*[1,1]$. The bar in ${\overline{D_{ }}}_{20}$ is merely a notational device signalling the fact that $D_4\simeq C_2\times C_2$ is special among dihedral groups in having automorphisms freely permuting the non-identity elements. See [5, p. 50 and footnote 3] for more. □

Remark 6. 

Conway and Smith describe in [5, Chapter 4.5] errors or omissions in previous catalogues of the isometry groups in ${\mathbb{E}}^4$. Perhaps the best known earlier enumeration of the groups is that of Du Val in [14, Sections 21–22]. Apparently, there are some redundancies to be found there.

□

5. Some Final Comments and Thanks

The grand antiprism has been examined elsewhere, generally in wider discussions of uniform 4-polyopes. We mention, for instance, [19,21] and [13]. This last paper employs subrootsystems for the group $H_4$, rather as in [17], but with a broader look that takes in other uniform polytopes, such as the snub 24-cell.

Our work on the grand antiprism is an offshoot of a more extensive investigation into abstract regular 4-polytopes whose automorphism groups are subgroups of low index in some orthogonal group $O(d,p,e)$ over a finite field $GF\left(p\right)$ [20]. (The parameter $e=\pm 1$ flags the Witt index for the corresponding d-dimensional orthogonal geometry.) For instance, $H_4\simeq O_1(4,5,+1)$, the subgroup generated by reflections whose roots have square spinor norm. From [20, Equation (14)], we find a similarly structured group and accompanying geometry whenever $p\equiv 1(mod4)$. Moreover, we get an abstract regular 4-polytope with tetrahedral facets and $p^3-p$ vertices. Inscribed in it, we must find a relative of the grand antiprism. But, of course, this abstract 4-polytope will not have a familiar convex realization.

We still do not properly understand the presentations of such orthogonal groups, when ones hands are tied (as they will be!) by using just 4 generating reflections. This in turn is necessary for an understanding of the universal regular polytopes whose facets are tetrahedra and whose vertex-figures are certain naturally occuring maps of type $\{3,p\}$ [20, Conjectures 1,2,3].

Finally, let me here thank Peter McMullen for his input, in particular for suggesting the use of the subgroup K in Section 2 as a way to more easily understand the structure of $G\left(\mathbf{A}\right)$.