Journal of Algebraic Combinatorics

, Volume 43, Issue 3, pp 665–691 | Cite as

Classification of tight regular polyhedra

Article

Abstract

A regular polyhedron of type \(\{p, q\}\) has at least 2pq flags, and it is called tight if it has exactly 2pq flags. The values of p and q for which there exist tight orientably regular polyhedra were previously known. We determine for which values of p and q there is a tight non-orientably regular polyhedron of type \(\{p, q\}\). Furthermore, we completely classify tight regular polyhedra in terms of their automorphism groups.

Keywords

Abstract regular polytope Tight polyhedron Tight polytope Flat polyhedron Flat polytope 

Mathematics Subject Classification

Primary 52B15 Secondary 51M20 05E18 52B70 

1 Introduction

Abstract polyhedra are combinatorial objects that generalize the face lattice of convex polyhedra. Those possessing the highest degree of symmetry are called regular polyhedra. The face lattices of platonic solids, known since antiquity, are all regular in this sense, and there are infinitely many more regular abstract polyhedra.

In [1], Marston Conder introduced the idea of a tight regular polyhedron: Any regular polyhedron with q-valent vertices and p-gons as faces has at least 2pq automorphisms, and the polyhedron is called tight if it has precisely this number of automorphisms. Tight polyhedra were also studied by the first author in [4]. In [2], Conder and the first author completely characterized the values of p and q of tight orientably regular polyhedra and further generalized to higher-dimensional analogues.

In the present paper we characterize the degree q of the vertices and the number p of edges in a face of tight non-orientably regular polyhedra. We also take the work from [2] one step further in the following direction. For many values of p and q, there are multiple non-isomorphic tight orientably regular polyhedra with p-gonal faces and q-valent vertices; here we determine the number of such polyhedra and describe their automorphism groups.

One of our main results is the following:

Theorem 1.1

There is a tight regular polyhedron of type \(\{p, q\}\) (that is, with p-gonal faces and q-valent vertices) if and only if one of the following is true:
  1. (a)

    p and q are both even.

     
  2. (b)

    p is odd and q is an even divisor of 2p.

     
  3. (c)

    q is odd and p is an even divisor of 2q.

     
  4. (d)

    \(p = 4\) and q is an odd multiple of 3.

     
  5. (e)

    \(q = 4\) and p is an odd multiple of 3.

     
In the second and third cases, there is one such polyhedron up to isomorphism, and it is orientably regular. In the fourth and fifth cases, there is one such polyhedron up to isomorphism, and it is non-orientably regular.

In Sects. 2 and 3, we review basic concepts and results on tight abstract regular polytopes and their automorphism groups. The classification of orientably regular and non-orientably regular polyhedra is obtained in Sects. 4 and 5, respectively. Theorem 1.1 follows directly from the results in these two sections.

2 Background

Our definitions are mostly taken from [8, Chs. 2,4], with some minor modifications.

2.1 Definition of an abstract polyhedron

Let \(\mathcal P\) be a ranked partially ordered set with elements of rank 0, called vertices, elements of rank 1, called edges and elements of rank 2, called faces. Let us say that two elements F and G are incident if \(F \le G\) or \(G \le F\). By a flag, we will mean a maximal chain (totally ordered set). The vertex-figure at a vertex \(F_0\) is \(\{G \mid G > F_0\}\). Then, \(\mathcal P\) is an (abstract) polyhedron if it satisfies the following properties:
  1. (1)

    Every flag of \(\mathcal P\) consists of a vertex, an edge and a face (all mutually incident).

     
  2. (2)

    Each edge is incident on exactly two vertices and two faces.

     
  3. (3)

    The graph determined by the vertex and edge sets is connected.

     
  4. (4)

    The vertex-figure at every vertex is isomorphic to the vertex and edge lattice of a connected 2-regular graph.

     
When considering finite polyhedra, the last property can be interpreted as vertex-figures being (finite) polygons, whereas the second and fourth properties imply that the faces are also polygons. As a consequence of the second and fourth properties above, given any flag \(\varPhi \) and \(i \in \{0, 1, 2\}\), there is a unique flag \(\varPhi ^i\) that differs from \(\varPhi \) only in its element of rank i. We say that \(\varPhi ^i\) is i-adjacent to \(\varPhi \) (or simply adjacent to \(\varPhi \) if the rank i is unimportant).

In the remainder of the paper, let us drop the qualifier “abstract” and simply refer to polyhedra.

Given a face of a polyhedron, if it is incident to p edges, then it must also be incident to p vertices. These edges and vertices occur in a single cycle, and we say that the face is a p-gon. Similarly, if a vertex is incident to q edges, then it is also incident to q faces, occurring in a single cycle. In this case we say that the vertex-figure is a q-gon. If \(\mathcal P\) is a polyhedron whose faces are all p-gons and whose vertex-figures are all q-gons, then we say that \(\mathcal P\) has Schläfli symbol\(\{p, q\}\) or that it is of type\(\{p, q\}\). A polyhedron that has a Schläfli symbol is said to be equivelar.

If \(\mathcal P\) is a polyhedron, then the dual of \(\mathcal P\), denoted \(\mathcal P^{\delta }\), is the polyhedron we obtain by reversing the partial order. If \(\mathcal P\) is of type \(\{p, q\}\), then \(\mathcal P^{\delta }\) is of type \(\{q, p\}\).

Given any convex polyhedron, the partially ordered set of its vertices, edges and faces, ordered by the usual geometric incidence, is an abstract polyhedron. Similarly, any face-to-face tessellation of the plane yields an (infinite) abstract polyhedron. Indeed, every abstract polyhedron with finite faces and vertex-figures corresponds to a face-to-face tiling of some surface, which may or may not be orientable. The tiling, also called a map, can be constructed by taking a topological p-gon (topological disk with its boundary divided in p segments) for each face F containing p edges. The p segments of the p-gon are labeled with the edges incident to F, in such a way that if two segments intersect in a point, the corresponding edges in the partial order have a vertex in common. The point of intersection is labeled by the common vertex. Since every edge belongs to two faces, it only remains to identify segments of the p-gons corresponding to the same edge in such a way that vertices with the same label are also identified.

On the other hand, some tilings fail to satisfy property (4) above, and therefore they do not correspond to abstract polyhedra. An example of this is the map \(\{4,4\}_{(1,1)}\) on the torus (see [3]).

2.2 Regularity and orientability

If \(\mathcal P\) and \(\mathcal Q\) are polyhedra, then a homomorphism from \(\mathcal P\) to \(\mathcal Q\) is a function that preserves incidence. We say that \(\mathcal P\)covers\(\mathcal Q\) if there is a surjective homomorphism \(\varphi \) from \(\mathcal P\) to \(\mathcal Q\) that also preserves rank and has the property that if flags \(\varPhi \) and \(\Psi \) are i-adjacent, then so are their images under \(\varphi \). An isomorphism from \(\mathcal P\) to \(\mathcal Q\) is an incidence- and rank-preserving bijection. An isomorphism from \(\mathcal P\) to itself is an automorphism of \(\mathcal P\), and the group of all automorphisms of \(\mathcal P\) is denoted by \(\varGamma (\mathcal P)\). There is a natural action of \(\varGamma (\mathcal P)\) on the flags of \(\mathcal P\), and due to the connectivity of \(\mathcal P\), the action of each automorphism is completely determined by its action on any given flag.

We say that \(\mathcal P\) is regular if the natural action of \(\varGamma (\mathcal P)\) on the flags of \(\mathcal P\) is transitive (and hence regular, in the sense of being sharply transitive). Indeed, for convex polyhedra, this definition is equivalent to any of the usual definitions of regularity.

Since each automorphism of \(\mathcal P\) is completely determined by its action on any particular flag, let us choose a base flag\(\varPhi \) of \(\mathcal P\). Then the automorphism group \(\varGamma (\mathcal P)\) is generated by the abstract reflections\(\rho _0, \rho _1, \rho _2\), where each \(\rho _i\) maps \(\varPhi \) to \(\varPhi ^i\). These generators satisfy (at least) the relations \(\rho _i^2 = 1\) for all i and \((\rho _0 \rho _2)^2 = 1\). A regular polyhedron must be equivelar, and if its type is \(\{p, q\}\), then \(\langle \rho _0, \rho _1 \rangle \) is dihedral of order 2p, and \(\langle \rho _1, \rho _2 \rangle \) is dihedral of order 2q. In other words, if \(\mathcal P\) is a regular polyhedron of type \(\{p, q\}\), then \(\varGamma (\mathcal P)\) is a smooth quotient of the string Coxeter group [pq], with presentation
$$\begin{aligned}{}[p, q] := \langle x, y, z \mid x^2 = y^2 = z^2 = 1, (xy)^p = (yz)^q = (xz)^2 = 1 \rangle . \end{aligned}$$
Let \(\varGamma = \langle \rho _0, \rho _1, \rho _2 \rangle \) be a group such that the generators \(\rho _i\) each have order 2 and such that \((\rho _0 \rho _2)^2 = 1\). Then we say that \(\varGamma \) is a string group generated by involutions of rank 3, which we will abbreviate to sggi. Now, for any \(I \subseteq \{0, 1, 2\}\), we define \(\varGamma _I = \langle \rho _i \mid i \in I \rangle \). We say that \(\varGamma \) is a string C-group of rank 3 if it satisfies the following intersection condition:
$$\begin{aligned} \varGamma _I \cap \varGamma _J = \varGamma _{I \cap J} \;\; \text { for all } I,J \subseteq \{0,1, 2\}. \end{aligned}$$
(1)
If \(\Gamma \) is the automorphism group of a polyhedron, the group \(\Gamma _I\) corresponds to the stabilizer under the automorphism group of the subset of the base flag consisting of elements with ranks not in I. In particular, \(\langle \rho _0, \rho _1 \rangle \) is the stabilizer of the base face, and \(\langle \rho _1, \rho _2 \rangle \) is the stabilizer of the base vertex. The intersection condition for \(\Gamma \) is a consequence of the definition of abstract polyhedron.

The automorphism group of regular polyhedron is a string C-group of rank 3. Furthermore, there is a natural way to reconstruct a regular polyhedron from its automorphism group and the generators \(\rho _i\). Indeed, regular polyhedra are in one-to-one correspondence with string C-groups of rank 3. Hence, every string C-group is the automorphism group of a (unique) regular polyhedron (see [8, Thm. 2E11]).

We will frequently encounter a group that is clearly an sggi, but where it is unclear whether it is a string C-group. The quotient criterion below is often useful ([8, Thm. 2E17]):

Proposition 2.1

Let \(\varGamma = \langle \rho _0, \rho _1, \rho _{2} \rangle \) be an sggi, and \(\varLambda = \langle \lambda _0, \lambda _1, \lambda _{2} \rangle \) a string C-group. If there is a homomorphism \(\pi : \varGamma \rightarrow \varLambda \) sending each \(\rho _i\) to \(\lambda _i\), and if \(\pi \) is one to one on the subgroup \(\langle \rho _0, \rho _{1} \rangle \) or the subgroup \(\langle \rho _1, \rho _{2} \rangle \), then \(\varGamma \) is a string C-group.

We next state another criterion to determine that some sggi’s are string C-groups:

Proposition 2.2

Let \(\varGamma = \langle \rho _0, \rho _1, \rho _2 \rangle \) be a string C-group. Let \(N = \langle (\rho _0 \rho _1)^k \rangle \) or \(\langle (\rho _1 \rho _2)^k \rangle \) for some \(k \ge 2\). If N is normal in \(\varGamma \), then \(\varGamma /N\) is a string C-group.

Proof

Let \(N = \langle (\rho _0 \rho _1)^k \rangle \) and suppose that N is normal. Let us write \(\overline{\rho _i}\) for the image of \(\rho _i\) under the canonical projection. Clearly \(\langle \overline{\rho _0}, \overline{\rho _1} \rangle \) and \(\langle \overline{\rho _1}, \overline{\rho _2} \rangle \) are both dihedral, and so by [8, Prop 2E16(a)], it suffices to show that \(\langle \overline{\rho _0}, \overline{\rho _1} \rangle \cap \langle \overline{\rho _1}, \overline{\rho _2} \rangle = \langle \overline{\rho _1} \rangle \). Consider an element in \(\langle \overline{\rho _0}, \overline{\rho _1} \rangle \cap \langle \overline{\rho _1}, \overline{\rho _2} \rangle \). We may write that element as \(\overline{g}\), where \(g \in \varGamma = \langle \rho _0, \rho _1, \rho _2 \rangle \). Then since \(\overline{g} \in \langle \overline{\rho _0}, \overline{\rho _1} \rangle \), it follows that \(g \in \langle \rho _0, \rho _1 \rangle N\), and the latter is the same as simply \(\langle \rho _0, \rho _1 \rangle \). Similarly, since \(\overline{g} \in \langle \overline{\rho _1}, \overline{\rho _2} \rangle \), it follows that \(g = h (\rho _0 \rho _1)^{mk}\) for some \(h \in \langle \rho _1, \rho _2 \rangle \) and some m. But then \(g (\rho _0 \rho _1)^{-mk} = h\) is an element of \(\langle \rho _0, \rho _1 \rangle \), and so h belongs to the intersection \(\langle \rho _0, \rho _1 \rangle \cap \langle \rho _1, \rho _2 \rangle \). Since \(\varGamma \) is a string C-group, this means that \(h \in \langle \rho _1 \rangle \). Finally, \(\overline{g} = \overline{h}\), so we see that \(\overline{g} \in \langle \overline{\rho _1} \rangle \). Therefore, \(\langle \overline{\rho _0}, \overline{\rho _1} \rangle \cap \langle \overline{\rho _1}, \overline{\rho _2} \rangle \) is contained in \(\langle \overline{\rho _1} \rangle \), and the reverse inclusion is obvious.

A dual argument proves the result if \(N = \langle (\rho _1 \rho _2)^k \rangle \). \(\square \)

Given a regular polyhedron \(\mathcal P\) with automorphism group \(\varGamma (\mathcal P) = \langle \rho _0, \rho _1, \rho _2 \rangle \), we define the abstract rotations\(\sigma _1 := \rho _0 \rho _1\) and \(\sigma _2 := \rho _1 \rho _2\). Then the subgroup \(\langle \sigma _1, \sigma _2 \rangle \) of \(\varGamma (\mathcal P)\) is denoted by \(\varGamma ^+(\mathcal P)\) and called the rotation subgroup of\(\mathcal P\). The index of \(\varGamma ^+(\mathcal P)\) in \(\varGamma (\mathcal P)\) is at most 2, and when the index is exactly 2, then we say that \(\mathcal P\) is orientably regular. Otherwise, if \(\varGamma ^+(\mathcal P) = \varGamma (\mathcal P)\), then we say that \(\mathcal P\) is non-orientably regular. Indeed, a regular polyhedron is orientably or non-orientably regular in accordance with whether the underlying surface is orientable or not (when viewing the polyhedron as a map). A regular polyhedron \(\mathcal P\) is orientably regular if and only if \(\varGamma (\mathcal P)\) has a presentation in terms of the generators \(\rho _0, \rho _1, \rho _2\) such that all of the relators have even length. As a consequence, we have the following:

Proposition 2.3

Let \(\mathcal P\) be a non-orientably regular polyhedron. If \(\mathcal P\) covers a regular polyhedron \(\mathcal Q\), then \(\mathcal Q\) is also non-orientably regular.

Proof

If \(\mathcal P\) is non-orientably regular, then some odd relation holds in \(\varGamma (\mathcal P)\), and the same relation must hold in \(\varGamma (\mathcal Q)\). \(\square \)

From the properties of the automorphism groups of regular polyhedra and the definitions of \(\sigma _1\) and \(\sigma _2\), it follows that the rotation subgroups of orientably regular polyhedra satisfy
$$\begin{aligned} (\sigma _1 \sigma _2)^2 = 1 \end{aligned}$$
(2)
and the intersection condition \(\langle \sigma _1 \rangle \cap \langle \sigma _2 \rangle = \{1\}\). Indeed, \(\langle \sigma _1 \rangle \), \(\langle \sigma _1 \sigma _2 \rangle \) and \(\langle \sigma _2 \rangle \) are the stabilizers in \(\Gamma ^+({{\mathcal {P}}})\) of the base face, base edge and base vertex of \({\mathcal {P}}\), respectively. Just as we can reconstruct a regular polyhedron from its automorphism group, we can also reconstruct an orientably regular polyhedron from its rotation subgroup and specified generators \(\sigma _1\) and \(\sigma _2\) [9, Thm. 1].

Let us say that \(\mathcal P\) has multiple edges if the underlying graph of \(\mathcal P\) has multiple edges with the same vertex set. (In other words, if there is a pair of vertices with more than one edge between them.) By regularity, if some pair of vertices has r edges between them, then every pair of vertices has either 0 or r edges between them. Polyhedra without multiple edges are particularly nice to work with combinatorially, in part because of the following property.

Proposition 2.4

If \({\mathcal {P}}\) is an orientably regular polyhedron with no multiple edges, then \(\varGamma ^+(\mathcal P)\) acts faithfully on the vertex set of \({\mathcal {P}}\).

Proof

Assume to the contrary that there is a non-trivial automorphism \(\gamma \) fixing each vertex. Since \({\mathcal {P}}\) has no multiple edges, \(\gamma \) must also fix every edge. In particular, \(\gamma \) fixes the base edge. Since \(\gamma \) fixes the base vertex and the base edge, that means that \(\gamma \in \langle \sigma _2 \rangle \cap \langle \sigma _1 \sigma _2 \rangle \), and by the intersection condition, it follows that \(\gamma \) is the identity. \(\square \)

Note that for \(p \ge 3\), the polyhedron with Schläfli type \(\{p,2\}\) has no multiple edges and the reflection \(\rho _2\) acts like the identity on the vertex set. These are the only polyhedra \({\mathcal {P}}\) with no multiple edges for which the full automorphism group \(\Gamma ({{\mathcal {P}}})\) does not act faithfully on the vertex set.

The dual of a regular polyhedron is itself regular. Furthermore, if \(\varGamma (\mathcal P) = \langle \rho _0, \rho _1, \rho _2 \rangle \) and \(\varGamma (\mathcal P^{\delta }) = \langle \rho _0', \rho _1', \rho _2' \rangle \), then to obtain the defining relations of \(\varGamma (\mathcal P^{\delta })\), we can simply change the defining relations of \(\varGamma (\mathcal P)\) by replacing each \(\rho _i\) with \(\rho _{2-i}'\). This also has the effect of replacing each \(\sigma _i\) with \((\sigma _{3-i}^{-1})'\).

2.3 Tight and flat polyhedra

It was shown in [4, Prop. 3.3] that a finite polyhedron of type \(\{p, q\}\) has at least 2pq flags. When it has exactly that many flags, the polyhedron is called tight (a term introduced by Marston Conder in [1]). Proposition 4.1 of [4] showed that every tight polyhedron is also flat: Every face is incident with every vertex. Furthermore, every flat polyhedron has a Schläfli symbol and is automatically tight as well.

A regular polyhedron \(\mathcal P\) with \(\varGamma (\mathcal P) = \langle \rho _0, \rho _1, \rho _2 \rangle \) is flat if and only if \(\varGamma (\mathcal P) = \langle \rho _0, \rho _1 \rangle \langle \rho _1, \rho _2 \rangle \) ([8, Prop. 4E4]). Equivalently, \(\mathcal P\) is flat if and only if \(\varGamma (\mathcal P) = \langle \sigma _1 \rangle \langle \rho _1 \rangle \langle \sigma _2 \rangle \). Let \(w \in \varGamma (\mathcal P)\) and suppose that
$$\begin{aligned} w = \sigma _1^i \rho _1^j \sigma _2^k = \sigma _1^{i'} \rho _1^{j'} \sigma _2^{k'}. \end{aligned}$$
Then
$$\begin{aligned} \sigma _1^{i-i'} \rho _1^j = \rho _1^{j'} \sigma _2^{k'-k}. \end{aligned}$$
Therefore, w is in \(\langle \rho _0, \rho _1 \rangle \) and \(\langle \rho _1, \rho _2 \rangle \), and by the intersection condition (Eq. 1), it follows that \(w \in \langle \rho _1 \rangle \). So \(\sigma _1^{i-i'} = 1\) and \(\sigma _2^{k'-k} = 1\), and thus \(\rho _1^j = \rho _1^{j'}\) as well. Therefore, the expression of w as \(\sigma _1^i \rho _1^j \sigma _2^k\) is essentially unique (except that we may, of course, change i to \(i+p\) and so on.)

In the remainder of the paper, we will find it useful to use the generating set \(\{\sigma _1, \rho _1, \sigma _2\}\) instead of \(\{\rho _0, \rho _1, \rho _2\}\). Let \(\varGamma = \langle \rho _0, \rho _1, \rho _2 \rangle \) be an sggi and let \(\sigma _1=\rho _0 \rho _1\), \(\sigma _2=\rho _1 \rho _2\). In analogy with regular polyhedra, let us say that the group \(\varGamma \) is tight if \(\varGamma = \langle \sigma _1 \rangle \langle \rho _1 \rangle \langle \sigma _2 \rangle \). If the order of \(\sigma _1\) is p and the order of \(\sigma _2\) is q, then we will say that the group \(\varGamma \) is of type\(\{p, q\}\).

The following results all help us determine when a group (or polyhedron) is tight.

Proposition 2.5

Suppose \(\varGamma = \langle \rho _0, \rho _1, \rho _2 \rangle \) is an sggi with \(\sigma _1=\rho _0 \rho _1\) and \(\sigma _2=\rho _1 \rho _2\) and with normal subgroup \(N = \langle \sigma _1^m \rangle \) or \(N = \langle \sigma _2^m \rangle \). If \(\varGamma / N\) is tight, then so is \(\varGamma \).

Proof

Without loss of generality, assume that \(N = \langle \sigma _1^m \rangle \). Let \(g \in \varGamma \), and let \(\varphi : \varGamma \rightarrow \varGamma /N\) be the canonical map. Let \(\overline{g} = \varphi (g)\). Then since \(\varGamma /N\) is tight, we may write \(\overline{g}\) as \(\overline{\sigma _1^i \rho _1^j \sigma _2^k}\) for some choice of i, j and k. Then g is in the coset \(N(\sigma _1^i \rho _1^j \sigma _2^k)\), and every element there is in \(\langle \sigma _1 \rangle \langle \rho _1 \rangle \langle \sigma _2 \rangle \). \(\square \)

Proposition 2.6

Let \(\varGamma = \langle \rho _0, \rho _1, \rho _2 \rangle \) be an sggi with \(\sigma _1=\rho _0 \rho _1\) and \(\sigma _2=\rho _1 \rho _2\). Then \(\varGamma \) is tight if and only if every expression of the form \(\sigma _2^i \sigma _1^j\) is equivalent to an expression of the form \(\sigma _1^{i'} \sigma _2^{j'}\) or of the form \(\sigma _1^{i'} \rho _1 \sigma _2^{j'}\).

Proof

The necessity is obvious. For sufficiency, we note that the assumption says that we may move any power of \(\sigma _1\) left past any power of \(\sigma _2\). Since we also have \(\rho _1 \sigma _1^i = \sigma _1^{-i} \rho _1\), we see that in any expression of a word in the generators of \(\varGamma \), we may move every \(\sigma _1\) to the left. Similarly, we may move every \(\sigma _2\) to the right (since \(\sigma _2^i \rho _1 = \rho _1 \sigma _2^{-i}\)), and so any element of \(\varGamma \) can be written as \(\sigma _1^i \rho _1^j \sigma _2^k\) for some i, j and k. \(\square \)

Proposition 2.7

If \(\mathcal P\) and \(\mathcal Q\) are polyhedra of type \(\{p, q\}\) such that \(\mathcal P\) covers \(\mathcal Q\), and if \(\mathcal P\) is tight, then \(\mathcal P\simeq \mathcal Q\).

Proof

Since \(\mathcal P\) is tight, it has 2pq flags, and thus \(\mathcal Q\) has at most 2pq flags. On the other hand, \(\mathcal Q\) itself has Schläfli symbol \(\{p, q\}\), and so it has at least 2pq flags. The result then follows. \(\square \)

3 Automorphism groups of tight regular polyhedra

Our goal is to find a complete classification of the tight regular polyhedra. In particular, we want to find, for each Schläfli symbol \(\{p, q\}\), how many tight regular polyhedra there are of that type (up to isomorphism), and provide presentations for their automorphism groups. We will proceed by showing that certain relations must hold, and then that these relations suffice to define the group.

We will frequently use the following simple result:

Proposition 3.1

Let \(\varGamma = \langle \rho _0, \rho _1, \rho _2 \rangle \) be an sggi with \(\sigma _1=\rho _0 \rho _1\) and \(\sigma _2=\rho _1 \rho _2\). Suppose that \(g_1 \cdots g_m = h_1 \cdots h_n\), where each \(g_i\) and \(h_i\) is in the set \(\{\sigma _1, \rho _1, \sigma _2\}\). Then \(g_m \cdots g_1 = h_n \cdots h_1\).

Proof

We note that conjugation by \(\rho _1\) inverts \(\sigma _1\) and \(\sigma _2\), and it fixes \(\rho _1\) (which is the same as inverting \(\rho _1\), since it is an involution). Therefore, conjugating the relation \(g_1 \cdots g_m = h_1 \cdots h_n\) by \(\rho _1\), we obtain \(g_1^{-1} \cdots g_m^{-1} = h_1^{-1} \cdots h_n^{-1}\). Inverting both sides then gives the desired result. \(\square \)

If \(\mathcal P\) is a tight regular polyhedron, then every element of \(\varGamma (\mathcal P)\) can be written uniquely in the form \(\sigma _1^i \sigma _2^j\) or \(\sigma _1^i \rho _1 \sigma _2^j\), with \(i \in \mathbb Z_p\) and \(j \in \mathbb Z_q\). In particular, \(\sigma _2^{-1} \sigma _1\) can be written this way. We make the following observation:

Proposition 3.2

Let \(\varGamma = \langle \rho _0, \rho _1, \rho _2 \rangle \) be an sggi with \(\sigma _1=\rho _0 \rho _1\) and \(\sigma _2=\rho _1 \rho _2\).
  1. (a)

    If \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \sigma _2^j\), then \(\sigma _1^{i+1}\) and \(\sigma _2^{j-1}\) are each inverted when conjugating by \(\rho _0\), \(\rho _1\) and \(\rho _2\). In particular, \(\langle \sigma _1^{i+1} \rangle \) and \(\langle \sigma _2^{j-1} \rangle \) are normal subgroups of \(\varGamma \).

     
  2. (b)

    If \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \rho _1 \sigma _2^j\), then \(\sigma _1^{i-2}\) is inverted when conjugating by \(\rho _0\), \(\rho _1\) and \(\sigma _2\), and commutes with \(\rho _2\), whereas \(\sigma _2^{j+2}\) is inverted when conjugating by \(\rho _1\) and \(\rho _2\) and \(\sigma _1\), and commutes with \(\rho _0\). In particular, \(\langle \sigma _1^{i-2} \rangle \) and \(\langle \sigma _2^{j+2} \rangle \) are normal subgroups of \(\varGamma \).

     

Proof

By Proposition 3.1, if \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \sigma _2^j\), then also \(\sigma _1 \sigma _2^{-1} = \sigma _2^j \sigma _1^i\). Therefore,
$$\begin{aligned} \sigma _2^{-1} \sigma _1^{i+1}&= \sigma _1^i \sigma _2^j \sigma _1^i \\&= \sigma _1^{i+1} \sigma _2^{-1}. \end{aligned}$$
Thus, \(\sigma _2^{-1}\) commutes with \(\sigma _1^{i+1}\). Since \(\sigma _2^{-1} = \rho _2 \rho _1\) and conjugation by \(\rho _1\) inverts \(\sigma _1^{i+1}\), it follows that conjugation by \(\rho _2\) also inverts \(\sigma _1^{i+1}\), and \(\langle \sigma _1^{i+1} \rangle \) is normal. A similar idea using \(\sigma _2^j \sigma _1^i \sigma _2^j\) instead of \(\sigma _1^i \sigma _2^j \sigma _1^i\) proves that \(\langle \sigma _2^{j-1} \rangle \) is normal.

For the second part, we use the elements \(\sigma _1^i \rho _1 \sigma _2^j \rho _1 \sigma _1^i\) and \(\sigma _2^j \rho _1 \sigma _1^i \rho _1 \sigma _2^j\) to show that \(\sigma _2 \sigma _1^{i-1} = \sigma _1^{-i+1} \sigma _2\) and that \(\sigma _1 \sigma _2^{-j-1} = \sigma _2^{j+1} \sigma _1^{-1}\), respectively. Using Eq. (2) it can now be verified that \(\sigma _2 \sigma _1^{i-2} = \sigma _1^{-i+2} \sigma _2\) and \(\sigma _1 \sigma _2^{-j-2} = \sigma _2^{j+2} \sigma _1\), and the statement follows. \(\square \)

Theorem 3.3

Let \(\mathcal P\) be a tight regular polyhedron of type \(\{p, q\}\). If \(\mathcal P\) is orientably regular, then for some i and j, the group \(\varGamma (\mathcal P)\) is the quotient of [pq] by the extra relation \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \sigma _2^j\). If \(\mathcal P\) is non-orientably regular, then for some i and j, the relation \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \rho _1 \sigma _2^j\) holds.

Proof

Let \(\varGamma \) be the quotient of [pq] by the relation \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \sigma _2^j\). By Proposition 3.2 (a), the subgroups \(\langle \sigma _1^{i+1} \rangle \) and \(\langle \sigma _2^{j-1} \rangle \) are normal. In the quotient of \(\varGamma \) by both of these subgroups, the relation \(\sigma _2^{-1} \sigma _1 = \sigma _1^{-1} \sigma _2\) holds, and by Proposition 3.1, the relation \(\sigma _1 \sigma _2^{-1} = \sigma _2 \sigma _1^{-1}\) also holds. Furthermore, \(\sigma _2 \sigma _1 = \sigma _1^{-1} \sigma _2^{-1}\) in the automorphism group of any regular polyhedron. Therefore
$$\begin{aligned} \sigma _2^r \sigma _1 = \sigma _2^{r-1} \sigma _1^{-1} \sigma _2^{-1} = \sigma _2^{r-2} \sigma _1 \sigma _2^{-2}, \end{aligned}$$
and it follows that for any r, \(\sigma _2^r \sigma _1 = \sigma _1^{(-1)^r} \sigma _2^{-r}\). Therefore, \(\sigma _2^r \sigma _1^s = \sigma _1^{s(-1)^r} \sigma _2^{r(-1)^s}\), and by Proposition 2.6, it follows that this quotient is tight. Then by two applications of Proposition 2.5, we see that \(\varGamma \) is itself tight. Furthermore, note that the relations of \(\varGamma \) are all even.

Now, let \(\mathcal P\) be a tight regular polyhedron of type \(\{p, q\}\). Then for some i and j, either the relation \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \sigma _2^j\) holds or the relation \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \rho _1 \sigma _2^j\) holds. If \(\mathcal P\) is orientably regular, it must be the former, since the latter relation is odd. The above analysis shows that this relation alone is enough to guarantee tightness, and so by Proposition 2.7, \(\varGamma (\mathcal P)\) must be this quotient of [pq]. On the other hand, if \(\mathcal P\) is non-orientably regular, then the relation \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \rho _1 \sigma _2^j\) must hold, since otherwise, \(\varGamma (\mathcal P)\) would be the group \(\varGamma \) above, all of whose relations are even. \(\square \)

So we see that for tight orientably regular polyhedra, their automorphism groups are single-relator quotients of string Coxeter groups. The same is not true, in general, of tight non-orientably regular polyhedra. However, two extra relations always suffice. We need the following lemmas.

Lemma 3.4

Let \(\mathcal P\) be a tight non-orientably regular polyhedron, with \(\varGamma (\mathcal P) = \langle \sigma _1, \rho _1, \sigma _2 \rangle \) and \(\varGamma (\mathcal P^{\delta }) = \langle \hat{\sigma }_1, \overline{\rho }_1, \hat{\sigma }_2 \rangle \). Then either \(\sigma _2^{-2} \sigma _1 = \sigma _1^a \sigma _2^b\) holds in \(\varGamma (\mathcal P)\) (for some a and b), or \(\hat{\sigma }_2^{-2} \hat{\sigma }_1 = \hat{\sigma }_1^a \hat{\sigma }_2^b\) holds in \(\varGamma (\mathcal P^{\delta })\).

Proof

Since \(\mathcal P\) is tight, either \(\sigma _2^{-2} \sigma _1 = \sigma _1^a \sigma _2^b\), or else \(\sigma _2^{-2} \sigma _1 = \sigma _1^a \rho _1 \sigma _2^b\). In the first case, we are done. Otherwise, consider \(\mathcal P^{\delta }\). Each relation of \(\varGamma (\mathcal P)\) yields a relation in \(\varGamma (\mathcal P^{\delta })\) by sending \(\rho _1\) to \(\overline{\rho }_1\) and \(\sigma _k\) to \(\hat{\sigma }_{3-k}^{-1}\). So if the relation \(\sigma _2^{-2} \sigma _1 = \sigma _1^a \rho _1 \sigma _2^b\) holds in \(\varGamma (\mathcal P)\), it follows that \(\hat{\sigma }_1^2 \hat{\sigma }_2^{-1} = \hat{\sigma }_2^{-a} \overline{\rho }_1 \hat{\sigma }_1^{-b}\) holds in \(\varGamma (\mathcal P^{\delta })\), and from this it follows (by Proposition 3.1) that \(\hat{\sigma }_2^{-1} \hat{\sigma }_1^2 = \hat{\sigma }_1^{-b} \overline{\rho }_1 \hat{\sigma }_2^{-a}\). Now, \(\mathcal P^{\delta }\) is also a tight non-orientably regular polyhedron, so Theorem 3.3 implies that the relation \(\hat{\sigma }_2^{-1} \hat{\sigma }_1 = \hat{\sigma }_1^i \overline{\rho }_1 \hat{\sigma }_2^j\) holds in \(\varGamma (\mathcal P^{\delta })\) for some i and j. Then, working in \(\varGamma (\mathcal P^{\delta })\) and using Proposition 3.2 (b), we get that:
$$\begin{aligned} \hat{\sigma }_2^{-2} \hat{\sigma }_1&= \hat{\sigma }_2^{-1} (\hat{\sigma }_2^{-1} \hat{\sigma }_1) \\&= \hat{\sigma }_2^{-1} \hat{\sigma }_1^i \overline{\rho }_1 \hat{\sigma }_2^j \\&= \hat{\sigma }_2^{-1} \hat{\sigma }_1^{i-2} \hat{\sigma }_1^2 \overline{\rho }_1 \hat{\sigma }_2^j \\&= \hat{\sigma }_1^{2-i} \hat{\sigma }_2^{-1} \hat{\sigma }_1^2 \overline{\rho }_1 \hat{\sigma }_2^j \\&= \hat{\sigma }_1^{2-i} \hat{\sigma }_1^{-b} \overline{\rho }_1 \hat{\sigma }_2^{-a} \overline{\rho }_1 \hat{\sigma }_2^j \\&= \hat{\sigma }_1^{2-i-b} \hat{\sigma }_2^{a+j}, \end{aligned}$$
and so a relation of the desired type holds in \(\varGamma (\mathcal P^{\delta })\). \(\square \)

Lemma 3.5

Let \(\varGamma = \langle \rho _0, \rho _1, \rho _2 \rangle \) be an sggi with \(\sigma _1 = \rho _0 \rho _1\) and \(\sigma _2 = \rho _1 \rho _2\). Suppose that \(\varGamma \) satisfies the relations \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \rho _1 \sigma _2^j\) and \(\sigma _2^{-2} \sigma _1 = \sigma _1^a \sigma _2^b\). Then \(\sigma _2^{-1} \sigma _1^2 = \sigma _1^{i-a} \rho _1 \sigma _2^{b-j-2}\). Furthermore, conjugation by \(\sigma _2\) inverts \(\sigma _1^{i-3-a}\) and \(\sigma _1\) commutes with \(\sigma _2^{b-2}\), and in particular, the subgroups \(\langle \sigma _1^{i-3-a} \rangle \) and \(\langle \sigma _2^{b-2} \rangle \) are normal.

Proof

By Proposition 3.2 (b), conjugation by \(\sigma _1\) inverts \(\sigma _2^{j+2}\). Therefore,
$$\begin{aligned} \sigma _2^{-1} \sigma _1^2&= \sigma _1^i \rho _1 \sigma _2^j \sigma _1 \\&= \sigma _1^i \rho _1 \sigma _2^{-2} \sigma _2^{j+2} \sigma _1 \\&= \sigma _1^i \rho _1 \sigma _2^{-2} \sigma _1 \sigma _2^{-(j+2)} \\&= \sigma _1^i \rho _1 \sigma _1^a \sigma _2^{b-j-2} \\&= \sigma _1^{i-a} \rho _1 \sigma _2^{b-j-2}. \end{aligned}$$
It follows from Proposition 3.1 that \(\sigma _1^2 \sigma _2^{-1} = \sigma _2^{b-j-2} \rho _1 \sigma _1^{i-a}\). Therefore,
$$\begin{aligned} \sigma _2^{-1} \sigma _1^2 \rho _1 \sigma _1^{i-a}&= \sigma _1^{i-a} \rho _1 \sigma _2^{b-j-2} \rho _1 \sigma _1^{i-a} \\&= \sigma _1^{i-a} \rho _1 \sigma _1^2 \sigma _2^{-1}. \end{aligned}$$
Then \(\sigma _2^{-1} \sigma _1^{2-i+a} = \sigma _1^{i-a-2} \sigma _2\). Therefore, by (2),
$$\begin{aligned} \sigma _2 \sigma _1^{3-i+a}&= \sigma _1^{-1} \sigma _2^{-1} \sigma _1^{2-i+a} = \sigma _1^{i-a-3} \sigma _2. \end{aligned}$$
It follows that \(\langle \sigma _1^{i-a-3} \rangle \) is normal. Finally, since \(\sigma _2^{-2} \sigma _1 = \sigma _1^a \sigma _2^b\), then also \(\sigma _1 \sigma _2^{-2} = \sigma _2^b \sigma _1^a\) (by Proposition 3.1). Therefore,
$$\begin{aligned} \sigma _2^{b-2} \sigma _1&= \sigma _2^b \sigma _1^a \sigma _2^b \\&= \sigma _1 \sigma _2^{b-2}, \end{aligned}$$
and so \(\sigma _2^{b-2}\) is normalized by \(\sigma _1\). \(\square \)

Theorem 3.6

Let \(\mathcal P\) be a tight non-orientably regular polyhedron of type \(\{p, q\}\). Then either \(\varGamma (\mathcal P)\) is the quotient of [pq] by the relations \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \rho _1 \sigma _2^j\) and \(\sigma _2^{-2} \sigma _1 = \sigma _1^a \sigma _2^b\) (for some choice of i, j, a and b), or \(\varGamma (\mathcal P^{\delta })\) is the quotient of [qp] by those relations.

Proof

Let us define \(\Delta (p,q)_{(i,j,a,b)}\) to be the quotient of [pq] by the relations \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \rho _1 \sigma _2^j\) and \(\sigma _2^{-2} \sigma _1 = \sigma _1^a \sigma _2^b\). Lemma 3.4 implies that either \(\varGamma (\mathcal P)\) is a quotient of \(\Delta (p,q)_{(i,j,a,b)}\) or that \(\varGamma (\mathcal P^{\delta })\) is a quotient of some \(\Delta (q,p)_{(i,j,a,b)}\), for some choice of (ijab). Without loss of generality, let us assume that the first is true. It remains to show that \(\varGamma (\mathcal P)\) is equal to \(\Delta (p,q)_{(i,j,a,b)}\) and not to a proper quotient. For that, it suffices to show that \(\Delta (p,q)_{(i,j,a,b)}\) is itself tight. In light of Proposition 2.5, we may take the quotient by any normal subgroup generated by a power of \(\sigma _1\) or \(\sigma _2\), and if that quotient is tight, then so is \(\Delta (p,q)_{(i,j,a,b)}\). There are several such normal subgroups; in particular, Proposition 3.2 (b) shows that \(\sigma _1^{i-2}\) and \(\sigma _2^{j+2}\) generate normal subgroups, and Lemma 3.5 shows that \(\sigma _1^{i-3-a}\) and \(\sigma _2^{b-2}\) generate normal subgroups. Taking the quotient by these subgroups yields the group \(\Delta (p', q')_{(2,-2,-1,2)}\) for some \(p'\) dividing p and some \(q'\) dividing q. Now, in this quotient, \(\sigma _2^{-2} \sigma _1 = \sigma _1^{-1} \sigma _2^2\), and therefore \(\sigma _2^{-3} \sigma _1 = \sigma _2^{-1} \sigma _1^{-1} \sigma _2^2 = \sigma _1 \sigma _2^3\). It follows that \(\langle \sigma _2^3 \rangle \) is normal. Similarly, the relation \(\sigma _2^{-1} \sigma _1 = \sigma _1^2 \rho _1 \sigma _2^{-2}\) holds, and so
$$\begin{aligned} \sigma _2^{-1} \sigma _1^4&= \sigma _1^2 \rho _1 \sigma _2^{-2} \sigma _1^3 \\&= \sigma _1^2 \rho _1 \sigma _1^{-1} \sigma _2^2 \sigma _1^2 \\&= \sigma _1^2 \rho _1 \sigma _1^{-1} \sigma _2 \sigma _1^{-1} \sigma _2^{-1} \sigma _1 \\&= \sigma _1^2 \rho _1 \sigma _1^{-1} \sigma _1^{-2} \rho _1 \sigma _2^2 \sigma _2^{-1} \sigma _1 \\&= \sigma _1^5 \sigma _2 \sigma _1 \\&= \sigma _1^4 \sigma _2^{-1}, \end{aligned}$$
and thus \(\langle \sigma _1^4 \rangle \) is normal as well. Taking the quotient by these subgroups yields \(\Delta (4,3)_{(2,-2,-1,2)}\). Using GAP [10], we can verify that this group is tight; in fact, it is the group of the hemicube. It follows that \(\Delta (p,q)_{(i,j,a,b)}\) is tight, proving the claim. \(\square \)

We see that the automorphism group of a tight regular polyhedron must be one of the groups in Theorem 3.3 or 3.6. Furthermore, the given groups are always tight, in the sense that \(\varGamma = \langle \sigma _1 \rangle \langle \rho _1 \rangle \langle \sigma _2 \rangle \). Two things remain to be determined for each family of groups. First, for which values of the parameters is the group a string C-group? Second, under what conditions is the group actually of type \(\{p, q\}\); in other words, when is there no collapse of the subgroups \(\langle \rho _0, \rho _1 \rangle \) and \(\langle \rho _1, \rho _2 \rangle \)? The answer to these questions is quite dissimilar in the orientable case versus the non-orientable case, and we require fairly different methods for the two cases.

4 Tight orientably regular polyhedra

We first consider the classification of tight, orientably regular polyhedra. Part of the classification was completed in [2, Thm. 3.4]:

Theorem 4.1

There is a tight orientably regular polyhedron of type \(\{p, q\}\) if and only if one of the following is true\(:\)
  1. (a)

    p and q are both even, or

     
  2. (b)

    p is odd and q is an even divisor of 2p, or

     
  3. (c)

    q is odd and p is an even divisor of 2q.

     

Furthermore, it was proved in [2, Thm. 3.3] that if p or q is odd, then there is at most one isomorphism type of tight orientably regular polyhedra of type \(\{p, q\}\). What remains to be determined is how many tight orientably regular polyhedra there are when p and q are both even, and to find presentations for their automorphism groups.

Let \(\varLambda (p,q)_{i,j}\) be the quotient of [pq] by the extra relation \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \sigma _2^j\). We determined in Theorem 3.3 that if \(\mathcal P\) is a tight orientably regular polyhedron of type \(\{p, q\}\), then it has automorphism group \(\varLambda (p,q)_{i,j}\) for some choice of i and j. For a given Schläfli symbol \(\{p, q\}\), we need to determine which values of i and j make \(\varLambda (p,q)_{i,j}\) the automorphism group of a tight orientably regular polyhedron of type \(\{p, q\}\).

Let us recall that if H is a subgroup of a group G, the largest subgroup of H which is normal in G is called the core of H in G, and we shall denote it \({\text {Core}}_{G}(H)\). If \({\text {Core}}_G(H)\) is trivial, then we say that H is core-free in G.

We are able to reduce our classification problem using the following result.

Proposition 4.2

\(\varLambda (p,q)_{i,j}\) is the automorphism group of a tight orientably regular polyhedron of type \(\{p, q\}\) if and only if there are values \(p'\) and \(q'\) such that
  1. (a)

    \(p'\) divides p and \(i+1\),

     
  2. (b)

    \(q'\) divides q and \(j-1\),

     
  3. (c)

    the group \(\varLambda (p,q')_{i,1}\) is the automorphism group of a tight orientably regular polyhedron of type \(\{p, q'\}\), with \(\langle \sigma _2 \rangle \) core-free in \(\varLambda (p,q')_{i,1}\), and

     
  4. (d)

    the group \(\varLambda (p',q)_{-1,j}\) is the automorphism group of a tight orientably regular polyhedron of type \(\{p', q\}\), with \(\langle \sigma _1 \rangle \) core-free in \(\varLambda (p',q)_{-1,j}\).

     

Proof

First, suppose that \(\varLambda (p,q)_{i,j}\) is the automorphism group of a tight orientably regular polyhedron of type \(\{p, q\}\). Let \(N := \langle \sigma _1^{p'} \rangle \) be the core of \(\langle \sigma _1 \rangle \) in \(\varLambda (p, q)_{i,j}\), with \(p'\) dividing p. Proposition 3.2 (a) shows that \(\langle \sigma _1^{i+1} \rangle \) is normal, and it follows that \(p'\) divides \(i+1\) as well. Then the quotient of \(\varLambda (p,q)_{i,j}\) by N is \(\varLambda (p',q)_{-1,j}\), and the latter has \(\langle \sigma _1 \rangle \) core-free. Since \(\sigma _2\) has order q in \(\varLambda (p,q)_{i,j}\) (by supposition), the same is true in \(\varLambda (p',q)_{-1,j}\); otherwise, we would have some \(\sigma _2^{q'} \in \langle \sigma _1^{p'} \rangle \), violating the intersection condition. Then Proposition 2.2 implies that \(\varLambda (p',q)_{-1,j}\) is a string C-group. Therefore, \(\varLambda (p',q)_{-1,j}\) is the automorphism group of a tight orientably regular polyhedron of type \(\{p', q\}\). An analogous argument, taking the quotient by the core of \(\langle \sigma _2 \rangle \) (which we set to be \(\langle \sigma _2^{q'} \rangle \)), shows that \(\varLambda (p, q')_{i,1}\) is the automorphism group of a tight orientably regular polyhedron of type \(\{p, q'\}\), where \(q'\) divides q and \(j-1\).

In the other direction, suppose that \(\varLambda (p,q')_{i,1}\) and \(\varLambda (p',q)_{-1,j}\) are automorphism groups of tight orientably regular polyhedra of types \(\{p, q'\}\) and \(\{p', q\}\), respectively, and suppose that \(p'\) divides p and \(i+1\), and that \(q'\) divides q and \(j-1\). It is clear from the presentations that \(\varLambda (p,q)_{i,j}\) covers \(\varLambda (p',q)_{i,j}\) and \(\varLambda (p,q')_{i,j}\). Since \(p'\) divides \(i+1\), it follows that \(i \equiv -1\) (mod \(p'\)), and thus \(\varLambda (p', q)_{i,j} = \varLambda (p', q)_{-1,j}\). Similarly, \(\varLambda (p,q')_{i,j} = \varLambda (p,q')_{i,1}\). So \(\varLambda (p,q)_{i,j}\) covers \(\varLambda (p,q')_{i,1}\) and \(\varLambda (p',q)_{-1,j}\). Since \(\varLambda (p,q')_{i,1}\) has type \(\{p, q'\}\), it follows that \(\sigma _1\) has order p in \(\varLambda (p,q)_{i,j}\), and since \(\varLambda (p',q)_{-1,j}\) has type \(\{p', q\}\), it follows that \(\sigma _2\) has order q in \(\varLambda (p,q)_{i,j}\). Finally, since the cover from \(\varLambda (p,q)_{i,j}\) to \(\varLambda (p',q)_{-1,j}\) is one to one on the facets and the latter is a string C-group, Proposition 2.1 implies that \(\varLambda (p,q)_{i,j}\) is also a string C-group. So \(\varLambda (p,q)_{i,j}\) is the automorphism group of an orientably regular polyhedron \(\mathcal P\) of type \(\{p, q\}\). Since the group \(\varLambda (p,q)_{i,j}\) is tight (by Theorem 3.3), it follows that \(\mathcal P\) is tight.

\(\square \)

There is a nice combinatorial interpretation of what it means for \(\langle \sigma _2 \rangle \) to be core-free in \(\varGamma (\mathcal P)\). We start by remarking that \(\langle \sigma _2 \rangle \) is core-free in \(\varGamma (\mathcal P)\) if and only if it is core-free in \(\varGamma ^+(\mathcal P)\), since \(\rho _1\) and \(\rho _2\) normalize any subgroup \(\langle \sigma _2^k \rangle \), and so if \(\sigma _1 (= \rho _0 \rho _1)\) normalizes such a subgroup, then so does \(\rho _0\). Therefore, we can work with \(\varGamma ^+(\mathcal P)\) instead. We start with a few simple results.

Proposition 4.3

Let \(\mathcal P\) be a tight polyhedron. Suppose \(\varphi \in \varGamma (\mathcal P)\) acts as a rotation at the vertex v; that is, \(\varphi \) fixes v while cyclically permuting the neighbors of v. If \(\varphi \) fixes some neighbor of v, then it fixes all neighbors of v.

Proof

Suppose that \(\varphi \) fixes u, and let w be another neighbor of v. Then there is some automorphism \(\psi \) that acts as a rotation at v and with the property that \(u \psi = w\). Furthermore, since \(\varphi \) and \(\psi \) both act as rotations at v, it follows that \(u \varphi \psi = u \psi \varphi \). Therefore,
$$\begin{aligned} w \varphi = u \psi \varphi = u \varphi \psi = u \psi = w, \end{aligned}$$
and so \(\varphi \) fixes every neighbor of v. \(\square \)

Corollary 4.4

Let \(\mathcal P\) be a tight orientably regular polyhedron. Let \(\varphi \in \varGamma ^+(\mathcal P)\), and suppose that \(\varphi \) fixes some vertex v and one of the neighbors of v. Then \(\varphi \) fixes all vertices of \(\mathcal P\).

Proof

Let \(\varphi \in \varGamma ^+(\mathcal P)\) and suppose that \(\varphi \) fixes some vertex v and one of its neighbors. Then by Proposition 4.3, \(\varphi \) fixes all of the neighbors of v. The stabilizer of each vertex is a dihedral group, and since \(\mathcal P\) is orientable and \(\varphi \in \varGamma ^+(\mathcal P)\), it follows that \(\varphi \) acts at a rotation at each of those neighbors. Therefore, \(\varphi \) fixes the neighbors of each neighbor of v. Continuing in this manner and using the connectivity of \(\mathcal P\), we see that \(\varphi \) fixes every vertex. \(\square \)

Proposition 4.5

Let \(\mathcal P\) be a tight orientably regular polyhedron of type \(\{p, q\}\), with \(\varGamma ^+(\mathcal P) = \langle \sigma _1, \sigma _2 \rangle \). Let v be the base vertex of \(\mathcal P\), and let \(q'\) be the smallest positive integer such that \(\sigma _2^{q'}\) fixes one of the neighbors of v. Then \(\langle \sigma _2^{q'} \rangle \) is the subgroup of \(\varGamma ^+(\mathcal P)\) that fixes every vertex of \(\mathcal P\), and \(\langle \sigma _2^{q'} \rangle = {\text {Core}}_{\varGamma ^+({{\mathcal {P}}})}(\langle \sigma _2 \rangle )\).

Proof

By Corollary 4.4, \(\sigma _2^{q'}\) must fix every vertex of \(\mathcal P\), from which it is immediate that \(\langle \sigma _2^{q'} \rangle \lhd \varGamma ^+(\mathcal P)\). Furthermore, if \(\varphi \in \varGamma ^+(\mathcal P)\) fixes every vertex of \(\mathcal P\), then it must lie in \(\langle \sigma _2 \rangle \), and by our choice of \(q'\) it follows that \(\langle \sigma _2^{q'} \rangle \) is the subgroup of \(\varGamma ^+(\mathcal P)\) that fixes every vertex.

Next we want to show that \(\langle \sigma _2^{q'} \rangle \) is the largest subgroup of \(\langle \sigma _2 \rangle \) that is normal in \(\varGamma ^+(\mathcal P)\). Suppose that \(\sigma _2^a \notin \langle \sigma _2^{q'} \rangle \). Then there is a vertex u such that \(u\sigma _2^a \ne u\). Since \(\mathcal P\) is tight, the base face is incident on every vertex, and thus we can find some \(b \in \mathbb {Z}\) such that \(v\sigma _1^b = u\). Then \(v \sigma _1^b\sigma _2^a \sigma _1^{-b} = u\sigma _2^a\sigma _1^{-b} \ne v\). Therefore, \(\sigma _1^b \sigma _2^a \sigma _1^{-b} \notin \langle \sigma _2 \rangle \) (since \(\sigma _2\) fixes v), and so \(\langle \sigma _2^a \rangle \) is not normal in \(\varGamma ^+({{\mathcal {P}})}\). \(\square \)

We are now ready to explain the connection between having multiple edges and the core of \(\langle \sigma _2 \rangle \).

Proposition 4.6

Let \(\mathcal P\) be a tight orientably regular polyhedron. Then \(\mathcal P\) has no multiple edges if and only if \(\langle \sigma _2 \rangle \) is core-free in \(\varGamma (\mathcal P)\).

Proof

First, suppose that \(\langle \sigma _2 \rangle \) has a nontrivial core. Then by Proposition 4.5, there is an automorphism \(\sigma _2^{q'}\) that fixes all vertices, and by Proposition 2.4, it follows that \(\mathcal P\) has multiple edges.

Conversely, suppose that \(\mathcal P\) has multiple edges, and let v be the base vertex. Let u be a neighbor of v, and let \(e_1\) and \(e_2\) be edges between u and v. By the regularity of \(\mathcal P\), there is some nontrivial even automorphism \(\varphi \) that sends the pair \((v, e_1)\) to \((v, e_2)\), and since \(\varphi \) fixes v, it follows that \(\varphi = \sigma _2^{q'}\) for some \(q'\). In order for \(\sigma _2^{q'}\) to send \(e_1\) to \(e_2\), it must be the case that \(\varphi \) fixes u. Then by Corollary 4.4, \(\sigma _2^{q'}\) fixes every vertex, which implies that \(\langle \sigma _2^{q'} \rangle \) is normal in \(\langle \sigma _2 \rangle \). Thus \(\langle \sigma _2 \rangle \) has a nontrivial core. \(\square \)

In light of Proposition 4.6, we may reinterpret Proposition 4.2 as follows:

Corollary 4.7

\(\varLambda (p,q)_{i,j}\) is the automorphism group of a tight orientably regular polyhedron of type \(\{p, q\}\) if and only if there are values \(p'\) and \(q'\) such that
  1. (a)

    \(p'\) divides p and \(i+1\),

     
  2. (b)

    \(q'\) divides q and \(j-1\),

     
  3. (c)

    \(\varLambda (p,q')_{i,1}\) is the automorphism group of a tight orientably regular polyhedron of type \(\{p, q'\}\), and with no multiple edges, and

     
  4. (d)

    \(\varLambda (p',q)_{-1,j}\) is the automorphism group of a tight orientably regular polyhedron of type \(\{p', q\}\), such that the dual has no multiple edges.

     

Our first step will then be to find all tight orientably regular polyhedra of type \(\{p, q\}\) with no multiple edges.

4.1 Tight orientably regular polyhedra with no multiple edges

There is exactly one orientably regular polyhedron \(\mathcal P\) with Schläfli type \(\{p,2\}\) for every \(p \ge 3\). This can be easily seen either by showing that the graph induced by the vertex and edge set must be connected and 2-regular, or by noting that the Coxeter group [p, 2] is isomorphic to \(D_p \times C_2\) and must cover \(\Gamma (\mathcal P)\). None of these polyhedra have multiple edges.

In what follows, we shall determine the remaining tight orientably regular polyhedra with no multiple edges.

In the results that follow, we will generally assume the following, which we call the usual setup (see Fig. 1). Let \({\mathcal {Q}}\) be a tight orientably regular polyhedron of type \(\{p, q\}\), with \(q \ge 3\), and suppose that \(\mathcal Q\) has no multiple edges. Let us fix a base face \(F_1\) and label the vertices with elements of \(\mathbb Z_p\) in such a way that \(i \sigma _1 = i + 1\). The flag \(\varPhi \) will consist of the vertex 1, the edge between 0 and 1, and the face \(F_1\). Let \(F_2\) be the other face containing the edge between 0 and 1, and let k be the other vertex of \(F_2\) that is adjacent to 1 (so that \(0 \sigma _2 = k\)).
Fig. 1

The usual setup for tight orientably regular polyhedra with no multiple edges

Lemma 4.8

Let \({\mathcal {Q}}\) be a tight orientably regular polyhedron of type \(\{p, q\}\) with \(q \ge 3\), no multiple edges, and with the usual setup. Then
  1. (a)

    \(p > q\).

     
  2. (b)

    p is even.

     
  3. (c)

    Every vertex of the dual of \({\mathcal {Q}}\) has exactly two neighbors.

     
  4. (d)

    k is even.

     
  5. (e)

    The vertices of \(F_2\), in clockwise order, are \((1, 0, -k+1, -k, -2k+1, -2k, \ldots , k+1, k)\).

     

Proof

A tight polyhedron of type \(\{p, q\}\) has p vertices, and each vertex has q neighbors; so in order to have no multiple edges, it must be that \(p > q\), proving part (a).

Recall that the base flag \(\varPhi \) consists of vertex 1, the edge between 0 and 1, and \(F_1\). The involutory automorphism \(\gamma = \rho _0 \sigma _1^k\) maps \(\varPhi \) to the flag \(\Psi \) consisting of vertex k, the edge between k and \(k+1\) and the face \(F_1\). Note that \(\gamma \) fixes \(F_1\) and maps, respectively, the vertices 0 and 1 to \(k+1\) and k. Then \(F_2 (= F_1\sigma _2)\) is mapped to a face \(F'\) sharing the edge between k and \(k+1\) with \(F_1\). Furthermore, \(\gamma \) fixes the edge between 1 and k (since it swaps their endpoints and there are no multiple edges), and since \(\gamma \) sends \(F_2\) to \(F'\), it follows that \(F'\) also contains that edge. Let \(\Upsilon \) be the flag containing vertex 1, the edge between 1 and k, and the face \(F_2\). Then \(\gamma \) maps \(\Upsilon \) to the flag containing vertex k, the edge between 1 and k, and \(F'\). If \(F' \ne F_2\), then \(\gamma \) maps \(\Upsilon \) to \(\Upsilon ^{0,2}\). But \({\mathcal {Q}}\) is orientable, and \(\gamma \) is an odd automorphism. So it must be that \(F' = F_2\) so that \(\gamma \) maps \(\Upsilon \) to \(\Upsilon ^0\). Hence \(F_2\) contains the edges between 0 and 1, between 1 and k, and between k and \(k+1\).

Now, the automorphism \(\sigma _1^k\) fixes \(F_1\), and it maps the edge between 0 and 1 to the edge between k and \(k+1\). Since \(F_1\) and \(F_2\) share both of those edges, it follows that \(\sigma _1^k\) also fixes \(F_2\). Therefore, \(F_2\) also contains the edge between \(k+1\) and 2k, since that is the image of the edge between 1 and k. Finally, an inductive procedure shows that \(F_2\) contains the edge between nk and \(nk+1\) and the edge between \(nk+1\) and \((n+1)k\) for every n. In particular, \(F_2\) shares every other edge with \(F_1\).

If p were odd, then \(F_2\) would have to share every edge with \(F_1\). Then there could only be two faces, which would imply that \(q = 2\). Since \(q \ge 3\), the parameter p must be even. Furthermore, this means that \(F_2\) shares half of its edges with \(F_1\), and half of its edges with some other face. By regularity, every face must share its edges with exactly two distinct faces, which means that in the dual of \(\mathcal Q\), every vertex has exactly two neighbors.

Just as \(F_2\) shares half of its edges with \(F_1\), the face \(F_1\) shares half of its edges with \(F_2\). If \(F_1\) shared two consecutive edges with \(F_2\), then it would have to share all of them (by regularity), and so it must share every other edge with \(F_2\). Since the two faces share the edge between 0 and 1, it follows that, for every i, they share the edge between 2i and \(2i+1\) but not the edge from 2i to \(2i-1\). Since they also share the edge between k and \(k+1\), it follows that k is even, proving part (d). Part (e) immediately follows. \(\square \)

Lemma 4.9

Let \({\mathcal {Q}}\) be a tight orientably regular polyhedron of type \(\{p, q\}\), with the usual setup. Then:
  1. (a)
    $$\begin{aligned} i\sigma _2 = \left\{ \begin{array}{ll} \frac{k(2-i)}{2}&{} \text{ if } i \text{ is } \text{ even }\\ 1+\frac{k(1-i)}{2}&{} \text{ if } i \text{ is } \text{ odd }. \end{array}\right. \end{aligned}$$
     
  2. (b)

    \((k/2)^2 \equiv 1\) modulo p / 2.

     

Proof

The automorphism \(\sigma _2\) sends vertex 0 to vertex k and fixes 1. Proceeding clockwise around \(F_1\) and applying \(\sigma _2\) gives us the vertices of \(F_2\) in clockwise order. From Lemma 4.8, the clockwise order of the vertices in \(F_2\) is \((1,0,-k+1,-k,-2k+1,-2k,\dots ,k+1,k)\), and part (a) follows.

For part (b), note first that since Lemma 4.8 says that every vertex of the dual of \({\mathcal {Q}}\) has exactly two neighbors, it follows that \({\text {Core}}_{\varGamma ^+({{\mathcal {Q}}})}(\langle \sigma _1 \rangle ) = \langle \sigma _1^2 \rangle \). In particular, this means that \(\sigma _2^{-1} \sigma _1^2 \sigma _2 = \sigma _1^{2s}\) for some s. Now, since \((\sigma _2 \sigma _1)^2 = 1\),
$$\begin{aligned} \sigma _1^2&= (\sigma _2 \sigma _1)^{-2} \sigma _1^2 (\sigma _2 \sigma _1)^2 \\&= (\sigma _1^{-1} \sigma _2^{-1} \sigma _1^{-1}) \sigma _1^{2s} (\sigma _1 \sigma _2 \sigma _1) \\&= \sigma _1^{-1} \sigma _2^{-1} \sigma _1^{2s} \sigma _2 \sigma _1 \\&= \sigma _1^{-1} \sigma _1^{2s^2} \sigma _1 \\&= \sigma _1^{2s^2}. \end{aligned}$$
Therefore \(2s^2 \equiv 2\) (mod p); that is, \(s^2 \equiv 1\) (mod p / 2). Now, \(F_1\sigma _2 = F_2\) and therefore, with the labeling of vertices as in Lemma 4.9, \(2\sigma _2 = 0\), \(1\sigma _2 =1\), \(0\sigma _2 = k\), \((p-1)\sigma _2 = k+1\) and so on. Then
$$\begin{aligned} 0 = 2\sigma _2 = 0\sigma _1^2 \sigma _2 = 0\sigma _2 \sigma _1^{2s} = k\sigma _1^{2s}, \end{aligned}$$
and so \(s = -k/2\). This implies that \(1 \equiv (-k/2)^2 \equiv (k/2)^2\) (mod p / 2). \(\square \)

Lemma 4.9 establishes the order of the vertices in \(F_2\), which is the face sharing the edge between 0 and 1 with \(F_1\) , and it determines the action of \(\sigma _2\) on the vertices. It follows from Proposition 2.4 that, in order to characterize all tight orientably regular polyhedra with no multiple edges, we only need to determine all possible values of p and k. (Note that the value of q plays no role on the expressions of \(\sigma _1\) and \(\sigma _2\).)

Lemma 4.9 (b) imposes a strong condition on the value of k. The following lemma suggests how restrictive this condition is. The proof is straightforward and omitted, but see [5, Section 1.2] for the number a(n) of solutions of \(x^2=1\) in \(\mathbb {Z}_n^*\).

Lemma 4.10

Let P be a prime and n a positive integer, and let \(X_{P,n}\) be the set of integers \(1 \le x \le P^{n-1}\) satisfying that \(x^2 \equiv 1\) modulo \(P^n\). Then
  1. (a)

    \(X_{2,1} = \{1\}\), \(X_{2,2} = \{1,3\}\) and \(X_{2,n} = \{1,2^{n-1}-1,2^{n-1}+1,2^n-1\}\) if \(n\ge 3\).

     
  2. (b)

    \(X_{P,n} = \{1,P^n-1\}\) if P is odd.

     

In general, if \(p = P_1^{\alpha _1} \cdots P_s^{\alpha _s}\) with \(P_1, \dots , P_s\) distinct primes and \(P_1=2\), then \((k/2)^2 \equiv 1\) (mod p / 2) if and only if \((k/2)^2 \equiv 1\) (mod \(2^{\alpha _1-1})\) and \((k/2)^2 \equiv 1\) (mod \(P_i^{\alpha _i}\)) for \(i\ge 2\).

We now obtain q from p and k. The value of q is the order of \(\sigma _2\), or alternatively, the smallest positive m such that \(2\sigma _2^m = 2\). An inductive procedure shows that for \(m \ge 2\),
$$\begin{aligned} 2\sigma _2^{m} = 2\left( k/2 - \left( k/2\right) ^2 + \left( k/2\right) ^3 - \cdots + \left( -1\right) ^{m}\left( k/2\right) ^{m-1}\right) . \end{aligned}$$
This implies that q is the smallest positive m satisfying that
$$\begin{aligned} 2 \equiv 2\left( k/2 - \left( k/2\right) ^2 + \left( k/2\right) ^3 - \cdots + \left( -1\right) ^{m-1}\left( k/2\right) ^{m}\right) \quad \text{(mod } p\text{) }, \end{aligned}$$
which by Lemma 4.9 (b) is equivalent to
$$\begin{aligned} 2 \sigma _2^m \equiv \left\{ \begin{array}{ll} \left( k-2\right) (m-1)/2 &{} \text{ if } m \text{ is } \text{ odd }\\ k+\left( k-2\right) (m-2)/2 &{} \text{ if } m \text{ is } \text{ even } \end{array} \right. \quad \text{(mod } p\text{). } \end{aligned}$$
(3)
Therefore q is the smallest positive m satisfying (3).

Now we are ready to state our main results about tight orientably regular polyhedra with no multiple edges.

Proposition 4.11

Let \({\mathcal {Q}}\) be a tight orientably regular polyhedron with no multiple edges and Schläfli type \(\{p,q\}\) with q odd. Then \(p=2q\). Furthermore, it is unique up to isomorphism.

Proof

From Eq. (3), we know that if q is odd, then \(2\equiv (k/2-1)(q-1)\) (mod p). Multiplying both sides by \((k/2+1)\) yields that \(k+2 \equiv ((k/2)^2-1)(q-1)\) (mod p), and since \((k/2)^2 \equiv 1\) (mod p / 2) and q is odd, it follows that \(k+2 \equiv 0\) (mod p); in other words, the only choice for k is \(k = p-2\). Substituting in Eq. (3), we obtain that \(-2(q-1) \equiv 2\) (mod p) and hence p divides 2q. Since p is even and \(p > q\) (by Lemma 4.8), it follows that \(p = 2q\). Such a polyhedron is unique up to isomorphism since k is determined by the value of p and by the fact that q is odd. \(\square \)

This result is also a consequence of [2, Thm. 3.4], since if p is a proper divisor of 2q, then \(p \le q\) and \({\mathcal {Q}}\) must have multiple edges.

Proposition 4.12

Let \({\mathcal {Q}}\) be a tight orientably regular polyhedron with no multiple edges and Schläfli type \(\{p,q\}\) with \(q \ge 4\) even. Let \(p=P_1^{\alpha _1} P_2^{\alpha _2} \cdots P_s^{\alpha _s}\) with \(P_i\) prime, \(P_i < P_{i+1}\) and \(\alpha _i \ge 1\) for all i. Then
  1. (a)

    \(P_1 = 2\);

     
  2. (b)

    the maximal power of 2 that divides q is either 2, 4 or \(2^{\alpha _1-1}\), and it is 4 only if \(\alpha _1 \ge 3\);

     
  3. (c)

    for every \(i \ge 2\), either q is coprime with \(P_i\) or \(P_i^{\alpha _i}\) divides q; and

     
  4. (d)

    q divides p.

     

Proof

Part (a) follows from Lemma 4.8 (b).

From Eq. (3) and Lemma 4.9 (b) we have that \((k-2)q/2 \equiv 0\) (mod p). This implies that \((k/2-1)q/2\equiv 0\) modulo p / 2. This is equivalent to
$$\begin{aligned} (k/2-1)q/2\equiv 0 \,\,\text{(mod } 2^{\alpha _1-1}\text{) } \quad \text{ and } \quad (k/2-1)q/2 \equiv 0 \,\,\text{(mod } P_i^{\alpha _i}\text{) } \text{ for } \text{ all } i \ge 2. \end{aligned}$$
(4)
If \(\alpha _1\ge 4\), Lemma 4.10 (a) implies that \(k/2-1\) is congruent to either 0, \(2^{\alpha _1-2}-2\), \(2^{\alpha _1-2}\) or \(-2\) (mod \(2^{\alpha _1-1}\)). First note that if \(k/2-1 \equiv 0\) (mod \(2^{\alpha _1-1}\)), then any even value of q satisfies the left-hand side of Eq. (4). On the other hand, if \(k/2-1 \equiv 2^{\alpha _1-2}-2\), (resp. \(k/2-1 \equiv 2^{\alpha _1-2}\), \(k/2-1 \equiv -2\)), then \(2^{\alpha _1-2}\) (resp. 2, \(2^{\alpha _1-2}\)) divides q / 2. Conversely, if \(2^{\alpha _1-2}\) (resp. 2, \(2^{\alpha _1-2}\)) divides q / 2, then the left part of Eq. (4) is satisfied whenever \(k/2-1 \equiv 2^{\alpha _1-2}-2\), (resp. \(k/2-1 \equiv 2^{\alpha _1-2}\), \(k/2-1 \equiv -2\)).

If \(\alpha _1 = 3\), then \(k/2-1\) is congruent to either 0 or 2 modulo 4. In the case when \(k/2-1 \equiv 0\) (mod 4), no restriction is imposed to q / 2, whereas if \(k/2-1 \equiv 2\) (mod 4), then q / 2 must be even. If \(\alpha _1 \in \{1,2\}\), then (4) is always satisfied and no restriction is imposed on q / 2.

From Lemma 4.10 (b), we know that \(k/2 \equiv \pm 1\) modulo \(P_i^{\alpha _i}\) for all \(i\ge 2\). Therefore \(k/2-1\) is congruent to either 0 or \(-2\) modulo \(P_i^{\alpha _i}\). Since 2 is coprime with \(P_i^{\alpha _i}\), we observe that if \(k/2-1 \equiv -2\) modulo \(P_i^{\alpha _i}\), then in order to satisfy the right-hand side of Eq. (4), \(P_i^{\alpha _i}\) must divide q / 2, and any even value of q with this property will work. Otherwise, if \(P_i^{\alpha _i}\) divides \(k/2-1\), then any even q satisfies the right-hand side of Eq. (4).

Since q / 2 is the smallest positive integer satisfying that \((k/2-1)q/2 \equiv 0\) modulo p / 2, the only factors of q / 2 are those required by the restrictions in the previous three paragraphs. In particular, if \(k/2 \equiv 1\) (mod \(2^{\alpha _1-1}\)) (resp. to \(2^{\alpha _1-2}-1\), \(2^{\alpha _1-2}+1\) or \(-1\)), then q / 2 is odd (resp. \(2^{\alpha _1-1}\), 4 or \(2^{\alpha _1-1}\) is the maximal power of 2 dividing q), implying (b). Furthermore, if \(i \ge 2\) and \(k/2 \equiv 1\) (mod \(P_i^{\alpha _i}\)), then \(P_i\) does not divide q, implying (c). Hence all factors of q are also factors of p and (d) holds. \(\square \)

We are now ready to fully characterize the tight orientably regular polyhedra with no multiple edges.

Theorem 4.13

Let \(p = P_1^{\alpha _1} P_2^{\alpha _2} \cdots P_s^{\alpha _s}\) with \(P_1 = 2\), \(P_1, \dots , P_s\) distinct primes and each \(\alpha _i\) a positive integer. For any even q with \(4 \le q < p\) satisfying (b), (c) and (d) of Proposition 4.12, there exists a tight orientably regular polyhedron with no multiple edges and type \(\{p,q\}\). The polyhedron is unique unless \(\alpha _1 \ge 4\) and \(2^{\alpha _1-1}\) divides q, in which case there are two such polyhedra. Moreover, every tight regular polyhedron with no multiple edges either has one of these types or has type \(\{2q,q\}\) for some odd q, or it corresponds to the map of type \(\{p,2\}\) on the sphere.

Proof

We already know by Proposition 4.11 that if q is odd, then \(p = 2q\) and \(k=-2\).

If \(q \ge 4\) is even, then we find \(k/2 \in \mathbb {Z}_{p/2}\) as a solution of the congruences
$$\begin{aligned} k/2 \equiv 1 \text{(mod } 2^{\alpha _1-1}\text{) }&\text{ if } q/2 \text{ is } \text{ odd } \text{( }\alpha _1 \ge 2\text{) },\\ k/2 \equiv 2^{\alpha _1-2}-1,-1 \text{(mod } 2^{\alpha _1-1}\text{) }&\text{ if } 2^{\alpha _1-2} \text{ divides } q/2 (\alpha _1 \ge 4),\\ k/2 \equiv 2^{\alpha _1-2}+1 \text{(mod } 2^{\alpha _1-1})&\text{ if } q/2 \text{ is } \text{ even } \text{ but } \text{ not } \text{ divisible } \text{ by } 4 (\alpha _1 \ge 2\text{), }\\ k/2 \equiv 1 \text{(mod } P_i^{\alpha _i}\text{) }&\text{ if } P_i \text{ is } \text{ odd } \text{ and } \text{ it } \text{ does } \text{ not } \text{ divide } q,\\ k/2 \equiv -1 \text{(mod } P_i^{\alpha _i}\text{) }&\text{ if } P_i \text{ is } \text{ odd } \text{ and } \text{ it } \text{ divides } q. \end{aligned}$$
This gives a unique solution (mod p / 2) unless \(\alpha _1 \ge 4\) and \(2^{\alpha _1-1}\) divides q, where there are two solutions. Multiplying by 2, we obtain k.

It remains to be shown that there exists a tight regular polyhedron for all such parameters p and k. Having chosen p, q and k, Lemma 4.8 describes the order of the vertices around \(F_2\), and Lemma 4.9 describes the action of \(\sigma _2\) on the vertices (and in particular, it describes the neighbors of vertex 1). We need to show that these choices actually yield a polyhedron. Arguing analogously to Lemma 4.8, it can be shown that if x and y are two consecutive neighbors of 1, then the order of the vertices in the face determined by these adjacencies is \((1,y,y-x+1,2y-x,2y-2x+1,3y-2x,\dots ,x-y+1,x)\). In other words, half of the edges go from a vertex i to \(i+y-1\), and half go from a vertex j to \(j-x+1\). With the q faces defined that way, it is easy to verify that every edge belongs to precisely two such faces, as a consequence of the fact that if x is a neighbor of 1, then so is \(2-x\) (by applying the automorphism \(\rho _1\)). It also follows that the order of the faces around neighboring vertices is the same, just reversing the orientation. This shows that these q faces suffice and that the diamond condition and strong flag connectivity hold. \(\square \)

We conclude by describing the automorphism groups of the polyhedra we have found. We determined earlier that the automorphism group of a tight orientably regular polyhedron with no multiple edges is \(\varLambda (p,q)_{i,1}\) for some choice of i. Labeling the vertices as usual (using some parameter k), we have that
$$\begin{aligned} 1 = 0\sigma _1 = k\sigma _2^{-1} \sigma _1 = k\sigma _1^i \sigma _2^j, \end{aligned}$$
which implies that \(k \sigma _1^i = 1\sigma _2^{-j} = 1\). Since also \(k \sigma _1^i = k+i\), it follows that \(i = -k+1\).

4.2 Full classification

We now return to the discussion of determining all tight orientably regular polyhedra of type \(\{p, q\}\). Corollary 4.7 implies that all such polyhedra cover tight orientably regular polyhedra with types \(\{p,q'\}\) and \(\{p',q\}\), with the property that the former and the dual of the latter have no multiple edges. On the other hand, there is only one tight orientably regular polyhedron of type \(\{p,q\}\) having such quotients.

Proposition 4.14

Let \({\mathcal {P}}\) and \({\mathcal {Q}}\) be tight orientably regular polyhedra of type \(\{p,q\}\) such that both cover a polyhedron of type \(\{p,q'\}\) with no multiple edges, and a polyhedron of type \(\{p',q\}\) whose dual has no multiple edges, for some \(q'\) dividing q and some \(p'\) dividing p. Then \({\mathcal {P}}\) and \({\mathcal {Q}}\) are isomorphic.

Proof

We know that \(\Gamma ({{\mathcal {P}}}) = \varLambda (p,q)_{i,j}\) and \(\Gamma ({{\mathcal {Q}}}) = \varLambda (p,q)_{i',j'}\) for some \(i, i', j, j'\). We need to show that \(i=i'\) and \(j=j'\).

Let \({{\mathcal {K}}}_1\) and \({{\mathcal {K}}}_2\) be the polyhedra with types \(\{p,q'\}\) and \(\{p',q\}\), respectively. Then \(\Gamma ({{\mathcal {K}}}_1) = \varLambda (p,q')_{i_1,1}\) and \(\Gamma ({{\mathcal {K}}}_2) = \varLambda (p',q)_{-1,j_2}\). Clearly \({{\mathcal {K}}}_1\) and \({{\mathcal {K}}}_2\) are quotients of \({\mathcal {P}}\) (and of \({\mathcal {Q}}\)) by \(\langle \sigma _2^{k_1}\rangle \) and by \(\langle \sigma _1^{k_2} \rangle \), respectively, for some \(k_1\) and \(k_2\). The relations \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \sigma _2^j\) and \(\sigma _2^{-1} \sigma _1 = \sigma _1^{i_1} \sigma _2\) of \(\varLambda (p,q)_{i,j}\) and \(\varLambda (p,q')_{i_1,1}\) imply that \(i_1 = i\); similarly, \(j_2 = j\). But the same is true for \(i'\) and \(j'\) and so \(\Gamma ({{\mathcal {P}}}) \equiv \Gamma ({{\mathcal {Q}}}) \equiv \varLambda (p,q)_{i_1,j_2}\). \(\square \)

Proposition 4.14 implies that, in order to determine all tight orientably regular polyhedra with type \(\{p,q\}\), we only need to determine all quotient maps satisfying the requirements of Corollary 4.7.

First, for each \(q'\) dividing q, we determine the values of i such that \(\varLambda (p,q')_{i,1}\) is the automorphism group of a tight orientably regular polyhedron of type \(\{p, q'\}\) with no multiple edges. Then, for each \(p'\) dividing p, we determine the values of j such that \(\varLambda (p', q)_{-1,j}\) is the automorphism group of a tight orientably regular polyhedron whose dual has no multiple edges. Finally, we determine which pairs of these polyhedra satisfy the conditions of Proposition 4.2 that \(p'\) divides \(i+1\) and \(q'\) divides \(j-1\).

Let us illustrate the procedure by determining all tight orientably regular polyhedra of type \(\{48, 32\}\). Proposition 4.12 implies that all possible values of \(q'\) (that yield a tight orientably regular polyhedron of type \(\{48, q'\}\) with no multiple edges) are 2, 4 or 8. Solving the congruences in the proof of Theorem 4.13, we find that \(k = -i+1\) is 2 when \(q' = 2\), is 26 when \(q'=4\), and is 14 or 38 if \(q'=8\). Since \(k = -i + 1\), this gives us that the values of i are \(-1\), 23, 35 and 11, respectively.

To find the tight orientably regular polyhedra of type \(\{p', 32\}\) such that \(\langle \sigma _1 \rangle \) is core-free, we will work with the dual polyhedron of type \(\{32, p'\}\). Then Proposition 4.12 implies that all possible values of \(p'\) are 2, 4 or 16. Solving the congruences in the proof of Theorem 4.13, we find that k is 2 when \(p' = 2\), is 18 when \(p' = 4\) and is 14 or 30 if \(p' = 16\). Therefore the values of i are \(-1\), 15, 19 and 3, respectively. To return to the dual polyhedron of type \(\{p', 32\}\), we note that the relation \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \sigma _2^j\) in \(\varGamma (\mathcal P)\) yields the relation \(\sigma _1 \sigma _2^{-1} = \sigma _2^{-i} \sigma _1^{-j}\) in \(\varGamma (\mathcal P^{\delta })\), and by Proposition 3.1, this is equivalent to the relation \(\sigma _2^{-1} \sigma _1 = \sigma _1^{-j} \sigma _2^{-i}\). So when considering the dual polyhedron with type \(\{p',32\}\), we must substitute i by \(-j\). This gives that the values of j are 1, 17, 13 and 29, respectively.

Examining all of the possible values for \(p'\), i, \(q'\) and j yields the following 10 groups whose parameters satisfy the conditions of Proposition 4.2:
$$\begin{aligned} \varLambda (48, 32)_{-1,1}&\varLambda (48, 32)_{-1,13} \\ \varLambda (48, 32)_{-1,17}&\varLambda (48, 32)_{-1,29} \\ \varLambda (48, 32)_{11,1}&\varLambda (48, 32)_{11,17} \\ \varLambda (48, 32)_{23,1}&\varLambda (48, 32)_{23,17} \\ \varLambda (48, 32)_{35,1}&\varLambda (48, 32)_{35,17} \\ \end{aligned}$$
Note that it is always possible to pick \(i = -1\) and \(j = 1\); indeed, this is the group named \(\varGamma (p, q)\) in [4], and it is the group of the polyhedron \(\{p, q \mid 2\}\) (see [8, p. 196]). It is clear that if there are many tight orientably regular polytopes of type \(\{p,q\}\), then the factorizations of p and q in primes have several factors in common. On the other hand, whenever p and q are relatively prime, the only tight orientably regular polytope of type \(\{2p,2q\}\) is \(\{2p, 2q \mid 2\}\) with group \(\varLambda (2p, 2q)_{-1, 1}\).

5 Tight non-orientably regular polyhedra

We now consider the classification of tight, non-orientably regular polyhedra.

In Theorem 3.6, we saw that every tight non-orientably regular polyhedron \(\mathcal P\) of type \(\{p, q\}\) has automorphism group \(\varGamma (\mathcal P) = \Delta (p,q)_{(i,j,a,b)}\) or its dual, where \(\Delta (p,q)_{(i,j,a,b)}\) is the quotient of [pq] by the extra relations \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \rho _1 \sigma _2^j\) and \(\sigma _2^{-2} \sigma _1 = \sigma _1^a \sigma _2^b\). It remains to determine which such groups actually appear as the automorphism group of a tight non-orientably regular polyhedron. First, we note the following:

Proposition 5.1

Let \(\mathcal P\) be a non-orientably regular polyhedron of type \(\{p, q\}\), with automorphism group \(\varGamma (\mathcal P) = \langle \sigma _1, \rho _1, \sigma _2 \rangle \). Then neither \(\langle \sigma _1^2 \rangle \) nor \(\langle \sigma _2^2 \rangle \) is normal.

Proof

Without loss of generality, let \(N = \langle \sigma _1^2 \rangle \) and suppose that N is normal in \(\varGamma (\mathcal P)\). Then by Proposition 2.2, \(\varGamma (\mathcal P) / N\) is a string C-group, and therefore, it is the automorphism group of a polyhedron \(\mathcal Q\) of type \(\{2, q\}\). Proposition 2.3 says that since \(\mathcal P\) is non-orientably regular, so is \(\mathcal Q\). But there is only a single polyhedron of type \(\{2, q\}\), and it is orientably regular. Thus, N cannot be normal after all. \(\square \)

We now work to find restrictions on the parameters (ijab). We start with several technical lemmas.

Lemma 5.2

Let \(\mathcal P\) be a tight non-orientably regular polyhedron of type \(\{p, q\}\) with \(\varGamma (\mathcal P) = \Delta (p,q)_{(i,j,a,b)}\). If q is odd, then \(b = 2\), and if q is even, then \(b = 2\) or \(2 + q/2\).

Proof

Lemma 3.5 established that \(\sigma _1 \sigma _2^{b-2} = \sigma _2^{b-2} \sigma _1\). Therefore,
$$\begin{aligned} \sigma _2^{b-2} (\sigma _2^{-1} \sigma _1) = (\sigma _2^{-1} \sigma _1) \sigma _2^{b-2}. \end{aligned}$$
On the other hand,
$$\begin{aligned} \sigma _2^{b-2} (\sigma _2^{-1} \sigma _1)&= \sigma _2^{b-2} (\sigma _1^i \rho _1 \sigma _2^j) \\&= (\sigma _1^i \rho _1 \sigma _2^j) \sigma _2^{2-b} \\&= (\sigma _2^{-1} \sigma _1) \sigma _2^{2-b}. \end{aligned}$$
It follows that \(\sigma _2^{b-2} = \sigma _2^{2-b}\), and thus \(b-2 \equiv 2-b\) (mod q). Therefore, \(2b \equiv 4\) (mod q), and the result then follows. \(\square \)

Lemma 5.3

Let \(\mathcal P\) be a tight non-orientably regular polyhedron of type \(\{p, q\}\) with \(\varGamma (\mathcal P) = \Delta (p,q)_{(i,j,a,b)}\). Then
  1. (a)

    \(a = 1 + p/2\)

     
  2. (b)

    If \(p = 4\), then \(j = 1\) or \(j = 1 + q/2\). If \(p \ne 4\), then \(j = 1 + q/2\).

     

Proof

In \(\Delta (p,q)_{(i,j,a,b)}\), the relations \(\sigma _2^{-1} \sigma _1 = \sigma _1^i \rho _1 \sigma _2^j\) and \(\sigma _2^{-2} \sigma _1 = \sigma _1^a \sigma _2^b\) both hold. Furthermore, Lemma 3.5 says that \(\sigma _2^{-1} \sigma _1^2 = \sigma _1^{i-a} \rho _1 \sigma _2^{b-j-2}\). Using these relations and the fact (from Proposition 3.2 (b)) that conjugation by \(\sigma _2\) inverts \(\sigma _1^{i-2}\), we get that
$$\begin{aligned} \sigma _1^a \sigma _2^b&= \sigma _2^{-2} \sigma _1 \\&= \sigma _2^{-1} \sigma _1^i \rho _1 \sigma _2^j \\&= \sigma _2^{-1} \sigma _1^{i-2} \sigma _1^2 \rho _1 \sigma _2^j \\&= \sigma _1^{2-i} \sigma _2^{-1} \sigma _1^2 \rho _1 \sigma _2^j \\&= \sigma _1^{2-i} \sigma _1^{i-a} \rho _1 \sigma _2^{b-j-2} \rho _1 \sigma _2^j \\&= \sigma _1^{2-a} \sigma _2^{2j-b+2}. \end{aligned}$$
Thus we see that \(\sigma _1^{2a-2} = \sigma _2^{2j-2b+2}\). Since \(\mathcal P\) is a polyhedron, \(\langle \sigma _1 \rangle \cap \langle \sigma _2 \rangle = \{ 1\}\), and it follows that \(2a - 2 \equiv 0\) (mod p) and that \((2j-2b+2) \equiv 0\) (mod q).
Now, without loss of generality, \(0 \le a \le p-1\). So, since \(2a \equiv 2\) (mod p), it follows that either \(a = 1\) or \(a = 1 + p/2\). If \(a = 1\), then we have that
$$\begin{aligned} \sigma _2^{-2} \sigma _1 = \sigma _1 \sigma _2^b, \end{aligned}$$
and \(\sigma _2^b\) is a conjugate of \(\sigma _2^{-2}\). Now, if q is odd, then \(b = 2\) (by Lemma 5.2), and it follows that \(\langle \sigma _2^2 \rangle \) is normal. On the other hand, if q is even, then \(\sigma _2^2\) has order q / 2 and so does \(\sigma _2^b\). This implies that b is even and again \(\langle \sigma _2^2 \rangle \) is normal. But by Proposition 5.1, that cannot happen. Therefore, it must be that \(a = 1 + p/2\).
Similarly, we have that \(2b \equiv 2j+2\) (mod q), and by Lemma 5.2, \(2b \equiv 4\) (mod q). Therefore, \(2j+2 \equiv 4\) (mod q) and so \(2j \equiv 2\) (mod q). Thus, either \(j = 1\) or \(j = 1 + q/2\). Now, if \(j = 1\), then Proposition 3.2 (b) says that \(N = \langle \sigma _2^3 \rangle \) is normal. In the quotient of \(\varGamma (\mathcal P)\) by N, the order of \(\sigma _1\) is still p, and we have that
$$\begin{aligned} \sigma _1^{1+p/2} \sigma _2^b&= \sigma _2^{-2} \sigma _1 \\&= \sigma _2 \sigma _1 \\&= \sigma _1^{-1} \sigma _2^{-1}. \end{aligned}$$
So \(\sigma _1^{2+p/2} = \sigma _2^{-b-1}\). Since \(\varGamma (\mathcal P)\) is a string C-group, Proposition 2.2 implies that \(\varGamma (\mathcal P) / N\) is a string C-group as well, and so again \(\langle \sigma _1 \rangle \cap \langle \sigma _2 \rangle = \{ 1\}\). So \(2 + p/2 \equiv 0\) (mod p), from which it follows that \(p = 4\). So if \(p \ne 4\), then \(j = 1 + q/2\). \(\square \)

Lemma 5.4

Let \(\mathcal P\) be a tight non-orientably regular polyhedron of type \(\{p, q\}\) with \(\varGamma (\mathcal P) = \Delta (p,q)_{(i,j,a,b)}\). Then the subgroups \(\langle \sigma _1^4 \rangle \) and \(\langle \sigma _2^6 \rangle \) are normal.

Proof

Proposition 3.2 (b) says that conjugation by \(\sigma _2\) inverts \(\sigma _1^{i-2}\), and Lemma 3.5 says that conjugation by \(\sigma _2\) inverts \(\sigma _1^{i-3-a}\). It follows that \(\sigma _2\) inverts \(\sigma _1^{a+1}\) and therefore, it also inverts \(\sigma _1^{2a+2}\). Since \(a = 1 + p/2\) (by Lemma 5.3), it follows that conjugation by \(\sigma _2\) inverts \(\sigma _1^4\) and so \(\langle \sigma _1^4 \rangle \) is normal.

For the other claim, Proposition 3.2 (b) says that \(\langle \sigma _1^{j+2} \rangle \) is normal, and so \(\langle \sigma _1^{2j+4} \rangle \) is also normal. By Lemma 5.3, \(j = 1\) or \(j = 1 + q/2\). In any case, \(2j+4 \equiv 6\) (mod q), and so, \(\langle \sigma _2^6 \rangle \) is normal. \(\square \)

Corollary 5.5

Let \(\mathcal P\) be a tight non-orientably regular polyhedron of type \(\{p, q\}\) with \(\varGamma (\mathcal P) = \Delta (p,q)_{(i,j,a,b)}\). Then p is divisible by 4 and q is divisible by 3.

Proof

Lemma 5.4 says that \(\langle \sigma _1^4 \rangle \) and \(\langle \sigma _2^6 \rangle \) are both normal, and Proposition 5.1 says that neither \(\langle \sigma _1^2 \rangle \) and \(\langle \sigma _2^2 \rangle \) is normal. It follows that \(\sigma _1^2 \notin \langle \sigma _1^4 \rangle \) and that \(\sigma _2^2 \notin \langle \sigma _2^6 \rangle \). Therefore, p is a multiple of 4 and q is a multiple of 3. \(\square \)

Lemma 5.6

Suppose that \(\mathcal P\) is a tight non-orientably regular polyhedron of type \(\{p, q\}\) with \(\varGamma (\mathcal P) = \Delta (p,q)_{(i,j,a,b)}\), If \(p/4 \equiv 3\) (mod 4), then \(i = p/4 - 1\), and otherwise \(i = 3p/4 - 1\).

Proof

Proposition 3.2 (b) says that \(\langle \sigma _1^{i-2} \rangle \) is normal, and Lemma 5.4 says that \(\langle \sigma _1^4 \rangle \) is normal. Because of Proposition 5.1, no subgroup of \(\langle \sigma _1 \rangle \) containing \(\langle \sigma _1^4 \rangle \) properly is normal, and thus \(\langle \sigma _1^{i-2} \rangle \) must be contained in \(\langle \sigma _1^4 \rangle \). It follows that \(i \equiv 2\) (mod 4).

Now, suppose we take the quotient of \(\varGamma (\mathcal P)\) by \(\langle \sigma _2^{j+2} \rangle \) and \(\langle \sigma _2^{b-2} \rangle \). This has the effect of replacing j with \(-2\) and b with 2 without changing i, a, or p. Then
$$\begin{aligned} \sigma _2 \sigma _1^4&= \sigma _1^{-1} \sigma _2^{-1} \sigma _1^3 \\&= \sigma _1^{i-1} \rho _1 \sigma _2^{-2} \sigma _1^2 \\&= \sigma _1^{i-1} \rho _1 \sigma _1^a \sigma _2^2 \sigma _1 \\&= \sigma _1^{i-1} \rho _1 \sigma _1^a \sigma _2 \sigma _1^{-1} \sigma _2^{-1} \\&= \sigma _1^{i-1} \rho _1 \sigma _1^{a-i} \rho _1 \sigma _2 \\&= \sigma _1^{2i-a-1} \sigma _2. \end{aligned}$$
Since \(a = 1 + p/2\), we have that \(\sigma _2 \sigma _1^4 \sigma _2^{-1} = \sigma _1^{2i-2-p/2}\). On the other hand, Lemma 5.4 says that \(\sigma _2 \sigma _1^4 \sigma _2^{-1} = \sigma _1^{-4}\), so it follows that \(2i - 2 - p/2 \equiv -4\) (mod p). Therefore, \(2i \equiv p/2 - 2\) (mod p), and thus \(i \equiv p/4 - 1\) (mod p / 2). Thus we see that \(i = p/4 - 1\) or \(i = 3p/4 - 1\). In order for \(i \equiv 2\) (mod 4), we need to pick \(i = p/4 - 1\) if \(p/4 \equiv 3\) (mod 4), and otherwise, we need to pick \(i = 3p/4 - 1\). \(\square \)

Lemma 5.7

Suppose that \(\mathcal P\) is a tight non-orientably regular polyhedron of type \(\{p, q\}\) with \(\varGamma (\mathcal P) = \Delta (p,q)_{(i,j,a,b)}\), Then \(b = 2\).

Proof

First, suppose that \(p = 4\). Then \(a = 3\) by Lemma 5.3 and \(i = 2\) by Lemma 5.6. Therefore,
$$\begin{aligned} \sigma _1^3 \sigma _2^b&= \sigma _2^{-2} \sigma _1 \\&= \sigma _2^{-1} \sigma _1^2 \rho _1 \sigma _2^j \\&= \sigma _2^{-1} \sigma _1^{-2} \rho _1 \sigma _2^j \\&= \sigma _1 \sigma _2 \sigma _1^{-1} \rho _1 \sigma _2^j \\&= \sigma _1 \sigma _1^2 \rho _1 \sigma _2^{-j} \rho _1 \sigma _2^j \\&= \sigma _1^3 \sigma _2^{2j}. \end{aligned}$$
It follows that \(b \equiv 2j\) (mod q). Since \(j = 1\) or \(j = 1 + q/2\), we must have that \(b \equiv 2\) (mod q), and since we can take \(0 \le b \le q-1\), it follows that \(b = 2\).

If \(p \ne 4\), we nevertheless have by Lemma 5.4 that \(\langle \sigma _1^4 \rangle \) is a normal subgroup. Taking the quotient by this subgroup forces \(p = 4\) without changing b, and so since \(b = 2\) in the quotient, it follows that \(b = 2\) in the original group. \(\square \)

Theorem 5.8

Let \(\mathcal P\) be a tight non-orientably regular polyhedron of type \(\{p, q\}\) such that \(\varGamma (\mathcal P) = \Delta (p,q)_{(i,j,a,b)}\). Then
  1. (a)

    p is an odd multiple of 4.

     
  2. (b)

    q is a multiple of 3. Furthermore, if \(p \ne 4\), then q is an odd multiple of 6.

     
  3. (c)

    If \(p/4 \equiv 3\) (mod 4), then \(i = p/4-1\), and otherwise \(i = 3p/4-1\).

     
  4. (d)

    If \(p \ne 4\), then \(j = 1 + q/2\), and if \(p = 4\) then either \(j = 1\) or \(j = 1 + q/2\).

     
  5. (e)

    \(a = 1 + p/2\).

     
  6. (f)

    \(b = 2\).

     

Proof

Parts (c) through (f) were proved in Lemmas 5.3 to 5.7. It remains to prove parts (a) and (b).

Corollary 5.5 tells us that p is a multiple of 4 and that q is a multiple of 3. Further, note that if q is odd, then \(j = 1\) (since \(j = 1 + q/2\) is impossible in this case). Then Lemma 5.3 tells us that \(p = 4\). So if \(p \ne 4\), it must be that q is a multiple of 6.

Now, suppose that 8 divides p. Then since \(\langle \sigma _1^4 \rangle \) is normal, so is \(\langle \sigma _1^8 \rangle \). Taking the quotient by this normal subgroup then yields a tight non-orientably regular polyhedron of type \(\{8, q\}\). In this quotient, \(\langle \sigma _2^6 \rangle \) is normal, and the quotient by this group yields a tight non-orientably regular polyhedron of type \(\{8, 6\}\) or \(\{8, 3\}\). But no such polyhedron exists (which we confirm by checking the Atlas of Small Regular Polytopes [7]). Therefore, 8 cannot divide p, and so p is an odd multiple of 4.

We have already established that if \(p \ne 4\), then q cannot be odd, and so it must be a multiple of 6. Suppose that q is a multiple of 12. Then since \(\langle \sigma _2^6 \rangle \) is normal (by Lemma 5.4), so is \(\langle \sigma _2^{12} \rangle \), and the quotient by this normal subgroup yields a tight non-orientably regular polyhedron of type \(\{p, 12\}\). Since \(p \ne 4\), Lemma 5.3 says that \(j = 7\). Now, \(\langle \sigma _2^6 \rangle \) is a normal subgroup of our quotient by \(\langle \sigma _2^{12} \rangle \), and in passing to the quotient by \(\langle \sigma _2^6 \rangle \), we may replace j with 1. In that case, Lemma 5.3 says that \(p = 4\) after all. Since \(p \ne 4\), it follows that q is not a multiple of 12. \(\square \)

Thus, with the exception of the case where \(p = 4\), there is only a single choice of parameters that (might) work, and in the case \(p = 4\), there are 2 choices. It remains to show that there really are tight non-orientably regular polyhedra of these types \(\{p, q\}\).

Lemma 5.9

Let r and k be odd, let \(p = 4r\) and let \(q = 6k\). Let i, j, a, and b satisfy the conditions of Theorem 5.8. Then \(\Delta (p,q)_{(i,j,a,b)}\) is a string C-group of type \(\{p, q\}\).

Proof

Since p is a multiple of 4 and q is a multiple of 6, the group \(\Delta (p,q)_{(i,j,a,b)}\) covers \(\Delta (4,6)_{(i,j,a,b)}\). In the latter, we may reduce i and a modulo 4, and we may reduce j and b modulo 6. The parameter i was chosen (in Lemma 5.6) such that \(i \equiv 2\) (mod 4), and since \(a = 1 + p/2 = 1 + 2r\) for some odd integer r, it follows that \(a \equiv 3\) (mod 4). The parameter j satisfies \(j \equiv 1\) (mod q / 2), and thus \(j \equiv 1\) (mod 3k), which implies that \(j \equiv 1\) (mod 3). Therefore, \(j \equiv 1\) or 4 (mod 6). Finally, \(b = 2\). It follows that \(\Delta (4,6)_{(i,j,a,b)} = \Delta (4,6)_{(2,1,3,2)}\) or \(\Delta (4,6)_{(2,4,3,2)}\). Using GAP [10], we can verify that these latter two groups are the automorphism groups of (non-isomorphic) tight polyhedra of type \(\{4, 6\}\), (\(\{4, 6\}*48b\) and \(\{4,6\}*48c\) in [7]), so in \(\Delta (p,q)_{(i,j,a,b)}\), the order of \(\sigma _1\) is divisible by 4 and the order of \(\sigma _2\) is divisible by 6.

Now, let \(G = \langle x, y \mid x^2 = y^2 = (xy)^r = 1 \rangle \). Then a small calculation shows that the function \(\varphi : \Delta (p,q)_{(i,j,a,b)} \rightarrow G\) that sends \(\rho _0\) to x, \(\rho _1\) to y, and \(\rho _2\) to 1 is a surjective group homomorphism. From this, it follows that the order of \(\sigma _1\) is divisible by r. Since the order of \(\sigma _1\) is also divisible by 4 and is a divisor of 4r, the order must be exactly 4r.

Similarly, let \(H = \langle y, z \mid y^2 = z^2 = (yz)^{3k} = 1 \rangle \). Then the function \(\varphi : \Delta (p,q)_{(i,j,a,b)} \rightarrow H\) sending \(\rho _0\) to 1, \(\rho _1\) to y, and \(\rho _2\) to z is a surjective group homomorphism. Thus, the order of \(\sigma _2\) is divisible by 3k, and since it is also divisible by 6 and a divisor of 6k, the order must be 6k.

We have established that \(\Delta (p,q)_{(i,j,a,b)}\) has type \(\{p, q\}\). To see that it is a string C-group, we note that \(\Delta (p,q)_{(i,j,a,b)}\) covers \(\Delta (4,q)_{(i,j,a,b)}\), which in turn covers \(\Delta (4,6)_{(i,j,a,b)}\). Since \(\Delta (4,6)_{(i,j,a,b)}\) is a string C-group, two applications of Proposition 2.1 show that so is \(\Delta (p,q)_{(i,j,a,b)}\). \(\square \)

It remains to show that there is a tight non-orientably regular polyhedron of type \(\{4, 3k\}\) whenever 3k is odd. This follows directly from [2, Thm. 5.1]. Combined with Theorem 5.8 and Lemma 5.9, we obtain the following result:

Theorem 5.10

There is a tight non-orientably regular polyhedron of type \(\{p, q\}\) if and only if
  1. (a)

    \(p = 4\) and \(q = 3k\), or

     
  2. (b)

    \(p = 4r\) and \(q = 6k\), with \(r > 1\) odd and k odd, or

     
  3. (c)

    \(q = 4\) and \(p = 3k\), or

     
  4. (d)

    \(q = 4r\) and \(p = 6k\), with \(r > 1\) odd and k odd.

     
Furthermore, in each case there is a unique such polyhedron up to isomorphism, except in the cases where \(p = 4\) and \(q = 6k\), or \(q = 4\) and \(p = 6k\), in which case there are two isomorphism types.

Now as in the case of tight orientably regular polyhedra, we consider the core of \(\langle \sigma _i \rangle \) in the automorphism groups of tight non-orientably regular polyhedra. Whenever \(\langle \sigma _1 \rangle \) and \(\langle \sigma _2 \rangle \) are not core-free in \(\Delta (p,q)_{i,j,a,b}\), we can take the quotients of \(\Delta \) by the two cores to obtain two tight non-orientably regular polyhedra. As in the orientable case, we could reconstruct \(\Delta (p,q)_{i,j,a,b}\) from these two quotients. The difficulty is that, unlike in the orientable case, some polyhedra with \(\langle \sigma _2 \rangle \) core-free might have multiple edges.

Let us show how this can happen. Corollary 4.4 showed that if \(\mathcal P\) is orientably regular and if some \(\varphi \in \langle \sigma _2 \rangle \) fixes one neighbor of v, then \(\varphi \) fixes every vertex. However, this may not be true in the non-orientable case. In particular, consider a polyhedron with double edges and with \(\varphi = \sigma _2^{q/2}\). Then it can happen that \(\varphi \) acts as a rotation at v but as a reflection through one of its neighbors u. This is illustrated in Fig. 2, where the two flags labeled \(\Psi \) are identified, and \(\varphi \) maps flag \(\varPhi \) into flag \(\Psi \) by a half-turn around v, but also by a reflection by a vertical line through u. This gives us a polyhedron with double edges, even though \(\langle \sigma _2 \rangle \) is core-free.
Fig. 2

Polyhedron with double edges but \(\langle \sigma _2 \rangle \) core-free

Examples of this situation are the duals of the polyhedra with automorphism groups \(\Delta (4,q)_{2,1,3,2}\) for any q divisible by 3. Figure 3 shows these polyhedra (represented as maps) for \(q \in \{3, 6, 9\}\); see [6] for an earlier appearance of these maps (see for example [6] for earlier appearance of these polyhedra as maps on surfaces). It is easy to see that in the polyhedron with group \(\Delta (4,q)_{2,1,3,2}\), each square face shares opposite edges with another square, as shown in the upper part of Fig. 3. This implies that the dual has double edges. On the other hand, it is easy to verify that \(\langle \sigma _1 \rangle \) is core-free in \(\Delta (4,q)_{2,1,3,2}\).
Fig. 3

Polyhedra with automorphism groups \(\Delta (4,q)_{2,1,3,2}\) for \(q \in \{3,6,9\}\)

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Copyright information

© Springer Science+Business Media New York 2015

Authors and Affiliations

  1. 1.University of Massachusetts BostonBostonUSA
  2. 2.National University of MexicoMoreliaMexico
  3. 3.Centro de Ciencias MatemáticasUniversidad Nacional Autónoma de MéxicoMoreliaMexico

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