# Classification of tight regular polyhedra

## Abstract

A regular polyhedron of type \(\{p, q\}\) has at least 2*pq* flags, and it is called *tight* if it has exactly 2*pq* 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 2*pq* 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**

*p*-gonal faces and

*q*-valent vertices) if and only if one of the following is true:

- (a)
*p*and*q*are both even. - (b)
*p*is odd and*q*is an even divisor of 2*p*. - (c)
*q*is odd and*p*is an even divisor of 2*q*. - (d)
\(p = 4\) and

*q*is an odd multiple of 3. - (e)
\(q = 4\) and

*p*is an odd multiple of 3.

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

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

- (2)
Each edge is incident on exactly two vertices and two faces.

- (3)
The graph determined by the vertex and edge sets is connected.

- (4)
The vertex-figure at every vertex is isomorphic to the vertex and edge lattice of a connected 2-regular graph.

*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.

*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 2

*p*, and \(\langle \rho _1, \rho _2 \rangle \) is dihedral of order 2

*q*. 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 [

*p*,

*q*], with presentation

*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*:

*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 \(\varGamma (\mathcal Q)\). \(\square \)

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 2*pq* 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.

*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 2*pq* flags, and thus \(\mathcal Q\) has at most 2*pq* flags. On the other hand, \(\mathcal Q\) itself has Schläfli symbol \(\{p, q\}\), and so it has at least 2*pq* 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**

- (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 \(\varGamma \).

- (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*

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 [*p*, *q*] 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*

*p*,

*q*] 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

*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 [*p*, *q*]. 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*

*i*and

*j*. Then, working in \(\varGamma (\mathcal P^{\delta })\) and using Proposition 3.2 (b), we get that:

**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*

**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 [*p*, *q*] 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 [*q*, *p*] by those relations.

*Proof*

*p*,

*q*] 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 (

*i*,

*j*,

*a*,

*b*). 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

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**

- (a)
*p*and*q*are both even, or - (b)
*p*is odd and*q*is an even divisor of 2*p*, or - (c)
*q*is odd and*p*is an even divisor of 2*q*.

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 [*p*, *q*] 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**

- (a)
\(p'\) divides

*p*and \(i+1\), - (b)
\(q'\) divides

*q*and \(j-1\), - (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

- (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*

*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,

*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**

- (a)
\(p'\) divides

*p*and \(i+1\), - (b)
\(q'\) divides

*q*and \(j-1\), - (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

- (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.

*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 \(0 \sigma _2 = k\)).

**Lemma 4.8**

- (a)
\(p > q\).

- (b)
*p*is even. - (c)
Every vertex of the dual of \({\mathcal {Q}}\) has exactly two neighbors.

- (d)
*k*is even. - (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 2*k*, 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 2*i* and \(2i+1\) but not the edge from 2*i* 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**

- (a)$$\begin{aligned} i\sigma _2 = \left\{ \begin{array}{ll} \frac{k(2-i)}{2}&{} \text{ if } i \text{ is } \text{ even }\\ 1+\frac{k(1-i)}{2}&{} \text{ if } i \text{ is } \text{ odd }. \end{array}\right. \end{aligned}$$
- (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.

*s*. Now, since \((\sigma _2 \sigma _1)^2 = 1\),

*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

*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**

*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

- (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 \(n\ge 3\).

- (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 \(i\ge 2\).

*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 \(m \ge 2\),

*q*is the smallest positive

*m*satisfying that

*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 2*q*. 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 2*q*, then \(p \le q\) and \({\mathcal {Q}}\) must have multiple edges.

**Proposition 4.12**

*i*. Then

- (a)
\(P_1 = 2\);

- (b)
the maximal power of 2 that divides

*q*is either 2, 4 or \(2^{\alpha _1-1}\), and it is 4 only if \(\alpha _1 \ge 3\); - (c)
for every \(i \ge 2\), either

*q*is coprime with \(P_i\) or \(P_i^{\alpha _i}\) divides*q*; and - (d)
*q*divides*p*.

*Proof*

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

*p*). This implies that \((k/2-1)q/2\equiv 0\) modulo

*p*/ 2. This is equivalent to

*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 \(k=-2\).

*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. \(\square \)

*i*. Labeling the vertices as usual (using some parameter

*k*), we have that

### 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.

*i*, \(q'\) and

*j*yields the following 10 groups whose parameters satisfy the conditions of Proposition 4.2:

*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 [*p*, *q*] 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 (*i*, *j*, *a*, *b*). 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*

*q*). Therefore, \(2b \equiv 4\) (mod

*q*), and the result then follows. \(\square \)

**Lemma 5.3**

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

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

*Proof*

*p*) and that \((2j-2b+2) \equiv 0\) (mod

*q*).

*p*), it follows that either \(a = 1\) or \(a = 1 + p/2\). If \(a = 1\), then we have that

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

*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

*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).

*j*with \(-2\) and

*b*with 2 without changing

*i*,

*a*, or

*p*. Then

*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*

*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**

- (a)
*p*is an odd multiple of 4. - (b)
*q*is a multiple of 3. Furthermore, if \(p \ne 4\), then*q*is an odd multiple of 6. - (c)
If \(p/4 \equiv 3\) (mod 4), then \(i = p/4-1\), and otherwise \(i = 3p/4-1\).

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

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

- (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 3*k*), 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 4*r*, the order must be exactly 4*r*.

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 3*k*, and since it is also divisible by 6 and a divisor of 6*k*, the order must be 6*k*.

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 3*k* 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**

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

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

*k*odd, or - (c)
\(q = 4\) and \(p = 3k\), or

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

*k*odd.

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.

*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.

*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 \(\Delta (4,q)_{2,1,3,2}\).

### References

- 1.Conder, M.: The smallest regular polytopes of any given rank. Adv. Math.
**236**, 92–110 (2013)MathSciNetCrossRefMATHGoogle Scholar - 2.Conder, M., Cunningham, G.: Tight orientably-regular polytopes. Ars Math. Contemp.
**8**, 68–81 (2015)MathSciNetMATHGoogle Scholar - 3.Coxeter, H.S.M., Moser, W.O.J.: Generators and relations for discrete groups. In: Ergebnisse der Mathematik und ihrer Grenzgebiete (Results in Mathematics and Related Areas) (4th ed.), vol. 14, Springer, Berlin (1980)Google Scholar
- 4.Cunningham, G.: Minimal equivelar polytopes. Ars Math. Contemp.
**7**(2), 299–315 (2014)MathSciNetMATHGoogle Scholar - 5.Finch, S., Sebah, P.: Squares and cubes modulo \(n\). http://arxiv.org/abs/math/0604465
- 6.Grek, A.S.: Regular polyhedra of simplest hyperbolic types (Russian). Ivanov. Gos. Ped. Inst. Ucen. Zap.
**34**, 27–30 (1963)MathSciNetGoogle Scholar - 7.Hartley, M.I.: An atlas of small regular abstract polytopes. Period. Math. Hung.
**53**, 149–156 (2006). doi:10.1007/s10998-006-0028-x MathSciNetCrossRefMATHGoogle Scholar - 8.McMullen, P., Schulte, E.: Abstract Regular Polytopes, Encyclopedia of Mathematics and its Applications, vol. 92. Cambridge University Press, Cambridge (2002)CrossRefMATHGoogle Scholar
- 9.Schulte, E., Weiss, A.I.: Chiral polytopes. applied geometry and discrete mathematics. In: DIMACS Ser. Discrete Math. Theoret. Comput. Sci., vol. 4, pp. 493–516. Amer. Math. Soc., Providence (1991)Google Scholar
- 10.The GAP Group: GAP – Groups, Algorithms, and Programming, Version 4.4.12 (2008)Google Scholar