On infinite 2-orbit polyhedra in classes $2_0$ and $2_2$

Abstract

We describe all 2-orbit skeletal polyhedra in classes $2_0$ and $2_2$ in $\mathbb {E}^3$. Those in class $2_2$ are continuous deformations of the three Petrie–Coxeter polyhedra and therefore are arranged in three families. Those in class $2_0$ are the Petrie duals of the chiral polyhedra in $\mathbb {E}^3$ and are classified in six families.

1 Introduction

Symmetry of polyhedra has delighted mathematicians since antiquity. This led to the discovery of the 5 Platonic solids and the 13 Archimedean solids many centuries ago. The requirement of convexity was implicit in their study and enumeration, and that resulted in relatively short lists of polyhedra with high degrees of symmetry.

Much later Kepler realized that many of the main properties of the Platonic solids held also for two star-shaped solids (with the right interpretation of edges and facets), and he regarded them as regular polyhedra. Kepler’s solids were the beginning of an abstraction of the properties that lead to interesting symmetric discrete objects with close relationship with the Platonic solids. This was continued by Poinsot, Petrie and Coxeter with the introduction of new regular polyhedra under broader definitions that still preserve most of the main features of the Platonic solids.

By comparison, there were fewer works about highly symmetric non-regular polyhedra (in some way the natural heirs of the Archimedean solids, prisms and pyramids). An example of these is the classification of uniform star polyhedra by Coxeter et al. (1954).

In the second half of the twentieth century, Grünbaum had the idea of removing membranes so that the facets become 1-dimensional objects: cycles or infinite paths. This allows them to be non-planar and even to be infinite (see Grünbaum 1977). A polyhedron then becomes a skeleton (embedding of a graph) equipped with distinguished subgraphs that play the role of facets. Grünbaum himself listed 47 regular polyhedra known by him at the time. A few years later Dress added one more and showed that the list of 48 regular polyhedra (under this skeletal approach) is complete (see Dress 1981, 1985).

Since then, much attention has been devoted to regular polyhedra and polytopes (generalizations to higher ranks of polygons and polyhedra) in various kinds of spaces (see for example Burgiel and Stanton 2000; Monson and Weiss 2000; Arocha et al. 2000; McMullen 2004).

The first works on skeletal non-regular polyhedra include two Ph.D. theses (Farris 1985; Eisenlohr 1990), both pursuing the classification of all fully-transitive polyhedra (transitive on vertices, edges and facets). The first thesis includes an enumeration of the finite fully-transitive polyhedra, and the second an enumeration of the infinite fully-transitive polyhedra that are planar. It became clear that the classification of all fully-transitive polyhedra is an ambitious project.

Fully-transitive non-regular polyhedra can be divided into 6 classes, depending on their symmetry types (see Sect. 2 for definitions). Four of those classes contain 2-orbit polyhedra, and the remaining two contain 4-orbit polyhedra. One of the classes of 2-orbit vertex-transitive polyhedra consists of the chiral ones; those which show all combinatorial rotations but no combinatorial reflections. The chiral polyhedra in $\mathbb {E}^3$ were enumerated by Schulte (2004, 2005). This was the first completely enumerated class of non-regular, fully-transitive polyhedra. The enumeration of all finite 2-orbit polyhedra was announced and will be carried out in Hubard and Miguel García-Velázquez (in preparation), where in particular all finite 2-orbit fully transitive polyhedra will appear.

Other works on non-regular skeletal polyhedra in Euclidean space include a list of vertex-transitive polyhedra obtained from Wythoffian operations on regular polyhedra (Schulte and Williams 2016), and the classification of finite 3-orbit polyhedra (Cunningham and Pellicer (in manuscript)).

In this paper we continue the analysis of 2-orbit vertex-transitive polyhedra. We enumerate the infinite ones in the classes $2_2$ and $2_0$.

The paper is organized as follows. We start in Sect. 2 with background, continuing in Sect. 3 with preliminary results on 2-orbit polyhedra. In Sect. 4 we present the first family of 2-orbit polyhedra in class $2_2$. The general analysis of the generators of the symmetry groups of polyhedra in class $2_2$ is carried out in Sect. 5; in that section most cases are ruled out, leaving only two possibilities. One of these two possibilities is discussed in detail in Sect. 6, where two more families of polyhedra in class $2_2$ are mentioned. The remaining three families of polyhedra arise in Sect. 7 from the second possibility of generators of symmetry groups. This accounts for the six families of polyhedra in class $2_2$. There are also six families of polyhedra in class $2_0$ and they are described in Sect. 8 through the application of Petrie duality to the chiral polyhedra. Finally, we present conclusions and open problems for this work in Sect. 9.

2 Background and definitions

While our approach in this paper derives much of its inspiration and techniques from the treatment in Schulte (2004), we instead follow Grünbaum (1977) for our basic definitions, though we will occasionally steal from Schulte (2004) and Hubard (2010).

We define a finite polygon $f:=[v_{1},v_{2},\ldots , v_{n}]$ in ${\mathbb {E}}^{3}$ to be the figure formed of the distinct points (vertices) $v_{i}$ together with the line segments (edges) formed by pairs of subsequent vertices $e_{i}:=[v_{i},v_{i+1}]$ for $i\in \{1,2,\ldots , n-1\}$ and $e_{n}:=[v_{n},v_{1}]$. Any two vertices must correspond to distinct points, and any two edges to distinct line segments, and therefore every polygon has at least 3 vertices and 3 edges. An infinite polygon is similarly determined by an infinite sequence of distinct points (vertices) $f:=[\ldots , v_{-1},v_{0},v_{1},v_{2},\ldots ]$ and the line segments (edges) of the form $e_{i}:=[v_{i},v_{i+1}]$ for $i\in {\mathbb {Z}}$, with the additional requirement that every compact subset of ${\mathbb {E}}^{n}$ intersects at most finitely many vertices and edges of f. Each edge is incident with each of the two vertices that define it; a polygon f is said to be incident to any vertex or edge used in its definition. A flag of a polygon is any incident pair (v, e) where v is a vertex, and e is an edge of the polygon f. A polyhedron $\mathcal {P}$ is any set of polygons (called the facets of $\mathcal {P}$) such that

• (P1) Each edge belongs to exactly two facets of $\mathcal {P}$.

• (P2) The set of polygons is connected; i.e., given any two edges e and $e'$ of $\mathcal {P}$ there exists a chain $e=e_{0},f_{1},e_{1},f_{2},e_{2},\ldots , f_{n},e_{n}=e'$ of edges and facets of $\mathcal {P}$ where each $f_{i}$ is incident with $e_{i-1}$ and $e_{i}$.

• (P3) Each compact subset of ${\mathbb {E}}^{3}$ meets only finitely many edges.

• (P4) Each vertex-figure (see below) forms a single polygon.

The vertex-figure $\mathcal {P}/v$ of $\mathcal {P}$ corresponding to a vertex $v\in \mathcal {P}$ consists of the vertices $v_{i}$ of $\mathcal {P}$ that share an edge of $\mathcal {P}$ with v, and the segments $[v_{i},v_{j}]$ when $[v,v_{i}]$ and $[v,v_{j}]$ are edges of $\mathcal {P}$ incident with one of the facets of $\mathcal {P}$. Due to the local finiteness requirement in the definition of a polyhedron, vertex-figures consist of only a finite number of vertices/edges. In any case each vertex of $\mathcal {P}/v$ is incident to 2 edges since each edge $[v,v_{i}]$ is incident with two facets of $\mathcal {P}$. In this treatment we will assume that each vertex-figure is a single polygon.¹ The faces of $\mathcal {P}$ are its vertices, edges, and facets.

The rank of a vertex is 0, an edge is 1, a polygon 2, and a polyhedron 3.

A flag of a polyhedron $\mathcal {P}$ is any triple $\Phi :=(v,e,f)$ consisting of a vertex v, edge e, and facet f of $\mathcal {P}$ that are mutually incident. We denote by $\mathcal {F}(\mathcal {P})$ the set of all flags of $\mathcal {P}$. Two flags are said to be adjacent if they differ by exactly one element (e.g., the flags formed from the same edge and facet, but the vertices at the two ends of the edge determine two adjacent flags); if $\Phi$ and $\Psi$ are adjacent flags that differ by a rank i element, then we write $\Psi =\Phi ^{i}$. The flags adjacent to a flag $\Phi$ are the neighbors of $\Phi$. We may associate with any polyhedron its connection group ${\textsf{Con}}(\mathcal {P})$ which acts on $\mathcal {F}(\mathcal {P})$ as a set of permutations on the left; the generators of this group are the maps $r_{i}:\mathcal {F}(\mathcal {P})\rightarrow \mathcal {F}(\mathcal {P})$ where $r_{i}(\Phi )=\Phi ^{i}$. Each $r_{i}$ is an involutory bijection on $\mathcal {F}(\mathcal {P})$, since each edge has two vertices, each vertex sits on exactly two edges of a facet of $\mathcal {P}$, and each edge belongs to exactly two facets of $\mathcal {P}$. Likewise, $r_{0}r_{2}=r_{2}r_{0}$.

The symmetry or isometry group ${\textsf{Iso}}(S)$ of a geometric figure S is the set of isometries of ${\mathbb {E}}^{3}$ that map the figure to itself. Elements of ${\textsf{Iso}}(S)$ will be denoted by capital latin letters. When considering the action of the symmetry group on polyhedra, we will take the action on the right. The actions of ${\textsf{Iso}}(\mathcal {P})$ and ${\textsf{Con}}(\mathcal {P})$ commute with each other, so one may evaluate the actions on flags in either order. A geometric polyhedron $\mathcal {P}$ is regular if ${\textsf{Iso}}(\mathcal {P})$ acts transitively on the flags of $\mathcal {P}$. A polyhedron $\mathcal {P}$ is a 2-orbit polyhedron if the action of ${\textsf{Iso}}(\mathcal {P})$ partitions the flags of $\mathcal {P}$ into precisely two non-empty orbits. A 2-orbit polyhedron is said to be chiral if it is not regular, with adjacent flags belonging to distinct orbits. The enumeration of chiral polyhedra in ${\mathbb {E}}^{3}$ was carried out by Schulte (2004, 2005). The automorphism group ${\textsf{Aut}}(\mathcal {P})$ is the set of combinatorial automorphisms (rank and adjacency preserving bijections) of $\mathcal {P}$ treated as a partially ordered set on the vertices, edges, and facets of $\mathcal {P}$. Elements of ${\textsf{Aut}}(\mathcal {P})$ also act on the right, and are denoted with lower case Greek letters and the actions of ${\textsf{Aut}}(\mathcal {P})$ and ${\textsf{Con}}(\mathcal {P})$ also commute, though the actions of ${\textsf{Aut}}(\mathcal {P})$ and ${\textsf{Iso}}(\mathcal {P})$ do not. There is a natural map from ${\textsf{Iso}}(\mathcal {P})$ to ${\textsf{Aut}}(\mathcal {P})$ via the action on the faces, but the groups need not be isomorphic (as the rectangle readily proves).

Whenever the affine span of the vertex set of $\mathcal {P}$ is ${\mathbb {R}}^{3}$, the only element of ${\textsf{Iso}}(\mathcal {P})$ fixing a flag is the identity. As we shall see, all polyhedra in classes $2_0$ and $2_2$ satisfy this property, and the following result follows easily for both ${\textsf{Iso}}(\mathcal {P})$ and ${\textsf{Aut}}(\mathcal {P})$.

Lemma 2.1

Let $\mathcal {P}$ be a non-planar 2-orbit polyhedron, let G be either ${\textsf{Iso}}(\mathcal {P})$ or ${\textsf{Aut}}(\mathcal {P})$ and let $\Phi \in \mathcal {F}(\mathcal {P})$. If $\Phi ^{i}$ is in the same orbit under G as $\Phi$ for some $i\in \{0,1,2\}$, then $\Psi$ and $\Psi ^{i}$ are in the same orbit for any $\Psi \in \mathcal {F}(\mathcal {P})$.

Let $\mathcal {P}$ be a 2-orbit polyhedron and let $I\subsetneq \{0,1,2\}$. Then we say $\mathcal {P}$ is in class $2_{I}$ if for every $i\in I$, each flag $\Phi$ and $\Phi ^{i}$ are in the same orbit, but $\Phi$ and $\Phi ^{j}$ are not for each $j\not \in I$.² Thus, the class $2_{\emptyset }$ (elsewhere denoted just by 2) is precisely the set of chiral polyhedra. For compactness of notation, we denote by $2_{i}$ the class $2_{\{i\}}$, and $2_{ij}$ the class determined by $I=\{i,j\}$ with $i<j$. A polyhedron is geometrically (combinatorially) fully transitive if ${\textsf{Iso}}(\mathcal {P})$ (${\textsf{Aut}}(\mathcal {P})$, resp.) acts transitively on the vertices, edges, and facets of $\mathcal {P}$ (regular polyhedra are always fully transitive). Hubard showed in Hubard (2010, Th. 5) that the only combinatorially fully transitive 2-orbit polyhedra are in classes $2_{\emptyset }, 2_{0}, 2_{1}$ and $2_{2}$, and so these are the only possibilities available for geometrically fully transitive polyhedra as well.

We will denote by $R_{i}$ the isometry taking the base flag $\Phi$ of $\mathcal {P}$ to its i-adjacent neighbor $\Phi ^{i}$ when $i\in I$; similarly $\rho _{i}$ denotes the corresponding automorphism. If we were working with regular polyhedra, these would be the only symmetries we would need to specify the generators of ${\textsf{Iso}}(\mathcal {P})$, but in the current work $\mathcal {P}$ is a 2-orbit polyhedron and so we will need to specify other generating symmetries of ${\textsf{Iso}}(\mathcal {P} )$. We will denote by $A_{ij}$ or $A_{ijk}$ the symmetries defined by $\Phi A_{ij}=\Phi ^{ij}=(\Phi ^{i})^{j}$ and $\Phi A_{ijk}=(\Phi ^{ij})^{k}$. We let $G_{I}=\{ R_{i}|i \in I\} \cup \{ A_{jk}|j,k\not \in I, j<k \} \cup \{ A_{jij}|i \in I, j\not \in I\text { and }|j-i|=1 \}$. We denote by $\mathcal {G}_{I}$ the corresponding set of generators of ${\textsf{Aut}}(\mathcal {P})$. In case the affine span of $\mathcal {P}$ is not $\mathbb {E}^3$ and we have multiple isometries mapping $\Phi$ to $\Phi ^i$, to $\Phi ^{ij}$ or to $\Phi ^{ijk}$ we may choose any such isometry as $R_i$, $A_{ij}$ or $A_{ijk}$, respectively; such a choice has no impact in the discussions of this paper.

In Hubard (2010) it was proved that ${\textsf{Aut}}(\mathcal {P})=\langle \mathcal {G}_{I}\rangle$ for any 2-orbit polyhedron in $2_{I}$. For some classes the the set $\mathcal {G}_{I}$ is not a minimal set of generators for polyhedra in class $2_{I}$. We note that if $\mathcal {P}$ is of type $2_{0}$ then a minimal generating set for ${\textsf{Aut}}(\mathcal {P})$ is $\{\rho _{0},\alpha _{12}\}$, and by duality, if $\mathcal {P}$ is of type $2_{2}$ then a minimal generating set is $\{\rho _{2},\alpha _{01}\}$. The following observation is easily shown.

Observation 2.2

Let $\mathcal {G}$ be a generating set for ${\textsf{Aut}}(\mathcal {P})$. Suppose there exists a set $G\subseteq {\textsf{Iso}}(\mathcal {P})$ such that for each $\gamma \in \mathcal {G}$, there is a $g\in G$ such that the action of g and $\gamma$ on the faces of $\mathcal {P}$ agree. Then for any $\tau \in {\textsf{Aut}}(\mathcal {P})$ there exists a corresponding isometry $T\in {\textsf{Iso}}(\mathcal {P})$ that has the same action on the faces of $\mathcal {P}$.

An immediate consequence of Observation 2.2 is that if we can find isometries of $\mathcal {P}$ corresponding to a minimal generating set for ${\textsf{Aut}}(\mathcal {P})$, then we have found a minimal generating set for ${\textsf{Iso}}(\mathcal {P})$.

A Petrie path $\pi$ of a polyhedron $\mathcal {P}$ is a sequence of edges $\ldots , e_{0}, e_{1}, \ldots$ in which for every i the edges $e_{i}$ and $e_{i+1}$ belong to the same facet, but $e_{i}$ and $e_{i+2}$ do not. Note that there is no requirement that each edge appear only once in the list, however, we will be very interested in the case when each Petrie path $\pi$ of $\mathcal {P}$ is a polygon, and in particular, when the set of vertices, edges, and Petrie paths of a polyhedron form a polyhedron; this is called the Petrial or Petrie dual of $\mathcal {P}$ and is denoted $\mathcal {P}^{\pi }$ (see Lemma 3.8).

Let $\mathcal {T}$ be the tiling of ${\mathbb {E}}^{3}$ by unit cubes with vertices at the integer coordinates. We will denote by $k\mathcal {T}$ any tiling of ${\mathbb {E}}^{3}$ by cubes with edges of length k.

3 Basic facts about 2-orbit polyhedra

In this section we will provide some basic information about 2-orbit polyhedra. Our treatment here relies heavily on the results in Hubard (2010). We denote the conjugation of a by b as $a^{b}=bab^{-1}$.

3.1 Type $2_{2}$

Hubard (2010) showed that every abstract 2-orbit polyhedron of type $2_{2}$ has an automorphism group generated by $\rho _{2}$ and $\alpha _{01}$. Thus, any geometrically 2-orbit polyhedron of this type must, by Observation 2.2, have a group of symmetries generated by corresponding isometries $R_{2}$ and $A_{01}$ with the same action on the base flag as $\rho _{2}$ and $\alpha _{01}$. (Here we are not excluding yet the possibility of planar polyhedra, in which case $R_2$ and $A_{01}$ would be isometries of the plane.)

We observe that $A_{121}=R_{2}^{A_{01}}$, since $\Phi A_{121}=\Phi ^{121}=r_{1}r_{2}r_{1}\Phi =r_{1}r_{0}r_{2}r_{0}r_{1}\Phi =r_{1}r_{0}r_{2}\Phi A_{01}^{-1}=r_{1}r_{0}\Phi R_{2}A_{01}^{-1}=\Phi A_{01}R_{2}A_{01}^{-1}$.

Lemma 3.1

Let $\mathcal {P}$ be a 2-orbit polyhedron under the action of ${\textsf{Iso}}(\mathcal {P})$ of type $2_{2}$, $\Phi =\{v_{0},e_{0},f_{0}\}$ a choice of base flag for $\mathcal {P}$, and $R_{2},A_{01}$ be the generators of ${\textsf{Iso}}(\mathcal {P})$ with respect to this choice of base flag. Then the stabilizer ${\textsf{Iso}}(\mathcal {P})_{v_{0}}$ of $v_{0}$ in ${\textsf{Iso}}(\mathcal {P})$ is generated by $\{R_{2},A_{121}.\}$.

Proof

By P3, every vertex-figure is finite. Since $\mathcal {P}$ is of type $2_{2}$, for each flag $\Psi \in \mathcal {P}/v_{0}$, the flag $\Psi ^{2}$ belongs to the same orbit, while the flag $\Psi ^{1}$ does not. Hence, the flags in the same orbit as $\Phi$ are precisely those in the orbit $\langle r_{2},r_{1}r_{2}r_{1}\rangle \Phi$. Recall that $A_{121}$ is the symmetry sending $\Phi$ to $r_{1}r_{2}r_{1} \Phi$. Thus if $\Psi \in \langle r_{2},r_{1}r_{2}r_{1}\rangle \Phi$, there exists an element $g\in \langle r_{2},r_{1}r_{2}r_{1}\rangle$ such that $\Psi =g\Phi$. Clearly $g=g'r_{2}$ or $g=g'a_{121}$ for some $g'\in \langle r_{2},r_{1}r_{2}r_{1}\rangle$ obtained by removing the last factor in $\{r_2, r_1 r_2 r_1\}$ of g. In the former case, $\Psi =g'(r_{2}\Phi )=g'(\Phi R_{2})=(g'\Phi )R_{2}$, while in the latter case $\Psi =g'(\alpha _{121}\Phi )=g'(\Phi A_{121})=(g'\Phi )A_{121}$ since the actions of ${\textsf{Con}}(\mathcal {P})$ and ${\textsf{Iso}}(\mathcal {P})$ commute. We may repeat this argument on $g'$ and hence, by finite induction, $\Psi \in \Phi \langle R_{2},A_{121}\rangle$. $\square$

Corollary 3.2

Let $\mathcal {P}$ be a geometrically 2-orbit polyhedron of type $2_{2}$, and let $v_{0}$ be the vertex of the base flag associated with the generators $R_{2},A_{01}$ of ${\textsf{Iso}}(\mathcal {P})$. Then $v_{0}$ lies in the intersection of the fixed point sets of $R_{2}$ and $A_{121}$.

Proof

It suffices to note that by definition, $v_{0}R_{2}=v_{0}$ and $v_{0}A_{121}=v_{0}$; and observe that by Lemma 3.1 any element of ${\textsf{Iso}}(\mathcal {P})$ that fixes $v_{0}$ is in $\langle R_{2},A_{121}\rangle$.$\square$

The next three results are valid for polyhedra even when the ambient space has dimension different from 3.

Lemma 3.3

There are no facet-transitive polyhedra whose facets consist of lines tiled by equal segments.

Proof

Assume that $\mathcal {P}$ is a counterexample. Given a facet f and an edge e of f there must exist another facet $f'$ containing e; but then $f'$ must lie on the unique line containing e. This forces f and $f'$ to lie on the same line $\ell$. Since there is only one tiling of $\ell$ by equal segments, one of which is e, we must have that $f = f'$ contradicting (P1). $\square$

Corollary 3.4

If $\mathcal {P}$ is facet-transitive and there exists $T \in {\textsf{Iso}}(\mathcal {P})$ that acts like a 1-step rotation on a facet then T is not a translation.

Lemma 3.5

Let $\mathcal {P}$ be a polyhedron in $\mathbb {E}^n$ with base flag $\Phi =\{v_{0},e_{0},f_{0}\}$ and $A_{01}, R_{2}\in {\textsf{Iso}}(\mathcal {P})$. Assume that there exists $T\in {\textsf{Iso}}(\mathbb {E}^{n})$ such that

• (1) $v_{0} T=v_{0}$,

• (2) $T^{2}=id$,

• (3) $T A_{01} T= A_{01}^{-1}$, and

• (4) $T R_{2} T= A_{121}$.

Then $\mathcal {P}$ is regular.

Proof

It suffices to prove that T is in ${\textsf{Iso}}(\mathcal {P})$. To see why, observe that if $\Phi =\{v_{0},e_{0},f_{0}\}$, then

$$\begin{aligned} \Phi T&=\{v_{0} T,[v_{0},v_{0} A_{01}] T,[v_{0},v_{0} A_{01}]\langle A_{01}\rangle T\}\\&=\{v_{0},[v_{0},v_{0} T A_{01}^{-1}],[v_{0},v_{0} A_{01}] T\langle A_{01}\rangle \}\\&=\{v_{0},[v_{0},v_{0} A_{01}^{-1}],[v_{0},v_{0} A_{01}^{-1}]\langle A_{01}\rangle \}\\&=\{v_{0},[v_{0},v_{0} A_{01}^{-1}],e_{0}\langle A_{01}\rangle \}. \end{aligned}$$

Observe that $[v_{0},v_{0} A_{01}^{-1}]$ is the other edge of $f_{0}$ containing $v_{0}$, and so $\Phi T=\Phi ^{1}$; in other words, T is an automorphism taking $\Phi$ to $\Phi ^{1}$. A similar calculation shows that $\Phi T A_{01}=\Phi ^{0}$ and so all of the flags adjacent to $\Phi$ are in its orbit under the action of ${\textsf{Iso}}(\mathcal {P})$, and $\mathcal {P}$ is regular.

To show that $T\in {\textsf{Iso}}(\mathcal {P})$, we consider the action of ${\textsf{Iso}}(\mathcal {P})$ on a vertex v of $\mathcal {P}$. Recall that $A_{121}\in \langle A_{01}, R_{2}\rangle \subseteq {\textsf{Iso}}(\mathcal {P})$. Since $v\in \mathcal {P}$, there exists $g\in {\textsf{Iso}}(\mathcal {P})$ such that $v=v_{0}g$, and $v T=v_{0}g T=v_{0} T g'$ for some $g'\in {\textsf{Iso}}(\mathcal {P})$ by (3) and (4) above. But $v_{0} T=v_{0}$, so $v T=v_{0}g'$ is a vertex of $\mathcal {P}$. We conclude that T permutes the vertex set of $\mathcal {P}$. Since $e_0 = [v_{0},v_{0} A_{01}]$ and $f_0 = [v_{0},v_{0} A_{01}]\langle A_{01}\rangle$, a similar argument shows that T permutes the edges and the facets of $\mathcal {P}$, and therefore $T \in {\textsf{Iso}}(\mathcal {P})$. $\square$

Observation 3.6

Let $\mathcal {P}$, $f_0$ and T be as in Lemma 3.5. Then ${f_0 T=f_0}$.

Proof

The edges of $f_0$ are $[v_{0},v_{0} A_{01}]\langle A_{01}\rangle$, and those of $f_0 T$ are

$$\begin{aligned}{}[v_{0},v_{0} A_{01}]\langle A_{01}\rangle T = [v_{0},v_{0} A_{01}] T \langle A_{01}\rangle = [v_{0},v_{0} A_{01}^{-1}]\langle A_{01}\rangle = [v_{0},v_{0} A_{01}]\langle A_{01}\rangle . \end{aligned}$$

$\square$

3.2 Wythoff style constructions

Since we are primarily concerned with the construction of geometric 2-orbit polyhedra, we must describe some of the basic features of their constructions.

Suppose we have a 2-orbit polyhedron $\mathcal {P}$ and its group of symmetries ${\textsf{Iso}}(\mathcal {P})$, and let us further suppose that ${\textsf{Iso}}(\mathcal {P})$ acts transitively on the vertices of $\mathcal {P}$. Further, let us suppose we have identified a set of generators G for ${\textsf{Iso}}(\mathcal {P})$ and an associated base flag $\Phi =\{v_{0},e_{0},f_{0}\}$. We wish to (re)construct the polyhedron $\mathcal {P}$ from the base vertex $v_{0}$ and the group ${\textsf{Iso}}(\mathcal {P})$. This will be our strategy for the enumeration of polyhedra in class $2_2$.

Since all polyhedra in class $2_2$ are fully transitive, we will be able to reconstruct them from their symmetry groups and a base vertex with a relatively simple version of the Wythoff construction (see for example Coxeter 1973, Section 5.7). Essentially, we do the obvious thing:

• 1. The set of vertices is $V=v_{0}\ {\textsf{Iso}}(\mathcal {P})=\{v_{0}\gamma :\gamma \in {\textsf{Iso}}(\mathcal {P})\}$.

• 2. The base edge is $e_{0}:=\{v_{0},v_{0}A_{01}\}$, and then the set of edges is $E:=e_{0}\ {\textsf{Iso}}(\mathcal {P})$.

• 3. The base facet is $f_{0}:=e_{0}\ \langle A_{01}\rangle$. Then the set of facets of $\mathcal {P}$ is $F:=f_{0}\ {\textsf{Iso}}(\mathcal {P})$.

Since $\mathcal {P}$ is fully transitive and items (2) and (3) above clearly yield the base edge and facet, the above construction will yield all the faces of $\mathcal {P}$. Thus, it suffices to identify the generating automorphisms and the location of the base vertex to construct the polyhedron $\mathcal {P}$ (though, of course, the generated group must have a discrete action on ${\mathbb {E}}^{3}$).

Regular polyhedra admit an even simpler version of Wythoff’s construction using the privileged generators $\langle R_{0}, R_{1}, R_{2}\rangle$ of ${\textsf{Iso}}(\mathcal {P})$:

• 1. The set of vertices is $V:=v_{0}{\textsf{Iso}}(\mathcal {P})$.

• 2. The base edge is $e_{0}=\{v_{0},v_{0}R_{0}\}$, and the set of edges is $E:=e_{0}{\textsf{Iso}}(\mathcal {P})$.

• 3. The base facet is $f_{0}=e_{0}\langle R_{0},R_{1}\rangle$, and the facets are $F:=f_{0}{\textsf{Iso}}(\mathcal {P})$.

3.3 Polyhedra obtained by Petrie duality

In Sects. 7 and 8 we shall enumerate certain sets of polyhedra through Petrie duality. In order to justify these enumerations we present the following results.

Lemma 3.7

Let T be an isometry of $\mathbb {E}^3$ and let $v \in \mathbb {E}^3$ be such that $v\langle T \rangle$ is discrete and has size at least 3. Then the vertex set $v\langle T \rangle$ together with the edge set $[v,vT] \langle T \rangle$ form a regular polygon.

Proof

Since $v\langle T \rangle$ has at least 3 points, T must have order at least 3. It follows that T is either a rotation, rotary reflection, translation, glide reflection or twist. When constructing the vertex set $v\langle T \rangle$ and edge set $[v,vT] \langle T \rangle$ from each of these isometries we obtain one of the polygons described in Grünbaum (1977, Section 2), and they are all regular. $\square$

More general details about orbits of points under powers of an isometry in Euclidean space can be found in Coxeter (1991, Section 1.5).

Lemma 3.8

Let $\mathcal {P}$ be a polyhedron whose Petrie paths have at least 3 distinct vertices and such that for each Petrie path $\pi$ there is a symmetry $T \in G(\mathcal {P})$ acting as a 1-step rotation on $\pi$. Then $\mathcal {P}^\pi$ is a polyhedron.

Proof

We need to show that $\mathcal {P}^\pi$ is a collection of polygons and that it satisfies conditions (P1) to (P4) in Sect. 2.

The facets of $\mathcal {P}^\pi$ are the Petrie paths of $\mathcal {P}$. It follows that the vertex and edge sets of each Petrie path of $\mathcal {P}^\pi$ are the orbits of a vertex and an incident edge under the group generated by some isometry (namely, the isometry T in the statement), and Lemma 3.7 implies that every Petrie path of $\mathcal {P}$ is a polygon; that is, every facet of $\mathcal {P}^\pi$ is a polygon.

From the definition of the Petrie paths it is easy to see that every edge belongs to precisely two of them; hence $\mathcal {P}^\pi$ satisfies (P1).

Let e and $e'$ be two edges of $\mathcal {P}^\pi$. We can construct a chain $e=e_0,f_1,e_1,\dots ,f_n, e_n=e'$ of edges and facets of $\mathcal {P}^\pi$ where $f_i$ is incident to $e_{i-1}$ and to $e_i$, from such a chain $e=e_0,\hat{f}_1,\hat{e}_1,\dots ,\hat{f}_m,e_m=e'$ of edges and facets of $\mathcal {P}$. It suffices to note that if $\hat{e}_{i-1}$ and $\hat{e}_i$ are incident to the facet $\hat{f}_i$ of $\mathcal {P}$ then there exist edges $x_1,\dots ,x_k$ of $\hat{f}_i$ such that $\hat{e}_{i-1}=x_1$, $\hat{e}_i=x_k$ and $x_{\ell -1}$ shares a vertex with $x_{\ell }$ for every $\ell$. In that situation we can replace the subchain $\hat{e}_{i-1},\hat{f}_i,\hat{e}_i$ with

$$\begin{aligned} \hat{e}_{i-1},\hat{g}_2,x_2,\hat{g}_3,x_3,\dots ,\hat{g}_{k},\hat{e}_i, \end{aligned}$$

where $\hat{g}_\ell$ is the unique Petrie path of $\mathcal {P}$ having $x_{\ell -1}$ and $x_\ell$ as consecutive edges. We conclude that $\mathcal {P}^\pi$ also satisfies (P2).

Conditions (P3) and (P4) follow from the fact that $\mathcal {P}^\pi$ has the same vertices, edges and vertex-figures as $\mathcal {P}$. $\square$

4 Two orbit polyhedra derived from $\{6,4|4\}$

In this section we describe a family of 2-orbit polyhedra in class $2_2$. They are all combinatorially isomorphic to the polyhedron $\{6,4|4\}$ found by Petrie (see Coxeter 1937).

There are many ways to define the polyhedron $\{6,4|4\}$. Perhaps the easiest one is as the vertex and edge set of the tiling of $\mathbb {E}^3$ by truncated octahedra (this is the tessellation $\#28$ in Grünbaum 1994), together with all the hexagons of that tessellation (but no squares) (see Fig. 1).

Fig. 1

The polyhedron $\{6,4|4\}$

Here it is convenient to give another description, using Wythoff’s construction as follows:

• Our base vertex will be $v_0 := (0,0,1)$.

• $R_0 : (x,y,z) \mapsto (-2+z,y,2+x)$.

• $R_1 : (x,y,z) \mapsto (y,x,2-z)$.

• $R_2 : (x,y,z) \mapsto (-x,y,z)$.

Then the base edge $e_0$ joins $v_0$ to $(-1,0,2)$, and the base facet $f_0$ is a hexagon with vertices listed in cyclic order:

$$\begin{aligned} (0,0,1), (-1,0,2), (-2,-1,2), (-2,-2,1), (-1,-2,0), (0,-1,0). \end{aligned}$$

The remaining facets come from the orbit of $f_0$ under ${\textsf{Iso}}(\{6,4|4\}) = \langle R_0, R_1, R_2 \rangle$.

To better visualize $\{6,4|4\}$, we may think of the cubic tessellation $2\mathcal {T}$ whose vertices have all entries in $2\mathbb {Z}$. Then $f_0$ is the convex hull of the midpoints of the edges of a Petrie polygon of the cube centered at $(-1,-1,1)$. The remaining facets are the images of $f_0$ under compositions of the reflections on the planes containing squares of $2\mathcal {T}$ (see Fig. 2). Each cube of $2\mathcal {T}$ contains only one hexagon of $\{6,4|4\}$.

Fig. 2

The polyhedron $\{6,4|4\}$ from the cubic tiling

The hexagons of $\{6,4|4\}$ can be divided into four classes, depending on their normal vector. Recall from McMullen and Schulte (2002, Section 6D) that the lattice $\Lambda _{(1,1,1)}$ consists of all vectors of $\mathbb {E}^3$ with integer coordinates such that either the three entries are even, or the three entries are odd. The hexagons in cubes whose centers are in $(-1,1,1)+2\Lambda _{(1,1,1)}$ have (1, 1, 1) as normal vector; similarly, those hexagons in cubes having centers in $(-1,-1,1)+2\Lambda _{(1,1,1)}$, in $(-1,1,-1)+2\Lambda _{(1,1,1)}$ and in $(-1,-1,-1)+2\Lambda _{(1,1,1)}$ have normal vectors $(1,-1,1)$, $(1,1,-1)$ and $(1,-1,-1)$, respectively.

We next show that $\{6,4|4\}$ can be thought of as a 2-orbit polyhedron in class $2_0$ with respect to the subgroup $\langle A_{01}, R_2 \rangle$.

Lemma 4.1

Let ${\textsf{Iso}}(\{6,4|4\}) = \langle R_0, R_1, R_2 \rangle$ and let $A_{01}$ be defined as in Sect. 2. Then $\langle A_{01},R_2 \rangle$ has index 2 in ${\textsf{Iso}}(\{6,4|4\})$.

Proof

It is well-known that every polyhedron with type $\{6,4\}$ is a quotient of the hyperbolic tessellation $\{6,4\}$ of hexagons meeting four around each vertex. As a consequence of this, ${\textsf{Iso}}(\{6,4|4\})$ is a quotient of the Coxeter group [6, 4], which is the symmetry group of the tessellation $\{6,4\}$. It is standard procedure to verify that if $[6,4] = \langle \rho _0,\rho _1,\rho _2 \rangle$ then $\langle \rho _1\rho _0,\rho _2 \rangle$ is an index 2 subgroup of [6, 4]. The image of $\langle \rho _1\rho _0,\rho _2 \rangle$ under the quotient to ${\textsf{Iso}}(\{6,4|4\})$ is precisely $\langle A_{01},R_2 \rangle$. Hence, the index of $\langle A_{01},R_2 \rangle$ in ${\textsf{Iso}}(\{6,4|4\})$ is at most 2.

It follows from McMullen and Schulte (2002, Section 7F) that ${\textsf{Iso}}(\{6,4|4\})$ can be obtained from the Coxeter group [6, 4] by adjoining the single extra relation $(R_0R_1R_2R_1)^4=id$. This relation has an even number of factors in $\{R_0, R_1\}$, which are the generators of ${\textsf{Iso}}(\{6,4|4\})$ that are not in $\langle A_{01},R_2 \rangle$. This implies that the coset $\langle A_{01},R_2 \rangle R_0$ equals $\langle A_{01},R_2 \rangle R_1$ and is distinct from $\langle A_{01},R_2 \rangle$. Therefore, $\langle A_{01},R_2 \rangle$ is an index 2 subgroup of ${\textsf{Iso}}(\{6,4|4\})$. $\square$

An important implication of Lemma 4.1 is that we can recover $\{6,4|4\}$ from the base vertex $v_0$ together with $\langle A_{01},R_2 \rangle$ by Wythoff’s construction. That is, the base edge $e_0$ has $v_0$ and $v_0 A_{01}$ as endpoints, the base facet $f_0$ has $e_0 \langle A_{01} \rangle$ as its edge set, and the remaining vertices, edges and facets are obtained as $v_0 \langle A_{01},R_2 \rangle$, $e_0 \langle A_{01},R_2 \rangle$ and $f_0 \langle A_{01},R_2 \rangle$, respectively.

In order to recover $\{6,4|4\}$ from ${\textsf{Iso}}(\{6,4|4\})$ it was necessary to choose $v_0$ in the intersection of the fixed sets of $R_1$ and $R_2$. (Recall that the stabilizer of $v_0$ is $\langle R_1, R_2 \rangle$.) That intersection consists only of the point (0, 0, 1), since the fixed set of $R_1$ is the line $\{(0,0,1) + \lambda (1,1,0) \, : \, \lambda \in \mathbb {R}\}$ while the fixed set of $R_2$ is the plane $x=0$.

On the other hand, if we want to recover $\{6,4|4\}$ from $\langle A_{01},R_2 \rangle$ without knowing beforehand the position of $v_0$, then the choice of base vertex is no longer unique. We know that $R_2$ and $A_{121}$ generate the stabilizer of $v_0$ in $\langle A_{01},R_2 \rangle$ (see Lemma 3.1), and hence $v_0$ is only required to belong to the intersection of the mirrors of these two isometries. As pointed out above, the mirror of $R_2$ is the plane $x=0$; on the other hand, the mirror of $A_{121}$ is the plane $y=0$. It follows that in principle the base vertex $v_0$ may be chosen as any point in the z-axis.

In what follows we shall construct a polyhedron $\mathcal {P}_{\{6,4\}}(s)$ through Wythoff’s construction from $\langle A_{01},R_2 \rangle$ starting with $v_0 = (0,0,s)$ as base vertex, for some values of s.

For any given choice (0, 0, s) of $v_0$ the base edge $e_0$ of $\mathcal {P}_{\{6,4\}}(s)$ has $(-s,0,2)$ as the other endpoint, and the vertices of the base facet $f_0$ of $\mathcal {P}_{\{6,4\}}(s)$ in cyclic order are

$$\begin{aligned} (0,0,s), (-s,0,2), (-2,-s,2), (-2,-2,2-s), (-2+s,-2,0), (0,-2+s,0). \end{aligned}$$

As in the regular case $s=1$, the remaining facets of $\mathcal {P}_{\{6,4\}}(s)$ are obtained as the images of $f_0$ by all products of conjugates of $R_2$; that is, they are obtained by reflecting $f_0$ through all planes containing squares of $2\mathcal {T}$.

Fig. 3

The 2-orbit realization of $\{6,4|4\}$ from $\langle A_{01},R_2 \rangle$ for $v_0 = (0,0,2/3)$

For every value of s different from 1 the facets of $\mathcal {P}_{\{6,4\}}(s)$ are not planar; they are regular antiprismatic skew hexagons (as defined in Grünbaum 1977) whose centers are still the centers of the cubes of $2\mathcal {T}$. The facet centered at $(-1,-1,-1)$ is shown on the left of Fig. 3 for the value of $s=2/3$. For the same value of s, the four hexagons of $\mathcal {P}_{\{6,4\}}(s)$ corresponding to those in Fig. 2 are shown on the right of Fig. 3.

We remark that each vertex of the hexagon f centered at some cube of $2\mathcal {T}$ lies on the line spanned by an edge of that cube. Figure 4 represents the cube centered at $(-1,-1,1)$ and indicates the lines on which the vertices of the hexagon $f_0$ move. The arrow indicates the direction of the movement when s increases.

Fig. 4

Axes where the vertices of the base facet $f_0$ lie

When $s<0$ or $s>2$ each hexagonal facet f of the resulting polyhedron is no longer bounded by the cube centered at the center of f. Nevertheless, every edge belongs to the plane $\Pi$ containing a certain square of that cube, so that it is invariant under the reflection about $\Pi$.

Not all values of s yield polyhedra. If s is an even integer then $v_0$ lies on the mirror $z=s$ of a reflection in $\langle A_{01},R_2 \rangle$, that does not belong to $\langle R_2, A_{121} \rangle$. As a consequence of this, the vertices collapse in pairs and every edge belongs to four polygons. As an example of this, if we choose $v_0$ to be (0, 0, 0) then $f_0$ becomes one of the Petrie polygons of the cube of $2\mathcal {T}$ centered at $(-1,-1,1)$, namely the one containing the vertices $(0,-2,0)$, (0, 0, 0) and (0, 0, 2), in that order. Its image under the reflection about the yz-plane is the Petrie polygon of the cube centered at $(1,-1,1)$ that also contains those three vertices. When reflecting with respect to the xy-plane we get two other hexagons containing the edge between $(0,-2,0)$ and (0, 0, 0). This edge now belongs to four hexagons, and therefore that structure is not a polyhedron as defined in Sect. 2. Since the only intersections of the z-axis with mirrors of elements in $\langle A_{01},R_2 \rangle$ are at points (0, 0, t) for even values of t, every choice of $s \in \mathbb {R} \setminus 2\mathbb {Z}$ yields a polyhedron $\mathcal {P}_{\{6,4\}}(s)$.

From the construction we know that $\mathcal {P}_{\{6,4\}}(1)$ coincides with $\{6,4|4\}$ and hence is regular. We claim that for other values of s the polyhedron $\mathcal {P}_{\{6,4\}}(s)$ is 2-orbit in class $2_2$. Indeed, the symmetry type of $\mathcal {P}_{\{6,4\}}(s)$ must be either regular or $2_2$ since its symmetry group contains $A_{01}$ and $R_2$. Assume now that $s \ne 1$ and $\mathcal {P}_{\{6,4\}}(s)$ is regular. Then it must also admit a symmetry R mapping the base flag to its 1-adjacent neighbor. If $f_0$ is non-planar then there is exactly one isometry preserving $v_0$ and $f_0$ but moving $e_0$. This isometry is the plane reflection about the bisector of the line segment between the two neighbors $(-s,0,2)$ and $(0,-2+s,0)$ of $v_0$. The normal vector of that plane is $(-s,0,2)-(0,-2+s,0) = (-s,2-s,2)$, which is not the normal vector of a plane reflection of an isometry of $2\mathcal {T}$ for any value of $s \notin \{0,2\}$. (Recall that for $s \in \{0,2\}$ the structure $\mathcal {P}(s)$ is not a polyhedron.) This implies that R does not preserve the set of centers of hexagons of $\mathcal {P}_{\{6,4\}}(s)$, and therefore it is not a symmetry of the polyhedron, a contradiction.

We have proved the following.

Theorem 4.2

For every $s \in \mathbb {R} \setminus 2\mathbb {Z}$ there is a polyhedron $\mathcal {P}_{\{6,4\}}(s)$ having $\langle A_{01},R_2 \rangle$ as a subgroup of index at most 2 of its symmetry group, with the properties that it contains (0, 0, s) as a one of its vertices, and it is combinatorially isomorphic to $\{6,4|4\}$. Furthermore, $\mathcal {P}_{\{6,4\}}(s)$ is geometrically regular if and only if $s=1$. If $s \ne 1$ then $\mathcal {P}_{\{6,4\}}(s)$ is a 2-orbit polyhedron in class $2_{2}$.

5 Analysis of fixed sets of generators of polyhedra in class $2_2$

Let $\mathcal {P}$ be a polyhedron in class $2_{2}$ with base flag $\Phi =\{v_{0},e_{0},f_{0}\}$. As noted in Sect. 2, ${\textsf{Iso}}(\mathcal {P})=\langle R_{2}, A_{01}\rangle$. Since $R_{2}$ fixes the vertex $v_{0}$ and edge $e_{0}$ of $\Phi$, and is an involution, $R_{2}$ must be either a half-turn about $e_{0}$ or a reflection through a plane containing $e_{0}$. Note also that for any flag $\Psi \in \mathcal {F}(\mathcal {P})$, $r_{0}\Psi$ and $r_{1}\Psi$ are in the other orbit, while $r_{2}\Psi$ is in the same orbit. Consequently, $r_{1}r_{2}r_{1}\Phi$ is in the same orbit as $\Phi$, and so there is an isometry $A_{121}:\Phi \mapsto \Phi ^{121}$. Since $R_{2}, A_{01}\in {\textsf{Iso}}(\mathcal {P})$, $A_{121}=R_{2}^{A_{01}}$, and so the fixed set of the other generator of the stabilizer ${\textsf{Iso}}(\mathcal {P})_{v_{0}}$ depends on the compositions of these actions.

5.1 Planar polyhedra

We first address the possibility of $\mathcal {P}$ lying in a plane. In that situation $R_2$ must be the reflection about the affine span of $e_0$, while $A_{01}$, not being an involution, must be either a rotation or a glide reflection. Note that translations are discarded by Corollary 3.4. In what follows we shall show that with such generators of ${\textsf{Iso}}(\mathcal {P})$, this polyhedron is forced to be regular.

Assume first that $A_{01}$ is a rotation about the point w. Since we are only considering infinite polyhedra, w cannot belong to the mirror of $R_2$. Then $A_{01} = R_m R_{m'}$, where m is the line through w and $v_0$, $m'$ is the line through w and the center u of $e_0$, while $R_m$ and $R_{m'}$ denote the reflections about m and $m'$, respectively (see Fig. 5 (I)). We set $T := R_m$, which trivially satisfies the first two items of Lemma 3.5. Furthermore,

$$\begin{aligned} TA_{01}T = R_m R_m R_{m'} R_m = R_{m'} R_m = A_{01}^{-1}, \end{aligned}$$

and so T also satisfies the third item. The fourth item follows from the fact that both $R_m$ and $A_{01}^{-1}$ map $e_0$ to the edge $e_{-1}$ of $f_0$ containing $v_0$ other than $e_0$, and therefore the conjugates of $R_2$ by both isometries are the reflection about the line spanned by $e_{-1}$. (Recall that $A_{121} = A_{01} R_2 A_{01}^{-1}$.) Now Lemma 3.5 implies that $\mathcal {P}$ is regular.

Fig. 5

Location of mirrors of symmetries if $\mathcal {P}$ is planar

We are left with the case where $A_{01}$ is a glide reflection with axis $\ell$. Then $A_{01} = R_m S_u$, where $R_m$ is the reflection about the line m through $v_0$ perpendicular to $\ell$ and $S_u$ is the half-turn about the midpoint u of $e_0$ (see Fig. 5 (II)). Here we again set $T := R_m$, which satisfies the first two items of Lemma 3.5. The verification that T satisfies the remaining two items is similar to that when $A_{01}$ was a rotation, and therefore $\mathcal {P}$ must be regular as well.

We summarize the previous discussion in the following result.

Theorem 5.1

There are no infinite 2-orbit polyhedra of type $2_2$ in the plane.

5.2 Basic aspects of 3-dimensional polyhedra

In the reminder of this section we assume that the affine span of $\mathcal {P}$ is the entire space $\mathbb {E}^3$.

There are four possibilities for the action of $A_{01}$: rotation about a line, rotary reflection, screw motion, and glide reflection. Again, translations are discarded by Corollary 3.4. Moreover, the fact that the affine span of $f_0$ cannot be 1-dimensional implies that if $A_{01}$ is a screw motion or glide reflection then $v_0$ does not belong to the invariant line(s) of $A_{01}$. Hence $A_{01}$ must be either rotation about a line (not containing $v_{0}$), a rotary reflection, a screw motion or a glide reflection with the restrictions above for $v_{0}$. Since a facet must be a polygon, $A_{01}$ must have order $\ge 3$, which eliminates half-turns from the possible rotations, as well as central inversions from those of rotary reflections for $A_{01}$. The set of fixed points of a rotation about a line m is that line m; a (non-trivial) rotary reflection has exactly one fixed point; a screw motion has no fixed points.

It turns out that many of the aforementioned possibilities always result in a regular polyhedron. This section is devoted to describing the arguments that demonstrate this fact. The remaining cases will be handled in Sects. 6 and 7.

With the foregoing information in hand, it is now possible to use Lemma 3.5 to show that some of these possibilities force $\mathcal {P}$ to be regular. In what follows we will find it notationally convenient to let $v_{i}:=v_{0}A_{01}^{i}$ and $e_{i}:=e_{0}A_{01}^{i}$.

5.3 $R_{2}$ is a half-turn

Suppose $R_{2}$ is a half-turn about the edge $e_{0}=\{v_{0},v_{0}A_{01}\}$.

5.3.1 $A_{01}$ is a screw motion

We consider first the case when $A_{01}$ is a screw-motion. Let $\ell$ be the angle bisector of $e_{0}$ and $e_{-1}$ (i.e., the two edges of $f_{0}$ containing $v_{0}$, see Fig. 6). Let T be the half-turn about $\ell$. It is easily seen that conditions (1) and (2) of Lemma 3.5 are satisfied for T, and that $f_{0}T=f_{0}$; in fact, $v_{i}T=v_{-i}$ for all i. Observe that $v_{0}TA_{01}T=v_{0}A_{01}T=v_{1}T=v_{-1}$, $v_{1}TA_{01}T=v_{-1}A_{01}T=v_{0}T=v_{0}$, and $v_{-1}TA_{01}T=v_{1}A_{01}T=v_{2}T=v_{-2}$. Since T and $A_{01}$ are isometries preserving $f_{0}$, so is their composition. Also, since an isometry of a helical polygon is determined by its action on three consecutive vertices, we conclude that $TA_{01}T=A_{01}^{-1}$. Thus (3) from Lemma 3.5 is satisfied.

Fig. 6

A helix whose generating screw motion $A_{10}$ rotates a quarter-turn per application. The base vertex $v_{0}$ and edge $e_{0}$, as well as the line $\ell$ are shown

What then is the relationship between the action of $TR_{2}T$ and that of $A_{121}$? The first thing to note is that for any isometry $I\in {\textsf{Iso}}(\mathcal {P})$, $IR_{2}I^{-1}$ is a half-turn around the image of the axis of rotation of $R_{2}$ under $I^{-1}$. Hence $A_{121}=A_{01}R_{2}A_{01}^{-1}$ is a half-turn around $e_{-1}$. Similarly, since $T^{-1}=T$, $TR_{2}T^{-1}$ is a half-turn around $e_{0}T=e_{-1}$. Thus (4) of Lemma 3.5 is satisfied. We have the following lemma.

Lemma 5.2

Let $\mathcal {P}$ be a polyhedron and let $R_{2},A_{01}\in {\textsf{Iso}}(\mathcal {P})$. If $R_{2}$ is a half-turn and $A_{01}$ is a screw motion, then $\mathcal {P}$ is regular.

5.3.2 $A_{01}$ is a rotary reflection

Every rotary reflection may be represented as the product of three reflections, two of which are in planes perpendicular to the third plane; the angle between the first two is $k\pi /n$ where n is the order of $A_{01}$ and k is coprime to n. Also note that since $f_{0}$ is a polygon, $n\ge 3$. Let $A_{01}=R_{\Pi _{a}}R_{\Pi _{b}} R_{\Pi _{c}}$, where $\Pi _{i}$ are the requisite planes of reflection (Fig. 7).

Fig. 7

A hexagon formed by a rotary reflection. Base vertex $v_{0}$ and edge $e_{0}$, as well as reflecting planes used in the decomposition of $A_{01}$ are labeled in the diagram

Let $\Pi$ be the plane containing $v_{0}$ and the rotation axis of $A_{01}$, then $\Pi$ contains the intersection $\ell$ of $\Pi _{a}$ and $\Pi _{b}$. Moreover, the rotation about $\ell$ determined by the composition of $R_{\Pi _{a}}$ and $R_{\Pi _{b}}$ is determined only by the angle between them, so we may choose our decomposition of $A_{01}$ so that $\Pi _{a}=\Pi$. If $R_{\Pi }$ is the reflection through the plane $\Pi$, then $R_{\Pi }$ clearly satisfies conditions (1) and (2) of Lemma 3.5. Observe also that $R_{\Pi }A_{01}R_{\Pi }^{-1}=R_{\Pi }R_{\Pi _{a}}R_{\Pi _{b}}R_{\Pi _{c}}R_{\Pi }$, but $R_{\Pi }=R_{\Pi _{a}}$, so $R_{\Pi }A_{01}R_{\Pi }^{-1}=R_{\Pi _{b}}R_{\Pi _{c}}R_{\Pi }$. Since $\Pi _{c}\perp \Pi$, we get $A_{01}^{R_{\Pi }}=R_{\Pi _{b}}R_{\Pi _{a}}R_{\Pi _{c}}=A_{01}^{-1}$, satisfying (3) of Lemma 3.5. Again using the fact that $R_{\Pi }R_{2}R_{\Pi }^{-1}$ is a half-turn about the image of the axis of $R_{2}$ around $e_{0}$ under $R_{\Pi }$, we see that $R_{2}^{R_{\Pi }}$ is a half turn around $e_{-1}$, agreeing with the action of $A_{121}$. We have the following observation, by Lemma 3.5.

Lemma 5.3

Let $\mathcal {P}$ be a polyhedron, and let $R_{2},A_{01}\in {\textsf{Iso}}(\mathcal {P})$. If $R_{2}$ is a half-turn and $A_{01}$ is a rotary reflection, then $\mathcal {P}$ is a regular polyhedron.

5.3.3 $A_{01}$ is a rotation about a line not containing $v_{0}$

When $A_{01}$ is a rotation, we can apply essentially the same argument as in Sect. 5.3.2, only we no longer require the plane $\Pi _{3}$ in the representation of $A_{01}$.

Lemma 5.4

Let $\mathcal {P}$ be a polyhedron, and let $R_{2},A_{01}\in {\textsf{Iso}}(\mathcal {P})$. If $R_{2}$ is a half-turn and $A_{01}$ is a rotation, then $\mathcal {P}$ is a regular polyhedron.

5.3.4 $A_{01}$ is a glide reflection

We consider the case when $A_{01}$ is a glide reflection. The orbit of every point under $A_{01}$ lies in a plane $\Lambda$ perpendicular to the invariant plane $\Omega$ of $A_{01}$, but $f_0$ does not belong to a line (see Lemma 3.3). It follows that half the vertices of $f_0$ lie on one side of $\Omega$ and the other half in the other side of $\Omega$ in an alternating fashion. We may again decompose $A_{01}$ into the product of three reflections: the reflection $R_{\Omega }$ through $\Omega$; the reflection $R_{\Pi }$ through the plane $\Pi$ perpendicular to $\Omega$ and to $\Lambda$ containing $v_{0}$; and the reflection $R_{\Sigma }$ through the midpoint of $e_{0}$ perpendicular to $\Omega$ and to $\Lambda$, then $A_{01}=R_{\Pi }R_{\Sigma }R_{\Omega }$. Let $T=R_{\Pi }$, then $TA_{01}T=R_{\Pi }R_{\Pi }R_{\Sigma }R_{\Omega }R_{\Pi }=R_{\Sigma }R_{\Pi }R_{\Omega }=A_{01}^{-1}$ since $\Pi$ and $\Omega$ are perpendicular, and $R_{\Sigma }R_{\Pi }$ is the inverse of the translational component of $A_{01}$. Also observe that both $A_{01}^{-1}$ and $R_\Pi$ map $e_0$ to $e_{-1}$, and so $R_{\Pi }R_{2}R_{\Pi }=A_{01}R_2A_{01}^{-1} =A_{121}$. Since $T=R_{\Pi }$ trivially satisfies conditions (1) and (2) of Lemma 3.5, we may again conclude that $\mathcal {P}$ is in fact regular (see Fig. 8). Note also that in this case all of the faces of $\mathcal {P}$ lie in the plane $\Lambda$, and hence $\mathcal {P}$ is planar. In summary.

Fig. 8

A base facet generated by a glide reflection $A_{01}$, with the base vertex $v_{0}$ and edge $e_{0}$ labeled, as well as the planes of reflection used in the decomposition of $A_{01}$

Lemma 5.5

Let $\mathcal {P}$ be a polyhedron and let $R_{2},A_{01}\in {\textsf{Iso}}(\mathcal {P})$. If $R_{2}$ is a half-turn and $A_{01}$ is a glide reflection then $\mathcal {P}$ is a regular polyhedron.

Corollary 5.6

Let $\mathcal {P}$ be a polyhedron (as above), and let $R_{2},A_{01}\in {\textsf{Iso}}(\mathcal {P})$. If $R_{2}$ is a half-turn then $\mathcal {P}$ is a regular polyhedron.

5.4 $R_{2}$ is a reflection

We now consider the cases when $R_{2}$ is a reflection in a plane $\Pi _2$ containing the edge $e_{0}=\{v_{0},v_{0}A_{01}\}$.

5.4.1 $A_{01}$ is a rotation about a line not containing $v_{0}$

Let $\ell$ be the line about which $A_{01}$ is a rotation, let $\Pi$ be a plane containing $v_{0}$ and $\ell$, and let $\Sigma$ be a plane containing the midpoint of $e_{0}$ and $\ell$. Then $A_{01}=R_{\Pi }R_{\Sigma }$, the product of the reflections through $\Pi$ and $\Sigma$. In the application of Lemma 3.5 to this setting, we will choose $T=R_{\Pi }$. Observe that $R_{\Pi }$ trivially satisfies conditions (1) and (2). Observe also that $TA_{01}T=R_{\Pi }R_{\Pi }R_{\Sigma }R_{\Pi }=A_{01}^{-1}$, so T satisfies condition (3).

Recall that $A_{121}=R_{2}^{A_{01}}=A_{01}R_{2}A_{01}^{-1}$, and so we have $A_{121}=R_{\Pi }R_{\Sigma }R_{2}R_{\Sigma }R_{\Pi }$. Note, however, that by construction $\Sigma$ and $\Pi _{2}$ are perpendicular since $e_{0}\subset \Pi _{2}$ and $e_{0}\not \subset \Sigma$ but $e_{0}$ is perpendicular to $\Sigma$. Thus $R_{\Sigma }$ and $R_{2}$ commute, hence, $A_{121}=R_{\Pi }R_{\Sigma }R_{\Pi }$, and $\mathcal {P}$ is regular.

5.4.2 $A_{01}$ is a glide reflection

We will use the notation for reflecting planes from Sect. 5.3.4. Hence, $A_{01}=R_{\Pi }R_{\Sigma }R_{\Omega }$ (note that $R_{\Omega }$ commutes with both $R_{\Sigma }$ and $R_{\Pi }$). Let $\ell$ be the line $\Pi \cap \Lambda$ passing through $v_{0}$, and let T be a half-turn about that line. Then T trivially satisfies (1) and (2) of Lemma 3.5. Since T is a half turn about the intersection of $\Lambda$ and $\Pi$, $R_{\Lambda }T=R_{\Pi }$. Note also that $\Lambda$ is perpendicular to each of the other planes, and $\Omega$ is perpendicular to $\Pi$ and $\Sigma$, so their associated reflections commute. Thus $TA_{01}T=R_{\Lambda }R_{\Pi }(R_{\Pi }R_{\Sigma }R_{\Omega })R_{\Lambda }R_{\Pi }=R_{\Sigma }R_{\Omega }R_{\Pi }=R_{\Omega }R_{\Sigma }R_{\Pi }=A_{01}^{-1}$, as desired (Fig. 9).

Fig. 9

A zigzag polygon with base vertex $v_{0}$ and base edge $e_{0}$. Also shown are the plane of reflection for the generating glide reflection $A_{01}$, the plane $\Pi$ used in the decomposition of the translational component of $A_{01}$ passing through $v_{0}$, the plane $\Pi _{2}$ corresponding to the reflection $R_{2}$, and the line $\ell$ fixed by the half-turn T

We know that the plane of reflection of $A_{121}$ is $\Pi _{2}A_{01}^{-1}$. We also know that the plane of reflection of $TR_{2}T$ is $\Pi _{2}T$. If we can show that these are the same plane, then we will have proven condition (4) of Lemma 3.5. Since both T and $A_{01}^{-1}$ map $e_{0}$ to $e_{-1}$, it suffices to show that both maps send a distinct line in $\Pi _{2}$ to the same line. Let ${\textbf {v}}$ be a vector rooted at the midpoint of $e_{0}$, which is contained in $\Pi _{2}\cap \Omega$. Thus $A_{01}^{-1}$ acts like a translation on ${\textbf {v}}$, sending it to a parallel vector rooted at the midpoint of $e_{0}$. On the other hand, since $\ell$ is perpendicular to $\Omega$, ${\textbf {v}}T$ is rooted at the midpoint of $e_{-1}$ and pointing in the opposite direction. Hence ${\textbf {v}}A_{01}^{-1}$ and ${\textbf {v}}T$ determine the same plane when combined with $e_{-1}$, so $TR_{2}T=A_{01}R_{2}A_{01}^{-1}$, as desired.

Lemma 5.7

Let $\mathcal {P}$ be a 2-orbit polyhedron of type $2_{2}$ with base flag $\Phi$, and let $R_{2},A_{01}\in {\textsf{Iso}}(\mathcal {P})$ be the associated generators. If $R_{2}$ is a reflection and $A_{01}$ is a glide reflection, then $\mathcal {P}$ is a regular polyhedron.

The previous analysis discards the possibility of $R_2$ being a half-turn, as well as the possibilities of $A_{01}$ being a rotation or glide reflection when $R_2$ is a plane reflection. In the next two sections we analyze the remaining two possibilities for the case that $R_2$ is a plane reflection. First we deal with the possibility of $A_{01}$ being a glide reflection, and then with that of $A_{01}$ being a screw motion.

6 Two orbit polyhedra derived from Petrie–Coxeter polyhedra

We now arrive at our first case where there are geometric 2-orbit polyhedra. As in Sect. 5.4, we will be concerned with polyhedra of type $2_{2}$, where $R_{2}$ is a reflection. As before, the plane of reflection of $A_{121}$ is $\Pi _{121}=\Pi _{2}A_{01}^{-1}$, and so $v_{0}$ may be any point in $\Pi _{121}\cap \Pi _{2}$, since these points remain fixed under the action of $\text {Stab}_{{\textsf{Iso}}(\mathcal {P})}\left( v_{0}\right) =\langle A_{121}, R_{2}\rangle$. Let $\chi =\{\Pi _{2}T\mid T\in {\textsf{Iso}}(\mathcal {P})\}$, the set of all reflection planes corresponding to conjugates of $R_{2}\in {\textsf{Iso}}(\mathcal {P})$. Note that ${\textsf{Iso}}(\mathcal {P})$ acts transitively on $\chi$. Let $G=\langle R_{\Pi }\mid \Pi \in \chi \rangle$. Note that G, being a discrete group acting on ${\mathbb {R}}^{3}$ generated by reflections, must be a Coxeter group. Furthermore, since $R_2$ fixes the base edge pointwise and $\mathcal {P}$ is edge-transitive, for each edge e of $\mathcal {P}$ there is a plane in $\chi$ that fixes e pointwise. It follows that $\chi$ is an infinite set.

What then can we say about the associated Coxeter diagram of G? The first thing to note is that $\chi$ divides ${\mathbb {E}}^{3}$ into convex cells bounded by members of $\chi$, and that to each cell C in the complement of $\chi$, the set of planes bounding that cell are sufficient (and necessary) to generate G.

We next present a property of the set $\chi$ just described.

Lemma 6.1

Let $\mathcal {P}$ be a polyhedron in class $2_2$ and let G and $\chi$ be defined as above. Then there is no direction vector contained in all the planes in $\chi$. In particular, the affine span of the orbit of any point under the action of G is ${\mathbb {E}}^{3}$.

Proof

From the findings of Sect. 5 we know that $R_2$ is a plane reflection and $A_{01}$ is either a rotary reflection or a screw motion.

Assume to the contrary that there is a direction vector contained in all planes of $\chi$. Since $v_0\in \Pi _2 \cap \Pi _{121}$, these two planes intersect on a line $\ell$ in a privileged direction of the planes of $\chi$. With the help of Lemma 3.5 we shall show that $\mathcal {P}$ is regular and not in class $2_2$, yielding the desired contradiction.

We define an isometry T as follows. If $A_{01}$ is a rotary reflection with fixed point o, then T is the reflection with respect to the plane spanned by o and $\ell$. (Note that $o \notin \ell$, since otherwise o would be a fixed point of every element in ${\textsf{Iso}}(\mathcal {P})$.) On the other hand, if $A_{01}$ is a screw motion then we define T as the half-turn about the line through $v_0$ perpendicular to the axis m of $A_{01}$.

The first two items of Lemma 3.5 are satisfied trivially, and the third one follows from the same arguments given in Sect. 5.3.2, if $A_{01}$ is a rotary reflection, and from Sect. 5.3.1, if $A_{01}$ is a screw motion.

To verify the fourth item of Lemma 3.5 it suffices to show that $\Pi _2 T = \Pi _{121}$, since these will be the reflection planes of $T R_2 T$ and $A_{121}$, respectively. Now, if $A_{01}$ is a rotary reflection then the distances from o to $\Pi _2$ and $\Pi _{121}$ are the same, since one is the image of the other under $A_{01}$. The fixed point set of T contains o, so it must be one of the two bisectors of $\Pi _2$ and $\Pi _{121}$, and we can conclude that $\Pi _2 T = \Pi _{121}$. If on the other hand $A_{01}$ is a screw motion, then the line m has two choices: either it is in the direction of $\ell$ or it is perpendicular to $\ell$ and the rotational component of $A_{01}$ is 2-fold (so that $A_{01}$ preserves that direction as all elements of ${\textsf{Iso}}(\mathcal {P})$ must). If m and $\ell$ are parallel then the distances from m to $\Pi _2$ and $\Pi _{121}$ are the same, since again one is an image of the other under $A_{01}$. In this case the axis of T is in a bisector of the planes $\Pi _2$ and $A_{01}$, and since this axis is also perpendicular to $\ell$, T maps $\Pi _2$ to $\Pi _{121}$. Finally, if the direction vector of m is perpendicular to that of $\ell$ we let $\Pi _\perp$ be the plane through $v_0$ perpendicular to $\ell$; then $A_{01}$ maps $\Pi _\perp \cap \Pi _2$ to $\Pi _\perp \cap \Pi _{121}$ and $A_{01}$ acts on $\Pi _\perp$ as a glide reflection. This forces the axis of T to the bisector in $\Pi _\perp$ of $\Pi _\perp \cap \Pi _2$ and $\Pi _\perp \cap \Pi _{121}$, and the planes $\Pi _2$ and $\Pi _{121}$ coincide. $\square$

Let C be a fundamental region for G, then since $\mathcal {P}$ is an infinite geometrically 3-dimensional polyhedron and in view of Lemma 6.1, C is bounded by at least 4 planes of reflection, and no more than 6. The possibilities for G thus correspond to the following Coxeter diagrams:

The following lemmas exclude most of the Coxeter diagrams for G.

Lemma 6.2

Let $\chi$ and G be as above. Then the Coxeter diagram of G does not include a component consisting of one dot.

Proof

Assume to the contrary that the diagram of G has one component that consists of a dot, and let $\Delta$ be the reflection plane corresponding to that component.

We first note that $\Delta \in \chi$, since the reflection about $\Delta$ is not generated by the remaining reflections in G. Now, $\Delta$ is perpendicular to all other planes in $\chi$. Transitivity of $\chi$ under ${\textsf{Iso}}(\mathcal {P})$ implies that every plane in $\chi$ is perpendicular to all other planes in $\chi$, forcing $\chi$ to have at most three planes, a contradiction. $\square$

Lemma 6.3

Let $\chi$ and G be as above. If the Coxeter diagram of G includes a component consisting of one link with label $\infty$ then it is diagram (3).

Proof

Assume that G has a component consistent of one link with label $\infty$. This means that there is a class $\mathcal {Y}$ of mirrors of reflections in G consisting of infinitely many parallel reflection planes, and that every plane in $\mathcal {Y}$ is perpendicular to every reflection plane of G that is not in $\mathcal {Y}$.

The reflections about planes in $\mathcal {Y}$ are not generated by the reflections in $\chi \setminus \mathcal {Y}$. This forces $\chi$ to have at least one plane in $\mathcal {Y}$. Transitivity of $\chi$ under ${\textsf{Iso}}(\mathcal {P})$ implies that every two planes in $\chi$ are either parallel or perpendicular, and this forces the diagram of G to be (3). $\square$

Lemma 6.4

Let $\chi$ and G be as above. Then the Coxeter diagram of G is either (3) or (12).

Proof

Lemmas 6.2 and 6.3 eliminate all possible diagrams save (3), (10), (11) and (12). To discard the possibilities of (10) and (11) we note that the group with diagram (10) is the symmetry group of the cubic tiling, while that with diagram (11) is an index 2 subgroup of the latter. In both cases they have two conjugacy classes of plane reflections: those about planes parallel to the coordinate planes and those about planes that join two opposite edges of some cubes.

Let $\mathcal {Y}$ denote the set of reflection planes of the cubic tiling parallel to coordinate planes and $\mathcal {Z}$ the set of reflection planes of the cubic tiling that join two opposite edges of some cubes. In any parallel class of planes in $\mathcal {Y}$ the distance between two planes is an integer multiple of the edge length of the tiling; in contrast, in any parallel class of planes in $\mathcal {Z}$ the distance between two planes is $\sqrt{2}/2$ times an integer multiple of the edge length of the tiling. It follows that there is no isometry of $\mathbb {E}^3$ mapping an infinite subset of $\mathcal {Y}$ to a subset of $\mathcal {Z}$. As a consequence, $\chi$ must be contained either in $\mathcal {Y}$ or in $\mathcal {Z}$.

However, the reflections about the planes in $\mathcal {Y}$ generate the group with diagram (3), while those about the planes in $\mathcal {Z}$ generate the group with diagram (12). Furthermore, the Coxeter group with diagram (3) does not contain reflections about two planes at an angle of $\pi /3$, while that with diagram (12) does not contain three reflections about mutually perpendicular planes (and so each of the groups with these two diagrams does not contain subgroups isomorphic to the group with the other diagram). We can thus conclude that the diagram of G cannot be (10) or (11). $\square$

Following McMullen and Schulte (2002), we shall denote the Coxeter diagrams (3) and (12) by $W_2\times W_2\times W_2$ and $P_4$, respectively. The next lemma follows naturally.

Lemma 6.5

Let $\mathcal {P}$ be a 2-orbit polyhedron of type $2_{2}$ with base flag $\Phi$, and let $R_{2},A_{01}\in {\textsf{Iso}}(\mathcal {P})$ be the associated generators. If $R_{2}$ is a reflection, then the associated Coxeter group generated by $\langle TR_{2}T^{-1}\mid T\in {\textsf{Iso}}(\mathcal {P})\rangle$ is the affine Coxeter group $P_4$ or the reducible affine Coxeter group $W_2\times W_2\times W_2$.

If the subgroup G is $P_4$, then it is generated by reflections corresponding to the convex hull of two adjacent vertices of a cube in the cubical tiling $\{4,3,4\}$ and two centers of cubes containing that edge and sharing a common square face (see Fig. 10). On the other hand, if G is the group $W_2\times W_2\times W_2$, then the fundamental region is a right parallelepiped. In fact, the fundamental region is a cube. To see why, let C be a fundamental region bounded by $\Pi _{2}$ and let $\Lambda$ denote the other bounding plane of C corresponding to reflection in the same parallel class as $R_{2}$; denote the associated reflection by $R_{\Lambda }$. Since ${\textsf{Iso}}(\mathcal {P})$ acts transitively on the planes of $\chi$, there is an isometry $T\in {\textsf{Iso}}(\mathcal {P})$ taking $\Pi _{2}$ to the plane $\Delta$ bounding the fundamental region C corresponding to one of the remaining nodes of the Coxeter diagram. Let $R_{\Delta }$ be the associated reflection to the plane $\Delta$. Since $\Delta$ bounds both C and CT, there exists a translation S in G leaving $\Delta$ invariant, and mapping CT to either C or $CR_{\Delta }$, thus either TS or $TSR_{\Delta }$ maps C to itself; without loss of generality, we will assume the former. Observe now that TS is distance preserving, and so the distance between $\Pi _{2}$ and $\Lambda$ must be the same as the distance between $\Delta$ and the other bounding plane of C in its parallel class.

Observation 6.6

If G is the group $W_2\times W_2\times W_2$, then (every/the) fundamental region of G is a cube.

Fig. 10

A fundamental region corresponding to the group $P_{4}$. Fundamental regions corresponding to $W_{2}\times W_{2}\times W_{2}$ are just the cubes of the tiling $\{4,3,4\}$

In this section we wish to describe the possible facets of $\mathcal {P}$ when $A_{01}$ is a rotary reflection. Throughout the discussion below we will denote the rotational axis of $A_{01}$ as $\ell$, and refer to the fixed point of $A_{01}$ as its center c. Together c and $\ell$ determine the plane of reflection $\Pi _{01}$ of $A_{01}$, so that $A_{01}$ is the composition of a rotation about $\ell$ with $R_{01}$, the reflection through the plane $\Pi _{01}$. Since ${\textsf{Iso}}(\mathcal {P})=\langle R_{2},A_{01}\rangle$, there are some restrictions on how $A_{01}$ interacts with $\chi$, and in particular, this places limitations on the relative positions of the center c and the axis $\ell$ given the location of $\Pi _{2}$. For example:

Observation 6.7

If the center of $A_{01}$ is on the plane $\Pi _{2}$, then ${\textsf{Iso}}(\mathcal {P})=\langle R_{2},A_{01}\rangle$ is a finite group, a contradiction.

6.1 $G=W_2\times W_2\times W_2$

When $G=W_2\times W_2\times W_2$ our set of planes $\chi$ contains the planes corresponding to the faces of the cubes in the cubic tiling $\mathcal {T}$ of ${\mathbb {R}}^{3}$ with Schläfli type $\{4,3,4\}$. For $A_{01}$ to be a rotary reflection and send $\chi$ to itself, $A_{01}$ must be either a rotary reflection of order 4 with $\ell$ parallel to an edge of $\mathcal {T}$, or of order 6 with $\ell$ a diagonal of a cubicle facet of $\mathcal {T}$. If $\ell$ is parallel to an edge of $\mathcal {T}$, and $\Pi _{2}$ contains a line parallel to $\ell$, then the plane $\Pi _{01}$ is invariant under the action of $R_{2}$ (and $A_{01}$ itself), so every element of $\langle R_{2},A_{01}\rangle$ contains lines parallel to $\ell$, a contradiction to Lemma 6.1. If on the other hand, $\Pi _{2}$ is perpendicular to $\ell$, then $\ell$ is preserved by both generators of ${\textsf{Iso}}(\mathcal {P})$, again contradicting Lemma 6.1. Hence, $A_{01}$ must be an order 6 rotary reflection.

Consider now the planes in $\xi =\{\Pi _{2}A_{01}^{k}|0\le k<6\}$, corresponding to the conjugates of $R_{2}$ by $A_{01}$. Let $x=\ell \cap \Pi _{2}$; then $xA_{01}$ is on $\ell$ and on the opposite side from c as x, but (naturally) $d(x,c)=d(xA_{01},c)$; the planes corresponding to even powers k will pass through x and the odd powers will pass through $xA_{01}$. Together, the planes of $\xi$ bound a cube C. This cube C induces a tiling $\mathcal {T}$ of ${\mathbb {R}}^{3}$ by cubes (see Fig. 11). By construction, this tiling is preserved by the actions of $A_{01}$ and $R_{2}$, and so is preserved by ${\textsf{Iso}}(\mathcal {P})$, so this choice for $A_{01}$ is admissible. In fact, any symmetry mapping a flag to its 2-adjacent neighbor must be a conjugate of $R_{2}$, and so all of the planes bounding the cells of $\mathcal {T}$ correspond to such symmetries. Since $R_{121}=R_{2}^{A_{01}}$, $\Pi _{121}=\Pi _{2}A_{01}^{-1}$ and $v_{0}\in \Pi _{2}\cap \Pi _{121}$. So, by definition, $v_{0}$ is going to lie on one of the lines determined by the edges of C that don’t intersect $\ell$, since the edges of C in $\Pi _{2}$ that do intersect $\ell$ do so at x, which is not contained in $\Pi _{121}$. Moreover, $R_{2}$ and $A_{121}$ generate the stabilizer of $v_{0}$, but these reflections correspond to perpendicular fixed planes and so $|\langle R_{2}, A_{121}\rangle |=4$, and there are exactly four faces around each edge, so the Schläfli type of $\mathcal {P}$ must be $\{6,4\}$. A special case of note is when $v_{0}$ is a midpoint of one of these edges, in which case the facet $f_{0}$ is a planar hexagon and $\mathcal {P}$ is the Petrie–Coxeter polyhedron $\{6,4|4\}$. Since the extensions of the edges of C that do not intersect $\ell$ also don’t intersect any of the axes of any of the other rotary reflections, $v_{0}$ may be positioned anywhere along the line of intersection of $\Pi _{2}$ and $\Pi _{121}$ as long as it doesn’t coincide with any vertex of a non-incident facet. This is accomplished by not placing $v_{0}$ at any of the vertices of $\mathcal {T}$; for example, if $v_{0}$ is a vertex of C, then the vertex-figure of $v_{0}$ is not a polygon; it is instead a square with doubled edges (arguably, two distinct squares sharing the same set of vertices; recall the first footnote regarding the definition of “polygon”). The polyhedra just described are precisely the polyhedra explained in Sect. 4.

Fig. 11

A portion of the tiling of ${\mathbb {R}}^{3}$ by cubes, with the green columns of cubes in front of the red, which are in turn in front of the blue. Black and gray dots correspond to intersection points of the lines preserved by the rotary reflections of order 6 for $\mathcal {P}$ with the corners of the cubes. The edges outlined in black correspond to the set of edges of a fundamental region C of $W_{2}\times W_{2}\times W_{2}$ that do not intersect the axis of rotary reflection associated with C (the dotted black line). Note that the lines extending these axes do not intersect any of the lines corresponding to the axes of rotation of the rotary reflections

6.2 $G=P_4$

When $G=P_{4}$ the planes in $\chi$ correspond to planes diagonally situated in the cubic tiling $\mathcal {T}$, with the fundamental regions so formed determined by the convex hull of adjacent vertices of $\mathcal {T}$ and the centers of two adjacent cubes containing the edge determined by them (see Fig. 10). Such a fundamental region $\mathcal {F}$ has two edges where the planes meet at a right angle, and the remaining four edges correspond to planes meeting at an angle of $\pi /3$; we will find it convenient to call the distinguished pair of edges of $\mathcal {F}$ blue edges, and the other four edges yellow edges.³ Let $\mathcal {S}$ denote the tiling by the fundamental regions of $P_{4}$, which we will often find it convenient to think of as a subdivision of the tiling $\mathcal {T}$. Since $v_{0}$ must lie at the intersection of the planes $\Pi _{2}$ corresponding to the generating reflection $R_{2}$, and the reflecting plane $\Pi _{121}$ corresponding to the reflection $A_{121}=R_{2}^{A_{01}}$, $v_{0}$ is at the intersection of two planes in $\chi$. As in Sect. 6.1, the only rotary reflections preserving the set of planes $\chi$ associated with $P_4$ are either (1) a four-fold rotary reflection around an axis of rotation parallel to the edges of the cubes, or (2) a six-fold rotary reflection around a diagonal of the cube. The centers in case (1) may lie at centers of cubical cells of $\mathcal {T}$ or (equivalently) vertices of $\mathcal {T}$ (these are the vertices of the tiling $\mathcal {S}$), the midpoint of a blue edge, or at the midpoint of a line connecting the midpoints of the blue edges of a region $\mathcal {F}$. The centers in case (2) have to lie on the boundary of $\mathcal {F}$, since there is no symmetry of $\mathcal {F}$ of order 6. Hence the centers in case (2) may lie at the vertices of $\mathcal {S}$ or the midpoints of yellow edges. Consequently, the facets of $\mathcal {P}$ will be either squares or hexagons, with vertices lying on the intersections of the planes in $\chi$. Note that for $P_{4}$ the planes in $\chi$ intersect in more than one way, with some of the lines of intersection involving only two planes meeting at blue edges in right angles, while other lines of intersection involve three planes, which meet along the diagonals of the cubicle cells of $\mathcal {T}$ (see Fig. 12) on yellow edges.

Fig. 12

On the left: Six copies of the fundamental region associated with $P_4$ under the action of a six-fold rotary reflection along a diagonal of the cubic lattice. On the right: Four copies of the fundamental region associated with $P_{4}$ under the action of a four-fold rotary reflection along a line parallel to an edge of the cubical lattice. The axis of rotation is shown in red, and the center is the black dot (the center of the rotary reflection could be either at a cube center as shown above, or at a vertex of the cube)

For the rest of this analysis we will assume that $\mathcal {F}$ is a fundamental region bounded by $\Pi _{2}$ and that $c\in \mathcal {F}$ (though possibly on the boundary). Note that if the rotocenter c of $A_{01}$ is a vertex of $\mathcal {F}$, then $c\not \in \Pi _{2}$, else $\langle R_{2},A_{01}\rangle$ fixes c, making ${\textsf{Iso}}(\mathcal {P})$ finite, a contradiction. Thus $\Pi _{2}$ must lie on a facet of $\mathcal {F}$ not containing c.

As in Sect. 6.1, our goal is to determine which choices of relative position of the rotocenter c, rotation axis $\ell$, and plane of reflection $\Pi _{2}$ with respect to the fundamental region $\mathcal {F}$ are admissible; in other words, for which of these choices for defining $A_{01}$ and $R_{2}$ is $\Pi _{2}\langle A_{01},R_{2}\rangle =\chi$. An important simplifying observation is, as we shall see, that some of these choices correspond to rescalings of other choices, and so will not generate all of $\chi$.

Consider first the case when $|A_{01}|=4$.

If $|A_{01}|=4$ and the center c of the rotary reflection is the midpoint of the line connecting the midpoints of two blue edges of $\mathcal {F}$, then the action of $A_{01}$ will send $\Pi _{2}$ to each of the other bounding planes of $\mathcal {F}$, and so this choice of $A_{01}$ is admissible. This is easily seen since this choice of $A_{01}$ permutes the four vertices of $\mathcal {F}$, and so permutes the planes determined by all but one of the four vertices of $\mathcal {F}$. The action of $A_{01}$ on the edges of $\mathcal {F}$ only has order 4 if $v_{0}$ is on a yellow edge, and there are six fundamental regions around each such edge, so for reasons analogous to those in Sect. 6.1, $\mathcal {P}$ is of type $\{4,6\}$. In fact, if the point $v_{0}$ is chosen to be a midpoint of a yellow edge, then $\mathcal {P}$ is the geometric Petrie–Coxeter polyhedron $\{4,6|4\}$ (see Fig. 13). Since all polyhedra obtained from this construction are geometric realizations of the abstract polyhedron $\{4,6|4\}$, when contemplating other possible locations of $v_{0}$ on $\Pi _{2}\cap \Pi _{01}$ it is again necessary to avoid placement where $v_{0}$ would coincide geometrically with a vertex of a non-incident facet of the abstract polyhedron. Consequently, $v_{0}$ may lie anywhere along $\Pi _{2}\cap \Pi _{01}$ except at a vertex or center of a cell C of $\mathcal {T}$. When $v_0$ is not a midpoint of a yellow edge of $\mathcal {F}$ then the faces of $\mathcal {P}$ are skew and $\mathcal {P}$ is in class $2_2$. The verification of these claims is analogous to that shown in Sect. 4 for the distinct realizations of $\{6,4|4\}$.

Fig. 13

A portion of the Petrie–Coxeter polyhedron $\{4,6|4\}$

Now suppose we reposition the center c to lie either at the midpoint of a blue edge, or at a vertex of $\mathcal {F}$, but keep $\ell$ parallel to the line connecting the midpoints of the two blue edges of $\mathcal {F}$. In either case the choice of $\Pi _{2}$ must be a face not containing c, but the angle of inclination of $\ell$ to $\Pi _{2}$ will be the same as in the previous admissible case. However, the fundamental region formed by $\Pi _{2}\langle A_{01},R_{2}\rangle$ will be either twice or four times as long, and so $\Pi _{2}\langle A_{01},R_{2}\rangle \ne \chi$. Thus, these are not admissible cases.

Suppose now that we orient $\ell$ so that it runs along a blue edge of $\mathcal {F}$. Then c may lie at either the midpoint of the edge or at a vertex. There are two equivalent choices for facet of $\mathcal {F}$ corresponding to $\Pi _{2}$ in the former case, and only one in the latter case. Inspection of the former case reveals that the images of $\Pi _{2}$ under the action of $A_{01}$ bound a region $\mathcal {R}$ homothetic to $\mathcal {F}$ with edges twice as long. By similarity, if c is a vertex of $\mathcal {F}$, the images of $\Pi _{2}$ will bound a region homothetic to $\mathcal {F}$ with edges four times as long. In either case $\Pi _{2}\langle A_{01},R_{2}\rangle$ is a strict subset of $\chi$.

The remaining possibilities involve c on a blue edge of $\mathcal {F}$, and $\ell$ parallel to the other blue edge (and so perpendicular to the choices for $\ell$ in both of the previously considered cases, see Fig. 14). Again, $\Pi _{2}$ is chosen to coincide with a face of $\mathcal {F}$ not incident with c. Here the line $\ell$ is parallel to all four images of $\Pi _{2}$, hence $R_{2}$ preserves the plane of reflection of $A_{01}$, and so then does all of ${\textsf{Iso}}(\mathcal {P})$. Thus all images of $\ell$ and $\Pi _{2}$ are parallel, another contradiction since this would mean that ${\textsf{Iso}}(\mathcal {P})$ would not act transitively on $\chi$.

Fig. 14

The three possibilities for the action of $A_{01}$ on a fundamental region of a cell of $P_{4}$ where the center of the rotary reflection is on a vertex of the fundamental region. On the left the axis of rotation is passing along the edge of the region containing the rotocenter, in the middle it is parallel to a line connecting the two blue edges of the fundamental region, and on the right it is parallel to the other blue edge. In each case, the plane $\Pi _{2}$ lies entirely outside the indicated cube of $\mathcal {T}$

Fig. 15

Here we have two different possible inclinations of a fundamental region to an axis of 6-fold rotary reflection (in red). On the left the fundamental region shares an edge with the axis of rotation, and on the right it does not. Up to enantiomorphism, these are the only two options. The plane of reflection for the rotary reflection is indicated by the outline of a disc in the plane. Planes of reflection in both cases are not incident to the rotocenter (in black)

We now consider the possibilities when $|A_{01}|=6$. Recall that in this case $\ell$ is a diagonal of the cube and c is on the boundary of $\mathcal {F}$, and up to enantiomorphism, there are essentially only two orientations of $\ell$ with respect to $\mathcal {F}$ (see Fig. 15).

We first consider the case when $\ell$ is a yellow edge of $\mathcal {F}$ and c is the midpoint of that edge. There are two equivalent choices for the plane $\Pi _{2}$, namely either of the faces of $\mathcal {F}$ not incident with c. Here we observe that the remaining faces of $\mathcal {F}$ are bounded by the planes obtained as images of $\Pi _{2}$ under the action of $A_{01}^{5},A_{01}^{4} R_{2}$ and $A_{01}^{2}R_{2}A_{01}^{2}$. Hence this is an admissible choice. The vertex $v_{0}$ must then lie on the line determined by the yellow edge $\hat{e}$ of $\mathcal {F}$ not incident with $\ell$, and there are six such cells around this edge, so for reasons analogous to those in Sect. 6.1, $\mathcal {P}$ is of type $\{6,6\}$. In fact, if $v_{0}$ is chosen to be the midpoint of $\hat{e}$, then $f_{0}$ is a planar hexagon and we obtain the Petrie–Coxeter polyhedron $\{6,6|3\}$ (see Fig. 16). As in the previous cases, the vertex $v_0$ must lie in the line spanned by $\hat{e}$, but not in any vertex or cube center of $\mathcal {T}$, since otherwise the stabilizer of $v_0$ in ${\textsf{Iso}}(\mathcal {P})$ is too large. For any choice of $v_0$ that yields a polyhedron, other than the midpoint of $\hat{e}$, the faces of $\mathcal {P}$ are skew and $\mathcal {P}$ is in class $2_2$. The verification of these claims is once more analogous to that shown in Sect. 4 for the distinct realizations of $\{6,4|4\}$.

Fig. 16

A portion of the polyhedron $\{6,6|3\}$

If we proceed as in the admissible case above, but instead position c at a vertex of $\mathcal {F}$, then there is now only one possible choice for $\Pi _{2}$, and $\Pi _{2}\langle A_{01}, R_{2}\rangle$ will bound a region $\mathcal {R}$ homothetic to $\mathcal {F}$ with edges twice as long, and so $\Pi _{2}\langle A_{01}, R_{2}\rangle \ne \chi$; hence this is not an admissible choice.

The remaining possibility is that c is a vertex of $\mathcal {F}$ and that $\ell$ is only incident to $\mathcal {F}$ at c. Again $\Pi _{2}$ must correspond to the only face of $\mathcal {F}$ that is not incident to c. Here the line $\ell$ is parallel to all six images of $\Pi _{2}$, hence $R_{2}$ preserves the plane of reflection of $A_{01}$, and so then does all of ${\textsf{Iso}}(\mathcal {P})$, the same contradiction we saw in the $|A_{01}|=4$ case.

We summarize the discussion of this section in the following theorem.

Theorem 6.8

Let $\mathcal {P}$ be a polyhedron in class $2_2$. Then it is (abstractly) isomorphic to one of the geometric regular polyhedra $\{6,4|4\}$, $\{4,6|4\}$ or $\{6,6|3\}$. Furthermore, every polyhedron in class $2_2$ isomorphic to a polyhedron $\mathcal {Q}$ among the three aforementioned can be obtained by Wythoff’s construction with respect to the subgroup $\langle A_{01},R_2 \rangle \le {\textsf{Iso}}(\mathcal {Q})$ for some specific choice of base vertex. For each $\mathcal {Q}$ the set $\mathcal {V}$ of valid base vertices to construct polyhedra in class $2_2$ consists of a union of open intervals in a line h. One point in $h\setminus \mathcal {V}$ is the one with which we recover $\mathcal {Q}$, while the structure constructed from any of the remaining points of $h\setminus \mathcal {V}$ by Wythoff’s construction has vertex-figures that are not simple polygons.

7 Two orbit polyhedra derived from Petrie duality

Some of the discussion in Sect. 6 can be used to analyze the case when $R_2$ is a plane reflection and $A_{01}$ is a screw motion. In particular, the orbit under ${\textsf{Iso}}(\mathcal {P})$ of the mirror $\Pi _2$ of $R_2$ must generate one of the Coxeter groups $W_2 \times W_2 \times W_2$ or $P_4$. However, using Petrie duality the enumeration of the polyhedra when $A_{01}$ is a screw motion can be derived from that when it is instead a rotary reflection.

Proposition 7.1

Let $\mathcal {P}$ be a polyhedron in class $2_2$ in $\mathbb {E}^3$. Then $\mathcal {P}^\pi$ is also a polyhedron in class $2_2$.

Proof

Lemma 3.8 will prove the proposition as long as we show that

• 1. if $\mathcal {P}$ is a polyhedron in class $2_2$ then its Petrie paths have at least 3 vertices,

• 2. for each Petrie path $\pi$ there is a symmetry acting like a 1-step rotation on $\pi$,

• 3. the symmetry type of the Petrial of a polyhedron in class $2_2$ is also in class $2_2$.

By definition, two consecutive edges of a Petrie path of a polyhedron $\mathcal {P}$ are two consecutive edges of a facet of $\mathcal {P}$. It follows that two consecutive edges of a Petrie path must involve precisely three vertices, and the length of each Petrie path is at least 3. This settles the first item.

Let $\mathcal {P}$ be in class $2_2$ and let $R_2$ and $A_{01}$ be the standard generators of ${\textsf{Iso}}(\mathcal {P})$. We claim that $R_2 A_{01}$ acts like a 1-step rotation on a Petrie path of $\mathcal {P}$. If $\Phi$ is the base flag then

$$\begin{aligned} \Phi R_2 A_{01} = r_2 \Phi A_{01}=r_2 r_1 r_0 \Phi . \end{aligned}$$

The application of the connection $r_2 r_1 r_0$ changes the vertex, then changes the edge, and finally it changes the facet. That means that starting on a flag $\Psi$ on a given edge e in a facet f, the flag $r_2 r_1 r_0 \Psi$ is in an edge $e'$ of f that contains the vertex of e not in $\Psi$, but $r_2 r_1 r_0 \Psi$ itself is in the facet different from f that contains $e'$. This property is preserved under symmetries, and therefore the edges of $\Phi (R_2 A_{01})^k$ and of $\Phi (R_2 A_{01})^{k+1}$ belong to a common facet $f_k$. On the other hand, the edge of $\Phi (R_2 A_{01})^{k+2}$ belongs to the facet $f_{k+1}$ that contains the edge of $\Phi (R_2 A_{01})^{k+1}$ other than $f_k$. It follows that $A_{01} R_2$ preserves that particular Petrie path $\pi$, and it behaves like a symmetry that rotates one step along it. Clearly, $R_2$ interchanges the two Petrie paths at the base edge. Since polyhedra in class $2_2$ are edge transitive, we can conclude that ${\textsf{Iso}}(\mathcal {P})$ acts transitively on Petrie paths. The conjugate $T^{-1} R_2 A_{01} T$ of $R_2 A_{01}$ will serve as 1-step rotation about the Petrie path $\pi T$, showing that for each Petrie path there is a symmetry acting like a 1-step rotation on it.

The last item was shown in Orbanić et al. (2010, Section 3). $\square$

In view of Proposition 7.1, the Petrie duals of the polyhedra described in Sect. 6 are also 2-orbit polyhedra in class $2_2$. The standard generators of the symmetry group of the latter are discussed next.

Lemma 7.2

Let $\mathcal {P}$ be a polyhedron in class $2_2$ and let $A_{01}$ and $R_2$ be defined as in Sect. 2. Then there exists a flag $\Phi$ of $\mathcal {P}^\pi$ such that $\Phi R_2 = \Phi ^2$ and $\Phi R_2 A_{01} = \Phi ^{01}$.

Proof

Let $\Psi = \{v_0, e_0, f_0\}$ be the base flag of $\mathcal {P}$, so that $\Psi R_2 = \Psi ^2$ and $\Psi A_{01} = \Psi ^{01}$, and let $\pi$ be the Petrie path of $\mathcal {P}$ that has the two edges $e_{-1}$ and $e_0$ incident to $v_0$ in $f_0$ as consecutive edges. We define the flag $\Phi$ of $\mathcal {P}^\pi$ as $\{v_0, e_0, \pi \}$.

When applying $R_2$ to $\pi$, we obtain a Petrie path $\pi '$ of $\mathcal {P}$ that still contains $v_0$ and $e_0$ (both are fixed by $R_2$), but not $e_{-1}$. To justify this we use the facts that every vertex is incident to at least 3 edges (this is a consequence of Axiom (P4)), and that $R_2$ must interchange $e_{-1}$ with the edge $e_{-1}'$ of $\pi$ at $v_0$ distinct to $e_0$ (see Fig. 17). It follows that $\Phi R_2 = \Phi ^2$.

Fig. 17

$R_2$ does not fix the Petrie path $\pi$

Since $R_2 A_{01} \in {\textsf{Iso}}(\mathcal {P})$, we have that $\pi R_2 A_{01}$ is a Petrie path of $\mathcal {P}$. Furthermore,

$$\begin{aligned} e_{-1}' R_2 A_{01}&= e_{-1} A_{01} = e_0,\\ e_0 R_2 A_{01}&= e_0 A_{01} = e_1, \\ v_0 R_2 A_{01}&= v_0 A_{01} = v_1, \end{aligned}$$

and hence $\pi R_2 A_{01} = \pi$ (we keep the notation of Sect. 5 and so $e_1$ is the edge in $f_0 \cap \pi$ other than $e_0$). From here it is immediate that $\Phi R_2 A_{01} = \Phi ^{01}$. $\square$

We are ready for the main result of this section.

Theorem 7.3

Let $\mathcal {P}$ be a 2-orbit polyhedron in class $2_2$. The standard generator $A_{01}$ of ${\textsf{Iso}}(\mathcal {P})$ mapping the base flag $\Phi$ to $\Phi ^{01}$ is a rotary reflection if and only if the standard generator $A_{01}'$ of ${\textsf{Iso}}(\mathcal {P}^\pi )$ mapping the base flag $\Psi$ to $\Psi ^{01}$ is a screw motion.

Proof

We already know from Proposition 7.1 that $\mathcal {P}$ is a polyhedron in class $2_2$ if and only if $\mathcal {P}^\pi$ is a polyhedron in class $2_2$. From Lemma 7.2 we conclude that $A_{01}' = R_2 A_{01}$, where $R_2$ is the standard generator mapping $\Phi$ to $\Phi ^2$ (and also $\Psi$ to $\Psi ^2$). Furthermore, in Sect. 5 it was shown that if $\mathcal {P}$ is a polyhedron in class $2_2$ then the symmetry $R_2$ is a plane reflection.

If $A_{01}'$ is a screw motion, then $A_{01} = R_2 A_{01}'$ must reverse orientation since $A_{01}'$ preserves orientation and $R_2$ reverses it. Similarly, if $A_{01}$ is a rotary reflection then $A_{01}' = R_2 A_{01}$ must preserve orientation. The analysis in Sect. 5 forces $A_{01}$ to be a rotary reflection in the first case and $A_{01}'$ a screw motion in the second case. $\square$

Remark 1

Theorem 7.3 can be proven without the analysis in Sect. 5 by using specific features of $A_{01}$, $A_{01}'$ and $R_2$. For example, to discard the possibility of $A_{01}$ being a glide reflection we must use that the edges of the base facet $f_0$ of $\mathcal {P}^\pi$ are neither parallel nor perpendicular to the axis of the screw motion $A_{01}'$, and that the mirror of $R_2$ contains one of the edges of $f_0$. It may be also necessary to show that if the mirror of $R_2$ contains the direction vector of the axis of $A_{01}'$ then the resulting structure is regular (which can be done by using Lemma 3.5).

Putting together Lemma 7.2 and Theorem 7.3 we get the following result.

Corollary 7.4

Let $\mathcal {P}$ be a 2-orbit polyhedron in class $2_2$ such that the symmetry $A_{01}$ is a screw motion. Then $\mathcal {P}$ is the Petrial of one of the polyhedra described in Sects. 4 and 6. In particular, it is combinatorially isomorphic to either $\{4,6|4\}^\pi$, $\{6,4|4\}^\pi$ or $\{6,6|3\}^\pi$.

In view of Corollary 7.4, each polyhedron $\mathcal {P}$ in class $2_2$ where $A_{01}$ is a screw motion is a modification of either $\{4,6|4\}^\pi$, $\{6,4|4\}^\pi$ or $\{6,6|3\}^\pi$, and is determined by the choice of base vertex $v_0$. The facets of $\mathcal {P}$ are helices, and using Lemma 3.7 we know that they are regular polygons; furthermore, they have the same axes as the corresponding regular polyhedron. When moving $v_0$ continuously the helices rotate around their axes with some translation factor along the direction of the axes; furthermore, the width of the helices increases or decreases. Figure 18 illustrates this phenomenon for the Petrie duals of the polyhedra with type $\{6,4|4\}$ with the choices of vertices (0, 0, 1) (the regular instance), (0, 0, 2/3) and (0, 0, 1/6).

Fig. 18

Movement of a helix from the regular choice $v_0 = (0,0,1)$ to the 2-orbit choices $v_0 = (0,0,2/3)$ and $v_0 = (0,0,1/6)$

8 The class $2_{0}$

The classification of 2-orbit polyhedra in class $2_0$ can be derived from Petrie duality together with the following result.

Proposition 8.1

The Petrial of every 2-orbit polyhedron in class $2_0$ is a polyhedron and it is chiral.

Proof

We proceed as in the proof of Proposition 7.1. The arguments are the same, just replacing $R_2 A_{01}$ by $A_{12} R_0$, so that $\Phi A_{12} R_0 = r_2 r_1 r_0 \Phi$, and using that $R_0$ interchanges the two Petrie paths through the base edge.

The fact that $\mathcal {P}^\pi$ is chiral was also shown in Orbanić et al. (2010, Section 3). $\square$

In view of Proposition 8.1, all 2-orbit polyhedra in class $2_0$ can be obtained by taking the Petrials of all the chiral polyhedra in $\mathbb {E}^3$. In what follows we describe the necessary aspects of the classification of chiral polyhedra in Schulte (2004, 2005).

All chiral polyhedra are infinite and are divided into 6 families. The polyhedra in each family have the same Schläfli type, and the generators of the stabilizers of the base vertex and the base facet are the same kind of isometry and have the same order. Each family is indexed with two parameters, although multiplication of both parameters by equal non-zero real numbers yields similar polyhedra. It follows that up to similarity there is only one parameter needed to index each family: a quotient of the two original parameters. Polyhedra in the last three families have infinite facets; they are helices. Table 1 summarizes the names, types and parameters of the six families. For the last three families, it also indicates the polygon over which the helical facets are constructed.

Table 1 Families of chiral polyhedra

In the column of parameters of Table 1 we forbid 1 and $-1$ for the first and fourth families, and 0 for the other four families. These parameters correspond to regular polyhedra related to each of the families. Admitting non-rational quotients as the parameters the first three families gives rise to non-discrete structures. When mentioning families of chiral polyhedra we shall assume that the parameters are only those in Table 1.

According to Schulte (2004, Theorems 5.18, 6.13), the chiral polyhedra in each of the first three families include infinitely many polyhedra that are mutually non-isomorphic. In fact, if $a, a', b, b', c, c', d, d' \in \mathbb {Z}$ are such that $a,c \ge 0$, a is coprime to b, $a'$ is coprime to $b'$, c is coprime to d and $c'$ is coprime to $d'$ then P(a, b) is combinatorially isomorphic to $P(a',b')$ if and only if $(a',b') \in \{(a,b),(b,a)\}$, whereas Q(c, d) (resp. $Q(c,d)^*$) is combinatorially isomorphic to $Q(c',d')$ (resp. $Q(c',d')^*$) if and only if $(c',d') \in \{(c,d),(-c,d)\}$. It was later shown in Pellicer and Ivić Weiss (2010) that all geometrically chiral polyhedra with finite facets are combinatorially chiral as well.

The story of the chiral polyhedra in the last three families is quite different to that of the polyhedra in the first three families. Within the same family, they are all combinatorially isomorphic, and they are all combinatorially regular. The chiral polyhedra $P_1(a,b)$ and $P_2(c,d)$ are respectively isomorphic to the regular polyhedra called $\{\infty ,3\}^{(a)}$ and $\{\infty ,3\}^{(b)}$ in McMullen and Schulte (1997). On the other hand, the chiral polyhedra $P_3(c,d)$ are isomorphic to a double cover of the polyhedron denoted by $\{\infty ,4\}_{\cdot ,*3}$ in McMullen and Schulte (1997) (see Pellicer and Ivić Weiss 2010).

Lemma 8.2

The Petrie dual of every chiral polyhedron is a polyhedron in class $2_0$.

Proof

This follows the lines of Lemma 3.8. The only difference is that if $\mathcal {P}$ is a chiral polyhedron then there is no symmetry of $\mathcal {P}$ fixing a Petrie path and acting like a 1-step rotation about it. With that in mind, it is enough to verify that the Petrie paths of every chiral polyhedron are polygons.

It was pointed out in McMullen and Schulte (2002, Section 7E) that the polyhedra $\{\infty ,3\}^{(a)}$ and $\{\infty ,3\}^{(b)}$ are Petrials of each other. This means that the Petrie paths of one of them are the facets of the other one, and therefore they are polygons. Since the polyhedra $P_1(a,b)$ and $P_2(c,d)$ are isomorphic to these two regular polyhedra, their Petrie paths are polygons.

For the polyhedra $\mathcal {P}$ of the remaining families we shall use that $G(\mathcal {P}) = \langle A_{01}, A_{12} \rangle$ (corresponding in Schulte 2004, 2005 to $S_1^{-1}$ and $S_2^{-1}$, respectively). If $\Phi$ is the base flag of $\mathcal {P}$ then

$$\begin{aligned} \Phi A_{01}^{-2} A_{12}^{-2}= & {} \Phi ^{10} A_{01}^{-1} A_{12}^{-2} = (\Phi A_{01}^{-1} A_{12}^{-2})^{10} = (\Phi A_{12}^{-2})^{1010} \\= & {} \Phi ^{21211010} = \Phi ^{212010} = \Phi ^{210210}. \end{aligned}$$

The discussion in the proof of Proposition 7.1 about the Petrie paths implies that $A_{01}^{-2} A_{12}^{-2}$ acts like a 2-step rotation along a Petrie walk of $\mathcal {P}$.

Using the descriptions of $A_{01}$ and $A_{12}$ in Schulte (2004, pp. 66, 84) and (2005, pp. 214, 215) we have that the polyhedra P(a, b), Q(c, d) and $P_3(c,d)$ have base vertex at (0, 0, 0), and that

$$\begin{aligned} (x,y,z) A_{01}^{-2} A_{12}^{-2} = \left\{ \begin{array}{ll} (x,-y,-z) + (-a-b, -a, b) &{}\quad \text{ if } \mathcal {P}\hbox { is }P(a,b),\\ (-y,-z,x) + (-c-d,-c+d,0) &{}\quad \text{ if } \mathcal {P}\hbox { is }Q(c,d),\\ (-y,z,-x) + (-c+d, -c-d, 0) &{}\quad \text{ if } \mathcal {P}\hbox { is }P_3(c,d).\\ \end{array}\right. \end{aligned}$$

When $\mathcal {P}$ is P(a, b), the symmetry $A_{01}^{-2} A_{12}^{-2}$ is a screw motion with 2-fold rotational component around the axis

$$\begin{aligned} \frac{1}{2}(0,-a,b) + \{\lambda (a+b,0,0) \, : \, \lambda \in \mathbb {R}\}, \end{aligned}$$

and translation component by vector $(-a-b,0,0)$. One of the neighbors of (0, 0, 0) in the base Petrie polygon is $(0,0,0)A_{01} = (a,0,b)$.

In the second case, $A_{01}^{-2} A_{12}^{-2}$ is a screw motion with 3-fold rotational component around the axis

$$\begin{aligned} \frac{1}{3}(-2d,-3c+d,0) + \{\lambda (-2d,2d,-2d) \, : \, \lambda \in \mathbb {R}\}, \end{aligned}$$

and translation component by vector $\frac{1}{3}(-2d,2d,-2d)$. One of the neighbors of (0, 0, 0) in the base Petrie polygon is $(0,0,0)A_{01} = (c,-c,d)$.

In the final case, $A_{01}^{-2} A_{12}^{-2}$ is a screw motion with 3-fold rotational component around the axis

$$\begin{aligned} \frac{1}{3}(2d,-3c-d,0) + \{\lambda (2d,-2d,-2d) \, : \, \lambda \in \mathbb {R}\}, \end{aligned}$$

and translation component by vector $\frac{1}{3}(2d,-2d,-2d)$. One of the neighbors of (0, 0, 0) in the base Petrie polygon is $(0,0,0)A_{01} = (c,-c,d)$.

It is routine to show that in the three cases $(0,0,0)A_{01}$ does not belong to $(0,0,0)\langle A_{01}^{-2} A_{12}^{-2}\rangle$. It follows that $A_{01}^{-2} A_{12}^{-2}$ induces two orbits on the vertices of the Petrie path. Hence, this path is a polygon. Since chiral polyhedra are edge-transitive and $A_{01}$ maps a Petrie path containing the base edge $e_0$ to the other Petrie path through $e_0$, we can conclude that chiral polyhedra are transitive on Petrie paths. This shows that in the three cases above, all Petrie paths are polygons.

It only remains to verify that the Petrie paths of $Q(c,d)^*$ are also polygons. As pointed out in Schulte (2004, p. 93), the base vertex of $Q(c,d)^*$ is $v^*=\frac{1}{2}(c,-c-d,-c+d)$, and the polyhedron can be constructed by Wythoff’s construction with respect to the generators $A_{12}$ and $A_{01}$ ($S_2^{-1}$ and $S_1^{-1}$, respectively, in the notation of Schulte (2004, 2005)). Then one of the Petrie paths has the edge between $v^*$ and $v^*A_{12}^{-1} = \frac{1}{2}(c-d,-c,c+d)$, and its vertex set consists of the orbits of these two vertices under $\langle A_{12}^2 A_{01}^2 \rangle = \langle A_{01}^{-2} A_{12}^{-2} \rangle$. A standard procedure can be used to verify that $v^*A_{12}^{-1} \notin v^*\langle A_{01}^{-2} A_{12}^{-2} \rangle$, implying that for the parameters in Table 1 the Petrie paths of $Q(c,d)^*$ are polygons as well. $\square$

As a direct consequence of Proposition 8.1 and Lemma 8.2 we have the following result.

Theorem 8.3

Every 2-orbit polyhedron in class $2_0$ is the Petrial of a chiral polyhedron in $\mathbb {E}^ 3$. Furthermore, the Petrial of every chiral polyhedron is a polyhedron in class $2_0$.

The facets of any 2-orbit polyhedron in class $2_0$ are helices of some sort. In general they are not regular. For instance, the base facet of $P(a,b)^\pi$ is a helix over the rectangle in the leftmost portion of Fig. 19. When a/b approaches to 1 the rectangle tends to a square, while if $b=0$ then it degenerates to a line segment (later we shall present another poyhedron whose facets are helices over such a base). The edges of the facets have all the same length, but they are inclined at two distinct angles with respect to the axis, so that they project into segments of distinct lengths. In particular, when $b=0$ then there are edges of the facet $f_0$ that are parallel to the axis of $f_0$ and edges that intersect it. The facets of $P_1(a,b)^\pi$ show a similar behavior.

Fig. 19

Bases of the helical facets of the Petrials of chiral polyhedra

For the remaining four families of polyhedra in class $2_0$ the facets are helices over hexagons. The projection of such helices to a plane perpendicular to the axis may look like the convex 2-orbit hexagon in the rightmost part of Fig. 19 or the star-shaped 2-orbit hexagon in the center part of Fig. 19, depending on the parameters c, d. A continuous change of these parameters induces a continuous change on the proportion between the two kinds of edges in the convex hull of the hexagon in the projection of the base facet.

It is beyond the scope of this paper to fully describe all polyhedra of type $2_0$. The geometric details of the chiral polyhedra are explained in Schulte (2004, 2005), and with that information the Petrial of each individual polyhedron can be derived directly.

We conclude this section with the description of the polyhedron $P_1(1,0)^\pi$. The geometric aspects of $P_1(1,0)$ were discussed in Pellicer (2017). Here we describe the 1-skeleton in another way that better suits our purposes.

The vertex set of $P_1(1,0)^\pi$ is that of the cubic tiling, say with vertex set $\mathbb {Z}^3$. The set of edges of $P_1(1,0)^\pi$ splits into three parallel classes:

• Edges parallel to the x-axis with vertices (x, y, z) and $(x+1,y,z)$ whenever x and z have the same parity.

• Edges parallel to the y-axis with vertices (x, y, z) and $(x,y+1,z)$ whenever x and y have the same parity.

• Edges parallel to the z-axis with vertices (x, y, z) and $(x,y,z+1)$ whenever y and z have the same parity.

The edges between the planes $z=0$ and $z=1$ for $0 \le x, y \le 3$ are shown in the left of Fig. 20. Similar layers of edges with respect to planes perpendicular to the y- and x-axes are shown in the center and right of Fig. 20, respectively.

Fig. 20

Edges of $P_1(1,0)^\pi$ in three layers with respect to the coordinate planes

When considering only the edge in the yz-plane we obtain infinitely many polygons, as shown in Fig. 21. All of them are translates by vectors in $\{(0,2m,0) \, : \, m \in \mathbb {Z}\}$. They are vertex-transitive 2-orbit polygons; the symmetry group of each one of them generated by a reflection through the bisector of a vertical edge $e_h$, and a half-turn with respect to the midpoint of a horizontal edge that shares a vertex with $e_h$. Clearly these polygons are not edge-transitive.

Fig. 21

Facets of $P_1(1,0)^\pi$ in the yz-plane

The facets of $P_1(1,0)^\pi$ are all polygons congruent to those in Fig. 21. They divide into three classes depending on the coordinate axis to which their axes are parallel. Any two facets in the same class are translates of each other by an element in $\Lambda _{(1,1,1)}$, which is the translation group of $P_1(1,0)^\pi$. These polygons can be understood as degenerate helices over rectangles, obtained when the length of the short side of the rectangle tends to 0. In that situation the edges of the polygon that were meant to project to the short edge of the rectangle become parallel to the axis of the helix.

We conclude this section by summarizing the findings of this section.

Theorem 8.4

There are six families of 2-orbit polyhedra in class $2_0$ in $\mathbb {E}^3$. They are the Petrials of the chiral polyhedra P(a, b), Q(c, d), $Q(c,d)^*$, $P_1(a,b)$, $P_2(c,d)$ and $P_3(c,d)$ in $\mathbb {E}^3$; the Schläfli types of the polyhedra in class $2_0$ of these families are $\{\infty ,6\}$, $\{\infty ,6\}$, $\{\infty ,4\}$, $\{\infty ,3\}$, $\{\infty ,3\}$ and $\{\infty ,4\}$, respectively.

The polyhedra in the families $P(a,b)^\pi$, $Q(c,d)^\pi$ and $(Q(c,d)^*)\pi$ are combinatorially in class $2_0$, while those in the families $P_1(a,b)^\pi$, $P_2(c,d)^\pi$ and $P_3(c,d)^\pi$ are combinatorially regular.

Proof

We only need to justify the second paragraph, which follows from Pellicer and Ivić Weiss (2010, Theorems 4.1, 4.3, 5.3, 5.6, 5.9) together with the fact that the dual or Petrial of a polyhedron $\mathcal {P}$ is combinatorially regular if and only if $\mathcal {P}$ is regular itself. $\square$

9 Conclusions and open problems

There is a notion of geometric duality consisting of taking as set of vertices the set of centers of old facets; as edges the line segments between (new) vertices corresponding to (old) facets that share an (old) edge, and as facets the cycles determined by the (old) facets around (old) vertices. This is valid, for example, for the Platonic solids. This notion has several limitations, including that it requires every facet to have a well-defined center, and the centers of any two facets to be distinct points. Nevertheless, we can try to apply this concept to the polyhedra in class $2_2$ determined in Sect. 6, since their facets are finite and the centers of any two of them are distinct. When doing that, the choice of initial vertex for Wythoff construction is irrelevant, since for all such choices the centers of the facets are the fixed points of the rotary reflections in their symmetry groups. These points remain invariant (along with the symmetry group itself) when changing the initial vertex, and therefore the structure obtained from all polyhedra in class $2_2$ with finite facets with the same Schläfli type is the same. Recall that for each symmetry group there is a unique choice of initial vertex that yields a regular polyhedron by Wythoff’s construction. The resulting polyhedra are the Petrie–Coxeter polyhedra $\{6,4|4\}$, $\{4,6|4\}$ and $\{6,6|3\}$. The notion of duality mentioned above does work for these polyhedra, where $\{6,4|4\}$ and $\{4,6|4\}$ are dual to each other while $\{6,6|3\}$ is self-dual. It follows that when we apply that notion of duality to a polyhedron in class $2_2$ with finite facets we recover one of the three regular Petrie–Coxeter polyhedra, and not a polyhedron in class $2_0$.

This paper is the third after Schulte (2004, 2005) addressing the enumeration of 2-orbit polyhedra in $\mathbb {E}^3$. Out of the 7 classes of such polyhedra, we now have the enumeration of 3 of them, namely classes 2 (chiral), $2_2$ and $2_0$. No polyhedron in any of these classes is planar. The polyhedra in classes 2 and $2_0$ can be divided in two: those that are combinatorially regular and those that are combinatorially 2-orbit. In contrast, the polyhedra in class $2_2$ are all combinatorially regular. In these three classes, every combinatorially regular polyhedron can be deformed to a geometrically regular polyhedron by a continuous motion that preserves the symmetry group at all times. This raises the following questions. (See Cunningham et al. 2015 for a proper definition in the combinatorial setting of the symmetry type of a polytope; that definition translates naturally to the geometric setting.)

Open problem 9.1

Is it possible to determine without performing the complete enumeration whether there exist (geometric) polyhedra of a given symmetry type $\tau$ that are combinatorially also of type $\tau$? What can we say about polyhedra of type $\tau$ that are combinatorially regular? What about polyhedra that are neither of type $\tau$ nor regular?

Open problem 9.2

Is there a polyhedron $\mathcal {P}$ that is combinatorially regular, but cannot be deformed to a regular polyhedron via a continuous motion that preserves ${\textsf{Iso}}(\mathcal {P})$ at all times?

The remaining classes of 2-orbit polyhedra are apparently very different to the ones considered so far. One of the main differences is that we know there exist planar polyhedra in classes $2_1$, $2_{01}$, $2_{02}$ and $2_{12}$. Examples of these are the tessellation by rhombi where every two rhombi are translates, the Archimedean tessellation (3.6.3.6) (see Grünbaum and Shephard 1987), the tessellation by rectangles, and the dual of the tessellation (3.6.3.6), respectively. This suggests that the enumeration of the polyhedra in these four classes will be more lengthy than the ones involving classes 2, $2_2$ and $2_0$.