Finite 3-orbit polyhedra in ordinary space, II

Abstract

We enumerate the 188 3-orbit skeletal polyhedra in ${\mathbb {E}}^3$ with irreducible symmetry group. The analysis is carried out by determining the polyhedra having each irreducible finite group of isometries as their symmetry group. Relevant information of every polyhedron is also organized in tables.

1 Introduction

The study of symmetries of polyhedra has a long history. Already the Greeks knew about all Platonic and Archimedean solids that in current terminology are the only vertex-transitive convex polyhedra whose facets are all regular, other than the prisms and antiprisms. Without stating proper definitions, the Greeks understood convexity as a defining condition for polyhedra.

Centuries later, a deeper understanding of the properties of symmetry of polyhedra led to modifications of the notion of ‘polyhedron’. First, the two stellated dodecahedra and then the great dodecahedron and great icosahedron were recognized as regular polyhedra (see Ref. [1]).

The twentieth century brought formal definitions and interactions among distinct areas of mathematics. This is the time of the appearance of the Petrie–Coxeter polyhedra (see Ref. [2]), and decades later of Grünbaum’s definition of what are known today as skeletal polyhedra (see Ref. [3]). Here, regularity is formally defined in terms of a group action.

Most of the study of symmetries of skeletal polyhedra has focused on the regular ones. Their full classification was given in the twentieth century in Refs. [4, 5]. Later generalizations to higher dimensions and other geometries were studied (see for example Refs. [6,7,8,9]).

Fewer studies have been performed on highly symmetric non-regular geometric polyhedra that are not convex. They include the enumeration of the starry uniform polyhedra with planar facets [10], the classification of chiral polyhedra [11, 12] and the polyhedra obtained by Wythoffian operations from regular ones [13, 14].

One of the natural ways to proceed when studying non-regular geometric polyhedra is to enumerate them according to the number of flag orbits (regular polyhedra have only one orbit). The enumeration of the finite 2-orbit polyhedra was announced [15]. The analysis of the infinite 2-orbit polyhedra is ongoing [16] and is a logical follow-up to the classification of the chiral polyhedra.

In Ref. [17], the authors enumerate the finite 3-orbit polyhedra with reducible symmetry group. In this paper, we complete the enumeration of finite 3-orbit polyhedra by studying those with irreducible symmetry group.

The paper is organized as follows. Section 2 is about basic concepts and results on skeletal polyhedra. The 7 finite irreducible groups of isometries of ${\mathbb {E}}^3$ and some of their properties are explained in Sect. 3. The approach we will follow towards the enumeration of the polyhedra is explained in Sect. 5, including the general theory of those polyhedra that admit a continuous movement that preserves the symmetry group and the combinatorics of the polyhedra, without being homotheties or orientation preserving isometries. The enumeration is carried out in Sects. 6, 7, 8, 9, and 10; each section deals with one symmetry group. We conclude with final remarks in Sect. 11, and information about the polyhedra is collected in tables in an appendix at the end of the paper.

2 Polygons and polyhedra

Here, we recall the main ideas and basic results on polygons, polyhedra and 3-orbit polyhedra. More details can be found in Ref. [17].

A finite skeletal polygon consists of a finite set of points in Euclidean space ${\mathbb {E}}^3$, joined by line segments in such a way that the underlying graph is simple, connected, and 2-regular. As a consequence of this, no polygon can have fewer than 3 edges. If two edges have no vertex in common then they may intersect only in an interior point. Polygons need not be convex nor planar. Even though in general polygons are allowed to be infinite, here we will only consider finite ones.

A symmetry of a polygon ${\mathcal {Q}}$ is an isometry of ${\mathbb {E}}^3$ preserving ${\mathcal {Q}}$ setwise. If ${\mathcal {Q}}$ has n vertices and its symmetry group has 2n elements acting on the vertex set as the standard actions of the 2n elements of the dihedral group $D_n$, we say that ${\mathcal {Q}}$ is a regular polygon. Regular polygons can be planar or skew. Planar regular polygons may be convex or star-shaped, whereas skew polygons are obtained from planar polygons by blending them with a line segment in the sense of Ref. [18, Chapter 5A]. For more details on regular polygons, see Ref. [3].

For a given collection ${\mathcal {X}}$ of skeletal polygons and a given vertex v of some polygon in ${\mathcal {X}}$, the vertex-figure at v is a graph whose vertices are the neighbors of v, two of which are joined by a line segment if and only if they are the neighbors of v in some polygon in ${\mathcal {X}}$.

A skeletal polyhedron ${\mathcal {P}}$ is a collection of polygons (also called facets) satisfying the following properties:

• Every compact subset of ${\mathbb {E}}^3$ contains finitely many vertices of (the facets of) ${\mathcal {P}}$.

• The graph induced by the vertices and edges of all polygons is connected.

• The vertex-figure at every vertex is a skeletal polygon.

As a consequence of the last item, every vertex of a skeletal polyhedron has degree at least 3, and every edge of a skeletal polyhedron belongs to precisely two facets. When convenient we may refer to the vertices, edges and facets of ${\mathcal {P}}$ by 0-faces, 1-faces and 2-faces, respectively. We note that elsewhere the facets are called ‘faces’; here, we keep the term ‘facet’ to avoid confusion. The 1-skeleton of ${\mathcal {P}}$ consists of the sets of vertices and edges of ${\mathcal {P}}$.

Skeletal polygons and polyhedra correspond, respectively, to faithful realizations of abstract polygons and abstract polyhedra in ${\mathbb {E}}^3$ as defined in Ref. [18, Chapter 5]. Unless explicitly stated, by ‘polygon’ and ‘polyhedron’, we shall understand ‘skeletal polygon’ and ‘skeletal polyhedron’, respectively. In this paper, all polyhedra that we consider are finite. For convenience, we assume that their center is at the origin o.

A flag of a polyhedron ${\mathcal {P}}$ is a triple consisting of incident vertex, edge and facet. Flags that differ in exactly one element are called adjacent, and i-adjacent if they differ precisely in the i-face. The axioms of polyhedra imply that for every $i \in \{0,1,2\}$ and for every flag $\varPhi$, there exists a unique i-adjacent flag of $\varPhi$, and we shall denote it by $\varPhi ^i$.

A polyhedron ${\mathcal {P}}$ is equivelar if there exist integers p and q such that all its facets are p-gons and all its vertices are q-valent. In such cases, the Schläfli type of ${\mathcal {P}}$ is defined as the ordered pair $\{p,q\}$.

A dual of ${\mathcal {P}}$ is a polyhedron ${\mathcal {P}}^\delta$ where for $i \in \{0,1,2\}$, there is a bijection between the set of i-faces of ${\mathcal {P}}$ and the set of $(2-i)$-faces of ${\mathcal {P}}^\delta$, where an i-face F is contained in a j-face G of ${\mathcal {P}}$ if and only if the $(2-j)$-face of ${\mathcal {P}}^\delta$ corresponding to G is contained in the $(2-i)$-face of ${\mathcal {P}}^\delta$ corresponding to F. In many instances, polyhedra have a natural dual that interchanges the role of vertices and facets.

A Petrie walk of a polyhedron ${\mathcal {P}}$ is a closed walk on the 1-skeleton of ${\mathcal {P}}$ where every two consecutive edges belong to a facet, but three consecutive edges never do. The Petrial of ${\mathcal {P}}$ is the collection ${\mathcal {P}}^\pi$ of Petrie walks of ${\mathcal {P}}$. Its sets of vertices, edges and vertex-figures are the same as those of ${\mathcal {P}}$, and so it may fail the definition of polyhedron only when the Petrie walks of ${\mathcal {P}}$ are not polygons. If ${\mathcal {P}}^\pi$ is a polyhedron, then $({\mathcal {P}}^\pi )^\pi ={\mathcal {P}}$.

A symmetry of a polyhedron ${\mathcal {P}}$ is an isometry preserving ${\mathcal {P}}$. We shall denote the group of symmetries of ${\mathcal {P}}$ by $G({\mathcal {P}})$. If the affine span of ${\mathcal {P}}$ has dimension 3 (as it will be the case for all polyhedra in this paper), then the flag stabilizers under $G({\mathcal {P}})$ must be trivial. If $G({\mathcal {P}})$ induces k orbits on flags then ${\mathcal {P}}$ is said to be a k-orbit polyhedron; 1-orbit polyhedra are called regular. We say that ${\mathcal {P}}$ is vertex-transitive if for every pair of vertices u, v there exists $T \in G({\mathcal {P}})$ such that $u = Tv$, and vertex-intransitive otherwise.

The symmetry type graph $T({\mathcal {P}})$ of ${\mathcal {P}}$ is a connected pre-graph (allowing semi-edges and multiple edges) with edges labeled in $\{0,1,2\}$. The vertex set of $T({\mathcal {P}})$ is the set of flag orbits of ${\mathcal {P}}$. Two vertices $X_1$ and $X_2$ of $T({\mathcal {P}})$ are joined by an edge labeled i whenever a flag in $X_1$ is i-adjacent to a flag in $X_2$. In addition, if two flags in the same orbit are i-adjacent then there is a semi-edge labeled i at the corresponding vertex.

Symmetry type graphs illustrate the local configuration of flags according to their flag orbits. The connected components of $T({\mathcal {P}})$ after removing the edges labeled i correspond to the distinct orbits of i-faces under $G({\mathcal {P}})$. Polyhedra with the same symmetry type graph are said to be in the same class.

According to Refs. [19, 20], there are three classes of 3-orbit polyhedra, called $3^{0,1}$, $3^1$ and $3^{1,2}$ in Ref. [19]; they correspond, respectively, to classes $3^2$, $3^{02}$ and $3^{0}$ in Ref. [20]. Their symmetry type graphs are those in Fig. 1.

Fig. 1

Symmetry type graphs of 3-orbit polyhedra

In what follows, we will develop terminology for discussing how symmetric the facets and vertices of a 3-orbit polyhedron are. Some care is required, because we will be interested in the number of flag orbits of a facet or vertex under its stabilizer in $G({\mathcal {P}})$, which may be different than the number of flag orbits under its own symmetry group. For example, it is possible to truncate a cube so that one obtains regular octagons as facets; however, the stabilizer of this octagon under [3, 4] has two orbits on the flags. Such a facet will be called 2-symmetric. Let us now fully describe our terminology.

We will say that an i-face F of a polyhedron ${\mathcal {P}}$ is 1-symmetric if for every $j \in \{0,1,2\}{\setminus } \{i\}$ there is a symmetry of ${\mathcal {P}}$ that maps a flag containing F to its j-adjacent flag. In the symmetry type graph of ${\mathcal {P}}$, 1-symmetric i-faces are those in a flag orbit represented by a vertex with semi-edges with labels in $\{0,1,2\}{\setminus } \{i\}$. In particular, every 3-orbit polyhedron has some 1-symmetric edges; polyhedra in class $3^{0,1}$ have some 1-symmetric vertices; and polyhedra in class $3^{1,2}$ have some 1-symmetric facets.

Under the assumption of trivial flag stabilizers under $G({\mathcal {P}})$, the stabilizer of a 1-symmetric i-face F is a dihedral group $D_k$ with 2k elements generated by involutions $T_j$, $j \in \{0,1,2\} {\setminus } \{i\}$ that fix F as well as the j-face of a given flag containing F. When $i=1$, then $k=2$ and hence the edge stabilizer of a 1-symmetric edge is isomorphic to ${\mathbb {Z}}_2 \times {\mathbb {Z}}_2$. If $i=0$, then F is a vertex and k is its degree, whereas if $i=2$, then F is a facet and k is its number of edges.

A 1-symmetric facet must be a regular polygon; it may be planar or skew. We will write $k_p$ to denote a planar k-gon and $k_s$ to denote a skew k-gon. When k is odd, then a regular k-gon is necessarily planar, and so we merely denote it by k.

In principle, it is possible to distinguish 1-symmetric facets even more finely by, for example, distinguishing convex polygons from star polygons. However, our notion of vt-equivalence that we will develop in Sect. 5 does not distinguish between the two types of polygons whenever the vertices are 3-valent and the symmetry group contains a plane reflection, which end up accounting for a large proportion of 3-orbit polyhedra in ${\mathbb {E}}^3$. Thus, we will not distinguish between different kinds of planar polygons.

If for every edge e of (the facet) F, there exists $T \in G({\mathcal {P}})$ preserving F while interchanging the endpoints of e, but there is no non-trivial symmetry of ${\mathcal {P}}$ preserving F and fixing one of its vertices, we say that F is 2-symmetric. A facet of this kind admits a symmetry acting like a 2-step rotation and hence it must have an even number of edges. The symmetries of ${\mathcal {P}}$ preserving F induce 2 orbits on the flags containing F. Among these symmetries, those that act on F like reflections are illustrated in Fig. 2a. The 2-symmetric facets of a 3-orbit polyhedron must be vertex-transitive, and they may be planar or skew. As for 1-symmetric facets, we will denote planar and skew facets with a subscript of p or s.

As pointed out in Ref. [17], besides the 2-symmetric facets described here, there are other possibilities of polygons and groups acting on them with 2 orbits on the flags, but those scenarios do not appear in the analysis of 3-orbit polyhedra.

We say that a facet F is 3-symmetric whenever the subgroup of $G({\mathcal {P}})$ preserving F induces on it 3 flag orbits. Such facets admit a symmetry of ${\mathcal {P}}$ acting like a 3-step rotation, but no symmetries acting like a 1- or 2-step rotation. This implies that the number of edges of F is divisible by 3. Furthermore, there are elements of $G({\mathcal {P}})$ preserving F that act on its vertices and edges in the same way a line reflection acts on the vertices and edges of a convex polygon; we shall refer to such symmetries by saying that they ‘act like reflections’. Some of the symmetries of F that act like reflections fix a vertex and some fix midpoints of edges. The symmetries acting like reflections on a 3-symmetric hexagon are illustrated in Fig. 2b.

The symmetry of a 3-symmetric facet that fixes a vertex must either be a plane reflection or a half-turn, while the symmetry that fixes the midpoint of an edge could be a plane reflection, half-turn, or central inversion. Of the six possible cases, we encountered three of them: where both symmetries are plane reflections, where both are half-turns, and where the symmetry that fixes a vertex is a reflection and the other symmetry is a half-turn. We will denote a 3-symmetric k-gon by $k_r$, $k_h$, or $k_{rh}$ according to which of these possibilities it realizes. Note that if k is odd, then $k_{rh}$ is not possible. Also, if $k = 3$ then the geometry of the facet does not depend on whether the generating symmetry is a reflection or half-turn, and so we merely denote the facet by 3.

Fig. 2

Reflections of 2-symmetric and 3-symmetric hexagons

We extend the definitions of 2-symmetric and 3-symmetric from facets to vertices by duality. In this way, a 2-symmetric vertex v has even degree and there are symmetries of ${\mathcal {P}}$ that fix v and any of the edges that contain v. On the other hand, 3-symmetric vertices are such that a symmetry of ${\mathcal {P}}$ acts like a 3-step rotation around them, but no element in $G({\mathcal {P}})$ acts like a 1-step or 2-step rotation around them. There are symmetries of ${\mathcal {P}}$ that fix a 3-symmetric vertex while reversing the cyclic order of the edges around it. The degree of a 3-symmetric vertex is divisible by 3.

Every 3-orbit polyhedron has two orbits on edges. One orbit consists of 1-symmetric edges, and the other edges containing flags in two distinct orbits. If ${\mathcal {P}}$ is a 3-orbit polyhedron in class $3^{0,1}$, then there is no symmetry swapping the endpoints of a 2-orbit edge, but there is a non-trivial symmetry fixing it pointwise. If ${\mathcal {P}}$ belongs to either of the other two classes then there are no non-trivial symmetries fixing 2-orbit edges pointwise, but for each edge, there is a symmetry that swaps its endpoints. For convenience, we shall refer to 2-orbit edges as 2-symmetric edges regardless of the class of 3-orbit polyhedra in question.

2.1 Class $3^{0,1}$

Polyhedra ${\mathcal {P}}$ in class $3^{0,1}$ have two orbits of vertices, and so they are vertex-intransitive. One orbit contains 1-symmetric vertices, and the other contains 2-symmetric vertices.

These polyhedra also have two orbits of edges. One of these orbits contains 2-symmetric edges joining a 1-symmetric vertex with a 2-symmetric vertex. The stabilizer of an edge e in this orbit has only one non-trivial element; this element swaps the two facets that contain e while fixing both endpoints.

The other orbit of edges contains half as many edges as the previous orbit. These edges are 1-symmetric and join two 2-symmetric vertices. The neighbors of a 2-symmetric vertex $v_2$ alternate between 1-symmetric and 2-symmetric vertices, implying the following results.

Lemma 1

Every 2-symmetric vertex of a polyhedron in class $3^{0,1}$ is incident to at least two 1-symmetric edges and its degree is an even number $d \ge 4$.

Proposition 2

[17, Proposition 3.3] Every skeletal polyhedron in class $3^{0,1}$ must have at least two 1-symmetric vertices and three 2-symmetric vertices.

Polyhedra in class $3^{0,1}$ are facet-transitive. The facets are 3-symmetric and, therefore, their number of edges is divisible by 3. One-third of the edges in each facet are 1-symmetric edges, and two consecutive 1-symmetric edges in a 3-symmetric facet are separated by two 2-symmetric edges. The next lemmas are consequences of the symmetries of 3-symmetric polygons.

Lemma 3

Let F be a 3-symmetric facet of a 3-orbit polyhedron in class $3^{0,1}$, and let e be an edge of F between two 2-symmetric vertices. Let T be a non-trivial involutory symmetry that fixes F and interchanges the two endpoints of e. Then, T does not fix any 2-symmetric vertex of F.

Proof

Suppose that F is a k-gon with vertices labeled $(1, 2, \ldots , k)$ and suppose that T fixes some vertex of F. Without loss of generality we may assume that T interchanges 1 and k. Then, it follows that it interchanges every i with $k+1-i$. In particular, if T fixes a vertex, then k must be odd. Furthermore, since the number of 2-symmetric vertices in F must be 2k/3 and the number of 1-symmetric vertices must be k/3, there is an even number of 2-symmetric vertices and an odd number of 1-symmetric vertices. It follows that the vertex fixed by T is 1-symmetric. $\square$

Given a 3-orbit polyhedron ${\mathcal {P}}$ in class $3^{0,1}$, we may preserve one orbit of vertices, while moving each of the vertices in the other orbit a fixed amount along the line that joins it with the center of ${\mathcal {P}}$. We say that two polyhedra obtained in this way are vi-equivalent (short for equivalent in the vertex intransitive case). We will study 3-orbit polyhedra in class $3^{0,1}$ up to vi-equivalence.

If ${\mathcal {P}}$ is in class $3^{0,1}$, then its facets are 3-symmetric a-gons, and it has 1-symmetric b-valent vertices and 2-symmetric c-valent vertices, for some a,b, and c. We note that moving one orbit of vertices may disrupt the planarity of a vertex-figure, so since we classify 3-orbit polyhedra in class $3^{0,1}$ up to vi-equivalence, we do not record whether the vertex-figures are planar or not. As described earlier, we will denote the facets as $a_r$, $a_h$, or $a_{rh}$ depending on the kind of symmetries that act like reflections. We will associate to each polyhedron in this class a modified Schläfli symbol like $\{a_r, (b, c)\}$.

2.2 Class $3^1$

Polyhedra in class $3^1$ are vertex- and facet-transitive. Their vertices and facets are 3-symmetric. They can be described by a Schläfli symbol like $\{9_r, 6_h\}$, where for both the facets and vertex-figures, we describe the generating involutions of their symmetry group.

They have two orbits on edges. One edge orbit contains 1-symmetric edges. The stabilizer of an edge e in the other edge orbit only contains the identity element and an element that swaps the two endpoints of e while interchanging the facets containing e.

They have two kinds of Petrie paths. There are some that contain only 2-symmetric edges, and some that alternate between 1-symmetric and 2-symmetric edges.

2.3 Class $3^{1,2}$

Polyhedra in class $3^{1,2}$ behave in a dual way to polyhedra in class $3^{0,1}$. They are vertex-transitive and all their vertices are 3-symmetric. They have two orbits on edges. One edge orbit contains 1-symmetric edges. The elements of the stabilizer of an edge e in the other orbit swap the endpoints of e and preserve the two facets containing e.

These polyhedra have two orbits of facets; one containing 1-symmetric facets and one containing 2-symmetric facets. The 1-symmetric edges belong to two 2-symmetric facets, whereas the 2-symmetric edges belong to a 1-symmetric facet and to a 2-symmetric facet.

The Petrie paths of polyhedra in class $3^{1,2}$ are all 3-symmetric. In each of them, every third edge is 1-symmetric, and two consecutive 1-symmetric edges are separated by two 2-symmetric edges.

If ${\mathcal {P}}$ is in class $3^{1,2}$, then its facets consist of 1-symmetric a-gons, 2-symmetric b-gons, and c-valent vertices. The vertices are described as in class $3^1$. The 1-symmetric facets and 2-symmetric facets are either planar or skew, and thus are denoted by $a_p$ or $a_s$ for example. If a is odd, then the facet is necessarily planar and we do not include a subscript. (On the other hand, b must be even, and so we always use its subscript.) We describe such a polyhedron with a Schläfli symbol $\{(a, b), c\}$ with the subscripts mentioned; e.g., $\{(3, 8_s), 6_{rh}\}$.

2.4 General results

Now, let us examine in more detail the geometric structure of the symmetry group of a 3-orbit polyhedron. We start with three useful results on the edges of 3-orbit polyhedra.

Lemma 4

Let ${\mathcal {P}}$ be a 3-orbit polyhedron with trivial flag stabilizers under $G({\mathcal {P}})$. Then, there exists an edge orbit ${\mathcal {O}}$ under $G({\mathcal {P}})$ such that for each $e\in {\mathcal {O}}$, the stabilizer in $G({\mathcal {P}})$ of e is generated by:

• an involution fixing e pointwise that interchanges the two facets containing e,

• an involution that interchanges the endpoints of e and preserves the two facets containing e.

Furthermore, these involutions commute.

Proof

As mentioned above, the three symmetry type graphs of 3-orbit polyhedra indicate the presence of 1-symmetric edges. The symmetry group of such an edge e acts transitively on the four flags containing e, and in particular there are symmetries that exchange a given one of these flags to its 2-adjacent flag and to its 0-adjacent flag. These symmetries act on the vertices and facets incident to e precisely as described in the items of the statement.

The commutativity follows from the facts that for every flag $\varPhi$ the flags $(\varPhi ^0)^2$ and $(\varPhi ^2)^0$ are equal. Indeed, trivial flag stabilizers imply that there is a unique symmetry mapping $\varPhi$ to $(\varPhi ^0)^2 = (\varPhi ^2)^0$. $\square$

Lemma 5

Let F be a facet of a 3-orbit polyhedron $\mathcal P$ with irreducible symmetry group. Consider a vertex v of F and its neighbors u and w in F, and suppose that the edge connecting u and v is 1-symmetric. Let T be the non-trivial symmetry of $\mathcal P$ that fixes u and v. Then, T does not fix w.

Proof

First, recall that every 1-symmetric edge does have a non-trivial symmetry that fixes u and v. This symmetry cannot fix F; otherwise it would fix a flag, and irreducibility of $G({\mathcal {P}})$ would imply that T is the identity. If T fixed w, then F and T(F) would both contain the vertices u, v, and w in the same order. But then the vertex-figure at v would contain a 2-cycle (with vertices u and w joined by a double edge, one per facet containing them) and be disconnected. $\square$

The Orbit–Stabilizer Theorem immediately implies the following:

Proposition 6

Let ${\mathcal {P}}$ be a 3-orbit polyhedron such that $|G({\mathcal {P}})| = N$. Then, ${\mathcal {P}}$ has precisely N/4 1-symmetric edges and N/2 2-symmetric edges.

The symmetry groups of 3-orbit polyhedra must contain many involutions, as illustrated by the following two results. The first one follows directly from the description of 1-, 2- and 3-symmetric vertices. The second one is a direct consequence of Ref. [19, Corollary 5.4].

Lemma 7

Let ${\mathcal {P}}$ be a 3-orbit polyhedron and v one of its vertices. Then, the stabilizer in $G({\mathcal {P}})$ of v contains a non-trivial involution. Furthermore, if ${\mathcal {P}}$ is in class $3^{0,1}$, then the stabilizer of v is generated by two involutions.

Proposition 8

If ${\mathcal {P}}$ is a 3-orbit polyhedron, then $G({\mathcal {P}})$ is generated by involutions.

Involutions shall play an important role in our analysis of 3-orbit polyhedra. In that analysis, we shall make use of the following obvious remark.

Remark 9

Let F be any facet of a 3-orbit polyhedron, and let T be a non-trivial involutory symmetry of F. If T does not fix any vertices of F, then F has an even number of vertices.

Now, we provide a condition on the number of vertices of a 3-orbit polyhedron in terms of the size of its symmetry group.

Proposition 10

If ${\mathcal {P}}$ is a 3-orbit polyhedron with symmetry group of order N, then the number V of vertices satisfies $V > \sqrt{3N/2}$.

Proof

A 3-orbit polyhedron with a group of order N has 3N flags and 3N/4 edges. Therefore, there must be at least 3N/4 pairs of vertices, which means $V(V-1) / 2 \ge 3N/4$. Then, $V(V-1) \ge 3N/2$; in particular $V^2 > 3N/2$ and the result follows. $\square$

In upcoming sections, we shall encounter graphs where all the vertices lie on some sphere ${\mathcal {S}}$. If there are no edges joining antipodal vertices, then we may project the edges to ${\mathcal {S}}$ and obtain an embedding of the graph on the sphere (possibly with edge crossings). In order to describe the facets of a polyhedron with such a 1-skeleton, we shall say that the facet F skips m edges at some vertex v with degree d if in some neighborhood of v on ${\mathcal {S}}$, the two edges of F itself leave m edges on one side and $d-m-2$ on the other. When doing so, we assume a fixed global orientation of ${\mathcal {S}}$, so that when tracing the facet we always skip edges on the same side.

For example, when describing the great dodecahedron (see Ref. [1]), we may say that we consider the 1-skeleton of the icosahedron and build the facets so that they skip an edge at every vertex. We could think of the Petrial of a Platonic solid ${\mathcal {P}}$ of degree d as the polyhedron with the 1-skeleton of ${\mathcal {P}}$ where the facets skip alternatingly 0 and $d-2$ edges.

When determining the candidate facets of a polyhedron with a given 1-skeleton, we must take into account that the resulting set of facets may not yield a polyhedron because the vertex-figure is not connected (failing the third condition to be a polyhedron). For example, if the 1-skeleton is that of the octahedron, then we cannot skip one edge at every vertex, since we would be left with the three equatorial squares; in this situation, each vertex-figure is a pair of line segments intersecting only in their midpoints.

2.5 Operations on polyhedra

If ${\mathcal {P}}$ has a dual, in some cases, it is possible to construct a canonical dual of ${\mathcal {P}}$ by taking as vertices of ${\mathcal {P}}^\delta$ the barycenters of the facets of ${\mathcal {P}}$. Then, the edges and facets are determined by the duality condition. With ${\mathcal {P}}^\delta$ arising from this construction, $G({\mathcal {P}}) = G({\mathcal {P}}^\delta )$, and we will simply call ${\mathcal {P}}^\delta$ ‘the dual of ${\mathcal {P}}$’. However, we may encounter polyhedra where two or more facets have the same barycenter, in which case the structures resulting from an attempt to find their duals in the way just explained would not satisfy our definition of polyhedron.

For most of the polyhedra ${\mathcal {P}}$ we will describe, there are pairs of facets that share more than one edge. Such a polyhedron cannot have a (geometric) dual, since such a dual would have a pair of vertices connected by multiple edges. In fact, if the facets of ${\mathcal {P}}$ are too big, then no dual can exist:

Proposition 11

Suppose ${\mathcal {P}}$ is a 3-orbit polyhedron with symmetry group of size N and with a dual. If the facets of ${\mathcal {P}}$ are all 3k-gons, then $k < \sqrt{N/6}$.

Proof

By Proposition 10, ${\mathcal {P}}$ must have more than $\sqrt{3N/2}$ facets. Each facet is part of 6k flags and there are 3N flags total, and the result is simple from there. $\square$

Proposition 11 is useful in filling out the tables in Sect. A, but we will not explicitly comment on when we are using it.

Note that if ${\mathcal {P}}$ has a dual ${\mathcal {P}}^{\delta }$, then $G({\mathcal {P}}) = G({\mathcal {P}}^{\delta })$. Similarly, if the Petrial of ${\mathcal {P}}$ (${\mathcal {P}}^{\pi }$) is a polyhedron, then $G({\mathcal {P}}) = G({\mathcal {P}}^\pi )$.

The three classes of 3-orbit polyhedra are related by duality and Petriality as follows.

Lemma 12

[17, Lemma 3.4] Let ${\mathcal {P}}$ be a 3-orbit polyhedron having a geometric dual ${\mathcal {P}}^\delta$.

• If ${\mathcal {P}}$ is in class $3^{0,1}$, then ${\mathcal {P}}^\delta$ is in class $3^{1,2}$.

• If ${\mathcal {P}}$ is in class $3^{1}$, then ${\mathcal {P}}^\delta$ is also in class $3^{1}$.

• If ${\mathcal {P}}$ is in class $3^{1,2}$, then ${\mathcal {P}}^\delta$ is in class $3^{0,1}$.

Lemma 13

[17, Lemma 3.5] Let ${\mathcal {P}}$ be a 3-orbit polyhedron whose Petrial ${\mathcal {P}}^\pi$ is a polyhedron.

• If ${\mathcal {P}}$ is in class $3^{0,1}$, then ${\mathcal {P}}^\pi$ is also in class $3^{0,1}$.

• If ${\mathcal {P}}$ is in class $3^{1}$, then ${\mathcal {P}}^\pi$ is in class $3^{1,2}$.

• If ${\mathcal {P}}$ is in class $3^{1,2}$, then ${\mathcal {P}}^\pi$ is in class $3^{1}$.

In addition to duality and Petriality, there are a few more operations on polyhedra that play an important role in our classification. The first is truncation. In the convex setting, we may think of ‘cutting off’ each vertex, obtaining a new polyhedron with two vertices on the relative interior of each original edge, and with two kinds of facets: those that correspond to the original vertex-figures and those that are truncations of the original facets, now with twice the number of edges as the original facets. More generally, truncation (and related operations) can be applied to skeletal polyhedra and indeed to abstract polyhedra; see Ref. [13] for a detailed description. The truncation of a regular polyhedron is in most cases a 3-orbit polyhedron, accounting for their prominence in our analysis. If ${\mathcal {P}}$ is an abstract polyhedron, we will use ${{\,\textrm{Tr}\,}}({\mathcal {P}})$ to denote the truncation of ${\mathcal {P}}$. We will also use $T_{[p,q]}$ to denote the 1-skeleton of the (geometric) truncation of $\{p,q\}$.

Dual to truncation is an operation known as the Kleetope operation. Applied to convex polyhedra, it can be thought of as attaching a pyramid to each facet. For example, the Kleetope of the cube is the tetrakis hexahedron, a 3-orbit polyhedron. For our purposes, we will define the Kleetope operation on abstract polyhedra only, defined by ${{\,\textrm{Kl}\,}}({\mathcal {P}}) = ({{\,\textrm{Tr}\,}}(\mathcal P^\delta ))^\delta$. We will also use $K_{[p,q]}$ to denote the 1-skeleton of the Kleetope of $\{p,q\}$.

The last operation is actually a family of three related operations, and we first describe the effect on a 1-skeleton S. The first operation is called $\zeta$ (see Ref. [21]), which replaces every edge $\{u, v\}$ with the edges $\{u, -v\}$ and $\{-u, v\}$. (Recall that the polyhedron has center at the origin o.) Since a 3-orbit polyhedron has two orbits of edges, we may also choose to perform this operation only to the 1-symmetric edges or only to the 2-symmetric edges. We will denote the operation that replaces all i-symmetric edges $\{u, v\}$ with $\{u, -v\}$ and $\{-u,v\}$ by $\zeta _i$. (A word of caution: $\zeta _i$ has a different definition in Ref. [21] but is used sparingly.) Note that $\zeta = \zeta _1 \zeta _2$.

In most cases we encounter, S will be centrally symmetric. However, we will also find it convenient to apply $\zeta _i$ to 1-skeleta (and polyhedra) that are not centrally symmetric, such as the truncated tetrahedron.

Definition 14

Let S be the 1-skeleton of a 3-orbit polyhedron with vertex set V and edge set $E_1 \cup E_2$ where $E_1$ contains the 1-symmetric edges and $E_2$ contains the 2-symmetric edges. Let $\overline{V} = V \cup -V$, and for $i = 1,2$ let

$$\begin{aligned} \overline{E_i} = \{\{u, -v\}: \{u, v\} \in E_i\}. \end{aligned}$$

Then,

• (a) $S^{\zeta _1}$ is one connected component of the 1-skeleton $(\overline{V}, \overline{E_1} \cup E_2)$.

• (b) $S^{\zeta _2}$ is one connected component of the 1-skeleton $(\overline{V}, E_1 \cup \overline{E_2})$.

• (c) $S^{\zeta }$ is one connected component of the 1-skeleton $(\overline{V}, \overline{E_1} \cup \overline{E_2})$.

We note that in the cases where we get a disconnected graph and have to take one connected component, these two components are congruent; in fact, they are the image of one another under a central inversion. For example, applying $\zeta _1$ to the 1-skeleton of a truncated cube yields two connected components, namely the truncations of the two regular tetrahedra that can be inscribed in the cube using its vertex set.

To extend these operations to polyhedra, we need to explain how the facets of ${\mathcal {P}}^{\zeta _i}$ should be obtained from the facets of ${\mathcal {P}}$. Informally, the idea is this: start with one typical facet from each orbit of facets. We may write the facet as a cyclic sequence of vertices $(v_1, v_2, \ldots , v_k)$. Then, depending on which $\zeta _i$ we are using and which kinds of edges the facet uses, some vertices $v_i$ are replaced with their antipode $-v_i$. In some cases (that will become clear), we need to apply this operation to the walk that goes twice around the facet: $(v_1, \ldots , v_k, v_1, \ldots , v_k)$. In any event, the effect of applying $\zeta _i$ now turns our original facet into a new cyclic walk $(v_1', \ldots , v_m')$, with $m = k$ or $m = 2k$. In some cases, we may have $m = k$ but the new cyclic walk actually has period m/2, in which case we just consider the new facet to be $(v_1', \ldots , v_{m/2}')$. In any case, we define the facets of ${\mathcal {P}}^{\zeta _i}$ to be the images of such cyclic walks under the symmetry group. We note that it does sometimes happen that the new facets are not polygons, in which case ${\mathcal {P}}^{\zeta _i}$ is not a polyhedron.

Now, let us describe the new facets in more detail.

Definition 15

Let $F = (v_1, \ldots , v_k)$ be a facet of a 3-orbit polyhedron.

• (a) If F is 1-symmetric (so that all the edges are 2-symmetric and $\zeta _1$ fixes F), then

  • (1a) If k is even, we define $F^{\zeta _2}$ as $(v_1, -v_2, v_3, \ldots , v_{k-1}, -v_k)$.

  • (1b) If k is odd, we define $F^{\zeta _2}$ as $(v_1, -v_2, v_3, \ldots , -v_{k-1}, v_k, -v_1, v_2, \ldots , -v_k)$.

• (b) If F is 2-symmetric (so that the edges alternate between 1-symmetric and 2-symmetric), without loss of generality, we assume that the edge from $v_i$ to $v_{i+1}$ is 1-symmetric when i is odd, and 2-symmetric when i is even. Then,

  • (2a) If k is divisible by 4, we define $F^{\zeta _1}$ as $(v_1, -v_2, -v_3, v_4 \ldots , -v_{k-1}, v_k)$ and $F^{\zeta _2}$ as $(v_1, v_2, -v_3, -v_4, v_5, \ldots , -v_{k-1}, -v_k)$.

  • (2b) If k is not divisible by 4, we define $F^{\zeta _1}$ as $(v_1, -v_2, -v_3, v_4, \ldots , v_{k-1}, -v_k, -v_1, v_2, \ldots , v_k)$ and $F^{\zeta _2}$ as $(v_1, v_2, -v_3, -v_4, v_5, \ldots , v_{k-1}, v_k, -v_1, -v_2, \ldots , -v_k)$.

• (c) If F is 3-symmetric, then without loss of generality, the 1-symmetric edges are $\{v_i, v_{i+1}\}$ with i divisible by 3 (and the remaining edges are 2-symmetric).

  • (3a) If k is even (and thus, divisible by 6), we define $F^{\zeta _1}$ as $(v_1, v_2, v_3, -v_4, -v_5, -v_6, v_7, \ldots , -v_k)$.

  • (3b) If k is odd, we define $F^{\zeta _1}$ as $(v_1, v_2, v_3, -v_4, -v_5, -v_6, v_7, \ldots , v_k, -v_1, -v_2, -v_3, \ldots , -v_k)$.

  • (3c) For all k, we define $F^{\zeta _2}$ as $(v_1, -v_2, v_3, v_4, -v_5, v_6, \ldots , -v_{k-1}, v_k)$.

We note that, strictly speaking, the description of $F^{\zeta _i}$ in Definition 15 depends on a particular ordering of the vertices of a facet, but since we are really only interested in how $\zeta _i$ acts on the whole orbit of facets, it does no harm to make an arbitrary choice of how we describe an individual facet.

The action of $\zeta _2$ on 1-symmetric facets has the same effect on them as applying $\zeta$ to a regular polygon, which is described in Ref. [21, Theorem 5.3]. In our analysis, the 1-symmetric facets are typically either planar facets whose affine hull does not contain the center of the polyhedron, or skew facets where the center of the facet coincides with the center of the polyhedron, and $\zeta _2$ interchanges these two possibilities.

In a similar way, both operations $\zeta _i$ will transform a planar 2-symmetric facet whose affine hull does not contain the center of the polyhedron into a skew facet whose center coincides with the center of the polyhedron.

Now, suppose F is a 3-symmetric facet, and fix a 1-symmetric edge $e_1$ from it. Consider the symmetry $T_1$ of F that interchanges the endpoints of $e_1$. Then, if ${\mathcal {P}}^{\zeta _2}$ is a polyhedron, the same symmetry $T_1$ fixes the induced facet $F^{\zeta _2}$ and interchanges the endpoints of $e_1$. Similarly, if $v_1$ is a 1-symmetric vertex of F, there is a symmetry $T_2$ of F that fixes $v_1$ while interchanging its neighbors, and the same symmetry will act in the same way on $F^{\zeta _2}$. Thus, we see that the symmetry group of a 3-symmetric facet remains unchanged under $\zeta _2$, and so the size and the type of the facet ($k_r$, $k_h$, or $k_{rh}$) remain unchanged as well.

We can now extend these operations to polyhedra.

Definition 16

Let P be a 3-orbit polyhedron. For $i = 1, 2$, we define a new structure $P^{\zeta _i}$ (which may or may not be a polyhedron) as follows:

• (a) If S is the 1-skeleton of P, then the 1-skeleton of $P^{\zeta _i}$ will be $S^{\zeta _i}$.

• (b) The facets of $P^{\zeta _i}$ are obtained from the facets of P in the manner described in Definition 15. We keep only those facets that lie in the connected component of $S^{\zeta _i}$ that we chose.

Let us note that if ${\mathcal {P}}$ is vertex-intransitive, then ${\mathcal {P}}^{\zeta _2}$ is vi-equivalent to ${\mathcal {P}}$, since up to vi-equivalence, we may scale one orbit of vertices independently of the other and so we can send, say, every 1-symmetric vertex v to $-v$. Thus, we will only use $\zeta _2$ on vertex-transitive polyhedra.

Example 17

Let P be the truncated cube (Fig. 3a). Then, the 1-skeleton of $P^{\zeta _1}$ has two connected components, each containing 12 vertices (Fig. 3b). The 1-skeleton of $P^{\zeta _2}$ is connected, and the facets are skew hexagons and skew octagons (Fig. 3c where only one 2-symmetric facet is shown).

Fig. 3

The truncated cube ${{\,\textrm{Tr}\,}}(\{4,3\})$ together with ${{\,\textrm{Tr}\,}}(\{4,3\})^{\zeta _1}$ and ${{\,\textrm{Tr}\,}}(\{4,3\})^{\zeta _2}$; 1-symmetric edges are dotted and 2-symmetric edges are solid

3 Finite irreducible groups of isometries of ${\mathbb {E}}^3$

In this section, we recall the finite irreducible groups of isometries of ${\mathbb {E}}^3$. Throughout we follow Ref. [22], although with a different notation.

All finite groups of isometries have a fixed point. A finite group G of isometries of ${\mathbb {E}}^3$ is said to be affinely reducible whenever there is a setwise fixed line, and hence also a setwise fixed plane (the orthogonal complement of the fixed line through the fixed point). Finite groups of isometries that are not affinely reducible are called affinely irreducible.

There are seven finite affinely irreducible groups of isometries of ${\mathbb {E}}^3$. They are tightly linked with the symmetry groups of the Platonic solids, and are described next.

The rotational tetrahedral group $[3,3]^+$ is the group of orientation preserving symmetries of the tetrahedron. It contains 12 elements and can be understood as the alternating group on the four vertices of the tetrahedron.

The full tetrahedral group [3, 3] is the group of all symmetries of the tetrahedron and contains 24 elements. It can be identified with the group of permutations of the vertices of the tetrahedron, and hence it is isomorphic to the symmetric group $S_4$.

The rotational octahedral group $[3,4]^+$ is the group of orientation preserving symmetries of the cube (and of the octahedron). It contains 24 elements and can be understood as the symmetric group on the four main diagonals of the cube.

The full octahedral group [3, 4] is the group of all symmetries of the cube and contains 48 elements. It is isomorphic to the direct product $S_4 \times C_2$ of the symmetric group on 4 elements with a cyclic group of order 2. The central element of [3, 4] is the central inversion with respect to the center of the cube.

The full octahedral group contains two index-two subgroups besides the rotational octahedral group. They can be understood through the two vertex-disjoint tetrahedra inscribed in the cube. In particular, the full tetrahedral group is the index-two subgroup of the full octahedral group consisting of all isometries preserving the two inscribed tetrahedra of the cube.

The group denoted by $[3,3]^*$ in Ref. [11] is the group of all orientation preserving isometries that fix the two tetrahedra of the cube together with all orientation reversing isometries that interchange these two tetrahedra. Since it includes the central inversion with respect to the center of the cube, it is isomorphic to $A_4 \times C_2$. It is commonly called the pyritohedral group.

The rotational icosahedral group $[3,5]^+$ is the group of orientation preserving symmetries of the dodecahedron (and of the icosahedron). It contains 60 elements and can be understood as the alternating group on the five cubes which can be inscribed in the dodecahedron.

Finally, the full icosahedral group [3, 5] is the group of all symmetries of the dodecahedron and contains 120 elements. It is isomorphic to the direct product $A_5 \times C_2$ of the alternating group on 5 elements with a cyclic group of order 2. The central element of [3, 5] is the central inversion with respect to the center of the dodecahedron.

The groups $[3,4]^+$ and $[3,5]^+$ contain no orientation reversing isometries, implying that polyhedra having these as their symmetry groups are invariant exclusively under orientation preserving isometries. It follows that all polyhedra with those symmetry groups are handed in the sense that a right polyhedron cannot be overlapped into a left polyhedron through a continuous motion. Nevertheless, right and left-handed versions of the same polyhedron are similar to each other, and such a polyhedron appears only once in our enumeration.

The following lemma will be used in subsequent sections. It can be proven by direct inspection.

Lemma 18

Let $G \in \{[3,3], [3,4], [3,5]\}$ and $T \in G$ of order $m>2$. Then, no power of T is a plane reflection.

4 Structure of the Appendix

Once we determine all the finite 3-orbit polyhedra in ${\mathbb {E}}^3$ with irreducible symmetry group, there are still some questions to consider, such as which polyhedra have a canonical geometric dual, and how the polyhedra are related by the Petrial operation. To help answer those questions and organize the information, we have compiled tables in Sect. A. Let us describe the format of those tables.

Each table consists of polyhedra with a fixed symmetry group. The 53 polyhedra with symmetry group [3, 4] are split into two tables; one for the vertex-transitive polyhedra and one for the vertex-intransitive polyhedra. The 114 polyhedra with symmetry group [3, 5] are split into four tables, with the first one containing the vertex-intransitive polyhedra.

Now, we explain the columns:

• (a) 1-skeleton: The polyhedra in each table are grouped together by their 1-skeleton. The names of these 1-skeleta are introduced in the appropriate section, and they are also summarized in Table 2.

• (b) #: We number the polyhedra with a given 1-skeleton so that other entries can cross-reference them.

• (c) Class: The class of 3-orbit polyhedra: $3^{0,1}$, $3^{1}$, or $3^{1,2}$.

• (d) Schläfli symbol: For each class of polyhedra, we describe the types of facets and vertices with an extended Schläfli symbol as described in Section 2.

• (e) Petrial: The Petrial of a given polyhedron must have the same 1-skeleton, so if the Petrial is also a polyhedron, then we indicate which by $\#N$ which means the Nth polyhedron with this same 1-skeleton. Otherwise we write N/A.

• (f) Dual: If the polyhedron has a canonical dual (see Sect. 2.5), then it is recorded here. We refer to the dual by listing the table it appears in, the name of its 1-skeleton, and its number. If it has no canonical dual, but its combinatorial dual is realizable, then we denote that in the same way but starting with “(NC)” meaning “non-canonical”. If the combinatorial dual is not realizable as a 3-orbit polyhedron, then we denote that with N/A.

• (g) Figure: If there is a figure that shows the types of facets, then we refer to it here.

• (h) Type: If there is an easy way to refer to the combinatorial type, we include it here. There are many pairs of polyhedra that are combinatorially equivalent, and if we have no other standard way to refer to the type, then we simply note here the combinatorially equivalent polyhedron by specifying the 1-skeleton and number. (For example, see the bottom two groups of Table 5.) If a polyhedron is combinatorially regular, we include that information in this column. If the combinatorial type does not have a common name, and if there are no other 3-orbit polyhedra with the same type, then again we simply name the type using the 1-skeleton and number.

  Note that by Ref. [20, Theorem 5.1], the combinatorial type of every trivalent 3-orbit polyhedron in class $3^{1,2}$ is the truncation of a regular map. Using the GAP package RAMP [23], we were often able to find a single possible candidate based on the Schläfli symbol, size, and Petrial. These regular maps are denoted in one of several possible ways:

  • (8a) The symbol $\{p, q \mid h\}_r$ denotes the universal regular map of type $\{p, q\}$ with Petrie paths of length r and 2-holes of length h. (A 2-hole of a polyhedron is a walk in the 1-skeleton that, upon entering a vertex, leaves using the second exit on the left.) We may omit either r or h when the other parameters are already sufficient to distinguish the polyhedron, giving us symbols like $\{p,q\}_r$ and $\{p,q \mid h\}$.

  • (8b) A symbol like $\{3,10\}*120b$ denotes a regular polyhedron in Ref. [24].

  • (8c) A symbol like N16.5 denotes a nonorientable regular map at https://www.math.auckland.ac.nz/~conder/RegularNonorientableMaps602.txt while a symbol like R3.8 denotes an orientable regular map in https://www.math.auckland.ac.nz/~conder/RegularOrientableMaps101.txt (see Ref. [25]).

For the reader who wants to double-check the information in the tables, here are some helpful hints:

• (a) Since the Petrial of ${\mathcal {P}}$ has the same 1-skeleton as ${\mathcal {P}}$, in order to find the isomorphism type of ${\mathcal {P}}^\pi$ there are usually only a few polyhedra to check. In many cases, a possible candidate for the Petrial of ${\mathcal {P}}$ has facets that share many vertices in a row in ${\mathcal {P}}$, and so it cannot be the Petrial.

• (b) Many 3-orbit polyhedra have a pair of facets that share multiple edges, which means that a geometric dual cannot exist. In particular, Proposition 11 rules out several possibilities. In other cases, it is enough to consider the possible Schläfli symbol of the dual and see that it does not occur among polyhedra with the appropriate symmetry group and class.

• (c) Note that $\zeta _2$ never changes the size of a 3-symmetric facet or indeed any 3-symmetric walk in a 1-skeleton. Therefore, if ${\mathcal {P}}$ is in class $3^{1,2}$, then the Petrie polygons of ${\mathcal {P}}^{\zeta _2}$ are the same size as those of ${\mathcal {P}}$. Thus, it often turns out that $({\mathcal {P}}^{\zeta _2})^\pi = ({\mathcal {P}}^{\pi })^{\zeta _2}$.

• (d) Perhaps the most time-consuming thing to verify is the finer structure of the facets; i.e., distinguishing $n_p$ from $n_s$ and distinguishing among $n_r$, $n_h$, and $n_{rh}$. The following observations may help:

  • (4a) If the symmetry group is $[3,4]^+$ or $[3,5]^+$, then the 3-symmetric facets must be type $n_h$.

  • (4b) A 3-symmetric facet with n odd must be either $n_r$ or $n_h$.

  • (4c) A 3-symmetric facet $n_r$ has its vertices contained in two parallel planes (or possibly a single plane for a certain choice of parameters in that equivalence class).

  • (4d) In many cases, Proposition 21 rules out type $n_h$ for a 3-symmetric facet.

• (e) When looking for combinatorially regular polyhedra, keep in mind that if ${\mathcal {P}}$ is combinatorially regular, then so is ${\mathcal {P}}^{\pi }$. This cuts the search space down considerably, since most 3-orbit polyhedra in class $3^1$ have Petrials where the two types of facets have different sizes.

5 Affinely irreducible 3-orbit polyhedra

In the following sections, we shall enumerate all 3-orbit polyhedra ${\mathcal {P}}$ in ${\mathbb {E}}^3$ where $G({\mathcal {P}})$ is irreducible. Since the vertex set spans the entire 3-dimensional space, the flag stabilizers in $G({\mathcal {P}})$ are trivial.

5.1 General results on symmetry groups of 3-orbit polyhedra

Not every finite irreducible group of isometries of ${\mathbb {E}}^3$ is the symmetry group of a 3-orbit polyhedron.

Theorem 19

There are no 3-orbit polyhedra ${\mathcal {P}}$ in ${\mathbb {E}}^3$ whose symmetry group $G({\mathcal {P}})$ is either $[3,3]^+$ or $[3,3]^*$.

Proof

This follows from Proposition 8, since neither $[3,3]^+$ nor $[3,3]^*$ is generated by its set of involutions. To see this, note that all involutions in these two groups fix each coordinate axis setwise. $\square$

Henceforth we may put aside the groups $[3,3]^+$and $[3,3]^*$. Furthermore, not every finite irreducible group that remains is the symmetry group of a vertex-intransitive 3-orbit polyhedron.

Theorem 20

There are no 3-orbit polyhedra in class $3^{0,1}$ whose symmetry group is $[3,4]^+$ or $[3,5]^+$.

Proof

By Lemma 7, each vertex of a polyhedron in class $3^{0,1}$ has a stabilizer that is generated by two involutions. When the symmetry group is $[3,4]^+$ or $[3,5]^+$, the only involutions are half-turns, and the only point that is stabilized by distinct half-turns is the center of the polyhedron, contradicting Proposition 2. $\square$

Let us discuss our general approach in the following sections. In each section, we will focus on a particular group G. We start by including a table with information about the group, which looks like the following:

Information about [3, 3]
Description	Symmetry group of a regular tetrahedron ${\mathcal {T}}$
Order	24
Admissible vertex orbit sizes	4, 6, 12
Involutions	6 plane reflections
	3 half-turns whose mirrors join midpoints of edges of ${\mathcal {T}}$

For the row ‘Admissible vertex orbit sizes’, we start by considering all possible vertex orbit sizes. This is done by starting with a fundamental region of the group and considering all possible vertex placements. It is also sometimes helpful to fix a convex polyhedron ${\mathcal {P}}$ such that $G({\mathcal {P}}) = G$, and then observe the behavior of G on points on the surface of ${\mathcal {P}}$. For example, [3, 3] is the symmetry group of a regular tetrahedron ${\mathcal {T}}$, and in addition to the unique fixed point, the orbit of a point under [3, 3] can have size 4, 6, 12, or 24 (corresponding, respectively, to the vertices of ${\mathcal {T}}$, the midpoints of edges of ${\mathcal {T}}$, points on edges of ${\mathcal {T}}$ other than midpoints, and generic points that do not lie on any reflection plane). Since the number of vertices in each orbit must be larger than 1 (by Proposition 2) and since each vertex is fixed by some involution (see Lemma 7), some vertex orbit sizes are not admissible and so they are excluded from the table.

For $G \in \{[3,3], [3,4], [3,5]\}$, where there is the possibility of polyhedra in class $3^{0,1}$, we then include a table that summarizes the basic combinatorics that are possible for such polyhedra:

Vertex-intransitive polyhedra with group [3, 3]
1-symmetric edges	6
2-symmetric edges	12
1-symmetric vertices	Four 3-valent
2-symmetric vertices	Four 6-valent or six 4-valent

Similarly, for vertex-transitive polyhedra, we start with the following table, which uses Proposition 6, Lemma 7, and Proposition 10:

Vertex-transitive polyhedra with group [3, 3]
1-symmetric edges	6
2-symmetric edges	12
Vertices	12 3-valent

In both cases, we are sometimes able to rule out other sizes of vertex orbit using further geometric arguments. Then, we identify the possible 1-symmetric and 2-symmetric edges by considering the sizes of orbits of pairs of points, and considering that we need the 1-skeleton to be connected. Finally, once we have identified a possible 1-skeleton, we determine the possible facets, and then check which choices of facets give us connected vertex-figures.

Some of the 1-skeleta that arise are twin graphs (short for graphs with twin vertices). We recall that a twin of a vertex v is a vertex $v'$ with the same neighborhood as v (see for example Ref. [26]); to construct the twin of a graph with vertex set $\{v_1,\ldots ,v_m\}$ we add vertices $v_1',\ldots , v_m'$ and an edge from $v_i'$ to each neighbor of $v_i$ for every $i \in \{1,\ldots ,m\}$. We can similarly define a twin 1-skeleton by taking an existing 1-skeleton and then making a copy of the vertices, but dilated so that the new vertices do not overlap the old ones; then we can add the line segments necessary so that the graph of the new 1-skeleton is a twin graph. We will use $Tw_{[p,q]}$ to denote the twin 1-skeleton of the polyhedron $\{p,q\}$.

When studying vertex-transitive polyhedra ${\mathcal {P}}$ with $G({\mathcal {P}}) \in \{[3,4],[3,5]\}$, we may prefer to visualize them as 2-dimensional objects. The vertex set is contained in a sphere ${\mathbb {S}}^2$, and the edges may be projected to geodesics of ${\mathbb {S}}^2$, as long as they do not join antipodal points. The combinatorics of ${\mathcal {P}}$ can then be completely described in terms of the set of points and projections of edges in ${\mathbb {S}}^2$, by establishing which cycles correspond to facets.

This visualization is particularly useful when ${\mathcal {P}}$ is centrally symmetric, since then we can project its immersion to ${\mathbb {S}}^2$ into the projective plane ${\mathbb {P}}^2$. There we represent the image of ${\mathcal {P}}$ under the projection as a set of points with a set of geodesics between them representing edges, and indicating which of those edges are in the same facet. As we shall see, the projections of k-gonal facets in ${\mathbb {S}}^2$ may transform into k-gonal polygons, (k/2)-gonal polygons, or degenerate polygons where some vertices (but not all) are visited twice, when projected to ${\mathbb {P}}^2$.

For our analysis, we will search for all i-symmetric polygons ($i \in \{1,2,3\}$) with perhaps multiple vertices in ${\mathbb {P}}^2$ in some given graphs with pre-established 1- and 2-symmetric edges. Each i-symmetric edge has two distinct lifts from ${\mathbb {P}}^2$ to ${\mathbb {S}}^2$ (one with endpoints u and v and the other with $-u$ and $-v$, say) and so each 1-symmetric polygon has two distinct lifts to a polygon in ${\mathbb {S}}^2$, while 2- and 3-symmetric polygons have each 4 lifts to ${\mathbb {S}}^2$.

5.2 Vertex-transitive polyhedra with non-rigid vertex sets

In the enumeration of 3-orbit polyhedra, we shall describe several vertex-transitive polyhedra whose vertices have one degree of freedom (up to similarity). Here, we develop the terminology and strategy to follow for such cases. We start by determining when this occurs.

Assume that ${\mathcal {P}}$ is a vertex-transitive 3-orbit polyhedron whose vertices have one degree of freedom (up to similarity). This means that we may move continuously a vertex $v_0$ along some specific trajectory to points $v_0'$, and move simultaneously every vertex $v_0 T$ of ${\mathcal {P}}$ to the points $v_0' T$, while carrying along edges and facets in the obvious way, in such a fashion that at every time (meaning, for every point $v_0'$ of the trajectory) the symmetry group remains unchanged. This occurs only if the vertex stabilizer consists only of a plane reflection and the identity, where the vertex may be chosen as any generic point in the reflection plane, say at distance r from o. In this situation, the trajectory is a section of circle centered at o of radius r contained in the reflection plane. If ${\mathcal {P}}$ is such a polyhedron, then $G({\mathcal {P}})$ must contain plane reflections, and therefore, $G({\mathcal {P}}) \in \{[3,3],[3,4],[3,5]\}$. Vertex-transitivity forces the number of vertices to be $|G({\mathcal {P}})|/2$, and it follows from Proposition 6 that every vertex has degree 3. We shall show next that if ${\mathcal {P}}$ is a vertex-transitive 3-orbit polyhedron with 3-valent vertices and $G({\mathcal {P}}) \in \{[3,3],[3,4],[3,5]\}$ then the vertex stabilizers are generated by a plane reflection; in so doing, we characterize the vertices having one degree of freedom in the sense explained above.

Proposition 21

Let ${\mathcal {P}}$ be a vertex-transitive 3-orbit polyhedron with 3-valent vertices such that $G({\mathcal {P}}) \in \{[3,3],[3,4],[3,5]\}$. Then,

• (a) The stabilizer of every vertex is of order 2 and is generated by a plane reflection.

• (b) The non-trivial element in the pointwise stabilizer of each 1-symmetric edge is a plane reflection.

Proof

Let v be a vertex of ${\mathcal {P}}$. From Lemma 7, we know that there is a non-trivial involution T fixing v. Furthermore, since v is contained in 3 edges, it is contained in 6 flags. Vertex-transitivity of ${\mathcal {P}}$ forces the stabilizer of v in G to consist only of T and the identity element. Besides plane reflections, the involutions in G are half-turns, and the central inversion if $G \ne [3,3]$. The axis of every half-turn is contained in a plane reflection (from the group under consideration), implying that T cannot be a half-turn, since otherwise the stabilizer of v would have more than 2 elements. On the other hand, the central inversion only fixes the center of ${\mathcal {P}}$, and hence it cannot stabilize v. This proves the first item.

The second item follows from the first, since the pointwise stabilizer of an edge must also stabilize both its vertices. $\square$

The plane reflection in Proposition 21 (a) can be specified also in the following alternative way.

Lemma 22

Let ${\mathcal {P}}$ be a vertex-transitive 3-orbit polyhedron such that each vertex stabilizer is generated by a plane reflection. Let u, v, w be vertices of ${\mathcal {P}}$ such that there are 2-symmetric edges $\{u,v\}$ and $\{v,w\}$. Then, the reflection plane of the generator of the stabilizer of v is the bisector of the line segment between the vertices u and w.

Proof

We know that there is a symmetry of ${\mathcal {P}}$ fixing v while swapping its two 2-symmetric edges. This must be the only non-trivial element in the stabilizer of v, which by hypothesis is a plane reflection. In order to interchange u and w, the fixed set (mirror) of the reflection must be the perpendicular bisector of the line segment between the vertices u, w. $\square$

In what follows, we describe in detail the impact of the degree of freedom of the vertices. We may codify the combinatorics of ${\mathcal {P}}$ through the action of $G({\mathcal {P}})$ on the vertex set as follows. The vertices of ${\mathcal {P}}$ are $v_0 G({\mathcal {P}})$, for any given vertex $v_0$ of ${\mathcal {P}}$. Each edge may be interpreted as its pair of endpoints, written in the form $\{v_0 T_1, v_0 T_2\}$ for some $T_1, T_2 \in G({\mathcal {P}})$. Finally, each facet may be written as its sequence of vertices $(v_0 T_1, v_0 T_2,\ldots , v_0 T_k)$. In this way, the vertex set consists of a list of images of $v_0$ under elements of $G({\mathcal {P}})$ (the orbit of $v_0$), the edge set consists of pairs of elements of $G({\mathcal {P}})$ applied to $v_0$, and the set of facets consists of vectors of elements of $G({\mathcal {P}})$ applied to $v_0$.

Let R be the generator of the stabilizer of $v_0$ and let $\varPi$ be its fixed set. If we replace the base vertex $v_0$ by some other point $w_0$ in $\varPi$ such that its stabilizer under $G({\mathcal {P}})$ is also $\langle R \rangle$, then we may use the codification above to recover a polyhedron ${\mathcal {P}}_w$ combinatorially isomorphic to ${\mathcal {P}}$. The vertex set of ${\mathcal {P}}_w$ is $w_0 G({\mathcal {P}})$. The edge set is now

$$\begin{aligned} \{\{w_0 T_1,w_0 T_2\} \,: \, \{v_0 T_1, v_0 T_2\} \, \text{ is } \text{ an } \text{ edge } \text{ of }\,{\mathcal {P}}\}, \end{aligned}$$

and the set of facets is

$$\begin{aligned} \{(w_0 T_1,\ldots ,w_0 T_k) \,: \, (v_0 T_1,\ldots , v_0 T_k) \, \text{ is } \text{ a } \text{ facet } \text{ of }\,{\mathcal {P}}\}. \end{aligned}$$

By following the convention that the distance from a vertex to the center o of ${\mathcal {P}}$ is some constant r, we are in practice admitting vertices in the intersection of $\varPi$ with the sphere S(o, r) centered at o of radius r. Clearly, a small movement of the vertex is reflected as a small movement of the entire structure, and so this can be understood as a continuous family of realizations of the same combinatorial polyhedron. Two polyhedra obtained in this fashion one from the other will be said to be vt-equivalent (the ‘vt’ standing for ‘vertex-transitive’).

When choosing the vertex $w_0$ in $\varPi$, we required that its stabilizer under $G({\mathcal {P}})$ is $\langle R \rangle$. If the stabilizer is larger, then the number of vertices will be smaller, implying that the resulting structure is no longer combinatorially isomorphic to ${\mathcal {P}}$. Therefore, vt-equivalent polyhedra cannot be taken continuously along all $\varPi \cap S(o,r)$; in general, there will be a finite number of points where the resulting structure is not isomorphic to the starting polyhedron.

Figure 4 illustrates the concept of vt-equivalence. We take as example the truncated cube, as shown in (b); the black dotted lines are the 1-symmetric edges; the solid black lines are the edges of the base facet, and the dark-gray solid lines are the edges of the remaining 7 triangles. The base vertex $v_0$ is assumed to be in the reflection plane shown in (a), and so it is allowed to belong to either one of two antipodal pair of edges, or to one of two antipodal diagonals of squares. The thin dotted line in (a) indicates the intersection of the upper facet of the underlying cube with some reflection plane $\Delta$, defined so that one of the neighbors of $v_0$ by a 2-symmetric edge is its image under the reflection about $\Delta$. Figure (c) shows the base triangle $f_0$ together with the three triangles that are joined to $f_0$ by 1-symmetric edges (again in black dotted lines). This realization can be thought of as flipping the two ends of each 1-symmetric edge from (b), and carrying the triangles along while flipping the vertices. Finally, (d) shows a realization where the base vertex is chosen in a diagonal of a square. The black triangle is again the base triangle $f_0$; for convenience, the diagonals of the squares where the vertices of $f_0$ lie are also indicated in that diagram. The three triangles joined to $f_0$ by 1-symmetric edges (thin dotted lines) are shown in dark-gray lines. The convex hull of the vertex set of the realizations where $v_0$ is in a diagonal of a square is a (possibly non-Archimedean) rhombicuboctahedron.

Fig. 4

Vt-equivalent realizations of the truncated cube

The previous discussion can be understood as a variation of Wythoff’s construction, described in Ref. [1, Sect. 5.7]. In that book, the construction is confined to polyhedra, and polytopes in general, constructed from groups generated by reflections. Later, it was expanded for more general polyhedra (see for example Refs. [10,11,12]).

5.3 Vertex-transitive polyhedra with 3-valent vertices

The aim of this subsection is to simplify the enumeration of vertex-transitive 3-valent polyhedra. Our first step is to show that the enumeration of vertex-transitive 3-valent polyhedra in class $3^1$ can be obtained from that of polyhedra in class $3^{1,2}$. That result is preceded by the following technical lemma.

Lemma 23

Let ${\mathcal {P}}$ be a vertex-transitive 3-orbit polyhedron with 3-valent vertices and such that $G({\mathcal {P}}) \in \{[3,3], [3,4], [3,5]\}$. Then, ${\mathcal {P}}$ is vt-equivalent to a polyhedron whose 1-symmetric edges do not contain its center.

Proof

Let e be a 1-symmetric edge and $\varPi$ the reflection plane of the non-trivial symmetry of ${\mathcal {P}}$ that stabilizes e pointwise (see Proposition 21).

Assume that the 1-symmetric edges of ${\mathcal {P}}$ contain the center o of ${\mathcal {P}}$. Then, there exists $R \in G({\mathcal {P}})$ that swaps the endpoints of e and that it is not the central inversion. In this situation the fixed set of R intersects $\varPi$ in a line m, and e is contained in the perpendicular $m'$ to m through o in $\varPi$. Any valid choice of $v_0$ in $(S(o,r) \cap \varPi ) {\setminus } m$ (see Sect. 5.2) yields a polyhedron vt-equivalent to ${\mathcal {P}}$, but whose 1-symmetric edges do not contain o. $\square$

Proposition 24

Let ${\mathcal {P}}$ be a vertex-transitive finite 3-orbit polyhedron in ${\mathbb {E}}^3$ with 3-valent vertices and symmetry group [3, 3], [3, 4] or [3, 5]. Then, ${\mathcal {P}}^\pi$ is a polyhedron.

Proof

We only need to show that all Petrie paths are polygons, since the 1-skeleta and vertex-figures of ${\mathcal {P}}$ and ${\mathcal {P}}^\pi$ are the same.

If some Petrie path $\pi = (x_1,\ldots ,x_m,x_1)$ repeats some vertex v, then it must repeat at least one of the three edges incident to v. Suppose that $\{v,w\}$ is such an edge. Given that ${\mathcal {P}}$ is either in class $3^{1}$ or in class $3^{1,2}$, its Petrie paths can be 1-, 2-, or 3-symmetric polygons as defined in Sect. 2. We shall split the discussion according to the kind of symmetry of $\pi$.

Suppose first that $\pi$ is 1-symmetric and let S be the 1-step (combinatorial) rotation along $\pi$. Then, S has order greater than 2, and therefore, it must be either a rotation or a rotatory reflection. In either case, the vertex set of $\pi$ is $v \langle S \rangle$ and the edge set is $\{v,w\} \langle S \rangle$, implying that $\pi$ is a polygon.

Now, suppose that $\pi$ is 2-symmetric and let S be the 2-step rotation along $\pi$. In this situation, the vertex set is $\{v,w\}\langle S \rangle$ and the edge set is $\{\{v,w\},\{w,vS\}\} \langle S \rangle$ (here we are assuming that S rotates v two steps in the direction of its neighbor w). The only way to have a repetition of v in $\pi$ is if $v = wS^j$ for some $j \in \{1,\ldots ,m-1\}$, but in that case vertices become 4-valent, which contradicts our hypothesis. (For example, v would be adjacent to w, to $w S^{-1}$, to $vS^j$ and to $vS^{j+1}$.)

Finally, suppose that $\pi$ is 3-symmetric and let S be the 3-step rotation along $\pi$. There are two possible ways on which the edge $\{v,w\}$ repeats depending on the order on which the vertices v, w appear in $\pi$:

• (A) $(\dots , v, w,\ldots , v, w,\ldots )$, or

• (B) $(\dots , v, w,\ldots , w, v,\ldots )$.

These possibilities are illustrated in Fig. 5, where 1-symmetric edges are indicated in gray.

Fig. 5

Possibilities of repetition of an edge in $\pi$ (gray edges are 1-symmetric)

If the repetition happens as in (A), then $\{v,w\}$ cannot be a 1-symmetric edge (see Fig. 5b), since otherwise a power of S would take one appearance of $\{v,w\}$ to the other while preserving $\pi$. This would force the non-trivial pointwise stabilizer of $\{v,w\}$ to be a power of S, contradicting Lemma 18.

We next assume still that the repetition happens as in (A), and that $\{v,w\}$ is a 2-symmetric edge (see Fig. 5a). In this situation, precisely one of the edges at v in $\pi$ must be 1-symmetric (or a power of S would take one appearance of $\{v,w\}$ to the other while preserving $\pi$, thus contradicting the fact that the pointwise stabilizer of a 2-symmetric edge is trivial), implying that at the other spot v appears between two 2-symmetric edges. Without loss of generality, assume that in its first appearance v is incident to a 1-symmetric edge and in its second one, it is incident to two 2-symmetric edges. Then, for some k, the first appearance of w is mapped by $S^k$ to the second appearance of v while preserving $\pi$. Furthermore, there exists $T \in G({\mathcal {P}})$ that preserves $\pi$ and v while interchanging its two neighbors in its second appearance (since it is incident to two 2-symmetric edges). It follows that $S^k T \in G({\mathcal {P}})$ swaps the vertices of the 2-symmetric edge $\{v,w\}$ while fixing $\pi$, a contradiction to the fact that the symmetry group of $\pi$ induces 3 orbits on its flags.

We are left with the case where $\pi$ is 3-symmetric and the repetition occurs as in (B). We can discard the possibility of $\{v,w\}$ being 2-symmetric (see Fig. 5d) as in the case where the repetition is as in (A). The only difference is that in the current situation it suffices a power of S (the symmetry T in the previous case plays no role). Therefore, we may assume that $\{v,w\}$ is 1-symmetric (see Fig. 5c).

If $\{v,w\}$ is 1-symmetric and $\pi$ does not repeat any 2-symmetric edge, then $\{v,w\}S^{m/6} = \{v,w\}$ (the order of S is m/3 and $\{v,w\}$ repeats precisely twice). It follows that the order of S is even. The order of S cannot be 2, since otherwise $\pi$ would be a hexagon of the form (v, w, y, w, v, z), and consecutive edges $\{w,y\}$ and $\{y,w\}$ are not possible in Petrie paths. The symmetry S cannot be a rotatory reflection whose order is twice an odd number, since otherwise $S^{m/6}$ is the central inversion with respect to the center o of ${\mathcal {P}}$ (this is the case for all such rotatory reflections in G), implying that the midpoint of the edge $\{v,w\}$ contains o, which by Lemma 23, we may assume not to be true.

Fig. 6

Possibilities for the lines containing 1-symmetric edges of $\pi$

The only remaining cases are when G is either [3, 3] or [3, 4], and S is a rotation or rotatory reflection of order 4. In that situation $S^2$ is a half-turn and its axis must contain the midpoints of the two 1-symmetric edges of $\pi$. Since [3, 3] is a subgroup of [3, 4], we may assume that the two 1-symmetric edges of $\pi$ lie on the two lines shown in either of the cubes in Fig. 6. Furthermore, if y is a vertex of $\pi$ that is not incident to a 1-symmetric edge of $\pi$, then by Lemma 22 the stabilizer of y must be the bisector of the line segment between its two neighbors $x_1$ and $x_2$. If $x_1$ and $x_2$ are endpoints of distinct edges in the cases illustrated in Fig. 6a or b, then their bisector contains no vertex in the orbit of $x_1$ under $G({\mathcal {P}})$ (recall that the stabilizer of $x_1$ has size 2, and therefore, $x_1$ cannot be a vertex or the midpoint of the edge of a cube in Fig. 6). This discards the possibility of S being a fourfold rotation. On the other hand, if S is a rotatory reflection and $x_1$ and $x_2$ are endpoints of distinct edges in the cases illustrated in Fig. 6c or d, then y must lie on the invariant plane of S, so that there is a symmetry fixing $\pi$ and y while swapping $x_1$ and $x_2$. In the case given by Fig. 6d, there is no vertex in the orbit of $x_1$ under $G({\mathcal {P}})$ in the invariant plane of S, while in the case given by Fig. 6c the vertices in the invariant plane of S that are in the orbit of $x_1$ under $G({\mathcal {P}})$ do not belong to the bisector of the line segment $x_1 x_2$. This discards the possibility of S being a rotatory reflection, and finishes the proof. $\square$

Corollary 25

The enumerations of all finite polyhedra with 3-valent vertices in class $3^1$ with symmetry groups [3, 3], [3, 4] and [3, 5] consist of the Petrials of all finite polyhedra with 3-valent vertices and with those symmetry groups in class $3^{1,2}$.

In view of Corollary 25, we may restrict our attention to polyhedra in class $3^{1,2}$. One kind of 3-orbit polyhedron in that class that frequently occurs is the truncation of a regular polyhedron. In what follows, we shall show that every 3-orbit polyhedron in class $3^{1,2}$ is vt-equivalent to either the truncation of a regular polyhedron, or to its image under the operation $\zeta _2$. Let ${\mathcal {P}}$ be a polyhedron in class $3^{1,2}$ with 3-valent vertices and symmetry group in $\{[3,3],[3,4],[3,5]\}$.

We continue the analysis of the position of the edges of ${\mathcal {P}}$ relative to the center of ${\mathcal {P}}$.

Lemma 26

Let ${\mathcal {P}}$ be a vertex-transitive 3-orbit polyhedron in class $3^{1,2}$ with 3-valent vertices and such that $G({\mathcal {P}}) \in \{[3,3], [3,4], [3,5]\}$. Then, the 2-symmetric edges do not contain the center of ${\mathcal {P}}$.

Proof

Let $f_1$ be a 1-symmetric facet containing the base vertex $v_0$. Then, there exists a rotation or rotatory reflection $R \in G({\mathcal {P}})$ whose order is at least 3, and acts like a 1-step rotation on the (2-symmetric) edges of $f_1$. We can conclude that the 2-symmetric edges do not intersect the center of ${\mathcal {P}}$, since otherwise it would not be possible to form $f_1$ from the orbit of such edges under $\langle R \rangle$. $\square$

It can be shown that the conclusion of Lemma 26 is also true for 1-symmetric edges when $G({\mathcal {P}}) \in \{[3,4], [3,5]\}$. However, if $G({\mathcal {P}}) = [3,3]$, there is a particular choice of $v_0$ where the 1-symmetric edges do contain the center of ${\mathcal {P}}$. For our purposes, we only need Lemma 23.

The next proposition is a direct consequence of Lemmas 26 and 23.

Proposition 27

Let ${\mathcal {P}}$ be a vertex-transitive 3-orbit polyhedron in class $3^{1,2}$ with 3-valent vertices and such that $G({\mathcal {P}}) \in \{[3,3], [3,4], [3,5]\}$. Then, ${\mathcal {P}}$ is vt-equivalent to a polyhedron, none of whose edges contains the center of ${\mathcal {P}}$.

We next analyze the 1-symmetric facets of ${\mathcal {P}}$. Each of them must be invariant under a rotation or a rotatory reflection of order at least 3 that permutes its edges cyclically.

Lemma 28

Let ${\mathcal {P}}$ be a polyhedron in class $3^{1,2}$ with 3-valent vertices and symmetry group in $\{[3,3],[3,4],[3,5]\}$. Let f be a 1-symmetric facet of ${\mathcal {P}}$, and let $R \in G({\mathcal {P}})$ have order at least 3 and cyclically permute the edges of f.

• (a) If R is a rotatory reflection, then ${\mathcal {P}}$ is vt-equivalent to a polyhedron with skew 1-symmetric facets.

• (b) If R is a rotation, then ${\mathcal {P}}$ is vt-equivalent to a polyhedron whose center is not in the affine span of any of its 1-symmetric facets.

Proof

Let v be a vertex of f and let $\varPi$ the reflection plane of $G({\mathcal {P}})$ containing v. Then, the reflection about $\varPi$ preserves f, since it is the only non-trivial symmetry of ${\mathcal {P}}$ that fixes v, whereas f is 1-symmetric. It follows that the axis $\ell$ of R is contained in $\varPi$.

If R is a rotatory reflection and f is planar or R is a rotation and the affine span of f contains the center of ${\mathcal {P}}$, then v is in the plane $\Delta$ perpendicular of $\ell$ at the center of ${\mathcal {P}}$, and we may choose v as a point of $\varPi$ outside $\Delta$ to obtain another polyhedron ${\mathcal {Q}}$ vt-equivalent to ${\mathcal {P}}$. If R is a rotatory reflection, ${\mathcal {Q}}$ has skew 1-symmetric facets; if R is a rotation then the affine span of the 1-symmetric facets of ${\mathcal {Q}}$ does not contain the center of ${\mathcal {Q}}$. $\square$

In view of Proposition 27 and Lemma 28, we say that a vertex-transitive 3-orbit polyhedron ${\mathcal {P}}$ in class $3^{1,2}$ with 3-valent vertices and symmetry group in $\{[3,3],[3,4],[3,5]\}$ is in general position whenever

• none of its edges contains the center of ${\mathcal {P}}$,

• either ${\mathcal {P}}$ has skew 1-symmetric facets, or it has planar 1-symmetric facets whose affine spans do not contain the center of ${\mathcal {P}}$.

In order to relate polyhedra with planar 1-symmetric facets with polyhedra with non-planar facets, we need the following result.

Proposition 29

Let ${\mathcal {P}}$ be a 3-orbit polyhedron in class $3^{1,2}$ in general position whose vertices are 3-valent, whose 1-symmetric facets are non-planar (resp. planar), and whose symmetry group is in $\{[3,3],[3,4],[3,5]\}$. Then, ${\mathcal {P}}^{\zeta _2}$ is a 3-orbit polyhedron in class $3^{1,2}$, and its 1-symmetric facets are planar (resp. non-planar).

Proof

By construction, ${\mathcal {P}}^{\zeta _2}$ has 1-symmetric and 2-symmetric facets; hence, it is in class $3^{1,2}$.

Now, if each 1-symmetric facet f of ${\mathcal {P}}$ lies on a plane $\varLambda$, then the 2-symmetric edge of ${\mathcal {P}}^{\zeta _2}$ induced by each edge of f has one endpoint in $\varLambda$ and the other in the image of $\varLambda$ under the central inversion at the center of ${\mathcal {P}}$ (which is not equal to $\varLambda$, by general position). Hence, the 1-symmetric facets of ${\mathcal {P}}^{\zeta _2}$ are skew. Similarly, if the vertices of f lie on two parallel planes (images of one another under the central inversion), then the 1-symmetric facets of ${\mathcal {P}}^{\zeta _2}$ are planar.

Suppose that we can show that the 2-symmetric facets are polygons. Since the 1-symmetric facets of ${\mathcal {P}}^{\zeta _2}$ are always polygons, we have that all its facets are indeed polygons. By definition, ${\mathcal {P}}^{\zeta _2}$ satisfies the first two items of our definition of polyhedron. Furthermore, each vertex v lies on the plane $\varPi$ fixing the 1-symmetric edge at v, and the reflection across that plane swaps the two 2-symmetric edges at v. It follows that in ${\mathcal {P}}^{\zeta _2}$ the 1-symmetric edge at v stays in $\varPi$ while the two other neighbors of w do not belong to that plane. From this, we conclude that the vertex-figures are triangles, and hence ${\mathcal {P}}^{\zeta _2}$ is a polyhedron.

It remains to prove that the 2-symmetric facets are polygons. Let F be a 2-symmetric facet of ${\mathcal {P}}$. If $F = (v_1, \ldots , v_k)$ then $F^{\zeta _2} = (v_1, v_2, -v_3, -v_4, v_5,\ldots )$, assuming that the 2-symmetric edges are the edges from $v_i$ to $v_{i+1}$ when i is even. The only way that $F^{\zeta _2}$ can fail to be a polygon is if some vertex is repeated and incident to two distinct 2-symmetric edges.

Therefore, suppose that $F^{\zeta _2}$ has 2-symmetric edges $\{u, v\}$ and $\{u, w\}$ with $v \ne w$. Then, there must have been 2-symmetric edges $\{u, -v\}$ and $\{u, -w\}$ among the edges of F and $-F$, and since these were both polygons, it follows that F contains just one of those edges; say $\{u, -v\}$. (In particular, this implies that $F \ne -F$.) Then, F and $-F$ have the vertex u in common. Since F is vertex-transitive and the central inversion commutes with all symmetries of F, it follows that F and $-F$ have all their vertices in common. Now, since ${\mathcal {P}}$ is 3-valent, every vertex is incident to a unique 1-symmetric edge. Thus, since F and $-F$ share all their vertices, they also share all their 1-symmetric edges.

Consider a vertex v of F and let $\{u, v\}$ be the 1-symmetric edge at v. Let $\{v, w\}$ be the 2-symmetric edge at v that F contains. If $-F$ contains $\{v, w\}$, then F and $-F$ would have two edges in a row in common, which would cause the vertex-figure at v in ${\mathcal {P}}$ to be non-polygonal. Therefore, $-F$ contains a different 2-symmetric edge at v, and since ${\mathcal {P}}$ is 3-valent, there is a unique other choice $\{v, w'\}$. Recall that F and $-F$ share all their vertices, and so $w'$ is a vertex of F as well. It follows that if we start at any vertex of F and follow any edge out of it, then we reach another vertex of F, so by the connectivity of ${\mathcal {P}}$ it follows that F (and $-F$) contain all the vertices of ${\mathcal {P}}$. If F is a 2k-gon, then ${\mathcal {P}}$ has 2k vertices, and the dihedral stabilizer of ${\mathcal {P}}$ acts vertex-transitively on them. However, no dihedral subgroup of [3, 3], [3, 4] or [3, 5] has an orbit of size 12, 24 or 60, respectively, and those are the required numbers of vertices of 3-valent 3-orbit polyhedra for each group. We may conclude that the 2-symmetric facets of ${\mathcal {P}}^{\zeta _2}$ are polygons, and ${\mathcal {P}}^{\zeta _2}$ itself is a polyhedron. $\square$

Lemma 30

Let ${\mathcal {P}}$ be a vertex-transitive 3-orbit polyhedron in class $3^{1,2}$ with planar 1-symmetric facets in general position such that $G({\mathcal {P}}) \in \{[3,3], [3,4], [3,5]\}$, and let e be an edge. Then, there exists a plane reflection $T \in G({\mathcal {P}})$ that preserves e while swapping its ends.

Proof

First, suppose that e is 1-symmetric. Then, by Lemma 23, there exists ${\mathcal {Q}}$ vt-equivalent to ${\mathcal {P}}$ whose 1-symmetric edges do not contain the center of ${\mathcal {Q}}$. In that situation, the three non-trivial symmetries preserving an edge of ${\mathcal {Q}}$ must be two plane reflections and a half-turn. Since $G({\mathcal {P}}) = G({\mathcal {Q}})$ and their symmetries act in the same way on their edges, we conclude that e is invariant under two plane reflections; one fixes it pointwise while the other swaps its endpoints.

Now, if e is 2-symmetric let f be the 1-symmetric facet that contains it. Then, there exists $T \in G({\mathcal {P}})$ that preserves f while interchanging the endpoints of e. Since T must fix the axis of the rotation or rotatory reflection that preserves f acting regularly on its edges, and T also must fix the midpoint of e, it must be a plane reflection. $\square$

Now, we are ready for the remaining two main theorems of this section.

Theorem 31

Let ${\mathcal {P}}$ be a vertex-transitive 3-orbit polyhedron in class $3^{1,2}$ in general position with 3-valent vertices and planar 1-symmetric facets such that $G({\mathcal {P}}) \in \{[3,3], [3,4], [3,5]\}$. Then, ${\mathcal {P}}$ is vt-equivalent to a truncation of a regular polyhedron.

Proof

Up to vt-equivalence, we may assume that the convex hulls of the 1-symmetric facets are small, and in particular that they do not intersect. Let $f_0$ be a 1-symmetric facet of ${\mathcal {P}}$. Recall that its edges are 2-symmetric. Our goal is to construct a regular polyhedron ${\mathcal {Q}}$ such that ${\mathcal {P}}$ is a truncation of ${\mathcal {Q}}$.

Since $f_0$ is 1-symmetric, for each of its vertices v there exists $T_v \in G({\mathcal {P}})$ that preserves $f_0$ and v, interchanging the two 2-symmetric edges incident to v. Since v is 3-valent, this means that $T_v$ fixes pointwise the 1-symmetric edge incident to v. From Proposition 21 (b), we conclude that $T_v$ is a plane reflection. Furthermore, if e is an edge of $f_0$ containing v, we know from Lemma 30 that the symmetry $T_e$ of ${\mathcal {P}}$ fixing e while swapping its endpoints is also a plane reflection. We conclude that the stabilizer of $f_0$ in $G({\mathcal {P}})$ is generated by the two plane reflections $T_v$ and $T_e$. As a consequence, there is a rotation $T_eT_v$ with axis $\ell _0$ mapping every edge of $f_0$ to the next one. This rotation cyclically permutes the 1-symmetric edges incident to the vertices of $f_0$. See Fig. 7.

Fig. 7

Local picture of the 1-symmetric facet $f_0$

Given any vertex w of $f_0$, we intend to extend the 1-symmetric edge $e_{w}$ at w until it meets with $\ell _0$. To justify that this can be assumed to occur, note first that both $e_{w}$ and $\ell _0$ are contained on the invariant plane of the reflection $T_w$ that fixes w while interchanging its two 2-symmetric edges. Now, if $\ell _0$ and $e_{w}$ were parallel, then our assumption of the small size of the convex hulls of the 1-symmetric facets implies that the 1-symmetric edges at vertices of $f_0$ would join $f_0$ with the 1-symmetric facet that winds around $\ell _0$ but on the other side of the center of ${\mathcal {P}}$. If $G({\mathcal {P}}) = [3,3]$, there is no such facet (the group does not act transitively on the rays of the threefold rotation axes), while if $G({\mathcal {P}}) \in \{[3,4], [3,5]\}$, then the two facets winding around $\ell _0$ and the 1-symmetric edges joining them would form a connected component of the 1-skeleton of ${\mathcal {P}}$ that is not invariant under the entire group, a contradiction. We conclude that $\ell _0$ intersects the line spanned by $e_{w}$. Finally, if we are in the situation where $\ell _0$ and $e_{w}$ intersect, then we may use vt-equivalence and move w by choosing it in the plane of the reflection $T_w$ as its image under the reflection by $\ell _0$ (in this plane), and adjust the remaining vertices accordingly. (See Fig. 8.) In that situation, the 1-symmetric edges will not intersect the rotation axes preserving 1-symmetric facets containing their endpoints. Therefore, it is possible to extend each 1-symmetric edge at both ends until it meets the rotation axes of the 1-symmetric facets containing its vertices. Let S be that graph.

Fig. 8

Adjusting the placement of w so that $\ell _0$ and $e_w$ do not intersect

Each 2-symmetric facet f of ${\mathcal {P}}$ has alternating 1- and 2-symmetric edges. Each 1-symmetric edge e of f has its representative $e'$ in the graph S; furthermore, two consecutive 1-symmetric edges $e_1$ and $e_2$ around f have endpoints in a 2-symmetric edge of ${\mathcal {P}}$, implying that $e_1'$ and $e_2'$ have an endpoint in common (namely, the common intersection of the affine spans of $e_1$ and $e_2$ with $\ell _0$). It follows that f induces a polygon $f'$ in S. It is easy to see that the union of the polygons $f'$ in S arising from 2-symmetric facets f of ${\mathcal {P}}$ form a polyhedron ${\mathcal {Q}}$. The symmetries of ${\mathcal {P}}$ all preserve ${\mathcal {Q}}$, and since the former has 3 times as many edges as the latter, we can conclude that ${\mathcal {Q}}$ is regular. Clearly, ${\mathcal {P}}$ is a truncation of ${\mathcal {Q}}$. $\square$

Theorem 32

Let ${\mathcal {P}}$ be a vertex-transitive 3-orbit polyhedron in class $3^{1,2}$ in general position with 3-valent vertices and non-planar 1-symmetric facets such that $G({\mathcal {P}}) \in \{[3,3], [3,4], [3,5]\}$. Then, ${\mathcal {P}}$ is vt-equivalent to $(Tr({\mathcal {Q}}))^{\zeta _2}$ for some finite regular polyhedron ${\mathcal {Q}}$.

Proof

Proposition 29 tells us that ${\mathcal {P}}^{\zeta _2}$ has planar 1-symmetric facets, and we may choose the base vertex so that the convex hulls of the 1-symmetric facets do not intersect. By Proposition 29, ${\mathcal {P}}^{\zeta _2}$ is a polyhedron. Since the 1-symmetric facets of ${\mathcal {P}}$ are not planar, they must split into two planar 1-symmetric facets of ${\mathcal {P}}^{\zeta _2}$, none of which contains the center of ${\mathcal {P}}$ in its convex hull. We can conclude that ${\mathcal {P}}^{\zeta _2}$ is in general position. An application of Theorem 31 concludes the proof. $\square$

Combining Theorems 31 and 32 with Corollary 25 and Proposition 29 yields the following corollary:

Corollary 33

The finite vertex-transitive 3-orbit polyhedra with 3-valent vertices and irreducible symmetry group consist of

• (a) The truncations of the finite regular polyhedra,

• (b) The polyhedra obtained by applying $\zeta _2$ to these truncations, and

• (c) The Petrials of all the above.

In particular, there are 72 such polyhedra.

According to Ref. [5], there are precisely 18 finite regular polyhedra in ${\mathbb {E}}^3$. In the notation of Ref. [1], they are the 5 platonic solids $\{3,3\}$, $\{3,4\}$, $\{4,3\}$, $\{3,5\}$, $\{5,3\}$ and the 4 Kepler–Poinsot polyhedra $\{5/2,3\}$, $\{5/2,5\}$, $\{3,5/2\}$, $\{5,5/2\}$, together with their Petrials. From Theorem 31, we know that there are precisely 18 polyhedra in class $3^{1,2}$ in general position with planar 1-symmetric facets and irreducible symmetry group: the truncations of the 18 finite regular polyhedra in ${\mathbb {E}}^2$.

The truncations of the Platonic solids are Archimedean solids and are well-understood. The truncations of $\{3,5/2\}$, $\{5,5/2\}$ have planar facets that can be made regular for a certain choice of base vertex, and therefore, are listed in Ref. [10] with the numbers 71 and 47, respectively. This is not the case for the truncations of $\{5/2,3\}$ and $\{5/2,5\}$, where the 2-symmetric facets are truncated pentagrams and are not regular for any choice of base vertex. The 2-symmetric facets of the Petrials of the Platonic solids and Kepler–Poinsot polyhedra are truncated skew polygons, and therefore, are non-planar. The truncations of these 18 polyhedra are all described in Ref. [14, Chapter 3] as the polyhedra $P^{0,1}$ listed in each subsection.

Applying $\zeta _2$ to each of the truncations transforms planar 1-symmetric facets into skew 1-symmetric facets and vice-versa, as long as we start with a truncation in general position. Furthermore, in all cases it returns a polyhedron. If ${\mathcal {P}}$ is a Platonic or Kepler–Poinsot polyhedron, then the 2-symmetric facets of ${{\,\textrm{Tr}\,}}({\mathcal {P}})$ are planar and the vertices of $({{\,\textrm{Tr}\,}}({\mathcal {P}}))^{\zeta _2}$ lie on two parallel planes; each 1-symmetric edge is contained in one of these planes while 2-symmetric edges have one endpoint in each plane. The 2-symmetric facets of the outcome of applying $\zeta _2$ to the truncations of the Petrials of the Platonic and Kepler–Poinsot solids are also non-planar with their vertices lying in two planes, but here every edge has one vertex in each of the planes.

Taking truncation of each of the finite regular polyhedra preserves the symmetry group. This is also the case when performing the $\zeta _2$ operation, with the following exception. If ${\mathcal {P}}$ is the tetrahedron or its Petrial, then ${{\,\textrm{Tr}\,}}({\mathcal {P}})$ is not centrally symmetric. Therefore, $({{\,\textrm{Tr}\,}}({\mathcal {P}}))^{\zeta _2}$ has twice as many vertices; furthermore, its symmetry group acquires the central inversion and it becomes [3, 4]. For all other choices of finite regular polyhedron ${\mathcal {P}}$, we have that the central inversion belongs to $G({\mathcal {P}})$, and the replacement of all 2-symmetric edges $\{u,v\}$ and $\{-u,-v\}$ of ${{\,\textrm{Tr}\,}}({\mathcal {P}})$ by $\{u,-v\}$ and $\{-u,v\}$ yields a connected graph. It follows that $G({\mathcal {P}}) = G(({{\,\textrm{Tr}\,}}({\mathcal {P}}))^{\zeta _2})$.

Some of the vt-equivalent realizations of the vertex-transitive 3-orbit polyhedra with 3-valent vertices and symmetry group [3, 3] will be illustrated in figures. This will not be repeated for the symmetry groups [3, 4] and [3, 5], where only one vt-equivalent realization will be illustrated for each polyhedron.

The vertex-transitive polyhedra with 3-valent vertices and symmetry group [3, 3], [3, 4], or [3, 5] are listed in Tables 3 (in the first four rows), 6 (in the first five groups), 9, and 10.

Finally, let us note that though the polyhedra in Tables 4 and 7 are not geometric truncations, they are combinatorially equivalent to truncated regular maps. Indeed, this follows from Ref. [20, Theorem 5.1].

6 Full tetrahedral group

Information about [3, 3]
Description	Symmetry group of a regular tetrahedron ${\mathcal {T}}$
Order	24
Admissible vertex orbit sizes	4, 6, 12
Involutions	6 plane reflections
	3 half-turns whose axes join midpoints of edges of ${\mathcal {T}}$

Throughout this section, we denote by ${\mathcal {T}}$ a regular tetrahedron.

There are two kinds of subgroups of [3, 3] isomorphic to ${\mathbb {Z}}_2 \times {\mathbb {Z}}_2$: the group consisting of the three half-turns and the identity, and the group generated by two reflections with perpendicular mirrors. Recall from Lemma 4 that for all classes of 3-orbit polyhedra with full tetrahedral symmetry group, there is an orbit of edges whose edge stabilizer is of this kind.

6.1 Vertex-intransitive cases

Vertex-intransitive polyhedra with group [3, 3]
1-symmetric edges	6
2-symmetric edges	12
1-symmetric vertices	Four 3-valent
2-symmetric vertices	Four 6-valent or six 4-valent

We first consider 3-orbit polyhedra in class $3^{0,1}$. Let ${\mathcal {P}}$ be one such polyhedron.

We first discard the possibility of ${\mathcal {P}}$ having 6 vertices that are 2-symmetric. Recall that the points whose orbit under [3, 3] have 6 elements are in the axes of the half-turns (aligned with the midpoints of edges of ${\mathcal {T}}$). An edge e between two of these vertices must be 1-symmetric, and such an edge is invariant under a subgroup of [3, 3] with 4 elements. However, each reflection plane of [3, 3] contains only one axis of a half-turn in [3, 3], which implies that the endpoints of e must belong to the same axis of a half-turn. This would imply that there is only one 1-symmetric edge incident to a 2-symmetric vertex, contradicting Lemma 1.

Now, we know that all 3-orbit polyhedra in class $3^{0,1}$ with full tetrahedral symmetry group have four 1-symmetric vertices, and four 2-symmetric vertices. The four vertices in each orbit must lie on the axes of the threefold rotations (each determined by a vertex and the center of the opposite triangle in ${\mathcal {T}}$).

Since the 2-symmetric vertices have degree 6, each such vertex v is adjacent to the other three 2-symmetric vertices and to the three 1-symmetric vertices that are not in the same axis as v. Up to vi-equivalence, we may assume that all vertices lie on a sphere, where vertices in the same rotation axis are at opposite sides of the center of ${\mathcal {P}}$. In this way, we get a graph vi-equivalent to the 1-skeleton of the convex Catalan solid called the triakis tetrahedron (see Fig. 9), which is the Kleetope of the tetrahedron and hence the dual of the truncated tetrahedron.

Fig. 9

Triakis tetrahedron; thick black edges belong to the underlying tetrahedron

In the 1-skeleton of the triakis tetrahedron, the only possible 3-symmetric facets are precisely the triangles of the triakis tetrahedron. If we assume otherwise, then we must skip either 2 or 4 edges at the 2-symmetric vertices. If u is a 1-symmetric vertex and its neighbors in a facet are v and w, by skipping 2 or 4 edges at v and w, we reach a vertex x adjacent by 1-symmetric edges in the same facet both to v and w, preventing the facet from being a polygon, or to be 3-symmetric (see Fig. 9, where the two possibilities of x are shown). We can conclude the following result.

Theorem 34

Up to similarity and vi-equivalence, the triakis tetrahedron is the only 3-orbit polyhedron in class $3^{0,1}$ in ${\mathbb {E}}^3$.

6.2 Vertex-transitive cases

Vertex-transitive polyhedra with group [3, 3]
1-symmetric edges	6
2-symmetric edges	12
Vertices	Twelve 3-valent

The only possibility is to have 12 3-valent vertices, and so the results in Sect. 5.3 imply the following.

Theorem 35

Up to similarity and vt-equivalence, the only vertex-transitive 3-orbit polyhedra in ${\mathbb {E}}^3$ with symmetry group [3, 3] are:

• the truncated tetrahedron ${{\,\textrm{Tr}\,}}(\{3,3\})$ in class $3^{1,2}$,

• the truncated hemi-cube ${{\,\textrm{Tr}\,}}(\{4,3\}_3)$ in class $3^{1,2}$,

• $({{\,\textrm{Tr}\,}}(\{3,3\}))^{\pi }$ in class $3^{1}$,

• $({{\,\textrm{Tr}\,}}(\{4,3\}_3))^{\pi }$ in class $3^{1}$.

Proof

We know from Theorems 31 and 32 that every polyhedron in class $3^{1,2}$ is either the truncation of a regular polyhedron, or the image under $\zeta _2$ of that truncation. The only regular polyhedra with symmetry group [3, 3] are the tetrahedron and the hemi-cube, yielding the first two items in the statement. However, as noted in Sect. 5.3, when applying $\zeta _2$ to polyhedra in the first two items the symmetry group increases to [3, 4]. We conclude that the only polyhedra in class $3^{1,2}$ are ${{\,\textrm{Tr}\,}}(\{3,3\})$ and ${{\,\textrm{Tr}\,}}(\{4,3\}_3)$. Corollary 25 is used directly to find all polyhedra in class $3^1$ from those in class $3^{1,2}$. $\square$

To conclude this section, we illustrate the two essentially different kinds of realizations of the 1-skeleton of the truncated tetrahedron (and hence of each of the vertex-transitive 3-orbit polyhedra with symmetry group [3, 3]). For sake of brevity, we shall not do the same for the groups [3, 4] or [3, 5].

The base vertex $v_0$ must belong to the mirror $\varPi$ of some plane reflection that fixes pointwise the 1-symmetric edge $e_0$ containing $v_0$. We may choose the base vertex as an interior point of the edge of ${\mathcal {T}}$ or of the edge of the dual of ${\mathcal {T}}$ contained in $\varPi$; in these cases, the convex hull is a truncated tetrahedron (possibly with non-regular hexagons) as in Fig. 10e. Alternatively, $v_0$ may be chosen as an interior point in the altitude of a triangle of ${\mathcal {T}}$ contained in $\varPi$ and which does not project to an edge of the dual of ${\mathcal {T}}$ from the center of ${\mathcal {T}}$. In this situation the convex hull is a cuboctahedron (possibly with rectangles instead of squares, and with two sets of triangles of distinct sizes) as in Fig. 10a. In both figures, ${\mathcal {T}}$ is shown in gray lines.

Fig. 10

Edges of the vertex-transitive 3-orbit polyhedra

The 1-symmetric edges are completely determined; they join a vertex with the unique other vertex in the same reflection plane of [3, 3]. One such edge e is illustrated in Fig. 10b, f for each case of vertex set. Note that the edge in Fig. 10b is not in the boundary of the convex hull of ${\mathcal {P}}$. The remaining 1-symmetric edges are the images of e under [3, 3]; each of them is parallel to an edge of ${\mathcal {T}}$. There are two possible choices of 2-symmetric edges depending on the choice of the symmetry that swaps their endpoints. They must be plane reflections, and up to conjugacy by elements in $G({\mathcal {P}})$ they can be chosen between a pair of perpendicular reflection planes of [3, 3]. A triangle for the two distinct choices is shown in Fig. 10c, d, g, h.

We summarize the enumeration in this section with the following theorem.

Theorem 36

There are 5 vertex-transitive 3-orbit polyhedra with symmetry group [3, 3], summarized in Table 3.

7 Rotational octahedral group

Information about $[3,4]^+$
Description	Orientation-preserving symmetries of a cube ${\mathcal {C}}$
Order	24
Admissible vertex orbit sizes	6, 12
Involutions	3 half-turns with mirrors parallel to the edges of ${\mathcal {C}}$
	6 half-turns with mirrors joining midpoints of opposite edges of ${\mathcal {C}}$.

Throughout this and the next section, we denote by ${\mathcal {C}}$ a cube.

Theorem 20 proves that ${\mathcal {P}}$ cannot be in class $3^{0,1}$, and so we will assume that ${\mathcal {P}}$ is vertex-transitive.

Vertex-transitive polyhedra with group $[3,4]^+$
1-symmetric edges	6
2-symmetric edges	12
Vertices	Twelve 3-valent vertices

The vertex set of ${\mathcal {P}}$ consists of the midpoints of the edges of ${\mathcal {C}}$. The convex hull of the vertex set is an Archimedean cuboctahedron. The 1-symmetric edges of ${\mathcal {P}}$ must be those joining two antipodal vertices (midpoints of edges of ${\mathcal {C}}$), since they are the only segments between pairs of points that are fixed pointwise by distinct non-trivial involutions in $[3,4]^+$.

To determine the 2-symmetric edges let $v_0$ be a vertex of ${\mathcal {P}}$, and let $e_0$ be the 1-symmetric edge incident to $v_0$ which, as we just established, must join $v_0$ to its antipode. A 2-symmetric edge $e_1$ at $v_0$ must be invariant under an involution $T_1 \in [3,4]^+$ that swaps its endpoints. We claim that this involution cannot be a half-turn about the axis $\ell$ of fourfold rotations. Indeed, those half-turns only preserve segments between midpoints of antipodal edges of ${\mathcal {C}}$ and altitudes of squares of ${\mathcal {C}}$ meeting $\ell$. It is easy to see that in these two situations the 1-skeleton of ${\mathcal {P}}$ is disconnected; in the latter case, it consists of three copies of a complete graph on 4 vertices (all lying in the same plane), whereas in the former case only of the six 1-symmetric edges.

Among the six axes of half-turns of $[3,4]^+$ that are not axes of fourfold rotations, one fixes pointwise $e_0$, one fixes $e_0$ while interchanging its endpoints, and the other four are equivalent under the stabilizer of $v_0$ in [3, 4], and therefore, yield isometric structures (in left- and right-handed versions). Figure 11a illustrates a vertex (black node) with its 1-symmetric edge (dashed line) and the four choices of 2-symmetric edges at that vertex.

Fig. 11

Edges and 1-symmetric facets of the vertex-transitive 3-orbit polyhedra

The orbit under $[3,4]^+$ of one of the candidates of 2-symmetric edges induces four triangles with disjoint vertex sets. The vertices of any of these triangles are the midpoints of three disjoint edges of a Petrie path of ${\mathcal {C}}$, and hence the center of every triangle is the center of ${\mathcal {C}}$. An orbit of 2-symmetric edges is illustrated in Fig. 11b, where the triangles are in distinct shades of gray.

The 1-skeleton of ${\mathcal {P}}$ is a connected, vertex-transitive cubic graph with girth 3 on 12 vertices. According to Ref. [27], it is isomorphic to the 1-skeleton of the truncated tetrahedron (the only graph with those properties). The facets of ${\mathcal {P}}$ are the only possible ones in the 1-skeleton of the truncated tetrahedron, described in Proposition 35, with the only difference being the way the truncated tetrahedron is realized. We will denote this 1-skeleton by $H_{[4,3]}$ and think of it as a realization of the 1-skeleton of the truncated hemi-cube; we will see later that the polyhedra with symmetry group $[3,5]^+$ can similarly be described in terms of the truncated hemi-dodecahedron and truncated hemi-icosahedron.

Theorem 37

There are four 3-orbit polyhedra in ${\mathbb {E}}^3$ with $G({\mathcal {P}})=[3,4]^+$, summarized in Table 4. In each case, the convex hull of the vertex set is an Archimedean cuboctahedron.

Fig. 12

3-symmetric and 2-symmetric facets of 3-orbit polyhedra with symmetry group $[3,4]^+$

To reinforce the visualization of the facets of the polyhedra in Theorem 37, Fig. 13 shows an Archimedean cuboctahedron with its vertices labeled. Vertices 1, 2, 3, 4 correspond to the upper square while vertices 9, 10, 11, 12 correspond to the lower square.

Fig. 13

Archimedean cuboctahedron

A sample 9-gon and a sample 12-gon of polyhedra in class $3^1$ with symmetry group $[3,4]^+$ are given by the lists of vertices (2, 7, 5, 11, 4, 10, 8, 3, 9) and (2, 7, 5, 4, 11, 1, 12, 6, 8, 10, 3, 9), respectively. A 2-symmetric 6-gon and a 2-symmetric 8-gon of polyhedra in class $3^{1,2}$ with symmetry group $[3,4]^+$ are given by the lists of vertices (5, 4, 10, 3, 9, 7) and (5, 4, 10, 8, 6, 12, 2, 7), respectively. These two last polygons can be constructed, respectively, from opposite triangles and opposite squares of the cuboctahedron.

8 Full octahedral group

Information about [3, 4]
Description	Symmetries of a cube ${\mathcal {C}}$
Order	48
Admissible vertex orbit sizes	6, 8, 12, 24
Involutions	Central inversion
	3 half-turns with mirrors parallel to the edges of ${\mathcal {C}}$
	6 half-turns with mirrors joining midpoints of opposite edges of ${\mathcal {C}}$
	3 reflections in mirrors that are parallel to the facets of a cube
	6 reflections in mirrors that contain pairs of opposite edges of a cube

8.1 Class $3^{0,1}$

Vertex-intransitive polyhedra with group [3, 4]
1-symmetric edges	12
2-symmetric edges	24
1-symmetric vertices	Six 4-valent or eight 3-valent
2-symmetric vertices	Six 8-valent, 8 6-valent, or 12 4-valent

We again set a reference cube ${\mathcal {C}}$ so that [3, 4] is the symmetry group of ${\mathcal {C}}$. We start by determining the graph induced by the 1-symmetric edges between 2-symmetric vertices.

If there are six 2-symmetric vertices, then each one is incident (that is, joined by an edge) to four 2-symmetric vertices. Thus, the 1-symmetric edges must form the 1-skeleton of an octahedron. If there are eight 2-symmetric vertices, each incident to three others, then the 1-symmetric edges either form the graph of a cube or of two disjoint tetrahedra inscribed in a cube. Finally, if there are 12 2-symmetric vertices, each incident to two others, then the 1-symmetric edges must belong to mirrors of plane reflections and, therefore, form three disjoint 4-cycles. We split our analysis according to the three possibilities of graph induced by the 2-symmetric vertices.

8.1.1 Octahedral graph

Our first case is when the 2-symmetric vertices induce the 1-skeleton of an octahedron. We split into cases depending on the number of 1-symmetric vertices.

Six 1-symmetric vertices Each 1-symmetric vertex is incident to four 2-symmetric vertices. Up to vi-equivalence, we may consider these vertices as the centers of facets of the dilated cube $2{\mathcal {C}}$, and the stabilizer of each 1-symmetric vertex has a single orbit of 2-symmetric vertices of size four, giving us the graph of a twin octahedron. We will denote the 2-symmetric vertices by 1 through 6 as in Fig. 14, and the 1-symmetric vertices will be denoted $1'$ through $6'$, with (for example) $1'$ being adjacent to the neighbors of 1.

Fig. 14

Labeling of vertices of the octahedron

Now, let F be a facet containing the 1-symmetric edge (2, 3). The non-trivial symmetry of $\mathcal P$ that fixes both endpoints is a reflection in a plane through vertices 2, 3, 4, and 5. Furthermore, this plane also contains vertices $2'$, $3'$, $4'$, and $5'$. By Lemma 5, a facet that starts with (2, 3) cannot continue to $2'$ or $4'$. Then, we may assume that F starts with $(2, 3, 1')$.

There are two symmetries that fix that edge while interchanging its endpoints: a reflection through the bisector of that edge, and a half-turn about a line through the center of that edge. One of these two symmetries must fix F. If the reflection fixes F, then F must contain the edge from $1'$ to 2 in addition to the edge from $1'$ to 3, and so we get triangular facets. The vertex-figure at 2 consists of two disjoint 4-cycles; the triangles $2 3 1', 2 5 1', 2 5 6', 2 3 6'$ induce one of the 4-cycles, while the other 4-cycle is induced by $2 1 5', 2 1 3', 2 6 3', 2 6 5'$. Therefore, we do not get a polyhedron this way. Thus, we may assume that a half-turn about the center of (2, 3) fixes F, and so F starts with $(6', 2, 3, 1')$. The image of this partial facet under the symmetry group of $\mathcal P$ includes a facet that starts $(6', 2, 5)$, one that starts $(1', 2, 5)$, and one that starts $(1', 2, 3)$. Therefore, again, we get a 4-cycle in the vertex-figure at 2, and so there are no polyhedra in this case.

Eight 1-symmetric vertices Up to vi-equivalence, we may consider the 1-symmetric vertices to be the vertices of the cube $\frac{1}{2} {\mathcal {C}}$. Each 1-symmetric vertex v is incident to three 2-symmetric vertices, and the stabilizer of v must act transitively on the neighbors of v. Thus, if v is connected to one of the nearest 2-symmetric vertices, then it must be connected to all three vertices at that distance; otherwise v will be connected to the three furthest 2-symmetric vertices instead. But these are vi-equivalent; switching every 1-symmetric vertex v with its opposite $-v$ induces the equivalence. Thus, we will assume that the 1-symmetric vertices are incident to the closest three 2-symmetric vertices. This gives us the 1-skeleton of a triakis octahedron.

We keep the labeling of the 2-symmetric vertices as in Fig. 14, and label the 1-symmetric vertices as in Fig. 15. In this way, 2-symmetric vertex 4 can be thought of as the center of the square determined by the 1-symmetric vertices c, d, h, g, and hence we assume that there is an edge from 4 to each of c, d, h and g. On the other hand, the edges in Fig. 15 are only a reference; they are not edges of the triakis octahedron.

Fig. 15

Labeling of 1-symmetric vertices

Let F be a facet containing the edge (2, 3). Up to symmetry, we may assume that the facet contains either (2, 3, b) or (2, 3, c). Suppose that the reflection through the middle of that edge fixes F. If F contains (2, 3, b), then this reflection gives us the edge between 2 and b, and so we get triangular facets (see Fig. 16a). This gives us the triakis octahedron. If instead, F contains (2, 3, c), then it also contains the edge (2, a). Then, the plane reflection that interchanges 2 with 3 and a with c is a symmetry of F, and by Lemma 3, this implies that the vertices 1 and 6 cannot be part of F. Considering the orbit of the partial facet (a, 2, 3, c) under the stabilizer of a, and throwing away those pieces that contain 1, 6, or 2 (since we have already used 2), we see that (a, 5, 4, c) must also be part of F. Thus, the facets are hexagons in the orbit of (a, 2, 3, c, 4, 5) (see Fig. 16b). It is straightforward to verify that the vertex-figures are all connected, and so this yields a polyhedron.

Finally, suppose that the half-turn through the middle of (2, 3) fixes F. Note that this half-turn does not fix any vertex, and so Remark 9 implies that F has an even number of vertices. Since 2/3 of the vertices of F must be 2-symmetric, and there are only six 2-symmetric vertices, that implies that F is a hexagon. Suppose F contains (2, 3, b). Then, it also contains the edge (2, f). Then, since b is only adjacent to vertices 1, 2, and 3, and 2 is already incident to two vertices of F, it follows that the edge (b, 1) is in F, and then applying the half-turn, we see that (f, 6) is in F. Therefore, F must contain (6, f, 2, 3, b, 1). Since F is a hexagon, this would have to be the whole facet—however, 1 is not adjacent to 6. Therefore, the last possibility is for F to contain (2, 3, c); applying the half-turn symmetry we get that F contains (e, 2, 3, c). Now, if F contains (c, 1), then it also contains (e, 6), and we would get that $F = (6, e, 2, 3, c, 1)$, which again doesn’t work since 1 and 6 are not adjacent. The only remaining possibility is $F = (5, e, 2, 3, c, 4)$ (see Fig. 16c). This is a hexagon, the vertex-figures are connected, and so we get a polyhedron. Note that the two polyhedra with hexagonal facets are not combinatorially equivalent; in the first case, each 8-valent vertex that is not contained in a given facet F is adjacent to either zero or two 3-valent vertices of F, whereas in the second case, each such 8-valent vertex is adjacent to one 3-valent vertex of F.

We have proven the following result.

Theorem 38

Up to similarity and vi-equivalence, there are three 3-orbit polyhedra ${\mathcal {P}}$ with $G({\mathcal {P}}) = [3,4]$ and with six 2-symmetric vertices. Their 1-skeleton is the triakis octahedron. One facet of each of them is illustrated in Fig. 16.

Fig. 16

Facets of 3-orbit polyhedra in class $3^{0,1}$ with six 2-symmetric vertices

8.1.2 Eight 2-symmetric vertices

Now, suppose that there are eight 2-symmetric vertices, each of degree 6. These vertices can be understood as the vertices of ${\mathcal {C}}$. Each vertex must be incident to three other vertices of this type, and so the 1-symmetric edges (which connect 2-symmetric vertices) must either form the skeleton of a cube, or else that of the stella octangula, that is, the two disjoint 1-skeleta of tetrahedra inscribed in the cube. Again, we split our analysis according on the number (6 or 8) of 1-symmetric vertices.

Six 1-symmetric vertices Each 1-symmetric vertex is incident to four 2-symmetric vertices. Up to vi-equivalence, we may consider these vertices as the centers of facets of $2{\mathcal {C}}$, each connected to the vertices of the corresponding facet of the cube. If the graph of 1-symmetric edges is a cube, then the full graph is the 1-skeleton of the tetrakis hexahedron. Otherwise, it can be thought as the 1-skeleton of the stella octangula where 6 vertices are added at the intersections of the edges, together with 24 edges from these 6 vertices along halves of the original edges of the stella octangula. We will denote the 2-symmetric vertices by 1 through 8, and the 1-symmetric vertices will be a through f (see Fig. 17), where d, e and f are opposite to b, c and a, respectively.

Fig. 17

Labeling of vertices

Let us start by assuming that the graph of 1-symmetric edges is a cube. Consider the 1-symmetric edge (1, 2), and let F be a facet containing this edge. By Lemma 5, the next vertex cannot be c. Then, without loss of generality, the facet contains (1, 2, a). If the reflection through the bisector of (1, 2) fixes F, then that implies that F is the triangle (1, 2, a). In this case, we get the tetrakis hexahedron. Now suppose that the half-turn at (1, 2) fixes F. Note that this symmetry does not fix any vertices, and so F must be either a hexagon or a 12-gon. Now, applying the half-turn to (1, 2, a), we see that F contains (b, 1, 2, a). In order for F to be a hexagon, there would need to be adjacent numbered vertices, different from 1 and 2, such that one of them is adjacent to a and the other to b. There is no such pair of vertices, and so F must be a 12-gon. There are two possibilities: one where the facets “pass through” the lettered vertices as (b, 1, 2, a, 4, 3, d, 8, 7, f, 5, 6), and one where the facets “turn” at the lettered vertices as (b, 1, 2, a, 3, 4, d, 8, 7, f, 6, 5). But “passing through” gives a disconnected vertex-figure at a.

The other choice gives us 12-gonal facets, while the vertex-figures at the 1-symmetric vertices are squares and those at the 2-symmetric vertices are hexagons.

Now, consider the graph represented on the right in Fig. 17, and the 1-symmetric edge (1, 3). Let F be a facet containing this edge. By Lemma 5, the next vertex cannot be a. Without loss of generality, F contains (1, 3, c). Suppose the reflection T through the bisector of (1, 3) fixes F. This reflection fixes vertices 2, 4, 6, and 8, and so Lemma 3 implies that none of these vertices are part of F. Now, applying T to (1, 3, c) shows us that F contains (b, 1, 3, c), and then the only possible facet is (b, 1, 3, c, 7, 5). This yields connected vertex-figures at every vertex and we get a polyhedron. Indeed, the vertex-figures at 1-symmetric vertices are squares, whereas those at 2-symmetric vertices are prismatic skew hexagons. Now, suppose instead that the half-turn $T'$ through the center of (1, 3) fixes F. In order to get connected vertex-figures at c, F must contain (1, 3, c, 2) or (1, 3, c, 7). In the first case, applying $T'$ gives us that F contains (4, e, 1, 3, c, 2). There must be a symmetry of F that sends the segment (4, e) to (3, c); indeed there are two such symmetries, both of order 2: the reflection through the bisector of (3, 4) and the half-turn through a line containing the center of (3, 4). Since these symmetries have order 2, it follows that F is the hexagon (4, e, 1, 3, c, 2), and the symmetry of F mapping (4, e) to (3, c) is in fact the reflection. This gives us connected vertex-figures (the same as in the case when the stabilizer of (1, 3) in G(F) is T instead of $T'$). The second case, where F contains (1, 3, c, 7), follows similarly and we get a polyhedron with a sample hexagonal facet (5, e, 1, 3, c, 7).

Let us show that the three polyhedra just produced are all combinatorially distinct, considering Fig. 18. We can distinguish (c) from the others by noting that the two 4-valent vertices of a facet have two common neighbors in (c) but none in common in Figs. 18d, e. Furthermore, we can distinguish (d) from (e) by noting that the four 6-valent vertices of a facet are all incident to a single 4-valent vertex in (d) but not in (e). (For example, the 6-valent vertices of the facet (1, 3, c, 2, 4, e) in (d) are all incident to vertex a, whereas in (e), there is no 4-valent vertex that is incident to all of the 6-valent vertices of the facet (1, 3, c, 7, 5, e).)

Fig. 18

Facets of 3-orbit polyhedra in class $3^{0,1}$ with eight 2-symmetric vertices

Eight 1-symmetric vertices Up to vi-equivalence, we may consider the 1-symmetric vertices to be the vertices of the cube $\frac{6}{5} {\mathcal {C}}$. Each 1-symmetric vertex v is incident to three 2-symmetric vertices. They can be either the ones corresponding to the neighbors of v in $\frac{6}{5} {\mathcal {C}}$, or the antipodes of the neighbors (corresponding to vertices at distance 2 from v in $\frac{6}{5} {\mathcal {C}}$).

If the graph induced by the 2-symmetric vertices is a cube, then we use the fact that antipodes belong to the same axis of threefold rotation, and therefore, these two choices yield vi-equivalent polyhedra. The 1-skeleton is isomorphic to the twin cube. On the other hand, if the graph induced by the 2-symmetric vertices is the union of the two tetrahedra inscribed in ${\mathcal {C}}$ then the three edges at a given 1-symmetric vertex join it with three 2-symmetric vertices in the same tetrahedron, and therefore, the 1-skeleton becomes disconnected (the union of two disjoint triakis tetrahedra). Hence, we can assume that the 1-skeleton is isomorphic to the twin cube.

Note here that the vertex-figures should all be polygons, as long as the two neighbors of a 2-symmetric vertex on a give facet belong to distinct axes of threefold rotation. The 1-symmetric vertices are all 3-valent, which implies that the vertex-figures are triangles. On the other hand, the 2-symmetric vertices are all 6-valent, which forces connectivity of the vertex-figures because of the way the different edge orbits are arranged around a 2-symmetric vertex.

We keep the labeling of 2-symmetric vertices as in the left of Fig. 17; then the 1-symmetric vertex twin to vertex j is called $j'$. Let F be a facet containing the edge (1, 2). By Lemma 5, since the reflection through the plane containing vertices 1, 2, 7, and 8 also fixes $1'$, the next vertex after 2 cannot be $1'$. Then, without loss of generality, F contains $(1, 2, 3')$. If the reflection that interchanges 1 and 2 is a symmetry of F, then F contains $(4', 1, 2, 3')$. Since this reflection does not fix any vertex, Remark 9 implies that F has an even number of sides. If F is a 12-gon, then the 1-symmetric edges of F consist of four disjoint edges in the orbit of (1, 2) under a cyclic subgroup of the symmetry group. The only such orbits are induced by either a quarter-turn about the center of the facet (2, 3, 6, 7) of the cube, or a fourfold rotatory reflection. In all cases the image of the partial facet $(4', 1, 2, 3')$ under the subgroup is disconnected. Therefore, F must be a hexagon. Then, there must be a symmetry of F that fixes $3'$ and $4'$ while sending (1, 2) to the other 1-symmetric edge of F; this must be the plane reflection through $3', 4', 5',$ and $6'$, which sends (1, 2) to (7, 8). Thus, $F = (4', 1, 2, 3', 7, 8)$.

If the half-turn is what fixes F while interchanging 1 and 2, then again Remark 9 implies that F has an even number of sides, and now F contains $(5', 1, 2, 3')$. An argument similar to the above shows that F cannot be a 12-gon. The only hexagon that contains $(5', 1, 2, 3')$ and is stabilized by the half-turn is $(5', 1, 2, 3', 7, 8)$.

To see that the two polyhedra we obtain are combinatorially distinct, consider all the neighbors of the 3-valent vertices of a facet. In the first type of facet, the remaining two 6-valent vertices are adjacent, but in the second type, they are not.

The polyhedra found in this case are enumerated in the following result.

Theorem 39

Up to similarity and vi-equivalence, there are seven 3-orbit polyhedra ${\mathcal {P}}$ in class $3^{0,1}$ with $G({\mathcal {P}}) = [3,4]$ and with eight 2-symmetric vertices. Two of them have the tetrakis hexahedron as their 1-skeleton, and three have as 1-skeleton the graph represented on the right of Fig. 17. Figure 18 shows in gray sample facets of each of these five polyhedra. The remaining two polyhedra have the twin cube as their 1-skeleton and a sample facet for each polyhedron is shown in Fig. 19, where the edges incident to the 1-symmetric vertices are omitted if they are not part of the facet.

Fig. 19

Facets of 3-orbit polyhedra in class $3^{0,1}$ with a twin cube as 1-skeleton

8.1.3 Twelve 2-symmetric vertices

Finally, suppose that there are 12 2-symmetric vertices, each of degree 4. These vertices can be understood as the midpoints of edges of ${\mathcal {C}}$ and we label them as in Fig. 20. The 1-symmetric edges must connect 2-symmetric vertices and have a stabilizer of order 4, and so the 1-symmetric edges form three disjoint 4-cycles (such as (a, c, k, i) in Fig. 20). We split the analysis on whether there are six or eight 1-symmetric vertices.

Six 1-symmetric vertices We assume that the 1-symmetric vertices are the centers of the squares of ${\mathcal {C}}$, labeled as in Fig. 20, with vertices 4, 5 and 6 opposite to 2, 3 and 1, respectively. Up to vi-equivalence, there are two graphs: we can either connect each 1-symmetric vertex with the edge midpoints of the same facet, or with the midpoints of the edges that are orthogonal to the facet.

Fig. 20

Labels of the 12 2-symmetric vertices

In the first case, consider the 1-symmetric edge (e, f). The 1-symmetric vertices that are adjacent to f are 2 and 3, both of which are fixed by the plane reflection through e, f, and g. Then, Lemma 5 implies that this graph is impossible for a 1-skeleton of a 3-orbit polyhedron.

In the second case, we get a disconnected graph. For example, one of the connected components consists of vertices 1 and 6 along with e, f, g, and h. In any case, we do not get a polyhedron with this vertex set.

Eight 1-symmetric vertices Now, the 1-symmetric vertices are assumed to be the vertices of ${\mathcal {C}}$, labeled as in Fig. 17. We must connect the eight 1-symmetric vertices each to three 2-symmetric vertices. Up to vi-equivalence, the only choice is to connect each vertex of the cube to the midpoints of the incident edges, since the remaining 2-symmetric vertices are in an orbit of size 6 under the stabilizer in [3, 4] of the given 1-symmetric vertex. We will denote this 1-skeleton by C.

The 1-symmetric vertices are 3-valent, and hence their vertex-figures are triangles. On the other hand, the vertex-figure of a 2-symmetric vertex is a quadrilateral, since the two edges incident to it on a facet do not belong to the same reflection plane of [3, 4] (one edge is 1-symmetric and the other is 2-symmetric). Therefore, any 3-symmetric polygon in this graph is a facet of a polyhedron in class $3^{0,1}$.

Without loss of generality, a facet F containing (e, f) contains (e, f, 2). Suppose that the symmetry that fixes F and (e, f) while interchanging e and f is the reflection through the plane containing a, c, and i. Then, by Lemma 3, F does not contain a, and so the facet must continue to b. In this case, the facet must be the hexagon (e, f, 2, b, d, 1).

Now, if the symmetry that fixes F and (e, f) while interchanging e and f is a half-turn, then since this half-turn fixes no vertices, Remark 9 implies that F has even length. Since F contains (e, f, 2), it must also contain the vertex 5, and then the only way to get a hexagon is with (e, f, 2, a, i, 5).

If F is a 12-gon, then its set of 1-symmetric edges must be the orbit under a cyclic subgroup of order 4 of (e, f) that contains four disjoint edges. There is only one such orbit; namely (e, f), (d, b), (g, h), and (j, l). The only possible facet that contains these and (5, e, f, 2) is the facet (5, e, f, 2, b, d, 4, h, g, 7, j, l). Finally, if F is an 18-gon, then the symmetry of F that sends (e, f) to one of the next 1-symmetric edges of F must be rotatory reflection of order 6. The conjugation of that symmetry by the half-turn that interchanges e with f is another symmetry of F; indeed it should be the inverse of the rotatory reflection. However, conjugation by a half-turn whose axis contains centers of two opposite squares does not invert any of the four sixfold rotatory reflections in [3, 4], and hence facets cannot have 18 vertices.

Note that the two polyhedra with hexagonal facets can be distinguished by considering whether the 3-valent vertices of a facet have a common neighbor or not.

The polyhedra with 12 2-symmetric vertices are enumerated in the next result.

Theorem 40

Up to similarity and vi-equivalence, there are three 3-orbit polyhedra ${\mathcal {P}}$ in class $3^{0,1}$ with $\varGamma ({\mathcal {P}}) = [3,4]$ and 12 2-symmetric vertices. The 1-skeleton, denoted C, is that of the cube but with subdivided edges on their halves, and both altitudes of each square. A sample facet for each polyhedron is shown in Fig. 21.

Fig. 21

Facets of 3-orbit polyhedra in class $3^{0,1}$ with 12 2-symmetric vertices

Let us summarize.

Theorem 41

Up to similarity and vi-equivalence, there are 13 vertex-intransitive 3-orbit polyhedra with symmetry group [3, 4], summarized in Table 5.

8.2 Vertex-transitive cases

If there are 24 vertices, then the stabilizer of each of them is generated by a plane reflection. Applying the results in Sect. 5.3, we obtain the following theorem. Recall that if the symmetry group of ${\mathcal {P}}$ is [3, 3], then that of ${\mathcal {P}}^{\zeta _2}$ must be [3, 4].

Vertex-transitive polyhedra with group [3, 4]
1-symmetric edges	12
2-symmetric edges	24
Vertices	Twelve 6-valent or 24 3-valent

Theorem 42

Up to similarity and vt-equivalence, there are 20 vertex-transitive 3-valent 3-orbit polyhedra in ${\mathbb {E}}^3$ with symmetry group [3, 4]:

• the truncations of the cube $\{4,3\}$, of the octahedron $\{3,4\}$, of $\{4,3\}^\pi$ and of $\{3,4\}^\pi$, all in class $3^{1,2}$;

• $(Tr(\{4,3\}))^{\zeta _2}$, $(Tr(\{3,4\}))^{\zeta _2}$, $(Tr(\{4,3\}^\pi ))^{\zeta _2}$, $(Tr(\{3,4\}^\pi ))^{\zeta _2}$, all in class $3^{1,2}$;

• $(Tr(\{3,3\}))^{\zeta _2}$, $(Tr(\{4,3\}_3))^{\zeta _2}$, both having as 1-skeleton that of $Tr(\{4,3\}^\zeta )$ and both in class $3^{1,2}$;

• the Petrials of each polyhedron in the previous items, all in class $3^1$.

For the remainder of the subsection, we assume that there are 12 vertices, corresponding to the midpoints of the edges of ${\mathcal {C}}$. These are the vertices of an Archimedean cuboctahedron $\mathcal{C}\mathcal{O}$. The 12 1-symmetric edges must correspond to diagonals of squares of the cuboctahedron. The 24 2-symmetric edges either coincide with the edges of the cuboctahedron, or they connect each vertex to the antipodes of its neighbors in the cuboctahedron. Thus, we obtain two possible 1-skeleta, and the operation $\zeta _2$ carries one to the other. As graphs, these are isomorphic, (verified in Sage [28]), although the isomorphism is not induced by $\zeta _2$ and we find it easier to relate the polyhedra with each skeleton via $\zeta _2$ rather than the abstract graph isomorphism. For convenience, we shall refer to the edges in the orbit of size 24 as ‘the edges of $\mathcal{C}\mathcal{O}$’, but bear in mind that they may or may not be the orbit of edges of the convex hull of the vertex set. The edges of $\mathcal{C}\mathcal{O}$ are solid, while those in the orbit with 12 elements are dashed in Fig. 22. We will denote the 1-skeleton containing the edges of the convex cuboctahedron as CO.

Fig. 22

The graph of CO. The solid edges are one edge orbit (also referred as ‘edges of $\mathcal{C}\mathcal{O}$’, and the dashed edges are another edge orbit. The top dashed edge continues as the bottom one, and the left dashed edge continues as the right one

To describe the 3-symmetric facets in both possibilities of 1-skeleton, we will use the graph in Fig. 22, which represents a distorted version of the 1-skeleton on a sphere. Since $\zeta _2$ fixes the size and type of 3-symmetric facet, this presents no difficulty when classifying the polyhedra with 1-skeleton $CO^{\zeta _2}$. In order to describe the 1-symmetric and 2-symmetric facets, we shall illustrate their images in the projective plane. With this understanding, the way the 2-symmetric edges lift depends completely on whether the 1-skeleton includes the edges of the cuboctahedron or not. The two possibilities for the lifts of the 1-symmetric edges must be considered; frequently one of them yields polygons whereas the other does not.

First, we consider the facet-transitive case (class $3^{1}$).

Each facet must use two solid edges in a row, followed by a dashed edge, two solid edges, etc. To specify the type of facets, it is enough to specify how many edges are skipped after traversing each edge on a given facet F. Whenever two consecutive solid edges of F are two edges of the same square of $\mathcal{C}\mathcal{O}$, then no matter which choice of dashed edge we make, the vertex-figures will be disconnected pairs of triangles. Any other configuration of solid-solid and solid-dashed in F yields connected vertex-figures, producing four possible types of facets: see Fig. 23. These four cases can also be derived by thinking of $\mathcal{C}\mathcal{O}$ as a convex cuboctahedron, noting that each choice (out of 2 possibilities) of consecutive solid edges together with each choice of dashed edges (also out of 2 possibilities) completely determines the non-trivial symmetry of F that fixes the vertex between two given consecutive dashed edges, as well as the non-trivial symmetry fixing a given dashed edge.

Fig. 23

The four possible facet-types for vertex-transitive, facet-transitive 3-orbit polyhedra with 12 vertices and symmetry group [3, 4], visualized on the sphere

These polyhedra are all distinct as abstract polyhedra. In the first but not the second, every pair of adjacent edges of a facet can be completed to a triangle in the graph. The third has a triangle of red vertices whereas the fourth does not. We note that though the last two polyhedra are equivelar of type $\{6,6\}$, there is no abstract regular polyhedron of type $\{6,6\}$ with 36 edges (see for example Ref. [24]). The first two are also not regular; the two facets meeting at a dashed edge share every third edge, whereas those meeting at a solid edge do not.

Now, let us consider the facet-intransitive polyhedra (in class $3^{1,2}$).

The argument used to discard two consecutive solid edges of a square of $\mathcal{C}\mathcal{O}$ in a facet works here as well. In the remaining choices the vertex-figures are connected. One type of facet uses only solid edges. If the 1-skeleton is CO, when we skip no edges we get triangles, and when we skip two edges, we get hexagons, as shown in Fig. 24a, b. The other type of facet has alternating edge types; a careful analysis gives us only 3 possible facets (see Fig. 24c–e). For clarity, Fig. 25 also shows the facet in Fig. 24e but visualized on the sphere. Each combination of 1-symmetric and 2-symmetric facet yields a polyhedron.

Fig. 24

The possible facets for a 3-orbit polyhedron in class $3^{1,2}$ with 12 vertices and symmetry group [3, 4], visualized in the projective plane. Above: the possible 1-symmetric facets. Below: the possible 2-symmetric facets

Fig. 25

One possible 2-symmetric facet for a 3-orbit polyhedron in class $3^{1,2}$ with 12 vertices and symmetry group [3, 4], visualized on a sphere

Two polyhedra with the same kind of vertex-figure, but distinct kinds of 4-gonal facets are non-isomorphic, since in the second kind of square there is a common neighbor of all vertices, but this does not occur in the first kind of square.

The details about the polyhedra with 1-skeleton $CO^{\zeta _2}$ follow similarly to those where the 1-skeleton is CO.

We summarize the discussion above in the following result.

Theorem 43

Up to similarity, there are 20 vertex-transitive 6-valent 3-orbit polyhedra in ${\mathbb {E}}^3$ with symmetry group [3, 4]. The 1-skeleton CO of 10 of them is that of the cuboctahedron together with the two diagonals of every square; the 1-skeleton of the remaining 10 is the image of CO under $\zeta _2$ (the 2-symmetric edges are those of the cuboctahedron). Four polyhedra with each 1-skeleton are in class $3^1$; a sample facet of each with CO as 1-skeleton is shown in Fig. 23. The remaining 12 are in class $3^{1,2}$; the 1-symmetric and 2-symmetric facets of those with CO as 1-skeleton are shown in Fig. 24.

Theorems 42 and 43 are summarized in the following result.

Theorem 44

Up to similarity and vt-equivalence, there are 40 vertex-transitive 3-orbit polyhedra with symmetry group [3, 4], summarized in Table 6.

9 Rotational icosahedral group

Information about $[3,5]^+$
Description	Orientation-preserving symmetry group of a regular dodecahedron ${\mathcal {D}}$
Order	60
Admissible vertex orbit sizes	30 (all points invariant under involutions are in the same orbit)
Involutions	15 half-turns with mirrors that join midpoints of opposite edges of ${\mathcal {D}}$

Throughout this and the next section, we denote by ${\mathcal {D}}$ a dodecahedron. Let ${\mathcal {P}}$ be a 3-orbit polyhedron with $G({\mathcal {P}})=[3,5]^+$.

By Theorem 20, ${\mathcal {P}}$ cannot be in class $3^{0,1}$, and so we will assume that ${\mathcal {P}}$ is vertex-transitive.

Vertex-transitive polyhedra with group $[3,5]^+$
1-symmetric edges	15
2-symmetric edges	30
Vertices	Thirty 3-valent

As in the situation when the symmetry group was $[3,4]^+$, the 1-symmetric edges must join a vertex with its antipode, since these are the only line segments between vertices of ${\mathcal {P}}$ that are fixed pointwise by an involution in $[3,5]^+$.

The 2-symmetric edges of ${\mathcal {P}}$ must be stabilized by involutions that swap their endpoints. Such involutions must be half-turns. Given a vertex $v_0$ of ${\mathcal {P}}$ there are two half-turns in $[3,5]^+$ that map it to its antipode and one that fixes it. The remaining 12 half-turns are divided into three conjugacy classes under the stabilizer of $v_0$ in [3, 5]. When constructing an edge with endpoints in $v_0$ and its image under a half-turn, two distinct choices of half-turns in the same class yield isometric collections of orbits of line segments under $[3,5]^+$, in a left- and right-handed version. Therefore, there are three essentially distinct possibilities for the 2-symmetric edges of ${\mathcal {P}}$. In Fig. 26a the square represents $v_0$, while the black dots are midpoints of edges in the intersections of the axes of representatives of the three classes of half-turns with ${\mathcal {D}}$. The corresponding 2-symmetric edges are illustrated in Fig. 26b in solid lines, whereas the dashed line indicates the 1-symmetric edge at $v_0$.

Fig. 26

Kinds of edges of vertex-transitive 3-orbit polyhedra with symmetry group isomorphic to $[3,5]^+$

Since the vertices of ${\mathcal {P}}$ are trivalent, the 2-symmetric edges of ${\mathcal {P}}$ form cycles. For one of the choices of half-turns stabilizing a 2-symmetric edge, these cycles are triangles; one such triangle is shown in Fig. 27a. For the remaining two choices of half-turns the cycles are pentagons; their vertices are the midpoints of alternate edges in some Petrie path of ${\mathcal {D}}$. One such pentagon is convex (see Fig. 27b), whereas the other one is star-shaped (see Fig. 27c). Note that the centers of all these polygons coincide with the center of ${\mathcal {D}}$.

Fig. 27

Polygons of 2-symmetric edges of vertex-transitive 3-orbit polyhedra with symmetry group isomorphic to $[3,5]^+$

It is possible to verify directly that if the 2-symmetric edges of ${\mathcal {P}}$ induce triangles, then the 1-skeleton of ${\mathcal {P}}$ is isomorphic to the truncated hemi-dodecahedron, whereas if these edges induce pentagons, then the 1-skeleton of ${\mathcal {P}}$ is isomorphic to the truncated hemi-icosahedron. These two graphs can be obtained by identifying antipodes of the 1-skeletons of the truncated dodecahedron and the truncated icosahedron, respectively. These two graphs are shown in Fig. 28 embedded in the projective plane (antipodes of the dotted circle are to be identified). The thick black lines represent the 1-symmetric edges, whereas the 2-symmetric ones are shown as thin gray segments.

Alternatively, the isomorphism of the 1-skeleton of ${\mathcal {P}}$ with the 1-skeleton of the truncated hemi-dodecahedron or truncated hemi-icosahedron can be verified by noting that the girth of the 1-skeleton must be 3 if the 2-symmetric edges form triangles, and 5 if these edges form pentagons (note that the two antipodes of the endpoints of an edge cannot be joined by an edge since $[3,5]^+$ does not contain the central inversion). According to Ref. [27], there is only one connected vertex-transitive cubic graph with 30 vertices and girth k for each $k \in \{3,5\}$; these two graphs cannot be other than the 1-skeleton of the truncated hemi-dodecahedron and that of the truncated hemi-icosahedron.

Note the similarity to what happened when the symmetry group was $[3,4]^+$; it was also the case there that the only possible 1-skeleton was isomorphic to the 1-skeleton of the truncated hemi-cube. It is not clear to us if there is a deeper reason for this coincidence. In particular, we cannot find a suitable geometric reason why this must be the case.

Fig. 28

The truncated hemi-dodecahedron and the truncated hemi-icosahedron

The 1-skeleton of the truncated hemi-dodecahedron has 120 automorphisms (this can be verified using Ref. [28]). They are the 60 automorphisms of the truncated hemi-dodecahedron as a map on the projective plane, together with another 60 automorphisms that transform the contractible 10-gons into non-contractible 10-gons. The two kinds of 10-gons correspond to the pentagons and Petrie paths of the hemi-dodecahedron. On the other hand, the 1-skeleton of the truncated hemi-icosahedron has 60 automorphisms; these correspond precisely to the automorphisms of the truncated hemi-icosahedron as a map on the projective plane.

Once we know that the geometry and combinatorics of the possible 1-skeleta of ${\mathcal {P}}$, it only remains to describe its facets. The 1-symmetric facets must be the triangles or pentagons shown in thin lines in Fig. 28; the possibilities for 2-symmetric and 3-symmetric facets are described next.

The 2-symmetric polygons of the truncated hemi-dodecahedron are necessarily 10-gons. They may be non-contractible or contractible, like those in Fig. 29a, b. The 2-symmetric polygons of the truncated hemi-icosahedron may be non-contractible hexagons or contractible decagons, like in Fig. 29c, d. Figure 30 shows the 3-symmetric facets in the 1-skeleta of the truncated hemi-dodecahedron and truncated hemi-icosahedron. In the first case they are 15-gons, and in the second they may be either 15-gons or 9-gons.

Fig. 29

2-symmetric facets

Fig. 30

3-symmetric facets

We have shown that for each symmetry type of vertex-transitive 3-orbit polyhedra there are three possible 1-skeleta, and for each choice of 1-skeleton there are two distinct choices of 2- or 3-symmetric facets. This yields 6 polyhedra in class $3^1$ and 6 more in class $3^{1,2}$.

Three consecutive edges of a polygon of each of the 6 polyhedra in class $3^1$ are shown in Fig. 31, where thin edges are 1-symmetric and fat edges are 2-symmetric. Figures (a) to (d) represent edges of 15-gons; the remaining edges of each polygon are obtained by rotating the 3 edges around the axis between the centers joining the front and back pentagons. Figures (e) and (f) represent edges of 9-gons; the remaining edges of each polygon are obtained by rotating the 3 edges around the axis between the fat vertex and its antipode.

Fig. 31

Three edges of four 3-symmetric 15-gons and two 3-symmetric 9-gons of polyhedra with symmetry group $[3,5]^+$

Now, consider the 6 polyhedra in class $3^{1,2}$. The 2-symmetric facets of four of them are 10-gons, whereas the remaining two have hexagonal 2-symmetric facets. Two consecutive edges of three of the 10-gons are illustrated in Fig. 32a–c; the remaining edges can be recovered through the fivefold rotation around the axis joining the centers of the front and back pentagons. The remaining 10-gon is shown in (d). The two hexagons are those in (e) and (f), where the fat vertex indicates the axis of threefold rotation. In all cases, fat edges are 2-symmetric and thin edges are 1-symmetric.

Fig. 32

Two edges of three 2-symmetric 10-gons, one 10-gon and two 6-gons of polyhedra with symmetry group $[3,5]^+$

Theorem 45

Up to similarity there are 12 3-orbit polyhedra in ${\mathbb {E}}^3$ with $G({\mathcal {P}}) = [3,5]^+$, summarized in Table 7. Each polyhedron occurs in a left-handed and right-handed form. In all cases, the convex hull of their vertex sets is an Archimedean icosidodecahedron.

10 Full icosahedral group

Information about [3, 5]
Description	Symmetry group of a regular dodecahedron ${\mathcal {D}}$
Order	120
Admissible vertex orbit sizes	12, 20, 30, 60
Involutions	Central inversion
	15 plane reflections
	15 half-turns with mirrors that join midpoints of opposite edges of ${\mathcal {D}}$

10.1 Class $3^{0,1}$

Vertex-intransitive polyhedra with group [3, 5]
1-symmetric edges	30
2-symmetric edges	60
1-symmetric vertices	Twelve 5-valent or 20 3-valent
2-symmetric vertices	Twelve 10-valent, 20 6-valent, or 30 4-valent

We first consider 3-orbit polyhedra in class $3^{0,1}$. Let ${\mathcal {P}}$ be one such polyhedron.

The 1-symmetric vertices either lie on the 6 axes of fivefold rotations or on the 10 axes of threefold rotations; in both cases, there are two vertices in each axis.

There cannot be 30 2-symmetric vertices, lying on the axes of twofold rotations, since in that case only the line segments between antipodal vertices would be invariant under a subgroup of [3, 5] with four elements. Then, there would be only one 1-symmetric edge incident to each 2-symmetric vertex, contradicting Lemma 1. Therefore, the 2-symmetric vertices are either 10-valent (12 of them) or 6-valent (20 of them).

10.1.1 2-symmetric vertices on fivefold axes

We first consider the case when the 2-symmetric vertices lie on the axes of fivefold rotations of [3, 5]. In this case, we may think that they are the centers of the facets of the dodecahedron ${\mathcal {D}}$. Five of the ten neighbors of each of these vertices are also 2-symmetric. These five neighbors must form an orbit under the action of the stabilizer in [3, 5] of the vertex. It follows that there are exactly two possible choices of 1-symmetric edges of ${\mathcal {P}}$, and the graph embedded in ${\mathbb {E}}^3$ induced by the 2-symmetric vertices is isometric to the 1-skeleton of either an icosahedron $\{3, 5\}$, or a great icosahedron $\{3, 5/2\}$. These are isomorphic as graphs, so for the moment we will assume that the graph is embedded as the 1-skeleton of an icosahedron.

For convenience, we label the vertices of an icosahedron as in the left of Fig. 33.

Fig. 33

Labels of the vertices and facets (‘front’ and ‘back’) of an icosahedron

If the 1-symmetric vertices lie on the axes of k-fold rotations of [3, 5], then they are k-valent and are the vertices of a dodecahedron if $k=3$, or of an icosahedron if $k=5$. In either situation, the neighbors of a 1-symmetric vertex v are an orbit under the stabilizer of v in [3, 5]. If $k=5$, there are two such choices, but they yield vi-equivalent polyhedra. Hence, we may assume that v is adjacent to the five 2-symmetric vertices that are neighbors of the 2-symmetric vertex in the same ray as v from the center of ${\mathcal {D}}$. In that case, we shall still use the numbers at the left of Fig. 33 for 1-symmetric vertices, but with a prime. On the other hand, if $k=3$, we shall locate the 1-symmetric vertices at the centers of the facets of the icosahedron induced by the 2-symmetric vertices, and use the labels in the center of Fig. 33 for the centers of the front facets, and those in the right of the same figure for the centers of the facets at the back. In that situation there are four choices, yielding two non-equivalent polyhedra: one has as 1-skeleton that of the triakis icosahedron, $K_{[3,5]}$, while the other has the graph $M_{[3,5]}$ obtained from the icosahedron by adding 20 trivalent vertices at the centers of the triangles, together with the 60 edges in the orbit of $\{a,4\}$ under [3, 5]. (The ‘M’ here stands for ‘modified Kleetope’; since the construction is similar to the ordinary Kleetope construction.) A given 1-symmetric vertex is connected to the vertices of a large equilateral triangle like the one shown in Fig. 34a. In other words, it suffices to find all polyhedra in class $3^{0,1}$ in the 1-skeleton of the triakis icosahedron $K_{[3,5]}$ and in $M_{[3,5]}$ if $k=3$, and in the twin 1-skeleton of the icosahedron $Tw_{[3,5]}$ if $k=5$. Each such polyhedron will have two realizations up to vi-equivalence; one where the 1-symmetric edges are those of an icosahedron and one where they are those of a great icosahedron.

Let us remark on the effect that $\zeta _1$ has on the 1-skeleta. For each 1-skeleton, $\zeta _1$ changes the 1-symmetric edges from the edges of an icosahedron to those of a great icosahedron, and vice-versa. Applying $\zeta _1$ to $Tw_{[3,5]}$ yields an isomorphic graph that can be understood as a twin great icosahedron $Tw_{[3,5/2]}$. Applying $\zeta _1$ to $K_{[3,5]}$ yields a 1-skeleton that is isomorphic as a graph to $M_{[3,5]}$; we will call the result $M_{[3,5/2]}$ since it is obtained from the 1-skeleton of $\{3, 5/2\}$ in the same way as $M_{[3,5]}$ is obtained from the 1-skeleton of $\{3,5\}$. Similarly, applying $\zeta _1$ to $M_{[3,5]}$ yields a 1-skeleton that is isomorphic as a graph to $K_{[3,5]}$ and which we will call $K_{[3,5/2]}$. Thus, we see that, for example, polyhedra with 1-skeleton $K_{[3,5]}$ also have a realization with 1-skeleton $M_{[3,5]}^{\zeta _1} = K_{[3,5/2]}$.

Fig. 34

1-symmetric vertex of ${\mathcal {P}}$

Recall that 3-symmetric facets have two non-equivalent automorphisms acting like reflections. Those preserving a vertex v must be plane reflections, since they must fix the center of the polyhedron, the vertex v, and the midpoint between the two neighbors of v in that facet. (Note that in the current circumstances those three points cannot be collinear.) On the other hand, those preserving a 1-symmetric edge while interchanging its endpoints may be plane reflections or half-turns. Let $f_0$ be a facet of ${\mathcal {P}}$, let $T_1 \in G({\mathcal {P}})$ fixing $f_0$ and a vertex $v_0$ of $f_0$, and let $T_2 \in G({\mathcal {P}})$ fixing an edge $e_0$ while interchanging its endpoints. Assume that $v_0$ is adjacent to a vertex of $e_0$ so that $T_1$ and $T_2$ generate the stabilizer of $f_0$.

The automorphism $T_1 T_2$ is the identity Id, or a rotation if $T_2$ is a plane reflection, or a rotatory reflection if $T_2$ is a half-turn.

Lemma 46

The symmetry $T_1 T_2$ is either the identity, a rotation of order 2 or a rotatory reflection of order 2 (that is, the central inversion).

Proof

Assume to the contrary that $T_1 T_2$ is none of the symmetries in the statement. The above arguments shows that $T_1 T_2$ is either a rotation of order k or a rotatory reflection of order 2k, for some $k \in \{3,5\}$. In the first case the facet $f_0$ invariant under $T_1$ and $T_2$ has k vertices that are 1-symmetric, and therefore, it must have 2k vertices that are 2-symmetric; furthermore, the 2-symmetric vertices are all in some plane perpendicular to the axis of $T_1 T_2$. However, there are no sets with 6 or 10 vertices of an icosahedron in the same plane perpendicular to a rotation axis, yielding the desired contradiction.

On the other hand, if $T_1 T_2$ is a rotatory reflection then $f_0$ must have 2k vertices that are 1-symmetric and 4k vertices that are 2-symmetric. In this situation the 2-symmetric vertices must be arranged in two parallel planes, and the contradiction arises again from the fact that there should be either 6 or 10 vertices of an icosahedron in the same plane, but that does not happen. $\square$

Fig. 35

3-symmetric 6-gons

If $T_1 T_2$ is a half-turn or the central inversion, then $f_0$ is a hexagon (of type $6_r$ or $6_{rh}$, respectively) and its 1-symmetric edges are antipodes of each other. (For the case of the half-turn, observe that the axis $\ell$ is the intersection of the mirrors of $T_1$ and $T_2$, so that all 2-symmetric vertices are in a plane perpendicular to $\ell$.) There is no such hexagon if the 1-skeleton of ${\mathcal {P}}$ is $K_{[3,5]}$, but there are two kinds if it is $Tw_{[3,5]}$ and two kinds if it is $M_{[3,5]}$. A sample hexagon of each kind with the labels of vertices given in Fig. 33 is illustrated in Fig. 35; (a) and (b) correspond to hexagons in $Tw_{[3,5]}$ while (c) and (d) are in $M_{[3,5]}$. (Note here that two 2-symmetric vertices of $M_{[3,5]}$ have either zero or two common 1-symmetric neighbors.) In each case, one of these two kinds is invariant under two plane reflections and the other is invariant under one plane reflection and a half-turn. In all four cases, the vertex-figures are polygons.

We claim that the four choices of ${\mathcal {P}}$ with hexagonal facets are mutually non-isomorphic (combinatorially). Those with $M_{[3,5]}$ as 1-skeleton have 3-valent 1-symmetric vertices while the other two have 5-valent 1-symmetric vertices. Furthermore, if the 1-skeleton of ${\mathcal {P}}$ is $Tw_{[3,5]}$ and the facets are invariant only under plane reflections (Fig. 35b), then the two 1-symmetric vertices of a facet (the only two of degree 5) have a common neighbor, while if the facets are invariant under a plane reflection and a half-turn (Fig. 35a), then the 1-symmetric vertices are at distance 3 in the 1-skeleton. On the other hand, if the 1-skeleton of ${\mathcal {P}}$ is $M_{[3,5]}$ and $f_0$ is invariant under a plane reflection and a half-turn (Fig. 35c), then the neighbors of the 1-symmetric vertices of $f_0$ (the only two vertices with degree 3) that are not in $f_0$ are antipodes in the icosahedron (and so they have no common neighbor), whereas if $f_0$ is invariant under two plane reflections (Fig. 35d), these neighbors themselves have common neighbors among the vertices of the icosahedron.

Finally, if $T_1 T_2$ is the identity, then the facets of ${\mathcal {P}}$ are all the triangles consisting of two 2-symmetric edges and one 1-symmetric edge. There are no such triangles if the 1-skeleton of ${\mathcal {P}}$ is $M_{[3,5]}$; a triangle for each of the other candidate 1-skeleta is shown in black in Fig. 36 with the notation in Fig. 33. If the 1-skeleton of ${\mathcal {P}}$ is $K_{[3,5]}$, then ${\mathcal {P}}$ is combinatorially isomorphic to the triakis icosahedron. Otherwise, the 1-skeleton is $Tw_{[3,5]}$ and ${\mathcal {P}}$ is combinatorially a triple cover of the icosahedron, still with polygonal vertex-figures.

Fig. 36

3-symmetric triangles

We have found six combinatorial types of polyhedra, each with two realizations, yielding 12 total. Let us explore the relationship between them by the operation $\zeta _1$. For most of the polyhedra we obtain with a given 1-skeleton, applying $\zeta _1$ to the polyhedron will give another polyhedron with the transformed 1-skeleton. Let us see why this does not always work. Let $f_0$ be a facet of ${\mathcal {P}}$ with symmetries $T_1$ and $T_2$ as above. Consider a facet $f_0'$ of ${\mathcal {P}}^{\zeta _1}$ that shares the fixed point of $T_1$ with $f_0$. Then, there are symmetries $T_1'$ and $T_2'$ of $f_0'$ that act analogously to $T_1$ and $T_2$. In particular, $T_1'$ is again a plane reflection. Now, $T_2$ switched the endpoints of some edge $\{u, v\}$ in ${\mathcal {P}}$, so the analogous symmetry $T_2'$ must switch the endpoints of an edge $\{u, -v\}$. It follows that $T_2'$ is the composition of $T_2$ with the central inversion. Therefore, $T_1' T_2'$ is the composition of $T_1 T_2$ with the central inversion.

Now, $T_1' T_2'$ is the identity (resp. the central inversion) if and only if $T_1 T_2$ is the central inversion (resp. the identity). Thus, $\zeta _1$ works to interchange triangular facets with facets $6_{rh}$. But if $T_1 T_2$ is a half-turn, then $T_1' T_2'$ is a plane reflection, which is impossible. Therefore, for example, applying $\zeta _1$ to the two polyhedra with 1-skeleton $M_{[3,5]}$ only produces one polyhedron with 1-skeleton $K_{[3,5/2]}$. Indeed, applying $\zeta _1$ to the polyhedron with facets as in Fig. 35d transforms the facet (9, d, 7, 5, a, 6) to the walk (9, d, 7, 9, k, 7) which is not a polygon.

The enumeration of the polyhedra discussed so far in this section is summarized next.

Proposition 47

Up to similarity and vi-equivalence, there are 12 3-orbit polyhedra ${\mathcal {P}}$ in class $3^{0,1}$ with $\varGamma ({\mathcal {P}}) = [3,5]$ where the 2-symmetric vertices are those of an icosahedron. The graph induced by the 2-symmetric vertices in six of them is the 1-skeleton of the icosahedron, and in the other six is the 1-skeleton of a great icosahedron.

The 1-skeleta of these 12 polyhedra are isomorphic to the triakis icosahedron, the graph $M_{[3,5]}$ defined above, or the twin icosahedron. Geometrically, there is one with each of the 1-skeleta $K_{[3,5]}$ and $K_{[3,5/2]}$, two with each of the 1-skeleta $M_{[3,5]}$ and $M_{[3,5/2]}$, and three with each of the 1-skeleta $Tw_{[3,5]}$ and $Tw_{[3,5/2]}$.

10.1.2 2-symmetric vertices on threefold axes

We move on to the case when the 2-symmetric vertices lie on the axes of threefold rotations of [3, 5], and we may assume that they are precisely the vertices of ${\mathcal {D}}$. Three of the neighbors of a 2-symmetric vertex v must be 2-symmetric, and must correspond to an orbit of vertices under the stabilizer of v in [3, 5]. The only two possibilities are that the three 2-symmetric neighbors of v in ${\mathcal {P}}$ are the three neighbors of v in ${\mathcal {D}}$, or that they are the antipodes of the neighbors of v in ${\mathcal {D}}$. It follows that the subgraph of the 1-skeleton of ${\mathcal {P}}$ induced by the 2-symmetric vertices is the 1-skeleton of either a dodecahedron, or a great stellated dodecahedron. These two are isomorphic as graphs, and for now we will assume that the 1-skeleton is a dodecahedron. We label the vertices of the dodecahedron as in the left of Fig. 37.

Fig. 37

Labels of the vertices and facets of a dodecahedron

Once again, the 1-symmetric vertices lie on the axes of k-fold rotations of [3, 5] for $k \in \{3,5\}$. If $k=3$, then they are trivalent and are the vertices of a dodecahedron, while if $k=5$ they are 5-valent and are the vertices of an icosahedron. For the case where $k=3$, we label the 1-symmetric and 2-symmetric vertices in the same ray from the center of ${\mathcal {P}}$ with the same letter, just distinguishing them by adding a prime to the 1-symmetric vertices. When considering the case where $k=5$, we label the 1-symmetric vertices as in the center and right of Fig. 37.

For the case $k=3$, there are two vertex orbits of ${\mathcal {D}}$ of size 3 under the stabilizer of a 1-symmetric vertex, and this indicates two choices of neighbors of a 1-symmetric vertex. However, these two choices are antipodal to each other and they yield vi-equivalent polyhedra, where the 1-skeleton is the twin dodecahedron $Tw_{[5,3]}$. On the other hand, if $k=5$, then there are four vertex orbits of ${\mathcal {D}}$ of size 5 under the stabilizer in [3, 5] of a 1-symmetric vertex. These give two possible 1-skeleta for ${\mathcal {P}}$ up to vi-equivalence: one is the 1-skeleton of the pentakis dodecahedron, denoted $K_{[5,3]}$, and the other is the 1-skeleton $M_{[5,3]}$ obtained from the 1-skeleton of the dodecahedron by adding 12 pentavalent vertices connected to the vertices of the large pentagons like the one shown in Fig. 34b. Hence, to describe all polyhedra in class $3^{0,1}$ with full icosahedral symmetry group it suffices to find all polyhedra in class $3^{0,1}$ with 1-skeleton $Tw_{[5,3]}$, $K_{[5,3]}$, and $M_{[5,3]}$. Each such polyhedron will have two realizations up to vi-equivalence; one where the 1-symmetric edges are those of a dodecahedron and one where they are those of a great stellated dodecahedron. Indeed, the 1-skeleta where the 1-symmetric edges are those of a great stellated dodecahedron can be thought of as $Tw_{[5/2,3]}$, $K_{[5/2,3]}$, and $M_{[5/2,3]}$, which as graphs are isomorphic to $Tw_{[5,3]}, K_{[5, 3]}$, and $M_{[5,3]}$, respectively. These two triples of graphs are related by the operation $\zeta _1$ so that $Tw_{[5,3]}^{\zeta _1}, K_{[5, 3]}^{\zeta _1}$, and $M_{[5,3]}^{\zeta _1}$ are isomorphic to $Tw_{[5/2,3]}$, $M_{[5/2,3]}$ and $K_{[5/2,3]}$, respectively. It follows that any polyhedron with 1-skeleton $Tw_{[5,3]}$ (resp. $K_{[5,3]}$, $M_{[5,3]}$) has another realization with 1-skeleton $Tw_{[5/2,3]}$ (resp. $M_{[5/2,3]}$, $K_{[5/2,3]}$).

As before, let $T_1$ and $T_2$ be symmetries of a facet $f_0$ of ${\mathcal {P}}$ preserving a vertex and an edge, respectively. Under the current situation, it is also true that $T_1$ is a plane reflection, whereas $T_2$ may be either a plane reflection or a half-turn.

The reasons explained in the proof of Lemma 46 still show that $\langle T_1 T_2 \rangle$ cannot contain a fivefold rotation, since there is no set of 10 coplanar vertices of a dodecahedron. However, now $\langle T_1 T_2 \rangle$ may contain a threefold rotation since if $k=3$, there are sets of 6 vertices in planes perpendicular to axes of threefold rotation.

If $T_1 T_2$ is a threefold rotation, then the three 1-symmetric edges of $f_0$ are coplanar. The only three such edges (up to symmetry) are shown in Fig. 38a, where the rotation axis goes through the center of the dodecahedron and the fat vertex.

Fig. 38

1-symmetric vertex of ${\mathcal {P}}$

There are two ways of joining these three edges to complete a 3-symmetric 9-gon of type $9_r$ invariant under $T_1 T_2$. One way is by making each of the 1-symmetric vertices adjacent to two non-adjacent consecutive vertices in the (hexagonal) convex hull of the edges, while the other is by making those vertices adjacent to opposite vertices in the convex hull of the edges. The first way can be achieved in $K_{[5,3]}$, $M_{[5,3]}$ and $Tw_{[5,3]}$, each with only one choice of 1-symmetric vertices. These are shown in that order in the first row of Fig. 39. The second way can only be achieved in the graph $M_{[5,3]}$, but with two essentially distinct choices of 1-symmetric vertices. (Note that two 2-symmetric vertices of $M_{[5,3]}$ that do not belong to the same pentagon of ${\mathcal {D}}$ have either two or zero common 1-symmetric neighbors.) These choices are illustrated in the second row of Fig. 39, where in each case only two consecutive 2-symmetric edges are added, indicating the 5-neighbors of the common 1-symmetric vertex.

Fig. 39

Possible 3-symmetric 9-gons of ${\mathcal {P}}$

Here, we point out that when considering the orbit of each such polygon under [3, 5] the vertex-figures arising from Fig. 39a, e become disconnected (each vertex-figure at a 2-symmetric vertex has three components, each being a line segment); in fact, Lemma 5 rules out the resulting structures. The vertex-figures arising from the other polygons are themselves polygons, and therefore, we get three 3-orbit polyhedra in class $3^{0,1}$ with 9-gonal facets and whose 2-symmetric vertices are in the axes of the threefold rotations.

These 3 polyhedra are mutually non-isomorphic. If the facets are (c), then the three 1-symmetric vertices of a facet (those with degree 3) have a neighbor in common, and this does not happen if the facets are (b) or (d). Moreover, in the 9-gon (d), every 1-symmetric vertex has a common neighbor with each vertex of its opposite 1-symmetric edge in the 9-gon, a property not satisfied by (b).

In the case that $T_1T_2$ is a rotatory reflection of order 6 then $f_0$ is an 18-gon. In this situation, $T_2$ must be a half-turn, and the axis of $T_1 T_2$ must belong to the mirror of $T_1$ and must be perpendicular to the axis of $T_2$. It follows that the centers of the six 1-symmetric edges must be coplanar and they must be like those in Fig. 38b. Since the midpoints of the 1-symmetric edges lie on a plane through the center of ${\mathcal {D}}$, the centers of all facets coincide. Let ${\mathcal {X}}$ be the circle containing the centers of all edges of $f_0$.

Fig. 40

Two 18-gons of polyhedra in class $3^{0,1}$

In order to construct a 3-symmetric 18-gon of type $18_{rh}$ invariant under $\langle T_1, T_2 \rangle$ and having the edges in Fig. 38b as its six 1-symmetric edges, any 1-symmetric vertex $v_0$ must be adjacent to two vertices in edges whose midpoints are consecutive in ${\mathcal {X}}$; furthermore, the two neighbors of a 1-symmetric vertex $v_0$ in $f_0$ must be in the same side of the plane spanned by ${\mathcal {X}}$. This can be done either by taking the endpoints of a 1-symmetric edge of ${\mathcal {P}}$ as the neighbors of $v_0$ in $f_0$, or by making $v_0$ adjacent to the two endpoints of a (non-trivial) diagonal of a pentagon. The first way cannot be achieved if the 1-skeleton is either $M_{[5,3]}$ or $Tw_{[5,3]}$, but it can if it is $K_{[5,3]}$ in the two different manners illustrated in Fig. 40. We still may discard Fig. 40a due to Lemma 5, but the vertex-figures arising from Fig. 40b are polygons and hence the orbit of such a facet constitutes a polyhedron. The second way could in principle be achieved in each of the three candidates of 1-skeleton, each in only one manner; these are shown in Fig. 41. However, Lemma 5 rules out the polygons in Fig. 41b. In the remaining two cases, the corresponding vertex-figures of ${\mathcal {P}}$ are polygons, yielding polyhedra. Once again, the two polyhedra constructed with these 18-gons are non-isomorphic since their 1-skeleta are distinct.

Fig. 41

Three 18-gons of polyhedra in class $3^{0,1}$

We claim that the 1-symmetric edges of $f_0$ are antipodes if $T_1 T_2$ is either a half-turn or the central inversion. This is clear in case $T_1 T_2$ the central inversion. If on the other hand, it is a half-turn then $T_2$ is a reflection and the four 2-symmetric vertices are coplanar, which is attained in ${\mathcal {D}}$ if and only if the 1-symmetric edges are antipodes. Hexagons whose opposite edges are antipodes in ${\mathcal {D}}$ do not exist if the 1-skeleton is either $K_{[5,3]}$ or $Tw_{[5,3]}$. However, there is one orbit under [3, 5] of each type in the graph $M_{[5,3]}$, shown in the left and center of Fig. 42. Both kinds of polygons induce connected vertex-figures.

Fig. 42

Two 6-gons and a triangle of polyhedra in class $3^{0,1}$

Finally, if $T_1 T_2$ is the identity, then $f_0$ is a triangle containing one 1-symmetric edge and one 1-symmetric vertex. This can only happen if the 1-skeleton of ${\mathcal {P}}$ is $K_{[5,3]}$, as shown in the right of Fig. 42, and the resulting polyhedron is the pentakis dodecahedron.

As in the previous subsection, applying $\zeta _1$ to each polyhedron will give another polyhedron except in the case where $T_1 T_2$ is a half-turn (and the facets are type $6_r$). The new two possibilities for $T_1 T_2$ (namely, a rotation of order 3 and a rotatory reflection of order 6) are interchanged by $\zeta _1$.

Proposition 48

Up to similarity and vi-equivalence, there are 18 3-orbit polyhedra ${\mathcal {P}}$ in class $3^{0,1}$ with $\varGamma ({\mathcal {P}}) = [3,5]$ where the 2-symmetric vertices are those of a dodecahedron. The graph induced by the 2-symmetric vertices in nine of them is the 1-skeleton of the dodecahedron, and in the other nine is the 1-skeleton of a great stellated dodecahedron.

The 1-skeleta of these polyhedra are isomorphic to the pentakis dodecahedron, the graph $M_{[5,3]}$ defined above, or the twin dodecahedron. Geometrically, there are three with each of the 1-skeleta $K_{[5,3]}$ and $K_{[5/2,3]}$, two with each of the 1-skeleta $Tw_{[5,3]}$ and $Tw_{[5/2,3]}$, and four with each of the 1-skeleta $M_{[3,5]}$ and $M_{[3,5/2]}$.

We summarize our previous discussion in the following theorem.

Theorem 49

Up to similarity and vi-equivalence there are 30 3-orbit polyhedra in class $3^{0,1}$ with $\varGamma ({\mathcal {P}}) = [3,5]$, summarized in Table 8.

10.2 Vertex-transitive case

Vertex-transitive polyhedra with group [3, 5]
1-symmetric edges	30
2-symmetric edges	60
Vertices	Twenty 9-valent, 30 6-valent, or 60 3-valent

Now, we assume that ${\mathcal {P}}$ is a vertex-transitive polyhedron with $G({\mathcal {P}})=[3,5]$.

We claim that ${\mathcal {P}}$ cannot have 30 vertices. Assume to the contrary that the vertices of ${\mathcal {P}}$ are in the midpoints of the edges of ${\mathcal {D}}$. Then, ${\mathcal {P}}$ has 30 6-valent vertices. Each vertex v of ${\mathcal {P}}$ must be incident to two 1-symmetric edges and to four 2-symmetric edges. Each 1-symmetric edge must be invariant under 4 distinct symmetries; however, the only segment between v and another midpoint of edge of ${\mathcal {D}}$ that has a stabilizer of order greater than 2 is the segment to its antipode. This contradicts the fact that there are two 1-symmetric edges incident to v.

Hence, we only need to consider polyhedra ${\mathcal {P}}$ with 20 and 60 vertices.

First, we consider the case where ${\mathcal {P}}$ has 60 vertices. In this situation the stabilizer of each vertex is generated by a plane reflection and the results of Sect. 5.3 apply. As a consequence, we have the following result.

Theorem 50

Up to similarity and vt-equivalence, there are 48 vertex-transitive 3-valent 3-orbit polyhedra in ${\mathbb {E}}^3$ with symmetry group [3, 5]:

• the truncations of the dodecahedron $\{5,3\}$, icosahedron $\{3,5\}$, great stellated dodecahedron $\{5/2,3\}$, great icosahedron $\{3,5/2\}$, great dodecahedron $\{5,5/2\}$, small stellated dodecahedron $\{5/2,5\}$ and the truncations of the Petrials of those 6 polyhedra (12 in total); all in class $3^{1,2}$;

• the image of the 12 polyhedra in the previous item under $\zeta _2$, all in class $3^{1,2}$;

• the Petrials of the 24 polyhedra in the previous two items, all in class $3^{1}$.

We are left with the case when ${\mathcal {P}}$ has 20 vertices. They must be located on the axes of threefold rotations of ${\mathcal {D}}$. Then, every vertex v is incident to six 2-symmetric edges and to three 1-symmetric edges. Each of these two sets of edges is an orbit under the vertex stabilizer in [3, 5]; recall that the latter is the dihedral group $D_3$ with 6 elements. As mentioned in Sect. 5, we may visualize ${\mathcal {P}}$ as a graph in ${\mathbb {S}}^2$ with prescribed facets (cycles), and project it to ${\mathbb {P}}^2$, since [3, 5] contains the central inversion. In the hemi-dodecahedron, there is precisely one orbit of vertices with 3 elements, and one of 6 elements under the stabilizer of v. The former consists of the neighbors of v and the latter of all vertices that are neither v nor adjacent to v. In particular, all edges of a 1-symmetric facet project to diagonals of pentagons in the hemi-dodecahedron.

Fig. 43

1- and 2-symmetric facets of ${\mathcal {P}}$

Fig. 44

3-symmetric facets of ${\mathcal {P}}$

An exhaustive search shows that in the graph described above all possible 1-symmetric facets are images under the symmetry group of the hemi-dodecahedron of those in black lines in Fig. 43a–c; all possible 2-symmetric facets are images of those in Fig. 43d, e; and all possible 3-symmetric facets are images of those in Fig. 44. The (projective) polygons in Fig. 44e, f are (degenerate) hexagons where an edge is traversed twice. When lifting them to ${\mathbb {S}}^2$, they become (proper) hexagons using one pair of antipodal 1-symmetric edges, and four 2-symmetric edges that are not paired by the central inversion (see Fig. 48).

To convince ourselves that we obtained all possible 3-symmetric facets $f_0$, we divided the cases as follows. There are three essentially distinct ways to choose two consecutive 2-symmetric edges of $f_0$, illustrated in Fig. 45a–c. There are also 3 essentially distinct ways to choose consecutive 1-symmetric and 2-symmetric edges of $f_0$, shown in Fig. 45d–f. Finally, we have a choice of whether the stabilizer of a 1-symmetric edge contains a reflection (r) or a half-turn (h). (Recall that the stabilizer of a vertex is always a reflection.) In this way, the facets in Fig. 44a–f correspond, respectively, to the choices $ae_r$, $ad_h$, $cd_r$, $cf_h$, $cf_r$, $ae_h$.

Fig. 45

Cases for choices of 1- and 2-symmetric edges of a 3-symmetric facet f; full and empty dots represent 1- and 2-symmetric vertices of f, respectively

Out of the 18 possible combinations in $\{a,b,c\} \times \{d,e,f\} \times \{r,h\}$ we still have to discard 12 of them. From Remark 9, we know that if the last entry is h then $f_0$ has an even number of vertices, and so it must be divisible by 6. In fact, that number must be 3k where k is the order of the 3-step rotation T around $f_0$. Since [3, 5] has no isometries of order 4, we find that T must be an involution, or a rotatory reflection of order 6. With this argument, we discard $bd_h$, $be_h$, $cd_h$ and $ce_h$ (T would be a rotatory reflection of order 10 in all these cases). We can further discard $af_h$, $bf_h$, $ad_r$, $be_r$ and $ce_r$, since traversing the corresponding path in the graph in the projective plane goes more than twice through a vertex (and so it cannot be lifted as a polygon in ${\mathbb {S}}^2$); in $ad_r$ all vertices belong to a pentagon appearing three times, while in the other four cases just mentioned one vertex is in the axis of the 3-step rotation T and this is a threefold rotation in the projective plane. The remaining three cases $af_r$, $bd_r$, $bf_r$ induce a degenerate 15-gon in the projective plane, with the vertices of a pentagon traversed twice and all the edges of this pentagon traversed once. It is not possible to lift such a path as a polygon in ${\mathbb {S}}^2$ since both 2-symmetric edges at a vertex should have a vertex on some lift of the special pentagon to ${\mathcal {D}}$, and so two lifts to ${\mathbb {S}}^2$ yield degenerate 15-gons that use the vertices of a pentagon twice while the other two lift into 30-gons.

The arguments in the enumeration of the 1- and 2-symmetric facets are simpler and we omit the details.

Recall that each edge in Figs. 43 and 44 can be lifted in four different ways to ${\mathbb {S}}^2$, by considering both antipodal pairs of vertices as candidates of endpoints. These four edges can be divided into pairs that are equivalent under [3, 5] and so we may think of only two possible ways of lifting the edges to ${\mathbb {S}}^2$.

Since 2- and 3-symmetric facets have two kinds of edges, each such facet in Figs. 43 and 44 can be lifted to four essentially different structures in ${\mathbb {S}}^2$, although they sometimes fail to be polygons. The corresponding polyhedra are related by the operations $\zeta$, $\zeta _1$ and $\zeta _2$.

Figures 46, 47 and 48 illustrate, respectively, the 1-, 2- and 3-symmetric facets obtained by lifting those in Figs. 43 and 44 in such a way that all edges are either edges of ${\mathcal {D}}$ or diagonals of its pentagons. We shall refer to the latter 1-skeleton by MD (standing for ‘modified dodecahedron’).

Fig. 46

Candidate 1-symmetric facets of 9-valent polyhedra in class $3^{1,2}$

Fig. 47

Candidate 2-symmetric facets of 9-valent polyhedra in class $3^{1,2}$

Fig. 48

Candidate 3-symmetric facets of 9-valent polyhedra in class $3^{1}$

Moreover, there is a hexagon obtained by lifting to ${\mathbb {S}}^2$ the polygon in Fig. 44f that has no valid representative in the 1-skeleton MD. The representative in $MD^{\zeta _1}$ is shown in Fig. 49.

Fig. 49

One additional candidate 3-symmetric facet for a 9-valent polyhedron in class $3^{1}$

As can be easily seen on the triangle and pentagons in Fig. 46, the operations $\zeta$, $\zeta _1$ and $\zeta _2$ do not induce isomorphisms on MD, since triangles or pentagons are transformed into hexagons or decagons. In fact, MD and $MD^{\zeta _1}$ each contain a clique with 5 vertices, whereas $MD^{\zeta }$ and $MD^{\zeta _2}$ do not.

According to Ref. [29] and to the software Sage [28], up to isomorphism there is only one 9-valent vertex-transitive graph on 20 vertices with girth 3, diameter 3, clique number 5 and containing 140 triangles. Both MD and $MD^{\zeta _1}$ satisfy those parameters, and hence they are isomorphic. Furthermore, the 2-symmetric edges can be distinguished as those that belong to only 4 triangles while the 1-symmetric edges belong to 6 of them. It follows that every polyhedron with a realization on the 1-skeleton MD also has a realization with 1-skeleton $MD^{\zeta _1}$ and vice-versa. Since $MD^\zeta = (MD^{\zeta _1})^{\zeta _2}$, we can conclude that $MD^\zeta$ and $MD^{\zeta _2}$ are also isomorphic and serve as 1-skeleton for the same abstract polyhedra. The automorphisms of graphs between MD and $MD^{\zeta _1}$ and between $MD^{\zeta _2}$ and $MD^\zeta$ are not induced by any of the operations $\zeta$, $\zeta _1$ or $\zeta _2$; part of the consequence is that the polygons projecting to the projective plane to those in Fig. 44e, f are related by the isomorphism between MD and $MD^{\zeta _1}$, but not by $\zeta _1$ itself. Here, we shall use Figs. 43, 44, 46, 47, 48 and 49 as references for the enumeration of the polyhedra on each 1-skeleton.

The previous discussion and a careful examination of the sizes of the preimages of the polygons shows the following.

• (i) The hexagon in Fig. 43d has four distinct polygonal preimages: the planar hexagon in Fig. 47a in MD, a skew hexagon in $MD^{\zeta }$, and two skew 12-gons in $MD^{\zeta _1}$ and $MD^{\zeta _2}$.

• (ii) The 6-gon in Fig. 43e has four distinct polygonal preimages: the skew 12-gon in Fig. 47b in MD, a planar non-convex hexagon in $MD^{\zeta _1}$, a skew hexagon in $MD^{\zeta _2}$ and another skew 12-gon in $MD^\zeta$.

• (iii) The triangle in Fig. 44a has four distinct polygonal preimages: the triangle in Fig. 48a in MD, another triangle in $MD^{\zeta _2}$, and two skew hexagons of type $6_{rh}$ in $MD^{\zeta _1}$ and $MD^{\zeta }$.

• (iv) The 9-gon in Fig. 44b has four distinct polygonal preimages: the 18-gon in Fig. 48b in MD, another 18-gon in $MD^{\zeta _2}$ (both of type $18_{rh}$), and two 9-gons of type $9_r$ in $MD^{\zeta _1}$ and $MD^\zeta$.

• (v) The 9-gon in Fig. 44c has four distinct polygonal preimages: the 9-gon in Fig. 48c in MD, another 9-gon in $MD^{\zeta _2}$ (both of type $9_{r}$), and two 18-gons of type $18_{rh}$ in $MD^{\zeta _1}$ and $MD^{\zeta }$.

• (vi) The triangle in Fig. 44d has four distinct polygonal preimages: the hexagon in Fig. 48d in MD, another hexagon in $MD^{\zeta _2}$ (both of type $6_{rh}$), and two triangles in $MD^{\zeta _1}$ and $MD^\zeta$.

• (vii) The degenerate hexagon in Fig. 44e has only two new preimages: the hexagon in Fig. 48e in MD and another hexagon in $MD^{\zeta _2}$ (both of type $6_{r}$). The remaining two preimages degenerate to unions of triangles like those in the Item (vi).

• (viii) The degenerate hexagon in Fig. 44f has only two new preimages: the hexagon in Fig. 49 in $MD^{\zeta _1}$ and another hexagon in $MD^\zeta$ (both of type $6_{r}$). The remaining two preimages degenerate to unions of triangles like those in Item (iii).

There are 20 polyhedra ${\mathcal {P}}$ in class $3^1$ with $G({\mathcal {P}})=[3,5]$ since in all cases the vertex-figures are polygons. They are described in Proposition 51. None of them are combinatorially regular; there are no regular polyhedra of type $\{3,9\}$, $\{6, 9\}$, $\{9,9\}$, or $\{18,9\}$ with 360 flags (see Ref. [24]).

The 1-symmetric facets of polyhedra in class $3^{1,2}$ only have 2-symmetric edges, implying that each such facet has only two non-congruent preimages in ${\mathbb {S}}^2$. By choosing the preimage of a 2-symmetric facet we automatically obtain the preimages of all 2-symmetric edges and hence also the preimages of the 1-symmetric facets.

The three choices of 1-symmetric facets together with the two choices of 2-symmetric facets give six possible structures. However, only four of them satisfy the conditions to be a polyhedron. Failure occurs when the 1-symmetric facets are like that in Fig. 43a and the 2-symmetric facets like that in Fig. 43d, or if the 1-symmetric facets are like that in Fig. 43c and the 2-symmetric facets like that in Fig. 43e; in those situations the vertex-figures are disconnected (union of 3 triangles). The remaining four cases indeed yield polyhedra.

Among the polyhedra in class $3^{1,2}$, only two are equivelar; in both cases, the 1-symmetric facets are the orbit of the image under $\zeta$ of the polygon in Fig. 46a. In one polyhedron the 2-symmetric facets are the orbit of the image under $\zeta$ of that in Fig. 47a, and in the other they are the orbit of the image under $\zeta _2$ of that in Fig. 47b. These polyhedra have type $\{6,9\}$, but they are not regular since there are no regular polyhedra with that type and 360 flags (see Ref. [24]).

Theorem 51

Up to similarity, there are 20 3-orbit polyhedra in class $3^{1}$ with $G({\mathcal {P}})=[3,5]$ and 20 vertices. The 1-skeleta are MD, $MD^{\zeta _2}$, $MD^{\zeta _1}$ and $MD^{\zeta }$, with 5 polyhedra in each 1-skeleton.

Up to similarity, there are 16 3-orbit polyhedra in class $3^{1,2}$ with $G({\mathcal {P}})=[3,5]$ and 20 vertices. Their 1-skeleta are MD, $MD^{\zeta _1}$, $MD^{\zeta _2}$ and $MD^{\zeta }$, with four polyhedra in each 1-skeleton.

Theorem 52

Up to similarity and vt-equivalence, there are 84 vertex-transitive 3-orbit polyhedra with symmetry group [3, 5], summarized in Tables 9, 10 and 11.

11 Summary of results

We summarize some of the interesting features of our data in the final theorem.

Theorem 53

There are 188 3-orbit polyhedra in ${\mathbb {E}}^3$ with irreducible symmetry group. There are 5 with symmetry group [3, 3], 4 with symmetry group $[3,4]^+$, 53 with symmetry group [3, 4], 12 with symmetry group $[3,5]^+$, and 114 with symmetry group [3, 5]. Furthermore,

• (a) Six are combinatorially regular; the rest are combinatorially 3-orbit.

• (b) There are 109 distinct combinatorial types: 30 that have a single 3-orbit geometric realization, and 79 that have two 3-orbit geometric realizations, including all the combinatorial types of polyhedra with symmetry group [3, 5].

• (c) There are 44 polyhedra in class $3^{0,1}$ (vertex-intransitive), 72 polyhedra in class $3^{1,2}$ (facet-intransitive), and 72 polyhedra in class $3^1$ (vertex- and facet-transitive).

• (d) 88 of the polyhedra are 3-valent.

• (e) There are 36 polyhedra (18 pairs) that have a canonical geometric dual. There are also two polyhedra with no canonical dual, but whose combinatorial dual is geometrically realizable as a 3-orbit polyhedron. None of the polyhedra is self-dual (not even combinatorially).

• (f) There are 130 polyhedra (65 pairs) such that the Petrial is also a polyhedron.

Appendix

See Tables 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11.

Table 1 The finite regular polyhedra in ${\mathbb {E}}^3$

Table 2 The 1-skeleta of 3-orbit polyhedra in ${\mathbb {E}}^3$ with irreducible symmetry group

Table 3 The five 3-orbit polyhedra with symmetry group [3, 3]

Table 4 The four 3-orbit polyhedra with symmetry group $[3, 4]^+$

Table 5 The 13 vertex-intransitive 3-orbit polyhedra with symmetry group [3, 4]

Table 6 The 40 vertex-transitive 3-orbit polyhedra with symmetry group [3, 4]

Table 7 The 12 3-orbit polyhedra with symmetry group $[3, 5]^+$; see Theorem 45

Table 8 The 30 vertex-intransitive 3-orbit polyhedra with symmetry group [3, 5]

Table 9 32 of the vertex-transitive 3-orbit polyhedra with symmetry group [3, 5] and 60 vertices

Table 10 The remaining 16 vertex-transitive 3-orbit polyhedra with symmetry group [3, 5] and 60 vertices

Table 11 The 36 vertex-transitive 3-orbit polyhedra with symmetry group [3, 5] and 20 vertices