Tight chiral polytopes

Abstract

A chiral polytope with Schläfli symbol $\{p_1, \ldots , p_{n-1}\}$ has at least $2p_1 \cdots p_{n-1}$ flags, and it is called tight if the number of flags meets this lower bound. The Schläfli symbols of tight chiral polyhedra were classified in an earlier paper, and another paper proved that there are no tight chiral n-polytopes with $n \ge 6$. Here we prove that there are no tight chiral 5-polytopes, describe 11 families of tight chiral 4-polytopes, and show that every tight chiral 4-polytope covers a polytope from one of those families.

1 Introduction

An abstract n-polytope is a partially ordered set that satisfies many of the properties of the face lattices of convex n-polytopes. The maximal chains (called flags) are analogous to the simplices in the barycentric subdivision of a convex polytope. Automorphisms are order-preserving bijections and are the combinatorial analogue of symmetries of convex polytopes.

The group of automorphisms of an abstract polytope acts semiregularly on the set of flags, and if the action is transitive (and thus regular), then the polytope is said to be regular. These polytopes are regarded as the most symmetric and have been extensively studied. The automorphism group of a regular polytope has a standard generating set, and it is possible to recover the polytope from a group in this form, making it possible to study regular polytopes completely in terms of their groups.

An abstract polytope is chiral whenever the automorphism group has two orbits on the flags such that flags that differ in only one element are in opposite orbits. This is the combinatorial analogue to having all symmetry by rotations but none by reflections. As with regular polytopes, the automorphism group of a chiral polytope has a standard form, and we can build a chiral polytope out of such a group. The study of chiral polytopes grew out from the study of chiral maps and twisted honeycombs (see [7, 8]), and while chiral 3-polytopes and chiral 4-polytopes are nowadays plentiful, constructing chiral n-polytopes with $n \ge 5$ seems to be much harder. To date, there is no known natural family of chiral n-polytopes with one polytope for each n (whereas there are many examples of families of regular n-polytopes, such as n-cubes). There is a construction, described in [21], that takes a chiral n-polytope as input and produces a chiral $(n+1)$-polytope, but the polytopes constructed this way are so large that their individual study is out of reach with the current computational means available.

How can we find small examples of chiral polytopes? One strategy is to specify part of the local structure (such as what kind of sub-units the polytope is built from) and then use that local structure to put a lower bound on the number of flags. This idea was used in [3] to find the smallest regular polytopes of each rank and in [11] to explore bounds in the size of chiral polytopes. A polytope is called tight if its number of flags is equal to some lower bound. For example, a chiral polyhedron (3-polytope) with p-gonal faces and q edges at each vertex must have at least 2pq flags, and so a tight chiral polyhedron has exactly 2pq flags (see [10]).

In [12], the first author determined the pairs (p, q) such that there is a tight chiral polyhedron with p-gonal faces and q edges at each vertex. Furthermore, the first author showed in [11] that there are no tight chiral n-polytopes with $n \ge 6$. In this work, we exhibit 11 families of tight chiral 4-polytopes (see Table 4) and show that every tight chiral 4-polytope covers one of the polytopes in these families. Furthermore, we prove the following theorem.

Theorem 1

There are no tight chiral 5-polytopes.

2 Background

In this section, we summarize relevant definitions and results.

2.1 Abstract polytopes

Regular abstract polytopes are a combinatorial generalization of the notion of (geometric) polyhedra explored by Petrie, Coxeter, Grünbaum and Dress in the $20\hbox {th}$ Century (see [6, 15, 16, 18]). In what follows, we recall the basic definitions. For further details see [19].

An abstract polytope $({\mathcal {P}},\le )$ of rank n is a partially ordered set satisfying the following four axioms.

• (I) It has a unique minimal element $F_{-1}$ and a unique maximal element $F_n$.

• (II) All maximal chains have precisely $n+2$ faces, including $F_{-1}$ and $F_n$. This induces a strictly increasing rank function ${{\,\mathrm{rank}\,}}: {\mathcal {P}}\rightarrow \{-1, \dots , n\}$ where ${{\,\mathrm{rank}\,}}(F_{-1})=-1$ and ${{\,\mathrm{rank}\,}}(F_n)=n$.

• (III) Diamond condition: Given two elements F, G with ${{\,\mathrm{rank}\,}}(G)={{\,\mathrm{rank}\,}}(F)+2$, there exist precisely two elements $H_1$ and $H_2$ with ${{\,\mathrm{rank}\,}}(H_1)={{\,\mathrm{rank}\,}}(H_2)={{\,\mathrm{rank}\,}}(F)+1$ such that $F \le H_i \le G$ for $i \in \{1,2\}$.

• (IV) Strong connectivity: For any pair of incident elements $\{F,G\} \subseteq {\mathcal {P}}$ with ${{\,\mathrm{rank}\,}}(G)-{{\,\mathrm{rank}\,}}(F) \ge 3$, the incidence graph of the open interval (F, G) is connected. (The incidence graph of a partially ordered set has the elements as vertices, and two are adjacent if and only if the corresponding elements are incident.)

Throughout this paper, we will encounter only abstract polytopes and we shall refer to them simply as ‘polytopes.’ Rank 2 and 3 polytopes are also called poylgons and polyhedra, respectively. For convenience, we refer to the polytope $({\mathcal {P}},\le )$ simply as ${\mathcal {P}}$. Two elements F, G of ${\mathcal {P}}$ are said to be incident if either $F \le G$ or $G \le F$.

The elements of ${\mathcal {P}}$ are called faces. Those of rank i are called i-faces. Following the tradition, the 0- 1- and $(n-1)$-faces are called vertices, edges and facets, respectively. For $i \in \{1, \dots , n-2\}$, we define the i-skeleton of ${\mathcal {P}}$ as the partially ordered set consisting of all the j-faces for $j \le i$. If $F_0$ is a vertex and $F_{n-1}$ is an incident facet, we say that the closed interval $[F_0,F_{n-1}]$ is a medial section of ${\mathcal {P}}$.

The closed intervals of a polytope (also called sections) satisfy the axioms of abstract polytopes. In particular, any medial section of a polytope is a polytope. The section $[F_0,F_n]$, where $F_0$ is a vertex, is called the vertex-figure at $F_0$. Every face F may be identified with the section $[F_{-1},F]$, and in this way it may be considered as an abstract polytope.

The maximal chains of ${\mathcal {P}}$ are called flags. Due to the diamond condition, for any flag $\varPhi$ and any rank $i \in \{0, \dots , n-1\}$ there exists a unique flag $\varPhi ^i$ that differs from $\varPhi$ precisely in the element of rank i. The flag $\varPhi ^i$ is called the i-adjacent flag of $\varPhi$. We extend this notation recursively in such a way that if w is a word on the alphabet $\{0,\dots , n-1\}$ and $i \in \{0, \dots , n-1\}$ then $(\varPhi ^w)^i = \varPhi ^{wi}$.

The dual ${\mathcal {P}}^\delta$ of a polytope ${\mathcal {P}}$ consists of the same elements as ${\mathcal {P}}$ with the partial order reversed. In this way, if F is an i-face of an n-polytope ${\mathcal {P}}$, then it is an $(n-i-1)$-face of ${\mathcal {P}}^\delta$.

An n-polytope is said to be flat whenever every vertex is incident to every facet. Given $0 \le k<m \le n$, we say that it is (k, m)-flat if every k-face is incident to every m-face.

There is a unique polytope of rank 0 and a unique polytope of rank 1. They correspond to the face lattices of a single point and of a line segment (with its two endpoints). For each integer $k \ge 2$, there is a unique polygon with k vertices, that corresponds to the face lattice of a convex k-gon. There is also a unique apeirogon with infinitely many vertices, corresponding to the face lattice of the tiling of the real line by unit intervals. Therefore the rank 2 sections of a polytope are all isomorphic to k-gons for some k or to apeirogons.

We say that a polytope is equivelar if, for every $i \in \{1, \dots , n-1\}$, all sections between an $(i-2)$-face and an incident $(i+1)$-face are $p_i$-gons for some numbers $p_i$, regardless of the choice of $(i-2)$-face and $(i+1)$-face. Regular and chiral polytopes defined below are examples of equivelar polytopes. The Schläfli type (or type for short) of an equivelar polytope is $\{p_1, \dots , p_{n-1}\}$.

We say that an n-polytope ${\mathcal {Q}}$ is a quotient of a polytope ${\mathcal {P}}$ whenever there exists a rank and adjacency preserving mapping from the faces of ${\mathcal {P}}$ to the faces of ${\mathcal {Q}}$. (We say that two i-faces are adjacent if they are incident to a common $(i-1)$-face and $(i+1)$-face.) In such cases, we say that ${\mathcal {P}}$ covers ${\mathcal {Q}}$.

An automorphism of ${\mathcal {P}}$ is an order-preserving bijection of its faces. The automorphism group is denoted by $\varGamma ({\mathcal {P}})$ and acts freely on the set of flags. It follows from the strong connectivity of ${\mathcal {P}}$ that all orbits of flags have the same size $|\varGamma ({\mathcal {P}})|$.

2.2 Regularity and chirality

In this subsection, we provide a general background on regular and chiral polytopes.

Our main interest in this paper is on chiral polytopes; hence, we shall follow the approach given in [22] to the study of the automorphism groups of these two classes of objects, and not the one in [19] for regular polytopes.

We say that an n-polytope ${\mathcal {P}}$ is regular whenever $\varGamma ({\mathcal {P}})$ acts transitively on the set of flags, and it is chiral whenever $\varGamma ({\mathcal {P}})$ induces two orbits on the flags in such a way that adjacent flags belong to distinct orbits. If ${\mathcal {P}}$ is regular or chiral we say that it is rotary.

For every $i \in \{0,\dots , n-1\}$ the automorphism group of a rotary polytope acts transitively on the i-faces. As a consequence, rotary polytopes are equivelar.

It is well-known that for every integers $p_1,\dots ,p_{n-1} \ge 2$ there is a regular polytope with type $\{p_1,\dots ,p_{n-1}\}$ (see [19, Chapter 3]. This is not the case for chiral polytopes, as shown by the following lemma.

Lemma 1

If the last entry of the type of a polytope ${\mathcal {P}}$ is 2 then ${\mathcal {P}}$ is not chiral.

Proof

If ${\mathcal {P}}$ is an n-polytope with a 2 as the last entry of its type then all $(n-3)$-faces belong to precisely two facets. By the diamond condition, also the $(n-2)$-faces belong to two facets. The connectivity of the $(n-2)$-skeleton shows that ${\mathcal {P}}$ has precisely two facets and all i-faces are incident to them for $i \le n-2$.

The function that fixes every i-face for $i \le n-2$ and interchanges the two $(n-1)$-faces is then an automorphism, and it maps every flag to its $(n-1)$-adjacent. Hence ${\mathcal {P}}$ is not chiral. $\square$

Every finite polygon is isomorphic to the face lattice of some convex regular polygon, and hence it is regular. Also the unique infinite 2-polytope is regular. Hence the rank of a non-regular polytope must be at least 3. Chiral polytopes exist in ranks 3 and higher (see [21]).

All sections of regular polytopes are regular. The facets and vertex-figures of a chiral n-polytope may be either regular or chiral; however, the $(n-2)$-faces must be regular (see [22, Proposition 9]). Note that chiral polytopes with chiral facets must have rank at least 4.

Much of the work on chiral polytopes has been done through a particular presentation of their automorphism groups that we explain next. For another useful presentation see for example [5].

Given a fixed base flag $\varPhi$ of a rotary n-polytope ${\mathcal {P}}$ there exist $\sigma _i \in \varGamma ({\mathcal {P}})$ for $i \in \{1,\dots ,n-1\}$ such that $\varPhi \sigma _i = \varPhi ^{i (i-1)}$. We shall denote the group $\langle \sigma _1, \dots , \sigma _{n-1}\rangle$ by $\varGamma ^+({\mathcal {P}})$ and call it the rotation group of ${\mathcal {P}}$. The automorphisms $\sigma _i$ are called standard generators of $\varGamma ^+({\mathcal {P}})$. If ${\mathcal {P}}$ has type $\{p_1,\dots , p_{n-1}\}$, then the order of $\sigma _i$ is $p_i$ and therefore $\varGamma ^+({\mathcal {P}})$ is a suitable quotient of the even subgroup $[p_1,\dots ,p_{n-1}]^+$ of the Coxeter group $[p_1,\dots ,p_{n-1}]$ (see for example [19, Chapter 3]).

If ${\mathcal {P}}$ is chiral then $\varGamma ({\mathcal {P}}) = \varGamma ^+({\mathcal {P}})$. Whenever ${\mathcal {P}}$ is regular, $\varGamma ^+({\mathcal {P}})$ has index at most 2 in ${\mathcal {P}}$; if the index is 2 we say that ${\mathcal {P}}$ is orientably regular, and it is non-orientably regular if $\varGamma ({\mathcal {P}})=\varGamma ^+({\mathcal {P}})$. In any of these cases, if F is an i-face and G is a j-face such that $F \le G$ and their ranks differ in at least 3 then $\varGamma ^+([F,G]) = \langle \sigma _{i+2}, \dots , \sigma _{j-1} \rangle$.

For a rotary polytope ${\mathcal {P}}$, the standard generators of $\varGamma ^+({\mathcal {P}})$ satisfy

$$\begin{aligned} (\sigma _i \dots \sigma _j)^2 = id\quad \text{ for } \text{ every } 1\le i < j \le n-1, \end{aligned}$$

(1)

as well as the intersection condition

$$\begin{aligned} A_I \cap A_J = A_{I \cup J} \quad \text{ for } \text{ every } I, J \subseteq \{0,\dots , n-1\}, \end{aligned}$$

(2)

where for $I \subseteq \{0,\dots ,n-1\}$ the set $A_I$ denotes the stabilizer in $\varGamma ^+({\mathcal {P}})$ of those faces $F_i$ of the base flag with ranks $i\in I$. If $I = \{1,\dots , n-1\} \setminus \{i,i+1,\dots ,j\}$ with $i<j$ then $A_I = \langle \sigma _{i+1}, \dots , \sigma _j \rangle$, which allows us to state the following lemma. For other sets, I the generating sets ${\mathcal {X}}_I$ of these stabilizers are more complicated (see [22, Sect. 3]).

Lemma 2

Let ${\mathcal {P}}$ be a rotary polytope with $\varGamma ^+({\mathcal {P}}) = \langle \sigma _1, \dots , \sigma _{n-1} \rangle$. If $j \le i+1 \le k$ then

$$\begin{aligned} \langle \sigma _1, \dots , \sigma _i \rangle \cap \langle \sigma _j,\dots , \sigma _{k} \rangle = \langle \sigma _j, \dots , \sigma _i \rangle . \end{aligned}$$

(3)

If ${\mathcal {P}}$ is chiral, we may choose the base flag in one or in the other flag orbit. These two choices produce non-equivalent sets of standard generators $\sigma _i$, in the sense that the defining relations for $\varGamma ^+({\mathcal {P}})$ will not be the same for the two sets. One may think of these two ways of looking at ${\mathcal {P}}$ as a left and right form of the same object; we can go from one to the other just by ‘reflecting’ our setting from the base flag into any of its adjacent flags. When doing this, we may take $\{\sigma _1^{-1},\sigma _1^2 \sigma _2, \sigma _3,\sigma _4,\dots ,\sigma _{n-1}\}$ as the new set of standard generators for $\varGamma ({\mathcal {P}})$. For a chiral polyhedron, another convenient new set of generators is $\{ \sigma _1^{-1}, \sigma _2^{-1} \}$. The enantiomorph of a chiral polytope ${\mathcal {P}}$ (with an implicit base flag chosen) consists of the same polytope but where we change the base flag to any of its adjacent flags. We denote the enantiomorph of ${\mathcal {P}}$ by ${\mathcal {P}}^*$. For more details about these forms see [23].

We mentioned that the rotation group of a rotary polytope is a group with a generating set satisfying (1) and the intersection condition (2). Conversely, a group with a generating set satisfying (1) and a suitable version of (2) is the rotation group of an orientable rotary polytope (that is, orientably regular or chiral).

The construction of the polytope from a group $\varGamma = \langle \sigma _1, \dots , \sigma _{n-1} \rangle$ is detailed in [22, Sect. 5]. It defines the i-face of the base flag as the subgroup of $\varGamma$ generated by the elements ${\mathcal {X}}_{\{i\}}$ of $A_{\{i\}}$ mentioned before Lemma 2. The remaining i-faces are the cosets of the base i-face under the right action of $\varGamma$. It also establishes that two faces are incident if they have non-empty intersection. In particular, the sets of facets may be identified with the right cosets of $\langle \sigma _1,\dots ,\sigma _{n-2} \rangle$ under $\varGamma$. Note that this construction can be performed even if the group does not satisfy the intersection condition. The output will still have well-defined flags, and it is possible to talk about regularity through the action of its automorphism group.

If ${\mathcal {P}}$ is non-orientably regular then that construction will produce the orientable double cover of ${\mathcal {P}}$. It follows that there is a one-to-one correspondence between orientable rotary polytopes and groups satisfying (1) together with some version of (2). For our purposes, we find convenient the following version of (2) that can be easily deduced from [22, Lemma 10].

Lemma 3

Let $\varGamma = \langle \sigma _1,\dots ,\sigma _{n-1} \rangle$ be a group where each $\sigma _i$ is nontrivial and the order of $\sigma _i \dots \sigma _j$ is 2, for every $1\le i < j \le n-1$. Then $\varGamma$ satisfies the intersection condition (2) if and only if

$$\begin{aligned} \langle \sigma _1, \ldots , \sigma _i \rangle \cap \langle \sigma _j, \ldots , \sigma _{i+1} \rangle = \langle \sigma _j, \ldots , \sigma _i \rangle , \end{aligned}$$

(4)

for every $2 \le j \le i+1 \le n-1$, where if $j = i+1$ then we interpret the right-hand side as being the trivial group.

If ${\mathcal {P}}$ is orientably regular (resp. chiral) with $\varGamma ^+({\mathcal {P}})= \langle \sigma _1,\dots ,\sigma _{n-1} \rangle$ then ${\mathcal {P}}^\delta$ is also orientably regular (resp. chiral) and, with respect to some flag, the i-th standard generator of $\varGamma ^+({\mathcal {P}}^\delta )$ is $\sigma _{n-1-i}^{-1}$, for $i \in \{1,\dots ,n-1\}$.

In upcoming sections, we will be interested in normal subgroups contained in $\langle \sigma _i \rangle$ for some i. In those situations the following result will prove useful.

Lemma 4

Let ${\mathcal {P}}$ be a rotary 4-polytope, and let $\varGamma ^+({\mathcal {P}}) = \langle \sigma _1, \sigma _2, \sigma _{3} \rangle$.

• (a) For every k, $\sigma _3 \sigma _1^k \sigma _3^{-1} = \sigma _2^{-1} \sigma _1^{-k} \sigma _2.$

• (b) If K is a subgroup of $\langle \sigma _1 \rangle$, then $\sigma _2^{-1} K \sigma _2 = K$ if and only if $\sigma _3^{-1} K \sigma _3 = K$.

Proof

We start with

$$\begin{aligned} \sigma _3 \sigma _1 = (\sigma _1 \sigma _2)^2 \sigma _3 \sigma _1 (\sigma _2 \sigma _3)^2 = \sigma _1 \sigma _2 (\sigma _1 \sigma _2 \sigma _3)^2 \sigma _2 \sigma _3 = \sigma _1 \sigma _2^2 \sigma _3. \end{aligned}$$

(5)

It follows that

$$\begin{aligned} \sigma _3 \sigma _1^k = (\sigma _1 \sigma _2^2)^k \sigma _3. \end{aligned}$$

Then

$$\begin{aligned} \sigma _3 \sigma _1^k \sigma _3^{-1} = (\sigma _1 \sigma _2^2)^k = (\sigma _2^{-1} \sigma _1^{-1} \sigma _2)^k = \sigma _2^{-1} \sigma _1^{-k} \sigma _2. \end{aligned}$$

That proves part (a). Part (b) follows since $K = \langle \sigma _1^k \rangle$ for some k. $\square$

2.3 Covers and quotients

From the definition of cover, we know that if ${\mathcal {P}}$ and ${\mathcal {Q}}$ are orientable rotary n-polytopes such that ${\mathcal {P}}$ covers ${\mathcal {Q}}$, then the flags of ${\mathcal {P}}$ in one orbit under $\varGamma ^+({\mathcal {P}})$ are mapped by the covering to the flags of ${\mathcal {Q}}$ in one orbit under $\varGamma ^+({\mathcal {Q}})$. As a consequence, there exists $N \triangleleft \varGamma ^+({\mathcal {P}})$ such that ${\mathcal {Q}}\cong {\mathcal {P}}/N$. In other words, the faces of ${\mathcal {Q}}$ can be taken as the orbits of faces of ${\mathcal {P}}$ under the action of N, and two of them are incident whenever an element in the orbit of one face is incident to some element in the orbit of the other face. (See [20, Example 2.15] for an example of a cover of polytopes that is not induced by the action of a group of automorphisms.)

Conversely, given $N \triangleleft \varGamma ^+({\mathcal {P}})$, the quotient ${\mathcal {P}}/N$ is a polytope if and only if $\varGamma ^+({\mathcal {P}}/N)$ satisfies (1) and the intersection condition (4) with respect to the generators $\{\sigma _i N\}_{i \in \{1,\dots , n-1\}}$.

Whenever ${\mathcal {P}}$ is chiral, there exists a normal subgroup $X({\mathcal {P}})$ of $\varGamma ({\mathcal {P}})$ satisfying that ${\mathcal {P}}/X({\mathcal {P}})$ is a regular structure (in the sense that all flags belong to the same orbit under $\varGamma ({\mathcal {P}}/X({\mathcal {P}})$) and that if $N \triangleleft \varGamma ({\mathcal {P}})$ is such that ${\mathcal {P}}/N$ is a regular structure then $N \ge X({\mathcal {P}})$. The group $X({\mathcal {P}})$ is called the chirality group of ${\mathcal {P}}$. Note that ${\mathcal {P}}$ is regular if and only if $X({\mathcal {P}})$ is trivial.

Elsewhere the chirality group has been introduced in other terms (see for example [1, 2] and [9]), but for our purposes the universal property of the chirality group mentioned here is more convenient.

The mix of two polytopes ${\mathcal {P}}$ and ${\mathcal {Q}}$ with base flags $\varPhi _{{\mathcal {P}}}$ and $\varPhi _{{\mathcal {Q}}}$, respectively, is the smallest structure ${\mathcal {P}}\lozenge {\mathcal {Q}}$ (which itself may or may not be a polytope) with well-defined ranks and adjacencies that covers simultaneously ${\mathcal {P}}$ and ${\mathcal {Q}}$, while mapping the base flag of ${\mathcal {P}}\lozenge {\mathcal {Q}}$ to $\varPhi _{{\mathcal {P}}}$ and $\varPhi _{{\mathcal {Q}}}$, respectively. As noted in [14, Section 3], the choice of base flags may be relevant when performing the mix of two chiral polytopes. This is often taken into account by choosing a base flag from which to construct the standard generators of the automorphism group.

If ${\mathcal {P}}$ and ${\mathcal {Q}}$ are orientable rotary polytopes with $\varGamma ^+({\mathcal {P}}) = \langle \sigma _1, \dots , \sigma _{n-1} \rangle$ and $\varGamma ^+({\mathcal {Q}}) = \langle \sigma _1', \dots , \sigma _{n-1}' \rangle$ then $\varGamma ^+({\mathcal {P}}\lozenge {\mathcal {Q}}) = \langle \tau _1, \dots , \tau _{n-1} \rangle \le \varGamma ^+({\mathcal {P}}) \times \varGamma ^+({\mathcal {Q}})$, where $\tau _i = (\sigma _i,\sigma _i')$. For convenience, we also denote $\varGamma ^+({\mathcal {P}}\lozenge {\mathcal {Q}})$ by $\varGamma ^+({\mathcal {P}}) \lozenge \varGamma ^+({\mathcal {Q}})$.

The mix of two orientably regular polytopes is orientably regular. However, the mix of an orientable rotary polytope with a chiral polytope may be either orientably regular or chiral.

The next lemma relates the notions of quotient and mix of orientable rotary polytopes.

Lemma 5

Let ${\mathcal {P}}$ be an orientable rotary polytope with base flag $\varPhi _0$ and let K, N be normal subgroups of $\varGamma ^+({\mathcal {P}})$. Then

$$\begin{aligned} {\mathcal {P}}/(K\cap N) \cong ({\mathcal {P}}/ K) \lozenge ({\mathcal {P}}/ N), \end{aligned}$$

where the base flags of ${\mathcal {P}}/K$ and ${\mathcal {P}}/N$ are taken as $\varPhi _0 \cdot K$ and $\varPhi _0 \cdot N$, respectively.

Proof

The regular structure ${\mathcal {P}}/(K\cap N)$ (which may or may not be a polytope) covers ${\mathcal {P}}/K$ mapping a face $F \cdot (K \cap N)$ to the face $F \cdot K$. Similarly, it covers ${\mathcal {P}}/N$. Hence ${\mathcal {P}}/(K\cap N)$ covers the mix $({\mathcal {P}}/ K) \lozenge ({\mathcal {P}}/ N)$.

Let $\varGamma ^+({\mathcal {P}})=\langle \sigma _1, \dots , \sigma _{n-1} \rangle$. Then there is a group epimorphism from $\varGamma ^+({\mathcal {P}}/(K \cap N))$ to $\varGamma ^+(({\mathcal {P}}/ K) \lozenge ({\mathcal {P}}/ N))$ mapping $\sigma _i \cdot (K \cap N)$ to $(\sigma _i \cdot K, \sigma _i \cdot N)$ for $i \in \{1,\dots , n-1\}$. This epimorphism sends the element $\sigma _{i_1} \cdots \sigma _{i_k} \cdot (K \cap N)$ to $(\sigma _{i_1} \cdots \sigma _{i_k} \cdot K, \sigma _{i_1} \cdots \sigma _{i_k} \cdot N)$. The latter is trivial if and only if $\sigma _{i_1} \cdots \sigma _{i_k} \in K \cap N$. Since the kernel of the epimorphism is trivial, the isomorphism holds. $\square$

Given a chiral polytope ${\mathcal {P}}$, there exists a smallest regular structure ${\mathcal {R}}$ with well-defined ranks and adjacencies of flags that covers ${\mathcal {P}}$ (even if this structure is not a polytope itself), in the sense that every regular polytope that covers ${\mathcal {P}}$ also covers ${\mathcal {R}}$. We shall call this structure the smallest regular cover of ${\mathcal {P}}$.

Sometimes the smallest regular cover of ${\mathcal {P}}$ is a polytope itself; for example, when the facets or the vertex-figures are regular (see [20, Corollary 7.5]). If the smallest regular cover of ${\mathcal {P}}$ is a polytope then it is elsewhere also called the minimal regular cover of ${\mathcal {P}}$; otherwise, ${\mathcal {P}}$ may have multiple polytopal regular covers that are minimal in the partial order given by the covering relation.

The smallest regular cover ${\mathcal {R}}$ of a chiral polytope ${\mathcal {P}}$ is the regular structure constructed (in the sense of [22]) from the group $\varGamma ({\mathcal {P}}) \lozenge \varGamma ({\mathcal {P}}^*)$, where ${\mathcal {P}}^*$ is the enantiomorph of ${\mathcal {P}}$ (see [20, Sect. 7]). We may assume that if $\varGamma ({\mathcal {P}}) = \langle \sigma _1, \dots , \sigma _{n-1} \rangle$ then

$$\begin{aligned} \varGamma ^+({\mathcal {R}}) = \langle (\sigma _1,\sigma _1^{-1}), (\sigma _2,\sigma _1^2 \sigma _2), (\sigma _3,\sigma _3), \ldots , (\sigma _{n-1},\sigma _{n-1})\rangle . \end{aligned}$$

We next relate the chirality group of a chiral polytope with its smallest regular cover. This is a direct consequence of [20, Remark 7.3].

Lemma 6

Let ${\mathcal {P}}$ be a chiral polytope and ${\mathcal {R}}$ its smallest regular cover. Then $X({\mathcal {P}})$ is isomorphic to the kernel of the quotient from $\varGamma ^+({\mathcal {R}})$ to $\varGamma ({\mathcal {P}})$.

The following result relates the smallest regular covers of chiral polytopes with that of one of its facets.

Lemma 7

Let ${\mathcal {P}}$ be a chiral polytope with chiral facets isomorphic to ${\mathcal {Q}}$. Then the facets of the smallest regular cover of ${\mathcal {P}}$ are isomorphic to the smallest regular cover of ${\mathcal {Q}}$.

Proof

Since the facets of ${\mathcal {P}}$ are chiral, ${\mathcal {P}}$ has rank $n \ge 4$.

Let $\varGamma ({\mathcal {P}})= \langle \sigma _1, \dots , \sigma _{n-1} \rangle$, let ${\mathcal {R}}_{{\mathcal {P}}}$ be the smallest regular cover of ${\mathcal {P}}$ and let ${\mathcal {R}}_{{\mathcal {Q}}}$ be the smallest regular cover of ${\mathcal {Q}}$. Then $\varGamma ^+({\mathcal {R}}_{{\mathcal {P}}})=\langle \sigma _1, \dots , \sigma _{n-1} \rangle \lozenge \langle \sigma _1^{-1}, \sigma _1^2 \sigma _2, \sigma _3, \dots , \sigma _{n-1}\rangle$ and $\varGamma ^+({\mathcal {R}}_{{\mathcal {Q}}}) = \langle \sigma _1, \dots , \sigma _{n-2} \rangle \lozenge \langle \sigma _1^{-1}, \sigma _1^2 \sigma _2, \sigma _3, \dots , \sigma _{n-2}\rangle$. Since the orientation preserving automorphism group of the facet of ${\mathcal {R}}_P$ is $\varGamma ^+({\mathcal {R}}_{{\mathcal {Q}}})$, the lemma holds. $\square$

We conclude this section with a result that relates the chirality group of a chiral polytope ${\mathcal {P}}$ with that of its facets.

Lemma 8

Let ${\mathcal {P}}$ be a chiral polytope with chiral facets isomorphic to ${\mathcal {Q}}$. Then $X({\mathcal {Q}}) \le X({\mathcal {P}})$.

Proof

Let ${\mathcal {R}}_{{\mathcal {P}}}$ be the smallest regular cover of ${\mathcal {P}}$, and let ${\mathcal {R}}_{{\mathcal {Q}}}$ be the smallest regular cover of ${\mathcal {Q}}$. Then, by Lemma 7, the facets of ${\mathcal {R}}_{{\mathcal {P}}}$ are isomorphic to ${\mathcal {R}}_{{\mathcal {Q}}}$. By Lemma 6, $X({\mathcal {Q}})$ is the kernel of the natural covering $\eta _{{\mathcal {Q}}}$ from $\varGamma ^+({\mathcal {R}}_{{\mathcal {Q}}})$ to $\varGamma ^+({\mathcal {Q}})$, whereas $X({\mathcal {P}})$ is the kernel of the natural covering $\eta _{{\mathcal {P}}}$ from $\varGamma ^+({\mathcal {R}}_{{\mathcal {P}}})$ to $\varGamma ^+({\mathcal {P}})$. Since the kernel of $\eta _{{\mathcal {Q}}}$ is contained in the kernel of $\eta _{{\mathcal {P}}}$, the result follows. $\square$

2.4 Tight polytopes

For the rest of the paper, all polytopes we deal with will be finite. A polytope of type $\{p_1, p_2, \ldots , p_{n-1}\}$ has at least $2p_1 p_2 \cdots p_{n-1}$ flags, and if it has exactly that many flags, we say it is tight [10, Prop. 3.3].

The first mention of the property of tightness occured in [3], while searching for the smallest regular polytopes of each rank. There it was proven that for $n \ge 4$, the regular n-polytopes with fewest flags are always tight. Their study was extended in [10] to equivelar polytopes that may not be regular. In particular, it was proven there that an equivelar polytope is tight if and only if every section of rank 3 is flat. It follows that every section of a tight polytope is itself tight. The following lemma is a natural consequence of this fact.

Lemma 9

Let ${\mathcal {P}}$ and ${\mathcal {Q}}$ be tight rotary polytopes with types $\{p,q\}$ and $\{q,r\}$, respectively. Suppose that $\varGamma ^+({\mathcal {P}})= [p,q]^+ /N_1$ and $\varGamma ^+({\mathcal {Q}}) = [q,r]^+ /N_2$ where $N_1 \triangleleft [p,q]^+$ and $N_2 \triangleleft [q,r]^+$ are subgroups induced by the sets of relations $R_1$ and $R_2$, respectively. Then a rotary 4-polytope with facets isomorphic to ${\mathcal {P}}$ and vertex-figures isomorphic to ${\mathcal {Q}}$ exists if and only if the group $[p,q,r]^+/ N_3$ has order pqr and satisfies the intersection condition (4), where $N_3$ is the subgroup induced by the relations in $R_1$ in the first two generators and the relations $R_2$ in the last two generators. Moreover, such a 4-polytope must be unique.

Tight regular and chiral polyhedra were studied more deeply in [4, 12] and [13]. We summarize relevant results on these polyhedra in Sect. 3. Some results on regular polytopes of higher ranks can be found in [4].

The next proposition summarizes Corollary 3.4 and Theorem 3.5 of [11].

Proposition 1

• (a) If ${\mathcal {P}}$ is a tight chiral 4-polytope then it has chiral facets or chiral vertex-figures (or both).

• (b) If ${\mathcal {P}}$ is a tight chiral 5-polytope then it has chiral facets, vertex-figures, and medial sections.

• (c) There are no tight chiral n-polytopes for $n \ge 6$.

Since we shall work with the automorphism groups of chiral polytopes in place of the polytopes themselves, it is useful to have a characterization of tightness that is entirely group-theoretic.

Proposition 2

Suppose that ${\mathcal {P}}$ is an orientable rotary n-polytope of type $\{p_1, \ldots , p_{n-1}\}$, with $\varGamma ^+({\mathcal {P}}) = \langle \sigma _1, \ldots , \sigma _{n-1} \rangle$. Then the following are equivalent:

• (a) ${\mathcal {P}}$ is tight.

• (b) $|\varGamma ^+({\mathcal {P}})| = p_1 \cdots p_{n-1}$.

• (c) $\varGamma ^+({\mathcal {P}}) = \langle \sigma _1 \rangle \cdots \langle \sigma _{n-1} \rangle$.

Proof

The equivalence of (a) and (b) follows from the fact that $|\varGamma ^+({\mathcal {P}})|$ is equal to half the number of flags.

Next we show that (b) and (c) are equivalent. For each $1 \le i \le n-1$, let

$$\begin{aligned} S_i = \langle \sigma _i \rangle \cdots \langle \sigma _{n-1} \rangle . \end{aligned}$$

Then $|S_{n-1}| = p_{n-1}$, and for $i < n-1$,

$$\begin{aligned} S_i = \langle \sigma _i \rangle S_{i+1}. \end{aligned}$$

Therefore,

$$\begin{aligned} |S_i| = \frac{|\langle \sigma _i \rangle | \cdot |S_{i+1}|}{|\langle \sigma _i \rangle \cap S_{i+1}|}, \end{aligned}$$

and since $\varGamma ^+({\mathcal {P}})$ satisfies the intersection condition (4), the intersection on bottom is trivial, and so

$$\begin{aligned} |S_i| = p_i \cdot |S_{i+1}|. \end{aligned}$$

It follows that $|S_1| = p_1 \cdots p_{n-1}$. This shows that (c) implies (b).

Conversely, if $|\varGamma ^+({\mathcal {P}})| = p_1 \cdots p_{n-1}$, then $\varGamma ^+({\mathcal {P}})$ has the same order as its subset $S_1$, which implies that $\varGamma ^+({\mathcal {P}}) = S_1$. $\square$

Note that (b) and (c) are equivalent only in the presence of the intersection condition.

In light of Proposition 2, we will say that the group $\varGamma = \langle \sigma _1, \ldots , \sigma _{n-1} \rangle$ is tight provided that $\varGamma = \langle \sigma _1 \rangle \cdots \langle \sigma _{n-1} \rangle$. Then $\varGamma$ is the rotation group of a tight orientable rotary polytope if and only if $\varGamma$ is tight, and it satisfies the intersection condition (4). The following result is immediate:

Proposition 3

If $\varGamma$ is tight, then any quotient of $\varGamma$ is tight. If ${\mathcal {P}}$ is a tight orientable rotary polytope then any quotient of ${\mathcal {P}}$ is tight.

Proposition 3 imposes a restriction on the quotients of tight orientable rotary polytopes. The contrapositive of the next proposition imposes another restriction to quotients of tight orientably regular polytopes, namely that tight regular polytopes do not have chiral quotients.

Proposition 4

If ${\mathcal {P}}$ is a tight orientable rotary n-polytope that covers a chiral n-polytope then ${\mathcal {P}}$ itself is chiral.

Proof

Let ${\mathcal {Q}}$ be a chiral quotient of ${\mathcal {P}}$. We proceed by induction over n. By Proposition 1 (c), it is only necessary to show the statement for $n \in \{3,4,5\}$.

The case when $n=3$ was proven in [12, Prop. 2.5]. If $n \in {4,5}$, then by Proposition 1 either the facets or the vertex-figures of ${\mathcal {Q}}$ are chiral $(n-1)$-polytopes. Since the facets and vertex-figures of ${\mathcal {Q}}$ are quotients of the facets and vertex-figures of ${\mathcal {P}}$, the inductive hypothesis implies that the facets or vertex-figures of ${\mathcal {P}}$ must be chiral. Hence ${\mathcal {P}}$ is chiral. $\square$

Propositions 3 and 4 have the following consequence. When taking polytopal quotients of a tight chiral polytope ${\mathcal {P}}$ by normal subgroups of $\varGamma ^+({\mathcal {P}})$, we obtain tight orientably regular or chiral polytopes, and if ${\mathcal {P}}$ is orientably regular then the quotients are tight and regular. This suggests to try to find successive proper quotients of tight chiral polytopes until we obtain tight regular polytopes. As we shall see, this is always possible. Proposition 6 gives a condition for such quotients to exist. Other conditions will be given in Sects. 4 and 5.

The chiral polytopes we will be interested in typically have a cyclic chirality group, generated by a power of some $\sigma _i$. The following result describes circumstances where this property is preserved when taking quotients.

Lemma 10

Let ${\mathcal {P}}$ be a tight chiral polytope with $\varGamma ({\mathcal {P}}) = \langle \sigma _1, \ldots , \sigma _n \rangle$ and ${\mathcal {Q}}$ a chiral quotient of ${\mathcal {P}}$ with $\varGamma ({\mathcal {Q}}) = \langle \sigma _1', \ldots , \sigma _n' \rangle$. If $X({\mathcal {P}}) \le \langle \sigma _2 \rangle$ then $X({\mathcal {Q}}) \le \langle \sigma _2' \rangle$.

Proof

Let $K \triangleleft \varGamma ({\mathcal {P}})$ such that ${\mathcal {Q}}= {\mathcal {P}}/K$, and let ${\mathcal {R}}= {\mathcal {P}}/K X(P)$. Then ${\mathcal {R}}$ is a quotient of ${\mathcal {P}}/ X({\mathcal {P}})$, and since the latter is regular, Propositions 3 and 4 imply that ${\mathcal {R}}$ is regular as well. Now, ${\mathcal {R}}$ is the quotient of ${\mathcal {Q}}$ by KX(P)/K, and since ${\mathcal {R}}$ is regular, that implies that $X({\mathcal {Q}})$ is contained in $KX({\mathcal {P}})/K$, which is the image of $X({\mathcal {P}})$ in $\varGamma ({\mathcal {Q}})$, and thus contained in $\langle \sigma _2' \rangle$. $\square$

Next, we describe useful structural properties of the normal subgroups of the rotation group of tight orientable rotary polytopes.

Lemma 11

Let ${\mathcal {P}}$ be a tight orientable rotary n-polytope with $\varGamma ^+({\mathcal {P}}) = \langle \sigma _1, \dots , \sigma _{n-1} \rangle$ and let $K \triangleleft \varGamma ^+({\mathcal {P}})$ such that ${\mathcal {P}}/K$ is a tight orientable rotary n-polytope. Then there exist nonnegative integers $\alpha _1, \dots , \alpha _{n-1}$ such that

$$\begin{aligned} K = \langle \sigma _1^{\alpha _1} \rangle \langle \sigma _2^{\alpha _2} \rangle \cdots \langle \sigma _{n-1}^{\alpha _{n-1}} \rangle . \end{aligned}$$

Moreover, ${\mathcal {P}}/K$ has type $\{\alpha _1,\dots ,\alpha _{n-1}\}$.

Proof

For $1 \le i \le n-1$, let $\alpha _i$ be the smallest positive integer such that $\sigma _i^{\alpha _i} \in K$, and let $H = \langle \sigma _1^{\alpha _1} \rangle \cdots \langle \sigma _{n-1}^{\alpha _{n-1}} \rangle$. Then clearly $H \subseteq K$. To show the reverse inclusion, let $\gamma \in K$. By Proposition 2, we may write $\gamma$ as $\sigma _1^{\beta _1} \cdots \sigma _{n-1}^{\beta _{n-1}}$ for some exponents $\beta _i$. Since $\gamma \in K$, we have that for every i,

$$\begin{aligned} K \sigma _1^{\beta _1} \cdots \sigma _i^{\beta _i} = K (\sigma _{i+1}^{\beta _{i+1}} \cdots \sigma _{n-1}^{\beta _{n-1}})^{-1}. \end{aligned}$$

Then, writing $\overline{\sigma _i}$ for the image of $\sigma _i$ in $\varGamma ^+({\mathcal {P}})/K$, we get that

$$\begin{aligned} \overline{\sigma _1^{\beta _1} \cdots \sigma _i^{\beta _i}} = \overline{(\sigma _{i+1}^{\beta _{i+1}} \cdots \sigma _{n-1}^{\beta _{n-1}})^{-1}}. \end{aligned}$$

Since $\varGamma ^+({\mathcal {P}})/K$ is the rotation group of a rotary polytope, Equation (3) implies that $\overline{\sigma _1^{\beta _1} \cdots \sigma _i^{\beta _i}} = 1$, which means that $\sigma _1^{\beta _1} \cdots \sigma _i^{\beta _i} \in K$ for every i. In particular, $\sigma _1^{\beta _1} \in K$, from which it follows that $\sigma _2^{\beta _2} \in K$ (since $\sigma _1^{\beta _1} \sigma _2^{\beta _2} \in K$), and continuing in this way it follows that each $\sigma _i^{\beta _i} \in K$. By our choice of exponents $\alpha _i$, that means that each $\beta _i$ is divisible by $\alpha _i$, and so $\gamma \in H$.

The type of ${\mathcal {P}}/K$ follows from Proposition 3 since K has order $p_1 \cdots p_{n-1} / \alpha _1 \cdots \alpha _{n-1}$. $\square$

Proposition 5

Suppose that ${\mathcal {P}}$ is a tight orientable rotary n-polytope with $\varGamma ^+({\mathcal {P}}) = \langle \sigma _1, \ldots , \sigma _{n-1} \rangle$, and let $N = \langle \sigma _1^{a_1} \rangle \langle \sigma _2^{a_2} \rangle \cdots \langle \sigma _{n-1}^{a_{n-1}} \rangle$ be a normal subgroup of $\varGamma ^+({\mathcal {P}})$. If N does not contain any generator $\sigma _i$, then $\varGamma ^+({\mathcal {P}}) / N$ is the rotation group of a tight orientable rotary polytope.

Proof

Let $\varGamma ^+({\mathcal {P}}) / N = \langle \overline{\sigma _1}, \ldots , \overline{\sigma _{n-1}} \rangle$. Since no generator $\sigma _i$ is in N, it follows that each $\overline{\sigma _i}$ has order at least 2. Then to prove that $\varGamma ^+({\mathcal {P}}) / N$ is the rotation group of an orientable rotary polytope, by Lemma 3 it suffices to show that

$$\begin{aligned} \langle \overline{\sigma _1}, \ldots , \overline{\sigma _i} \rangle \cap \langle \overline{\sigma _j}, \ldots , \overline{\sigma _{i+1}} \rangle = \langle \overline{\sigma _j}, \ldots , \overline{\sigma _i} \rangle , \end{aligned}$$

for all i and j such that $2 \le j \le i+1 \le n-1$. (In fact, it suffices to show that the subgroup on the left is included in the subgroup on the right, since the reverse inclusion is obvious.) Tightness will then follow from Proposition 3.

Consider an element of $\varGamma ^+({\mathcal {P}}) / N$ that lies in

$$\begin{aligned} \langle \overline{\sigma _1}, \ldots , \overline{\sigma _i} \rangle \cap \langle \overline{\sigma _j}, \ldots , \overline{\sigma _{i+1}} \rangle . \end{aligned}$$

We may write this element as $\overline{\varphi _1} = \overline{\varphi _2}$, where

$$\begin{aligned} \varphi _1 \in \langle \sigma _1, \ldots , \sigma _i \rangle \end{aligned}$$

and

$$\begin{aligned} \varphi _2 \in \langle \sigma _j, \ldots , \sigma _{i+1} \rangle . \end{aligned}$$

Then $\varphi _1 = \gamma \varphi _2$ for some $\gamma \in N$. Since ${\mathcal {P}}$ is tight, Proposition 2(c) says that we may write $\gamma = \sigma _1^{b_1} \cdots \sigma _{n-1}^{b_{n-1}}$. Setting $\gamma _1 = \sigma _1^{b_1} \cdots \sigma _{j-1}^{b_{j-1}}$ and $\gamma _2 = \sigma _j^{b_j} \cdots \sigma _{n-1}^{b_{n-1}}$, we have that by definition $\gamma _1$ and $\gamma _2$ both lie in N. Now,

$$\begin{aligned} \gamma _1^{-1} \varphi _1 = \gamma _2 \varphi _2, \end{aligned}$$

and it follows that

$$\begin{aligned} \gamma _1^{-1} \varphi _1 \in \langle \sigma _1, \ldots , \sigma _i \rangle \cap \langle \sigma _j, \ldots , \sigma _{n-1} \rangle . \end{aligned}$$

Then since $\varGamma ^+({\mathcal {P}})$ satisfies the intersection condition, it follows from Lemma 2 that $\gamma _1^{-1} \varphi _1 \in \langle \sigma _j, \ldots , \sigma _i \rangle$. And since $\gamma _1 \in N$, this implies that $\overline{\varphi _1} \in \langle \overline{\sigma _j}, \ldots , \overline{\sigma _i} \rangle$, which is what we wanted to show. $\square$

When considering $\varGamma ^+({\mathcal {P}})$ as a group acting on the set of i-faces of ${\mathcal {P}}$ for some i, the kernel of this action is a natural normal subgroup of ${\mathcal {P}}$ to consider. (Recall that the kernel of the action of a group $\varGamma$ on a set X is the subgroup of $\varGamma$ fixing X pointwise.) The next results give sufficient conditions for the kernel of the action on the vertex set to be nontrivial.

Lemma 12

Let ${\mathcal {P}}$ be a tight orientable rotary polyhedron. If $\gamma \in \varGamma ^+({\mathcal {P}})$ fixes a vertex and one of its neighbors then it fixes all vertices of ${\mathcal {P}}$.

Proof

Let $u_0$ be the base vertex of ${\mathcal {P}}$. Let $\varGamma ^+({\mathcal {P}}) = \langle \sigma _1, \sigma _2 \rangle$, and let $\gamma \in \varGamma ^+({\mathcal {P}})$ such that it fixes $u_0$ and one of its neighbors $v_0$.

Since the stabilizer of $u_0$ is $\langle \sigma _2 \rangle$ then $\gamma = \sigma _2^a$ for some a. Now, if $\sigma _2^a$ fixes $v_0$ then it must fix all neighbors of $u_0$, since all of them are images of $v_0$ under $\langle \sigma _2 \rangle$. Since the choice of base vertex is arbitrary, we have proven that if $\gamma$ fixes a vertex u and one of its neighbors then it fixes all neighbors of u.

The result then follows from the connectivity of the 1-skeleton of ${\mathcal {P}}$. $\square$

The fact that the base facet of a tight polytope ${\mathcal {P}}$ contains all vertices of ${\mathcal {P}}$ implies the following corollary.

Corollary 1

Let ${\mathcal {P}}$ be a tight orientable rotary n-polytope with $\varGamma ^+({\mathcal {P}}) = \langle \sigma _1, \dots , \sigma _{n-1} \rangle$. If $\sigma _2^a$ fixes a neighbor of the base vertex, then it fixes all vertices of ${\mathcal {P}}$.

Corollary 2

Let ${\mathcal {P}}$ be a tight orientable rotary n-polytope with type $\{p_1,\dots , p_{n-1}\}$ with $p_1 \le p_2$. Then the kernel of the action of $\varGamma ^+({\mathcal {P}})$ on the vertex set is nontrivial.

Proof

If ${\mathcal {P}}$ is a tight polytope of type $\{p_1, \ldots , p_{n-1}\}$, then it has $p_1$ vertices. The automorphism $\sigma _2$ fixes the base vertex while permuting the remaining $p_1 - 1$. If $p_1 \le p_2$, then each neighbor of the base vertex must have a nontrivial stabilizer under $\langle \sigma _2 \rangle$, since the group has order $p_2$, which is larger than the largest possible orbit. $\square$

Now we are ready to exhibit a proper normal subgroup N of $\varGamma ^+({\mathcal {P}})$ that is a key element in discussions in Sects. 4 and 5.

Proposition 6

Let ${\mathcal {P}}$ be a tight orientable rotary n-polytope with $n \ge 3$ with type $\{p_1, \dots , p_{n-1}\}$ satisfying that $p_1 \ge p_2$, and rotation group $\varGamma ^+({\mathcal {P}})= \langle \sigma _1, \dots , \sigma _{n-1} \rangle$. Then there exists an integer k such that $\langle \sigma _1^k \rangle$ is a nontrivial normal subgroup of $\varGamma ^+({\mathcal {P}})$.

Proof

By the dual version of Corollary 2 the group $\langle \sigma _1, \sigma _2 \rangle$ has a nontrivial kernel when acting on the 2-faces of the base 3-face of $\mathcal P$. These 2-faces correspond to cosets of $\langle \sigma _1 \rangle$ in $\langle \sigma _1, \sigma _2 \rangle$. Then there exists $k \in \{1,\dots , p_1-1\}$ such that $\langle \sigma _1 \rangle \sigma _2^\ell \sigma _1^k = \langle \sigma _1 \rangle \sigma _2^\ell$ for every $\ell$. In particular, when $\ell =-1$ this implies that $\sigma _2^{-1} \sigma _1^k \sigma _2 \in \langle \sigma _1 \rangle$. Since the latter group is cyclic, we have that $\langle \sigma _1^k \rangle$ is normal in $\langle \sigma _1, \sigma _2 \rangle$. The result follows from Lemma 4 and commutativity of $\sigma _1^k$ with $\sigma _i$ for every $i \ge 4$. $\square$

3 Tight orientable rotary polyhedra and 4-polytopes

Much of the discussion on tight chiral n-polytopes for $n \ge 4$ in Sects. 4, 5 and 6 is based on what we know about tight orientable rotary polyhedra. In this section, we summarize some important facts about them.

We start with a simple result related to Lemma 1, and one of its consequences for tight orientable rotary polyhedra.

Lemma 13

For every $p \ge 2$, there is a unique polyhedron of type $\{p,2\}$ and it is regular.

Proof

Let ${\mathcal {P}}$ be a polyhedron with type $\{p,2\}$. Then every vertex of ${\mathcal {P}}$ is incident with precisely two edges and precisely two facets. Since adjacent vertices belong to the same two facets, the connectivity of ${\mathcal {P}}$ forces ${\mathcal {P}}$ itself to have only two facets. It follows that ${\mathcal {P}}$ is isomorphic to the face lattice of the map on the sphere whose 1-skeleton is an equatorial p-gon, and its two facets are the northern and southern hemispheres. Clearly ${\mathcal {P}}$ is regular. $\square$

Lemma 14

If ${\mathcal {P}}$ is an orientable rotary polyhedron with $\varGamma ^+({\mathcal {P}})=\langle \sigma _1, \sigma _2 \rangle$ and $\langle \sigma _1 \rangle \triangleleft \varGamma ^+({\mathcal {P}})$, then ${\mathcal {P}}$ has type $\{p,2\}$ for some p. In particular, ${\mathcal {P}}$ is regular.

Proof

If $\sigma _2^{-1} \langle \sigma _1 \rangle \sigma _2 = \langle \sigma _1 \rangle$ then $\sigma _2^{-1} \sigma _1 \sigma _2 = \sigma _1^k$ for some k. Now, $\sigma _2^{-1} \sigma _1 \sigma _2= \sigma _2^{-1} \sigma _2^{-1} \sigma _1^{-1}$, implying that $\sigma _2^{-2} = \sigma _1^{k+1}$. The intersection condition (4) tells us that $\sigma _2$ has order 2, and hence, the type of ${\mathcal {P}}$ is $\{p,2\}$ for some p. Lemma 13 implies the regularity of ${\mathcal {P}}$. $\square$

The rotation groups of tight orientable rotary polyhedra have many normal subgroups contained in the vertex or facet stabilizer. In the next result, we describe some of these normal subgroups.

Proposition 7

Suppose ${\mathcal {P}}$ is a chiral or orientable rotary polyhedron of type $\{p, q\}$, with $\varGamma ^+({\mathcal {P}}) = \langle \sigma _1, \sigma _2 \rangle$. If $\langle \sigma _2^a \rangle \triangleleft \varGamma ^+({\mathcal {P}})$, then $\sigma _2^a \sigma _1 = \sigma _1 \sigma _2^{sa}$ for some s such that $s^2 \equiv 1$ (mod q/a). In particular, $\sigma _1^2$ commutes with $\sigma _2^a$, and if p is odd, then $\sigma _2^a$ is central.

Proof

Without loss of generality, we may assume that a is a positive divisor of q. The subgroup $\langle \sigma _2^a \rangle$ is normal if and only if $\sigma _1^{-1} \sigma _2^a \sigma _1 = \sigma _2^{sa}$ for some s. Furthermore, we note that

$$\begin{aligned} \sigma _2^a&= (\sigma _1 \sigma _2)^{-2} \sigma _2^a (\sigma _1 \sigma _2)^2 \\&= (\sigma _2^{-1} \sigma _1^{-1} \sigma _2^{-1}) \sigma _2^{sa} (\sigma _2 \sigma _1 \sigma _2) \\&= \sigma _2^{-1} \sigma _1^{-1} \sigma _2^{sa} \sigma _1 \sigma _2 \\&= \sigma _2^{-1} \sigma _2^{s^2 a} \sigma _2 \\&= \sigma _2^{s^2 a}, \end{aligned}$$

so that $a \equiv s^2 a$ (mod q), and thus $s^2 \equiv 1$ (mod q/a). It is now clear then that $\sigma _1^2$ commutes with $\sigma _2^a$, and if p is odd, then $\langle \sigma _1^2 \rangle = \langle \sigma _1 \rangle$ so that $\sigma _1$ commutes with $\sigma _2^a$ as well. $\square$

Lemma 15

Let ${\mathcal {P}}$ be a tight regular polyhedron with $\varGamma ^+({\mathcal {P}})=\langle \sigma _1,\sigma _2 \rangle$ and $\langle \sigma _2 \rangle$ core-free. Then $\langle \sigma _1^2 \rangle \triangleleft \varGamma ^+({\mathcal {P}})$.

Proof

First, [13, Theorem 3.3] says that $\varGamma ^+({\mathcal {P}})$ is the quotient of $[p, q]^+$ by the extra relation $\sigma _2^{-1} \sigma _1 = \sigma _1^i \sigma _2^j$ for some i and j. By [13, Proposition 3.2(a)], the subgroup $\langle \sigma _2^{j-1} \rangle$ is normal, so since we are supposing that $\langle \sigma _2 \rangle$ is core-free, we need $j = 1$. Then taking the relation $\sigma _2^{-1} \sigma _1 = \sigma _1^i \sigma _2$ and multiplying on the left by $\sigma _1^{-1}$ and then rewriting $\sigma _1^{-1} \sigma _2^{-1}$ as $\sigma _2 \sigma _1$ gives us $\sigma _2 \sigma _1^2 = \sigma _1^{i-1} \sigma _2$. Furthermore, i must be odd; Sect. 4 of [13] uses a parameter k which is shown at the end of the section to satisfy $i = 1 - k$, and k and p are both even by [13, Lemma 4.8]. (Note that Lemma 4.8 requires the polyhedron to have no multiple edges; this is equivalent to asking for $\langle \sigma _2 \rangle$ to be core-free, by [13, Proposition 4.6].) Thus $\sigma _2$ normalizes $\langle \sigma _1^2 \rangle$ and thus this subgroup is normal. $\square$

Tight orientably regular polyhedra with no multiple edges were classified in [13, Theorem 4.13]. The next theorem is a direct consequence.

Theorem 2

The types of the tight orientably regular polyhedra with no multiple edges are:

• (a) $\{p,2\}$ for some $p \ge 2$,

• (b) $\{2q,q\}$ for some odd integer $q\ge 3$,

• (c) $\{p,q\}$ with $p = 2^{\alpha _1}P_2^{\alpha _2} \cdots P_k^{\alpha _k}$ for some $\alpha _1 > 0$, some distinct odd primes $P_2,\dots ,P_k$, and q a proper even divisor of p satisfying that

  • the maximal power of 2 dividing q is either 2, 4 or $2^{\alpha _1-1}$, and if it is 4 then $\alpha _1 \ge 3$,

  • for $i\in \{2,\dots ,k\}$, either $P_i^{\alpha _i}$ divides q or $P_i$ is coprime with q.

In [12], an atomic chiral polyhedron was defined as a tight chiral polyhedron with type $\{p,q\}$ that covers no chiral polyhedron of type $\{p',q\}$ or of type $\{p,q'\}$ for $p'$ a proper divisor of p and $q'$ a proper divisor of q. It is easy to see that every tight chiral polyhedron covers an atomic chiral polyhedron. Furthermore, we will see in Corollary 4 that a stronger condition is true: atomic chiral polyhedra do not cover any chiral polyhedra.

The atomic chiral polyhedra were classified in [12, Lemma 4.10, Theorem 4.11, Theorem 4.14]. Here we summarize and slightly simplify this classification (see [12, Theorem 4.15]).

Table 1 The atomic chiral polyhedra. An atomic chiral polyhedron with name ${\mathcal {P}}(p,q)_t$ has an automorphism group that is a quotient of $[p,q]^+$ by the relation(s) given in the “Extra relations” column, and the subscript indicates the name of an additional parameter

Theorem 3

Every atomic chiral polyhedron ${\mathcal {P}}$ is one of the polyhedra in Table 1, with chirality group $X({\mathcal {P}})$ and enantiomorph ${{\mathcal {P}}}^*$ as described in the table.

Proof

First we will prove the claim for atomic chiral polyhedra of type $\{2m, m^{\alpha }\}$ and $\{m^{\alpha }, 2m\}$. We start by noting that for any rotation group $\langle \sigma _1, \sigma _2 \rangle$ and for all t, the relation $\sigma _2^{-1} \sigma _1 = \sigma _1^3 \sigma _2^t$ is equivalent to $\sigma _2 \sigma _1^2 = \sigma _1^2 \sigma _2^t$, since:

$$\begin{aligned} \sigma _2^{-1} \sigma _1&= \sigma _1^3 \sigma _2^t&\text {Multiply both sides by}\, \sigma _1^{-1}\hbox { on the left} \\ \sigma _1^{-1} \sigma _2^{-1} \sigma _1&= \sigma _1^2 \sigma _2^t&\text {Use}\, \sigma _1^{-1} \sigma _2^{-1} = \sigma _2 \sigma _1 \\ \sigma _2 \sigma _1^2&= \sigma _1^2 \sigma _2^t. \end{aligned}$$

Similarly, for all t, the relation $\sigma _2 \sigma _1^{-1} = \sigma _1^{-3} \sigma _2^t$ is equivalent to $\sigma _1^2 \sigma _2 = \sigma _2^{-t} \sigma _1^2$, since:

$$\begin{aligned} \sigma _2 \sigma _1^{-1}&= \sigma _1^{-3} \sigma _2^t&\text {Multiply both sides by}\, \sigma _1 \text {on the left } \\ \sigma _1 \sigma _2 \sigma _1^{-1}&= \sigma _1^{-2} \sigma _2^t&\text {Use }\sigma _1 \sigma _2 = \sigma _2^{-1} \sigma _1^{-1} \\ \sigma _2^{-1} \sigma _1^{-2}&= \sigma _1^{-2} \sigma _2^t&\text {Invert both sides} \\ \sigma _1^2 \sigma _2&= \sigma _2^{-t} \sigma _1^2. \end{aligned}$$

Now, suppose that ${\mathcal {P}}$ is the atomic chiral polyhedron of type $\{2m, m^{\alpha }\}$ whose group is the quotient of $[2m, m^{\alpha }]^+$ by the relations $\sigma _2^{-1} \sigma _1 = \sigma _1^3 \sigma _2^{1+km^{\alpha -1}}$ and $\sigma _2 \sigma _1^{-1} = \sigma _1^{-3} \sigma _2^{-1+km^{\alpha -1}}$. (See [12, Theorem 4.11].) Then the above discussion shows that this group is equivalent to the quotient of $[2m, m^{\alpha }]^+$ by the relations $\sigma _2 \sigma _1^2 = \sigma _1^2 \sigma _2^{1+km^{\alpha -1}}$ and $\sigma _1^2 \sigma _2 = \sigma _2^{1-km^{\alpha -1}} \sigma _1^2$. Furthermore, the second of those relations is superfluous, since if $\sigma _2 \sigma _1^2 = \sigma _1^2 \sigma _2^{1+km^{\alpha -1}}$ then

$$\begin{aligned} \sigma _2^{1-km^{\alpha -1}} \sigma _1^2 = \sigma _1^2 \sigma _2^{(1-km^{\alpha -1})(1+km^{\alpha -1})} = \sigma _1^2 \sigma _2. \end{aligned}$$

So $\varGamma ({\mathcal {P}})$ may be written in the form as it appears in Table 1.

Next, the proof of [12, Theorem 3.6] shows that $\langle \sigma _2^{m^{\alpha -1}} \rangle$ is normal and that the quotient of ${\mathcal {P}}$ by this normal subgroup is regular. Thus $X({\mathcal {P}})$ is a nontrivial subgroup of $\langle \sigma _2^{m^{\alpha -1}} \rangle$, and since the latter has prime order m, this implies that $X({\mathcal {P}}) = \langle \sigma _2^{m^{\alpha -1}} \rangle$.

To find a presentation for $\varGamma ({{\mathcal {P}}}^*)$, we may change the defining relations of $\varGamma ({\mathcal {P}})$ by replacing $\sigma _1$ with $\sigma _1^{-1}$ and replacing $\sigma _2$ with $\sigma _2^{-1}$. This yields:

$$\begin{aligned} \sigma _2^{-1} \sigma _1^{-2}&= \sigma _1^{-2} \sigma _2^{-1-km^{\alpha -1}}&\text {Invert both sides} \\ \sigma _1^2 \sigma _2&= \sigma _2^{1+km^{\alpha -1}} \sigma _1^2. \end{aligned}$$

From this, we obtain $\sigma _1^2 \sigma _2^{1-km^{\alpha -1}} = \sigma _2^{(1+km^{\alpha -1})(1-km^{\alpha -1})} \sigma _1^2 = \sigma _2 \sigma _1^2$. Thus, the enantiomorph replaces the parameter k with $-k$ (or equivalently, $m-k$).

A presentation for the dual of ${\mathcal {P}}$ (with respect to the same base flag as ${\mathcal {P}}$) is obtained by changing each defining relation, replacing $\sigma _1$ with $\sigma _2^{-1}$ and $\sigma _2$ with $\sigma _1^{-1}$. Applying this to the relation $\sigma _2 \sigma _1^2 = \sigma _1^2 \sigma _2^{1+km^{\alpha -1}}$ and then inverting both sides yields $\sigma _2^2 \sigma _1 = \sigma _1^{1+km^{\alpha -1}} \sigma _2^2$, matching the second row of Table 1.

This finishes the proof for atomic chiral polyhedra of type $\{2m, m^{\alpha }\}$ and their duals. The proof for the remaining polyhedra is analogous (referencing [12, Theorems 3.7 and 3.8]), except that for type $\{2^{\beta -1}, 2^{\beta }\}$ and its dual, it is not possible to simplify the presentation in the same way that we can for the other two cases. $\square$

Corollary 3

Let ${\mathcal {P}}$ be an atomic chiral polyhedron with type $\{p,q\}$ and $p\ge q$. Then p is a prime power.

It turns out that the atomic chiral polyhedra satisfy a stronger condition than their definition would seem to imply.

Corollary 4

If ${\mathcal {P}}$ is an atomic chiral polyhedron, then it does not properly cover any chiral polyhedron.

Proof

Suppose that ${\mathcal {P}}$ is an atomic chiral polyhedron of type $\{p, q\}$, and without loss of generality, assume that $p \ge q$ so that p is a prime power $m^{\alpha }$ (where we could have $m = 2$). By the definition of atomic, ${\mathcal {P}}$ does not properly cover any chiral polyhedra of type $\{p, q'\}$ or $\{p', q\}$. Furthermore, if ${\mathcal {Q}}$ is an orientable rotary polyhedron of type $\{p', q'\}$ where $p'$ is a proper divisor of p, then the kernel of the natural map from $\varGamma ({\mathcal {P}})$ to $\varGamma ({\mathcal {Q}})$ contains $\langle \sigma _1^{m^{\alpha -1}} \rangle$ (see Table 1), and since that is the chirality group of ${\mathcal {P}}$, it follows that ${\mathcal {Q}}$ is regular. $\square$

In light of Corollary 4, let us now make a (harmless) redefinition of what it means to be atomic, while simultaneously generalizing the definition to higher rank.

Definition 1

A chiral polytope is atomic if it is tight and it does not properly cover any chiral polytopes.

The following result is an immediate consequence of the definition of atomicity and [12, Corollary 4.3], which states that every tight chiral polyhedron of type $\{p, q\}$ covers a tight orientable rotary polyhedron of type $\{p', q\}$ or $\{p, q'\}$.

Proposition 8

If ${\mathcal {P}}$ is a tight chiral polyhedron of type $\{p, q\}$ that is not atomic, then it covers a tight chiral polyhedron of type $\{p', q\}$ or $\{p, q'\}$.

When mixing tight orientable rotary polyhedra we may not get a tight structure, as shown next.

Proposition 9

Let ${\mathcal {P}}$ and ${\mathcal {Q}}$ be distinct atomic chiral polyhedra of types $\{p, q\}$ and $\{p, q'\}$, respectively, with $q'$ a divisor of q (not necessarily proper). Then ${\mathcal {P}}\lozenge {\mathcal {Q}}$ is not tight, regardless of the choice of base flags.

Proof

The mix of ${\mathcal {P}}$ and ${\mathcal {Q}}$ with respect to any choice of base flags must have type $\{p,q\}$, and if it were tight then it should be isomorphic to ${\mathcal {P}}$ and have ${\mathcal {Q}}$ as a proper quotient. This is not possible since ${\mathcal {P}}$ is atomic. $\square$

We conclude this section with some technical lemmas that allow us to find polytopal quotients of tight orientable rotary 4-polytopes.

Lemma 16

Let ${\mathcal {P}}$ be a tight chiral polyhedron with type $\{p,q\}$ and $\varGamma ({\mathcal {P}}) = \langle \sigma _1,\sigma _2 \rangle$. Then $\langle \sigma _i^2 \rangle$ is not normal in $\varGamma ({\mathcal {P}})$.

Proof

Let ${\mathcal {Q}}$ be an atomic chiral polyhedron covered by ${\mathcal {P}}$ with automorphism group $\langle \tau _1, \tau _2 \rangle$. If we assume that $\langle \sigma _i^2 \rangle \triangleleft \varGamma ({\mathcal {P}})$, then by the correspondence theorem in group theory we must also have that $\langle \tau _i^2 \rangle \triangleleft \varGamma ({\mathcal {Q}})$.

It was proven in [12, Proposition 4.1] that if ${\mathcal {Q}}$ has type $\{p',q'\}$ and $p'>q'$ then $\langle \tau _1 \rangle$ has a proper subgroup normal in $\varGamma ({\mathcal {Q}})$. On the other hand, it is shown in [12, Proposition 4.5] that either $\langle \tau _1 \rangle$ or $\langle \tau _2 \rangle$ is core-free in $\varGamma ({\mathcal {Q}})$. Up to duality, we may assume that $p'>q'$, and hence, we only need to show that $\langle \tau _1^2 \rangle$ is not normal in $\varGamma ({\mathcal {Q}})$.

Now, using the classification of atomic chiral polyhedra we see that if $\{p,q\} = \{m^{\alpha }, 2m\}$ then $\langle \tau _1^2 \rangle = \langle \tau _1 \rangle$ and this is not normal in $\varGamma ({\mathcal {Q}})$ (see Lemma 14). On the other hand, if p and q are powers of 2, the dual version of [12, Lemma 4.13] tells us that the core of $\langle \tau _1 \rangle$ is $\langle \tau _1^4 \rangle$. Hence, $\langle \tau _1^2 \rangle$ is not normal in $\varGamma ({\mathcal {Q}})$. $\square$

Lemma 17

Let ${\mathcal {P}}$ be a chiral 4-polytope with chiral facets and let K be the kernel of the action of $\varGamma ({\mathcal {P}})$ on the vertex set. Then $\sigma _i \notin K$ for $i \in \{1,2,3\}$.

Proof

The group K is a normal subgroup of $\varGamma ({\mathcal {P}})$ that is contained in $\langle \sigma _2, \sigma _3 \rangle$ since it fixes the base vertex. The intersection condition (3) implies that $\sigma _1 \not \in K$.

If $\sigma _2 \in K$, then also $\sigma _1 \sigma _2 \sigma _1^{-1} \in K$, which implies that $\sigma _2^{-1} \sigma _1^{-2} \in K$ and so $\sigma _1^2 \in K$. Since $\langle \sigma _1 \rangle$ has trivial intersection with K (again by the intersection condition), this implies that $\sigma _1^2 = id$, which contradicts Lemma 1.

Similarly, if $\sigma _3 \in K$, then also $\sigma _2 \sigma _3 \sigma _2^{-1} \in K$, which implies that $\sigma _3^{-1} \sigma _2^{-2} \in K$ and so $\sigma _2^2 \in K$. It follows that $\sigma _1^{-1} \sigma _2^2 \sigma _1 \in K$. Then $\sigma _1^{-1} \sigma _2^2 \sigma _1$ lies in the intersection of $\langle \sigma _1, \sigma _2 \rangle$ with $\langle \sigma _2, \sigma _3 \rangle$, and so it must lie in $\langle \sigma _2 \rangle$. This implies that $\langle \sigma _2^2 \rangle$ is normal in $\langle \sigma _1, \sigma _2 \rangle$, contradicting Lemma 16. $\square$

Lemma 18

Let ${\mathcal {P}}$ be a tight orientable rotary 4-polytope, with $\varGamma ^+({\mathcal {P}}) = \langle \sigma _1, \sigma _2, \sigma _{3} \rangle$. Let K be the kernel of the action of $\varGamma ^+({\mathcal {P}})$ on the vertex set. Then:

• (a) There are integers a and b such that $K = \langle \sigma _2^{a} \rangle \langle \sigma _3^{b} \rangle$.

• (b) ${\mathcal {P}}/ K$ is a tight orientable rotary 4-polytope.

Proof

Let a be the smallest positive integer such that $\sigma _2^a \in K$, and let b be the smallest positive integer such that $\sigma _3^b \in K$. (We allow the possibility that $\sigma _2^a = id$ or $\sigma _3^b = id$.) Let $N = \langle \sigma _2^a \rangle \langle \sigma _3^b \rangle$. Then clearly N is contained in K. To prove the first part, it remains to show that K is contained in N.

Let $H = \langle \sigma _2, \sigma _{3} \rangle$, and suppose that the order of $\sigma _1$ is p. Since ${\mathcal {P}}$ is tight, it has p vertices, which we can identify with the cosets $H, H \sigma _1, \ldots , H \sigma _1^{p-1}$. The action of each automorphism on the vertices is by multiplication on the right. Now, suppose that $\varphi \in K$, which in particular implies that $\varphi \in \langle \sigma _2, \sigma _{3} \rangle$. Since ${\mathcal {P}}$ is tight, Proposition 2 implies that we may write $\varphi = \sigma _2^{c} \sigma _{3}^{d}$. Since $\sigma _2^c \sigma _3^d$ fixes all vertices, it follows that the action of $\sigma _2^{c}$ on vertices is the same as the action of $\sigma _3^{-d}$ on vertices. Note that $\sigma _3^{-1}$ fixes the neighbor of the base vertex in the base edge, namely,

$$\begin{aligned} H \sigma _1^{-1} \sigma _3^{-1} = H (\sigma _3 \sigma _1)^{-1} = H (\sigma _1 \sigma _2^2 \sigma _3)^{-1} = H \sigma _1^{-1}. \end{aligned}$$

It follows that $\sigma _3^{-d}$ fixes that vertex, and thus so does $\sigma _2^{c}$. However, by Corollary 1, if a power of $\sigma _2$ fixes a neighbor of the base vertex, then it fixes all vertices. Therefore, $\sigma _2^{c} \in K$, from which it follows that $\sigma _3^{d} \in K$. Then by our choice of a and b, it follows that $\varphi \in N$.

The second part follows from the first along with Lemma 17. $\square$

4 Atomic chiral 4-polytopes with chiral facets and vertex-figures

To understand the structure of tight chiral 4-polytopes, we use a strategy similar to what was done with tight chiral polyhedra. Recall that a tight chiral 4-polytope is atomic if it does not properly cover any chiral polytopes. It is clear that every tight chiral polytope covers an atomic chiral polytope. Our goal will be to classify the atomic chiral 4-polytopes.

By Proposition 1 (a), the facets or the vertex-figures of an atomic chiral 4-polytope must be chiral. In this section, we classify all atomic chiral 4-polytopes that have chiral facets and chiral vertex-figures, leaving the case when one of them is regular for Sect. 5. We will show in Theorem 4 that an atomic chiral 4-polytope with chiral facets and chiral vertex-figures must have atomic chiral facets and atomic chiral vertex-figures. The classification of atomic chiral polyhedra will be then used to find all atomic chiral 4-polytopes with chiral facets and vertex-figures.

4.1 The structure of atomic chiral 4-polytopes with chiral facets and vertex-figures

Now we study atomic chiral 4-polytopes with chiral facets and chiral vertex-figures. We find several restrictions on atomic chiral 4-polytopes, culminating in Theorem 4.

Proposition 10

Let ${\mathcal {P}}$ be an atomic chiral 4-polytope of type $\{p, q, r\}$, with chiral facets and vertex-figures, and with $\varGamma ({\mathcal {P}}) = \langle \sigma _1, \sigma _2, \sigma _3 \rangle$. Then:

• (a) $\langle \sigma _1 \rangle$ and $\langle \sigma _3 \rangle$ are core-free in $\varGamma ({\mathcal {P}})$.

• (b) $q > p$ and $q > r$.

Proof

By duality, for the first part it suffices to prove that $\langle \sigma _1 \rangle$ is core-free. Suppose that ${\mathcal {P}}$ is a tight chiral 4-polytope with chiral facets and vertex-figures and suppose that $\langle \sigma _1 \rangle$ is not core-free. In other words, there is a nontrivial normal subgroup $N = \langle \sigma _1^a \rangle$ of $\varGamma ({\mathcal {P}})$. If $\sigma _1 \in N$, then $\langle \sigma _1 \rangle$ is normal in $\langle \sigma _1, \sigma _2 \rangle$, and by Lemma 14, this implies that the facets are regular, contradicting our assumptions. So $\sigma _1 \not \in N$. Then the dual of Proposition 5 shows that $\varGamma ({\mathcal {P}}) / N$ is the rotation group of a tight rotary polytope ${\mathcal {Q}}$. Since $\langle \sigma _2, \sigma _3 \rangle$ has trivial intersection with N, the vertex-figures of ${\mathcal {Q}}$ must be isomorphic to the vertex-figures of ${\mathcal {P}}$, which are chiral. Thus ${\mathcal {Q}}$ is chiral, which means that ${\mathcal {P}}$ is not atomic. This proves part (a).

By Proposition 6, if $p \ge q$ then there exists a proper divisor k of p such that $\langle \sigma _1^k \rangle \triangleleft \varGamma ({\mathcal {P}})$ contradicting part (a). A dual argument follows if $r \ge q$. $\square$

Proposition 11

Let ${\mathcal {P}}$ be an atomic chiral 4-polytope of type $\{p, q, r\}$, with chiral facets and vertex-figures, and with $\varGamma ({\mathcal {P}}) = \langle \sigma _1, \sigma _2, \sigma _3 \rangle$. Then:

• (a) The chirality group $X({\mathcal {P}})$ is $\langle \sigma _2^{q'} \rangle$ for some $q'$ with $q/q'$ prime,

• (b) The chirality groups of the base facet and vertex-figure are isomorphic to $X({\mathcal {P}})$.

Proof

Let H and K be the kernels of the actions of $\varGamma ({\mathcal {P}})$ on the vertices and on the facets of ${\mathcal {P}}$, respectively. By Proposition 10(b) together with Corollary 2 and its dual form, $H \le \langle \sigma _2, \sigma _3 \rangle$ and $K \le \langle \sigma _1, \sigma _2 \rangle$ are nontrivial normal subgroups of $\varGamma ({\mathcal {P}})$. Therefore $H \cap K$ is a normal subgroup of $\varGamma ({\mathcal {P}})$ that by the intersection condition is contained in $\langle \sigma _2 \rangle$.

Now, Lemma 18 and its dual show that ${\mathcal {P}}/H$ and ${\mathcal {P}}/K$ are polytopes, and since H and K are nontrivial and ${\mathcal {P}}$ is atomic, ${\mathcal {P}}/H$ and ${\mathcal {P}}/K$ are regular. Moreover, by Lemma 5, ${\mathcal {P}}/(H\cap K) \cong {\mathcal {P}}/H \lozenge {\mathcal {P}}/K$ is also regular, implying that $H \cap K$ is nontrivial.

Since ${\mathcal {P}}/(H\cap K)$ is regular, $X({\mathcal {P}}) \le H\cap K = \langle \sigma _2^m \rangle$ for some m. If q/m is not prime, then $\langle \sigma _2^{mk} \rangle \triangleleft \varGamma ({\mathcal {P}})$ for any k, in particular, for some k such that q/mk is prime. By atomicity of ${\mathcal {P}}$, its quotient by $\langle \sigma _2^{mk} \rangle$ is regular and, since it is a maximal quotient, $X({\mathcal {P}}) = \langle \sigma _2^{mk} \rangle$. This concludes part (a).

Part (b) follows from Part (a) and Lemma 8. $\square$

Proposition 12

Let ${\mathcal {P}}$ be an atomic chiral 4-polytope of type $\{p,q,r\}$ with chiral facets and vertex-figures. If q is a prime power then the facets and vertex-figures of ${\mathcal {P}}$ are atomic chiral polyhedra.

Proof

Suppose ${\mathcal {P}}$ has facets isomorphic to ${\mathcal {Q}}_1$ and vertex-figures isomorphic to ${\mathcal {Q}}_2$. By Proposition 11, $X({\mathcal {P}}) = X({\mathcal {Q}}_1)=X({\mathcal {Q}}_2)$, and these groups are cyclic of prime order. If q is a prime power then $X({\mathcal {P}})$ is contained in all proper subgroups of $\langle \sigma _2 \rangle$, and so ${\mathcal {Q}}_1$ does not cover any tight chiral polyhedra of type $\{p, q'\}$ with $q'$ a proper divisor of q. Proposition 10 says that $\langle \sigma _1 \rangle$ is core-free, and so also ${\mathcal {Q}}_1$ does not cover any tight chiral polyhedra of type $\{p', q\}$ with $p'$ a proper divisor of p. It follows that ${\mathcal {Q}}_1$ is atomic, and a dual argument proves that ${\mathcal {Q}}_2$ is atomic as well. $\square$

We are now ready to prove the main necessary condition for a tight chiral 4-polytope with chiral facets and chiral vertex-figures to be atomic.

Theorem 4

If ${\mathcal {P}}$ is an atomic chiral 4-polytope with chiral facets and chiral vertex-figures, then the facets and vertex-figures are atomic chiral polyhedra.

Proof

Assume that ${\mathcal {P}}$ has type $\{p,q,r\}$. The facets and vertex-figures of ${\mathcal {P}}$ are isomorphic to some chiral polyhedra ${\mathcal {Q}}_1$ and ${\mathcal {Q}}_2$, respectively. Proposition 11 (b) tells us that $X({\mathcal {P}}) = \langle \sigma _2^{q'} \rangle$ with $q/q'$ prime and that $X({\mathcal {P}}) = X({\mathcal {Q}}_1) = X({\mathcal {Q}}_2)$.

Assume to the contrary that ${\mathcal {Q}}_i$ is not atomic for some $i \in \{1,2\}$. Then, by Proposition 12, q must have at least two distinct prime factors, which by Corollary 3 implies that neither ${\mathcal {Q}}_1$ nor ${\mathcal {Q}}_2$ is atomic. Let $m = q/q'$, which is prime (but not necessarily odd). Then $q = m^{\alpha } t$ for some $\alpha$ and some t not divisible by m.

Since ${\mathcal {Q}}_1$ is not atomic there exists $N_1 \triangleleft \langle \sigma _1,\sigma _2 \rangle$ such that ${\mathcal {Q}}_1/N_1$ is an atomic chiral polyhedron. By Lemma 11, there exist a and b such that $N_1 = \langle \sigma _1^{a} \rangle \langle \sigma _2^{b} \rangle$ and ${\mathcal {Q}}_1 / N_1$ has type $\{a,b\}$.

Now $X({\mathcal {Q}}_1)=X({\mathcal {P}})$, and therefore, $\sigma _2^{q'} \notin N_1$. It follows that q/b divides t and $m^{\alpha }$ divides b. Lemma 10 implies that $X({\mathcal {Q}}_1/N_1)$ is contained in the subgroup generated by the second standard generator of $\varGamma ({\mathcal {Q}}_1/N_1)$. Since ${\mathcal {Q}}_1/N_1$ is atomic, we can conclude that $a < b$ by the classification of atomic chiral polyhedra. Corollary 3 now tells us that $b = m^{\alpha }$.

We proceed in a dual manner to observe that there exists $N_2= \langle \sigma _2^{b'}\rangle \langle \sigma _3^{c} \rangle \le \langle \sigma _2,\sigma _3 \rangle$ such that ${\mathcal {Q}}_2/N_2$ is an atomic chiral polyhedron with type $\{b',c\} = \{m^{\alpha },c\}$. In particular, $b=b'$.

Let $K= \langle \sigma _1^{a} \rangle \langle \sigma _2^{b} \rangle \langle \sigma _3^{c} \rangle$. We claim that $K\triangleleft \varGamma ({\mathcal {P}})$. To see this, note that

$$\begin{aligned} \sigma _2^{-1}(\sigma _1^{k_1 a} \sigma _2^{k_2 b} \sigma _3^{k_3 c})\sigma _2 = (\sigma _2^{-1} \sigma _1^{k_1 a} \sigma _2^{k_2 b} \sigma _2)(\sigma _2^{-1} \sigma _3^{k_3 c} \sigma _2) \in (\langle \sigma _1^{a} \rangle \langle \sigma _2^{b} \rangle )(\langle \sigma _2^{b} \rangle \langle \sigma _3^{c} \rangle ), \end{aligned}$$

and as noted in the proof of Lemma 4,

$$\begin{aligned} \sigma _3(\sigma _1^{k_1 a} \sigma _2^{k_2 b} \sigma _3^{k_3 c})\sigma _3^{-1} = (\sigma _2^{-1}(\sigma _1^{-k_1 a})\sigma _2)(\sigma _3 \sigma _2^{k_2 b}\sigma _3^{k_3 c} \sigma _3^{-1}) \in (\langle \sigma _1^{a} \rangle \langle \sigma _2^{b} \rangle )(\langle \sigma _2^{b} \rangle \langle \sigma _3^{c} \rangle ).\end{aligned}$$

A dual argument shows that K is invariant under conjugation by $\sigma _1$.

Now, Lemma 17 and Proposition 5 imply that ${\mathcal {P}}/K$ is a polytope, and since ${\mathcal {P}}$ is atomic this polytope must be regular of type $\{a, b, c\}$. In particular, this implies that the facets are regular polyhedra of type $\{a, b\}$. On the other hand, the facets must be a quotient of ${\mathcal {Q}}_1/N_1$, which is a tight chiral polyhedron of type $\{a, b\}$. But no tight polyhedron properly covers another polyhedron of the same type, and so we have a contradiction. $\square$

Now let us show that the conditions in Proposition 10 and Theorem 4 suffice if we want to build an atomic chiral 4-polytope.

Corollary 5

A tight chiral 4-polytope ${\mathcal {P}}$ with chiral facets and vertex-figures is atomic if and only if

• (a) The facets and vertex-figures are atomic, and

• (b) $\langle \sigma _1 \rangle$ and $\langle \sigma _3 \rangle$ are core-free in $\varGamma ({\mathcal {P}})$.

Proof

Theorem 4 and Proposition 10 prove that the conditions are necessary. Now, suppose that ${\mathcal {P}}$ satisfies the conditions. If ${\mathcal {Q}}$ is a proper chiral quotient of ${\mathcal {P}}$, then ${\mathcal {Q}}$ is still tight, and so Proposition 1 says that either the facets or vertex-figures are chiral. Without loss of generality, suppose that the facets of ${\mathcal {Q}}$ are chiral. The facets of ${\mathcal {P}}$ cover the facets of ${\mathcal {Q}}$, and since the facets of ${\mathcal {P}}$ are atomic, this implies that ${\mathcal {Q}}$ has the same facets. In particular, if ${\mathcal {P}}$ has type $\{p, q, r\}$, then ${\mathcal {Q}}$ has type $\{p, q, r'\}$ for some $r'$ dividing r. By tightness, $|\varGamma ({\mathcal {P}})| = pqr$ and $|\varGamma ({\mathcal {Q}})| = pqr'$, and so since ${\mathcal {Q}}$ is a proper quotient of ${\mathcal {P}}$, we have $r' \ne r$. Furthermore, $\varGamma ({\mathcal {Q}}) = \varGamma ({\mathcal {P}}) / \langle \sigma _3^{r'} \rangle$. But this contradicts that $\langle \sigma _3 \rangle$ is core-free in $\varGamma ({\mathcal {P}})$. $\square$

4.2 Classification of atomic chiral 4-polytopes with chiral facets and vertex-figures

In light of Lemma 9, once we know the possible types of facets and vertex-figures of an atomic chiral 4-polytope, all we need to do is try amalgamating the compatible pairs and see which ones give us a group of the proper size that satisfies the intersection condition. Theorem 4 implies that the facets and vertex-figures must appear on Table 1. Combined with Proposition  10, we find that the automorphism group of an atomic chiral 4-polytope with chiral facets and vertex-figures must be one of the groups in Table 2. For simplicity, we avoid including the various parameters (such as m, $\alpha$, and $k_1$) in the names of the groups. The “extra relations” show how to define the group as a quotient of the given parent group.

Table 2 The candidate groups of atomic chiral 4-polytopes with chiral facets and vertex-figures

Using GAP [17], we verified that $\varGamma _2, \varGamma _3,$ and $\varGamma _4$ have the correct order and satisfy the intersection condition for $\beta = 5$ and $\beta = 6$, and for all four choices of $(\epsilon _1, \epsilon _2)$. Thus, for these parameter values, the group is the automorphism group of a tight chiral polytope. We similarly verified that $\varGamma _1$ is the automorphism group of a tight chiral polytope for $m = 3, \alpha \in \{2,3\}, k_1 = k_2 \in \{1,2\}$ and for $m = 5, \alpha = 2, k_1 = k_2 \in \{1, \ldots , 4\}$. Furthermore, for these values of m and $\alpha$, we verified that $\varGamma _1$ does not have the proper order when $k_1 \ne k_2$, and so does not define the automorphism group of a tight chiral polytope.

For the group $\varGamma _1$, we will show that we do in fact need $k_1 = k_2$. Then, for each group we will describe a permutation representation of the group. There is a standard strategy that we used to determine the permutation representation, based on the following facts. If ${\mathcal {P}}$ is a tight chiral 4-polytope of type $\{p, q, r\}$, then the cosets of $\langle \sigma _1 \rangle$ are of the form $\langle \sigma _1 \rangle \sigma _2^b \sigma _3^c$, and $\varGamma ({\mathcal {P}})$ acts on the set of cosets by right multiplication. Furthermore, since $\varGamma ({\mathcal {P}})$ is tight, then for every i we can rewrite $\langle \sigma _1 \rangle \sigma _2^b \sigma _3^c \sigma _i$ as $\langle \sigma _1 \rangle \sigma _2^{b'} \sigma _3^{c'}$ for some $b'$ and $c'$. So for each i, we determined how $b'$ and $c'$ depend on b and c. We then encode the coset $\langle \sigma _1 \rangle \sigma _2^b \sigma _3^c$ as the pair $(b,c) \in {\mathbb {Z}}_q \times {\mathbb {Z}}_r$ and write down a description of the multiplication.

Once we have a permutation representation, the following lemma will show that we indeed have found the group of a tight chiral polytope.

Lemma 19

Suppose that ${\mathcal {P}}$ is a tight orientable rotary polyhedron and that ${\mathcal {Q}}$ is a tight chiral polyhedron. Let $\varGamma = \langle \sigma _1, \sigma _2, \sigma _3 \rangle = [p,q,r]^+/N_3$, the amalgamation of $\varGamma ^+({\mathcal {P}})$ with $\varGamma ^+({\mathcal {Q}})$ as defined in Lemma 9. Suppose that there is a permutation group $G = \langle \pi _1, \pi _2, \pi _3 \rangle$ on ${\mathbb {Z}}_q \times {\mathbb {Z}}_r$ such that the function that sends each $\sigma _i$ to $\pi _i$ determines a group epimorphism. Further, suppose that:

• (a) $\pi _1$ fixes (0, 0).

• (b) There is some point (b, c) such that the smallest power of $\pi _1$ that fixes (b, c) is $\pi _1^p$.

• (c) $(b, 0) \pi _2 = (b+1, 0)$ for all b.

• (d) $(b,c) \pi _3 = (b, c+1)$ for all b and c.

Then $\varGamma \cong G$, and $\varGamma$ is the rotation group of a tight chiral polytope of type $\{p, q, r\}$.

Proof

First, note that since G is a quotient of $\varGamma$, then $\pi _1^p = \pi _2^q = \pi _3^r = id$. The given conditions then imply that no smaller powers of any $\pi _i$ will equal the identity. Now, since $\varGamma$ is a tight quotient of $[p,q,r]^+$, it follows that $|G| \le |\varGamma | \le pqr$. If we can show that G satisfies the intersection condition, then Proposition 2 will imply that $|G| = pqr$ and thus that $G \cong \varGamma$ and that $\varGamma$ is the rotation group of a tight orientable rotary polytope of type $\{p, q, r\}$. Furthermore, by Lemma 9, such a polytope will have chiral vertex-figures isomorphic to ${\mathcal {Q}}$ and thus it will be chiral itself.

To show that G satisfies the intersection condition, we first need to show that

$$\begin{aligned} \langle \pi _1 \rangle \cap \langle \pi _2 \rangle = \{ id\} = \langle \pi _2 \rangle \cap \langle \pi _3 \rangle . \end{aligned}$$

If $\varphi = \pi _1^a = \pi _2^b$, then

$$\begin{aligned} (0, 0) = (0, 0) \pi _1^a = (0, 0) \pi _2^b = (b, 0), \end{aligned}$$

and so $b \equiv 0$ (mod q), which implies that $\varphi$ is trivial. Similarly, if $\varphi = \pi _2^b = \pi _3^c$, then

$$\begin{aligned} (b, 0) = (0, 0) \pi _2^b = (0, 0) \pi _3^c = (0, c), \end{aligned}$$

which implies that $\varphi$ is trivial. Finally, we need to show that

$$\begin{aligned} \langle \pi _1, \pi _2 \rangle \cap \langle \pi _2, \pi _3 \rangle . \end{aligned}$$

Consider $\varphi$ in this intersection. Since G is a quotient of the tight group $\varGamma$, we may write $\varphi = \pi _1^a \pi _2^b = \pi _2^{b'} \pi _3^c$ for some $a, b, b', c$. We have

$$\begin{aligned} (0, 0) \pi _1^a \pi _2^b = (0, 0) \pi _2^b = (b, 0) \end{aligned}$$

and

$$\begin{aligned} (0, 0) \pi _2^{b'} \pi _3^c = (b', 0) \pi _3^c = (b', c). \end{aligned}$$

It follows that $c \equiv 0$ (mod r) and thus that $\pi _3^c = id$. So $\varphi = \pi _2^{b'} \in \langle \pi _2 \rangle$, as desired. $\square$

Theorem 5

The group $\varGamma _1$ is the automorphism group of an atomic chiral 4-polytope of type $\{2m, m^{\alpha }, 2m\}$ if and only if $k_1 = k_2$.

Proof

First let us show that $k_1 = k_2$. Note that

$$\begin{aligned} \sigma _1 \sigma _2^{k_1 m^{\alpha -1}}&= \underline{\sigma _1 \sigma _2^{-1}} \sigma _2^{1+k_1 m^{\alpha -1}} \\&= \sigma _2^{1-k_1 m^{\alpha -1}} \underline{\sigma _1^3 \sigma _2^{1+k_1 m^{\alpha -1}}} \\&= \sigma _2^{1-k_1 m^{\alpha -1}} \sigma _2^{-1} \sigma _1 \\&= \sigma _2^{-k_1 m^{\alpha -1}} \sigma _1 \\ \end{aligned}$$

Thus, conjugation by $\sigma _1$ inverts $\sigma _2^{k_1 m^{\alpha -1}}$, and since $1 \le k_1 \le m-1$ and m is prime, this implies that conjugation by $\sigma _1$ inverts $\sigma _2^{m^{\alpha -1}}$. A similar argument shows that conjugation by $\sigma _3$ inverts $\sigma _2^{m^{\alpha -1}}$. Then, using Lemma 4(a) we see that

$$\begin{aligned} \sigma _3 \underline{\sigma _2^{-1} \sigma _1}&= \underline{\sigma _3 \sigma _1^3} \sigma _2^{1+k_1 m^{\alpha -1}} \\&= \sigma _2^{-1} \sigma _1^{-3} \underline{\sigma _2 \sigma _3 \sigma _2} \sigma _2^{k_1 m^{\alpha -1}} \\&= \sigma _2^{-1} \sigma _1^{-3} \underline{\sigma _3^{-1} \sigma _2^{k_1 m^{\alpha -1}}} \\&= \underline{\sigma _2^{-1} \sigma _1^{-3}} \sigma _2^{-k_1 m^{\alpha -1}} \sigma _3^{-1} \\&= \sigma _1^{-1} \sigma _2^{1-2k_1 m^{\alpha -1}} \sigma _3^{-1}. \end{aligned}$$

On the other hand,

$$\begin{aligned} \underline{\sigma _3 \sigma _2^{-1}} \sigma _1&= \sigma _2^{1+k_2 m^{\alpha -1}} \underline{\sigma _3^3 \sigma _1} \\&= \sigma _2^{k_2 m^{\alpha -1}} \underline{\sigma _2 \sigma _1 \sigma _2} \sigma _3^{-3} \sigma _2^{-1} \\&= \underline{\sigma _2^{k_2 m^{\alpha -1}} \sigma _1^{-1}} \sigma _3^{-3} \sigma _2^{-1} \\&= \sigma _1^{-1} \sigma _2^{-k_2 m^{\alpha -1}} \underline{\sigma _3^{-3} \sigma _2^{-1}} \\&= \sigma _1^{-1} \sigma _2^{1 - 2k_2 m^{\alpha -1}} \sigma _3^{-1}. \\ \end{aligned}$$

Thus $\sigma _2^{1-2k_1 m^{\alpha -1}} = \sigma _2^{1-2k_2 m^{\alpha -1}}$, and since $k_1$ and $k_2$ are defined modulo m (which is an odd prime), it follows that $k_1 = k_2$.

Now, fix $k_1 = k_2 = k$. Let $D = k m^{\alpha -1}$. For $b \in {\mathbb {Z}}_{m^{\alpha }}$, we define ${\overline{b}} = \displaystyle -b + \frac{b(b-1)}{2} D$. Then we define permutations of ${\mathbb {Z}}_{m^{\alpha }} \times {\mathbb {Z}}_{2m}$ as follows:

$$\begin{aligned} (b, c) \pi _1&= {\left\{ \begin{array}{ll} ({\overline{b}} + \frac{c}{2} D, -c), &{} \text { if }c\text { is even}, \\ ({\overline{b}} + 2 - \frac{c-1}{2} D, 2-c), &{} \text { if }c\text { is odd,} \end{array}\right. } \\ (b, c) \pi _2&= {\left\{ \begin{array}{ll} (b + 1 + \frac{c}{2} D, c), &{} \text { if }c\text { is even}, \\ (b - 1 - \frac{c-1}{2} D, c-2), &{} \text { if }c\text { is odd,} \end{array}\right. } \\ (b, c) \pi _3&= (b, c+1). \end{aligned}$$

We want to show that $\langle \pi _1, \pi _2, \pi _3 \rangle$ satisfies the defining relations of $\varGamma _1$. Here are several intermediate calculations; the first three formulas help verify the fourth and fifth.

• (a) $\overline{b} D = -bD$

• (b) $\overline{b + tD} = \overline{b} - tD$

• (c) $\overline{\overline{b}} = b(1+D)$

• (d) $(b, c) \pi _1 \pi _2 = (\overline{b}+1, -c)$

• (e) $(b, c) \pi _1^2 = {\left\{ \begin{array}{ll} (b(1+D) - cD, c) &{} \text { if }c\text { is even}, \\ (b(1-D) + cD, c) &{} \text { if }c\text { is odd}. \end{array}\right. }$

• (f) $(b, c) \pi _2^m = {\left\{ \begin{array}{ll} (b+m, c), &{} \text { if }c\text { is even}, \\ (b-m, c), &{} \text { if }c\text { is odd}. \end{array}\right. }$

From the above, it is straightforward to show that

$$\begin{aligned} (b, c) \pi _1^{2t} = {\left\{ \begin{array}{ll} (b(1+tD) - tcD, c) &{} \text { if }c\text { is even}, \\ (b(1-tD) + tcD, c) &{} \text { if }c\text { is odd}. \end{array}\right. } \end{aligned}$$

Then $(b, c) \pi _1^{2m} = (b,c)$ since $mD \equiv 0$ (mod $m^{\alpha }$). We note that the action of $\pi _1$ on the second coordinate makes it clear that $\pi _1$ has even order, and for $1 \le t \le m-1$ we have $(1, 0) \pi _1^{2t} = (1+tD, 0) \ne (1, 0)$. So $\pi _1$ has order 2m (and not a proper divisor).

From the sixth calculation above, it is clear that $\pi _2^{m^{\alpha }} = id$. It’s also clear that $\pi _3$ has order 2m.

Next, we want to show that $(\pi _1 \pi _2)^2 = (\pi _1 \pi _2 \pi _3)^2 = id$. Since $(b, c) \pi _1 \pi _2 = (\overline{b} + 1, -c)$, we have:

$$\begin{aligned} (b, c) (\pi _1 \pi _2)^2&= (\overline{b} + 1, -c) \pi _1 \pi _2 \\&= (\overline{\overline{b} + 1} + 1, c), \\ \end{aligned}$$

and

$$\begin{aligned} \overline{\overline{b} + 1} + 1&= -(\overline{b}+1) + \frac{(\overline{b}+1)\overline{b}}{2}D + 1 \\&= -\overline{b} + \frac{(-b+1)(-b)}{2}D \\&= b - \frac{b(b-1)}{2}D + \frac{(-b+1)(-b)}{2}D \\&= b. \end{aligned}$$

So $(b, c) (\pi _1 \pi _2)^2 = (b, c)$. Essentially the same proof shows that $(b, c) (\pi _1 \pi _2 \pi _3)^2 = (b, c)$. Verifying that $(b, c) (\pi _2 \pi _3)^2 = (b,c)$ is straightforward.

Finally, verifying that $\pi _2 \pi _1^2 = \pi _1^2 \pi _2^{1+D}$ and $\pi _3^2 \pi _2 = \pi _2^{1+D} \pi _3^2$ is relatively straightforward with the hints above. Lemma 19 and Corollary 5 then finish the proof. $\square$

Theorem 6

The group $\varGamma _2$ is the automorphism group of an atomic chiral 4-polytope of type $\{8, 2^{\beta }, 8\}$ for all four choices of $(\epsilon _1, \epsilon _2)$ and for every $\beta \ge 5$.

Proof

Let $D = 2^{\beta -3}$. For $b \in {\mathbb {Z}}_{2^{\beta }}$, we define ${\overline{b}} = -b + b(b-1)D\epsilon _1$. Then we define permutations of ${\mathbb {Z}}_{2^{\beta }} \times {\mathbb {Z}}_8$ as follows:

$$\begin{aligned} (b, c) \pi _1&= {\left\{ \begin{array}{ll} ({\overline{b}} + D \epsilon _2 c, -c), &{} \text { if }c\text { is even}, \\ ({\overline{b}} + 2 - D \epsilon _2 (c-1), 2-c), &{} \text { if }c\text { is odd,} \end{array}\right. } \\ (b, c) \pi _2&= {\left\{ \begin{array}{ll} (b + 1 + D \epsilon _2 c, c), &{} \text { if }c\text { is even}, \\ (b - 1 - D \epsilon _2 (c-1), c-2), &{} \text { if }c\text { is odd,} \end{array}\right. } \\ (b, c) \pi _3&= (b, c+1). \end{aligned}$$

The following intermediate calculations can be used to verify that there is a well-defined epimorphism from $\varGamma _2$ to $\langle \pi _1, \pi _2, \pi _3 \rangle$ sending each $\sigma _i$ to $\pi _i$.

• (a) $4D \equiv 2^{\beta -1}$ and $8D \equiv 0$ (mod $2^{\beta }$)

• (b) If $\beta = 5$, then $D^2 \equiv 2^{\beta -1}$ (mod $2^{\beta }$), and if $\beta \ge 6$ then $D^2 \equiv 0$ (mod $2^{\beta }$).

• (c) $\overline{b} D = -bD$

• (d) $\overline{b+2tD} = \overline{b} - 2tD$ for all t

• (e) $\overline{\overline{b}} = b(1+2D \epsilon _1)$

• (f) $(b, c) \pi _1 \pi _2 = (\overline{b}+1, -c)$

• (g) $(b, c) \pi _1^2 = {\left\{ \begin{array}{ll} (b(1+2D \epsilon _1) -2D \epsilon _2 c, c) &{} \text { if }c\text { is even}, \\ (b(1-2D \epsilon _1) + 2D \epsilon _1 + 2D \epsilon _2(c-1), c) &{} \text { if }c\text { is odd}. \end{array}\right. }$

• (h) $(b, c) \pi _2^8 = {\left\{ \begin{array}{ll} (b+8, c), &{} \text { if }c\text { is even}, \\ (b-8, c), &{} \text { if }c\text { is odd}. \end{array}\right. }$

We omit the details of showing that $\langle \pi _1, \pi _2, \pi _3 \rangle$ satisfies the defining relations of $\varGamma _2$. Lemma 19 and Corollary 5 then finish the proof. $\square$

Theorem 7

The group $\varGamma _3$ is the automorphism group of an atomic chiral 4-polytope of type $\{2^{\beta -1}, 2^{\beta }, 2^{\beta -1}\}$ for all four choices of $(\epsilon _1, \epsilon _2)$ and for every $\beta \ge 5$.

Proof

Let $D = 2^{\beta -3}$. For $b \in {\mathbb {Z}}_{2^{\beta }}$, we define

$$\begin{aligned} {\overline{b}} = {\left\{ \begin{array}{ll} b(1+ D \epsilon _1) &{} \text { if }b\text { is even}, \\ (b-1)(1-D\epsilon _1) - 1 &{} \text { if }b\text { is odd.} \end{array}\right. } \end{aligned}$$

Then we define permutations of ${\mathbb {Z}}_{2^{\beta }} \times {\mathbb {Z}}_{2^{\beta -1}}$ as follows:

$$\begin{aligned} (b, c) \pi _1&= {\left\{ \begin{array}{ll} ({\overline{b}} + 2c + D \epsilon _2 c, c(D+1)) &{} \text { if }c\text { is even,} \\ ({\overline{b}} + 2c - D \epsilon _2 (c-1), c(D+1)-D) &{} \text { if }c\text { is odd,} \\ \end{array}\right. } \\ (b, c) \pi _2&= {\left\{ \begin{array}{ll} (b + 1 - 2c + D \epsilon _2 c, c(D-1)) &{} \text { if }c\text { is even,} \\ (b + 1 - 2c - D \epsilon _2 (c-1), c(D-1)-D) &{} \text { if }c\text { is odd,} \\ \end{array}\right. } \\ (b, c) \pi _3&= (b, c+1). \end{aligned}$$

The following intermediate calculations can be used to verify that there is a well-defined epimorphism from $\varGamma _3$ to $\langle \pi _1, \pi _2, \pi _3 \rangle$ sending each $\sigma _i$ to $\pi _i$.

• (a) $4D \equiv 2^{\beta -1}$ and $8D \equiv 0$ (mod $2^{\beta }$)

• (b) If $\beta = 5$, then $D^2 \equiv 2^{\beta -1}$ (mod $2^{\beta }$), and if $\beta \ge 6$ then $D^2 \equiv 0$ (mod $2^{\beta }$).

• (c) $(b, c) \pi _1 \pi _2 = (\overline{b}+1, -c)$

• (d) $\overline{\overline{b}+1} = b - 1$

• (e) $(b, c) \pi _2^4 = {\left\{ \begin{array}{ll} (b+4, c) &{} \text { if }c\text { is even } \\ (b+4 + 4D, c) &{} \text { if }c\text { is odd}. \end{array}\right. }$

• (f) $(b, c) \pi _2^8 = (b+8, c)$.

• (g) $(b, c) \pi _1^2 = {\left\{ \begin{array}{ll} (b(1+2D \epsilon _1) + 2c(2+D \epsilon _1), c) &{} \text { if }b\text { is even}, \\ (b(1-2D \epsilon _1) + 2c(2-D \epsilon _1) + 4D - 4, c) &{} \text { if }b\text { is odd.} \\ \end{array}\right. }$

• (h) $(b, c) \pi _1^4 = {\left\{ \begin{array}{ll} (b + c(4D+8), c) &{} \text { if }b\text { is even}, \\ (b + (c-1)(4D+8), c) &{} \text { if }b\text { is odd.} \end{array}\right. }$

• (i) $(b, c) \pi _1^{2^{\beta -2}} = (b + 4D(b+c), c) = {\left\{ \begin{array}{ll} (b, c) &{} \text { if }b\text { and }c\text { have the same parity}, \\ (b + 4D, c) &{} \text { if }b\text { and }c\text { have opposite parity.} \end{array}\right. }$

Here we give more details on how to verify that $\langle \pi _1, \pi _2, \pi _3 \rangle$ satisfies the extra relations from Table 2. To verify the relation $\pi _2^{-1} \pi _1 = \pi _1^{-1+2^{\beta -2}} \pi _2^{-3 + \epsilon _1 2^{\beta -2}}$, we rewrite it:

$$\begin{aligned} \pi _2^{-1} \pi _1&= \pi _1^{-1+2^{\beta -2}} \pi _2^{-3 + \epsilon _1 2^{\beta -2}}&\text {Multiply by }\pi _1^{-1}\text { on the left} \\ \pi _1^{-1} \pi _2^{-1} \pi _1&= \pi _1^{-2+2^{\beta -2}} \pi _2^{-3 + \epsilon _1 2^{\beta -2}}&\pi _1^{-1} \pi _2^{-1} = \pi _2 \pi _1 \\ \pi _2 \pi _1^2&= \pi _1^{-2+2^{\beta -2}} \pi _2^{-3 + \epsilon _1 2^{\beta -2}}&\text {Multiply by }\pi _1^2\text { on the left and } \pi _2^4\text { on the right} \\ \pi _1^2 \pi _2 \pi _1^2 \pi _2^4&= \pi _1^{2^{\beta -2}} \pi _2^{1 + \epsilon _1 2^{\beta -2}}&\end{aligned}$$

Then we can show that both sides send (b, c) to

$$\begin{aligned} {\left\{ \begin{array}{ll} (b + 4Db + 2D\epsilon _1 + 1 - 2c + D\epsilon _2 c, c(D-1)) &{} \text { if }c\text { is even,} \\ (b + 4Db - 2D\epsilon _1 + 1 - 2c - D\epsilon _2 (c-1), c(D-1) - D) &{} \text { if }c\text { is odd.} \end{array}\right. } \end{aligned}$$

After showing that that relation holds, we can use it to rewrite the second relation into a form that is easier to verify:

$$\begin{aligned} \pi _2 \pi _1^{-1}&= \pi _1^{1+2^{\beta -2}} \pi _2^{3+\epsilon _1 2^{\beta -2}}&\text {Multiply by }\pi _1^{-1}\text { on the left} \\ \pi _1^{-1} \pi _2 \pi _1^{-1}&= \pi _1^{2^{\beta -2}} \pi _2^{3+\epsilon _1 2^{\beta -2}}&\text {Rewrite using first relation} \\ \pi _2^{3-\epsilon _1 2^{\beta -2}} \pi _1^{-2^{\beta -2}}&= \pi _1^{2^{\beta -2}} \pi _2^{3+\epsilon _1 2^{\beta -2}}&\text {Multiply by }\pi _2\text { on the left and right} \\ \pi _2^{4-\epsilon _1 2^{\beta -2}} \pi _1^{-2^{\beta -2}} \pi _2&= \pi _2 \pi _1^{2^{\beta -2}} \pi _2^{4+\epsilon _1 2^{\beta -2}} \end{aligned}$$

Then we can show that both sides send (b, c) to

$$\begin{aligned} {\left\{ \begin{array}{ll} (b + 4Db - 2D\epsilon _1 + 5 - 2c + D\epsilon _2 c, c(D-1)) &{} \text { if }c\text { is even,} \\ (b + 4Db - 2D\epsilon _1 + 5 - 2c - D\epsilon _2 (c-1), c(D-1) - D) &{} \text { if }c\text { is odd.} \end{array}\right. } \end{aligned}$$

To verify the third relation, we rewrite it as $\pi _2 \pi _3^{-1} \pi _2 = \pi _2^{4+2 \epsilon _2 D} \pi _3^{1+2D}$. Then we can show that both sides send (b, c) to

$$\begin{aligned} {\left\{ \begin{array}{ll} (b+4+2 \epsilon _2 D, c+1+2D) &{} \text { if }c\text { is even,} \\ (b+4-2 \epsilon _2 D, c+1+2D) &{} \text { if }c\text { is odd.} \end{array}\right. } \end{aligned}$$

To verify the fourth relation, we multiply both sides by $\pi _2^4$ on the left and $\pi _2$ on the right to obtain $\pi _2^4 \pi _3 = \pi _2^{1+2\epsilon _2 D} \pi _3^{-1+2D} \pi _2$. Then we can show that both sides send (b, c) to

$$\begin{aligned} {\left\{ \begin{array}{ll} (b+4, c+1) &{} \text { if }c\text { is even,} \\ (b+4+4D, c+1) &{} \text { if }c\text { is odd.} \end{array}\right. } \end{aligned}$$

Lemma 19 and Corollary 5 then finish the proof. $\square$

Theorem 8

The group $\varGamma _4$ is the automorphism group of an atomic chiral 4-polytope of type $\{8, 2^{\beta }, 2^{\beta -1}\}$ for all four choices of $(\epsilon _1, \epsilon _2)$ and for every $\beta \ge 5$.

Proof

We use the same permutation representation as Theorem 7, except that we now define ${\overline{b}} = -b + b(b-1) D \epsilon _1$ as in Theorem 6. Note that since the relations of $\varGamma _4$ that involve only $\sigma _2$ and $\sigma _3$ are the same as the relations in $\varGamma _3$, and the permutation representation for those two elements is the same, the only relations that need to be verified are those that include $\sigma _1$. Here are some intermediate calculations:

• (a) $(b, c) \pi _1^2 = {\left\{ \begin{array}{ll} (b(1+2D \epsilon _1), c) &{} \text { if }c\text { is even}, \\ (b(1-2D \epsilon _1) + 2D\epsilon _1, c) &{} \text { if }c\text { is odd.} \\ \end{array}\right. }$

• (b) $(b, c) \pi _1^4 = {\left\{ \begin{array}{ll} (b(1+4D \epsilon _1), c) &{} \text { if }c\text { is even}, \\ (b(1-4D \epsilon _1) - 4D, c) &{} \text { if }c\text { is odd.} \\ \end{array}\right. }$

• (c) $(b, c) \pi _1 \pi _2 = (\overline{b}+1, -c)$. (Note that since this calculation and the definition of $\overline{b}$ is the same as in Theorem 6, it follows at once that $\pi _1 \pi _2$ and $\pi _1 \pi _2 \pi _3$ have order 2.)

• (d) $(b, c) \pi _2^8 = (b+8, c)$. (This follows from the same calculation in Theorem 7.)

Lemma 19 and Corollary 5 then finish the proof. $\square$

Table 4 includes information on all of the atomic chiral 4-polytopes with chiral facets and vertex-figures.

5 Atomic chiral 4-polytopes with regular facets and chiral vertex-figures

Now we switch our attention to atomic chiral 4-polytopes with regular facets and chiral vertex-figures. The goal is to show that the vertex-figures are atomic chiral polyhedra and then use the classifications in Sect. 3 to find all atomic chiral 4-polytopes with regular facets.

5.1 The structure of atomic chiral 4-polytopes with regular facets and chiral vertex-figures

As in the previous section, we start by studying normal subgroups of the rotation group of atomic chiral 4-polytopes, in this case with regular facets.

Lemma 20

Let ${\mathcal {P}}$ be an atomic chiral 4-polytope with regular facets, chiral vertex-figures and type $\{p,q,r\}$. If $\varGamma ({\mathcal {P}})=\langle \sigma _1,\sigma _2, \sigma _3 \rangle$ then

• (a) $\langle \sigma _1 \rangle$ is core-free,

• (b) q is even,

• (c) $p<q$.

Proof

If $\langle \sigma _1^k \rangle \triangleleft \varGamma ({\mathcal {P}})$, then by Proposition 5, ${\mathcal {P}}/\langle \sigma _1^k \rangle$ is a tight polytope with vertex-figures isomorphic to those of ${\mathcal {P}}$. The chirality of the vertex-figures of ${\mathcal {P}}$ contradicts atomicity, proving part (a).

To prove part (b), assume to the contrary that q is odd. Since $\langle \sigma _1 \rangle$ is core-free, the type of the facets of ${\mathcal {P}}$ must be the dual of one of the types listed in Theorem 2. The only possibility for q being odd is if the facets of ${\mathcal {P}}$ have type $\{2,q\}$. This contradicts Lemma 1.

Part (c) follows from part (1) and Proposition 6. $\square$

Lemma 21

Let ${\mathcal {P}}$ be an atomic chiral 4-polytope with regular facets and chiral vertex-figures. If ${\mathcal {P}}$ has type $\{p,q,r\}$, then the vertex-figures of ${\mathcal {P}}$ do not cover a chiral polyhedron with type $\{q,r'\}$ for $r'<r$.

Proof

Assume to the contrary that the vertex-figures of ${\mathcal {P}}$ cover a chiral polyhedron ${\mathcal {Q}}$ with type $\{q,r'\}$ with $r'<r$. Then $\langle \sigma _3^{r'} \rangle \triangleleft \langle \sigma _2,\sigma _3 \rangle$, and $\varGamma ({\mathcal {Q}})=\langle \sigma _2,\sigma _3 \rangle / \langle \sigma _3^{r'}\rangle$. In particular, $\langle \sigma _3^{r'} \rangle$ is normalized by conjugation by $\sigma _2$. The dual version of Lemma 4 implies that it is also normalized by conjugation by $\sigma _1$ and hence $\langle \sigma _3^{r'} \rangle \triangleleft \varGamma ({\mathcal {P}})$.

By Proposition 5, ${\mathcal {P}}/\langle \sigma _3^{r'} \rangle$ is a 4-polytope. Furthermore, its vertex-figures are isomorphic to ${\mathcal {Q}}$, which is chiral. This contradicts the atomicity of ${\mathcal {P}}$. $\square$

Lemma 22

Let ${\mathcal {P}}$ be an atomic chiral 4-polytope with type $\{p,q,r\}$, regular facets and chiral vertex-figures. Then the vertex-figures of ${\mathcal {P}}$ do not cover a chiral polyhedron with type $\{q',r\}$ with either $q'$ an even divisor of q or $q' < q/2$.

Proof

Let $\varGamma ({\mathcal {P}}) = \langle \sigma _1, \sigma _2, \sigma _3 \rangle$.

Assume first that the vertex-figures of ${\mathcal {P}}$ cover a chiral polyhedron ${\mathcal {Q}}$ with type $\{q',r\}$ with $q'$ even. Then $\langle \sigma _2^{q'} \rangle \triangleleft \langle \sigma _2,\sigma _3 \rangle$. By the dual version of Lemma 15, $\langle \sigma _2^2 \rangle \triangleleft \langle \sigma _1,\sigma _2 \rangle$. Since $\langle \sigma _2^{q'} \rangle \le \langle \sigma _2^2 \rangle$ and the latter is cyclic, we have that $\langle \sigma _2^{q'} \rangle \triangleleft \varGamma ({\mathcal {P}})$. Then Proposition 5 shows that ${\mathcal {P}}/\langle \sigma _2^{q'} \rangle$ is a 4-polytope whose vertex-figures are isomorphic to ${\mathcal {Q}}$, contradicting atomicity of ${\mathcal {P}}$.

Now, if $q'$ is odd and $q' < q/2$ then $\langle \sigma _2^{2q'} \rangle$ is invariant under conjugation by all generators $\sigma _i$ and hence it is a proper normal subgroup of $\varGamma ({\mathcal {P}})$. It follows that ${\mathcal {P}}/\langle \sigma _2^{2q'} \rangle$ is a proper quotient of ${\mathcal {P}}$ whose vertex-figures cover ${\mathcal {Q}}$. Proposition 4 implies that ${\mathcal {P}}/\langle \sigma _2^{2q'} \rangle$ is a chiral quotient of ${\mathcal {P}}$, again contradicting atomicity of ${\mathcal {P}}$. $\square$

We are now ready to prove the main necessary condition for a tight chiral 4-polytope with regular facets and chiral vertex-figures to be atomic.

Theorem 9

If ${\mathcal {P}}$ is an atomic chiral 4-polytope with regular facets and chiral vertex-figures then the vertex-figures are atomic chiral polyhedra.

Proof

Let $\varGamma ({\mathcal {P}})=\langle \sigma _1,\sigma _2,\sigma _3 \rangle$ and assume that the facets and vertex-figures of ${\mathcal {P}}$ are isomorphic to ${\mathcal {Q}}_1$ and ${\mathcal {Q}}_2$, respectively. We shall abuse notation and write $\varGamma ^+({\mathcal {Q}}_1) = \langle \sigma _1,\sigma _2 \rangle$ and $\varGamma ({\mathcal {Q}}_2) = \langle \sigma _2,\sigma _3 \rangle$.

Assume to the contrary that ${\mathcal {Q}}_2$ is not atomic. Lemmas 21 and 22 imply that ${\mathcal {Q}}_2$ covers no chiral polyhedron with type $\{q,r'\}$ with $r'<r$ and that the only chiral polyhedron covered by ${\mathcal {Q}}_2$ with type $\{q',r\}$ for $q'<q$ is such that $q'=q/2$. Furthermore, q/2 must be odd.

First, we show that ${\mathcal {Q}}_2 / \langle \sigma _2^{q/2} \rangle$ is atomic. Note that since $\sigma _2^{q/2}$ has order 2 and $\langle \sigma _2^{q/2} \rangle \triangleleft \varGamma ({\mathcal {Q}}_2)$, $\sigma _2^{q/2}$ is central in $\varGamma ({\mathcal {Q}}_2)$. Let $\varGamma ({\mathcal {Q}}_2 / \langle \sigma _2^{q/2} \rangle ) = \langle {\hat{\sigma }}_2,{\hat{\sigma }}_3 \rangle$. By Lemma 22, ${\mathcal {Q}}_2 / \langle \sigma _2^{q/2} \rangle$ does not cover a chiral polyhedron with type $\{q'',r\}$ for $q''<q/2$. On the other hand, if ${\mathcal {Q}}_2 / \langle \sigma _2^{q/2} \rangle$ covers a chiral polyhedron with type $\{q/2,r'\}$ for $r'<r$ then ${\hat{\sigma }}_2^{-1} {\hat{\sigma }}_3^{r'} {\hat{\sigma }}_2 = {\hat{\sigma }}_3^{ar'}$ for some integer a. Lifting this relation to $\varGamma ({\mathcal {Q}}_2)$, we have that $\sigma _2^{-1} \sigma _3^{r'} \sigma _2 = \sigma _2^{\epsilon q/2}\sigma _3^{ar'}$ for some $\epsilon \in \{0,1\}$; however, by Lemma 21$\langle \sigma _3^{r'} \rangle$ is not normal in $\varGamma ({\mathcal {Q}}_2)$ and hence $\epsilon =1$. Then, conjugation by $\sigma _2$ interchanges the subgroups $\langle \sigma _3^{r'} \rangle$ and $\langle \sigma _2^{q/2} \sigma _3^{r'} \rangle$, implying that $\sigma _2^{-\ell } \sigma _3^{r'} \sigma _2^\ell \in \langle \sigma _3^{r'} \rangle$ if and only if $\ell$ is even. It follows that $\sigma _3^{r'} = \sigma _2^{q/2} \sigma _3^{r'} \sigma _2^{q/2} \notin \langle \sigma _3^{r'} \rangle$, a contradiction. Therefore ${\mathcal {Q}}_2 / \langle \sigma _2^{q/2} \rangle$ is atomic.

Since ${\mathcal {Q}}_2 / \langle \sigma _2^{q/2} \rangle$ is atomic and has type $\{q/2,r\}$ with q/2 odd, we have that $q/2 = m^\beta$ and $r=2m$ for some odd prime m and positive integer $\beta$. In particular $q=2m^\beta$ and, by Theorem 2, p must be the odd prime power $m^\beta$. Furthermore, by [12, Prop. 3.2 and Thm. 3.6], the atomic chiral polyhedron of type $\{m^{\beta }, 2m\}$ covers a tight regular polyhedron of type $\{m, 2m\}$, and so $\langle \sigma _2^m \rangle$ is normal in $\langle \sigma _2, \sigma _3 \rangle / \langle \sigma _2^{q/2} \rangle$ and indeed in $\langle \sigma _2, \sigma _3 \rangle$ itself. Then, since the dual version of Lemma 15 tells us that $\langle \sigma _2^2 \rangle$ is normal in $\langle \sigma _1, \sigma _2 \rangle$, it follows that $\langle \sigma _2^{2m} \rangle$ is normal in $\varGamma ({\mathcal {P}})$.

Abusing notation let $\varGamma ^+({\mathcal {Q}}_2/\langle \sigma _2^{m} \rangle )=\langle \sigma _2',\sigma _3 \rangle$. Since m is odd and ${\mathcal {Q}}/\langle \sigma _2^{m} \rangle$ is regular, Proposition 7 and the dual version of Lemma 15 imply that $\sigma _3^2$ is central in $\varGamma ^+({\mathcal {Q}}_2/\langle \sigma _2^{m} \rangle )$. If $\varGamma ^+({\mathcal {P}}/\langle \sigma _2^{2m} \rangle ) = \langle \sigma _1,\sigma _2'',\sigma _3 \rangle$ then

$$\begin{aligned} \sigma _2'' \sigma _3^2 (\sigma _2'')^{-1} = \sigma _3^2 (\sigma _2'')^{\varepsilon m} \end{aligned}$$

for some $\varepsilon \in \{0,1\}$. Now, $(\sigma _2'')^{m}$ generates a normal subgroup of order 2, and is thus central. Then

$$\begin{aligned} id= \sigma _2'' \sigma _3^{2m} (\sigma _2'')^{-1} = (\sigma _3^2 (\sigma _2'')^{\varepsilon m})^{m} = \sigma _3^{2m} (\sigma _2'')^{\varepsilon m^2} = (\sigma _2'')^{\varepsilon m^2}. \end{aligned}$$

Since $\sigma _2''$ has order 2m and $m^2$ is odd, it follows that $\varepsilon = 0$, so in fact, $\sigma _3^2$ commutes with $\sigma _2''$. Then, by (5) we have that

$$\begin{aligned} \sigma _1^{-1} \sigma _3^2 \sigma _1 = ((\sigma _2'')^2 \sigma _3)^2 = \sigma _2'' \sigma _3^{-2} (\sigma _2'')^{-1} = \sigma _3^{-2}, \end{aligned}$$

and so conjugation by $\sigma _1$ inverts $\sigma _3^2$. Since $p=m^\beta$ is odd, this implies that $\sigma _3^2 = \sigma _3^{-2}$, and so 2m (the order of $\sigma _3$) divides 4, which is impossible. $\square$

Corollary 6

A tight chiral 4-polytope ${\mathcal {P}}$ of type $\{p, q, r\}$ with regular facets and vertex-figures is atomic if and only if

• (a) The vertex-figures are atomic,

• (b) q is even, and

• (c) $\langle \sigma _1 \rangle$ is core-free in $\varGamma ({\mathcal {P}})$.

Proof

Theorem 9 and Lemma 20 prove that the conditions are necessary. To prove that they suffice, suppose that ${\mathcal {P}}$ satisfies the three conditions and suppose that ${\mathcal {P}}$ properly covers a chiral 4-polytope ${\mathcal {Q}}$. Then the facets of ${\mathcal {Q}}$ are covered by the regular facets of ${\mathcal {P}}$, and by Proposition 4, the facets of ${\mathcal {Q}}$ are regular. Then by Proposition 1, the vertex-figures of ${\mathcal {Q}}$ are chiral. These vertex-figures are covered by the vertex-figures of ${\mathcal {P}}$, which are atomic, and so ${\mathcal {Q}}$ has the same vertex-figures as ${\mathcal {P}}$. In particular, ${\mathcal {Q}}$ has type $\{p', q, r\}$ for some $p'$ dividing p. By tightness, $|\varGamma ({\mathcal {Q}})| = p'qr$ and $|\varGamma ({\mathcal {P}})| = pqr$, and since ${\mathcal {Q}}$ is a proper quotient of ${\mathcal {P}}$, we have $p' < p$. Now, the kernel of the natural epimorphism from $\varGamma ({\mathcal {P}})$ to $\varGamma ({\mathcal {Q}})$ includes $\sigma _1^{p'}$. On the other hand, $|\langle \sigma _1^{p'} \rangle | = p/p'$ so that $|\varGamma ({\mathcal {P}})| = |\varGamma ({\mathcal {Q}})| \cdot |\langle \sigma _1^{p'} \rangle |$. It follows that $\sigma _1^{p'}$ generates a nontrivial normal subgroup of $\varGamma ({\mathcal {P}})$, contradicting that $\langle \sigma _1 \rangle$ is core-free. So ${\mathcal {P}}$ must be atomic. $\square$

5.2 Classification of atomic chiral 4-polytopes with regular facets and chiral vertex-figures

If ${\mathcal {P}}$ is an atomic chiral 4-polytope with regular facets and chiral vertex-figures, then Lemma 20 implies that the facets must be the dual of one of the polyhedra in Theorem 2, and Theorem 9 implies that the vertex-figures must be one of the polyhedra in Theorem 3 or its dual. The dual of Lemma 1 implies that the facets cannot be type $\{2, q\}$. Then, after some manipulation of the relations in [13, Sect. 4] we have the following lemma:

Lemma 23

The facets of an atomic chiral 4-polytope with regular facets must be one of the following:

• (a) Type $\{m, 2m\}$ for an odd prime m, with rotation group $[m, 2m]^+ / (\sigma _2^2 \sigma _1 = \sigma _1 \sigma _2^2)$.

• (b) Type $\{4, 8\}$, with rotation group $[4, 8]^+ / (\sigma _2^2 \sigma _1 = \sigma _1 \sigma _2^2)$.

• (c) Type $\{4, 2^{\beta }\}$ for some $\beta \ge 5$, with rotation group $[4, 2^{\beta }]^+ / (\sigma _2^{-1} \sigma _1 = \sigma _1^{-1} \sigma _2^{1+2^{\beta -1}})$.

• (d) Type $\{2^{\beta -1}, 2^{\beta }\}$ for some $\beta \ge 5$, with rotation group $[2^{\beta -1}, 2^{\beta }]^+ / (\sigma _2 \sigma _1^{-1} = \sigma _1 \sigma _2^{3-\epsilon 2^{\beta -1}})$, with $\epsilon \in \{0, 1\}$.

Table 3 The candidate groups of atomic chiral 4-polytopes with regular facets and chiral vertex-figures

Now there are eight possibilities for the automorphism group of an atomic chiral 4-polytope with regular facets and chiral vertex-figures; see Table 3.

We will show that the first three groups do correspond to atomic chiral 4-polytopes, whereas the remaining groups do not.

Theorem 10

The group $\varLambda _1$ is the automorphism group of an atomic chiral polytope of type $\{m, 2m, m^{\alpha }\}$, for each k satisfying $1 \le k \le m-1$.

Proof

Let $D = k m^{\alpha -1}$. Then we define permutations of ${\mathbb {Z}}_{2m} \times {\mathbb {Z}}_{m^{\alpha }}$ as follows:

$$\begin{aligned} (b, c) \pi _1&= {\left\{ \begin{array}{ll} (b + 2c, c + \frac{c(c-1)}{2} D) &{} \text { if }b\text { is even}, \\ (b + 2c - 2, c + \frac{c(c-1)}{2} D) &{} \text { if }b\text { is odd}, \\ \end{array}\right. } \\ (b, c) \pi _2&= (b + 1 - 2c, -c + \frac{c(c-1)}{2} D), \\ (b, c) \pi _3&= (b, c+1). \end{aligned}$$

Here are a few intermediate calculations.

• (a) For all n, $(b,c) \pi _1^n = {\left\{ \begin{array}{ll} (b + 2nc, c + n\frac{c(c-1)}{2} D) &{} \text { if }b\text { is even}, \\ (b + 2nc - 2n, c + n\frac{c(c-1)}{2} D) &{} \text { if }b\text { is odd}. \\ \end{array}\right. }$

• (b) $(b,c) \pi _2^2 = (b+2, c(1+D))$

• (c) $(b,c) \pi _1 \pi _2 = {\left\{ \begin{array}{ll} (b+1, -c) &{} \text { if }b\text { is even }, \\ (b-1, -c) &{} \text { if }b\text { is odd. } \end{array}\right. }$.

From these, it is routine to show that there is a well-defined epimorphism from $\varLambda _1$ to $\langle \pi _1, \pi _2, \pi _3 \rangle$ sending each $\sigma _i$ to $\pi _i$. Lemma 19 and Corollary 6 then finish the proof. $\square$

Theorem 11

The group $\varLambda _2$ is the automorphism group of an atomic chiral polytope of type $\{4, 8, 2^{\beta }\}$.

Proof

Let $D = 2^{\beta -3}$. For $b \in {\mathbb {Z}}_{8}$, we define

$$\begin{aligned} {\overline{b}} = {\left\{ \begin{array}{ll} b &{} \text { if }b\text { is even}, \\ b-2 &{} \text { if }b\text { is odd.} \end{array}\right. } \end{aligned}$$

Then we define permutations of ${\mathbb {Z}}_{8} \times {\mathbb {Z}}_{2^{\beta }}$ as follows:

$$\begin{aligned} (b, c) \pi _1&= {\left\{ \begin{array}{ll} ({\overline{b}} - 2c, c(1+D\epsilon )) &{} \text { if }c\text { is even,} \\ ({\overline{b}} - 2c + 4, c(1 - D \epsilon ) + D\epsilon ) &{} \text { if }c\text { is odd,} \\ \end{array}\right. } \\ (b, c) \pi _2&= {\left\{ \begin{array}{ll} (b + 1 - 2c, c(-1 + D \epsilon )) &{} \text { if }c\text { is even,} \\ (b + 1 - 2c, c(-1 - D \epsilon ) + D\epsilon ) &{} \text { if }c\text { is odd,} \\ \end{array}\right. } \\ (b, c) \pi _3&= (b, c+1). \end{aligned}$$

We note that

$$\begin{aligned} (b,c) \pi _2^2 = {\left\{ \begin{array}{ll} (b+2, c(1-2D\epsilon ) &{} \text { if }c\text { is even}, \\ (b+2, c(1+2D\epsilon ) &{} \text { if }c\text { is odd}. \end{array}\right. } \end{aligned}$$

Then it is routine to show that there is a well-defined epimorphism from $\varLambda _2$ to $\langle \pi _1, \pi _2, \pi _3 \rangle$ sending each $\sigma _i$ to $\pi _i$. Lemma 19 and Corollary 6 then finish the proof. $\square$

Theorem 12

The group $\varLambda _3$ is the automorphism group of an atomic chiral polytope of type $\{4, 2^{\beta -1}, 2^{\beta }\}$.

Proof

Let $D = 2^{\beta -3}$. For $b \in {\mathbb {Z}}_{2^{\beta -1}}$, we define

$$\begin{aligned} {\overline{b}} = {\left\{ \begin{array}{ll} b(-1-D) &{} \text { if }b\text { is even}, \\ b(-1-D)+D &{} \text { if }b\text { is odd.} \end{array}\right. } \end{aligned}$$

Then we define permutations of ${\mathbb {Z}}_{2^{\beta -1}} \times {\mathbb {Z}}_{2^{\beta }}$ as follows:

$$\begin{aligned} (b, c) \pi _1&= {\left\{ \begin{array}{ll} ({\overline{b}} - cD, c(-1+D\epsilon )) &{} \text { if }c\text { is even,} \\ ({\overline{b}} - (c-1)D + 2, (1-c)(1+D\epsilon ) + 1) &{} \text { if }c\text { is odd,} \\ \end{array}\right. } \\ (b, c) \pi _2&= {\left\{ \begin{array}{ll} (b + 1 + cD, c(1+D\epsilon )) &{} \text { if }c\text { is even,} \\ (b - 1 + (c-1)D, (c-1)(1-D\epsilon )-1 &{} \text { if }c\text { is odd,} \\ \end{array}\right. } \\ (b, c) \pi _3&= (b, c+1). \end{aligned}$$

Here are some intermediate calculations:

• (a) If a is even, then $\overline{a+b} = \overline{a} + \overline{b}$.

• (b) $\overline{{\overline{b}}} \equiv b(1+2D)$ (mod $2^{\beta -1}$).

• (c) $(b, c) \pi _1^2 = {\left\{ \begin{array}{ll} (b(1+2D), c(1-2D\epsilon )) &{} \text { if }c\text { is even,} \\ (b(1+2D)-2D, (c-1)(1+2D\epsilon )+1) &{} \text { if }c\text { is odd,} \\ \end{array}\right. }$

• (d) $(b, c) \pi _2^8 = {\left\{ \begin{array}{ll} (b+8, c), &{} \text { if }c\text { is even}, \\ (b-8, c-16), &{} \text { if }c\text { is odd.} \end{array}\right. }$

• (e) $(b,c) \pi _1 \pi _2 = (\overline{b}+1, -c)$

Let us rewrite the first extra relation of $\varLambda _3$ as $\sigma _2 \sigma _1^2 = \sigma _1^2 \sigma _2^{1+2D}$ (see the proof of Theorem 3, noting that $\sigma _1^{-1} = \sigma _1^3$). Similarly, we rewrite the second extra relation by multiplying both sides on the left by $\sigma _2$, and the third relation by multiplying both sides on the right by $\sigma _2$. Then one can check that there is a well-defined epimorphism from $\varLambda _3$ to $\langle \pi _1, \pi _2, \pi _3 \rangle$ sending each $\sigma _i$ to $\pi _i$. Lemma 19 and Corollary 6 then finish the proof. $\square$

In order to rule out the remaining cases, we will use the following lemma.

Lemma 24

Suppose that $\varLambda$ is a quotient of $[p, q, r]^+$ satisfying

$$\begin{aligned} \sigma _2^2 \sigma _1&= \sigma _1 \sigma _2^{2t}, \\ \sigma _3^{-1} \sigma _2&= \sigma _2^a \sigma _3^c, \\ \sigma _3 \sigma _2^{-1}&= \sigma _2^b \sigma _3^{-c}, \end{aligned}$$

and suppose that b is odd. Then

$$\begin{aligned} \sigma _2^{2t+2} \sigma _3 = \sigma _3 \sigma _2^{-t(b+1)-1+a}. \end{aligned}$$

Proof

First, we note that

$$\begin{aligned} \sigma _1^{-1} \sigma _2^2 \underline{\sigma _3 \sigma _1} = \sigma _1^{-1} \underline{\sigma _2^2 \sigma _1} \sigma _2^2 \sigma _3 = \sigma _2^{2t+2} \sigma _3. \end{aligned}$$

On the other hand,

$$\begin{aligned} \sigma _1^{-1} \underline{\sigma _2^2 \sigma _3} \sigma _1&= \sigma _1^{-1} \underline{\sigma _2 \sigma _3^{-1}} \sigma _2^{-1} \sigma _1 \\&= \sigma _1^{-1} \sigma _3^c \underline{\sigma _2^{-b-1} \sigma _1} \\&= \underline{\sigma _1^{-1} \sigma _3^c \sigma _1} \sigma _2^{-t(b+1)} \\&= \underline{\sigma _2 \sigma _3^{-c}} \sigma _2^{-t(b+1)-1} \\&= \sigma _3 \sigma _2^{-t(b+1)-1+a}. \end{aligned}$$

$\square$

Proposition 13

The groups $\varLambda _4$ and $\varLambda _6$ satisfy

$$\begin{aligned} \sigma _3^{-1} \sigma _2 = \sigma _2^{-1 - \epsilon 2^{\beta -2}} \sigma _3^{-3} \end{aligned}$$

and

$$\begin{aligned} \sigma _3 \sigma _2^{-1} = \sigma _2^{1 - \epsilon 2^{\beta -2}} \sigma _3^3. \end{aligned}$$

Proof

First:

$$\begin{aligned} \sigma _2 \sigma _3^2&= \sigma _3^2 \sigma _2^{1 + \epsilon 2^{\beta -2}}&\text {Rewrite }\sigma _2 \sigma _3\text { as }\sigma _3^{-1} \sigma _2^{-1} \\ \sigma _3^{-1} \sigma _2^{-1} \sigma _3&= \sigma _3^2 \sigma _2^{1 + \epsilon 2^{\beta -2}}&\text {Multiply on the left by }\sigma _3 \\ \sigma _2^{-1} \sigma _3&= \sigma _3^3 \sigma _2^{1 + \epsilon 2^{\beta -2}}&\text {Invert both sides} \\ \sigma _3^{-1} \sigma _2&= \sigma _2^{-1 - \epsilon 2^{\beta -2}} \sigma _3^{-3}&\end{aligned}$$

Next, we note that since $\sigma _2 \sigma _3^2 = \sigma _3^2 \sigma _2^{1 + \epsilon 2^{\beta -2}}$, it follows that

$$\begin{aligned} \sigma _2^{1 - \epsilon 2^{\beta -2}} \sigma _3^2 = \sigma _3^2 \sigma _2^{(1 - \epsilon 2^{\beta -2})(1 + \epsilon 2^{\beta -2})} = \sigma _3^2 \sigma _2. \end{aligned}$$

Finally:

$$\begin{aligned} \sigma _3^2 \sigma _2&= \sigma _2^{1 - \epsilon 2^{\beta -2}} \sigma _3^2&\text {Rewrite }\sigma _3 \sigma _2\text { as }\sigma _2^{-1} \sigma _3^{-1} \\ \sigma _3 \sigma _2^{-1} \sigma _3^{-1}&= \sigma _2^{1 - \epsilon 2^{\beta -2}} \sigma _3^2&\text {Multiply on the right by }\sigma _3 \\ \sigma _3 \sigma _2^{-1}&= \sigma _2^{1 - \epsilon 2^{\beta -2}} \sigma _3^3.&\end{aligned}$$

$\square$

Theorem 13

The groups $\varLambda _4$ and $\varLambda _5$ are not the automorphism groups of tight chiral 4-polytopes.

Proof

In $\varLambda _4$ and $\varLambda _5$, the relation $\sigma _2^{-1} \sigma _1 = \sigma _1^{-1} \sigma _2^{1+2^{\beta -1}}$ is equivalent to $\sigma _2 \sigma _1^{-1} = \sigma _1 \sigma _2^{-1+2^{\beta -1}}$ (see [13, Proposition 3.1]), and this implies that $\sigma _2^2 \sigma _1 = \sigma _2 \sigma _1^{-1} \sigma _2^{-1} = \sigma _1 \sigma _2^{2(-1+2^{\beta -2})}$. Then Proposition 13 and Lemma 24 prove that both groups satisfy $\sigma _2^{2^{\beta -1}} \sigma _3 = \sigma _3$, and so $\sigma _2$ does not have order $2^{\beta }$ as required. $\square$

Theorem 14

The groups $\varLambda _6$ and $\varLambda _7$ are not the automorphism groups of tight chiral 4-polytopes.

Proof

If either group is the automorphism group of a tight chiral 4-polytope, then Proposition 6 implies that $\langle \sigma _1, \sigma _2 \rangle$ and $\langle \sigma _2, \sigma _3 \rangle$ both have a normal subgroup of the form $\langle \sigma _2^k \rangle$. It follows that $\langle \sigma _2^{2^{\beta -1}} \rangle$ is normal in $\langle \sigma _1, \sigma _2, \sigma _3 \rangle$, which means that $\sigma _2^{2^{\beta -1}}$ is central.

Now, the relation $\sigma _2 \sigma _1^{-1} = \sigma _1 \sigma _2^{3-\epsilon _1 2^{\beta -1}}$ implies that $\sigma _2^2 \sigma _1 = \sigma _1 \sigma _2^{2(1+\epsilon _1 2^{\beta -2})}$. Proposition 13 implies that we may use Lemma 24, which then implies that in $\varLambda _6$, $\sigma _2^{4+\epsilon _1 2^{\beta -1}} \sigma _3 = \sigma _3 \sigma _2^{-4-\epsilon _1 2^{\beta -1}}$. So conjugation by $\sigma _3$ inverts $\sigma _2^{4+\epsilon _1 2^{\beta -1}}$, and since $\sigma _2^{2^{\beta -1}}$ is central, this implies that conjugation by $\sigma _3$ inverts $\sigma _2^4$. Now, $\sigma _3^2 \sigma _2 = \sigma _3 \sigma _2^{-1} \sigma _3^{-1} = \sigma _2^{1+\epsilon _2 2^{\beta -2}} \sigma _3^2$, and it follows that $\sigma _3^4 \sigma _2 = \sigma _2^{1+\epsilon _2 2^{\beta -1}} \sigma _3^4$. Then $\sigma _3^4 \sigma _2^2 = \sigma _2^{2(1+\epsilon _2 2^{\beta -1})} \sigma _3^4 = \sigma _2^2 \sigma _3^4$, and since $\sigma _3$ has order 8, this implies that $\sigma _2^2 = \sigma _3^4 \sigma _2^2 \sigma _3^4$. So:

$$\begin{aligned} \sigma _3 \sigma _2^4&= \sigma _2^{-1} \underline{\sigma _3^{-1} \sigma _2} \sigma _2^2 \\&= \sigma _2^{-2+\epsilon _2 2^{\beta -2}} \underline{\sigma _3^{-3} \sigma _2^2} \\&= \sigma _2^{-2+\epsilon _2 2^{\beta -2}} \underline{\sigma _3 \sigma _2^2} \sigma _3^4 \\&= \sigma _2^{-2+\epsilon _2 2^{\beta -2}} \sigma _2^{-2+\epsilon _2 2^{\beta -2}} \sigma _3^{-3} \sigma _3^4 \\&= \sigma _2^{-4+\epsilon _2 2^{\beta -1}} \sigma _3. \end{aligned}$$

Since we also have that conjugation by $\sigma _3$ inverts $\sigma _2^4$, this implies that $\sigma _2^{2^{\beta -1}} = id$, and so $\sigma _2$ does not have the desired order.

In $\varLambda _7$, Lemma 24 implies that $\sigma _2^{4+\epsilon _1 2^{\beta -1}} \sigma _3 = \sigma _3 \sigma _2^{4+\epsilon _1 2^{\beta -1}}$. Since $\sigma _2^{2^{\beta -1}}$ is central, this implies that $\sigma _3$ commutes with $\sigma _2^4$. However,

$$\begin{aligned} \sigma _3 \sigma _2^{-4-\epsilon _2 2^{\beta -2}}&= \underline{\sigma _3 \sigma _2^{-1}} \sigma _2^{-3-\epsilon _2 2^{\beta -2}} \\&= \sigma _2^{-3+\epsilon _2 2^{\beta -2}} \underline{\sigma _3^{-1+2^{\beta -2}} \sigma _2^{-3-\epsilon _2 2^{\beta -2}}} \\&= \sigma _2^{-4+\epsilon _2 2^{\beta -2}} \sigma _3. \end{aligned}$$

Then $\sigma _2^{-2^{\beta -2}} = \sigma _2^{2^{\beta -2}}$, which implies that $\sigma _2^{2^{\beta -1}} = id$, and again $\sigma _2$ does not have the desired order. $\square$

Theorem 15

The group $\varLambda _8$ is not the automorphism group of tight chiral 4-polytope.

Proof

If $\varLambda _8$ were the automorphism group of a tight chiral 4-polytope, then $\langle \sigma _2 \rangle$ would be core-free in $\langle \sigma _2, \sigma _3 \rangle$ (see Proposition 6 and [12, Proposition 4.5]). We will show that in fact, $\sigma _3$ normalizes a nontrivial subgroup of $\langle \sigma _2 \rangle$.

From the relation $\sigma _2 \sigma _1^{-1} = \sigma _1 \sigma _2^{3-\epsilon _1 2^{\beta -2}}$, it follows that $\sigma _2^2 \sigma _1 = \sigma _1 \sigma _2^{2-\epsilon _1 2^{\beta -2}}$, and thus, for each k, $\sigma _2^{2k} \sigma _1 = \sigma _1 \sigma _2^{(1-\epsilon _1 2^{\beta -3})2k}$. Then

$$\begin{aligned} \sigma _1^{-1} \sigma _3 \underline{\sigma _2^2 \sigma _1}&= \underline{\sigma _1^{-1} \sigma _3 \sigma _1} \sigma _2^{2 - \epsilon _1 2^{\beta -2}} \\&= \sigma _2^2 \sigma _3 \sigma _2^{2 - \epsilon _1 2^{\beta -2}}. \end{aligned}$$

On the other hand,

$$\begin{aligned} \sigma _1^{-1} \underline{\sigma _3 \sigma _2^2} \sigma _1&= \sigma _1^{-1} \sigma _2^{-2+2^{\beta -2}} \underline{\sigma _3^{-3+\epsilon _2 2^{\beta -2}} \sigma _1} \\&= \underline{\sigma _1^{-1} \sigma _2^{-2+2^{\beta -2}} \sigma _1} \sigma _2 \sigma _3^{3-\epsilon _2 2^{\beta -2}} \sigma _2^{-1} \\&= \sigma _2^{-1+2^{\beta -2}+\epsilon _1 2^{\beta -2}} \sigma _3^{3-\epsilon _2 2^{\beta -2}} \sigma _2^{-1} \\&= \sigma _2^{-1+2^{\beta -2}+\epsilon _1 2^{\beta -2}} \underline{\sigma _3^{3- \epsilon _2 2^{\beta -2}} \sigma _2^{1-2^{\beta -2}}} \sigma _2^{-2+2^{\beta -2}} \\&= \sigma _2^{-1+2^{\beta -2}+\epsilon _1 2^{\beta -2}} \sigma _2^{-1} \sigma _3 \sigma _2^{-2+2^{\beta -2}}. \end{aligned}$$

Putting these together, we find that $\sigma _2^{-4+2^{\beta -2} + \epsilon _1 2^{\beta -2}} \sigma _3 = \sigma _3 \sigma _2^{4 - 2^{\beta -2} - \epsilon _1 2^{\beta -2}}$, and so $\sigma _3$ normalizes a nontrivial subgroup of $\langle \sigma _2 \rangle$. $\square$

Table 4 summarizes the atomic chiral 4-polytopes with chiral vertex figures. The duals of the first three rows yield atomic chiral 4-polytopes with regular vertex-figures, and the last two rows correspond to a dual pair of chiral 4-polytopes. In total, there are 11 families of atomic chiral 4-polytopes. Thus we have shown:

Theorem 16

Every tight chiral 4-polytope covers one of the polytopes in Table 4 or its dual.

Proposition 14

If ${\mathcal {P}}$ is an atomic chiral 4-polytope with regular facets, then $X({\mathcal {P}})$ is contained in $\langle \sigma _3 \rangle$.

Proof

An atomic chiral 4-polytope ${\mathcal {P}}$ with regular facets has automorphism group $\varLambda _1$, $\varLambda _2$, or $\varLambda _3$. In each case, the chirality group of the vertex-figures is a cyclic group of prime order of the form $\langle \sigma _3^c \rangle$ that is normal in $\langle \sigma _2, \sigma _3 \rangle$ (see Table 1), and thus in $\varGamma ({\mathcal {P}})$ (by the dual of Lemma  4). The quotient of $\varGamma ({\mathcal {P}})$ by this normal subgroup is a polytope, by Proposition 5, and thus, it is regular (by atomicity). Therefore, the chirality group of the vertex-figures contains the chirality group of ${\mathcal {P}}$, and since the former has prime order and the latter is nontrivial, it follows that the two coincide, proving the claim. $\square$

Table 4 The atomic chiral 4-polytopes with chiral vertex-figures

6 Tight chiral 5-polytopes

Recall that a tight chiral 5-polytope must have chiral facets and chiral vertex-figures (see Proposition 1 (c)). In this section, we prove Theorem 1, that is, that no such polytope exists.

Recall that a tight chiral 5-polytope ${\mathcal {P}}$ with $\varGamma ({\mathcal {P}})= \langle \sigma _1,\sigma _2,\sigma _3,\sigma _4 \rangle$ is atomic if it does not properly cover any tight chiral polytope. Clearly, every tight chiral 5-polytope covers an atomic chiral 5-polytope.

We start by giving properties that atomic chiral 5-polytopes must satisfy, should they exist.

Lemma 25

Let ${\mathcal {P}}$ be a tight chiral 5-polytope with type $\{p,q,r,s\}$ where $q \ge r$. Then the kernel of the action of $\varGamma ({\mathcal {P}})$ on the chains containing a 3-face and a facet is nontrivial.

Proof

The stabilizer of the chain containing the base 3-face and the base facet is $\varDelta = \langle \sigma _1, \sigma _2 \rangle$. The remaining chains can be associated to right cosets of $\varDelta$. Proposition 6 implies that there is a nontrivial subgroup $\langle \sigma _2^k \rangle$ that is normal in $\langle \sigma _2, \sigma _3, \sigma _4 \rangle$. Then it follows that for all a and b we have $(\langle \sigma _1, \sigma _2 \rangle \sigma _3^a \sigma _4^b) \sigma _2^k = \langle \sigma _1, \sigma _2 \rangle \sigma _3^a \sigma _4^b$, and so $\sigma _2^k$ fixes all chains containing a 3-face and a facet. $\square$

Lemma 26

Let ${\mathcal {P}}$ be an atomic chiral 5-polytope with $\varGamma ({\mathcal {P}}) = \langle \sigma _1, \sigma _2, \sigma _3, \sigma _4\rangle$ and type $\{p, q,r, s\}$. If $q \ge r$ then

• (a) $X({\mathcal {P}})$ is $\langle \sigma _2^{q'} \rangle$ for some $q'$ satisfying that $q/q'$ is prime,

• (b) The chirality groups of the base facet and the base vertex-figure are also $\langle \sigma _2^{q'} \rangle$.

Proof

Let H and K be the kernels of the actions of $\varGamma ({\mathcal {P}})$ on the vertices and on the chains consisting of a 3-face and a 4-face, respectively. By Corollary 2 and Lemma 25, H and K are nontrivial. Therefore $H \cap K$ is a normal subgroup of $\varGamma ({\mathcal {P}})$ that by the intersection condition is contained in $\langle \sigma _2 \rangle$. The rest of the proof is as in Proposition 11. $\square$

Now we can prove Theorem 1.

Proof of Theorem 1

It suffices to show that there are no atomic chiral 5 polytopes. Suppose to the contrary that ${\mathcal {P}}$ is an atomic chiral 5-polytope. Up to duality, we may assume that $q \ge r$. Let ${\mathcal {K}}$ be the base facet. It must be a tight chiral 4-polytope, by Proposition 1(b), and since the facets of the facets of a chiral polytope are always regular, ${\mathcal {K}}$ has regular facets. Now, Lemma 26 says that ${\mathcal {K}}$ has chirality group contained in $\langle \sigma _2 \rangle$. Let ${\mathcal {K}}'$ be an atomic chiral 4-polytope that is covered by ${\mathcal {K}}$, with $\varGamma ({\mathcal {K}}') = \langle \sigma _1', \sigma _2', \sigma _3' \rangle$. Then Lemma 10 says that $X({\mathcal {K}}')$ is contained in $\langle \sigma _2' \rangle$, which contradicts Proposition 14. $\square$

7 Concluding remarks

The study of tight chiral polytopes was originated in the search for chiral polytopes with a small number of flags. In ranks 3 and 4, the atomic chiral polytopes are now classified; this constitutes the first step for a full classification of tight chiral 3- and 4-polytopes. However, the techniques used to classify tight regular polyhedra fail in the chiral setting, and the full classification seems to require several more steps.

The nonexistence of tight chiral n-polytopes for $n \ge 5$ strengthens the general belief that for each $n \ge 5$ the chiral n-polytopes with the fewest flags have considerably more flags than the regular n-polytopes with the fewest flags. See also [11, Theorem 5.5].