Internal and External Invariance of Abstract Polytopes

A dissertation presented by

Gabe Cunningham

The Department of Mathematics

In partial fulﬁllment of the requirements for the degree of Doctor of Philosophy

in the ﬁeld of

Mathematics

Northeastern University Boston, Massachusetts May, 2012

1 Internal and External Invariance of Abstract Polytopes

Gabe Cunningham

ABSTRACT OF DISSERTATION

Submitted in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy in Mathematics in the Graduate School of Science of Northeastern University, May 2012

2 Abstract

In addition to the usual symmetries by reﬂections and rotations, abstract polytopes can have external symmetries such as self-duality and self-Petriality. In this dissertation, we describe and study a way of measuring the invariance of abstract polytopes under such external operations. We then present methods for constructing abstract polytopes with speciﬁed external symmetries. In particular, we describe how to construct polyhedra that are self-dual and self-Petrie, and how to construct polytopes that are self-dual and chiral.

3 Acknowledgements

I would like to thank the members of my committee for their time and attention.

I would like to thank Professor Schulte for all of the help he has given me over the years. He has helped proofread my papers, given me many interesting articles to read, and invited me to rewarding seminars and conferences. He has been instrumental in getting my career started.

Finally, I would like to thank my wife, Nomeda. She inspires me to work hard, not only by her words, but by her example. Furthermore, if it hadn’t been for her urging, there is a good chance I would never have even applied to graduate school and found out how rewarding it could be.

4 Contents

Abstract 2

Acknowledgements 4

Contents 5

1 Background 7

1.1 Introduction ...... 7

1.2 Abstract polytopes ...... 9

1.3 Regularity ...... 17

1.4 Direct regularity ...... 28

1.5 Chirality ...... 30

1.6 Duality ...... 34

1.7 Petrie duality ...... 36

2 The mixing operation 39

2.1 Mixing ﬁnitely generated groups ...... 39

5 2.2 Variance groups and the structure of the mix ...... 45

2.3 Mixing polytopes ...... 51

2.4 Polytopality of the mix ...... 55

3 Measuring invariance 66

3.1 External and internal invariance ...... 66

3.2 Variance of the mix ...... 71

4 Constructing self-dual and self-Petrie polytopes 80

4.1 Self-dual polytopes ...... 81

4.2 Self-dual, self-Petrie polyhedra ...... 83

4.3 Self-dual, self-Petrie covers of universal polyhedra ...... 87

5 Constructing chiral polytopes 92

5.1 Building chiral polytopes of high rank ...... 92

5.2 Self-dual chiral polytopes ...... 96

6 Appendix 103

Bibliography 105

6 Chapter 1

Background

1.1 Introduction

The study of abstract polytopes is a growing ﬁeld, uniting combinatorics with geometry and group theory. Abstract polytopes are combinatorial structures with a distinctly geometric ﬂavor, generalizing the face-lattices of convex polytopes. The result is a rich theory that encompasses not only polytopes, but also face-to-face tilings of the plane, tessellations of space-forms, and other fascinating structures.

As in the classical theory, the abstract polytopes that have been most extensively stud- ies are the regular polytopes: those whose automorphism group acts transitively on the set of their flags. These polytopes possess maximal symmetry by abstract reflections. Closely related are the chiral polytopes, which possess two flag-orbits under the action of the automorphism group. Intuitively, a chiral polytope has full rotational symmetry, but it occurs in two mirror-image forms. Compared to regular polytopes, relatively little is known about chiral polytopes. In particular, we know very few examples of chiral polytopes of high rank (dimension).

7 In addition to the study of polytopes with a high degree of internal symmetry – that is, polytopes with a large number of polytope automorphisms – the study of external symmetries of polytopes and maps on surfaces has been gaining attention; for example, see [15, 32]. The idea is to consider operations that transform one polytope into another, and to then look for (or construct) polytopes that are invariant under this transformation. The classic example is self-duality, but there are many others.

In this dissertation, we focus on external symmetries arising from certain outer automorphisms of groups. In particular, we consider outer automorphisms of the automorphism group of the universal polytope of rank n, a polytope that covers all others of the same rank. There are three principal external symmetries this gives rise to: duality, Petrie duality, and mirror symmetry. Our goal is then to measure to what extent a polytope is invariant under these symmetries.

Our main tool for constructing new polytopes is the mixing construction, which creates the minimal common cover of two polytopes [24]. The mix of a polytope with its dual is self-dual, and in general, the mix of a polytope with all of its images under a group of transformations is invariant under that group of transformations. The result, however, is not always a polytope; we must check that separately.

We start by giving background on (abstract) polytopes, including regularity, chirality, duality, and Petrie duality, in Chapter 1. In Chapter 2, we describe the mixing construction for groups and for polytopes, and we examine in detail the structure of the mix of two polytopes. In particular, we describe several circumstances that guarantee that the mix of two polytopes will again be a polytope. Next, we describe how to use mixing to measure the invariance of a polytope under external symmetries in Chapter 3. In Chapter 4, we focus on duality and Petrie duality, and we explicitly describe many self-dual, self-Petrie polyhedra. Finally, in Chapter 5, we construct chiral polytopes that are self-dual – in other

8 words, self-dual polytopes that lack mirror symmetry.

1.2 Abstract polytopes

Abstract polytopes are essentially generalizations of the face-lattice of a convex polytope, and they retain much of the geometric flavor of convex polytopes. As we build up the definition, given in [23], we will give some geometric justification for the properties we require. Some of our later constructions produce objects that are not quite polytopes, and so we start with these more general definitions.

Deﬁnition 1.1. A ﬂagged poset of rank n is a partially ordered set P such that there is a unique maximal element, a unique minimal element, and such that each maximal chain of

P has n + 2 elements. We use Fn to denote the maximal element, and F−1 to denote the minimal element.

We call the elements of a flagged poset faces, and we call the maximal chains flags. Since each flag of a flagged poset contains the same number of faces, every flagged poset has a natural rank function. In particular, we define the rank of F−1 to be −1 and the rank of Fn to be n, with the rank of any other face obtained in the natural way from its position in any

flag. A face of rank j is called a j-face. The faces F−1 and Fn are called the improper faces of P; the remaining faces are the proper faces. If F and G are faces of a flagged poset P such that F ≤ G, then we define the section G/F to be the poset consisting of those faces H such that F ≤ H ≤ G. In fact, if F is a j-face and G is a k-face, then the section G/F is a flagged poset of rank k − j − 1, where G plays the role of the maximal face, and F plays the role of the minimal face.

We think of the rank of a face as its dimension, informally speaking. Thus, in analogy with convex polytopes, we refer to 0-faces as vertices and 1-faces as edges. Faces of rank

9 n−1 are called facets. The requirement that a ﬂagged poset of rank n must contain a unique n-face and a unique (−1)-face is analogous to the fact that a convex n-polytope is viewed to have a single face of dimension n (namely, the whole polytope) and a single face of dimension −1 (namely, the empty set).

There is a natural identification between each j-face F and its section F/F−1, and we occasionally blur the distinction between these two concepts. If F is a k-face, then the co- k-face at F (or sometimes simply the co-face at F ) is the section Fn/F . In particular, if F is a vertex, then we call the co-face at F a vertex-figure. If G is a facet and F is a vertex, then the section G/F is called a medial section. Note that a medial section is a facet of the vertex-figure Fn/F and also a vertex-figure of the facet G/F−1.

The requirement that all ﬂags of a ﬂagged poset have the same length ensures that we get no “holes” and that all maximal proper faces have the same rank. Essentially, we are forcing the set of proper faces to form a polytopal complex [13] where the maximal polytopes have the same dimension (except, of course, our polytopes will be abstract instead of convex).

We note that there is one flagged poset of rank −1 (where the minimal and maximal element coincide), and one flagged poset of rank 0 (where the only two elements are the minimal and maximal elements). In rank 1, there are already infinitely many flagged posets.

Flagged posets are really too general to have much geometric ﬂavor. The notion of a pre-polytope is much closer to what we want.

Definition 1.2. A flagged poset of rank n is a pre-polytope of rank n (also called an n-pre- polytope) if it satisfies the following “diamond condition”: whenever F < G where F is a (j − 1)-face and G is a (j + 1)-face (for any j = 0, 1, . . . , n − 1), then there are exactly two j-faces H such that F < H < G.

The diamond condition is a strong restriction on the types of structures we investigate. In

10 particular, we see that each 1-face (edge) of an n-pre-polytope contains two 0-faces (vertices), and that each (n − 2)-face is contained in two (n − 1)-faces. This again parallels the case with convex polytopes. As with ﬂagged posets, we note that every section of a pre-polytope is again a pre-polytope.

There is a way to formulate the diamond condition in terms of flags instead of faces. Given two flags of a flagged poset, we say that they are adjacent if they differ in exactly one face. If that face is an i-face, we say that the flags are i-adjacent.

Proposition 1.3. Let P be a flagged poset of rank n. Then P has the diamond property (and is therefore a pre-polytope) if and only if each flag of P has a unique i-adjacent flag for each i = 0, . . . , n − 1.

Proof. The claim is trivially true for n ≤ 0. Suppose n ≥ 1, and ﬁx an i between 0 and

n − 1. Let Fi+1 be a face of rank i + 1 and Fi−1 a face of rank i − 1 such that Fi−1 < Fi+1.

Consider a ﬂag Φ = {F−1,...,Fi−1,Fi,Fi+1,...,Fn}. Then any ﬂag that is i-adjacent to Φ

yields a proper face in the section Fi+1/Fi−1, and any proper face other than Fi in Fi+1/Fi−1 yields a ﬂag that is i-adjacent to Φ. Therefore, the section Fi+1/Fi−1 has two proper faces if and only if Φ has a unique i-adjacent ﬂag.

For a flag Φ in a pre-polytope, we denote the unique i-adjacent flag by Φi. We then inductively define Φi1,...,ik to be (Φi1,...,ik−1 )ik . Note that for any i,Φi,i = Φ, and if |i − j| ≥ 2, then Φi,j = Φj,i.

If P and Q are n-pre-polytopes, then we can identify their respective maximal elements and minimal elements to obtain another n-pre-polytope. This essentially amounts to taking the disjoint union of P and Q. Working with disconnected pre-polytopes does not add much to the theory, so we will restrict ourselves to connected pre-polytopes.

11 Deﬁnition 1.4. Let P be a ﬂagged poset of rank n. We say that P is connected if n ≤ 1 or if the Hasse diagram of P\{F−1,Fn} is a connected graph. In other words, P is connected if for every pair of proper faces G and H, there is a chain of proper faces

G = G0 ≤ G1 ≥ G2 ≤ G3 ≥ · · · ≥ Gk−1 ≤ Gk = H.

Every section of a pre-polytope is again a pre-polytope, but the sections of a connected pre-polytope need not be connected. For example, consider the “dihedral bowtie”: two triangular dihedra joined at a vertex. It has 5 vertices, 6 edges, and 4 triangular faces: see Figure 1.1. As we see, this is a connected pre-polytope. However, if we take the section

F jj 3 UUU jjj | CC UUU jjj | C UUUU jjjj || CC UUU jjj || CC UUUU jjj || C UUU jjjj 0 UUU 0 T1 T2 T W T p NN ggggp 2 OOWWWWW o 1 PP pp NN gggg pp OOO WWWWooo PPP ppp gggggNNN ppp OO ooWWWW PP pp gggg ppNN ooOoO WWWWW PPP pp gggg pp NN oo OO WWWW PPP pppgggg ppp NN ooo OOO WWWW PP gpggg ac p N 0 o 0 WWW 0 0 ab TT UUU bc b c i a c i a b ?? TTTT UUU DD z @@ iiii AA iiii | ? TTT UUUU DD zz @@ iiii AA iiii | ?? TTTT UUU D z iii@i iAii || ?? TTT UUUU DD zz iii @@ iiii AA || ? TTT UUU DD zz iiii @ iii A || TTT UUU ziiii 0 iiii 0 a WWW b P c b gg a WWWW PPP nnn gggg WWWWW PP nn ggggg WWWW PPP nnn ggggg WWWWW PP nn gggg WWWW PPP nnn ggggg WWWWWP nnggggg F−1 g

Figure 1.1: Hasse diagram of the dihedral bowtie

F3/c, we get the disjoint union of two digons; see Figure 1.2. Thus, we see that working with connected pre-polytopes is a little unsatisfying. In order to build them from smaller pieces, we may have to consider pre-polytopes that are not connected.

We dodge this diﬃculty in the simplest possible way: by deciding not to deal with such structures. We say that a ﬂagged poset is strongly connected if all of its sections are connected

12 F3 o @PP ooo~ @PPP oo ~~ @@ PPP ooo ~~ @@ PPP ooo ~~ @@ PPP ooo ~ 0 PP 0 T1 T2 T T @ 2 A 1 @@ ~~ AA }} @~@~ A}A} ~ @@ }} AA ~~ @ }} A ac~ 0 0 QQ bc b cm a c QQQ CC { mmm QQQ CC {{ mmm QQQ CC {{ mmm QQQCC {{mmm QQ c {mmm

Figure 1.2: Hasse diagram of a section of the dihedral bowtie

(including the whole ﬂagged poset itself). Finally, we come to our desired deﬁnition:

Deﬁnition 1.5. An (abstract) polytope of rank n (also called an (abstract) n-polytope) is a strongly connected n-pre-polytope. That is, an n-polytope is a partially ordered set P satisfying the following four properties:

(1) P has a unique maximal element Fn and a unique minimal element F−1.

(2) Every ﬂag of P has n + 2 elements.

(3) (Diamond condition): whenever F < G where F is a (j−1)-face and G is a (j+1)-face, then there are exactly two j-faces H such that F < H < G.

(4) P is strongly connected.

There is only a single polytope in each of the ranks −1, 0, and 1. In rank 2, the polytopes are the face-lattices of k-gons, for 2 ≤ k ≤ ∞. In general, the face-lattice of a convex polytope is indeed an abstract polytope, though there are also many fascinating new structures.

A slightly stronger notion of connectivity proves quite useful. The adjacency relation between flags gives rise to the flag graph of a polytope in a natural way; namely, the vertices of the flag graph are the flags of the flagged poset, and two flags are connected by an edge

13 if they are adjacent (that is, if they differ in just a single element). We say that a flagged poset is flag-connected if its flag graph is connected. That is, if for every pair of flags Φ and Ψ, there is a sequence

Φ = Φ0, Φ1,..., Φk = Ψ,

such that for each i,Φi is adjacent to Φi+1. Every ﬂag-connected ﬂagged poset is also connected.

Note that the dihedral bowtie, though connected, is not ﬂag-connected. For example,

0 there is no sequence of adjacencies that will take a ﬂag containing T1 to a ﬂag containing T1.

Like connected pre-polytopes, flag-connected pre-polytopes have the unfortunate property that their sections may not be flag-connected. Nevertheless, we shall often encounter flag-connected pre-polytopes with flag-connected facets and vertex-figures. As such, we make the following definition:

Definition 1.6. A semi-polytope of rank n (also called an n-semi-polytope) is a flag- connected n-pre-polytope such that the facets and vertex-figures are both flag-connected.

Note that for n ≤ 3, all semi-polytopes are in fact polytopes.

If every section of a flagged poset is flag-connected (including the whole poset itself), then we say that the poset is strongly flag-connected. Even though a flagged poset can be connected without being flag connected, Proposition 2A1 in [23] shows that a flagged poset is strongly connected if and only if it is strongly flag-connected.

Let P be an n-pre-polytope such that its sections of rank 2 are all connected. Then for any (i − 2)-face F and (i + 1)-face G such that F ≤ G, the section G/F is a k-gon for some

k (2 ≤ k ≤ ∞). Having ﬁxed P, we deﬁne pi(F,G) to be the number of i-faces in the section

G/F ; in other words, if G/F is a k-gon, then pi(F,G) = k. If every pi(F,G) depends only

on i (and not on the particular choice of F and G), then we deﬁne pi := pi(F,G) and we say

14 that P has Schläflisymbol {p1, . . . , pn−1} or that P is of type {p1, . . . , pn−1}. A pre-polytope that has a Schläfli symbol is said to be equivelar.

Any section of an equivelar pre-polytope is itself equivelar. In particular, if P is equivelar of type {p1, . . . , pn−1}, and if F is an i-face and G is a j-face such that F ≤ G, then G/F is equivelar of type {pi+2, . . . , pj−1}. If P is an equivelar pre-polytope such that the facets are all isomorphic to K and the vertex-ﬁgures are all isomorphic to L, then we say that P is of type {K, L}. (In [23], P is said to be in the class hK, Li.)

We now move on to functions between pre-polytopes. Let P and Q be ﬂagged posets and ϕ : P → Q a function. We call ϕ a (polytope) homomorphism if whenever F ≤ G in P we also have F ϕ ≤ Gϕ in Q. (Following the convention of [23], polytope homomorphisms always act on the right.) Naturally, if ϕ is bijective and its inverse is a homomorphism, then we call ϕ an isomorphism, and if ϕ is an isomorphism from P to itself, we call it an automorphism (of P). The group of automorphisms of a ﬂagged poset P is denoted Γ(P).

The automorphism group of a flagged poset has a natural action on the flags. For any automorphism ϕ of a flagged poset P, if the flags Φ and Φ0 are i-adjacent, then so are Φϕ and Φ0ϕ. Thus, if P is a pre-polytope, Φiϕ = (Φϕ)i for each flag Φ. We can extend this result as follows:

Proposition 1.7. Let P be a pre-polytope. Let ϕ ∈ Γ(P), and let Φ be a ﬂag of P. Then

Φi1,...,ik ϕ = (Φϕ)i1,...,ik .

In particular, if P is ﬂag-connected, then ϕ is determined by the action on Φ.

15 Proof. We see that

Φi1,...,ik ϕ = (Φi1,...,ik−1 )ik ϕ

= (Φi1,...,ik−1 ϕ)ik .

By induction, the ﬁrst claim is proved. Now, suppose P is ﬂag-connected and that Φϕ is

i1,...,ik known. If Ψ is any other ﬂag of P, then Ψ = Φ for some sequence i1, . . . , ik, and thus the action of ϕ on Ψ is known. Since we know how ϕ acts on every ﬂag, we know where it takes every face, and thus ϕ is completely determined.

Corollary 1.8. Let P be a flag-connected pre-polytope, and let ϕ ∈ Γ(P). If ϕ fixes any flag Φ of P, then ϕ = , the identity automorphism.

Proof. By Proposition 1.7, ϕ is uniquely determined by where it sends Φ. Since the automorphism also ﬁxes Φ, we must have that ϕ = .

In practice, we rarely work with homomorphisms in full generality. We will usually be interested in isomorphisms and automorphisms. However, there is one other type of homomorphism that we will often use.

Definition 1.9. Let P and Q be flagged posets, and let ϕ : P → Q be a homomorphism. Then ϕ is a covering if it satisfies the following properties:

(1) The function ϕ is surjective.

(2) If F is a face of P, then the rank of F ϕ is equal to the rank of F .

0 0 (3) Let Φ and Φ be ﬂags of P that are i-adjacent. Let Fi be the i-face of Φ and let Fi be

0 0 0 the i-face of Φ . Then Fiϕ 6= Fi ϕ. That is, Φϕ and Φ ϕ are also i-adjacent ﬂags; in particular, they do not coincide.

16 If there is a covering from P to Q, we say that P covers Q. Note that if P covers Q, then P and Q must have the same rank.

Proposition 1.10. Let P be a pre-polytope, let Q be a flag-connected pre-polytope, and let ϕ : P → Q be a homomorphism. If ϕ satisfies properties 2 and 3 in Definition 1.9, then ϕ is a covering.

Proof. It suﬃces to show that ϕ is surjective. Let Φ be a ﬂag of P, and let Ψ = Φϕ. Since Q

i1,...,ik is ﬂag-connected, we can write any ﬂag of Q as Ψ for some sequence (i1, . . . , ik). Since

ϕ preserves ﬂag-adjacency, we get that Φi1,...,ık ϕ = Ψi1,...,ik , and thus ϕ is surjective.

1.3 Regularity

Classically, the polytopes that have received the most attention are the regular ones: those having the highest degree of symmetry. There are several equivalent definitions of regular convex polytopes, and it takes some care to determine which of these definitions generalizes nicely for abstract polytopes. Furthermore, whatever definition of regularity we choose, we would like for the face lattice of a regular convex polytope to be a regular abstract polytope.

For example, the face lattice of a regular k-gon should be a regular abstract polygon. Of course, the face lattice of any other convex k-gon is the same. Thus we see that every abstract polyhedron has regular faces and regular vertex-figures. If we were to define regularity inductively – say, by declaring an n-polytope regular if it has regular facets and vertex-figures – then every polytope would be regular! Even if we restrict the definition to additionally require the facets to all be isomorphic and the vertex-figures to all be isomorphic as well, we get a fairly permissive definition. Instead, we use the following definition:

Deﬁnition 1.11. A ﬂagged poset P is regular if the action of its automorphism group Γ(P)

17 on its ﬂags is transitive.

Though we define regularity for flagged posets, we will usually only work with regular polytopes or regular semi-polytopes, for reasons that will become clear in a moment. Given a flagged poset P, we usually fix a flag Φ that we call the base flag. If P is regular, then any two choices of base flag are equivalent, since there is an automorphism of P that carries the first choice to the second. If P is not regular, however, then choosing a base flag amounts to choosing an orbit of the action of Γ(P) on the flags.

Now, let P be a regular semi-polytope. Since P is a pre-polytope, every ﬂag has a unique i-adjacent ﬂag for each i = 0, . . . , n − 1. By the regularity of P, there are automorphisms

i i i ρ0, . . . , ρn−1 (called abstract reﬂections) such that Φρi = Φ . Recalling that (Φ )ϕ = (Φϕ)

ik,...,i1 for any ϕ ∈ Γ(P), it follows that Φρi1 ··· ρik = Φ . Now, since P is ﬂag-connected, any automorphism is determined by where it sends Φ, so that if Φϕ = Φik,...,i1 we must

have ϕ = ρi1 ··· ρik . Thus we see that the abstract reﬂections ρ0, . . . , ρn−1 generate Γ(P).

i,i 2 i,j j,i Furthermore, since Φ = Φ we must have ρi = , and since Φ = Φ whenever |i − j| ≥ 2,

2 we must have (ρiρj) = whenever |i − j| ≥ 2.

If P is a polytope (and not just a semi-polytope), then the group Γ(P) satisﬁes the following “intersection property” [23]:

hρi | i ∈ Ii ∩ hρj | j ∈ Ji = hρi | i ∈ I ∩ Ji.

Semi-polytopes do not necessarily satisfy the intersection property, but they do satisfy some special cases of it.

Proposition 1.12. Let P be a regular n-semi-polytope, with automorphism group Γ = hρ0, . . . , ρn−1i. For each i = 0, . . . , n − 1, let Γi = hρj | j 6= ii. Then Γ satisﬁes the

18 following “empty intersection property”:

Γ0 ∩ · · · ∩ Γn−1 = hi,

and the following “weak intersection property”:

\ Γj = hρii, for each i ∈ {0, . . . , n − 1}. j6=i

Furthermore, the action of Γ on the ﬂags of P is simply transitive.

Proof. Let Φ = {F0,...,Fn−1} be the base flag of P. (We frequently omit the improper faces when listing the elements of a flag.) For each i, the subgroup Γi stabilizes Fi, and thus the group Γ0 ∩ · · · ∩ Γn−1 stabilizes Φ. Suppose that ϕ ∈ Γ0 ∩ · · · ∩ Γn−1, so that it stabilizes Φ. Then by Corollary 1.8, ϕ = , and this proves the empty intersection property. Furthermore, this shows that the action of Γ on the flags of P is simply transitive, since if Φϕ = Φψ, then ϕψ−1 stabilizes Φ, and so ϕ = ψ.

T Now, the group j6=i Γj stabilizes every face of Φ except for Fi. It is clear that hρii is contained in this stabilizer. Since P is a pre-polytope, there is exactly one ﬂag that is i-adjacent to Φ. Then by simple transitivity, ρi is the only automorphism that sends Φ to

i Φ , and we see that the stabilizer is equal to hρii.

We thus make the following deﬁnitions:

Definition 1.13. Let Γ = hρ0, . . . , ρn−1i be a finitely generated group such that the order of each ρi is 2 and such that whenever |i − j| ≥ 2, the order of ρiρj is 2. Then we say that Γ is a string group generated by involutions (or sggi for short.) If Γ also satisfies the empty intersection property and the weak intersection property, then we call Γ a string pre-C-group.

19 If in fact Γ satisﬁes the full intersection property

hρi | i ∈ Ii ∩ hρj | j ∈ Ji = hρi | i ∈ I ∩ Ji, then we call Γ a string C-group.

The ‘C’ in “string C-group” stands for Coxeter, since string C-groups share many of the properties of string Coxeter groups. In particular, every string Coxeter group is a string C-group. However, the class of string C-groups is much more general. Note that by our preceding discussion, the automorphism group of a regular semi-polytope is a string pre-C- group, and the automorphism group of a regular polytope is a string C-group.

We will soon see how to build regular semi-polytopes and regular polytopes out of string pre-C-groups and string C-groups, respectively. In order to use these constructions effec- tively, we need to know when an sggi satisfies the various intersection properties. Section 2E of [23] gives some general results for when an sggi satisfies the (full) intersection property. Here we present two similar results for the weak intersection property.

Proposition 1.14. Let Γ = hρ0, . . . , ρn−1i be an sggi that satisﬁes the empty intersection property. Then for each i in {0, . . . , n − 1},

Γ0 ∩ · · · ∩ Γi−1 = hρi, . . . , ρn−1i and

Γi+1 ∩ · · · ∩ Γn−1 = hρ0, . . . , ρii.

Proof. We will prove the ﬁrst claim by induction on i; the proof of the second is analogous and follows by a ‘dual’ argument, replacing each ρi by ρn−1−i. The claim is obvious for i = 1.

Suppose that Γ0 ∩ · · · ∩ Γi−1 = hρi, . . . , ρn−1i and consider Γ0 ∩ · · · ∩ Γi. This is equal to

20 hρi, . . . , ρn−1i∩Γi. Now, Γi = hρ0, . . . , ρi−1, ρi+1, . . . , ρn−1i. The subgroups hρ0, . . . , ρi−1i and hρi+1, . . . , ρn−1i commute, so

Γi = hρ0, . . . , ρi−1ihρi+1, . . . , ρn−1i.

Therefore,

Γ0 ∩ · · · ∩ Γi = hρi, . . . , ρn−1i ∩ hρ0, . . . , ρi−1ihρi+1, . . . , ρn−1i.

Now, let w be in this intersection. Then w ∈ hρ0, . . . , ρi−1ihρi+1, . . . , ρn−1i, and so we can write w = uv with u ∈ hρ0, . . . , ρi−1i and v ∈ hρi+1, . . . , ρn−1i. We also have that

−1 w ∈ hρi, . . . , ρn−1i. Thus we see that u = wv ∈ hρi, . . . , ρn−1i, so that in fact u ∈ hρ0, . . . , ρi−1i ∩ hρi, . . . , ρn−1i. Therefore, u ∈ Γ0 ∩ · · · ∩ Γn−1, and since Γ satisﬁes the empty intersection property, u = . So w = v ∈ hρi+1, . . . , ρn−1i, which shows that

Γ0 ∩ · · · ∩ Γi ≤ hρi+1, . . . , ρn−1i.

The other containment is obvious, and so the claim is proved.

Corollary 1.15. Let Γ be an sggi that satisﬁes the empty intersection property. Then Γ is

a string pre-C-group if and only if for each i we have hρ0, . . . , ρii ∩ hρi, . . . , ρn−1i = hρii.

Proof. From Proposition 1.14, we see that if Γ satisﬁes the empty intersection property, T then j6=i Γj = hρ0, . . . , ρii ∩ hρi, . . . , ρn−1i. Since Γ is a string pre-C-group if and only if T j6=i Γj = hρii, the claim follows.

Let P be a regular n-semi-polytope, with base ﬂag Φ = {F−1,...,Fn}. Then for −1 ≤ i < j ≤ n, the automorphism group of Fj/Fi can be identiﬁed with the stabilizer of the

21 chain {F−1,...,Fi,Fj,...,Fn} in Γ(P). If P is a polytope, then the stabilizer of Fk is

Γk := hρi | i 6= ki.

Therefore,

Γ(Fj/Fi) = (Γ0 ∩ · · · ∩ Γi) ∩ (Γj ∩ · · · ∩ Γn−1)

= hρi+1, . . . , ρj−1i.

If P is just a semi-polytope, however, the situtation is somewhat more complicated. In order for Γk to be the stabilizer of Fk, the sections Fn/Fk and Fk/F−1 must both be ﬂag-connected:

Proposition 1.16. Let P be a regular n-semi-polytope, with base ﬂag Φ = {F−1,...,Fn}.

Then the stabilizer of Fk in Γ := Γ(P) is Γk if and only if the sections Fn/Fk and Fk/F−1 are regular and ﬂag-connected.

Proof. First, suppose that the stabilizer of Fk is Γk. Let Φ1 = {F−1,...,Fk} and let Φ2 =

{Fk,...,Fn}. That is, Φ1 is the part of Φ that lies in Fk/F−1, and Φ2 is the part of Φ that lies in Fn/Fk. Now, to prove that Fk/F−1 is regular and flag-connected, it suffices to show that for any other flag Ψ1 of Fk/F−1, there is an automorphism of Fk/F−1 that sends Φ1 to

Ψ1. Fix a particular choice of Ψ1. We can extend it to Ψ, a ﬂag of P. Since P is regular, there is an automorphism α sending Φ to Ψ. Now, Φ and Ψ both contain the k-face Fk, so

α is in the stabilizer of Fk. Therefore, α ∈ Γk = hρi | i 6= ki. Now, we can decompose Γk as

Γk = hρ0, . . . , ρk−1ihρk+1, . . . , ρn−1i.

Therefore, we can write α = α1α2, where α1 ∈ hρ0, . . . , ρk−1i and α2 ∈ hρk+1, . . . , ρn−1i. We can identify α1 with an automorphism of Fk/F−1, and this automorphism sends Φ1 to Ψ1,

22 as desired. So Fk/F−1 is regular and ﬂag-connected. The proof for Fn/Fk is analogous.

Now suppose that the sections Fn/Fk and Fk/F−1 are regular and ﬂag-connected. It is

clear that Γk stabilizes Fk, so we only need to show that Γk is the entire stabilizer. Since Γ

acts simply transitively on the ﬂags of P, it suﬃces to show that Γk acts transitively on all

flags containing Fk. In fact, it suffices to show that for any flag Ψ containing Fk, there is

an automorphism α ∈ Γk such that Φα = Ψ. Let Φ1 be the part of Φ in Fk/F−1. Since Ψ

contains Fk as its single k-face, we can also take Ψ1 to be the part of Ψ in Fk/F−1. Now,

Φ1 and Ψ1 are ﬂags of Fk/F−1, so by supposition, there is an automorphism α1 of Fk/F−1 that sends Φ1 to Ψ1. Furthermore, we can identify the generators of Γ(Fk/F−1) with the generators {ρ0, . . . , ρk−1} of Γ. Now, in a similar fashion, we can take Φ2 and Ψ2 to be the parts of the ﬂags Φ and Ψ in Fn/Fk, and we obtain an automorphism α2 of Fn/Fk that sends

Φ2 to Ψ2. This automorphism can be written in terms of the generators {ρk+1, . . . , ρn−1} of Γ. Now, set α = α1α2. Then α ∈ Γk and α sends Φ to Ψ, which is what we wanted to prove.

Corollary 1.17. Let P be a regular n-semi-polytope. Then the stabilizer of F0 is hρ1, . . . , ρn−1i, and the stabilizer of Fn−1 is hρ0, . . . , ρn−2i.

Let P be a regular n-semi-polytope. Clearly, any regular semi-polytope is equivelar; that is, P has a Schläfli symbol. In fact, P is of type {p1, . . . , pn−1}, where pi is the order of ρi−1ρi. It is also clear that the facets of P must all be isomorphic, and that the vertex-figures must all be isomorphic as well, so that P is of type {K, L} for some regular (n − 1)-pre-polytopes K and L. (The facets and vertex-figures might not be semi-polytopes; i.e., might not have flag-connected facets and vertex-figures themselves.)

For a given Schl¨aﬂi symbol {p1, . . . , pn−1}, there is a regular, universal polytope of that type that covers all semi-polytopes of that type. We use {p1, . . . , pn−1} to refer to this universal polytope. Whenever this universal polytope is the face-lattice of a regular convex

23 polytope, the name used here is the same as the usual Schl¨aﬂi symbol for the polytope (see

[7]). The automorphism group of {p1, . . . , pn−1} is the string Coxeter group [p1, . . . , pn−1].

All polytopes of rank n are quotients of the universal polytope of rank n

Un := {∞,..., ∞}. | {z } n−1

This polytope has automorphism group

2 2 2 Wn := [∞,..., ∞] = hρ0, . . . , ρn−1 | ρ0 = ··· = ρn−1 = , (ρiρj) = when |i − j| ≥ 2i. (1.1)

There is a natural action of Wn on the ﬂags of any regular n-semi-polytope P with base ﬂag

j1,...,jk i,j1,...,jk Φ, given by Φ ρi = Φ . That is, ρi from Wn acts in the same way as ρi from Γ(P), justifying our choice of notation. Now, if w ∈ Wn stabilizes Φ, then for any i ∈ {0, . . . , n−1},

i Φρiwρi = Φ wρi

i = (Φw) ρi

i = Φ ρi

= Φ.

Therefore, StabWn (Φ) is normal in Wn. Furthermore, if w stabilizes Φ, then it must in fact stabilize all ﬂags of P. Therefore, Wn/StabWn (Φ) acts simply transitively on the ﬂags of P.

In fact, Γ(P) ' Wn/StabWn (Φ).

In [23], the authors show how to build a regular polytope from a string C-group. In fact, the construction can be applied to any sggi to obtain a ﬂagged poset. The construction is as follows. Let Γ = hρ0, . . . , ρn−1i be an sggi. Let Γj = hρi | i 6= ji, as before. We deﬁne a

24 poset P(Γ) by setting the j-faces of P(Γ) to be the cosets Γjϕ, where ϕ ∈ Γ. Then we say

that Γjϕ ≤ Γkψ if j ≤ k and Γjϕ ∩ Γkψ 6= ∅. Lemma 2E6 in [23] shows that this deﬁnes a partial order.

We note that for any sggi Γ, there is a natural group homomorphism f :Γ → Γ(P(Γ))

sending ψ to fψ, where fψ acts on P(Γ) by

(Γjϕ)fψ = Γj(ϕψ).

Several of the results in [23, Ch. 2E] apply not just to string C-groups, but to sggi’s or string pre-C-groups. We reproduce the statements here (with minor modiﬁcation in some cases).

Lemma 1.18 (2E8). Let Γ be an sggi, and let K ⊆ {0, 1, . . . , n − 1}. Then Γ is transitive on all chains of the form {Γiϕi | i ∈ K}.

Lemma 1.19 (2E9). Let Γ be a string pre-C-group. Then the stabilizer of the base ﬂag is trivial, and for any j ∈ {0, . . . , n − 1}, the stabilizer of the chain {Γ0,..., Γj−1, Γj+1,..., Γn−1} is hρji.

Lemma 1.20 (2E10). The natural homomorphism f :Γ → Γ(P(Γ)) is injective.

Now we are able to expand [23, Thm. 2E11] to sggi’s and string pre-C-groups.

Theorem 1.21. Let Γ be an sggi. Then P(Γ) is a ﬂag-connected ﬂagged poset. If Γ is also a string pre-C-group, then P(Γ) is a regular semi-polytope with Γ(P(Γ)) ' Γ.

Proof. First, suppose that Γ is an sggi. We have Γ−1 = Γn = Γ, and these play the role of the minimal and maximal face, respectively. Now, let Ω be a chain in P(Γ); we want to show that it extends to a chain with n + 2 elements. By Lemma 1.18, we can write Ω as ΦK ϕ for

25 some ϕ ∈ Γ, where ΦK = {Γi | i ∈ K}. Then Ω extends to the flag Φϕ. Thus, we see that P(Γ) is a flagged poset. In fact, since the action of Γ on P(Γ) is transitive on flags (again by Lemma 1.18), we see that P(Γ) is flag-connected.

Now, suppose Γ is also a string pre-C-group. Then Lemma 1.19 shows that the base flag has a unique j-adjacent flag for each j in {0, . . . , n − 1}. Since the action of Γ is transitive on flags, this implies that every flag has a unique j-adjacent flag for each j in {0, . . . , n − 1}. Therefore, P(Γ) is a flag-connected pre-polytope.

We now need to show that the facets and vertex-figures of P(Γ) are flag-connected. Fix a facet Fn−1 of P(Γ). By Corollary 1.17, the stabilizer of the facet is Γn−1 = hρ0, . . . , ρn−2i, and we can naturally identify this stabilizer with the automorphism group of the facet. Now, given two flags in the facet, they naturally extend to flags Φ and Ψ of P(Γ), both containing Fn−1. Since Γ acts transitively on the flags, there are elements ϕ and ψ of Γ such

−1 that Φ = (Γ0ϕ, . . . , Γn−1ϕ) and Ψ = (Γ0ψ, . . . , Γn−1ψ). Then ϕ ψ sends Φ to Ψ. Since

−1 Γn−1ϕ = Γn−1ψ = Fn−1, it follows that ϕ ψ ∈ Γn−1. Therefore, there is an automorphism of the facet sending Φ to Ψ, which shows that the facets are flag-connected. The same argument works to show that the vertex-figures are flag-connected as well.

It remains to show that P(Γ) is regular and that Γ(P(Γ)) ' Γ. By Lemma 1.20, the homomorphism f :Γ → Γ(P(Γ)) is injective. I claim that it is surjective as well. Let

α ∈ Γ(P(Γ)); we want to ﬁnd some ψ ∈ Γ such that fψ = α. By Lemma 1.18, there is an element ψ ∈ Γ such that fψ sends the base ﬂag Γ0 < ··· < Γn−1 to the same place that α

−1 −1 sends it. Therefore, fψα fixes the base flag. Then by Corollary 1.8, this means that fψα is the identity, and thus fψ = α. So f is an isomorphism. Finally, this means that Γ(P(Γ)) is transitive on the flags of P(Γ), and thus P(Γ) is regular.

Thus, just as we have a one-to-one correspondence between string C-groups of rank n

26 on a distinguished set of generators and regular n-polytopes, so too we have a one-to-one correspondence between string pre-C-groups of rank n on a distinguished set of generators and regular n-semi-polytopes. In particular, for any regular n-semi-polytope Q, the semi- polytope P(Γ(Q)) is naturally isomorphic to Q.

This correspondence also respects the natural covering relations:

0 0 0 Proposition 1.22. Let Γ = hρ0, . . . , ρn−1i and Γ = hρ0, . . . , ρn−1i be string pre-C-groups.

0 0 0 If Γ covers Γ via the epimorphism sending each ρi to ρi, then P(Γ) covers P(Γ ).

Proof. Write π for the natural epimorphism from Γ to Γ0. We deﬁne a function π : P(Γ) →

0 0 P(Γ ) by (Γjϕ)π = Γj(ϕπ). To check that this is a well-deﬁned function, we note that if

0 0 0 Γj = Γjϕ, then ϕ ∈ Γj = hρi | i 6= ji. Therefore, ϕπ ∈ Γj and thus Γj = Γj(ϕπ). Next, we

need to show that π preserves the order relation. If Γjϕ ≤ Γkψ, then by deﬁnition there is

0 0 some α ∈ Γ such that α ∈ Γjϕ ∩ Γkψ. It follows that απ ∈ Γj(ϕπ) ∩ Γk(ψπ), and therefore

0 0 Γj(ϕπ) ≤ Γk(ψπ). So π is a polytope homomorphism.

Now we need to show that π is a covering. It is clear that π preserves rank, so by Proposition 1.10, it suffices to show that it preserves flag adjacency. Let Φ be a flag of P(Γ).

j We can write it as Φ = (Γ0ϕ, . . . , Γn−1ϕ) for some ϕ ∈ Γ. The j-adjacent ﬂag Φ can be

j 0 0 written Φ = (Γ0ρjϕ, . . . , Γn−1ρjϕ). Then π sends Φ to (Γ0(ϕπ),..., Γn−1(ϕπ)), and since

0 j 0 0 0 0 (ρjϕ)π = (ρjπ)(ϕπ) = ρj(ϕπ), it follows that π sends Φ to (Γ0ρj(ϕπ),..., Γn−1ρj(ϕπ)), which is j-adjacent to the image of Φ. Therefore, π preserves adjacency, and thus it is a covering.

In light of the correspondence between string pre-C-groups and regular semi-polytopes, we view regular semi-polytopes P and Q (with chosen base ﬂags) as being equal if and only if the stabilizer in Wn of the base ﬂag is the same in each case. That is, if Γ(P) = Wn/M

and Γ(Q) = Wn/K for some normal subgroups M and K of Wn, then P = Q if and only if

27 M = K.

We also use the correspondence to characterize covering relations between regular semi- polytopes:

Proposition 1.23. Let P and Q be regular n-semi-polytopes with Γ(P) ' Wn/M and

Γ(Q) ' Wn/K. Then Q covers P if and only if K ≤ M.

Proof. Suppose ϕ : Q → P is a covering. Since P and Q are regular, we can compose any covering with automorphisms of P and Q to obtain a covering that sends the base ﬂag Ψ of Q to the base ﬂag Φ of P. Thus, without loss of generality, suppose that ϕ sends Ψ to Φ.

ik,...,i1 Let ψ = ρi1 ··· ρik ∈ Wn stabilize Ψ. Then Ψ = Ψ. Now, since ϕ preserves adjacency,

it must send Ψ = Ψik,...,i1 to Φik,...,i1 . Thus we see that Φik,...,i1 = Φ, so that ψ stabilizes Φ as well. Since K is the stabilizer of Ψ and M is the stabilizer of Φ, we see that K ≤ M.

Conversely, if K ≤ M, then Γ(Q) naturally covers Γ(P), and by Proposition 1.22, P(Γ(Q)) covers P(Γ(P)). Since the former is naturally isomorphic to Q and the latter is naturally isomorphic to P, the claim is proved.

1.4 Direct regularity

The directly regular polytopes form an important subclass of the regular polytopes. Direct regularity is essentially an orientability condition; intuitively, a regular polytope is directly regular if there is no rotation that carries the polytope to its mirror image. In order to make this intuitive idea precise, we need to deﬁne rotations.

Let P be a regular semi-polytope with automorphism group Γ(P) = hρ0, . . . , ρn−1i. Then

+ the abstract rotations σi := ρi−1ρi (for i = 1, . . . , n−1) generate the rotation subgroup Γ (P) of Γ(P), which has index at most 2. If the index is 2, then it is not possible to write any of

28 the abstract reﬂections ρi as a product of abstract rotations. This is exactly what we want for direct regularity.

Deﬁnition 1.24. Let P be a regular ﬂag-connected n-pre-polytope. Then P is directly regular if n ≤ 1 or if Γ+(P) has index 2 in Γ(P).

Note that all universal polytopes {p1, . . . , pn−1}, including the regular convex polytopes, are directly regular.

The ﬂag-connected sections of a directly regular semi-polytope are themselves directly regular. In particular, the rotation subgroup of hρi, . . . , ρji is hσi+1, . . . , σji, and this must have index 2 since hσ1, . . . , σn−1i has index 2 in hρ0, . . . , ρn−1i.

The rotation subgroups of sggi’s inherit many nice properties, including analogues of the intersection properties. Let P be a directly regular semi-polytope, and let Γ+ =

+ + + hσ1, . . . , σn−1i = Γ (P). We deﬁne Γ0 = hσ2, . . . , σn−1i,Γn−1 = hσ1, . . . , σn−2i, and for

+ + 1 ≤ i ≤ n − 2, we deﬁne Γi = hσ1, . . . , σi−1, σiσi+1, σi+2, . . . , σn−1i. Then Γi is the stabilizer of the i-face in the base ﬂag. From this we get the following analogue of the empty intersection property:

+ + Γ0 ∩ · · · ∩ Γn−1 = hi, and the following analogue of the weak intersection property:

\ + Γj = hi. j6=i

Note that the latter condition simply says that no element of Γ+ sends the base ﬂag to an i-adjacent ﬂag; this is required in order for P to be directly regular. In any case, we see that the empty intersection property is redundant when dealing with the rotation subgroups of string pre-C-groups, and we need only consider the weak intersection property.

29 Now suppose that P is a directly regular polytope (not just a semi-polytope), and let Γ+ be as before. For 1 ≤ i ≤ j ≤ n − 1 we deﬁne

τi,j := σiσi+1 ··· σj,

and for 0 ≤ i ≤ n let τ0,i := τi,n := 1; then τi,i = σi for i 6= 0, n. For I ⊆ {0, . . . , n − 1} we deﬁne

+ ΓI := hτi,j | i ≤ j and i − 1, j ∈ Ii.

Then we get the following analogue of the intersection property [30]:

+ + + ΓI ∩ ΓJ = ΓI∩J for I,J ⊆ {0, . . . , n − 1}.

In light of these properties, we classify rotation subgroups in a way similar to sggi’s (see Deﬁnition 1.13):

+ 2 Definition 1.25. Let Γ = hσ1, . . . , σn−1i be a finitely generated group such that (σi ··· σj) = for all i < j. Then we call Γ+ a string rotation group. If Γ+ also satisfies the weak intersection property (for rotation groups), we call Γ+ a string pre-chi-group, and if Γ+ satisfies the full intersection property (for rotation groups), we call Γ+ a string chi-group. (The use of “chi” here emphasizes the connection these groups have with chiral polytopes, which will be defined in the next section.)

1.5 Chirality

Chirality is a phenomenon that has been studied in many subjects, including chemistry, biology, and physics. Intuitively, an object is said to be chiral if it is impossible to bring

30 the object to coincide with its mirror image using only rotations and translations. The prototypical example is the human hand; indeed, we often refer to a chiral object as being “left-handed” or “right-handed”.

In the study of abstract polytopes, we take a somewhat more specific definition of chirality. We will get to the technical definition in a moment, but the idea is that a chiral polytope has full rotational symmetry, but that there is no automorphism that brings the left-handed form to the right-handed form. Compare this to the definition of a directly regular polytope, where there is an automorphism that brings the left-handed form to the right-handed form, but no such rotation.

If P is a directly regular semi-polytope, then the abstract rotation σi := ρi−1ρi is an automorphism of P that sends the base ﬂag Φ to Φi,i−1. On the other hand, if P is chiral,

we should have an automorphism that acts the same way as the rotation σi, whereas no

automorphism acts as the reﬂection ρi. We make this precise below.

Definition 1.26. Let P be a flag-connected n-pre-polytope with base flag Φ. Then P is chiral if there is no automorphism that sends Φ to any of its adjacent flags Φi, and if there

i,i−1 are automorphisms σ1, . . . , σn−1 such that for each i, Φσi = Φ .

As an immediate consequence of the definition, a chiral semi-polytope has two flag-orbits under the action of Γ(P), and adjacent flags are always in different orbits.

In analogy with directly regular polytopes, we call the elements σi abstract rotations. In fact, these elements generate the entire group Γ(P), so that everything that was said in the previous section about rotation subgroups of directly regular semi-polytopes also applies to the automorphism groups of chiral semi-polytopes. In particular, the automorphism group of a chiral polytope is a string chi-group, and the automorphism group of a chiral semi-polytope is a string pre-chi-group. For convenience, we deﬁne Γ+(P) := Γ(P) whenever P is chiral.

31 If Γ+ is the rotation subgroup of a directly regular semi-polytope P, then conjugation by

+ −1 2 ρ0 (in Γ) is a group automorphism of Γ that sends σ1 to σ1 , σ2 to σ1σ2, and ﬁxes every other generator. Similar group automorphisms are induced by conjugation with any ρi. On the other hand, if Γ+ is the automorphism group of a chiral semi-polytope, then Γ+ admits no automorphism that moves the generators this way.

+ If σ1, . . . , σn−1 is the set of generators of Γ corresponding to the base ﬂag Φ of a chiral

−1 2 + semi-polytope P, then σ1 , σ1σ2, σ3, . . . , σn−1 is the set of generators of Γ corresponding to the base flag Φ0 that is 0-adjacent to Φ. The choice of base flag is essentially a choice of orientation; we say that a chiral semi-polytope occurs in two enantiomorphic (mirror image) forms. We write P for the enantiomorphic form of P, with base flag Φ0 where Φ is the base flag of P.

+ Let Γ be generated by σ1, . . . , σn−1, and let w be a reduced word in the free group on these generators. We deﬁne the enantiomorphic (or mirror image) word w of w to be the

−1 2 word obtained from w by replacing every occurrence of σ1 by σ1 and σ2 by σ1σ2, while

+ + keeping all σj with j ≥ 3 unchanged. Then if Γ = Γ (P) for a directly regular semi- polytope P, the elements of Γ+(P) corresponding to w and w are conjugate in Γ(P). Note that w = w for all words w.

The flag-connected sections of a chiral n-semi-polytope are either chiral or directly regular. Furthermore, the (n − 2)-faces and the co-faces at edges must be directly regular [30]. This feature makes it difficult to find new examples of chiral semi-polytopes, which is part of what makes them so interesting. Given a regular n-polytope P, there are typically many ways to build a regular (n + 1)-polytope with facets isomorphic to P (for example, see [27]). On the other hand, if P is a chiral n-polytope with chiral facets, then it is impossible to build a chiral (n + 1)-polytope with facets isomorphic to P.

The simplest examples of chiral polytopes are the torus maps {4, 4}(b,c), {3, 6}(b,c) and

32 {6, 3}(b,c), with b, c 6= 0 and b 6= c (see [8]). Many other examples of chiral 3-polytopes are known, and they mostly come to us from the study of regular irreﬂexible maps. A fair amount of progress has been made in ﬁnding chiral 4-polytopes as well, primarily by extending the known chiral 3-polytopes. In ranks 5 and higher, however, only a handful of examples of chiral polytopes are known [5]. Indeed, it was only recently that we knew for certain that there are chiral polytopes in every rank [28].

Chiral semi-polytopes, like regular semi-polytopes, are always equivelar. A chiral n-

semi-polytope P is of type {p1, . . . , pn−1}, where pi is the order of the generator σi. Chiral semi-polytopes also have the property that their faces are all isomorphic, and so are there vertex-ﬁgures, so a chiral semi-polytope P is of type {K, L} for some directly regular or chiral semi-polytopes K and L.

+ The rotation subgroup Wn of Wn (see Equation 1.1) has presentation

+ + 2 Wn := [∞,..., ∞] = hσ1, . . . , σn−1 | (σi ··· σj) = for i < ji. (1.2)

+ Analogous to the case with regular semi-polytopes, there is a natural action of Wn on the ﬂags of any chiral or directly regular n-semi-polytope P with base ﬂag Φ. In fact,

+ + Γ (P) ' W /Stab + (Φ). n Wn

There is also a way to build a semi-polytope from a string rotation group that satis-

+ ﬁes a suitable intersection property. In particular, let Γ = hσ1, . . . , σn−1i. Recall that

+ + + we deﬁned Γ0 = hσ2, . . . , σn−1i,Γn−1 = hσ1, . . . , σn−2i, and for 1 ≤ i ≤ n − 2, Γi =

+ hσ1, . . . , σi−1, σiσi+1, σi+2, . . . , σn−1i. Then we build a poset P(Γ ) by setting the j-faces to

+ + + + + be the cosets of Γj , and where again we say that Γj ϕ ≤ Γk ψ if j ≤ k and if Γj ϕ ∩Γk ψ 6= ∅.

In [30], the authors prove that when we apply the preceding construction to a string chi- group, we get a polytope that is either directly regular or chiral. As before, the construction

33 works more generally, and the arguments used with sggi’s and string pre-C-groups can be (somewhat tediously) adapted for string rotation groups and string pre-chi-groups, with the help of some results from [30]. We are able to conclude the following.

Proposition 1.27. Let Γ+ be a string rotation group. Then P(Γ+) is a ﬂagged poset.

Proposition 1.28. Let Γ+ be a string pre-chi-group. Then P(Γ+) is a semi-polytope with Γ+(P(Γ+)) = Γ+. Furthermore, P(Γ+) is directly regular if and only if there is an involutory

−1 2 group automorphism that sends σ1 to σ1 , σ2 to σ1σ2, and σi to σi for i ≥ 3. Otherwise, P(Γ+) is chiral.

Thus, there is a natural correspondence between string pre-chi-groups and semi-polytopes that are chiral or directly regular. Using this correspondence, we can characterize covers of chiral and directly regular polytopes in terms of their rotation groups. The proof of the following proposition is essentially the same as for Proposition 1.23.

Proposition 1.29. Let P and Q be chiral or directly regular n-semi-polytopes with Γ+(P) '

+ + + Wn /M and Γ (Q) ' Wn /K. Then Q covers P if and only if K ≤ M.

1.6 Duality

The usual geometric operation of dualizing a convex polytope has a natural counterpart for abstract polytopes. For any polytope P, we obtain the dual of P (denoted Pδ) by simply reversing the partial order. A duality from P to Q is an anti-isomorphism; that is, a bijection δ between the face sets such that F < G in P if and only if δ(F ) > δ(G) in Q. If a polytope is isomorphic to its dual, then it is called self-dual.

If P is of type {K, L}, then Pδ is of type {Lδ, Kδ}. Therefore, in order for P to be self-dual, it is necessary (but not suﬃcient) that K is isomorphic to Lδ (in which case it is

34 also true that Kδ is isomorphic to L).

A self-dual regular polytope always possesses a duality that fixes the base flag. For chiral polytopes, this may not be the case. If a self-dual chiral polytope P possesses a duality that sends the base flag to another flag in the same orbit (but reversing its direction), then there is a duality that fixes the base flag, and we say that P is properly self-dual [16]. In this case, the groups Γ+(P) and Γ+(Pδ) have identical presentations. If a self-dual chiral polytope has no duality that fixes the base flag, then every duality sends the base flag to a flag in the other orbit, and P is said to be improperly self-dual. In this case, the groups Γ+(P) and Γ+(Pδ) have identical presentations instead.

δ δ If P is a regular polytope with Γ(P) = hρ0, . . . , ρn−1i, then the group of P is Γ(P ) =

0 0 0 + hρ0, . . . , ρn−1i, where ρi = ρn−1−i. If P is a directly regular or chiral polytope with Γ (P) =

δ + δ 0 0 0 −1 hσ1, . . . , σn−1i, then the rotation group of P is Γ (P ) = hσ1, . . . , σn−1i, where σi = σn−i. Equivalently, if Γ+(P) has presentation

hσ1, . . . , σn−1 | w1, . . . , wki

then Γ+(Pδ) has presentation

0 0 hσ1, . . . , σn−1 | δ(w1), . . . , δ(wk)i, where if w = σ ··· σ , then δ(w) := (σ0 )−1 ··· (σ0 )−1. i1 ij n−i1 n−ij

+ + + δ Suppose P is a chiral or directly regular polytope with Γ (P) = Wn /M. Then Γ (P ) =

+ + + δ Wn /δ(M), where δ(M) = {δ(w) | w ∈ M}. If δ(M) = M, then Γ (P) = Γ (P ), so P is properly self-dual.

δ If P is a chiral polytope, then Pδ is naturally isomorphic to P . Indeed, if w is a word

35 + in the generators σ1, . . . , σn−1 of Γ (P), then

−1 δ(w) = (σ1σ2 ··· σn−1)δ(w)(σ1σ2 ··· σn−1) ,

δ so we see that the presentation for Pδ is equivalent to that of P . In particular, if Γ+(P) =

+ Wn /M, then δ(M) = δ(M) := {δ(w) | w ∈ W }

+ since M is a normal subgroup of Wn , and thus δ(δ(M)) = M.

1.7 Petrie duality

There is a second duality operation that is defined on abstract polyhedra. We start with the idea of the Petrie polygons of a polyhedron. The definition usually used is that a Petrie polygon is a maximal edge-path (closed if the polyhedron is finite) such that every two successive edges lie on a common face, but no three successive edges do. For most polyhedra, this definition works well, but it is slightly problematic for polyhedra having at least one vertex of degree 2. In particular, it is not always possible for three successive edges to avoid lying on a common face. We thus make the following more precise definition:

Definition 1.30. Let P be a flag-connected pre-polyhedron, and let E = (...,E1,E2,...) be a (possibly finite) sequence of edges of P. Then E is a Petrie polygon of P if there is a sequence of flags (..., Φ1, Φ2,...) of P such that:

(1) For each i, Ei ∈ Φi.

(2) For i 6= j, Φi 6= Φj.

0,1,2 (3) For each i, Φi+1 = Φi .

36 0,1,2 (4) If E is ﬁnite, say E = (E1,...,Ek), then Φ1 = Φk .

A Petrie polygon in this sense does have the property that every two successive edges share a common face, and no three successive edges share a common face unless it is unavoidable.

We can now describe the second duality operation. Given a polyhedron P, its Petrie dual Pπ consists of the same vertices and edges as P, but its faces are the Petrie polygons of P. Taking the Petrie dual of a polyhedron also forces the old faces to be the new Petrie polygons, so that Pππ 'P. If P is isomorphic to Pπ, then we say that P is self-Petrie.

In some cases, the Petrie dual of a polyhedron is not a polyhedron, or even a pre- polyhedron. For example, the Petrie dual of the toroidal polyhedron {3, 6}(1,1) is not a pre- polyhedron [23]. Nevertheless, our later results are able to steer around such complications.

When talking about polyhedra, we will frequently consider their Petrie polygons, so we expand some of our earlier terminology. If P is a regular polyhedron, then its Petrie polygons all have the same length, and that length is the order of ρ0ρ1ρ2 in Γ(P). A regular polyhedron of type {p, q} and with Petrie polygons of length r is also said to be of type {p, q}r. If P is of type {p, q}r and it covers every other polyhedron of type {p, q}r, then we call it the universal polyhedron of type {p, q}r and we denote it by {p, q}r. The automorphism group of {p, q}r is denoted by [p, q]r, and this group is the quotient of [p, q] by the single extra

r relation (ρ0ρ1ρ2) = . We will also extend our notation and use [p, q]r for the group with presentation

2 2 2 p 2 q r hρ0, ρ1, ρ2 | ρ0 = ρ1 = ρ2 = (ρ0ρ1) = (ρ0ρ2) = (ρ1ρ2) = (ρ0ρ1ρ2) = i,

even when there is no universal polyhedron of type {p, q}r.

The operations δ and π form a group of order 6, isomorphic to S3. A regular polyhedron that is self-dual and self-Petrie is invariant under all 6 operations; such a polyhedron must

37 δ be of type {n, n}n for some n ≥ 2. In general, if P is of type {p, q}r, then P is of type

π {q, p}r, and P is of type {r, q}p. Furthermore, the dual and the Petrie dual of a universal polyhedron is again universal.

38 Chapter 2

The mixing operation

The mixing operation on polytopes ([23], [24]) is analogous to the parallel product of groups [33], the tensor product of graphs, and the join of maps and hypermaps [4]. It gives us a natural way to ﬁnd the minimal common cover of two regular or chiral polytopes. The basic method is to ﬁnd the parallel product of the automorphism groups (or rotation groups) of two polytopes, and then to build a poset from the resulting group. There are two main challenges. First, we want to determine how the structure of the mix depends on the two component polytopes. Second, we want to know when the mix of two polytopes is a polytope, and not just a poset. Neither issue is solved in full generality here, and it seems unlikely that a fully general and satisfactory solution exists to either problem. Nevertheless, in a wide variety of cases, it is possible to easily determine the structure and polytopality of the mix.

2.1 Mixing ﬁnitely generated groups

0 0 0 Let Γ = hx1, . . . , xni and Γ = hx1, . . . , xni be ﬁnitely generated groups on n generators. Then

0 0 0 the elements zi := (xi, xi) ∈ Γ × Γ (for i = 1, . . . , n) generate a subgroup of Γ × Γ that we

39 call the mix of Γ and Γ0, denoted ΓΓ0 (see [23, Ch.7A]). In other words, ΓΓ0 is the diagonal subgroup of Γ × Γ0, consisting of elements that are equivalent to (x ··· x , x0 ··· x0 ) for i1 ik i1 ik

some sequence i1, . . . , ik.

The following proposition follows immediately.

0 0 0 0 Proposition 2.1. Let Γ = hx1, . . . , xni and Γ = hx1, . . . , xni, and let zi = (xi, xi) for each i ∈ {1, . . . , n}. Then for 1 ≤ i ≤ j ≤ n,

0 0 hzi, . . . , zji = hxi, . . . , xji hxi, . . . , xji.

Let us consider some simple examples of the mix of two groups.

r 0 0 0 s 0 Example 2.2. Let Γ = hx1 | x1 = i and let Γ = hx1 | (x1) = i. Then Γ Γ has a single

0 generator z1 = (x1, x1), and the order of z1 is the least common multiple of r and s.

Example 2.3. Let

2 2 r Γ = hx1, x2 | x1 = x2 = (x1x2) = i, and

0 0 0 0 2 0 2 0 0 s Γ = hx1, x2 | (x1) = (x2) = (x1x2) = i.

0 0 0 Then the generators z1 = (x1, x1) and z2 = (x2, x2) also have order 2, and thus Γ Γ is a dihedral group. The order of z1z2 is the least common multiple of r and s.

Example 2.4. For any n-generator group Γ, the group Γ Γ is naturally isomorphic to Γ.

There is another way to characterize the mix of two groups which is in many ways more useful:

0 0 0 Proposition 2.5. Let Γ = hx1, . . . , xni and Γ = hx1, . . . , xni. Let Λ = hy1, . . . , yni, and suppose that there is an epimorphism f :Λ → Γ with f(yi) = xi and an epimorphism

40 0 0 0 0 0 0 0 0 f :Λ → Γ with f (yi) = xi. Let K = ker f and K = ker f . Then Γ Γ ' Λ/(K ∩ K ).

Proof. Consider the homomorphism g :Λ → Γ × Γ0 deﬁned by g(w) = (f(w), f 0(w)). The

0 image of yi is (xi, xi), and since Λ is generated by the elements yi, the image of g is generated

0 0 by the diagonal elements (xi, xi), and so Im g = Γ Γ . Now, w ∈ ker g if and only if w ∈ ker f ∩ ker f 0 = K ∩ K0. Therefore,

Γ Γ0 = Im g ' Λ/ ker g = Λ/(K ∩ K0).

Note that we can always ﬁnd such a group Λ; for instance, take Λ to be the free group

+ on n generators. In our applications, we will usually use Λ = Wn or Λ = Wn .

Our next result, and many later results, use the idea of a natural cover of a group. We

0 0 0 say that a group Γ = hx1, . . . , xni naturally covers Γ = hx1, . . . , xni if there is a well-deﬁned

0 0 epimorphism f :Γ → Γ deﬁned by f(xi) = xi. If Λ is a group on n speciﬁed generators and Γ = Λ/M and Γ0 = Λ/M 0, then Γ naturally covers Γ0 if and only if M ≤ M 0.

0 0 0 0 Corollary 2.6. Let Γ = hx1, . . . , xni and Γ = hx1, . . . , xni. Then Γ Γ is the minimal group that naturally covers Γ and Γ0. That is, if Λ naturally covers both Γ and Γ0, then it also naturally covers Γ Γ0.

Dual to the mix is the comix of two groups. We deﬁne the comix Γ Γ0 to be the amal- gamated free product that identiﬁes the generators of Γ with the corresponding generators

0 0 0 0 of Γ . That is, if Γ has presentation hx1, . . . , xn | Ri and Γ has presentation hx1, . . . , xn | Si, then Γ Γ0 has presentation

0 0 −1 0 −1 0 hx1, x1, . . . , xn, xn | R, S, x1 x1, . . . , xn xni.

0 0 Equivalently, we can just add the relations from Γ to those of Γ, replacing each xi with xi.

41 r 0 0 0 s Example 2.7. Let Γ = hx1 | x1 = i and let Γ = hx1 | (x1) = i. Then

0 r s Γ Γ = hx1 | x1 = x1 = i.

Therefore, the group of Γ Γ0 is cyclic, and its order is the greatest common divisor of r and s.

Example 2.8. Let

4 3 2 Γ = hσ1, σ2 | σ1 = σ2 = (σ1σ2) = i,

the rotation subgroup of the cube, and let

0 0 0 0 3 0 3 0 0 2 Γ = hσ1, σ2 | (σ1) = (σ2) = (σ1σ2) = i,

the rotation subgroup of the tetrahedron. Then a presentation for Γ Γ0 is given by

0 4 3 3 2 Γ Γ = hσ1, σ2 | σ1 = σ1 = σ2 = (σ1σ2) = i.

4 3 3 2 Since σ1 = σ1, the generator σ1 must be trivial. Therefore, σ2 = σ2 = , and σ2 must be trivial as well. So Γ Γ0 is the trivial group. Example 2.9. For any group Γ on n generators, the group Γ Γ is naturally isomorphic to Γ.

As with the mix, there is a simple way to characterize the comix using quotients.

0 0 0 Proposition 2.10. Let Γ = hx1, . . . , xni and Γ = hx1, . . . , xni be groups on n generators.

Suppose Λ = hy1, . . . , yni such that there is an epimorphism f :Λ → Γ with f(yi) = xi

0 0 0 0 0 and an epimorphism f :Λ → Γ with f(yi) = xi. Let K = ker f and K = ker f . Then Γ Γ0 ' Λ/KK0. 42 Proof. Since Γ Γ0 satisﬁes all the relations of Γ and of Γ0, it is clear that it is a quotient of both. Thus we get an epimorphism g from Λ to Γ Γ0. Set N = ker g, so that Γ Γ0 ' Λ/N. Since there are natural epimorphisms from Γ to Γ Γ0 and from Γ0 to Γ Γ0, this means that K ≤ N and K0 ≤ N, so that KK0 ≤ N. Now, K is the normal closure in Λ of the relators in R and K0 is the normal closure in Λ of the relators in S, while N is the normal closure of the relators in both. Since K and K0 are both normal in Λ, so is KK0, and thus the normal closure of the relators in R and S must be contained in KK0; i.e., N ≤ KK0. Thus N = KK0 and the claim is proved.

0 0 0 0 Corollary 2.11. Let Γ = hx1, . . . , xni and Γ = hx1, . . . , xni. Then Γ Γ is the maximal group that is naturally covered by Γ and Γ0. That is, if Λ is naturally covered by both Γ and Γ0, then it is also naturally covered by Γ Γ0.

The mixing and comixing operations on groups are commutative, associative, and idempotent in a natural way:

0 0 0 00 00 00 Proposition 2.12. Let Γ = hx1, . . . , xni, Γ = hx1, . . . , xni, and Γ = hx1, . . . , xni be groups on n speciﬁed generators. Then

(1) Γ Γ0 ' Γ0 Γ

(2) Γ Γ0 ' Γ0 Γ (3) (Γ Γ0) Γ00 ' Γ (Γ0 Γ00)

(4) (Γ Γ0) Γ00 ' Γ (Γ0 Γ00) (5) Γ Γ ' Γ Γ ' Γ,

where the isomorphisms are all given by sending each generator of the ﬁrst group to the corresponding generator of the second group.

43 Our next result is obvious and we have already used it implicitly in our earlier examples.

0 0 0 0 Proposition 2.13. Let Γ = hx1, . . . , xni and Γ = hx1, . . . , xni. Let zi = (xi, xi) so that

0 Γ Γ = hz1, . . . , zni. Then for each word w = zi1 ··· zik , the order of w is the least common multiple of the orders of x ··· x and x0 ··· x0 . i1 ik i1 ik

Corollary 2.14. The mix of two sggi’s is an sggi, and the mix of two string rotation groups is a string rotation group.

0 0 0 Proof. Suppose Γ = hx1, . . . , xni and Γ = hx1, . . . , xni are sggi’s; that is, the order of each

0 0 0 xi and xi is 2, and the orders of xixj and xixj are also both 2 whenever |i − j| ≥ 2. Let

0 0 zi = (xi, xi) so that Γ Γ = hz1, . . . , zni. Then by Proposition 2.13, each zi has order 2, and

0 zizj has order 2 when |i − j| ≥ 2; therefore, Γ Γ is an sggi. The proof for string rotation groups is analogous.

Corollary 2.15. The mix of two string pre-chi-groups is a string pre-chi-group.

0 0 0 Proof. Let Γ = hσ1, . . . , σn−1i and Γ = hσ1, . . . , σn−1i be string pre-chi-groups. In particular, they are string rotation groups, and Corollary 2.14 tells us that Γ Γ0 is a string rotation group. Now we just have to show that ΓΓ0 satisﬁes the weak intersection property. For each

0 0 T i, let βi = (σi, σi), and write Λ for Γ Γ = hβ1, . . . , βn−1i. Now, if βi1 ··· βik ∈ j6=i Λj, then that means that σ ··· σ ∈ T Γ and σ0 ··· σ0 ∈ T Γ0 . Since Γ and Γ0 both satisfy the i1 ik j6=i j i1 ik j6=i j weak intersection property, both of these intersections are trivial, and thus βi1 ··· βik = (, ), and we see that Λ = Γ Γ0 satisﬁes the weak intersection property as well.

We note that analogous statements about string pre-C-groups, string C-groups, and string chi-groups are not true. For example, [4, 3, 3] [3, 3, 4] is not a string C-group, and [4, 3, 3]+ [3, 3, 4]+ is not a string chi-group. In fact, we will see later that the mix of two string C-groups need not even be a string pre-C-group.

44 2.2 Variance groups and the structure of the mix

We observed earlier that the canonical projections from Γ × Γ0 to its factors restrict to epimorphisms from Γ Γ0 to its factors. It is natural, then, to study the kernel of these epimorphisms as a way of determining the structure of Γ Γ0.

0 0 0 0 Deﬁnition 2.16. Let Γ = hx1, . . . , xni and Γ = hx1, . . . , xni. Let f :Γ Γ → Γ and f 0 :Γ Γ0 → Γ0 be the natural epimorphisms. We denote ker f by X(Γ0|Γ) and call it the variance group of Γ0 with respect to Γ. Similarly, we denote ker f 0 by X(Γ|Γ0) and call it the variance group of Γ with respect to Γ0. In other words, X(Γ0|Γ) consists of the elements of Γ Γ0 of the form (1, w) with w ∈ Γ0, and X(Γ|Γ0) consists of the elements of Γ Γ0 of the form (w, 1) with w ∈ Γ.

Example 2.17. For any group Γ on n generators, the variance group X(Γ|Γ) is trivial.

Example 2.18. Suppose Γ is cyclic of order r and Γ0 is cyclic of order s, with r and s coprime. Then X(Γ|Γ0) is isomorphic to Γ and X(Γ0|Γ) is isomorphic to Γ0.

As with the mix of two groups, representing Γ and Γ0 as quotients of a group Λ gives us a natural way to determine the structure of X(Γ|Γ0).

Proposition 2.19. Let Γ = Λ/K and Γ0 = Λ/K0 be ﬁnitely generated groups, where Λ is a group with n speciﬁed generators. Then:

(1) X(Γ|Γ0) ' K0/(K ∩ K0) ' KK0/K and X(Γ0|Γ) ' K/(K ∩ K0) ' KK0/K0.

(2) Let g :Γ → Γ Γ0 and g0 :Γ0 → Γ Γ0 be the natural epimorphisms. Then ker g ' X(Γ|Γ0) and ker g0 ' X(Γ0|Γ). In particular, X(Γ|Γ0) and X(Γ0|Γ) can be viewed as normal subgroups of Γ and Γ0, respectively.

45 Proof. (1) We prove the claim for X(Γ|Γ0); the other case is analogous. Consider

f 0 :Γ Γ0 → Γ0.

We can rewrite this using quotients of Λ to

f 0 :Λ/(K ∩ K0) → Λ/K0,

and X(Γ|Γ0) is the kernel of f 0. Since f 0 sends a coset w(K ∩ K0) to the coset wK0, the kernel consists of those cosets w(K ∩ K0) with w ∈ K0. Thus the variance group X(Γ|Γ0) is K0/(K ∩ K0). Furthermore, since K and K0 are both normal in Λ, this is isomorphic to KK0/K.

(2) Again, we rewrite the domain and codomain of g using quotients of Λ, so that

g :Λ/K → Λ/KK0.

This homomorphism sends a coset wK to a coset wKK0, so the kernel consists of those cosets wK with w ∈ KK0. Thus we see that the kernel of g is KK0/K, and by the previous part, this is isomorphic to X(Γ|Γ0).

0 0 0 0 Corollary 2.20. Let Γ = hx1, . . . , xni and Γ = hx1, . . . , xni. Then X(Γ |Γ) is trivial if and

0 0 only if there is a well-deﬁned homomorphism π :Γ → Γ sending xi to xi.

Proof. Write Γ = F/K and Γ0 = F/K0, where F is the free group on n generators and K and K0 are the necessary normal subgroups. Then by Proposition 2.19, the group X(Γ0|Γ) is isomorphic to K/(K ∩ K0). Now, the required homomorphism from Γ to Γ0 exists if and

46 only if K ≤ K0, and in this case, K/(K ∩ K0) is trivial.

The fact that f 0 :Γ Γ0 → Γ0 and g :Γ → Γ Γ0 have isomorphic kernels is what allows us to use the comix of two polytopes to derive information about the mix. The following properties are immediate:

0 0 0 Proposition 2.21. Let Γ = hx1, . . . , xni and Γ = hx1, . . . , xni be ﬁnite. Then:

(1) Γ Γ0 is ﬁnite, and

|Γ Γ0| = |X(Γ|Γ0)| · |Γ0| = |X(Γ0|Γ)| · |Γ|.

(2) Γ Γ0 is ﬁnite, and |Γ| |Γ0| |Γ Γ0| = = . |X(Γ|Γ0)| |X(Γ0|Γ)|

(3) |Γ Γ0| · |Γ Γ0| = |Γ| · |Γ0|.

(4) |X(Γ|Γ0)| |Γ| = |X(Γ0|Γ)| |Γ0|

Intuitively speaking, the group X(Γ0|Γ) tells us something about how much Γ0 and Γ overlap. If Γ covers Γ0, then as we saw above, X(Γ0|Γ) is trivial. At the other extreme, if Γ Γ0 = Γ × Γ0, then Γ and Γ0 have no overlap, and X(Γ0|Γ) = Γ0.

In general, we will be much more interested in the mix of two groups than the comix. However, as we see in Proposition 2.21, knowing some information about the comix gives us information about the mix. This is important because it is often diﬃcult to calculate information about Γ Γ0 directly. Even if Γ and Γ0 have only a few hundred elements – a

47 fairly modest size for the automorphism groups of regular polytopes – their mix can have tens or hundreds of thousands of elements, and calculating the exact number using Todd- Coxeter enumeration can be practically impossible. On the other hand, for the groups we deal with, Γ Γ0 tends to be quite small and easily calculated by a computer or even by hand. Proposition 2.21 is thus invaluable for determining the size of Γ Γ0.

Even when Γ and Γ0 are both infinite, it often happens that Γ Γ0 is trivial. Just as with finite groups, the mix Γ Γ0 is equal to Γ × Γ0 whenever Γ Γ0 is trivial, even if Γ and Γ0 are infinite. In order to prove this, though, it is helpful to write Γ Γ0 as an explicit subgroup of Γ × Γ0.

0 0 0 0 Proposition 2.22. Let Γ = hx1, . . . , xni and Γ = hx1, . . . , xni. Let N = X(Γ|Γ ) and

0 0 0 0 0 0 N = X(Γ |Γ), and let g :Γ/N → Γ /N be the isomorphism sending xiN to xiN . Then

Γ Γ0 = {(u, v) ∈ Γ × Γ0 | g(uN) = vN 0}.

In particular, X(Γ|Γ0) × X(Γ0|Γ) is a normal subgroup of Γ Γ0.

Proof. First of all, by Proposition 2.19,

Γ/N ' Γ Γ0 ' Γ0/N 0,

so that g really is an isomorphism. Let f :Γ Γ0 → Γ and f 0 :Γ Γ0 → Γ0 be the natural epimorphisms, so that N 0 = ker f and N = ker f 0 (see Deﬁnition 2.16). Then the ﬁrst part follows directly by Lemma A.1 (Goursat’s Lemma; see the Appendix). For the last part, note that if u ∈ N and v ∈ N 0, then uN = N and vN 0 = N 0. Therefore, g(uN) = g(N) = N 0 = vN 0, so that (u, v) ∈ Γ Γ0. Then we see that N × N 0 is a subgroup of Γ Γ0, and since N is a normal subgroup of Γ and N 0 is a normal subgroup of Γ0, it

48 immediately follows that N × N 0(= X(Γ|Γ0) × X(Γ0|Γ)) is normal in Γ Γ0.

Now we are able to reﬁne Proposition 2.21 to include inﬁnite groups.

0 0 0 0 Theorem 2.23. Let Γ = hx1, . . . , xni and Γ = hx1, . . . , xni. If Γ Γ is ﬁnite of order k, then the index of Γ Γ0 in Γ × Γ0 is k.

Proof. If Γ and Γ0 are finite, this follows immediately from the third equation of Proposi- tion 2.21. Now let Γ and Γ0 be of arbitrary size, and set N = X(Γ|Γ0) and N 0 = X(Γ0|Γ), as in Proposition 2.22. If Γ Γ0 is finite of order k, then N 0 has index k in Γ0. For each fixed u ∈ Γ, Proposition 2.22 says that the element (u, v) is in Γ Γ0 if and only if g(uN) = vN 0 in Γ Γ0. In other words, having fixed u we can pick any v that lies in the same (corresponding) coset. Then since N 0 has index k in Γ0, the set of (u, v) ∈ Γ Γ0 for a fixed u has “index” k in {u} × Γ0. Therefore, letting u range over all elements of Γ, we see that Γ Γ0 has index k in Γ × Γ0.

Corollary 2.24. If Γ Γ0 is trivial, then Γ Γ0 = Γ × Γ0.

We have now examined the structure of Γ Γ0 in some detail. The next thing we will examine is how the variance group of the mix or comix depends on the component groups.

Proposition 2.25. Let Γ, ∆, and ∆0 be groups with n speciﬁed generators. Then Γ ∆0 naturally covers Γ∆ if and only if X(∆0|Γ) (viewed as a subgroup of Γ∆0) naturally covers X(∆|Γ) (viewed as a subgroup of Γ ∆).

Proof. Let F be the free group on n generators. Let M, K, and K0 be the normal subgroups of F such that Γ = F/M, ∆ = F/K, and ∆0 = F/K0. Then X(∆0|Γ) = MK0/K0 ' M/(M ∩ K0) and X(∆|Γ) = MK/K ' M/(M ∩ K), by Proposition 2.19. Furthermore, Γ ∆0 = F/(M ∩ K0), and Γ ∆ = F/(M ∩ K), by Proposition 2.5. Thus we see that Γ ∆0

49 naturally covers Γ ∆ if and only if M ∩ K0 ≤ M ∩ K, and this is true if and only if X(∆0|Γ) naturally covers X(∆|Γ).

Corollary 2.26. Let Γ, Γ0, and ∆ be groups with n speciﬁed generators. Then

X(∆ Γ0|Γ Γ0) ' X(∆|Γ Γ0).

Proof. Since (∆ Γ0) (Γ Γ0) = ∆ (Γ Γ0), each of these mixes naturally covers the other, and Proposition 2.25 tells us that the given variance groups naturally cover each other. Therefore they must be isomorphic.

Proposition 2.27. Let Γ, ∆, and ∆0 be groups with n speciﬁed generators. Then X(∆∆0|Γ) is isomorphic to a subgroup of X(∆|Γ) × X(∆0|Γ).

Proof. Let F be the free group on n generators, and let M, K, and K0 be the normal subgroups of F such that Γ = F/M, ∆ = F/K, and ∆0 = F/K0. Then by Proposition 2.19,

X(∆0|Γ) = MK0/K0 ' M/M ∩ K0,

X(∆|Γ) = MK/K ' M/M ∩ K,

and X(∆ ∆0|Γ) = M(K ∩ K0)/(K ∩ K0) ' M/(M ∩ K ∩ K0).

Now, consider the group homomorphism ϕ : M(K ∩ K0) → MK/K × MK0/K0 sending each element w to (wK, wK0). This is clearly well-deﬁned, and the kernel is K ∩ K0. Therefore, M(K ∩ K0)/(K ∩ K0) is isomorphic to the image of ϕ, which is what we wanted to show.

Proposition 2.28. Let Γ, Γ0, and ∆ be groups with n speciﬁed generators. Then X(∆|ΓΓ0) is a normal subgroup of X(∆|Γ) ∩ X(∆|Γ0).

50 Proof. Let F be the free group on n generators, and let M, M 0, and K be the normal subgroups of F such that Γ = F/M,Γ0 = F/M 0, and ∆ = F/K. Then we have X(∆|ΓΓ0) = (M ∩ M 0)K/K, X(∆|Γ) = MK/K, and X(∆|Γ0) = M 0K/K. It is clear that (M ∩ M 0)K is a subgroup of MK ∩ M 0K, and since (M ∩ M 0)K is a normal subgroup of F , it is also a normal subgroup of MK ∩ M 0K. Therefore, (M ∩ M 0)K/K is a normal subgroup of MK/K ∩ M 0K/K = (MK ∩ M 0K)/K.

Proposition 2.29. Let Γ, Γ0, ∆, and ∆0 be groups with n speciﬁed generators. Then X(∆ ∆0|Γ Γ0) is isomorphic to a subgroup of X(∆|Γ) × X(∆0|Γ0).

Proof. First, by Proposition 2.27, the group X(∆ ∆0|Γ Γ0) is isomorphic to a subgroup of X(∆|Γ Γ0) × X(∆0|Γ Γ0). Then Proposition 2.28 shows that X(∆|Γ Γ0) ≤ X(∆|Γ) and that X(∆0|Γ Γ0) ≤ X(∆0|Γ0).

Proposition 2.30. Let Γ, Γ0, and ∆ be groups with n speciﬁed generators. Then X(∆ Γ0|Γ Γ0) is isomorphic to a normal subgroup of X(∆|Γ).

Proof. First, Corollary 2.26 says that X(∆ Γ0|Γ Γ0) ' X(∆|Γ Γ0); then Proposition 2.28 tells us that this is a normal subgroup of X(∆|Γ).

2.3 Mixing polytopes

There are two ways we can apply our deﬁnition of mixing to semi-polytopes. First, suppose P and Q are regular n-semi-polytopes. Then by Corollary 2.14, the mix of Γ(P) with Γ(Q) is an sggi. Thus, we can build a regular ﬂagged poset from this group (as detailed in Section 1.3), and we call this poset the mix of P and Q and denote it P Q. Similarly, if P and Q are chiral or directly regular n-semi-polytopes, then Γ+(P) Γ+(Q) is a string pre-chi-group by

51 Corollary 2.15. Therefore, we can build a chiral or directly regular semi-polytope from this group, and we again call this semi-polytope the mix of P and Q and denote it P Q.

We similarly deﬁne the comix of P and Q to be the poset built from Γ(P) Γ(Q) or Γ+(P) Γ+(Q). However, these groups often have trivial generators, which makes it problematic to build a semi-polytope out of them.

Whenever P and Q are directly regular semi-polytopes, there are two ways we can deﬁne their mix. The following proposition says that these two deﬁnitions are compatible. As

0 0 + usual, we let Γ(P) = hρ0, . . . , ρn−1i, Γ(Q) = hρ0, . . . , ρn−1i,Γ (P) = hσ1, . . . , σn−1i, and

+ 0 0 Γ (Q) = hσ1, . . . , σn−1i.

Proposition 2.31. Let P and Q be directly regular n-semi-polytopes. Then (Γ(P)Γ(Q))+ = Γ+(P) Γ+(Q).

+ + Proof. Taking Γ(P) Γ(Q) = hα0, . . . , αn−1i and Γ (P) Γ (Q) = hβ1, . . . , βn−1i as above

0 0 + (where αi = (ρi, ρi) and βi = (σi, σi)), we see that (Γ(P)Γ(Q)) is generated by the elements

0 0 0 0 0 αi−1αi = (ρi−1, ρi−1)(ρi, ρi) = (ρi−1ρi, ρi−1ρi) = (σi, σi) = βi, where i = 1, . . . , n − 1.

Thus we see that whether we build P Q from Γ(P) Γ(Q) or Γ+(P) Γ+(Q), we get the same poset.

For our purposes, it is generally more useful to work with mixing via rotation groups, but we will occasionally need to mix via full automorphism groups. As we will see, the results governing the mixing of both types of groups are very similar.

We now consider a few examples.

Example 2.32. Let P = {p} and Q = {q}. Then P Q = {r}, where r is the least common multiple of p and q.

52 Example 2.33. Let P = {4, 4}(1,2), the chiral torus map with 5 square faces meeting 4 at each vertex [8]. Let Q = {4, 4}(2,1), the mirror image of P. Then P Q = {4, 4}(5,0), which is a regular torus map with 25 square faces meeting 4 at each vertex [1].

To apply the results of the previous section, we want to look at X(Γ+(P)|Γ+(Q)) if P and Q are chiral or directly regular, and at X(Γ(P)|Γ(Q)) if P and Q are regular. In the case where P and Q are both directly regular, we can use either one:

Proposition 2.34. Let P and Q be directly regular n-semi-polytopes. Then

X(Γ+(P)|Γ+(Q)) = X(Γ(P)|Γ(Q)).

Proof. Recall that for any chiral or directly regular n-semi-polytope P, we can write Γ+(P) =

+ + Wn /M for some normal subgroup M of Wn (see Equation 1.2). Similarly, for any regular n-semi-polytope P, we can write Γ(P) = Wn/M (Equation 1.1). Let M and K be the normal

+ subgroups of Wn such that Γ(P) = Wn/M and Γ(Q) = Wn/K. Then we must have Γ (P) =

+ + + + + Wn /M and Γ (Q) = Wn /K. Then by Proposition 2.19, we see that X(Γ (P)|Γ (Q)) ' MK/M ' X(Γ(P)|Γ(Q)).

Therefore, we make the following deﬁnition:

Definition 2.35. Let P and Q be n-semi-polytopes. If P and Q are both regular, we define X(P|Q) := X(Γ(P)|Γ(Q)). If instead P and Q are both chiral or directly regular, we define X(P|Q) := X(Γ+(P)|Γ+(Q)).

The following two propositions follow directly from Propositions 2.5 and 2.10.

Proposition 2.36. Let P and Q be chiral or directly regular n-semi-polytopes, with Γ+(P) =

+ + + + + + + Wn /M and Γ (Q) = Wn /K. Then Γ (P Q) ' Wn /(M ∩ K) and Γ (P) Γ (Q) ' + Wn /MK.

53 Proposition 2.37. Let P and Q be regular n-semi-polytopes, with Γ(P) = Wn/M and

Γ(Q) = Wn/K. Then Γ(P) Γ(Q) ' Wn/(M ∩ K) and Γ(P) Γ(Q) ' Wn/MK.

The mixing and comixing operations on polytopes are commutative, associative, and idempotent, just as they are on groups:

Proposition 2.38. Let P and Q be n-semi-polytopes such that they are both regular or they are both chiral or directly regular. Then:

(1) P Q = Q P

(2) P Q = Q P (3) (P Q) R = P (Q R)

(4) (P Q) R = P (Q R) (5) P P = P P = P.

Of course, even if P'Q, it may be the case that P Q 6' P. For example, if P is a chiral polytope, then P' P, but P P is not isomorphic to P.

Proposition 2.39. Let P1 and P2 both be regular n-polytopes, or both be chiral or directly regular n-polytopes. If P1 is of type {K1, L1} and P2 is of type {K2, L2}, then P1P2 is of type

{K1K2, L1L2}. Furthermore, if P1 is of type {p1, . . . , pn−1} and P2 is of type {q1, . . . , qn−1}, then P1 P2 is of type {`1, . . . , `n−1}, where `i = lcm(pi, qi) for i ∈ {1, . . . , n − 1}.

Proof. The ﬁrst part follows from Proposition 2.1, and the second part follows from Propo- sition 2.13.

54 2.4 Polytopality of the mix

Our main goal is to use the mixing operation to construct new polytopes. We have already seen that the mix of chiral or directly regular semi-polytopes is a semi-polytope. What about the mix of two regular semi-polytopes? In other words, suppose Γ and Γ0 are string pre-C-groups, and let Λ = Γ Γ0. Under what conditions can Λ fail to itself be a string pre-C-group?

We start by noting that Corollary 2.14 says that Λ is an sggi. Now, for each i, the group

0 Λi is a subgroup of Γi × Γi, and therefore

0 0 Λ0 ∩ · · · ∩ Λn−1 ≤ (Γ0 × Γ0) ∩ · · · ∩ (Γn−1 × Γn−1)

0 0 = (Γ0 ∩ · · · ∩ Γn−1) × (Γ0 ∩ · · · ∩ Γn−1)

= hi × h0i.

Thus we see that Λ has the empty intersection property. We can apply a similar argument to conclude that

\ 0 Λj ≤ hρii × hρii. j6=i Therefore, the only way that Λ can fail to be a string pre-C-group is if for some i, the

T 0 intersection j6=i Λj is equal to hρii × hρii.

0 Let αi = (ρi, ρi) so that Λ = hα0, . . . , αn−1i. By Proposition 1.14,

\ Λj = hα0, . . . , αii ∩ hαi, . . . , αn−1i. j6=i

Suppose that Λ does not have the weak intersection property. Then we can ﬁx a value for i

55 such that

(ρi, ) ∈ hα0, . . . , αii ∩ hαi, . . . , αn−1i.

0 0 Now, since (ρi, ) ∈ hα0, . . . , αii, there is a word (ρj1 ··· ρjm , ρj1 ··· ρjm ) that is equal to (ρi, ),

with 0 ≤ jk ≤ i for each jk. If m is even, then Γ has a relator of odd length. Similarly, if m is odd, then Γ0 has a relator of odd length. So either P(Γ) or P(Γ0) (or possibly both) must not be directly regular.

Of course, this is not new information; we already knew that the mix of two directly regular semi-polytopes is again a directly regular semi-polytope. But we can conclude something even stronger. Assuming again that m is even, then we in fact get a relator of odd length in hρ0, . . . , ρii. So not only is P(Γ) not directly regular, but in fact, its (i + 1)-faces are not directly regular. In general, in order for Λ to not have the weak intersection property, there

0 0 must be some i such that either hρ0, . . . , ρii or hρ0, . . . , ρii has a relator of odd length, and

0 0 such that hρi, . . . , ρn−1i or hρi, . . . , ρn−1i has a relator of odd length. In particular, since a string pre-C-group on 2 generators is dihedral, having only relators of even length, we must have 2 ≤ i ≤ n − 3, so that in particular, n ≥ 5.

Example 2.40. Let Γ0 = [2, 3, 3, 2], and let Γ be the quotient of [3, 4, 4, 3] with the extra

3 3 relations (ρ0ρ1ρ2) = (ρ4ρ3ρ2) = . A calculation with GAP [12] conﬁrms that Γ is a string C-group of order 576; in fact, Γ is the automorphism group of a regular polytope that is not directly regular. Another calculation shows that the intersection hα0, . . . , α2i ∩ hα2, . . . , α4i has 4 elements, and thus Γ Γ0 does not have the weak intersection property.

We now move on to determining when the mix of two polytopes is a polytope. In fact, in some cases, we can mix a polytope with a semi-polytope and still get a polytope:

Proposition 2.41. Let P be a chiral or directly regular n-polytope with facets isomorphic to K. Let Q be a chiral or directly regular n-semi-polytope with facets isomorphic to K0. If

56 K covers K0, then P Q is polytopal. The same is true if both P and Q are regular instead.

Proof. Since K covers K0, the facets of P Q are isomorphic to K. Therefore, the canonical projection from Γ+(P) Γ+(Q) → Γ+(P) is one-to-one on the subgroup of the facets, and by [3, Lemma 3.2], the group Γ+(P) Γ+(Q) has the intersection property. Therefore, P Q is a polytope.

In general, when we mix P and Q, we have to verify the full intersection property. But as we shall see, some parts of the intersection property are automatic. Recall that for a subset

I of {0, . . . , n − 1} and a string rotation group Γ = hσ1, . . . , σn−1i, we deﬁne

ΓI = hτi,j | i ≤ j and i − 1, j ∈ Ii,

where τi,j = σi ··· σj.

Proposition 2.42. Let I,J ⊂ {0, . . . , n − 1}. Suppose Λ = Γ Γ0, where Γ and Γ0 are string

0 chi-groups. Then ΛI ∩ ΛJ ⊆ ΓI∩J × ΓI∩J .

0 0 Proof. Let γ ∈ ΛI ∩ ΛJ . Write γ = (γ1, γ2). Then γ1 ∈ ΓI ∩ ΓJ and γ2 ∈ ΓI ∩ ΓJ . Now,

0 0 0 0 since Γ and Γ are string chi-groups, ΓI ∩ ΓJ = ΓI∩J and ΓI ∩ ΓJ = ΓI∩J . Therefore,

0 γ ∈ ΓI∩J × ΓI∩J .

0 0 Corollary 2.43. Let Λ = hβ1, . . . , βn−1i = Γ Γ , where Γ = hσ1, . . . , σn−1i and Γ =

0 0 hσ1, . . . , σn−1i are string chi-groups. Let i ≤ j < k ≤ `. Then hβi, . . . , βji∩hβk, . . . , β`i = hi.

Proof. Noting that Λ{i−1,...,j} = hβi, . . . , βji, we get the desired result by applying Proposi- tion 2.42 with I = {i − 1, . . . , j} and J = {k − 1, . . . , `}.

We can apply these same methods to string C-groups to obtain the following:

57 0 0 0 Proposition 2.44. Let Γ = hρ0, . . . , ρn−1i and Γ = hρ0, . . . , ρn−1i be string C-groups. Let

0 0 αi = (ρi, ρi), so that Γ Γ = hα0, . . . , αn−1i. Then

0 hαi | i ∈ Ii ∩ hαj | j ∈ Ji ≤ hρi | i ∈ I ∩ Ji × hρi | i ∈ I ∩ Ji.

In particular, if I and J are disjoint, then hαi | i ∈ Ii ∩ hαj | j ∈ Ji = hi.

Corollary 2.45. Let P and Q be chiral or directly regular polyhedra. Then P Q is a chiral or directly regular polyhedron.

Proof. In order for P Q to be a polyhedron (and not just a pre-polyhedron), it must satisfy the intersection property. For polyhedra, the only requirement is that hσ1i ∩ hσ2i = hi, which holds by Corollary 2.43.

Corollary 2.46. Let P and Q be regular polyhedra. Then P Q is a regular polyhedron.

Proof. We will show just one of the conditions for the intersection property; the others are

0 veriﬁed similarly. Proposition 2.44 says that hα0, α1i ∩ hα1, α2i ≤ hρ1i × hρ1i. Now, hα0, α1i is the group of a regular polygon, and in particular, it is directly regular. So neither (ρ1, )

0 nor (, ρ1) is in hα0, α1i. Thus we see that hα0, α1i ∩ hα1, α2i = hα1i, as desired.

Corollary 2.45 and Corollary 2.46 are extremely useful. In addition to telling us that the mix of any two polyhedra is a polyhedron, it makes it simpler to verify the polytopality of the mix of 4-polytopes, since the facets and vertex-ﬁgures of the mix are guaranteed to be polytopal.

There is another simple way of seeing that the mix of two polyhedra is polyhedron. A semi-polytope of rank 3 is always a polyhedron, so as long as ΓΓ0 is a string pre-C-group or a string pre-chi-group, the poset P(Γ Γ0) will be a polyhedron. Corollary 2.15 tells us that

58 the mix of two string pre-chi-groups is a string pre-chi-group, so that case is easily settled. The mix of string C-groups need not be a string pre-C-group in general, as we saw earlier, but in order to fail, we must have n ≥ 5. Thus the mix of two string pre-C-groups of rank 3 is again a string pre-C-group of rank 3.

We now prove some general results that work for polytopes in any rank. We start by presenting part of [3, Theorem 9.1], noting that the requirement that P and Q are ﬁnite are not needed for this part of the theorem.

Proposition 2.47. Let P be a chiral or directly regular n-polytope of type {p1, . . . , pn−1},

and let Q be a chiral or directly regular n-polytope of type {q1, . . . , qn−1}. If pi and qi are relatively prime for each i = 1, . . . , n − 1, then P Q is a chiral or directly regular n-polytope

+ + + of type {p1q1, . . . , pn−1qn−1}, and Γ (P Q) = Γ (P) × Γ (Q).

Note that Example 2.40, in which we mixed a directly regular polytope with a regular polytope that is not directly regular, shows that direct regularity is essential to Proposi- tion 2.47.

Example 2.48. Let P = {3, 4}, the octahedron, and let Q = {4, 3}, the cube. Then P Q is a polyhedron of type {12, 12} with Γ+(P) Γ+(Q) = Γ+(P) × Γ+(Q).

We now work on a generalization of Proposition 2.47.

0 0 0 Lemma 2.49. Let Γ = hσ1, . . . , σn−1i and Γ = hσ1, . . . , σn−1i be string chi-groups. Let

0 0 βi = (σi, σi), so that Γ Γ = hβ1, . . . , βn−1i. Then

0 0 hβ1, . . . , βn−2i ∩ hβ2, . . . , βn−1i ≤ hσ2, . . . , σn−2i × hσ2, . . . , σn−2i, with equality if

0 0 hβ1, . . . , βn−2i = hσ1, . . . , σn−2i × hσ1, . . . , σn−2i

59 and

0 0 hβ2, . . . , βn−1i = hσ2, . . . , σn−1i × hσ2, . . . , σn−1i.

Proof. We have that

hβ1, . . . , βn−2i ∩ hβ2, . . . , βn−1i

0 0 0 0 = h(σ1, σ1),..., (σn−2, σn−2)i ∩ h(σ2, σ2),..., (σn−1, σn−1)i

0 0 0 0 ≤ (hσ1, . . . , σn−2i × hσ1, . . . , σn−2i) ∩ (hσ2, . . . , σn−1i × hσ2, . . . , σn−1i)

0 0 0 0 = (hσ1, . . . , σn−2i ∩ hσ2, . . . , σn−1i) × (hσ1, . . . , σn−2i ∩ hσ2, . . . , σn−1i)

0 0 = hσ2, . . . , σn−2i × hσ2, . . . , σn−2i, where the last line follows from the fact that Γ and Γ0 are string chi-groups.

Applying the same technique to string C-groups yields the following result.

0 0 0 Lemma 2.50. Let Γ = hρ0, . . . , ρn−1i and Γ = hρ0, . . . , ρn−1i be string C-groups. Let

0 0 αi = (ρi, ρi), so that Γ Γ = hα0, . . . , αn−1i. Then

0 0 hα0, . . . , αn−2i ∩ hα1, . . . , αn−1i ≤ hρ1, . . . , ρn−2i × hρ1, . . . , ρn−2i, with equality if

0 0 hα0, . . . , αn−2i = hρ0, . . . , ρn−2i × hρ0, . . . , ρn−2i and

0 0 hα1, . . . , αn−1i = hρ1, . . . , ρn−1i × hρ1, . . . , ρn−1i.

Proposition 2.51. Let P be a chiral or directly regular n-polytope of type {K, L}, and let Q be a chiral or directly regular n-polytope of type {K0, L0}. Suppose K K0 and L L0 are polytopes. Further, suppose that K has vertex-ﬁgures M and that K0 has vertex ﬁgures M0,

60 and suppose that Γ+(M M0) = Γ+(M) × Γ+(M0). Then P Q is polytopal. The same is true if we suppose that P and Q are both regular and that Γ(M M0) = Γ(M) × Γ(M0).

Proof. We start with the case where P and Q are chiral or directly regular. Let Γ+(P) =

+ 0 0 0 + hσ1, . . . , σn−1i and let Γ (Q) = hσ1, . . . , σn−1i. Let βi = (σi, σi), so that Γ (P Q) =

0 0 hβ1, . . . , βn−1i. The facets of P Q are K K , and the vertex ﬁgures are L L , both of which are polytopal. Thus, to verify the polytopality of P Q, it suﬃces to check that

hβ1, . . . , βn−2i ∩ hβ2, . . . , βn−1i = hβ2, . . . , βn−2i ([30, Lemma 10]). From Lemma 2.49 we get that

0 0 hβ1, . . . , βn−2i ∩ hβ2, . . . , βn−1i ≤ hσ2, . . . , σn−2i × hσ2, . . . , σn−2i.

The right hand side is just Γ+(M) × Γ+(M0), and since this is equal to Γ+(M M0) = hβ2, . . . , βn−2i by assumption, the result follows.

When P and Q are both regular, then to verify polytopality it suﬃces to check that hα0, . . . , αn−2i ∩ hα1, . . . , αn−1i = hα1, . . . , αn−2i ([23, 2E16(a)]). Then the argument used above, along with Lemma 2.50, proves that the mix is polytopal.

Theorem 2.52. Let P be a chiral or directly regular n-polytope of type {K, L}. Let Q be a chiral or directly regular n-polytope of type {K0, L0}. Let M be the vertex-ﬁgures of K, and let M0 be the vertex-ﬁgures of K0. Suppose that K K0 and L L0 are polytopal, and suppose that Γ+(K K0) = Γ+(K) × Γ+(K0) and Γ+(L L0) = Γ+(L) × Γ+(L0). Then P Q is polytopal if and only if Γ+(M M0) = Γ+(M) × Γ+(M0). The same is true if P and Q are both regular, and if all rotation groups are replaced with full automorphism groups.

+ + 0 0 0 Proof. Let Γ (P) = hσ1, . . . , σn−1i and let Γ (Q) = hσ1, . . . , σn−1i. Let βi = (σi, σi), so that

+ + 0 + + 0 Γ (PQ) = hβ1, . . . , βn−1i. Proposition 2.51 proves that if Γ (MM ) = Γ (M)×Γ (M ),

61 then P Q is polytopal. Conversely, suppose P Q is polytopal. Since P Q is polytopal,

+ 0 Γ (M M ) = hβ2, . . . , βn−2i = hβ1, . . . , βn−2i ∩ hβ2, . . . , βn−1i.

0 0 Now, by Lemma 2.49, the right hand side is equal to hσ2, . . . , σn−2i × hσ2, . . . σn−2i, which is Γ+(M) × Γ+(M0). The proof for regular polytopes is essentially the same.

These propositions show us how P Q can fail to be polytopal, even if it has polytopal facets and vertex-ﬁgures. In particular, if Γ+(M M0) is “small” compared to Γ+(M) × Γ+(M0), then we need the inequality in the second line to be “loose”. We will make these notions more precise later.

We are now able to generalize Proposition 2.47.

Theorem 2.53. Let P be a chiral or directly regular n-polytope of type {p1, . . . , pn−1}, and let Q be a chiral or directly regular n-polytope of type {q1, . . . , qn−1}. If pi and qi are relatively prime for each i = 2, . . . , n − 2 (but not neccessarily for i = 1 or i = n − 1), then P Q is a chiral or directly regular n-polytope. Furthermore, if n ≥ 4, then Γ+(P) Γ+(Q) is a subgroup of index 4 or less in Γ+(P) × Γ+(Q).

Proof. We prove the claim by induction. The claim is trivially true for n ≤ 2, and Corol- lary 2.45 establishes the case n = 3. Now, suppose the claim is true for (n − 1)-polytopes, and let P and Q be n-polytopes satisfying the given conditions. Suppose that P is of type {K, L} with the vertex-ﬁgures of K being M. Similarly, suppose that Q is of type {K0, L0} with the vertex-ﬁgures of K0 being M0. Now, the facets of P Q are isomorphic to K K0.

0 The facet K has type {p1, . . . , pn−2}, and K has type {q1, . . . , qn−2}. Now, pi and qi are relatively prime for each i = 2, . . . , n − 2, so by the inductive hypothesis, K K0 is polytopal. Similarly, L L0 is polytopal. Then by Proposition 2.51, to prove polytopality of P Q,

62 it suﬃces to show that Γ+(M M0) = Γ+(M) × Γ+(M0). We have that M is of type

0 {p2, . . . , pn−2}, and M is of type {q2, . . . , qn−2}. Then Proposition 2.47 tells us that since

+ 0 + + 0 each pi is relatively prime to qi, we do in fact have that Γ (M M ) = Γ (M) × Γ (M ). Therefore P Q is polytopal.

For the last part, note that if n ≥ 4, then the generators σ2, . . . , σn−2 are all trivial in

+ + 0 + + Γ (P) Γ (Q); in fact, σi = σi in Γ (P) Γ (Q), and thus each σi has order dividing pi 2 2 and qi. Combined with the relations (σ1σ2) = and (σn−2σn−1) = , this forces σ1 and

+ + σn−1 to have order dividing 2. Therefore, Γ (P) Γ (Q) has order at most 4, and thus Γ+(P) Γ+(Q) has index at most 4 in Γ+(P) × Γ+(Q).

Example 2.54. Let P = {3, 3, 3}, the 4-simplex, and let Q = {3, 4, 3}, the 24-cell. Then P Q is a directly regular polytope. Furthermore, Γ+(P) Γ+(Q) is trivial, and thus Γ+(P) Γ+(Q) = Γ+(P) × Γ+(Q).

We conclude this section with two more general results.

Theorem 2.55. Let P be a chiral or directly regular n-polytope of type {K, L}. Let Q be a chiral or directly regular n-polytope of type {K0, L0}. Let M be the vertex-ﬁgures of K, and let M0 be the vertex-ﬁgures of K0. If

|Γ+(M M0)| > |Γ+(K K0)| · |Γ+(L L0)|, then P Q is not polytopal.

Proof. First, note that

Γ+(M) × Γ+(M0) = (Γ+(K) × Γ+(K0)) ∩ (Γ+(L) × Γ+(L0)).

63 Suppose P Q is polytopal. Then we know from Theorem 2.52 that

Γ+(K K0) ∩ Γ+(L L0) = Γ+(M M0).

Now, by Theorem 2.23 and Proposition A.3 in the Appendix,

|Γ+(M M0)| = [Γ+(M) × Γ+(M0):Γ+(M M0)] = [(Γ+(K) × Γ+(K0)) ∩ (Γ+(L) × Γ+(L0)) : Γ+(K K0) ∩ Γ+(L L0)]

≤ [Γ+(K) × Γ+(K0):Γ+(K K0)] · [Γ+(L) × Γ+(L0):Γ+(L L0)]

= |Γ+(K K0)| · |Γ+(L L0)|.

Proposition 2.56. Let P and Q be chiral or directly regular n-polytopes. Suppose P is of type {p1, . . . , pn−1}, and that Q is of type {q1, . . . , qn−1}. Let ri = gcd(pi, qi) for i ∈

{1, . . . , n − 1}. If there is an integer m ∈ {2, . . . , n − 2} such that rm−1 = rm+1 = 1 and rm ≥ 3, then P Q is not a polytope.

+ + 0 0 0 Proof. Let Γ (P) = hσ1, . . . , σn−1i,Γ (Q) = hσ1, . . . , σn−1i, and βi = (σi, σi) for each i ∈ {1, . . . , n − 1}. To show that P Q is not polytopal, it suﬃces to show that

hβm−1, βmi ∩ hβm, βm+1i= 6 hβmi.

Now, since pm−1 and qm−1 are relatively prime, there is an integer k such that kpm−1 ≡ 1

0 (mod qm−1). Then since the order of σm−1 is pm−1 and the order of σm−1 is qm−1, we see that

kpm−1 kpm−1 0 kpm−1 0 βm−1 = (σm−1 , (σm−1) ) = (, σm−1),

64 and therefore

kpm−1 2 2 0 0 2 2 (βm−1 βm) = (σm, (σm−1σm) ) = (σm, ),

0 0 2 2 since we have (σiσi+1) = for any i ∈ {1, . . . , n − 2}. Thus, (σm, ) ∈ hβm−1, βmi. Similarly,

0 0 there is an integer k such that k pm+1 ≡ 1 (mod qm+1), and thus

0 k pm+1 2 2 0 0 2 2 (βmβm+1 ) = (σm, (σmσm+1) ) = (σm, ).

2 Therefore, (σm, ) ∈ hβm, βm+1i as well. So we see that

2 (σm, ) ∈ hβm−1, βmi ∩ hβm, βm+1i.

k 2 0 k Now, since rm ≥ 3, there is no integer k such that σm = σm and (σm) = . Therefore,

2 (σm, ) 6∈ hβmi, and that proves the claim.

65 Chapter 3

Measuring invariance

In this chapter, we develop the theory of internal and external invariance of polytopes; the distinction is similar to that between inner and outer automorphisms of a group. Our framework provides a uniﬁed way to measure the extent to which a polytope is chiral, self- dual, or self-Petrie. Our goal is then to understand how the variance of P Q depends on P and Q, and to use this knowledge to build polytopes with speciﬁed symmetries.

3.1 External and internal invariance

Our study of invariance starts with the symmetries of Wn = hρ0, . . . , ρn−1i, the automorphism group of the universal n-polytope. Let P be a regular n-semi-polytope with base ﬂag Φ.

j1,...,jk i,j1,...,jk Recall that Wn acts on the ﬂags of P by Φ ρi = Φ . If M is the stabilizer of the base ﬂag Φ under this action, then M is normal in Wn and Γ(P) = Wn/M.

ϕ Suppose ϕ is in Aut(Wn), the group of group automorphisms of Wn, and deﬁne P to

ϕ be the ﬂagged poset built from Wn/ϕ(M). If ϕ ﬁxes M, then P and P are naturally isomorphic, and as before we shall consider them equal. On the other hand, if ϕ(M) 6= M,

66 then the polytopes P and Pϕ are distinct, and they need not be isomorphic, even though ϕ induces an isomorphism of their automorphism groups.

Similarly, if P is a chiral or directly regular n-semi-polytope with base ﬂag Φ, then

+ j1,...,jk i,i−1,j1,...,jk Wn = hσ1, . . . , σn−1i acts on that ﬂags of P by Φ σi = Φ . If M is the stabilizer

+ + + of the base ﬂag under this action, then M is normal in Wn and Γ (P) = Wn /M. Now,

+ ϕ + taking ϕ ∈ Aut(Wn ), we similarly deﬁne P to be the ﬂagged poset built from Wn /ϕ(M).

Deﬁnition 3.1. Let P be a regular or chiral n-semi-polytope. Let ϕ be a group automorphism

+ ϕ of Wn or Wn (whichever is appropriate), and let P be deﬁned as above.

(1) If P = Pϕ, we say that P is internally ϕ-invariant.

(2) If P= 6 Pϕ, we say that P is internally ϕ-variant.

(3) If P'Pϕ, we say that P is externally ϕ-invariant.

(4) If P 6' Pϕ, we say that P is externally ϕ-variant.

Of course, if a semi-polytope is internally ϕ-invariant, it must also be externally ϕ- invariant. Similarly, if a semi-polytope is externally ϕ-variant, it must also be internally ϕ-variant.

As usual, there are two ways to work with a directly regular polytope: via its automorphism group or via its rotation group. These are compatible in a natural way:

+ + + + Proposition 3.2. Let ϕ ∈ Aut(Wn) such that ϕ(Wn ) = Wn . Let ϕ ∈ Aut(Wn ) be the automorphism induced by ϕ. If P is a directly regular semi-polytope, then Pϕ = Pϕ+ .

+ + ϕ Proof. Let Γ (P) = Wn /M. Since P is directly regular, Γ(P) = Wn/M. Now, P is the directly regular semi-polytope with group Wn/ϕ(M). This polytope has rotation group

+ + + ϕ+ Wn /ϕ(M) = Wn /ϕ (M), which is the rotation group of P .

67 We now consider several applications.

Example 3.3. Let P be a regular semi-polytope with Γ(P) = Wn/M, and let χw ∈ Aut(Wn)

be conjugation by w ∈ Wn. Then since M is normal in Wn, χw ﬁxes M. Therefore, every

regular polytope is internally χw-invariant.

+ −1 2 Example 3.4. Let χ be the automorphism of Wn that sends σ1 to σ1 , σ2 to σ1σ2, and ﬁxes

all other generators σi. This is the automorphism induced by conjugation by ρ0 in Wn. If P is directly regular, then it is internally χ-invariant. If P is chiral instead, then Pχ is the enantiomorphic form of P (that is, Pχ = P). Since Pχ 'P but Pχ 6= P, we see that chiral semi-polytopes are externally χ-invariant, but internally χ-variant.

Example 3.5. Let δ be the automorphism of Wn that sends each ρi to ρn−i−1, and let P be a regular n-semi-polytope. Then Pδ is the dual of P (and indeed, our notation for the dual was chosen in anticipation of this fact). The semi-polytope P is externally δ-invariant if and only if it is self-dual. Every regular self-dual semi-polytope has a duality that ﬁxes the base ﬂag while reversing the order [23], and therefore if P is regular and self-dual, the

δ polytopes P and P have the same ﬂag-stabilizer in Wn. Thus we see that a regular self-dual semi-polytope is always internally δ-invariant.

+ + −1 Example 3.6. Let δ be the automorphism of Wn that sends each σi to σn−i. This is the automorphism induced by δ in the previous example (and by an abuse of notation, we

+ frequently use δ to denote this automorphism of Wn as well). Then a directly regular or chiral semi-polytope P is externally δ+-invariant if and only if it is self-dual. If P is properly self- dual (i.e., if there is a duality that ﬁxes the base ﬂag), then it is internally δ+-invariant; otherwise it is internally δ+-variant.

Example 3.7. Let π be the automorphism of W3 that sends ρ0 to ρ0ρ2 and ﬁxes every other

π ρi. Let P be a regular semi-polytope. Then P is the Petrie dual of P.

68 We note here that, in general, Wn has an automorphism ψ that sends ρ2 to ρ0ρ2 while

ﬁxing every other ρi [18]. In rank 3, that automorphism is δπδ, and when applied to a polyhedron it produces a polyhedron except in rare cases. On the other hand, in ranks 4

ψ p and greater, P frequently fails to be a polytope. To see why, consider the relation (ρ2ρ3) =

p ψ in Γ(P), which induces the relation (ρ0ρ2ρ3) = in Γ(P ). Since ρ0 commutes with ρ2ρ3,

−p p p we get that ρ0 = (ρ2ρ3) . If p is odd, then we get that ρ0 = (ρ2ρ3) , a clear violation of the intersection property. In fact, even the empty intersection property is violated, so we do not even get a semi-polytope. Due to this and other diﬃculties, we shall not treat the automorphism ψ here.

We will now explore the connection between invariance and polytope covers.

Proposition 3.8. Let P be a chiral or regular n-semi-polytope, and let ϕ be an automorphism

+ of Wn or Wn , as appropriate. Suppose Q is a internally ϕ-invariant n-semi-polytope that covers P. Then Q covers Pϕ.

Proof. Suppose P and Q are both regular; the proof is essentially the same in the other cases. We have Γ(P) = Wn/M and Γ(Q) = Wn/K for some normal subgroups M and K of Wn. Since Q covers P, we get that K ≤ M, by Proposition 1.23. Then ϕ(K) ≤ ϕ(M) as well, and since Q is internally ϕ-invariant, ϕ(K) = K. Therefore K ≤ ϕ(M), and so Q covers Pϕ.

Corollary 3.9. Let P be a chiral or regular n-semi-polytope, and let ϕ be an automorphism

+ of Wn or Wn , as appropriate. Suppose that ϕ has ﬁnite order k, and that Q is an internally ϕ-invariant n-semi-polytope that covers P. Then Q covers P Pϕ · · · Pϕk−1 .

Proof. Repeated application of Proposition 3.8 shows that Q covers Pϕ, Pϕ2 ,..., and Pϕk−1 . Therefore, it covers their mix.

69 As we shall see shortly, the mix P Pϕ · · · Pϕk−1 is actually the minimal internally ϕ-invariant cover of P. As such, we make the following deﬁnition.

Deﬁnition 3.10. Let P be a chiral or regular n-semi-polytope, and let ϕ be an automorphism

+ ϕ ϕ ϕk−1 of Wn or Wn (as appropriate) of ﬁnite order k. Then we deﬁne P to be P P · · ·P .

Proposition 3.11. Let P be a chiral or regular n-semi-polytope, and let ϕ be an automor-

+ ϕ phism of Wn or Wn (as appropriate) of order ﬁnite k. Then P is the minimal internally ϕ-invariant cover of P.

Proof. Since Pϕk = P, it is clear that (Pϕ)ϕ = Pϕ. So Pϕ is internally ϕ-invariant. By Corollary 3.9, every internally ϕ-invariant cover of P must cover Pϕ as well. Thus it follows that Pϕ is minimal.

In the rest of this chapter, in order to avoid duplication, we will usually assume that ϕ is

+ an automorphism of Wn , and that any polytopes we deal with are chiral or directly regular. Note, however, that the deﬁnitions below all still make sense if we work with automorphisms of Wn instead and assume that our polytopes are regular.

+ ϕ Given an automorphism ϕ of Wn , we can consider the variance groups X(P|P ) and

ϕ + + X(P |P). By Proposition 2.19, if Γ (P) = Wn /M, then the former is isomorphic to Mϕ(M)/M, and the latter is isomorphic to Mϕ(M)/ϕ(M). Since M ' ϕ(M), the groups X(P|Pϕ) and X(Pϕ|P) are isomorphic. We make the following deﬁnition:

Deﬁnition 3.12. Let P be a chiral or directly regular semi-polytope of rank n. Let ϕ ∈

+ Aut(Wn ). We deﬁne

ϕ + + ϕ Xϕ(P) := X(P|P ) := X(Γ (P)|Γ (P )), and we call this the (internal) ϕ-variance group of P; we usually omit the word ‘internal’.

70 + + ϕ In other words, Xϕ(P) is the kernel of the natural epimorphism from Γ (P) Γ (P ) to Γ+(Pϕ) (and also the kernel of the natural epimorphism from Γ+(P) to Γ+(P) Γ+(Pϕ)). + + ϕ + ϕ Example 3.13. If P is internally ϕ-invariant, then Γ (P) Γ (P ) ' Γ (P ), and Xϕ(P) is trivial.

Example 3.14. If P is a chiral polytope, then its chirality group X(P) is deﬁned to be the kernel of the natural epimorphism from Γ+(P) Γ+(P) to Γ+(P) [9]. Taking ϕ = χ (as deﬁned in Example 3.4), we see that the chirality group is the same thing as the χ-invariance

group Xχ(P).

ϕ The group Xϕ(P) gives us a measure of how diﬀerent P is from P . At one extreme,

Xϕ(P) might be trivial, in which case P is internally ϕ-invariant. At the other extreme,

+ Xϕ(P) might coincide with the whole group Γ (P); in that case, we say that P is totally (internally) ϕ-variant. (Again, we usually drop the word ‘internally’ for brevity.)

3.2 Variance of the mix

Using the tools developed in the previous section, our goal now is to determine how Xϕ(PQ)

depends on Xϕ(P) and Xϕ(Q). One of the primary motivations is to determine when P Q is chiral, and indeed, many of the results presented here generalize results from [9] and [3]. We start with a simple result.

Proposition 3.15. Let P and Q be chiral or directly regular n-semi-polytopes, and let ϕ ∈

+ ϕ ϕ ϕ Aut(Wn ). Then (P Q) = P Q .

+ + + + + ϕ + Proof. Write Γ (P) = Wn /M and Γ (Q) = Wn /K. Then Γ ((P Q) ) = Wn /ϕ(M ∩ K),

+ ϕ ϕ + + while Γ (P Q ) = Wn /(ϕ(M) ∩ ϕ(K)). Since ϕ ∈ Aut(Wn ), the subgroups ϕ(M ∩ K) and ϕ(M) ∩ ϕ(K) are equal, which proves the result.

71 Lemma 3.16. Let P and Q be chiral or directly regular n-semi-polytopes, and let ϕ ∈

+ ϕ Aut(Wn ) have ﬁnite order. Then P Q is internally ϕ-invariant if and only if it covers P and Qϕ.

Proof. If P Q is internally ϕ-invariant, then by Corollary 3.9, it covers (P Q)ϕ. Further- more, by Proposition 3.15, the latter polytope is equal to Pϕ Qϕ, which covers both Pϕ and Qϕ. Conversely, if P Q covers Pϕ and Qϕ, then it covers (P Q)ϕ, which itself covers P Q. Then we must have that (P Q)ϕ = P Q; that is, P Q must be internally ϕ-invariant.

We now give the main theorem of this section, from which we will be able to derive many results on the ϕ-invariance of P Q.

Theorem 3.17. Let P and Q be chiral or directly regular n-semi-polytopes, and let ϕ ∈

+ Aut(Wn ) have ﬁnite order. Suppose P Q is internally ϕ-invariant. Then there is a natural epimorphism from Γ+(P)Γ+(Q) to Γ+(P)Γ+(Pϕ), and it restricts to an epimorphism from

ϕ ϕ + + X(Q|P) to Xϕ(P ) = X(P |P). Similarly, the natural epimorphism from Γ (P) Γ (Q)

+ + ϕ ϕ ϕ to Γ (Q) Γ (Q ) restricts to an epimorphism from X(P|Q) to Xϕ(Q ) = X(Q |Q).

+ + + + Proof. Let Γ (P) = Wn /M and Γ (Q) = Wn /K. By Lemma 3.16, since P Q is internally ϕ-invariant, it covers Pϕ, which covers P Pϕ. Therefore, Γ+(P) Γ+(Q) naturally covers Γ+(P) Γ+(Pϕ). Since

+ + + Γ (P) Γ (Q) = Wn /(M ∩ K) and

+ + ϕ + Γ (P) Γ (P ) = Wn /(M ∩ ϕ(M)), this means that M ∩ K ≤ M ∩ ϕ(M). Thus, the group M/(M ∩ K) naturally covers M/(M ∩ϕ(M)). By Proposition 2.19, the former is the subgroup X(Q|P) of Γ+(P)Γ+(Q),

72 and the latter is the subgroup X(Pϕ|P) of Γ+(P) Γ+(Pϕ). The result then follows by symmetry.

When ϕ is an involution, the converse of Theorem 3.17 also holds:

Theorem 3.18. Let P and Q be chiral or directly regular n-semi-polytopes. Let ϕ ∈

+ ϕ Aut(Wn ) have order 2. If X(Q|P) naturally covers Xϕ(P ) and X(P|Q) naturally cov-

ϕ + ers Xϕ(Q ) (as quotients of Wn ), then P Q is internally ϕ-invariant.

ϕ + + + Proof. Suppose X(Q|P) naturally covers Xϕ(P ). Writing Γ (P) = Wn /M and Γ (Q) =

+ + + Wn /K, we again get that M ∩ K ≤ M ∩ ϕ(M). Therefore, Γ (P) Γ (Q) naturally covers Γ+(P) Γ+(Pϕ). So P Q covers P Pϕ, which is equal to Pϕ since ϕ has order 2. Similarly, P Q naturally covers Q Qϕ = Qϕ. Then by Lemma 3.16, P Q is internally ϕ-invariant.

+ Example 3.19. The automorphism χ of Wn (Example 3.4) has order 2, and a chiral or directly regular semi-polytope P is directly regular if and only if it is internally χ-invariant. Therefore, Theorem 3.17 and Theorem 3.18 tell us that P Q is directly regular if and only if it covers P Pχ and Q Qχ (i.e., P P and Q Q).

Now we will see several ways to apply Theorem 3.17.

Theorem 3.20. Let P and Q be ﬁnite chiral or directly regular n-semi-polytopes, and let

+ ϕ ∈ Aut(Wn ) have ﬁnite order. If P Q is internally ϕ-invariant, then |Xϕ(P)| divides

+ + + + + + |Γ (Q)|/|Γ (P) Γ (Q)| and |Xϕ(Q)| divides |Γ (P)|/|Γ (P) Γ (Q)|.

ϕ Proof. If P Q is internally ϕ-invariant, then by Theorem 3.17, X(Q|P) covers Xϕ(P ). If P

ϕ ϕ and Q are both ﬁnite, then so are X(Q|P) and Xϕ(P ), and thus |Xϕ(P )| divides |X(Q|P)|.

ϕ + + Furthermore, |Xϕ(P )| = |Xϕ(P)| and, by Proposition 2.21, |X(Q|P)| = |Γ (Q)|/|Γ (P) Γ+(Q)|. The result then follows by symmetry.

Corollary 3.21. Let P and Q be chiral or directly regular n-semi-polytopes, and let ϕ ∈

+ Aut(Wn ) have finite order. Suppose that P has infinite ϕ-variance group Xϕ(P) and that Q is finite. Then P Q is internally ϕ-variant (that is, not internally ϕ-invariant).

Proof. Since Q is ﬁnite, so is X(Q|P), which is isomorphic to a subgroup of Q. Then there

ϕ is no epimorphism from the ﬁnite group X(Q|P) to the inﬁnite group Xϕ(P ) ' Xϕ(P), and thus by Theorem 3.17, P Q must be internally ϕ-variant.

Corollary 3.22. Let P and Q be chiral or directly regular n-semi-polytopes, and let ϕ ∈

+ Aut(Wn ) have ﬁnite order. Suppose P is internally ϕ-variant and that Q has a rotation

+ + group Γ (Q) that is simple. If Xϕ(P) is not isomorphic to Γ (Q), then P Q is internally ϕ-variant.

ϕ Proof. Theorem 3.17 says that if PQ is internally ϕ-invariant, then X(Q|P) covers Xϕ(P ).

ϕ + Now, since P has a nontrivial ϕ-variance group, Xϕ(P ) is nontrivial, and since Γ (Q) is simple, the normal subgroup X(Q|P) of Γ+(Q) is either trivial or the whole group Γ+(Q).

ϕ + The only way for X(Q|P) to cover Xϕ(P ) is for X(Q|P) to be Γ (Q), and then the only

ϕ nontrivial group it covers is itself. Therefore, if Xϕ(P) (and thus Xϕ(P )) is not isomorphic

+ ϕ to Γ (Q), then X(Q|P) cannot cover Xϕ(P ), and the mix P Q is internally ϕ-variant by Theorem 3.17.

Corollary 3.22 has a natural analogue using regular semi-polytopes, where Q has a full automorphism group that is simple. Both forms are particularly useful since there are a number of papers that address which simple groups arise as the automorphism groups or rotation subgroups of abstract polytopes. For a small sampling, see [10, 11, 14, 22].

74 We see that for ﬁnite polytopes, there are several simple tests that we can apply to determine whether P Q is internally ϕ-invariant. We would like to extend the results to the mix of three or more polytopes. In order to do that, however, we need to know more

ϕ about the size of Xϕ(P Q) = X(P Q|(P Q) ). We start with some simple observations.

Proposition 3.23. Let P and Q be chiral or directly regular n-polytopes, and let ϕ ∈

+ Aut(Wn ) have ﬁnite order. Then Xϕ(P Q) is isomorphic to a subgroup of Xϕ(P)×Xϕ(Q).

ϕ ϕ Proof. Apply Proposition 2.29, noting that Xϕ(P Q) = X(P Q|P Q ), Xϕ(P) =

ϕ ϕ X(P|P ), and Xϕ(Q) = X(Q|Q ).

Proposition 3.24. Let P and Q be chiral or directly regular n-polytopes, and let ϕ ∈

+ Aut(Wn ) have ﬁnite order. Suppose that Q is internally ϕ-invariant. Then Xϕ(P Q) is a

normal subgroup of Xϕ(P).

Proof. Since Q is internally ϕ-invariant, Γ+(Qϕ) = Γ+(Q). Therefore,

+ + + ϕ + Xϕ(P Q) = X(Γ (P) Γ (Q)|Γ (P ) Γ (Q)),

and we can apply Proposition 2.30 to get the desired result.

Next we generalize Theorem 3.20 to ﬁnd a lower bound for |Xϕ(P Q)|.

Theorem 3.25. Let P and Q be ﬁnite chiral or directly regular n-semi-polytopes, and let ϕ ∈

+ Aut(Wn ) have ﬁnite order. Then |Xϕ(P Q)| is an integer multiple of |Xϕ(P)|/|X(Q|P)|.

Proof. Since P Q Pϕ Qϕ = (P Q) (P Q)ϕ,

75 the latter covers P Pϕ. Therefore, |Γ+(P) Γ+(Pϕ)| divides |Γ+(P Q) Γ+((P Q)ϕ)|.

+ The former has size |Γ (P)| · |Xϕ(P)|, by Proposition 2.21, while the latter has size

+ + |Γ (P Q)| · |Xϕ(P Q)| = |Γ (P)| · |X(Q|P)| · |Xϕ(P Q)|.

Thus we see that |Xϕ(P)| divides |X(Q|P)| · |Xϕ(P Q)|, and thus |Xϕ(P Q)| is an integer multiple of |Xϕ(P)|/|X(Q|P)|.

Thus we see that, for instance, if P has a large ϕ-variance group Xϕ(P) and if Q is fairly

small (which forces X(Q|P) to be small), then Xϕ(P Q) is still large.

A careful reﬁnement lets us make a similar statement about inﬁnite ϕ-variance groups:

Theorem 3.26. Let P and Q be chiral or directly regular n-semi-polytopes, and let ϕ ∈

+ Aut(Wn ) have finite order. If X(Q|P) is finite and Xϕ(P) is infinite, then |Xϕ(P Q)| is infinite.

Proof. Consider the commutative diagram below, where the maps are all the natural epimorphisms:

f1 Γ+(P Pϕ Q Qϕ) / Γ+(P Pϕ)

f2 g1 Γ+(P Q) / Γ+(P) g2

Then ker(g1◦f1) = ker(g2◦f2). Now, since Xϕ(P) = ker g1 is inﬁnite by assumption, it follows

−1 that ker(g1 ◦f1) is infinite. We have that ker(g2 ◦f2) = f2 (ker g2). Now, ker f2 = Xϕ(P Q), and ker g2 = X(Q|P), which is finite by assumption. If Xϕ(P Q) is finite, then so is the

+ −1 preimage of any ﬁnite subgroup of Γ (P Q) under f2. Therefore, f2 (ker g2) is ﬁnite, and

−1 since f2 (ker g2) = ker(g2 ◦ f2) = ker(g1 ◦ f1), the group Xϕ(P Q) must be inﬁnite.

Let us apply this result to extend Theorem 3.20 in a diﬀerent way.

76 Theorem 3.27. Let P, Q, and R be ﬁnite chiral or directly regular n-semi-polytopes, and

+ + let ϕ ∈ Aut(Wn ) have ﬁnite order. If P Q R is internally ϕ-invariant, then |Γ (R)| is

+ an integer multiple of |Xϕ(P)|/|Γ (Q)|.

Proof. Theorem 3.20 tells us that if (P Q) R is internally ϕ-invariant, then |Xϕ(P

+ of |Xϕ(P)|/|X(Q|P)|, and since |X(Q|P)| divides |Γ (Q)|, this is an integer multiple of

+ |Xϕ(P)|/|Γ (Q)|. The result follows.

Thus we see that it is possible (though a bit unwieldy) to extend Theorem 3.20 to the mix of three and indeed any number of polytopes. In practice, however, we usually mix polytopes two at a time and try to ﬁnd the exact ϕ-variance group of the mix. The following results illustrate a few cases where we can easily do so.

Theorem 3.28. Let P and Q be ﬁnite chiral or directly regular n-semi-polytopes, and let

+ ϕ ∈ Aut(Wn ) have ﬁnite order. Suppose P is internally ϕ-variant, with Xϕ(P) simple,

+ and suppose that Q is internally ϕ-invariant. If |Xϕ(P)| does not divide |Γ (Q)|, then

Xϕ(P Q) = Xϕ(P).

Proof. By Theorem 3.20, the mix P Q is internally ϕ-variant. Proposition 3.24 says that

Xϕ(PQ) is a normal subgroup of Xϕ(P), which is simple. Since PQ is internally ϕ-variant,

Xϕ(P Q) must be nontrivial, and therefore Xϕ(P Q) = Xϕ(P).

Theorem 3.29. Let P and Q be ﬁnite chiral or directly regular n-semi-polytopes, and let

+ ϕ ∈ Aut(Wn ) have ﬁnite order. Suppose that P and Q are internally ϕ-variant and that

+ + Γ (Q) is simple. If Xϕ(P) is not isomorphic to Γ (Q), then P Q is internally ϕ-variant.

Proof. Suppose P Q is internally ϕ-invariant. Then by Lemma 3.16, Γ+(P) Γ+(Q) naturally covers Γ+(P) Γ+(Pϕ). Thus, we get the following commutative diagram, where the

77 maps are all the natural epimorphisms:

f1 Γ+(P) Γ+(Q) / Γ+(P) Γ+(Pϕ) OOO nn OO nnn OO nng f2 OO nn OO' wnnn Γ+(P)

Therefore, ker f2 = ker(g ◦ f1), and in particular, ker f1 is normal in ker f2. Now, we can

+ view ker f2(= X(Q|P)) as a normal subgroup of Γ (Q), which is simple. Then ker f2 is

simple (possibly trivial), and ker f1 is likewise simple (possibly trivial). Suppose ker f1 is

trivial. Then ker f2 ' ker g = Xϕ(P). Since P is internally ϕ-variant, ker g is nontrivial.

+ + Therefore, ker f2 is nontrivial, so we must have ker f2 ' Γ (Q). But then Xϕ(P) ' Γ (Q),

+ violating our assumptions. Now, suppose instead that ker f1 ' Γ (Q). Then we must

+ have ker f2 ' Γ (Q) as well. Then since ker f2 = ker(g ◦ f1), we must have that ker g is trivial, contradicting that P is internally ϕ-variant. Therefore, P Q must be internally ϕ-variant.

Theorem 3.30. Let P and Q be ﬁnite chiral or directly regular n-semi-polytopes and let

+ + + ϕ + + ϕ ϕ ∈ Aut(Wn ) have ﬁnite order. Suppose that Γ (P) Γ (P ) and Γ (Q) Γ (Q ) have

+ + + no nontrivial common quotients. Then Γ (P Q) = Γ (P) × Γ (Q) and Xϕ(P Q) =

Xϕ(P) × Xϕ(Q).

Proof. Since Γ+(P) is a quotient of Γ+(P) Γ+(Pϕ) and Γ+(Q) is a quotient of Γ+(Q) Γ+(Qϕ), the groups Γ+(P) and Γ+(Q) also have no nontrivial common quotient. Therefore, Γ+(P) Γ+(Q) is trivial, and Γ+(P Q) = Γ+(P) × Γ+(Q), by Proposition 2.21. Now, on

78 the one hand,

|Γ+(P) Γ+(Pϕ) Γ+(Q) Γ+(Qϕ)| = |(Γ+(P) Γ+(Q)) (Γ+(P) Γ+(Q))ϕ|

+ + = |Γ (P) Γ (Q)| · |Xϕ(P Q)|

+ + = |Γ (P)| · |Γ (Q)| · |Xϕ(P Q)|.

On the other hand,

|Γ+(P) Γ+(Pϕ) Γ+(Q) Γ+(Qϕ)| = |(Γ+(P) Γ+(Pϕ)) (Γ+(Q) Γ+(Qϕ))|

+ + = |Γ (P)| · |Xϕ(P)| · |Γ (Q)| · |Xϕ(Q)|, again by Proposition 2.21. Therefore,

|Xϕ(P Q)| = |Xϕ(P)| · |Xϕ(Q)|.

By Proposition 3.23, Xϕ(P Q) is isomorphic to a subgroup of Xϕ(P) × Xϕ(Q), and thus it follows that Xϕ(P Q) = Xϕ(P) × Xϕ(Q).

79 Chapter 4

Constructing self-dual and self-Petrie polytopes

We will now start applying the results of Chapter 3 to the operations of duality and Petrie duality. Recall that the dual of P is denoted Pδ, and it is obtained from P by reversing the partial order. The Petrie dual of a polyhedron P is denoted Pπ, and it is obtained from P by keeping the vertices and edges, but replacing the faces with the set of Petrie polygons. Each of these operations also has a corresponding group automorphism; the automorphism

+ −1 δ of Wn = hσ1, . . . , σn−1i sends each σi to σn−i, and the automorphism π of W3 = hρ0, ρ1, ρ2i sends ρ0 to ρ0ρ2 while ﬁxing ρ1 and ρ2.

Our ﬁrst goal is to use the results from Chapter 3 to construct self-dual covers of a given polytope and to determine when such covers are polytopal. Then we will see how to construct polyhedra that are simultaneously self-dual and self-Petrie, and we will give several examples of such polyhedra.

80 4.1 Self-dual polytopes

By mixing a polytope P with its dual, we get a properly self-dual cover Q of P; that is, Q is internally δ-invariant, so Q = Qδ. Our ﬁrst question is: under what conditions is this cover a polytope? Since Corollary 2.45 tells us that the mix of two polyhedra is a polyhedron, we focus on ranks 4 and higher. We start by applying Proposition 2.41 and Theorem 2.53 to the mix of P with Pδ.

Proposition 4.1. Let P be a chiral or regular polytope of type {K, L}, and suppose Lδ covers K. Then P Pδ is a self-dual polytope.

Proof. The facets of Pδ are Lδ, and the facets of P are K. Since Lδ covers K, Proposition 2.41 says that P Pδ is a polytope, and Proposition 3.11 tells us that P Pδ is internally δ- invariant, i.e., properly self-dual.

Proposition 4.2. Let P be a chiral or directly regular n-polytope, with n = 2m+1. Suppose

P is of type {p1, . . . , pn−1}, and that pi is coprime to pn−i for every i ∈ {2, . . . , m}. Then

δ δ P P is a self-dual polytope. Furthermore, if p1 is coprime to pn−1, then Γ(P P ) = Γ(P) × Γ(Pδ).

Proof. The result follows directly from Theorem 2.53 and Proposition 2.47.

Clearly, if n is even, then it is impossible to apply Theorem 2.53 to the mix of P with Pδ, since in that case, P and Pδ will agree in the middle entry of their Schl¨aﬂisymbol. In fact, when n is even, having certain numbers pi relatively prime to pn−i is actually an impediment to polytopality.

Proposition 4.3. Let P be a chiral or directly regular n-polytope, with n = 2m. Suppose P is of type {p1, . . . , p2m−1}, that pm−1 and pm+1 are relatively prime, and that pm ≥ 3. Then P Pδ is not a polytope.

81 Proof. Apply Proposition 2.56 to P and Pδ.

Example 4.4. If P is the 4-cube, which is the universal polytope of type {4, 3, 3}, then P Pδ is not a polytope.

We will brieﬂy explore a way to build self-dual regular polytopes of high rank. Given any polytope P of type {p1, . . . , pn−1}, there is a polytope {P, 2} of type {p1, . . . , pn−1, 2} and with facets isomorphic to P. We build this polytope by inserting two new faces between the maximal element and the facets of P, and connecting both new faces to every facet of P. It is then easy to show that if P is regular, so is {P, 2}, and Γ({P, 2}) = Γ(P) × hρni. This extension also works nicely with direct regularity:

Proposition 4.5. Let P be a directly regular n-polytope and let Q = {P, 2}. Then Q is directly regular, and Γ+(Q) ' Γ(P).

Proof. Since P is directly regular, all of the relators in Γ(P) have even length. The only

2 2 relators that we add in extending P to Q are ρn and relators of the form (ρiρn) , all of which are even. Therefore, Q is still directly regular.

+ Now, let w ∈ Γ (Q) ≤ Γ(Q) = hρ0, . . . , ρni. Since Γ(Q) = Γ(P) × hρni, we can write w

i uniquely as w = vρn, where i = 0 or 1, and where v ∈ Γ(P).

Now, consider the canonical projection from Γ(Q) = Γ(P)×hρni to Γ(P), and let f be its

+ i restriction to Γ (Q). That is, f(vρn) = v. We want to show that f is an isomorphism from

+ + i Γ (Q) to Γ(P). First, let w ∈ Γ (Q), and suppose that f(w) = . Then writing w = vρn,

i + we see that v = . Therefore, w = ρn. Then since w ∈ Γ (Q), we must have w = , and thus f is injective. To prove surjectivity, we note that if v ∈ Γ+(P), then v ∈ Γ+(Q), and

+ + f(v) = v, while if v ∈ ρ0Γ (P), then ρnv ∈ Γ (Q), and f(ρnv) = v. Thus we see that f :Γ+(Q) → Γ(P) is an isomorphism.

82 Theorem 4.6. Let P be a directly regular n-polytope of type {p1, . . . , pn−1} with every pi odd.

Then there is a self-dual directly regular (2n−1)-polytope of type {2p1,... 2pn−1, 2pn−1,... 2p1},

δ 2n−4 with n-faces that cover P and with rotation group Γ(P) × Γ(P ) × C2 ' Γ(P) × Γ(P) ×

2n−4 C2 .

Proof. We start by extending P n − 1 times as above to obtain a (2n − 1)-polytope Q of

n−1 type {q1, . . . , q2n−2} := {p1, . . . , pn−1, 2,... 2}, with group Γ(Q) = Γ(P) × C2 . Then by

+ n−2 δ Proposition 4.5, Γ (Q) ' Γ(P) × C2 . Now, let Q be the dual of Q. If i < n, then

δ qi is odd and q2n−1−i = 2, so by Proposition 4.2, we get that Q Q is a self-dual directly

+ δ regular polytope of type {2p1,... 2pn−1, 2pn−1,... 2p1} and with rotation group Γ (QQ ) =

+ + δ δ 2n−4 Γ (Q) × Γ (Q ) = Γ(P) × Γ(P ) × C2 .

For example, if we apply Theorem 4.6 to the n-simplex, then we obtain a (2n−1)-polytope

2n−4 of type {6,..., 6} with rotation group Sn+1 × Sn+1 × C2 .

4.2 Self-dual, self-Petrie polyhedra

Regular polyhedra that are self-dual and self-Petrie have a high degree of “external” symmetry in addition to the maximal “internal” symmetry that regularity measures. Closely related are the self-dual and self-Petrie maps on surfaces, which have been studied in [19, 20, 29]. Many of these results can be adapted to abstract polyhedra. We do so and present several new results here.

Suppose we want to construct the minimal self-dual, self-Petrie cover of a regular polyhedron P. Our ﬁrst instinct might be to mix P with its dual to construct its self-dual cover Q, and then to mix Q with its Petrial to construct its self-Petrie cover R. However, there is no guarantee that R is self-dual. Constructing a polyhedron that is simultaneously self-dual

83 and self-Petrie requires a slightly subtler approach.

In [18], the author proves that the automorphisms δ and π (of W3) generate a symmetric group of order 6. Therefore, a regular polyhedron P has at most 6 distinct images under these operations. Deﬁne P∗ to be the mix of these 6 polyhedra; i.e.,

P∗ := P Pδ Pπ Pδπ Pπδ Pδπδ.

∗ ∗ Then P is a regular polyhedron. Now, if Γ(P) = W3/M, then Γ(P ) is the quotient of W3 by M ∩ δ(M) ∩ π(M) ∩ δπ(M) ∩ πδ(M) ∩ δπδ(M), and since this subgroup is ﬁxed by both δ and π, it follows that P∗ is self-dual and self-Petrie. In fact, it is the minimal self-dual, self-Petrie cover of P.

δ π If P if of type {p, q}r, then P is of type {q, p}r, and P is of type {r, q}p. Therefore, a polyhedron that is self-dual and self-Petrie must be of type {k, k}k for some k ≥ 2. In

∗ particular, if P is of type {p, q}r, then P is of type {k, k}k, where k is the least common multiple of p, q, and r.

Determining the global structure of Γ(P∗), such as its size or isomorphism type, is quite challenging. In general, the problem is completely intractable; the mix of 6 groups is simply too large for the usual algorithms in combinatorial group theory to work with. Therefore, we turn our attention to ﬁnding ways to avoid such direct calculation. Usually, such results will hinge on calculating the comix of certain polyhedra, building on Theorem 2.23.

Theorem 4.7. Let P be a regular self-Petrie polyhedron of type {p, q}p. Suppose p is odd and that p and q are coprime. Then

Γ(P∗) = Γ(P) × Γ(Pδ) × Γ(Pδπ).

84 δ δ Proof. First, we note that P is of type {q, p}p. In Γ(P) Γ(P ), the order of ρ0ρ1 divides p and q, and since p and q are coprime, we get ρ0ρ1 = ; that is, ρ0 = ρ1. Similarly, ρ1ρ2 = , and so ρ1 = ρ2. Now, the order of ρ0ρ1ρ2 divides p. On the other hand, ρ0ρ1ρ2 = ρ0, and so the order divides 2 as well. Therefore, we must have that ρ0ρ1ρ2 = . This forces all of the generators to be trivial, and thus Γ(P) Γ(Pδ) is trivial and Γ(P) Γ(Pδ) = Γ(P) × Γ(Pδ), by Corollary 2.24.

δ δπ Now, P P is of type {pq, pq}p, and P is of type {p, p}q. Then in their comix, we get that ρ0ρ1ρ2 has order dividing p and q, and thus ρ0ρ1ρ2 is trivial. Thus ρ0 = ρ1ρ2, and so

2 p (ρ1ρ2) = . On the other hand, we also have that (ρ1ρ2) = in the comix, and since p is odd, this means that ρ1ρ2 is trivial. So ρ0 is trivial, and ρ1 = ρ2. Similarly, ρ2 = ρ0ρ1,

2 p and thus (ρ0ρ1) = . But again, we also have that (ρ0ρ1) = , and thus ρ0ρ1 = . So

ρ0 = ρ1 = ρ2 = and we see that the comix is trivial. Therefore, the mix is the direct product Γ(P) × Γ(Pδ) × Γ(Pδπ), again by Corollary 2.24.

Finally, we note that since P is self-Petrie, it follows that Pπ = P, Pπδ = Pδ, and Pπδπ = Pδπ. Therefore, the self-dual, self-Petrie cover of P consists of just the three distinct polyhedra we have mixed.

Corollary 4.8. Let P be a regular polyhedron of type {p, q}r. Suppose p and r are odd and both coprime to q. Let Q = P Pπ. Then

Γ(P∗) = Γ(Q) × Γ(Qδ) × Γ(Qδπ).

Proof. We start by noting that P∗ = Q∗. The polyhedron Q is a self-Petrie polyhedron of type {`, q}`, where ` is the least common multiple of p and r. Since p and r are both odd and coprime to q, ` is also odd and coprime to q. Then we can apply Theorem 4.7 to Q and the result follows.

85 The condition in Theorem 4.7 that p is odd is essential. When p is even, we cannot tell from the type alone whether Γ(P) Γ(Pδ) has order 1 or 2. However, if P is the universal ∗ polyhedron of type {p, q}p, then we can still determine |Γ(P )|.

Theorem 4.9. Let P = {p, q}p, the universal polyhedron of type {p, q}p. Suppose p is even and that p and q are coprime. Then Γ(P∗) is a group of index 8 in Γ(P) × Γ(Pδ) × Γ(Pδπ). In particular, if P is ﬁnite, then

|Γ(P∗)| = |Γ(P)|3/8.

Proof. Since p and q are coprime, a presentation for Γ(P) Γ(Pδ) is given by

2 2 2 2 p hρ0, ρ1, ρ2 | ρ0 = ρ1 = ρ2 = ρ0ρ1 = (ρ0ρ2) = ρ1ρ2 = (ρ0ρ1ρ2) = i, and direct calculation shows that this is a group of order 2. Therefore, Γ(P) Γ(Pδ) has index 2 in Γ(P) × Γ(Pδ).

δ δπ Now we consider (Γ(P)Γ(P )) Γ(P ). In this group, the order of ρ0ρ1ρ2 divides both p and q, and thus ρ0ρ1ρ2 = . Therefore, ρ0ρ1 = ρ2, which forces ρ0ρ1 to have order dividing

2, and similarly ρ0 = ρ1ρ2, which forces ρ1ρ2 to have order dividing 2. Therefore, the comix is a (not necessarily proper) quotient of the group [2, 2]1, a group of order 4. Now, since

δπ δπ Γ(P ) = [p, p]q and p is even, we see that Γ(P ) covers [2, 2]1. We similarly see that Γ(P)

δ δ covers [2, 1]1 and that Γ(P ) covers [1, 2]1, and thus Γ(P) Γ(P ) covers [2, 1]1 [1, 2]1, which is equal to [2, 2]1. Thus we see that the group [2, 2]1 is the maximal group covered by both Γ(P) Γ(Pδ) and Γ(Pδπ), and therefore it is their comix. Therefore, (Γ(P) Γ(Pδ)) Γ(Pδπ) has index 4 in (Γ(P) Γ(Pδ)) × Γ(Pδπ), which itself has index 2 in (Γ(P) × Γ(Pδ)) × Γ(Pδπ), and the claim is proved.

86 We note here that the arguments used can be generalized to give bounds on |Γ(P∗)| even when we cannot calculate the exact value. In the next section, however, we will only consider cases where we can calculate |Γ(P∗)| exactly.

4.3 Self-dual, self-Petrie covers of universal polyhedra

In this section, we will consider several of the ﬁnite universal polyhedra P = {p, q}r and calculate |Γ(P∗)|, the group order for the minimal self-dual, self-Petrie, regular cover of P. The results are summarized in Table 4.1. In most cases, the sizes are easily calculated by applying Corollary 4.8 to P or one of its images under the duality operations. A few others can be found by applying Theorem 4.9 or by calculating the size directly using GAP [12]. We cover the remaining cases here.

We start with P = {2, 2s}. The order of ρ0ρ1ρ2 in Γ(P) is 2s, and therefore P = {2, 2s}2s, which has an automorphism group of order 8s. If s = 1, then P is already self-dual and self-Petrie. For any s, there are only 3 distinct images of P under the duality operations, and thus P∗ = P Pδ Pπ.

δ Now, P P is a self-dual regular polyhedron of type {2s, 2s}2s. We note that for any s,

δ the group Γ(P) Γ(P ) is equal to [2, 2]2s. In fact, this group is equal to [2, 2]2, the group of order 8 generated by 3 commuting involutions (i.e., the direct product of three cyclic groups of order 2). Then by Proposition 2.21, we see that |Γ(P) Γ(Pδ)| = 8s2. In order to determine whether P Pδ is self-Petrie, we would like to determine the group Γ(P) Γ(Pδ) explicitly. A computation with GAP [12] suggests that the group is always the quotient of

2 [2s, 2s]2s by the extra relation (ρ1ρ0ρ1ρ2) = . Since this extra relation also holds in Γ(P) and Γ(Pδ), this quotient must cover Γ(P) Γ(Pδ). Therefore, to prove that this is in fact

87 the mix, it suﬃces to show that this quotient has order 8s2.

Start by considering the Cayley graph G of

2 2 2 2 2 hρ0, ρ1, ρ2 | ρ0 = ρ1 = ρ2 = (ρ0ρ2) = (ρ1ρ0ρ1ρ2) = i,

determined by the generators ρ0, ρ1, and ρ2. Starting at a vertex and building out from it, we see that the Cayley graph of this group is the graph of the uniform tiling 4.8.8 of the plane by squares and octagons. Figure 4.1 gives a local picture of G, and shows how this tiling can be thought of as being built from square tiles ﬁtting together edge-to-edge.

2 2 2 0 0 0 0 0 0 0 2 2 2 ?? ?? ?? ??1 1 ??1 1 ??1 1 ?? ?? ?? ? ? ? 2 2 2 0 0 0 0 0 0 2 2 2 ?? ?? ?? ?? ?? ?? 1 1 ?? 1 1 ?? 1 1 ?? ? ? ? 2 2 2 0 0 0 0 0 0 0 2 2 2 ?? ?? ?? ??1 1 ??1 1 ??1 1 ?? ?? ?? ? ? ? 2 2 2 0 0 0 0 0 0 2 2 2 ?? ?? ?? ?? ?? ?? 1 1 ?? 1 1 ?? 1 1 ?? ? ? ? 2 2 2 0 0 0 0 0 0 0 2 2 2 ?? ?? ?? ??1 1 ??1 1 ??1 1 ?? ?? ?? ? ? ? 2 2 2 0 0 0 0 0 0 2 2 2 ?? ?? ?? ?? ?? ?? 2 1 1 ?? 2 1 1 ?? 2 1 1 ?? ? ? ?

Figure 4.1: Local picture of the Cayley graph G

2s Now, let us see what happens when we introduce the remaining relations (ρ0ρ1) = and

2s (ρ1ρ2) = . We note that the components of G induced by edges labeled 0 and 1 are vertical

88 zig-zags, while the components of G induced by edges labeled 1 and 2 are horizontal zig-zags.

2s Therefore, adding the relation (ρ0ρ1) = forces an identiﬁcation between points that are

2s s square tiles away vertically, and adding the relation (ρ1ρ2) = forces an identiﬁcation between points that are s square tiles away horizontally. Therefore, the Cayley graph of Γ(P Pδ) consists of an s × s grid of square tiles, with opposite sides identiﬁed. Each square tile has 8 vertices of the Cayley graph, so there are a total of 8s2 vertices, which shows that the given group has 8s2 elements. Therefore the given group is indeed the mix.

Resuming our discussion of whether P Pδ is self-Petrie, we consider (Γ(P) Γ(Pδ)) π 2 Γ(P ), which is the quotient of [2s, 2s]2 by the extra relation (ρ1ρ0ρ1ρ2) = . Put another

δ 2 way, we get this group as the quotient of Γ(P) Γ(P ) by the extra relation (ρ0ρ1ρ2) = .

2 Using the Cayley graph from before, we see that the relation (ρ0ρ1ρ2) = forces us to identify each square tile with the square tiles that touch it at a corner. When s is odd, this forces us to identify every pair of square tiles, leaving us with a Cayley graph with 8 vertices. When s is even, we instead get 2 distinct square tiles, and the Cayley graph has 16 vertices. Therefore, |Γ(P) Γ(Pδ) Γ(Pδπ)| = 8s3 if s is odd, and 4s3 if s is even. In other words, if P = {2, 2s} and s = 2k + 1, then |Γ(P∗)| = 8(2k + 1)3, while if s = 2k, then |Γ(P∗)| = 32k3.

Now suppose that P = {2, 2k +1}. This polyhedron is covered by Q = {2, 4k +2} (which

δ δ is equal to {2, 4k+2}4k+2, as we observed earlier); therefore, QQ covers P P . Calculating the size of the mix, we ﬁnd that |Γ(Q) Γ(Qδ)| = |Γ(P) Γ(Pδ|, and thus these two groups must be equal. From this it easily follows that Q∗ = P∗, and thus |Γ(P∗)| = 8(2k + 1)3.

2 Next we consider the polyhedra {4, 4}2s, which are the torus maps {4, 4}(s,s) with 16s

π ﬂags [8]. First, suppose that s = 2k, so that P = {4, 4}4k. Then Γ(P) Γ(P ) = [4, 4]4, which has order 64. Therefore, by Proposition 2.21,

|Γ(P) Γ(Pπ)| = |Γ(P)| · |Γ(Pπ)|/64 = 64k4.

89 π πδ Now, P P is of type {4k, 4}4k, and P is the universal polyhedron of type {4, 4k}4. Thus

π πδ (Γ(P) Γ(P )) Γ(P ) is [4, 4]4 or a proper quotient. Since Γ(P) covers [4, 4]4, so does π πδ π πδ Γ(P) Γ(P ). Then since Γ(P ) also covers [4, 4]4, so does (Γ(P) Γ(P )) Γ(P ), and thus the comix is the whole group [4, 4]4. Thus we see that

|Γ(P) Γ(Pπ) Γ(Pπδ)| = |Γ(P) Γ(Pπ)| · |Γ(Pπδ)|/64 = 64k6.

π Now suppose that s = 2k + 1, so that P = {4, 4}4k+2. Then Γ(P) Γ(P ) = [2, 4]2, which has size 8. Therefore,

|Γ(P) Γ(Pπ)| = |Γ(P)| · |Γ(Pπ)|/8 = 32(2k + 1)4.

π πδ Now, P P is of type {8k + 4, 4}8k+4 and P is the universal polyhedron {4, 4k + 2}4.

π πδ πδ Thus (Γ(P) Γ(P )) Γ(P ) is [4, 2]4 or a proper quotient. Clearly, Γ(P ) covers [4, 2]4. π Furthermore, Γ(P) covers [4, 4]2 and Γ(P ) covers [2, 4]4; therefore, their mix covers [4, 4]2

[2, 4]4, which covers [4, 2]4. Therefore, the comix is the whole group [4, 2]4 of order 16, and we see that

|Γ(P) Γ(Pπ) Γ(Pπδ)| = |Γ(P) Γ(Pπ)| · |Γ(Pπδ)|/16 = 32(2k + 1)6.

It would be natural here to consider the torus maps {3, 6}2s = {3, 6}(s,0). However, in this case there are 6 distinct polyhedra under the duality operations, and the problem seems to be intractible.

Finally, we note that the self-dual, self-Petrie covers of {3, 3}4 and {3, 4}6 have groups of the same order. In fact, since {3, 4}6 covers {3, 4}3, these two polyhedra have the same self-dual, self-Petrie cover, of type {12, 12}.

90 P |Γ(P)| |Γ(P∗)| Method 3 {2, 2k + 1}4k+2 4(2k + 1) 8(2k + 1) Hand 3 {2, 4k}4k 16k 32k Hand 3 {2, 4k + 2}4k+2 8(2k + 1) 8(2k + 1) Hand 3 {3, 3}4 24 (24) Thm. 4.7 3 {3, 4}6 48 (48) /8 GAP 3 {3, 5}5 60 (60) Thm. 4.7 3 {3, 5}10 120 (120) GAP 3 {3, 6}6 108 (108) /216 GAP 6 {3, 7}8 336 (336) /8 Cor. 4.8 6 {3, 7}9 504 (504) Cor. 4.8 6 {3, 7}13 1092 (1092) Cor. 4.8 6 {3, 7}15 12180 (12180) Cor. 4.8 6 {3, 7}16 21504 (21504) /8 Cor. 4.8 3 {3, 8}8 672 (672) /8 Thm. 4.9 6 {3, 8}11 12144 (12144) /8 Cor. 4.8 3 {3, 9}9 3420 (3420) GAP 6 {3, 9}10 20520 (20520) /216 Cor. 4.8 2 6 {4, 4}4k 64k 64k Hand 2 6 {4, 4}4k+2 16(2k + 1) 32(2k + 1) Hand 3 {4, 5}5 160 (160) Thm. 4.7 6 {4, 5}9 6840 (6840) /8 Cor. 4.8

Table 4.1: Self-dual, self-Petrie covers of several ﬁnite polyhedra {p, q}r

91 Chapter 5

Constructing chiral polytopes

Our ﬁnal goal is the construction of chiral polytopes. Though we have many examples of chiral polyhedra and maps (see [6, 34]) and chiral 4-polytopes (see [25, 26]), only a handful of examples are known in ranks 5 and higher (see [5, 28]). Mixing cannot, by itself, construct a chiral n-polytope from a chiral (n−1)-polytope. However, once we have a chiral n-polytope, we can use mixing to create many more chiral n-polytopes. Results like Theorem 3.20, Corollary 3.21, and Corollary 3.22 give us simple ways to determine when the mix of a chiral polytope with another polytope is chiral.

5.1 Building chiral polytopes of high rank

+ Recall that, in Example 3.4, we described an automorphism χ of Wn = hσ1, . . . , σn−1i

−1 2 that sends σ1 to σ1 , sends σ2 to σ1σ2, and ﬁxes σ3, . . . , σn−1. As noted earlier, if P is a chiral or directly regular polytope, then it is always externally χ-invariant. Furthermore, P is internally χ-invariant if and only if it is directly regular (and therefore it is internally χ-variant if and only if it is chiral).

92 For the rest of the section, we will change our terminology slightly to ﬁt with earlier accounts on the measurement of chirality (see [2, 3, 9]). We will continue to say that a polytope is directly regular or chiral instead of using the terms “internally χ-invariant” and “internally χ-variant”. For Pχ we will use P, and we call this the enantiomorphic form of

χ P. Finally, the group Xχ(P), which is equal to X(P|P ) = X(P|P), is denoted by X(P) and called the chirality group of P.

In [23], it is shown that for each n ≥ 4, there are inﬁnitely many n-toroids of type {4, 3n−3, 4}. In particular, for each s ≥ 2 and k = 1, 2, or n − 1, there is a regular toroid

n−3 n+k−2 n−1 {4, 3 , 4}(sk,0n−k−1) that has a group of order 2 (n − 1)!s . These toroids provide a rich source of polytopes to mix with chiral polytopes, and in many cases, the mixing will yield another chiral polytope.

Theorem 5.1. Let n ≥ 4, and let P be a chiral n-polytope of type {p1, . . . , pn−1} such that p1 and pn−1 are odd and such that p2, . . . , pn−2 are not divisible by 3. If X(P) is inﬁnite, or if the largest prime factor of |X(P)| is greater than n − 1, then there are inﬁnitely many chiral

+ n−3 + n-polytopes of type {4p1, 3p2,..., 3pn−2, 4pn−1}, with groups Γ (P) × [4, 3 , 4](sk,0n−k−1).

n−3 Proof. Let Q(s, k) = {4, 3 , 4}(sk,0n−k−1). Proposition 2.47 establishes that P Q(s, k) is polytopal and that its type is {4p1, 3p2,..., 3pn−2, 4pn−1}. When X(P) is inﬁnite, we can appeal to Corollary 3.21 to see that the mix is chiral for any choice of s and k. Otherwise, let m be the largest prime factor of |X(P)|. In order for P Q(s, k) to be directly regular,

n−3 + we must have that |X(P)| divides |[4, 3 , 4](sk,0n−k−1)|, by Theorem 3.20. In particular, m must divide 2n+k−2(n−1)!sn−1. Since m is prime, if m > n−1, the only way this can happen is when m divides s. Therefore, for any k and any s such that m - s, the mix P Q(s, k) is chiral.

Another general way of producing new chiral polytopes is by mixing chiral polytopes

93 with chiral or directly regular polytopes that have a simple rotation group. Here is one particularly nice result of this type.

Theorem 5.2. Let n ≥ 4, and let P be a chiral n-polytope of type {p1, . . . , pn−1} such that no pi is divisible by 3. Suppose that X(P) is not isomorphic to An+1. Let Q be the n-simplex.

+ Then P Q is a chiral polytope of type {3p1,..., 3pn−1} with group Γ (P) × An+1.

Proof. The n-simplex Q = {3, 3,..., 3} (with n − 1 threes) has rotation group An+1, which is simple for n ≥ 4. Now, Proposition 2.47 establishes that P Q is polytopal, that its type

+ is {3p1,..., 3pn}, and that its group is Γ (P) × An+1. Furthermore, Corollary 3.22 tells us that P Q is chiral.

What we would really like is a way to build a chiral (n + 1)-polytope from a given chiral n-polytope in some way. In [31], the authors proved that for any chiral n-polytope K with directly regular facets, there is a universal chiral (n+1)-polytope U(K) with facets isomorphic to K. This polytope covers all other chiral (n + 1)-polytopes with facets isomorphic to K. In many cases, this extension has an inﬁnite chirality group:

Lemma 5.3. Let K be a chiral n-polytope with directly regular facets such that Γ+(K) Γ+(K) is ﬁnite. (For example, the latter holds if Γ+(K) is itself ﬁnite.) Let P = U(K). Let

+ + + + Γ (P) = hσ1, . . . , σni so that Γ (K) = hσ1, . . . , σn−1i. If σn−1 = in Γ (K) Γ (K), or if + + σn−1 = σn−2 in Γ (K) Γ (K), then X(P) is inﬁnite.

Proof. Since X(P) is the kernel of the natural epimorphism from Γ+(P) to Γ+(P) Γ+(P) (see Proposition 2.19) and Γ+(P) is infinite, it suffices to show that Γ+(P) Γ+(P) is finite. + + + + Now, suppose that σn−1 = in Γ (K) Γ (K). Then σn−1 = in Γ (P) Γ (P) as 2 + + 2 well. Furthermore, the relation (σn−1σn) = also holds in Γ (P) Γ (P), so σn = + + + + in Γ (P) Γ (P). Suppose instead that σn−1 = σn−2 in Γ (K) Γ (K) (and thus in

94 + + 2 2 2 2 Γ (P) Γ (P)). Then (σn−2σn−1σn) = , and thus (σn−1σn) = = (σn−1σn) . Therefore, 2 2 σn−1σnσn−1 = σn, and so = (σn−1σn) = σn. In any case, we see that σn has order dividing 2 in Γ+(P) Γ+(P).

2 2 Now, σn commutes with σi if i < n − 2. Since (σn−1σn) = and σn = , it follows that

−1 2 σnσn−1 = σn−1σn. Finally, since (σn−2σn−1σn) = , we see that

2 (σn−2σn−1)(σn−2σn−1σn) (σn−1σn) = (σn−2σn−1)(σn−1σn)

2 σnσn−2 = σn−2σn−1σn.

+ + Therefore, given any word w in Γ (P) Γ (P), we can bring every factor of σn in w to k the right and write w = uσn, where k is 0 or 1 and where u contains only the generators

+ + + + σ1, . . . , σn−1, and is therefore a word in Γ (K) Γ (K). Therefore, Γ (P) Γ (P) is ﬁnite; in particular, it is at most twice as large as Γ+(K) Γ+(K). Theorem 5.4. Let K be a ﬁnite chiral n-polytope with directly regular facets and group

+ Γ (K) = hσ1, . . . , σn−1i. Let P = U(K), and let Q be a ﬁnite directly regular (n+1)-polytope.

+ + If σn−1 = or σn−1 = σn−2 in Γ (K) Γ (K), then P Q is a chiral n-semi-polytope with inﬁnite chirality group X(P Q).

Proof. By Lemma 5.3, the conditions given on K suffice to ensure that P has an infinite chirality group. Then Corollary 3.21 applies to show that P Q also has an infinite chirality group.

2 2 For example, let K be the chiral polyhedron {4, 4}(b,c) with p = b + c an odd prime.

+ + Then Γ (K) Γ (K) = [4, 4](1,0), the group of the torus map{4, 4}(1,0), in which σ1 = σ2 [1]. Therefore, Theorem 5.4 says that we can mix U(K) with any ﬁnite directly regular polytope to obtain a polytope with inﬁnite chirality group.

95 Note that in Theorem 5.4, we had to choose K to have directly regular facets. Since the (n − 2)-faces of a chiral polytope must be directly regular, there is no universal polytope U(P) if P has chiral facets. In particular, there is no universal polytope U(U(K)). However, using mixing, we can build a chiral polytope with directly regular facets K K:

Theorem 5.5. Let K be a ﬁnite chiral polytope with directly regular facets. Let P = U(K), and let Q = {K K, 2}. If X(K) = Γ+(K), then P Q is a chiral polytope with directly regular facets.

Proof. Since X(K) = Γ+(K), Γ+(K) Γ+(K) is trivial, by Proposition 2.19. Furthermore, the polytope Q is ﬁnite, since K is ﬁnite. Then by Theorem 5.4, the mix PQ is chiral. Moreover, the facets of Q cover the facets of P, so by Proposition 2.41, the mix is polytopal.

Unfortunately, Theorem 5.5 cannot be repeatedly applied, because it requires that we start with a ﬁnite chiral polytope, and it produces an inﬁnite chiral polytope.

Finally, we note that Lemma 5.3, Theorem 5.4, and Theorem 5.5 all still apply when P is any inﬁnite chiral polytope with facets isomorphic to K – the universality of U(K) is not necessary.

5.2 Self-dual chiral polytopes

We now see how to construct chiral polytopes that are self-dual. This problem ties together many of the diﬀerent features we have developed so far. The construction itself is simple; we start with a chiral polytope and mix it with its dual. This procedure always yields something self-dual, but it may not be chiral or polytopal. We dealt with the issue of polytopality in Chapter 4, so all that remains is to determine when P Pδ is chiral.

96 The results in this section all mix the chiral polytope P with its dual Pδ, yielding something that is properly self-dual (that is, internally δ-invariant). If we replace Pδ with Pδ, then the results all still hold, but they yield polytopes that are improperly self-dual instead (that is, externally δ-invariant but internally δ-variant).

Proposition 5.6. Let P be a ﬁnite chiral polytope. If |Γ+(P) Γ+(P)| < |Γ+(P) Γ+(Pδ)|, then P Pδ is chiral.

By taking into account the Schl¨aﬂi symbol of P, we obtain a slightly stronger result.

Theorem 5.7. Let P be a finite chiral n-polytope of type {p1, . . . , pn−1}. Define ì = lcm(pi, pn−i) for i = 1, . . . , n − 1, and let ` = lcm(`1/p1, . . . , `n−1/pn−1). If

|Γ+(P) Γ+(P)| < ` |Γ+(P) Γ+(Pδ)|,

then P Pδ is chiral.

Proof. Suppose P Pδ is directly regular. Then Γ+(P) Γ+(Pδ) covers Γ+(P) Γ+(P), by Lemma 3.16. Let π be the corresponding natural epimorphism. Now, P Pδ is of type

+ δ {`1, . . . , `n−1}, while P P is of type {p1, . . . , pn−1}. Let Γ (P P ) = hσ1, . . . , σn−1i. Then

pi p1 pn−1 pi σi ∈ ker π for each i = 1, . . . , n − 1. So hσ1 , . . . , σn−1 i ≤ ker π. Now, the order of σi in

+ δ Γ (P P ) is ì/pi since the order of σi is ì and pi divides ì. Then ker π contains elements

of order `i/pi for i = 1, . . . , n − 1, and thus it has size at least ` = lcm(`1/p1, . . . , `n−1/pn−1).

97 Now, |Γ+(P) Γ+(Pδ)| = | ker π||Γ+(P) Γ+(P)|, and therefore

|Γ+(P) Γ+(P)| = | ker π||Γ+(P) Γ+(Pδ)| ≥ ` |Γ+(P) Γ+(Pδ)|,

proving the desired result.

We now illustrate this theorem by looking at a few broad classes of examples where polytopality and chirality of the mix are guaranteed.

Theorem 5.8. Let P be a ﬁnite chiral polyhedron of type {p, q}. Let `1 = lcm(p, q), and

+ + 2 + + δ δ suppose that |Γ (P) Γ (P)| < `1/pq |Γ (P) Γ (P )|. Then P P is a properly self-dual chiral polyhedron of type {`1, `1}.

Proof. From Corollary 2.45, we know that P Pδ is a chiral or directly regular polyhedron.

2 δ Now, we apply Theorem 5.7. We have that ` = lcm(`1/p, `1/q) = `1/pq, and therefore, P P is chiral.

Theorem 5.9. Let P be a ﬁnite chiral polytope of odd rank of type {p1, . . . , pn−1}. Suppose

that pi and pn−i are coprime for i ∈ {1, . . . , n − 1}, and suppose that

+ + |Γ (P) Γ (P)| < lcm(p1, . . . , pn−1).

δ Then P P is a properly self-dual chiral polytope of type {p1pn−1, p2pn−2, . . . , pn−1p1}, and with group Γ+(P) × Γ+(Pδ).

Proof. With the given conditions, Proposition 4.2 applies to show us that P Pδ is a polytope

+ + δ with group Γ (P) × Γ (P ). To prove chirality, we apply Theorem 5.7, noting that `i =

+ + δ pipn−i, ` = lcm(p1, . . . , pn−1), and |Γ (P) Γ (P )| = 1.

98 Theorem 5.10. Let P be a chiral polyhedron of type {p, q} with simple automorphism group, and suppose that p 6= q. Then P Pδ is a properly self-dual chiral polyhedron.

Proof. Since the chirality group X(P) is a normal subgroup of the simple group Γ+(P), and P is chiral, it follows that X(P) = Γ+(P). Therefore, Γ+(P) Γ+(P) is trivial. Since p 6= q, Theorem 5.8 then tells us that P Pδ is a properly self-dual chiral polyhedron.

We note that there are many examples of chiral polyhedra with simple automorphism groups. For example, in [4], the authors give several examples of chiral polyhedra whose

automorphism group is the Mathieu group M11.

Next we consider the simplest chiral polyhedra: the torus maps. Since the torus map

{4, 4}(b,c) is already (improperly) self-dual, we work only with {3, 6}(b,c) and its dual. Let

2 2 P = {3, 6}(b,c), where m := b + bc + c is a prime and m ≥ 5. (The primality of m is not essential, but it makes some of our calculations easier.) We have that |Γ+(P)| = 6m, |X(P)| = m, and |Γ+(P) Γ+(P)| = 6 [1]. Unfortunately, Theorem 5.8 isn’t enough to establish the chirality of P Pδ. We need the following result, which relies on the fact that δ Pδ = P .

Theorem 5.11. Let P be a ﬁnite chiral polytope, and suppose that

2 + + + δ + δ |X(P)| > Γ (P) Γ (P) Γ (P ) Γ (P ) .

Then P Pδ is chiral.

99 Proof. Suppose that P Pδ is directly regular. Then (P Pδ)(P Pδ) = P Pδ. Therefore,

|Γ+(P) Γ+(Pδ)| = |Γ+(P) Γ+(Pδ) Γ+(P) Γ+(Pδ)|

δ = |(Γ+(P) Γ+(P)) (Γ+(Pδ) Γ+(P ))| δ |Γ+(P) Γ+(P)| · |Γ+(Pδ) Γ+(P )| = δ |(Γ+(P) Γ+(P)) (Γ+(Pδ) Γ+(P ))| |Γ+(P) Γ+(P)|2 = δ , |(Γ+(P) Γ+(P)) (Γ+(Pδ) Γ+(P ))| where the third line follows from Proposition 2.21. Rearranging, we get that

+ + 2 δ |Γ (P) Γ (P)| |(Γ+(P) Γ+(P)) (Γ+(Pδ) Γ+(P ))| = |Γ+(P) Γ+(Pδ)| |Γ+(P) Γ+(P)|2 ≥ |Γ+(P)|2 = |X(P)|2,

where the last line again follows from Proposition 2.21. The result then follows immediately.

Corollary 5.12. Let P be a chiral polytope of type {p1, . . . , pn−1}, and suppose that

2 + + |X(P)| > |[p1, . . . , pn−1] [pn−1, . . . , p1] |.

Then P Pδ is chiral.

δ δ Proof. Since P is of type {p1, . . . , pn−1}, so are P and P P. Similarly, P P is of

+ + + δ + δ type {pn−1, . . . , p1}. Therefore, (Γ (P) Γ (P)) (Γ (P ) Γ (P )) is a quotient of + + [p1, . . . , pn−1] [pn−1, . . . , p1] , and the result follows from Theorem 5.11.

+ + Now, returning to P = {3, 6}(b,c), we see that |X(P)| = m and that |[3, 6](b,c) [6, 3](b,c)| 100 is at most 12 (the order of [3, 3]+). Since m ≥ 5, we can conclude that P Pδ is chiral using Corollary 5.12.

It is instructive to calculate the full structure of P Pδ in this case. To do so requires that we ﬁrst calculate the size of Γ+(P) Γ+(Pδ) directly. Since m is prime, b and c must be coprime, and in particular, at least one of them must be odd. We can assume that b is odd

+ + δ by changing from P = {3, 6}(b,c) to P = {3, 6}(c,b) if necessary. Now, in Γ (P) Γ (P ), we have the relation

−1 −1 b −1 −1 c (σ1σ2 σ1 σ2) (σ2σ1σ2 σ1 ) = .

2 3 3 Now, using the facts that (σ1σ2) = σ1 = σ2 = and that b is odd, we see that

−1 −1 b b (σ1σ2 σ1 σ2) = (σ1σ1σ2σ2)

−1 −1 b = (σ1 σ2 )

b = (σ2σ1)

= σ2σ1.

Therefore,

−1 c σ2σ1(σ2σ1 σ2) =

−1 c −1 σ2(σ2σ1(σ2σ1 σ2) )σ2 =

−1 −1 c −1 (σ2(σ2σ1)σ2 )(σ2(σ2σ1 σ2) σ2 ) =

−1 −1 −1 −1 c σ2 σ1σ2 (σ2 σ1 ) = ,

101 and thus

−1 −1 σ2 σ1σ2σ1 = if c is odd,

−1 −1 σ2 σ1σ2 = if c is even.

2 In the ﬁrst case, we see that σ1σ2 = σ2σ1, and since we also have (σ1σ2) = , we see

−1 −1 that σ1 = σ2 . In the second case, we also directly get that σ1 = σ2 , and therefore

σ1σ2 = σ2σ1. In any case, the extra relation from {6, 3}(b,c) is rendered redundant, and we

+ + δ see that Γ (P) Γ (P ) is generated by σ1 and has order 3. We can now determine the full structure of PPδ. Since |Γ+(P)Γ+(Pδ)| = |Γ+(P)|2/3 = 12m2, the polyhedron P Pδ has 24m2 ﬂags. Furthermore, P Pδ is of type {6, 6}, and thus must have 2m2 vertices, 6m2 edges, and 2m2 2-faces. This polyhedron corresponds to a map on an orientable surface of genus m2 + 1.

102 Chapter 6

Appendix

Goursat’s Lemma characterizes the subgroups of a direct product; it appears as an exercise on page 75 of [21], and a proof is given below.

Lemma A.1 (Goursat’s Lemma). Let G1 and G2 be groups, and let H be a subgroup of

G1 ×G2 such that the restrictions of the canonical projections π1 : H → G1 and π2 : H → G2 are surjective. Let N1 = ker π1 and N2 = ker π2. Then we can identify N2 as a normal subgroup of G1 and N1 as a normal subgroup of G2. Furthermore, there is an isomorphism

ϕ : G1/N2 → G2/N1 such that

H '{(g1, g2) | g1 ∈ G1, g2 ∈ G2, ϕ(g1N2) = g2N1}.

Proof. We ﬁrst show that N2 is normal in G1 × 1. It is clear that N2 is a subgroup of G1 × 1.

Let (g1, 1) ∈ G1 × 1. Since π1 is surjective, there exists an element h = (g1, g2) ∈ H for some

−1 g2 ∈ G2. Now, N2 is a normal subgroup of H, so hN2h = N2. In other words, given an

−1 −1 element (n, 1) ∈ N2, it follows that h(n, 1)h ∈ N2, which is to say that (g1ng1 , 1) ∈ N2.

Thus we see that N2 is normal in G1 × 1, and it can be identiﬁed with a normal subgroup of

103 G1. Similarly, N1 can be identiﬁed with a normal subgroup of G2.

Now, consider the homomorphism π : H → G1/N2 × G2/N1 deﬁned as π(g1, g2) =

0 0 0 (g1N2, g2N1). Let h = (g1, g2) ∈ H and h = (g1, g2) ∈ H and suppose that g1N2 =

0 −1 0 −1 0 −1 0 0−1 g1N2. Then g1 g1 ∈ N2 which means that (g1 g1, 1) ∈ H. Then h(g1 g1, 1)h , which is

0−1 0−1 0 equal to (1, g2g2 ), is also an element of H, so that g2g2 ∈ N1 and thus g2N1 = g2N1.

0 0 Conversely, if g2N1 = g2N1, then g1N2 = g1N2 by similar reasoning. Therefore, given any coset g1N2, there is exactly one coset g2N1 such that (g1N2, g2N1) ∈ π(H). So there is a well- defined homomorphism ϕ : G1/N2 → G2/N1 given by ϕ(g1N2) = g2N1, where (g1N2, g2N1) is the unique element of π(H) that has g1N2 as its first coset. By the previous reasoning, this is clearly one-to-one, and it follows immediately that it is a group homomorphism. Furthermore, it is clear by the definition of ϕ that H is as desired.

Proposition A.2. Let H and H0 be subgroups of a group G. Let K be a subgroup of H. Then [H ∩ H0 : K ∩ H0] ≤ [H : K].

Proof. Proposition 4.8 in [17] tells us that if A and B are subgroups of C, then [B : B ∩A] ≤ [C : A]. Now, taking A = K, B = H ∩ H0, and C = H yields the desired result.

Proposition A.3. Let H and H0 be subgroups of a group G. Let K be a subgroup of H, and K0 a subgroup of H0. Then [H ∩ H0 : K ∩ K0] ≤ [H : K][H0 : K0].

Proof. We have that [H ∩ H0 : K ∩ K0] = [H ∩ H0 : K ∩ H0][K ∩ H0 : K ∩ K0]. Applying Proposition A.2 to each term on the right-hand side yields the desired result.

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