UNIVERSITY OF CALGARY

Separation in and X-Raying of Convex Bodies

by

Ryan Gordon Trelford

A THESIS

SUBMITTED TO THE FACULTY OF GRADUATE STUDIES

IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE

DEGREE OF DOCTOR OF PHILOSOPHY

DEPARTMENT OF MATHEMATICS AND STATISTICS

CALGARY, ALBERTA

SEPTEMBER, 2014

c© Ryan Gordon Trelford 2014

UNIVERSITY OF CALGARY

FACULTY OF GRADUATE STUDIES

The undersigned certify that they have read, and recommend to the Faculty of GraduateStudies for acceptance, a thesis entitled “Separation in and X-Raying of Convex Bodies”submitted by Ryan Gordon Trelford in partial fulfillment of the requirements for thedegree of DOCTOR OF PHILOSOPHY.

Supervisor, Dr. Ted BisztriczkyDept. of Mathematics and Statistics

Dr. Thi DinhDept. of Mathematics and Statistics

Dr. Wayne EberlyDept. of Computer Science

Co–Supervisor, Dr. Karoly BezdekDept. of Mathematics and Statistics

Dr. Joseph LingDept. of Mathematics and Statistics

Dr. Barry MonsonDept. of Mathematics and Statistics

University of New Brunswick

Date

Abstract

In this thesis, we introduce an algorithm to compute 6-dimensional totally-sewn neigh-

bourly polytopes. We apply this algorithm to construct a class of semi-cyclic 6-polytopes,

and give a complete description of their faces. We then confirm the Separation Conjec-

ture for these semi-cyclic 6-polytopes. Next, we verify the X-ray Conjecture in the plane

by showing that triangles are the only planar convex sets that cannot be X-rayed along

two lines. We use this result to show that any 3-dimensional convex body exhibiting

mirror symmetry also satisfies the X-ray conjecture.

ii

Acknowledgements

Although having just one author, this thesis is truly the result of the contributions from

many. I surely cannot remember everyone who assisted me during the writing of this

dissertation and I apologize for any oversights.

I would like to thank my supervisor Dr. Ted Bisztriczky for his never-ending support

during my time here. I cannot express how grateful I am for the countless hours he spent

helping me to first understand my research, and to later properly present it in this thesis.

Additionally, the funding that he supplied, particularly during my last year, ensured that

I was able to survive through to the completion of this degree. Although I was not always

the most receptive student, I am truly grateful for, and will never forget, the knowledge

of both mathematics and life that he shared with me.

My thanks also go to my co-supervisor Dr. Karoly Bezdek, whose passion for math-

ematics is truly contagious. His seemingly unending knowledge of discrete mathematics

truly inspired me to learn more about this field. I enjoyed our many meeting and discus-

sions about mathematics, and the entertaining stories he would tell. I hope that I have

the opportunity to work with him again in the future.

I owe a special thank you to my teaching mentor Dr. Thi Dinh. Teaching alongside

him was a remarkable experience, and his advice on how to present material to students

(particularly discrete mathematics) played a major role in the writing of this thesis. He

instilled a sense of confidence in me, and I am a stronger person because of him. I am

also indebted to him for the time he spent listening to me discuss my ideas and results,

and for the comments and suggestions that he made.

Thank you to Dr. Viktor Vıgh, who helped me take my first steps in the vast world

of convex polytopes. Our research collaboration and friendship were both a great expe-

rience, and I learned so much from him.

iii

Also, I must thank my officemates, past and present: Karin Arikushi, Pooyan Shir-

vani Ghomi, Stacey Wynn Lamont, Victoria Labute, Samuel Reid, Brian Chan, Sarah

Couzens and Muhammad Khan. Each of you made my time here more bearable, and

your friendship and support of me is appreciated more than you know.

I cannot forget all of my other mathematical friends that have come and gone dur-

ing my time here: Marie-Andree B.Langlois, Sarah Chisholm, Aaron Christie, Matthew

Couch, Diane Fenton, Azamed Gezahagne, Jan Hubicka, Cameron Hodgins, Lyudmila

Korobenko, Maximiliano Liprandi, Matthew Musson, Jason Nicholson, Mike Pors, Ander

Steele and Colin Weir. You have all supported, challenged and inspired me in so many

different ways, and I am a better person for having known each of you.

I would like to express my gratitude to the administrative staff who on so many

occasions went out of their way to help me and make my life here more bearable. Thank

you to the three Graduate Program Administrators - Joanne Mellard, Carissa Matthews

and Yanmei Fei. They truly care about students and always ensure that our programs

stay on track. Thank you also to the Graduate Program Director Dr. Renate Scheidler

for all of her hard work, particularly to secure funding for many grad students to attend

conferences. She has succeeded in making the graduate program an extremely enjoyable

experience.

Finally, thank you to my family for always being a constant source of support and

encouragement throughout my life.

iv

Dedication

I would like to dedicate this thesis to my wife Yang for her endless support, understanding,

encouragement and patience. Without her, this dissertation would not have been possible.

v

Table of Contents

Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viList of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Neighbourly Polytopes and the Sewing Construction . . . . . . . . . . . . 16

2.1 The Sewing Construction . . . . . . . . . . . . . . . . . . . . . . . . 162.2 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.3 Sewing in Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.4 Keeping Track of Universal Faces . . . . . . . . . . . . . . . . . . . 27

3 Semi-Cyclic 6 -Polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.1 The Facets of Sp(n) . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.2 The Universal Faces of Sp(n) . . . . . . . . . . . . . . . . . . . . . . 463.3 Quotient Polytopes of Sp(n) . . . . . . . . . . . . . . . . . . . . . . 51

4 Separation in Semi-Cyclic 6 -Polytopes . . . . . . . . . . . . . . . . . . . 644.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.2 The Case F ∈ D2(m− 1) . . . . . . . . . . . . . . . . . . . . . . . . 70

4.3 The Case F ∈ D3(m− 1) . . . . . . . . . . . . . . . . . . . . . . . . 784.3.1 The Case j − i Even . . . . . . . . . . . . . . . . . . . . . . 784.3.2 The Case j − i Odd . . . . . . . . . . . . . . . . . . . . . . . 87

4.4 The Case F ∈ D1(m− 1) . . . . . . . . . . . . . . . . . . . . . . . . 924.4.1 The Case j Even . . . . . . . . . . . . . . . . . . . . . . . . 934.4.2 The Case j Odd . . . . . . . . . . . . . . . . . . . . . . . . . 1024.4.3 The Case F = [x1, V2, Vi, xm−1], i = 4, 5, 6 . . . . . . . . . . . 105

4.5 The Case m ∈ {9, 10} . . . . . . . . . . . . . . . . . . . . . . . . . . 1065 X-Raying Mirror Symmetric Convex Bodies in R3 . . . . . . . . . . . . . 109

5.1 Illumination and X-raying . . . . . . . . . . . . . . . . . . . . . . . 1095.2 The X-Ray Conjecture in the Plane . . . . . . . . . . . . . . . . . . 1125.3 X-Raying Convex Bodies of R3 Symmetric about a Plane . . . . . . 116

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

vi

List of Tables

2.1 The dependancies among the steps in Algorithm 2. . . . . . . . . . . . . 32

3.1 The facets of S9(12). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

vii

List of Figures

1.1 An example of using quotient polytopes to indicate separation. . . . . . . 101.2 A cyclic polytope with a non-cyclic subpolytope. . . . . . . . . . . . . . . 15

2.1 Beyond and beneath facets . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 The case i = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.1 Sp(n)/[x1, Vi, xn] for 3 ≤ i ≤ p− 4. . . . . . . . . . . . . . . . . . . . . . 523.2 Sp(n)/[Vi, Vn−1] for 3 ≤ i ≤ p− 4. . . . . . . . . . . . . . . . . . . . . . . 533.3 Sp(n)/[V1, V3]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.4 Sp(n)/[V1, Vn−1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.5 Sp(n)/[x1,Wp−4, xn]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.6 Sp(n)/[x1,Wp−3, xn]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.7 Sp(n)/[x1,Wp−2, xn]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.8 Sp(n)/[x1,Wi, xn] for p− 1 ≤ i ≤ n− 4 and i− p even. . . . . . . . . . . 563.9 Sp(n)/[x1,Wi, xn] for p− 1 ≤ i ≤ n− 4 and i− p odd. . . . . . . . . . . . 573.10 Sp(n)/[x1,Wi, xn] for p− 4 ≤ i ≤ n− 4. . . . . . . . . . . . . . . . . . . . 583.11 Sp(n)/[Wp−4, Vn−1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.12 Sp(n)/[Wp−3, Vn−1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.13 Sp(n)/[Wp−2, Vn−1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.14 Sp(n)/[Wi, Vn−1] for p− 1 ≤ i ≤ n− 4 and i− p even. . . . . . . . . . . . 603.15 Sp(n)/[Wi, Vn−1] for p− 1 ≤ i ≤ n− 4 and i− p odd. . . . . . . . . . . . 613.16 Sp(n)/[Wi, Vn−1] for p− 4 ≤ i ≤ n− 4. . . . . . . . . . . . . . . . . . . . 613.17 Sp(n)/[Vi, Vn−1] for p− 3 ≤ i ≤ n− 4. . . . . . . . . . . . . . . . . . . . . 623.18 Sp(n)/[Vn−3, Vn−1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.19 Sp(n)/[x1, V2, xn]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.20 Sp(n)/[V2, Vn−1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.1 From separation to strict separation. . . . . . . . . . . . . . . . . . . . . 664.2 S(m)/[x1,Wi, xm] and the hyperplanes H1, H2 and H3. . . . . . . . . . . 714.3 S(m)/[Wi, Vm−1] and the hyperplanes H2, H3 and H4. . . . . . . . . . . . 724.4 S(m)/[V1, Vm−1] and the hyperplanes H2, H5 and H6. . . . . . . . . . . . 734.5 S(m)/[V1, V3] and the hyperplane H7. . . . . . . . . . . . . . . . . . . . . 744.6 S(m)/[x1,Wi−1, xm] and the hyperplane H8. . . . . . . . . . . . . . . . . 754.7 S(m)/[x1,Wi, xm] and hyperplanes H1, H2, H3, and H4. . . . . . . . . . 794.8 S(m)/[V1, Vm−1], j − i even, and the hyperplane H6. . . . . . . . . . . . . 804.9 S(m)/[x1,Wj−1, xm], j − i even, and the hyperplanes H7, H8, H9, and H10. 814.10 S(m)/[Vj, Vm−1], j − i even, and the hyperplanes H5, H11 and H12. . . . . 824.11 S(m)/[V1, Vm−1], j − i odd, and the hyperplane H6. . . . . . . . . . . . . 884.12 S(m)/[x1,Wj−1, xm], j − i odd, and the hyperplanes H7, H

′8, H9 and H ′10. 89

4.13 S(m)/[Vj, Vm−1], j − i odd, and the hyperplanes H5, H′11 and H ′12. . . . . 90

4.14 S(m)/[x1, Vi, xm] and the hyperplanes H1, H2, H3 and H4. . . . . . . . . . 944.15 S(m)/[V1, Vm−1] and the hyperplane H6. . . . . . . . . . . . . . . . . . . 95

viii

4.16 S(m)/[x1,Wj−1, xm], j even, and the hyperplanes H7, H8, H9 and H10. . . 964.17 S(m)/[Vj, Vm−1], j even, and the hyperplanes H5, H11 and H12. . . . . . . 974.18 S(m)/[x1,Wj−1, xm], j odd, and the hyperplanes H7, H

′8, H9 and H ′10. . . 103

4.19 S(m)/[Vj, Vm−1], j odd, and the hyperplanes H5, H′11 and H ′12. . . . . . . 104

5.1 The planar convex body C and the supporting lines `xv1, `xv2

, ` pv1and ` qv2

. 115

ix

1

Introduction

Discrete Geometry is a field of mathematics concerned with arrangements of geometric

objects and their combinatorial properties. Among the many well-known topics in this

field (see [2] [19]), covering problems and sphere packings are two of the more widely

studied. Discrete Geometry has also become a more active area of research with the

increased use of computers to aid with difficult computations.

One of the famous conjectures in Discrete Geometry is the Gohberg-Markus-Hadwiger

(G-M-H) Covering Conjecture. It is that every d-dimensional convex body K may be

covered by at most 2d smaller homothetic copies of K. Formulated independently by

Gohberg and Markus [14] and by Hadwiger [16], it has been verified in the plane by Levi

[18], but remains open in dimensions greater than two.

In an effort to verify the Covering Conjecture, several equivalent formulations have

been introduced. The first formulation deals with illumination. A point x on the bound-

ary of a d-dimensional convex body K is said to be illuminated by a non-zero vector

v if the open ray emanating from x with direction v intersects the interior of K. A

collection of directions v1,v2, . . . ,vn is said to illuminate K if every boundary point

of K is illuminated by one of v1,v2, . . . ,vn. The Illumination Conjecture, formulated

by Boltyanski and Hadwiger, states that every d-dimensional convex body can be illu-

minated by at most 2d directions, and its equivalence to the Covering Conjecture was

shown by Boltyanski in [6].

A second equivalent formulation of the Covering Conjecture is the Separation Con-

jecture. For a convex body K and a point O in the interior of K, a hyperplane H is

said to strictly separate O from a face F of K if F belongs to one of the open halfspaces

determined by H and O lies in the other. The Separation Conjecture asserts that for

any d-dimensional convex body K and any point O in the interior of K, at most 2d hy-

2

perplanes are necessary to strictly separate any face of K from O. In [1], Bezdek showed

that if a convex body K is illuminated by n directions, then any interior point O of K∗,

the dual of K, is strictly separated from any face of K∗ by one of at most n hyperplanes.

Thus, the Separation Conjecture is a dual formulation of the Covering Conjecture. We are

interested in the Separation Problem: given any d-dimensional convex body K with O in

the interior of K, find the minimum number of hyperplanes, s(O,K), required to strictly

separate any face of K from O. We define s(K) = max{s(O,K) |O in the interior of K}.

If s(K) ≤ 2d then the Separation Conjecture is verified for K.

One of the main objects studied in this thesis are neighbourly d-dimensional polytopes

for d ≥ 4. Described by Grunbaum [15] as a “rather freakish family of polytopes,”

they are those d-dimensional polytopes P with the property that every bd/2c vertices

determine a face of P . They were first studied by Caratheodory [8, 9] in his work

on cyclic polytopes, and now play a prominent role in polytope theory, most notably

in McMullen’s proof of the famous Upper Bound Theorem [20]. In 1982 Shemer [21]

introduced the Sewing Construction, allowing one to add or “sew” a vertex onto an even

dimensional neighbourly polytope with n vertices to obtain a neighbourly polytope with

n+1 vertices. Shemer showed also that the number of combinatorial types of neighbourly

polytopes grows super-exponentially with the number of vertices.

We are interested in solving the Separation Problem for neighbourly polytopes. This

study began in 1997, when Bezdek and Bisztriczky [3] proved the following:

Theorem 1 (Cyclic Separation Theorem). Let C ⊂ Rd be a cyclic d-polytope, d ≥ 3.

Then there is a number s(d) ≤ 2d such that s(C) ≤ s(d). In particular, s(3) = 6 and for

m ≥ 2,

s(2m) = s(2m− 1) +

(2m− 1

m

)and

s(2m+ 1) = 2s(2m).

3

In 1999, Talata [24] showed that if C is a cyclic 2m-polytope, m ≥ 2, then s(C) =

(m+ 1)2. Since then, all results concerning neighbourly d-polytopes and the Separation

Problem have been been obtained only for d = 4. Finbow-Singh and Oliveros [12]

obtained results for 4-dimensional semi-cyclic polytopes, and recently Bisztriczky and

Fodor (to appear) extended their result to totally-sewn 4-polytopes. In this thesis, we

introduce a class of semi-cyclic 6-polytopes P and examine the Separation Problem for

this class. We show that s(P ) ≤ 29, and thus verify the Separation Conjecture for

these semi-cyclic 6-polytopes. These are the first results of the Separation Problem for

non-cyclic neighbourly 6-polytopes.

After introducing the necessary terminology and concepts in Chapter 1, we consider

Shemer’s sewing construction in Chapter 2. Let P be a neighbourly 6-polytope with n

vertices and P+ be a neighbourly 6-polytope with n + 1 vertices obtained by sewing a

vertex x+ onto P . We introduce a recursive algorithm that returns the list of facets of P+

given the list of facets of P , and then extend the algorithm to return the list of universal

faces of P+ given the list of universal faces of P . We use this algorithm in Chapter 3 to

give a complete description of a class of semi-cyclic 6-polytopes. We construct first the

list of facets of these polytopes, and then describe the universal edges and universal 3-

faces. Finally, we give representations of several quotient polytopes used in the following

chapter.

Chapter 4 verifies the Separation Conjecture for the semi-cyclic 6-polytopes. We show

that if P is a semi-cyclic 6-polytope with O in the interior of P , then there is a collection

H of at most twenty-nine hyperplanes such that O is strictly separated from any face of

P by some H ∈ H.

Chapter 5 deals with X-raying, a variant of illumination. Let K be a d-dimensional

convex body, x be a boundary point of K, and v be a non-zero vector. We say that x

is X-rayed along a line with direction v if the line passing through x with direction v

4

intersects the interior of K, that is, if either v or −v illuminate x. The convex body K

is X-rayed by a collection of lines with directions v1,v2, . . . ,vn if every boundary point

x of K is X-rayed along one of the lines. The X-ray Conjecture (Bezdek and Zamfirescu

[4]) asserts that every d-dimensional convex body K is X-rayed along at most 3 · 2d−2

lines. We verify the X-ray Conjecture in the plane, and show that 3 lines are necessary

if, and only if, the planar convex body is a triangle. We then use this result to show that

any 3-dimensional convex body that is symmetric about a plane may be X-rayed along

at most 6 lines.

5

Chapter 1

Background

We begin with a brief introduction to the basic tools and concepts of convex polytopes

that the reader is assumed to know in order to understand this thesis. A more detailed

exposition appears in [7], [15], [21], and [26].

Let R denote the set of real numbers, and let Rd = {(a1, . . . , ad) | ai ∈ R, i = 1, . . . , d}

denote the real d-dimensional space. An element x ∈ Rd will be refered to as either a

point or a vector. We equip Rd with the standard inner product 〈· , ·〉, and norm || · ||,

and for a set A ⊂ Rd, we assume the reader is familiar with the closure cl(A) of A, the

interior int(A) of A, and the boundary bd(A) of A.

Let X = {x1, . . . , xn} be a collection of n ≥ 1 distinct vectors in Rd. A linear

combination of X is a vector of the form λ1x1 + · · ·+λnxn where λi ∈ R for i = 1, . . . , n.

The set X is called linearly independent when λ1x1 + · · · + λnxn is the zero vector if,

and only if, λ1 = · · · = λn = 0. We say that X is linearly dependent if X is not linearly

independent. The set of all linear combinations of X will be denoted by span(X).

A set L ⊂ Rd is called a linear subspace if every linear combination of points from L is

contained in L. A basis for a linear subspace L is any maximal linearly independent set of

L, that is, any linearly independent set B ⊂ L with L = span(B). The dimension of L,

dim(L), is the cardinality of any basis of L. The intersection of any two linear subspaces

is again a linear subspace, and for a given set M ⊂ Rd, span(M) is the intersection of

all the linear subspaces that contain M .

For X = {x1, . . . , xn} ⊂ Rd, an affine combination of X is any linear combination

λ1x1 + · · · + λnxn of X where λi ∈ R for i = 1, . . . , n, and λ1 + · · · + λn = 1. The

set X is called affinely independent when both λ1x1 + · · · + λnxn is the zero vector and

6

λ1 + · · · + λn = 0 if, and only if, λ1 = · · · = λn = 0. We call X affinely dependent if

it is not affinely independent. We note that X is affinely independent if, and only if,

{x2− x1, x3− x1, . . . , xn− x1} is linearly independent. The set of all affine combinations

of X will be denoted by aff(X), or by 〈X〉.

A set A ⊂ Rd is called an affine subspace or a flat if every affine combination of

points from A is contained in A. An affine basis for an affine subspace A is an affinely

independent set B ⊂ A, with A = 〈B〉. The dimension of A, dim(A), is given by

dim(A) = |B| − 1 for any affine basis B of A. If A = ∅, then dim(A) = −1 by definition.

The intersection of any two affine subspaces is an affine subspace, and for any M ⊂ Rd,

〈M〉 is the intersection of all affine subspaces that contain M , and the dimension of M

is dim(M) = dim(〈M〉).

A set K ⊂ Rd is called convex if, and only if, for any x, y ∈ K, λx + (1 − λ)y ∈ K

with 0 ≤ λ ≤ 1. For X = {x1, . . . , xn} ⊂ Rd, a convex combination of X is any point of

the form λ1x1 + · · ·+ λnxn with λ1 + · · ·+ λn = 1 and 0 ≤ λi ≤ 1 for i = 1, . . . , n. The

convex hull of X is the set of all convex combinations of X, and is denoted by conv(X)

or [X]. The intersection of convex sets is a convex set, and for a set M ⊂ Rd, [M ] is the

intersection of all convex sets that contain M . The convex set K is called a convex body

if K is closed, bounded, and dim(K) = d.

A set H ⊂ Rd is called a hyperplane if there is a y ∈ Rd, y 6= 0, and an a ∈ R such that

H = {x ∈ Rd | 〈x, y〉 = a}. We note that H is an affine (d− 1)-dimensional subspace. A

hyperplane H induces two open half-spaces, {x ∈ R | 〈x, y〉 < a} and {x ∈ R | 〈x, y〉 > a},

as well as two closed half-spaces, {x ∈ R | 〈x, y〉 ≤ a} and {x ∈ R | 〈x, y〉 ≥ a}.

For a given set X ⊂ Rd, a hyperplane H is called a supporting hyperplane of X if

H ∩X 6= ∅, and X is contained in one of the closed halfspaces determined by H. If H is

a supporting hyperplane of X then the closed halfspace determined by H that contains

X is called a supporting halfspace of X. For any Y ⊂ Rd, a hyperplane H separates X

7

and Y if X lies in one of the closed half-spaces determined by H, while Y lies in the

other, and H strictly separates X and Y if X lies in one of the open half-spaces induced

by H and Y lies in the other open half-space.

Let X = {x1, . . . , xn} ⊂ Rd. The set P = [X] is called a convex polytope, or simply a

polytope, and if dim(P ) = d, then we refer to P as a d-polytope. Let P be a d-polytope,

and let F ⊂ P . We call F a face of P if F = ∅, F = P , or if there exists a supporting

hyperplane H of P such that F = H ∩ P . If F = P , then F is an improper face of P ;

otherwise, F is a proper face of P . Note that each proper face F of P is itself a convex

polytope, and that P has finitely many faces. We call a 0-dimensional face a vertex, a

1-dimensional face an edge, and a (d− 1)-dimensional face a facet. We denote the set of

all vertices of P by V(P ), the set of all edges of P by E(P ), and the set of all facets of

P by F(P ). More generally, Fj(P ) denotes the set of all j-dimensional faces (or simply

j-faces) of P for j = −1, 0, . . . , d. For Y ⊂ V(P ), we call P ′ = [Y ] a subpolytope of P .

Remark 1.1.1. Let P ⊂ Rd be a d-polytope. Then

1. If F is a face of P and G is a face of F , then G is a face of P .

2. If F1, . . . , Fk are faces of P then⋂kj=1 Fj is a face of P .

3. Every proper face of P is the intersection of the facets of P that contain it. In

particular, every (d− 2)-face of P is contained in exactly two facets of P .

4. If Fi ∈ Fi(P ) and Fk ∈ Fk(P ) with Fi ⊂ Fk and −1 ≤ i < k ≤ d, then for

j = i+1, . . . , k−1, there are faces Fj ∈ Fj(P ) so that Fi ⊂ Fi+1 ⊂ · · · ⊂ Fk−1 ⊂ Fk.

5. If G is a k-face of P and P ′ is a subpolytope of P with V(G) ⊆ V(P ′), then G is a

k-face of P ′.

A k-polytope is called a k-simplex if it has k+ 1 vertices (the vertices are necessarily

affinely independent). A d-polytope is called simplicial if all of its facets are simplices.

8

The boundary complex B(P ) of a d-polytope P is the set of all proper faces of P ;

that is, B(P ) =⋃d−1j=−1Fj(P ). We define (L(P ),⊆), where L(P ) = B(P ) ∪ {P}, to be

the face lattice of P : the set of all faces of P partially ordered by inclusion. Remark

1.1.1(3) guarantees that the face lattice of P is determined entirely by F(P )∪{P}. Two

polytopes P and Q are combinatorially equivalent if there exists an inclusion preserving

one-to-one correspondence between F(P ) and F(Q), that is, if (L(P ),⊆) and (L(Q),⊆)

are isomorphic, and in this case we write P ∼= Q. We call P and Q dual to one another if

there exists an inclusion reversing one-to-one correspondence between F(P ) and F(Q),

that is, if (L(P ),⊆) and (L(Q),⊆) are anti-isomorphic.

With reference to Remark 1.1.1(4), we note that {G ∈ L(P ) |Fi ⊆ G ⊆ Fk} is a

sublattice of L(P ). It is denoted by Fk/Fi and it is called the quotient polytope determined

by Fi and Fk. It is known (cf. [15]) that Fk/Fi is the face lattice of a (k − i − 1)-

polytope. What is of importance to us are realizations of Fk/Fi. Let P ⊂ Rd be a

d-polytope, x ∈ V(P ), and H be a hyperplane that strictly separates x and V(P ) \ {x}.

The vertex figure of P at x is the (d−1)-polytope P ∩H, and it is a realization of P/{x}

(which we denote as P/x). For each z ∈ V(P ) \ {x}, define {z∗} = [z, x] ∩ H. Then

P/x ∼= conv{z∗ | z ∈ V(P ) \ {x}}. Note that z∗ ∈ P ∩H is a vertex of P ∩H if, and only

if, [z, x] ∈ E(P ). As a rule, we denote any such P ∩H by P/x.

Let P be a simplicial d-polytope, [x, z] ∈ E(P ), and d ≥ 2. Then P ∗ = P/x is a

(d − 1)-polytope, and we denote by w∗ the vertex of P ∗ that corresponds to [x,w] ∈

E(P ). Then z∗ ∈ V(P ∗), and we consider P ∗∗ = P ∗/z∗ = (P/x)/z∗. We denote by

y∗∗ the vertex of P ∗∗ that corresponds to the edge [y∗, z∗] in P ∗, and hence, the 2-face

[x, y, z] in P . In fact, we note that P ∗∗ = (P/x)/z∗ is a realization of P/[x, z] and

P ∗∗ = conv{y∗∗ | y ∈ V(P ) \ {x, z}}.

Let H be a hyperplane that strictly separates x and V(P ) \ {x}, and H∗ be an affine

(d−2)-space in H that strictly separates z∗ from V(P ∗)\{z∗}. Then for any 2-face [x, y, z]

9

of P , y∗∗ = H∗∩ [y∗, z∗] = H∗∩ [H∩ [y, x], H∩ [z, x]] = H∗∩ [H∩ [x, y, z]] = H∗∩ [x, y, z],

and 〈x,H∗〉 separates y and z.

For a proper k-face G = [x1, . . . , xk+1] of a simplicial d-polytope P , a realization

of P/G is given by (P/[x1, . . . , xk])/xk+1∼= P/[x1, . . . , xk+1] and henceforth, we assume

that each realization is obtained as above by iteration. We summarize this construction

below.

Theorem 1.1.2. Let P be a simplicial d-polytope, G = [y0, y1, . . . , yk] be a k-face for

some k = 0, . . . , d− 1, and {z1, . . . , z`} ⊂ V(P ) \ V(G). For j = d− k − 1, . . . , d− 1, let

Hj be a j-flat such that Hd−1 ⊃ Hd−2 ⊃ · · · ⊃ Hd−k−1 and P/[y0, y1, . . . , yj−1] is realized

in Hd−j. Then

1. P = P/G is a (d− k − 1)-polytope in Hd−k−1.

2. z ∈ V(P ) if, and only if, there is a z ∈ V(P ) such that {z} = 〈G, z〉 ∩Hd−k−1 and

[G, z] is a (k + 1)-face of P .

3. [z1, . . . , z`] is a face of P/G if, and only if, [G, z1, . . . , z`] is a face of P .

4. A (d− k − 2)-flat H separates y and z in Hk+1 if, and only if, 〈G,H〉 separates y

and z in Rd.

Example 1.1.3. Let P be a simplicial d-polytope and G be a (d − 3)-face of P . Then

P/G is a polygon, and x∗ ∈ P/G is a vertex of P/G if, and only if, [G, x] is a (d − 2)

face of P , and two vertices y∗ and z∗ of P/G determine an edge of P/G if, and only if,

[G, y, z] is a (d− 1)-face of P . In Figure 1.1a, we see that the (d− 3)-face G is contained

in five (d − 2)-faces: [G, y] for y ∈ {xi, xj, xk, xl, xm}, and in exactly five (d − 1)-faces:

[G, y, z] for {y, z} ∈ {{xi, xj}, {xj, xk}, {xk, xl}, {xl, xm}, {xi, xm}}. Furthermore, Figure

1.1b shows that the line L∗ = 〈x∗k, x∗m〉 strictly separates a∗ and b∗ in H2, and thus

〈G, xk, xm〉 strictly separates a and b in Rd.

10

(a) The quotient polytope P/G ⊂ H2. (b) Separation in P/G.

Figure 1.1: An example of using quotient polytopes to indicate separation.

Example 1.1.3 shows how we use quotient polytopes to gain information about the

facial structure of a polytope, as well as showing when a hyperplane spanned by vertices

of the polytope separates two points. We can view the realizations of a vertex figure as

a projection of the polytope onto a given hyperplane.

Remark 1.1.4. We define a projected quotient polytope to be a quotient polytope ob-

tained via a sequence of vertex figures such that each vertex figure is realized as the convex

hull of the intersection of the polytope and a hyperplane that separates the given vertex

from all of the other vertices. Henceforth, all quotient polytopes are assumed to be pro-

jected.

Let A ⊂ Rd. The convex set A◦ = {x ∈ Rd | 〈x, y〉 ≤ 1 for every y ∈ A} is the polar

set of A. If the origin of Rd is contained in int(A), then A◦ is bounded, and if A is

bounded, then the origin is in the interior of A◦. If P is a d-polytope containing the

origin of Rd, then Q = P ◦ is also a d-polytope, and Q is dual to P .

The following class of polytopes play an important role in the theory of polytopes,

and they are central to this thesis. Let P be a d-polytope. For k = 1, . . . , d, we say that

P is k-neighbourly if any k vertices of P determine a proper face of P .

11

Theorem 1.1.5 (Grunbaum [15]). Let P be a k-neighbourly d-polytope.

1. Every k vertices of P form an affinely independent set.

2. P is j-neighbourly for j = 1, . . . , k.

3. For any W ⊆ V(P ) with |W | ≥ k + 1, [W ] is k-neighbourly.

4. If k > bd/2c, then P is a d-simplex.

5. For 0 ≤ j ≤ k, |Fj(P )| =(|V(P )|j + 1

).

6. Every face of P with dimension at most 2k − 1 is a simplex.

Of particular interest are those d-polytopes that are bd/2c-neighbourly. Such poly-

topes are referred to as neighbourly d-polytopes. For d even, Theorem 1.1.5(6) states

that P is simplicial.

Let P be a neighbourly d-polytope, and S ⊂ V(P ). We call [S] a missing face of P if

[S] is not a face of P , but every subpolytope of [S] is a face of P . We know (cf. Shemer

[21]) that the missing faces of P are bd/2c-simplices.

The most well-known class of neighbourly polytopes are the cyclic polytopes. In Rd,

the moment curve Md is the curve x(t) = (t, t2, . . . , td) for t ∈ R; d ≥ 2. It is known

that a hyperplane H of Rd contains at most d points from Md(R). The convex hull of

any distinct n ≥ d+ 1 points from Md(R) is (cf. [13]) combinatorially equivalent to the

convex hull of any other distinct n points from Md(R). We denote such a polytope by

C(n, d). From above, we note that C(n, d) is simplical.

Let n ≥ d + 1, t1 < t2 < · · · < tn ∈ R, xi = x(ti), and C(n, d) = [x1, . . . , xn]. Then

V(C(n, d)) = {x1, . . . , xn} and we observe that C(d + 1, d) is a simplex. We define a

total ordering on the vertices of C(n, d) by xi < xj if ti < tj. We refer to this ordering

as the vertex array of C(n, d). A vertex xk is between xi and xj if either xi < xk < xj or

xj < xk < xi.

12

Theorem 1.1.6 (Gale’s Evenness Condition [13]). Let V(C(n, d)) = {x1, . . . , xn}, n ≥

d+ 1 and x1 < x2 < · · · < xn. Then a set W of d vertices of C(n, d) determines a facet

of P if, and only if, there are an even number of vertices in W between any two vertices

of V(C(n, d)) \W .

As an example with d = 6 and n = 9, Gale’s Evenness Condition states that the

facets of C(9, 6) are:

[x1, x2, x3, x4, x5, x6], [x1, x2, x3, x4, x6, x7], [x1, x2, x3, x4, x7, x8], [x1, x2, x3, x4, x8, x9],

[x1, x2, x4, x5, x6, x7], [x1, x2, x4, x5, x7, x8], [x1, x2, x4, x5, x8, x9], [x1, x2, x5, x6, x7, x8]

[x1, x2, x5, x6, x8, x9], [x1, x2, x6, x7, x8, x9], [x1, x2, x3, x4, x5, x9], [x1, x2, x3, x5, x6, x9],

[x1, x2, x3, x6, x7, x9], [x1, x2, x3, x7, x8, x9], [x1, x3, x4, x5, x6, x9], [x1, x3, x4, x6, x7, x9],

[x1, x3, x4, x7, x8, x9], [x1, x4, x5, x6, x7, x9], [x1, x4, x5, x7, x8, x9], [x1, x5, x6, x7, x8, x9],

[x2, x3, x4, x5, x6, x7], [x2, x3, x4, x5, x7, x8], [x2, x3, x4, x5, x8, x9], [x2, x3, x5, x6, x7, x8],

[x2, x3, x5, x6, x8, x9], [x2, x3, x6, x7, x8, x9], [x3, x4, x5, x6, x7, x8], [x3, x4, x5, x6, x8, x9],

[x3, x4, x6, x7, x8, x9], [x4, x5, x6, x7, x8, x9].

Finally, a d-polytope P is cyclic if P ∼= C(n, d) for some n ≥ d+1. Thus, a d-polytope

P is cyclic if, and only if, P is simplicial and there is a vertex array of P so that P satisfies

Gale’s Evenness Condition.

We now state some results from the theory of neighbourly polytopes that we use in

subsequent chapters.

Theorem 1.1.7 (Shemer [22]). Let C be a cyclic 2m-polytope with V(C) =

{x1, . . . , x2m+2}, and vertex array x1 < x2 < · · · < x2m+2. Then C has two missing

faces: [x1, x3, . . . , x2m+1] and [x2, x4, . . . , x2m+2].

13

Let U be a k-face of a neighbourly 2m-polytope P . Then U is a universal face of P

if [U, S] is a face of P for all S ⊂ V(P ) such that |S| ≤ b(2m − k − 1)/2c. It follows

from the definition that U is a universal k-face of P if, and only if, P/U is a neighbourly

(2m − k − 1)-polytope. The set of all universal k-faces of P are denoted by Uk(P ). It

follows from Remark 1.1.1(5) that if U is a universal face of P , then U is a universal face

of any subpolytope P ′ of P such that V(U) ⊂ V(P ′).

It is known that every edge of a 2m-simplex is universal, and for a cyclic polytope with

2m + 2 vertices and vertex array x1 < x2 < · · · < x2m+2, E = [xi, xj] is a universal edge

if, and only if, i− j is odd. The following theorem, which follows from Gale’s Evenness

Condition, characterizes all odd dimensional universal faces for cyclic 2m-polytopes with

at least 2m+ 3 vertices.

Theorem 1.1.8. Let C be a cyclic 2m-polytope with n ≥ 2m + 3 vertices and vertex

array x1 < · · · < xn, and G ∈ F2k−1(C) for some k = 1, . . . ,m. Then G ∈ U2k−1(C) if,

and only if, V(G) satisfies Gale’s Evenness Condition.

In particular, an edge E of a cyclic 2m polytope with n ≥ 2m+ 3 vertices is universal

if, and only if, E = [x1, xn], or E = [xi, xi+1] for some i = 1, . . . , n − 1. Thus, a cyclic

2m-polytope with n ≥ 2m + 3 vertices has exactly n universal edges, with each vertex

lying in exactly two universal edges. It was shown (cf. Shemer [23]) that a neighbourly

2m-polytope with n ≥ 2m + 3 vertices is cyclic if, and only if, it has n universal edges,

and that a neighbourly non-cyclic 2m-polytope has at least 2m+ 4 vertices and at most

n− 2 universal edges.

Theorem 1.1.9 (Gale [13]). Let P be a neighbourly 2m-polytope. If |V(P )| ≤ 2m + 3,

then P is cyclic.

Theorem 1.1.10 (Shemer [21]). If U1 and U2 are disjoint universal faces of a neighbourly

2m-polytope P , then [U1, U2] is a universal face of P provided |V(U1)|+ |V(U2)| ≤ 2m+1.

14

Theorem 1.1.11 (Shemer [22]). Let P be a neighbourly 2m-polytope with at least 2m+2

vertices, and H be any hyperplane of R2m spanned by 2m vertices of P . Let E be a

universal edge of P whose vertices do not lie in H. Then E is contained in one of the

open halfspaces determined by H.

Proof. Let H be a hyperplane that is spanned by 2m vertices of P , and let E = [xi, xj] be

a universal edge of P such that xi, xj /∈ H. Consider the subpolytope P ′ of P determined

by xi, xj, and the 2m vertices of P that span H. Then P ′ is a neighbourly 2m-polytope

with 2m + 2 vertices, and so P ′ is cyclic. We relable the vertices of P ′ as y1, . . . , y2m+2

so that they satisfy the vertex array y1 < y2 < · · · < y2m+2. By Theorem 1.1.7, the

missing faces of P ′ are G1 = [y1, y3, . . . , y2m+1] and G2 = [y2, y4, . . . , y2m+2]. Since E is a

universal edge of P ′, we have that every m-dimensional subpolytope of P ′ that contains

E is a face of P ′, and so E 6⊂ G1, E 6⊂ G2, and E = [yk, yk+`] with ` odd. Hence by Gale’s

Evennness Condition, F = [y1, . . . , yk−1, yk+1, . . . , yk+`−1, yk+`+1, . . . , y2m+2] is a facet of

P ′, H = 〈F 〉 and E is contained in one of the open halfspaces determined by H.

The facets of a neighbourly d-polytope necessarily determine the facets of any of its

subpolytopes only if d is even (cf. Shemer [21]). In odd dimensions, this need not happen.

For example, let P be a 3-polytope of type C(8, 3) with vertex array x1 < · · · < x8,

and let Q be the polytope depicted in Figure 1.2a. Then P and Q are combinatorially

equivalent, with the inclusion preserving isomorphism satisfying xi 7−→ yi for i = 1, . . . , 8.

The subpolytope P ′ = [x1, . . . , x7] of P is a 3-polytope of type C(7, 3), however, the

subpolytope Q′ = [y1, . . . , y7] of Q (see Figure 1.2b) is not simplicial and is thus (despite

satisfying Gale’s Evenness Condition) not cyclic. This observation led to neighbourly

2m-poltopes being heavily studied.

In Chapter 3, we give a detailed construction of a class of neighbourly 6-polytopes

which we refer to as semi-cyclic 6-polytopes. Then in Chapter 4, we consider the problem

15

(a) The 3-polytope Q is cyclic. (b) The subpolytope Q′ of Q is not cyclic.

Figure 1.2: A cyclic polytope with a non-cyclic subpolytope.

of finding a collection of hyperplanes H such that a given interior point O of a semi-cyclic

6-polytope P is strictly separated from every facet of P by at least one of the hyperplanes

in H. We accomplish this by locating O∗ in polygons P/G for some 3-faces G of P .

16

Chapter 2

Neighbourly Polytopes and the Sewing Construction

In Chapter 1, we introduced neighbourly polytopes, and constructed the cyclic polytopes

as an example of neighbourly polytopes. From the definition of neighbourly polytopes,

it is clear that all 2- and 3-polytopes are neighbourly. However, it was believed until the

1950’s that for d ≥ 4, the class of neighbourly polytopes was exactly the class of cyclic

polytopes. In 1982, Shemer [21] introduced a process called sewing, whereby one adds

a vertex to a given neighbourly 2m-polytope with n vertices and obtains a neighbourly

2m-polytope with n+ 1 vertices. Shemer showed that the cyclic polytopes are obtained

in this way, and that the number of combinatorial types of neighbourly 2m-polytopes

with n vertices grows super-exponentially as n grows. In this Chapter, we introduce the

sewing construction for neighbourly 2m-polytopes, and develop an algorithm to recover

the universal faces of the sewn polytope.

2.1 The Sewing Construction

We define the sewing process introduced by Shemer in [21]. Let P denote a neighbourly

2m-polytope with n ≥ 2m + 3 vertices, m ≥ 2. Let x1, x2, . . . , xm and y1, y2, . . . , ym be

distinct vertices of P , and define Φ1 = [x1, y1], Φj = [Φj−1, xj, yj] for 2 ≤ j ≤ m, and

Φ0 = ∅. Then Φj is a (2j−1)-polytope and T = {Φ1, . . . ,Φm} is a universal tower in P if

Φj ∈ U2j−1(P ) for j = 1, . . . ,m. We denote by Fj, the set of facets of P that contain Φj,

with the additional convention that Fj = ∅ if j > m and F0 = F(P ). Furthermore, let

C(T ) = (F1 \F2)∪ (F3 \F4)∪ · · · . Let F be a facet of P . We say that a point x /∈ 〈F 〉

is beyond F (with respect to P ) if 〈F 〉 separates P and x; otherwise, x is beneath F . A

17

point x is exactly beyond C(T ) if it is beyond F for every F ∈ C(T ), and beneath F for

every F ∈ F0 \ C(T ). Lemma 4.4 in [21] states that for every universal tower T , there

exists a point x+ = x+(T ) that lies exactly beyond C(T ).

Figure 2.1 should clarify the position of x+ with respect to the facets of P . In Category

I are those facets of P that do not contain the sewing edge Φ1; x+ is beneath these facets.

In the Category II are those facets of P that contain the sewing edge Φ1, but do not

contain the sewing 3-face Φ2; x+ is beyond these facets, and so on. The existence of x+

is fundamental to the sewing construction as it guarantees that the obtained polytope is

also neighbourly.

In this section, we assume that P is a neighbourly 2m-polytope that contains a

universal tower T defined in terms of x1, x2, . . . , xm and y1, y2, . . . , ym as above. The

importance of the following theorem lies in the fact that it allows one to construct infinite

families of neighbourly polytopes.

Φ0 beneath Category I

Φ1 beyond Category II

Φ2 beneath Category III

...

Figure 2.1: Beyond and beneath facets

Theorem 2.1.1 (Shemer [21], Theorem 4.6). Let T be a universal tower of P , and x+

lie exactly beyond C(T ). Then P+ = [P, x+] is a neighbourly 2m-polytope, and V(P+) =

18

V(P ) ∪ {x+}. We say that P+ is obtained by sewing the vertex x+ onto the polytope P

through T .

By varying the universal tower, the combinatorial structure of P+ may change. Thus,

understanding the behaviour of the universal faces is very important. The following

theorem indicates which elements of T are universal in P+, and also yields some new

universal faces of P+.

Theorem 2.1.2 (Shemer [21], Theorem 4.6). Let P+ = [P, x+] be the neighbourly 2m-

polytope obtained by sewing x+ onto P through the tower T , and let 1 ≤ j ≤ m.

1. If j < m, then Φj is a face of P+.

2. Φj is a universal face of P+ if, and only if, j is even.

3. [Φj−1, xj, x+] is a universal (2j − 1)-face of P+.

Shemer [21] also described all of the universal faces of a neighbourly 2m-polytope in

terms of its missing faces.

Definition 2.1.3 (Shemer [21], Definition 4.1). If G is a proper face of P and M ⊆

V(P ) \ V(G), then we say that [M ] is a G-missing face of P if [M,G] is not a face of P ,

but [S,G] is a face of P for every proper subset S of M . We define

M(P/G) = {[M ] | [M ] is a G-missing face of P},

M(P ) =M(P/∅).

We note that M(P ) is the set of missing faces of P as defined in Chapter 1.

To simplify the notation, we do not distinguish between a missing face (or a G-missing

face) and its vertices when it is clear from the context which interpretation is implied.

Thus if M is a missing face, M may refer to the missing face itself, or M may refer to the

19

set of vertices of the missing face. The following lemma allows us to recover the missing

faces of P+, using the quotient polytopes (cf. Remark 1.1.4) of P with respect to the

sewing faces.

Lemma 2.1.4 (Shemer [21], Lemma 4.7). Let P+ = [P, x+] be obtained by sewing x+

onto P through the tower T . If M ⊂ V(P+), then M ∈M(P+) if, and only if, either

1. M =(⋃ j

i=1{x2i−1, y2i−1})∪ A for some integer 0 ≤ j ≤ (m + 1)/2 and some

A ⊆ V(P ) such that A∗ ∈M(P/Φ2j), or

2. M =(⋃ j

i=1{x2i, y2i})∪ B ∪ {x+} for some integer 0 ≤ j ≤ m/2 and some B ⊆

V(P ) such that B∗ ∈M(P/Φ2j+1).

(Recall that Φm+1 = P and M(P/P ) = {∅}).

Example 2.1.5. To help illustrate Lemma 2.1.4, consider the following. Let C = C(n, 6)

be a cyclic 6-polytope with n ≥ 9 vertices and vertex array x1 < x2 < · · · < xn.

Then by Lemma 1.1.8, Φ1 = [x1, xn] ∈ U1(C), Φ2 = [x1, x2, xn−1, xn] ∈ U3(C), and

Φ3 = [x1, x2, x3, xn−2, xn−1, xn] ∈ U5(C). Let C+ = [C, x+] be obtained by sewing x+ onto

C through T = {Φ1,Φ2,Φ3}. It follows from Gale’s Evenness Condition (Theorem 1.1.6)

that C/Φ1 is a 4-polytope of type C(n − 2, 4) with vertex array x∗2 < x∗3 < · · · < x∗n−1,

C/Φ2 is a polygon with n − 4 vertices and edges [x∗∗3 , x∗∗4 ], [x∗∗4 , x

∗∗5 ], . . . , [x∗∗n−2, x

∗∗3 ], and

C/Φ3 is a 0-dimensional polytope with vertex x∗∗∗4 = x∗∗∗5 = · · · = x∗∗∗n−3. We note that

G = [x1, x3, x5, x7] ∈M(C) as any subpolytope of G is a face of C by neighbourliness and

there are no distinct xi, xj ∈ V(C)\{x1, x3, x5, x7} such that [x1, x3, x5, x7, xi, xj] satisfies

Gale’s Evenness Condition. By Lemma 2.1.4(1) with j = 0, we see that [x1, x3, x5, x7] ∈

M(C+). In fact, all missing faces of C are missing faces of C+. For similar reasons,

[x∗2, x∗4, x∗7] ∈ M(C/Φ1), and Lemma 2.1.4(2) with j = 0 shows that [x2, x4, x7, x

+] ∈

M(C+). The missing faces of C/Φ2 are exactly of the form [x∗∗i , x∗∗j ] such that [x∗∗i , x

∗∗j ] /∈

E(C/Φ2). In this case, Lemma 2.1.4(1) with j = 1 states that [x1, xi, xj, xn] ∈ M(C+).

20

Since C/Φ3 = [x∗∗∗i ] for i = 4, . . . , n−3, x∗∗∗i is not a proper face and so it is a missing face

of C/Φ3 by definition. Lemma 2.1.4(2) with j = 1 shows that [x2, xi, xn−1, x+] ∈M(C+)

for i = 4, . . . , n− 3. Finally, Φ4 = C by definition, so C/Φ4 = ∅, and thus has no missing

faces. Lemma 2.1.4(1) with j = 2 yields that [x1, x3, xn−2, xn] ∈M(C+).

Finally, we present the connection between universal faces and missing faces.

Proposition 2.1.6 (Shemer [21], Proposition 4.2(8)). U is a universal (2k − 1)-face of

a neighbourly 2m-polytope P if, and only if, |M ∩ U | ≤ k for every M ∈M(P ).

For completeness, we include also a result from [15]:

Theorem 2.1.7 (Grunbaum [15], Theorem 5.2). Let P = [P, x] for some x /∈ P with x

not in the affine hull of any facet of P . For all facets F of P , exactly one of the following

holds:

(a ) F is a facet of P , that is, x is beneath F .

(b ) F = [G, x] where G is a (2m− 2)-face of P such that there exist facets F1, F2 of P

with G = F1 ∩ F2, such that x is beneath F1 and beyond F2. In this case, we say

that G has the beyond-beneath property (BBP ).

We refer to the facets above as type (a ) facets and type (b ) facets.

2.2 Main Result

The remaining sections of this chapter contain results obtained in collaboration with

Viktor Vıgh for neighbourly 2m-polytopes [25]. Since this thesis is concerned with neigh-

bourly 6-polytopes, we state the results for m = 3.

Proposition 2.2.1. Let P be a neighbourly 6-polytope and U be a universal face of P

with U ⊂ V for some face V of P . Then V is a universal face of P if, and only if, V/U

is a universal face of P/U .

21

Proof. Let U ∈ Uk(P ), and V ∈ Fj(P ) with 0 ≤ k < j ≤ 5. Then V ∈ Uj(P ) if, and

only if, [V, S] is a face of P for any S ⊂ V(P ) with |S| ≤ b(5 − j)/2c. Since V/U is a

(j − k− 1)-face of the (5− k)-polytope P/U , it follows that V/U ∈ Uj−k−1(P/U) if, and

only if, [V/U,R] is a face of P/U for any R ⊂ V(P/U) with

|R| ≤⌊

(5− k)− (j − k − 1)− 1

2

⌋=

⌊5− j

2

⌋.

Let ` = b(5 − j)/2c. Now U ∈ Uk(P ) implies that for any v1, v2, . . . , v` ∈ V(P ),

[V, v1, v2, . . . , v`] is a face of P if, and only if, [V/U, v∗1, v∗2, . . . , v

∗` ] is a face of P/U . This

completes the proof.

Let P be a neighbourly 6-polytope, T = {Φ1,Φ2,Φ3} be a universal tower of P with

Φj a universal (2j − 1)-face of P for j = 1, 2, 3, and P+ = [P, x+] be obtained by sewing

x+ onto P through T . Let i ∈ {1, 2} and consider the quotient polytope P/Φi, which

is a neighbourly (6 − 2i)-polytope. We sew a vertex z+i onto P/Φi through a particular

universal tower, so that the sewn polytope [P/Φi, z+i ] helps to understand the facial

structure of P+. To do this, we define first the universal tower of P/Φi that corresponds

to the universal tower T of P . Since Φi is universal in P , there corresponds a vertex v∗

of P/Φi to every vertex v ∈ V(P ) \ V(Φi). Let n > i, and consider Φn ⊃ Φi and the

face Φ∗n of P/Φi that corresponds to Φn. From Proposition 2.2.1, it follows that Φ∗n is a

universal face of P/Φi. We denote by

T ∗ =

{Φ∗2,Φ∗3} if i = 1

{Φ∗3} if i = 2

the universal tower of P/Φi corresponding to the universal tower T of P . Let (P/Φi)+ =

[P/Φi, z+i ] be obtained by sewing z+i onto P/Φi through T ∗.

Proposition 2.2.2. With the notation above, let F be a facet of P such that Φi ⊂ F ,

and F ∗ be the corresponding facet of P/Φi with i ∈ {1, 2}.

22

(1 ) z+1 is beneath (beyond ) F ∗ ∈ F(P/Φ1) if, and only if, x+ is beyond (beneath ) F ,

(2 ) z+2 is beneath (beyond ) F ∗ ∈ F(P/Φ2) if, and only if, x+ is beneath (beyond ) F .

Proof. Let F ∈ F(P ) with Φ1 ⊂ F . Then z+1 is beyond F ∗ ∈ F(P/Φ1) if, and only if,

Φ∗2 ⊂ F ∗ and F ∗ 6= Φ∗3, or equivalently, Φ2 ⊂ F and Φ3 6= F , but the latter is valid if,

and only if, x+ is beneath F .

Let F ∈ F(P ) with Φ2 ⊂ F . Then z+2 is beyond F ∗ ∈ F(P/Φ2) if, and only if,

F ∗ = Φ∗3, or equivalently, F = Φ3, which is valid if, and only if, x+ is beyond F .

Note that from Theorem 2.1.2(3), [Φi−1, xi, x+] and [Φi−1, yi, x

+] are universal faces

of P+ for i ∈ {1, 2}. We are now in position to state our main theorem.

Theorem 2.2.3. With the notation above;

(P/Φ2)+ = [P/[x1, y1, x2, y2], z

+2 ] ∼= [P, x+]/[x1, y1, x2, x

+] = P+/[Φ1, x2, x+] (2.1)

and

(P/Φ1)+ = [P/[x1, y1], z

+1 ] ∼= [P, x+]/[x1, x

+] = P+/[x1, x+] (2.2)

Proof. Let P be a neighbourly 6-polytope with n ≥ 9 vertices with universal tower

T = {Φ1,Φ2,Φ3} with Φ1 = [x1, y1], Φ2 = [Φ1, x2, y2], and Φ3 = [Φ2, x3, y3]. We deal

with equation (2.1) first. Since Φ2 ∈ U2(P ), P/Φ2 is a polygon with n − 4 vertices and

the edge [x∗3, y∗3]. Label the remaining n − 6 vertices of V(P ) \ V(Φ3) as v1, v2, . . . , vn−6

so that

[y∗3, v∗1], [v∗1, v

∗2], . . . , [v∗n−6, x

∗3]

are edges of P/Φ2. Since Φ∗3 = [x∗3, y∗3] and T ∗ = {Φ∗3} is the corresponding universal

tower of P/Φ2, we sew z+2 onto P/Φ2 by placing it beyond the edge [x∗3, y∗3] of P/Φ2 and be-

neath all other edges of P/Φ2 to obtain (P/Φ2)+ (see Figure 2.2a). Now, with the conven-

tion that vn−5 = x3 and v0 = y3, and since [v∗j , v∗j+1] is an edge of P/Φ2 for j = 0, . . . , n−6,

23

it follows that [x1, y1, x2, y2, vj, vj+1] ∈ F(P ). Since {vj, vj+1} 6⊂ {x3, y3}, x+ is be-

neath [x1, y1, x2, y2, vj, vj+1], and since P is simplicial, [x1, y1, x2, vj, vj+1] ∈ F4(P ). Thus,

there is a w ∈ V(P ) \ {x1, y1, x2, y2, vj, vj+1} such that [x1, y1, x2, w, vj, vj+1] ∈ F(P ).

Since w 6= y2, it follows that x+ is beyond [x1, y1, x2, w, vj, vj+1] and [x1, y1, x2, vj, vj+1]

has the BBP; that is, [x1, y1, x2, vj, vj+1, x+] ∈ F(P+), and [v∗∗j , v

∗∗j+1] is an edge of

P+/[Φ1, x2, x+]. Since both (P/Φ2)

+ and P+/[Φ1, x2, x+] are polygons with n − 3 ver-

tices, the correspondance v∗ 7→ v∗∗ for v ∈ V(P )\V(Φ2) and z+2 7→ y∗∗2 is a combinatorial

equivalence between (P/Φ2)+ and P+/[Φ1, x2, x

+] (see Figure 2.2b).

(a) (P/Φ2)+

(b) P+/[Φ1, x2, x+]

Figure 2.2: The case i = 2.

To prove equation (2.2), we note that (P/Φ1)+ and P+/[x1, x

+] are simplicial 4-

polytopes. It follows that it is sufficient to prove that [v∗1, v∗2, v∗3, v∗4] ∈ F((P/Φ1)

+) if,

and only if, [v∗∗1 , v∗∗2 , v

∗∗3 , v

∗∗4 ] ∈ F(P+/[x1, x

+]) with the convention that z+1 7→ y∗∗1 . We

start by examining the facets of P+/[x1, x+]. Now [v∗∗1 , v

∗∗2 , v

∗∗3 , v

∗∗4 ] ∈ F(P+/[x1, x

+]) if,

and only if, [x1, v1, v2, v3, v4, x+] ∈ F([P, x+]). All of these facets of [P, x+] are of type

(b) (see Theorem 2.1.7(b )). This means that [x1, v1, v2, v3, v4, x+] ∈ F([P, x+]) if, and

only if, [x1, v1, v2, v3, v4] ∈ F4(P ) has the BBP.

In what follows, we prove that [x1, v1, v2, v3, v4] ∈ F4(P ) has the BBP if, and only if,

[v∗1, v∗2, v∗3, v∗4] ∈ F((P/Φ1)

+).

24

Case 1. y1 /∈ {v1, v2, v3, v4}. In this case, we claim that [x1, v1, v2, v3, v4] uniquely cor-

responds to a type (a) facet of (P/Φ1)+. First note that [x1, v1, v2, v3, v4] is

contained in exactly two facets of P . Since [x1, v1, v2, v3, v4] has the BBP, we

assume that x+ is beyond [x1, y1, v1, v2, v3, v4] and beneath [x1, w, v1, v2, v3, v4]

for some w ∈ V(P ) \ {x1, y1, v1, v2, v3, v4}. Similarly, for every facet F =

[y1, x1, w1, w2, w3, w4] of P with x+ beyond F , we have that F ′ =

[x1, w1, w2, w3, w4] ∈ F4(P ) with the BBP. In summary, [x1, v1, v2, v3, v4] ∈

F4(P ) with the BBP if, and only if, x+ is beyond [x1, y1, v1, v2, v3, v4]. Since all

facets of P/[x1, y1] that z+1 is beneath correspond to type (a) facets of (P/Φ1)+

(see Theorem 2.1.7(a )), Proposition 2.2.2 implies that if y1 /∈ {v1, v2, v3, v4},

then [x1, v1, v2, v3, v4] ∈ F4(P ) has the BBP if, and only if, [v∗1, v∗2, v∗3, v∗4] is a

type (a) facet of (P/Φ1)+.

Case 2. y1 ∈ {v1, v2, v3, v4}. We prove that [x1, y1, v1, v2, v3] ∈ F4(P ) has the

BBP if, and only if, [v∗1, v∗2, v∗3, z

+1 ] is a type (b) facet of (P/Φ1)

+. Since

[x1, y1, v1, v2, v3] ∈ F4(P ) has the BBP, there exist v, w ∈ V(P ) such that

x+ is beneath [v, x1, y1, v1, v2, v3] ∈ F(P ), and beyond [w, x1, y1, v1, v2, v3] ∈

F(P ). Using Proposition 2.2.2, we obtain that z+1 is beyond exactly one

of [v∗, v∗1, v∗2, v∗3], [w∗, v∗1, v

∗2, v∗3] ∈ F(P/[x1, y1]) and beneath the other. This

implies that [v∗1, v∗2, v∗3] ∈ F2(P/[x1, y1]) has the BBP, and from Theorem

2.1.7(b ), it follows that [v∗1, v∗2, v∗3, z

+1 ] is a type (b) facet of (P/Φ1)

+.

The importance of Theorem 2.2.3 is that it allows us to reduce a sewing in 6-

dimensions to a trivial sewing in 2-dimensions. The key observation is that while the

sewings on the right-hand sides of (2.1) and (2.2) are in 6-dimensions, the sewings on the

left-hand sides are only in two and four dimensions.

25

2.3 Sewing in Practice

In this section, we present an algorithm for sewing in practice. If the dimension is fixed,

then this algorithm is the best possible; that is, it has linear running time in the number

of facets of P . Since P+ has more facets than P , we do not expect better than this. The

algorithm is based on the special case of Theorem 2.2.3 with

(P/Φ1)+ ∼= P+/[x1, x

+] ∼= P+/[y1, x+]. (2.3)

In this section, we assume that we have a list of facets for the initial polytope P . It

is either the list of facets of a cyclic polytope obtained using Gale’s Evenness Condition,

or it is the list of facets obtained from a previous sewing. We make some preliminary

remarks. First, the algorithm takes the list of facets of P as input, and outputs the list

of the facets of P+. Second, if F ∈ F(P ) such that F does not contain Φ1, then x+

is beneath F by the sewing and F ∈ F(P+) (a type (a) facet). Third, each type (b)

facet F of P+ contains x+, and hence F = [G, x+] for some G ∈ F4(P ) with the BBP.

It follows that there is an F2 ∈ F(P ) such that G ⊂ F2 and Φ1 ⊂ F2. This implies that

G ∩ Φ1 6= ∅, and thus, F ∩ Φ1 6= ∅. This means that all the type (b) facets of P+ may

be read from the list of facets of P+/[x1, x+] and P+/[y1, x

+]. Since the list of facets of

P/Φ1 and P/Φ2 are easily derived by checking which facets of P contain Φ1 and Φ2, we

assume also that the list of facets of P/Φ1 and P/Φ2 are given.

The algorithm is based on the fact that by using (2.3), we reduce the 6-dimensional

sewing of P to a 4-dimensional sewing of P/Φ1, which is then further reduced to a

2-dimensional sewing of (P/Φ1)/Φ∗2∼= P/Φ2. We then recover (P/Φ1)

+ from the 2-

dimensional sewn polytope (P/Φ2)+, and then we recover P+ from the 4-dimensional

sewn polytope (P/Φ1)+. To aid in the notation, if U is a universal face of P , and v is a

vertex of P that is not a vertex of U , then the vertex v∗ of P/U will also be denoted as

26

v. Thus, unless otherwise stated, the vertex w of P/U corresponds to the vertex w of P .

Algorithm 1

Input: The facets of P , the facets (3-faces) of P/Φ1, and the facets (edges) of P/Φ2.

Step 1. Sew z+2 onto P/Φ2 to obtain the list of the edges of (P/Φ2)+ = [P/Φ2, z

+2 ].

(This is a trivial sewing in the plane: the edges of (P/Φ2)+ are [x3, z

+2 ], [y3, z

+2 ],

and all edges of P/Φ2 with the exception of [x3, y3].)

Step 2. (i) For each 3-face G of P/Φ1: if z+1 is beneath G, then add G to the list of

the 3-faces of (P/Φ1)+ = [P/Φ1, z

+1 ]. (This generates all of the type (a )

3-faces of (P/Φ1)+.)

(ii) For each edge E of (P/Φ2)+: add G = [E, x2, z

+1 ] to the list of the 3-

faces of (P/Φ1)+. We note that the vertex z+2 of (P/Φ2)

+ corresponds to

the vertex y2 of (P/Φ1)+. (This generates all of the type (b ) 3-faces of

(P/Φ1)+ that contain x2.)

(iii) For each edge E of P/Φ2: if z+2 is beneath E then add G = [E, y2, z+1 ] to

the list of the 3-faces of (P/Φ1)+. (This generates the remaining type (b )

3-faces of (P/Φ1)+: those that contain y2, but not x2.)

Step 3. (i) For each facet F of P : if x+ is beneath F , then add F to the list of the

facets of P+ = [P, x+]. (This generates all of the type (a ) facets of P+.)

(ii) For each 3-face G of (P/Φ1)+: add F = [G, x1, x

+] to the list of the facets

of P+. We note that the vertex z+1 of (P/Φ1)+ corresponds to the vertex

y1 of P+. (This generates all of the type (b ) facets of P+ that contain x1.)

(iii) For each 3-face G of P/Φ1: if z+1 is beneath G then add F = [G, y1, x+] to

the list of the facets of P+. (This generates the remaining type (b ) facets

of P+: those that contain y1, but not x1.)

27

Output: The facets of P+.

The correctness of the algorithm follows from (2.3), Theorem 2.1.7, and our prelimi-

nary remarks.

2.4 Keeping Track of Universal Faces

In this section, we give a complete picture of the odd dimensional universal faces using

Theorem 2.2.3. With these results, Algorithm 1 is extended so that it keeps track of the

universal faces during the sewing process. Note that the list of the facets contains all of

the combinatorial information of the polytope, but it is very time consuming to list all

universal faces of a polytope from the list of its facets.

Proposition 2.4.1. Let C = C(2` + 2, 2`) be a cyclic 2`-polytope with 2` + 2 vertices.

Then any set X of 2k − 1 distinct vertices of C contains at least k vertices of a missing

face of C for 1 ≤ k ≤ `+ 1.

Proof. By Theorem 1.1.7, C(2` + 2, 2`) has exactly two missing faces; both have ` + 1

vertices which together partition V(C). If X contained at most k−1 vertices from either

missing face of C, then X would have at most 2k − 2 vertices, a contradiction.

Lemma 2.4.2. Any U ∈ U2k−1(P+), with x+ ∈ U , intersects the sewing edge Φ1 for

k ∈ {1, 2, 3}.

Proof. For k = 3, let x+ ∈ U ∈ U5(P+) = F(P+). Then U is a type (b) facet of P+, and

U = [G, x+] with G ∈ F4(P ) such that there are F1, F2 ∈ F(P ) with x+ beneath F1 and

beyond F2. It follows that Φ1 ⊂ F2, so |G ∩ {x1, y1}| ≥ 1 and U ∩ Φ1 6= ∅.

For k = 2, let x+ ∈ U ∈ U3(P+). We suppose that U∩Φ1 = ∅ and seek a contradiction.

Let U = [v1, v2, v3, x+]. Then Proposition 2.1.6 implies that

|U ∩M | ≤ 2 (2.4)

28

for all M ∈M(P+). From Lemma 2.1.4(2), it follows that

{x+} ∪B ∈M(P+) (2.5)

for every B ⊂ V(P ) such that B∗ ∈M(P/Φ1). From (2.4) and (2.5), we obtain that

|B∗ ∩ {v∗1, v∗2, v∗3}| ≤ 1 (2.6)

for every B∗ ∈M(P/Φ1). Consider any distinct w∗1, w∗2, w

∗3 ∈ V(P/Φ1) \ {v∗1, v∗2, v∗3}, and

letQ∗ = [v∗1, v∗2, v∗3, w

∗1, w

∗2, w

∗3]. SinceQ∗ is a cyclic 4-polytope with 6 vertices, Proposition

2.4.1 yields that there exists an X∗ ∈M(Q∗) such that |X∗ ∩ {v∗1, v∗2, v∗3}| ≥ 2. This is a

contradiction to (2.6) since Q∗ is a subpolytope of P/Φ1 and X∗ ∈M(P/Φ1).

For k = 1, an analogous proof shows that any U ∈ U1(P+) with x+ ∈ U intersects

Φ1, which yields that [x1, x+] and [y1, x

+] are the only universal edges of P+ that contain

x+. This fact also follows from Theorem 4.10(1) of [21].

Lemma 2.4.3. Let U be a universal face of P . Then U is not a universal face of P+ in

each of the following cases:

(1) U ∈ F(P ) and x+ is beyond U .

(2) U ∈ F3(P ) and Φ1 ⊂ U 6= Φ2.

(3) U = Φ1.

Proof. Let U be a universal face of P . Note that (1) follows from x+ being sewn onto P

through T = {Φ1,Φ2,Φ3}.

For (2), let U ∈ F3(P ) such that Φ1 ⊂ U 6= Φ2, and suppose that U ∈ U3(P+). From

Lemma 2.1.4(1), we have that

|M ∩ U | ≤ 2 (2.7)

for every M ∈M(P+). Lemma 2.1.4(1) shows that

A ∪ {x1, y1} ∈ M(P+) (2.8)

29

for every A ⊂ V(P ) such that A∗ ∈M(P/Φ2). Since Φ1 ⊂ U , (2.7) and (2.8) imply that

A ∩ (V(U) \ {x1, y1}) = ∅ (2.9)

for every A∗ ∈M(P/Φ2). Since Φ2 6⊂ U , there is a v ∈ V(U)\V(Φ2). Let {w1, w2, w3} ⊆

V(P ) \ (V(Φ2) ∪ {v}). The subpolytope Q∗ = [v∗, w∗1, w∗2, w

∗3] of P/Φ2 is a quadrilateral,

and so v∗ determines a missing face of Q∗ with exactly one of w∗1, w∗2, w

∗3, say, w∗1. Since

P/Φ2 is neighbourly, X∗ = {v∗, w∗1} ∈ M(P/Φ2), and so {X, x1, y1} ∈ M(P+) by Lemma

2.1.4(1). Since v ∈ U , v ∈ X ∩ (V(U) \ {x1, y1}), a contradiction to (2.9).

If U = Φ1, then U /∈ U1(P+) follows from Theorem 2.1.2(2), so (3) holds.

We now state the main result of this section.

Theorem 2.4.4. Let U be a universal face of P . Then

(1) U ∈ U1(P+) if, and only if, U 6= Φ1 and [U, u, x+] ∈ U3(P+) for any u ∈ {x1, y1} \

V(U).

(2) U ∈ U3(P+) if, and only if, U = Φ2 or, Φ1 6⊆ U and [U, u, x+] ∈ F(P+) for any

u ∈ {x1, y1} \ V(U).

Proof. Let U ∈ U1(P+). Then from Lemma 2.4.3 it follows that U 6= Φ1. From Lemma

2.1.2(3), we obtain that [x1, x+] ∈ U1(P+) and [y1, x

+] ∈ U1(P+), and it follows from

Theorem 1.1.10 that [U, u, x+] ∈ U3(P+) for u ∈ {x1, y1} \ V(U).

Let U 6= Φ1 and [U, u, x+] ∈ U3(P+). By definition, U ∈ U1(P+) if [U, v1, v2] is a face

of P+ for any v1, v2 ∈ V(P+). Since [U, u, x] ∈ U3(P+), it follows that [U, u, x+, v] is a

face of P+ for every v ∈ V(P+). This implies that [U, u, v] and [U, x+, v] are faces of

P+, so we assume that {v1, v2} ∩ {u, x+} = ∅. Since U ∈ U1(P ), we have G = [U, v1, v2]

is a face of P . Suppose that x+ is beyond every facet F of P that contains G. Then

F ∈ F 1, and Φ1 = [x1, y1] ⊂ F . Since G is the intersection of all facets that contain it,

30

we obtain that u ∈ G, and hence, u ∈ U , a contradiction. So there is a F ∈ F(P ) such

that G ⊂ F ∈ F(P+), and G = [U, v1, v2] is a face of P+. Thus, [U, v1, v2] is a face of P+

for any v1, v2 ∈ V(P+), and U ∈ U3(P+).

Let U ∈ U3(P ). If U = Φ2, then (2) follows from U ∈ U3(P+) by Lemma 2.1.2(2),

and [U, x3, x+] ∈ U5(P+) and [U, y3, x

+] ∈ U5(P+) by Lemma 2.1.2(3).

Let U 6= Φ2. If U ∈ U3(P+), then the proof that Φ1 6⊂ U and [U, u, x+] ∈ U5(P+) for

any u ∈ {x1, y1} \ V(U) is analogous to (1).

Let [U, u, x+] ∈ U5(P+) = F(P+) such that Φ1 6⊆ U . By definition, U ∈ U3(P+) if

[U, v] is a face of P+ for any v ∈ V(P+). Since [U, u, x+] ∈ F(P+), it follows that [U, u]

and [U, x+] are faces of P+, so we assume that v /∈ {u, x+}. We now argue as above that

[U, v] is a face of P for any v ∈ V(P+).

The list of universal faces of a sewn polytope follows from Lemma 2.1.4 and Proposi-

tion 2.1.6. Lemma 2.1.4 characterizes the missing faces of a sewn polytope, and Propo-

sition 2.1.6 describes the universal faces of the sewn polytope in terms of the missing

faces. However, this involves determining |V(F ) ∩ V(M)| for every face F of P+ and

every missing face M of P+. Lemma 2.4.2 and Theorem 2.4.4 together provide a way

to keep track of the universal faces during the sewing process. Algorithm 2 below is

an extension of Algorithm 1 that keeps track of not only the facets of a sewn polytope,

but also all odd-dimensional universal faces. The algorithm receives as input the list of

all odd-dimensional universal faces of a 6-dimensional neighbourly polytope P that is

either cyclic or that has been obtained from a previous sewing, and returns the list of

all odd-dimensional universal faces of P+. As in Algorithm 1, we assume that the list

of facets of P/Φ1 and P/Φ2 are given, from which it follows by Lemma 2.2.1 that the

odd-dimensional universal faces of P/Φ1 and P/Φ2 are also given. As in Algorithm 1, we

drop the “∗” notation used for the vertices of a quotient polytope.

31

Algorithm 2

Input: The odd-dimensional universal faces of P , P/Φ1, and P/Φ2.

Step 1. Sew z+2 onto P/Φ2 to obtain the list of the edges of (P/Φ2)+ = [P/Φ2, z

+2 ].

Step 2. (i) Obtain the list of the 3-faces of (P/Φ1)+ = [P/Φ1, z

+1 ] using Step 2 of

Algorithm 1.

(ii) (a) For each universal edge E of P/Φ1: if E 6= [x2, y2] and either [E, x2, z+1 ]

is a 3-face of (P/Φ1)+ or [E, y2, z

+1 ] is a 3-face of (P/Φ1)

+, then add

E to the list of universal edges of (P/Φ1)+. (This generates all of the

universal edges of (P/Φ1)+ that are also universal edges of P/Φ1.)

(b) Add [x2, z+1 ] and [y2, z

+1 ] to the list of universal edges of (P/Φ1)

+.

(This generates the only two universal edges of (P/Φ1)+ that are cre-

ated during the sewing process.)

Step 3. (i) Obtain the list of facets of P+ = [P, x+] using Step 3 of Algorithm 1.

(ii) (a) For each universal 3-face G of P : if G = Φ2, then add G to the

list of universal 3-faces of P+. If Φ1 6⊂ G 6= Φ2, then choose u ∈

{x1, y1} \ V(G). If [G, u, x+] is a facet of P+, then add G to the list

of universal 3-faces of P+. (This generates all of the universal 3-faces

of P+ that are also universal 3-faces of P .)

(b) For each universal edge E of (P/Φ1)+: add [E, x1, x

+] to the list of

universal 3-faces of P+. Note that the vertex z+1 of (P/Φ1)+ cor-

responds to the vertex y1 of P+. (This generates all of the universal

3-faces of P+ that are created during the sewing process which contain

x1.)

(c) For each universal edge E of (P/Φ1)+: if z+1 is not a vertex of E,

then add [E, y1, x+] to the list of universal 3-faces of P+. (This gen-

32

erates the remaining universal 3-faces of P+ that are created during

the sewing process: those that contain y1, but not x1.)

(iii) (a) For each universal edge E of P : if E 6= Φ1 and at least one of

[E, x1, x+] and [E, y1, x

+] is a universal 3-face of P+, then add E to

the list of universal edges of P+. (This generates all of the universal

edges of P+ that are also universal edges of P .)

(b) Add [x1, x+] and [y1, x

+] to the list of universal edges of P+. (This

generates the only two universal edges of P+ that are created during

the sewing process.)

Output: The odd-dimensional universal faces of P+.

The numbers in Table 2.1 show the order of the steps in Algorithm 2, and the arrows

show dependency. The correctness of Algorithm 2 follows from Lemma 2.4.2 and Theorem

2.4.4.

(P/Φ2)+ (P/Φ1)

+ P+

U5( · ) Step 3(i)

↗ ↓U3( · ) Step 2(i) Step 3(ii)

↗ ↓ ↗ ↓U1( · ) Step 1 Step 2(ii) Step 3(iii)

Table 2.1: The dependancies among the steps in Algorithm 2.

33

Chapter 3

Semi-Cyclic 6 -Polytopes

Let P denote a neighbourly 2m-polytope, m ≥ 2. Theorem 1.1.9 states that if |V(P )| ≤

2m + 3, then P is cyclic. It is known (Shemer [21]) that any cyclic 2m-polytope is

obtained by a sequence of sewings, and thus there is a natural ordering of its vertices.

Let |V(P )| = n ≥ 2m+3. We say that P is totally-sewn if there is a sequence {Pk}nk=2m+3

of subpolytopes of P = Pn such that P2m+3 = [x1, x2, . . . , x2m+3] is cyclic with the vertex

array x1 < x2 < · · · < x2m+3, |V(Pk)| = k and for k ≥ 2m+3, Pk+1 is obtained by sewing

xk+1 onto Pk. Then P = Pn = [x1, x2, . . . , xn] and we say that P has the vertex array

x1 < x2 < · · · < xn. In particular, cyclic polytopes are totally sewn.

Let C(p) = C(p, 6) denote a cyclic 6-polytope with p ≥ 9 vertices and vertex array

x1 < x2 < · · · < xp. It follows from Theorem 1.1.8 that

Ep = [x1, xp] ∈ U1(C(p)),

Gp = [x1, x2, xp−1, xp] = [Ep, x2, xp−1] ∈ U3(C(p)),

Fp = [x1, x2, xp−3, xp−2, xp−1, xp] = [Gp, xp−3, xp−2] ∈ U5(C(p)) = F(C(p)),

showing that Tp = {Ep, Gp, Fp} is a universal tower of C(p). Let Sp(p+ 1) = [C(p), xp+1]

denote the neighbourly 6-polytope obtained by sewing xp+1 onto C(p) through Tp. It

follows from Theorem 2.1.2(3) that

Ep+1 = [x1, xp+1] ∈ U1(Sp(p+ 1)),

Gp+1 = [x1, x2, xp, xp+1] = [Ep+1, x2, xp] ∈ U3(Sp(p+ 1)),

Fp+1 = [x1, x2, xp−2, xp−1, xp, xp+1] = [Gp+1, xp−2, xp−1] ∈ U5(Sp(p+ 1)) = F(Sp(p+ 1)).

Then Tp+1 = {Ep+1, Gp+1, Fp+1} is a universal tower of Sp(p + 1). Let Sp(p + 2) =

34

[Sp(p+1), xp+2] denote the neighbourly 6-polytope obtained by sewing xp+2 onto Sp(p+1)

through Tp+1.

More generally, let m ≥ p and let Sp(m+ 1) = [Sp(m), xm+1] denote the neighbourly

6-polytope with m+1 vertices obtained by sewing xm+1 onto Sp(m) through the universal

tower Tm = {Em, Gm, Fm} with

Em = [x1, xm], Gm = [Em, x2, xm−1], and Fm = [Gm, xm−3, xm−2].

We say that the totally-sewn neighbourly 6-polytope Sp(m) is a semi-cyclic polytope and

define Sp(p) = C(p).

Remark 3.0.1. Semi-cyclic 4-polytopes were defined by Finbow-Singh and Oliveros in

[12]. To be consistent with their definition, a semi-cyclic 6-polytope is any neighbourly

polytope constructed in the above manner with Fm being any facet containing Gm, m ≥ p.

Thus, our above construction generates a specific class of semi-cyclic 6-polytopes which

we denote by Sp(n) for n ≥ p ≥ 9.

In this chapter, we give a complete description of the faces of Sp(n) for n ≥ p, and

give representations of the quotient polytopes of Sp(n) that will be utilized in Chapter

4. We first introduce the follow sets of vertices of Sp(n):

V1, V2, . . . , Vn−1,

T1, T2, . . . , Tn−1, and

Wp−4,Wp−3, . . . ,Wn−2,

with Vi = {xi, xi+1} (successive pairs), Ti = {xi, xi+1, xi+2} (successive triples), Wp−4 =

{x3, xp−2}, and Wi = {xi, xi+2} (separated pairs). In [. . . , Vi, Vj, . . .], we assume that

i + 2 ≤ j when no conditions on i and j are present. A similar convention holds for

successive triples, separated pairs, and singletons. For example, in [x2, Ti, Vj], we assume

that 3 ≤ i ≤ n− 4 and that i+ 3 ≤ j ≤ n− 1.

35

3.1 The Facets of Sp(n)

In this section, we give a complete description of the facets of Sp(n) for n ≥ p.

Proposition 3.1.1. Every subpolytope of Sp(n) with 6 vertices is a 5-simplex.

Proof. Let F be a subpolytope of Sp(n) with |V(F )| = 6. Then dim(F ) ≤ 5. By Theorem

1.1.5, F is a 3-neighbourly polytope. Since 3 > bdim(F )/2c, Theorem 1.1.5 yields that

F is a simplex. Since F has 6 vertices, it follows that F is a 5-simplex.

To aid in the description of the facets of Sp(n), we define the following families of

5-dimensional subpolytopes of Sp(n). Let

R1,p(n) = {[V1, Vi, Vj] | 3 ≤ i ≤ p− 4}⋃

{[V2, Vi, Vj] | 4 ≤ i ≤ p− 3}⋃

{[Vi, Vj, Vk] | 3 ≤ i ≤ r}

with r = p− 5 if n = p, and r = p− 4 otherwise,

R2,p(n) = {[V1,Wi, Vj] | p− 4 ≤ i ≤ n− 4},

D1,p(n) = {[x1, V2, Vi, xn]} | 4 ≤ i ≤ p− 3}⋃

{[x1, Vi, Vj, xn] | 3 ≤ i ≤ p− 4} and

D2,p(n) = {[V1, Ti, xn] | p− 3 ≤ i ≤ n− 3}.

For n ≥ p+ 1, let

R3,p(n) = {[x2, Ti, Vj] | p− 3 ≤ i ≤ n− 4} and

D3,p(n) = {[x1,Wi, Vj, xn] | p− 4 ≤ i ≤ n− 5},

and finally, when n ≥ p+ 2, let

R4,p(n) = {[Wi, Vj, Vk] | p− 4 ≤ i ≤ n− 6}.

36

For simplicity, we define R3,p(p) = R4,p(p) = R4,p(p + 1) = D3,p(p) = ∅, and for n ≥ p,

we define

Rp(n) = R1,p(n) ∪R2,p(n) ∪R3,p(n) ∪R4,p(n) and

Dp(n) = D1,p(n) ∪D2,p(n) ∪D3,p(n).

Remark 3.1.2. Let 1 ≤ ` ≤ 4 and m ≥ p. We note that R`,p(m) ⊂ R`,p(m + 1),

from which it follows that Rp(m) ⊂ Rp(m + 1), and if X ∈ Rp(m + 1) \ Rp(m), then

xm+1 ∈ X. Furthermore, note that for 1 ≤ ` ≤ 3, D`,p(m)∩D`,p(m+ 1) = ∅ implies that

Dp(m) ∩ Dp(m+ 1) = ∅.

In this section, we show that F(Sp(n)) = Rp(n) ∪ Dp(n). It is known (see [15]) that

a neighbourly 6-polytope with n vertices has(n− 3

3

)+

(n− 4

2

)=n(n− 4)(n− 5)

6

facets. We first show that |Rp(n) ∪ Dp(n)| = n(n− 4)(n− 5)/6.

Lemma 3.1.3. Let n ≥ p ≥ 9. Then Rp(n)∩Dp(n) = ∅, Dp(n)∩F(Sp(n+ 1)) = ∅, and

Rp(n) ∩ F(Sp(n)) ⊂ F(Sp(n+ 1)).

Proof. Recall that Tn = {En, Gn, Fn}, with En = [x1, xn], Gn = [En, x2, xn−1], and

Fn = [Gn, xn−3, xn−2], and that F ∈ F(Sp(n)) ∩ F(Sp(n + 1)) if, and only if, En 6⊂ F or

Gn ⊂ F 6= Fn.

Let X ∈ Dp(n). Then En ⊂ X. If X ∈ D1,p(n), then xn−1 /∈ X, and if X ∈ D3,p(n),

then x2 /∈ X. In either case, Gn 6⊂ X. If X ∈ D2,p(n) and Gn ⊂ X, then X = Fn. Thus

if X ∈ Dp(n), then either En ⊂ X and Gn 6⊂ X, or X = Fn. (3.1)

It follows that if X ∈ F(Sp(n)), then xn+1 is beyond X, and X /∈ F(Sp(n + 1)).

Moreover, since xn+1 /∈ X, it follows that X /∈ F(Sp(n + 1)) \ F(Sp(n))). Hence,

Dp(n) ∩ F(Sp(n+ 1)) = ∅.

37

Let X ∈ Rp(n). If En ⊂ X, then X ∈ {[V1, Vi, Vn−1] | 3 ≤ i ≤ p − 4} ⊂ R1,p(n) or

X ∈ {[V1,Wi, Vn−1] | p − 4 ≤ i ≤ n − 4} ⊂ R2,p(n). In either case, Gn ⊂ X, and since

Vn−3 6⊂ X, it follows that X 6= Fn. Thus,

if X ∈ Rp(n), then En 6⊂ X or Gn ⊂ X 6= Fn. (3.2)

It follows that if X ∈ Rp(n)∩F(Sp(n)), then xn+1 is beneath X, and X ∈ F(Sp(n+ 1)).

Finally, from (3.1) and (3.2), it follows that Rp(n) ∩ Dp(n) = ∅.

Lemma 3.1.4. Let n ≥ p ≥ 9. Then R1,p(n), R2,p(n), R3,p(n), and R4,p(n) are pairwise

disjoint.

Proof. Let n ≥ p + 2 and X ∈ Rp(n). If X ∈ R1,p(n), then X contains three disjoint

successive pairs, and X /∈ R2,p(n) ∪ R3,p(n) ∪ R4,p(n). If X ∈ R2,p(n), then x1 ∈ X,

and X /∈ R3,p(n) ∪ R4,p(n). If X ∈ R3,p(n), then x2 ∈ X, and X /∈ R4,p(n). Thus,

R1,p(n), R2,p(n), R3,p(n), and R4,p(n) are pairwise disjoint. Then it follows from Remark

3.1.2 that the result holds for n ∈ {p, p+ 1}.

Lemma 3.1.5. Let n ≥ p ≥ 9. Then D1,p(n), D2,p(n), and D3,p(n) are pairwise disjoint.

Proof. Let n ≥ p, and X ∈ Dp(n). If X ∈ D1,p(n), then X ∩ {xp−1, . . . , xn−1} = ∅,

and X /∈ D2,p(n) ∪ D3,p(n). Let X ∈ D2,p(n). Then x2 ∈ X, and X /∈ D3,p(n). Hence,

D1,p(n), D2,p(n), and D3,p(n) are pairwise disjoint.

Lemma 3.1.6. Let n ≥ p ≥ 9. Then

|Rp(n) ∪ Dp(n)| = n(n− 4)(n− 5)

6

Proof. From Lemmata 3.1.3, 3.1.4, and 3.1.5, R1,p(n), R2,p(n), R3,p(n), R4,p(n), D1,p(n),

D2,p(n), and D3,p(n) partition Rp(n)∪Dp(n). Thus, we compute the cardinalities of the

partitioning sets, and add the results. Recall that R1,p(n) = A ∪B ∪ C, with

A = {[V1, Vi, Vj] | 3 ≤ i ≤ p− 4},

38

B = {[V2, Vi, Vj] | 4 ≤ i ≤ p− 3} and

C = {[Vi, Vj, Vk] | 3 ≤ i ≤ r}

for r = p− 5 if n = p, and r = p− 4 if n ≥ p + 1. Since A,B, and C partition R1,p(n),

|R1,p(n)| = |A|+ |B|+ |C|. To calculate A, we choose a Vi with 3 ≤ i ≤ p− 4, and since

i+ 2 ≤ j ≤ n− 1, we then have n− i− 2 choices for Vj. Summing over i, and using the

fact that∑m

i=1 i = m(m+ 1)/2, we obtain

|A| =p−4∑i=3

(n− i− 2)

=

p−4∑i=3

(n− 2)−p−4∑i=3

i

= (n− 2)(p− 4− 2)−(

(p− 4)(p− 3)

2− 3

)= n(p− 6)− 2p+ 12− p2

2+

7p

2− 3

= n(p− 6)− p2

2+

3p

2+ 9. (3.3)

Similarly,

|B| =p−3∑i=4

(n− i− 2)

=

p−3∑i=4

(n− 2)−p−3∑i=4

i

= (n− 2)(p− 3− 3)−(

(p− 3)(p− 2)

2− 6

)= n(p− 6)− 2p+ 12− p2

2+

5p

2+ 3

= n(p− 6)− p2

2+p

2+ 15, (3.4)

and recalling that∑m

i=1 i2 = m(m+ 1)(2m+ 1)/6, we find

|C| =r∑i=3

n−3∑j=i+2

(n− j − 2)

39

=r∑i=3

(n−3∑j=i+2

(n− 2)−n−3∑j=i+2

j

)

=r∑i=3

((n− 2)(n− 3− i− 1)−

((n− 3)(n− 2)

2− (i+ 1)(i+ 2)

2

))=

r∑i=3

((n− 2)(n− 4)− (n− 2)i− (n− 3)(n− 2)

2+i2 + 3i+ 2

2

)=

r∑i=3

(n− 2

2(2(n− 4)− (n− 3)) + 1 +

i2

2+

(7− 2n)i

2

)=

r∑i=3

n2 − 7n+ 12

2+

1

2

r∑i=3

i2 +7− 2n

2

r∑i=3

i

=n2 − 7n+ 12

2(r − 2) +

1

2

(r(r + 1)(2r + 1)

6− 5

)+

7− 2n

2

(r(r + 1)

2− 3

)=n2 − 7n+ 12

2(r − 2) +

1

12(2r3 + 3r2 + r)− 5

2+

7− 2n

4(r2 + r − 6)

=r3

6+

(7− 2n

4+

1

4

)r2 +

(n2 − 7n+ 12

2+

1

12+

7− 2n

4

)r

− 2(n2 − 7n+ 12)

2− 5

2− 6(7− 2n)

4

=r3

6+

(8− 2n

4

)r2 +

(6n2 − 42n+ 72 + 1 + 21− 6n

12

)r

− 4n2 − 28n+ 48 + 10 + 42− 12n

4

=r3

6+

(4− n

2

)r2 +

(3n2 − 24n+ 47

6

)r − n2 + 10n− 25. (3.5)

If n = p, then r = p− 5 and (3.5) simplifies to

|C| = (p− 5)3

6+

(4− p

2

)(p− 5)2 +

(3p2 − 24p+ 47

6

)(p− 5)− p2 + 10p− 25

=p3 − 15p2 + 75p− 125

6− p3 − 14p2 + 65p− 100

2+

3p3 − 39p2 + 167p− 235

6

− p2 + 10p− 25

=p3

6− 3p2 +

107p

6− 35. (3.6)

Otherwise, r = p− 4 and (3.5) becomes

|C| = (p− 4)3

6+

(4− n

2

)(p− 4)2 +

(3n2 − 24n+ 47

6

)(p− 4)− n2 + 10n− 25

40

=p3 − 12p2 + 48p− 64

6+

4p2 − 32p+ 64− (p2 − 8p+ 16)n

2

+47p− 188 + (3p− 12)n2 − (24p− 96)n

6− n2 + 10n− 25.

=n2

2(p− 6)− n

2(p2 − 36) +

p3

6− p

6− 35. (3.7)

For n = p, we use equations (3.3), (3.4) and (3.6) to compute |R1,p(n)|.

|R1,p(p)| = |A|+ |B|+ |C|

= p(p− 6)− p2

2+

3p

2+ 9 + p(p− 6)− p2

2+p

2+ 15

+p3

6− 3p2 +

107p

6− 35

=p3

6− 2p2 +

47p

6− 11,

while for n ≥ p+ 1, |R1,p(p)| is computed using equations (3.3), (3.4) and (3.7).

|R1,p(p)| = |A|+ |B|+ |C|

= n(p− 6)− p2

2+

3p

2+ 9 + n(p− 6)− p2

2+p

2+ 15 +

n2

2(p− 6)

− n

2(p2 − 36) +

p3

6− p

6− 35

=n2

2(p− 6)− n

2(p2 − 4p− 12) +

p3

6− p2 +

11p

6− 11.

The computations of the remaining cardinalities are omitted, being that they are similar

(and ridiculously easier) to the computation of |R1,p(n)|. We obtain

|R2,p(n)| =n−4∑i=p−4

(n− i− 3)

=n2

2− n

2(2p− 3) +

p2

2− 3p

2+ 1,

|D1,p(n)| = (p− 6) +

p−4∑i=3

(n− i− 3)

= n(p− 6)− p2

2+

3p

2+ 9,

|D2,p(n)| = n− p+ 1.

41

For n ≥ p+ 1, we have

|R3,p(n)| =n−4∑i=p−3

(n− i− 3)

=n2

2− n

2(2p− 1) +

p2

2− p

2,

|D3,p(n)| =n−5∑i=p−4

(n− i− 4)

=n2

2− n

2(2p− 1) +

p2

2− p

2,

and for n ≥ p+ 2, we obtain

|R4,p(n)| =n−6∑i=p−4

n−3∑j=i+3

(n− j − 2)

=n3

6− n2p

2+n

6(3p2 − 1)− p3

6+p

6.

Recalling that R3,p(p) = R4,p(p) = R4,p(p+ 1) = D3,p(p) = ∅, basic algebra yields that

|Rp(n)| = |R1,p(n)|+ |R2,p(n)|+ |R3,p(n)|+ |R4,p(n)|

=n3

6− 2n2 +

47n

6− 10,

|Dp(n)| = |D1,p(n)|+ |D2,p(n)|+ |D3,p(n)|

=n2

2− 9n

2+ 10,

which in turn leads to

|Rp(n) ∪ Dp(n)| = |Rp(n)|+ |Dp(n)| = n(n− 4)(n− 5)

6.

We argue by induction that F(Sp(n)) = Rp(n) ∪ Dp(n), the base case occuring for

n = p.

Lemma 3.1.7. Let p ≥ 9. Then F(Sp(p)) = Rp(p) ∪ Dp(p).

42

Proof. Recall that Sp(p) = C(p), Rp(p) = R1,p(p) ∪ R2,p(p) and Dp(p) = D1,p(p) ∪

D2,p(p). Using Gale’s Evenness Condition, we partition the facets of C(p) into two

sets; F even(C(p)) = {[Vi, Vj, Vk]}, and Fodd(C(p)) = {[x1, Vi, Vj, xp]} (cf. page 34). By

definition, we have that R1,p(p) ⊂ F even(C(p)). In fact,

F even(C(p)) \R1,p(p) = {[V1, Vp−3, Vp−1]} = {[V1, Tp−3, xp]} = D2,p(p).

Similarly, D1,p(p) ⊂ Fodd(C(p)), and we have

Fodd(C(p)) \D1,p(p) = {[V1,Wp−4, Vp−1]} = R2,p(p).

Hence, F(C(p)) = Rp(p) ∪ Dp(p).

We now prove our main result.

Theorem 3.1.8. Let n ≥ p ≥ 9. Then F(Sp(n)) = Rp(n) ∪ Dp(n).

Proof. We argue by induction; the case n = p being proven in Lemma 3.1.7. Assume that

Rp(m)∪Dp(m) = F(Sp(m)) for some m ≥ p. We first show thatRp(m+1)∪Dp(m+1) ⊆

F(Sp(m + 1)). Recall that Sp(m + 1) = [Sp(m), xm+1] is obtained be sewing xm+1 onto

Sp(m) through T = {Em, Gm, Fm}. From the proof of Lemma 3.1.3, we know that xm+1

is beneath any facet of Sp(m) contained in Rp(m), while xm+1 is beyond any facet of

Sp(m) contained in Dp(m).

Let F ∈ Rp(m+ 1). If F ∈ Rp(m), then Rp(m) ⊂ F(Sp(m)) and Lemma 3.1.3 yield

that F ∈ F(Sp(m + 1)). Let F /∈ Rp(m). Then Remark 3.1.2 yields that xm+1 ∈ F .

Let F = [G, xm+1]. By Theorem 2.1.7 and Lemma 3.1.3, it is sufficient to determine

F ∈ Rp(m) and F ∈ Dp(m) such that G = F ∩ F . From Lemma 3.1.4, it follows that

Rp(m+ 1) \ Rp(m) =⋃4i=1 (Ri,p(m+ 1) \Ri,p(m)).

Case 1. F ∈ R1,p(m+ 1) \R1,p(m).

43

Subcase 1.1. F = [V1, Vi, Vm] for some 3 ≤ i ≤ p− 4. Then F = [V1, Vi, Vm−1], and

F =

[x1, V2, V4, xm] , for i = 3,

[x1, V2, Vi, xm] , otherwise.

Subcase 1.2. F = [V2, Vi, Vm] for some 4 ≤ i ≤ p − 3. Then F = [V2, Vi, Vm−1], and

F = [x1, V2, Vi, xm].

Subcase 1.3. F = [Vi, Vj, Vm] for some 3 ≤ i ≤ p− 4. If m = p, then

F =

[Vp−5, Vp−3, Vp−1] , for i = p− 4,

[Vi, Vp−3, Vp−1] , for j = p− 2 and i < p− 4,

[V1, Vj, Vp−1] , otherwise.

If m ≥ p+ 1, then

F =

[Vi, Vm−3, Vm−1] , for j = m− 2,

[Vi, Vj, Vm−1] , otherwise.

For any m ≥ p, F = [x1, Vi, Vj, xm].

Case 2. F ∈ R2,p(m+ 1) \R2,p(m). Then F = [V1,Wi, Vm] for some p− 4 ≤ i ≤ m− 3.

If m = p, then

F =

[V1, Vp−3, Vp−1] , for i = p− 3,

[V1,Wi, Vp−1] , otherwise.

If m ≥ p+ 1, then m− 5 ≥ p− 4. It follows that

F =

[V1,Wm−5, Vm−1] , for i = m− 3,

[V1,Wi, Vm−1] , otherwise.

44

For any m ≥ p,

F =

[x1, V2, Vp−1, xm] , for i = p− 4,

[V1, Ti, xm] , otherwise.

Case 3. F ∈ R3,p(m+ 1) \R3,p(m). Then F = [x2, Ti, Vm] for some p− 3 ≤ i ≤ m− 3.

If m = p, then F = [V2, Vp−3, Vp−1]. Otherwise, m ≥ p+ 1, and m− 4 ≥ p− 3.

Then

F =

[x2, Tm−4, Vm−1] , for i = m− 3,

[x2, Ti, Vm−1] , otherwise.

For any m ≥ p, F = [V1, Ti, xm].

Case 4. F ∈ R4,p(m + 1) \ R4,p(m). Then m ≥ p + 1, and F = [Wi, Vj, Vm] for

some p − 4 ≤ i ≤ m − 5. If m = p + 1, then F = [V3, Vp−2, Vp] (recall that

Wp−4 = {x3, xp−2}). If m = p+ 2, then

F =

[Wp−4, Vp−1, Vp] , for i = p− 4 and j = p− 1,

[Wp−4, Vp−1, Vp+1] , for i = p− 4 and j = p,

[Vp−4, Vp−1, Vp+1] , for i = p− 3 and j = p.

For m ≥ p+ 3, we have m− 7 ≥ p− 4, and

F =

[Wm−7, Vm−3, Vm−1] , for i = m− 5,

[Wi, Vm−3, Vm−1] , for i < m− 5 and j = m− 2,

[Wi, Vj, Vm−1] , otherwise.

For any m ≥ p+ 1, F = [x1,Wi, Vj, xm] ∈ D3,p(m).

45

Thus, F ∈ F(Sp(m+ 1)), and Rp(m+ 1) ⊂ F(Sp(m+ 1)). We turn our attention to the

case when F ∈ Dp(m + 1). Since it is clear that xm+1 ∈ F , we let F = [G, xm+1], and

again seek F ∈ Rp(m) and F ∈ Dp(m) with G = F ∩ F .

Case 5. F ∈ D1,p(m+ 1).

Subcase 5.1. F = [x1, V2, Vi, xm+1] for some 4 ≤ i ≤ p− 3. Then

F =

[V1, V3, V5] , for i = 4,

[V1, V3, Vi] , otherwise,

and F = [x1, V2, Vi, xm].

Subcase 5.2. F = [x1, Vi, Vj, xm+1] for some 3 ≤ i ≤ p− 4. Then F = [V1, Vi, Vj], and

F =

[x1, Vi, Vm−2, xm] , for j = m− 1,

[x1, Vi, Vj, xm] , otherwise.

Case 6. F ∈ D2,p(m + 1). Then F = [V1, Ti, xm+1] for some p − 3 ≤ i ≤ m − 2. It

follows that

F =

[V1, Vp−4, Vp−2] , for i = p− 3,

[V1,Wi−2, Vi+1] , otherwise,

and

F =

[V1, Tm−3, xm] , for i = m− 2,

[V1, Ti, xm] , otherwise.

Case 7. F ∈ D3,p(m + 1). Then F = [x1,Wi, Vj, xm+1] for some p − 4 ≤ i ≤ m − 4,

and F = [V1,Wi, Vj]. If m = p, then F = [x1, V3, Vp−2, xp] (recall that Wp−4 =

46

{x3, xp−2}). If m = p+ 1, then

F =

[x1,Wp−4, Vp−1, xp+1] , for i = p− 4 and j = p− 1,

[x1,Wp−4, Vp−1, xp+1] , for i = p− 4 and j = p,

[x1, Vp−4, Vp−1, xp+1] , for i = p− 3 and j = p.

If m ≥ p+ 2, then m− 6 ≥ p− 4 and

F =

[x1,Wm−6, Vm−2, xm] , for i = m− 4,

[x1,Wi, Vm−2, xm] , for i < m− 4 and j = m− 1,

[x1,Wi, Vj, xm] , otherwise.

Thus, F ∈ F(Sp(m+1)) and Dp(m+1) ⊂ F(Sp(m+1)). Combined with the first part of

the proof, we have shown that for n ≥ p ≥ 9, Rp(n)∪Dp(n) ⊆ F(Sp(n)). Equality follows

from Lemma 3.1.6 and the fact that Sp(n) is a 6-dimensional neighbourly polytope with

n vertices, and as such, |F(Sp(n))| = n(n− 4)(n− 5)/6.

Note that since the neighbourly 6-polytopes Sp(n) are not cyclic, there is no ordering

on the vertices that will allow one to use Gale’s Evenness Condition to list the facets.

However, in light of Theorem 3.1.8, one may use the descriptions for Ri,p(n) for i =

1, 2, 3, 4 and Dj,p(n) for j = 1, 2, 3 (see pages 34–36) to generate the list of facets of

Sp(n). Table 3.1 contains such a list for the case p = 9 and n = 12.

3.2 The Universal Faces of Sp(n)

In this section, we use Algorithm 2 (cf. page 31) to describe the universal 3-faces and

universal edges of Sp(n). Following the same convention used in Algorithm 2, we do not

use the “∗”-notation when labelling vertices of quotient polytopes. Since En and Gn

47

R1,9(12)

[x1, x2, x3, x4, x5, x6] [x1, x2, x3, x4, x6, x7] [x1, x2, x3, x4, x7, x8] [x1, x2, x3, x4, x8, x9] [x1, x2, x3, x4, x9, x10]

[x1, x2, x3, x4, x10, x11] [x1, x2, x3, x4, x11, x12] [x1, x2, x4, x5, x6, x7] [x1, x2, x4, x5, x7, x8] [x1, x2, x4, x5, x8, x9]

[x1, x2, x4, x5, x9, x10] [x1, x2, x4, x5, x10, x11] [x1, x2, x4, x5, x11, x12] [x1, x2, x5, x6, x7, x8] [x1, x2, x5, x6, x8, x9]

[x1, x2, x5, x6, x9, x10] [x1, x2, x5, x6, x10, x11] [x1, x2, x5, x6, x11, x12] [x2, x3, x4, x5, x6, x7] [x2, x3, x4, x5, x7, x8]

[x2, x3, x4, x5, x8, x9] [x2, x3, x4, x5, x9, x10] [x2, x3, x4, x5, x10, x11] [x2, x3, x4, x5, x11, x12] [x2, x3, x5, x6, x7, x8]

[x2, x3, x5, x6, x8, x9] [x2, x3, x5, x6, x9, x10] [x2, x3, x5, x6, x10, x11] [x2, x3, x5, x6, x11, x12] [x2, x3, x6, x7, x8, x9]

[x2, x3, x6, x7, x9, x10] [x2, x3, x6, x7, x10, x11] [x2, x3, x6, x7, x11, x12] [x3, x4, x5, x6, x7, x8] [x3, x4, x5, x6, x8, x9]

[x3, x4, x5, x6, x9, x10] [x3, x4, x5, x6, x10, x11] [x3, x4, x5, x6, x11, x12] [x3, x4, x6, x7, x8, x9] [x3, x4, x6, x7, x9, x10]

[x3, x4, x6, x7, x10, x11] [x3, x4, x6, x7, x11, x12] [x3, x4, x7, x8, x9, x10] [x3, x4, x7, x8, x10, x11] [x3, x4, x7, x8, x11, x12]

[x3, x4, x8, x9, x10, x11] [x3, x4, x8, x9, x11, x12] [x3, x4, x9, x10, x11, x12] [x4, x5, x6, x7, x8, x9] [x4, x5, x6, x7, x9, x10]

[x4, x5, x6, x7, x10, x11] [x4, x5, x6, x7, x11, x12] [x4, x5, x7, x8, x9, x10] [x4, x5, x7, x8, x10, x11] [x4, x5, x7, x8, x11, x12]

[x4, x5, x8, x9, x10, x11] [x4, x5, x8, x9, x11, x12] [x4, x5, x9, x10, x11, x12] [x5, x6, x7, x8, x9, x10] [x5, x6, x7, x8, x10, x11]

[x5, x6, x7, x8, x11, x12] [x5, x6, x8, x9, x10, x11] [x5, x6, x8, x9, x11, x12] [x5, x6, x9, x10, x11, x12]

R2,9(12)[x1, x2, x3, x7, x8, x9] [x1, x2, x3, x7, x9, x10] [x1, x2, x3, x7, x10, x11] [x1, x2, x3, x7, x11, x12] [x1, x2, x6, x8, x9, x10]

[x1, x2, x6, x8, x10, x11] [x1, x2, x6, x8, x11, x12] [x1, x2, x7, x9, x10, x11] [x1, x2, x7, x9, x11, x12] [x1, x2, x8, x10, x11, x12]

R3,9(12)[x2, x6, x7, x8, x9, x10] [x2, x6, x7, x8, x10, x11] [x2, x6, x7, x8, x11, x12] [x2, x7, x8, x9, x10, x11] [x2, x7, x8, x9, x11, x12]

[x2, x8, x9, x10, x11, x12]

R4,9(12) [x3, x7, x8, x9, x10, x11] [x3, x7, x8, x9, x11, x12] [x3, x7, x9, x10, x11, x12] [x6, x8, x9, x10, x11, x12]

D1,9(12)

[x1, x2, x3, x4, x5, x12] [x1, x2, x3, x5, x6, x12] [x1, x2, x3, x6, x7, x12] [x1, x3, x4, x5, x6, x12] [x1, x3, x4, x6, x7, x12]

[x1, x3, x4, x7, x8, x12] [x1, x3, x4, x8, x9, x12] [x1, x3, x4, x9, x10, x12] [x1, x3, x4, x10, x11, x12] [x1, x4, x5, x6, x7, x12]

[x1, x4, x5, x7, x8, x12] [x1, x4, x5, x8, x9, x12] [x1, x4, x5, x9, x10, x12] [x1, x4, x5, x10, x11, x12] [x1, x5, x6, x7, x8, x12]

[x1, x5, x6, x8, x9, x12] [x1, x5, x6, x9, x10, x12] [x1, x5, x6, x10, x11, x12]

D2,9(12) [x1, x2, x6, x7, x8, x12] [x1, x2, x7, x8, x9, x12] [x1, x2, x8, x9, x10, x12] [x1, x2, x9, x10, x11, x12]

D3,9(12)[x1, x3, x7, x8, x9, x12] [x1, x3, x7, x9, x10, x12] [x1, x3, x7, x10, x11, x12] [x1, x6, x8, x9, x10, x12] [x1, x6, x8, x10, x11, x12]

[x1, x7, x9, x10, x11, x12]

Table 3.1: The facets of S9(12).

are universal faces of Sp(n), the quotient polytope Sp(n)/En is a neighbourly 4-polytope

with n− 2 vertices and the quotient polytope Sp(n)/Gn is a polygon with n− 4 vertices.

From Theorem 3.1.8, it follows that (with abuse of notation)

F3(Sp(n)/En) = R1,p(n)/En ∪R2,p(n)/En ∪D1,p(n)/En ∪D2,p(n)/En

with

R1,p(n)/En :={[x2, Vi, xn−1] | 3 ≤ i ≤ p− 4},

R2,p(n)/En :={[x2,Wi, xn−1] | p− 4 ≤ i ≤ n− 4},

D1,p(n)/En :={[V2, Vi] | 4 ≤ i ≤ p− 3} ∪ {[Vi, Vj] | 3 ≤ i ≤ p− 4},

D2,p(n)/En :={[x2, Ti] | p− 3 ≤ i ≤ n− 3},

48

and that

E(Sp(n)/Gn) =R1,p(n)/Gn ∪R2,p(n)/Gn ∪D1,p(n)/Gn

with

R1,p(n)/Gn :={[Vi] | 3 ≤ i ≤ p− 4},

R2,p(n)/Gn :={[Wi] | p− 4 ≤ i ≤ n− 4},

D2,p(n)/Gn :={[Vn−3]}.

For Step 1 of Algorithm 2, sew a vertex z+2 onto Sp(n)/Gn through the tower T /Gn :=

{Fn/Gn} = {xn−3, xn−2} = {[Vn−3]}. This amounts to choosing z+2 so that it is beyond

[Vn−3] and beneath all other edges of Sp(n)/Gn. Let (Sp(n)/Gn)+ = [Sp(n)/Gn, z+2 ]. The

edges of (Sp(n)/Gn)+ are

E((Sp(n)/Gn)+) = R1,p(n)/Gn ∪R2,p(n)/Gn ∪ {[xn−3, z+2 ], [xn−2, z+2 ]}.

For Step 2(i) of Algorithm 2, sew a vertex z+1 onto Sp(n)/En through the tower

T /En := {Gn/En, Fn/En} to obtain (Sp(n)/En)+ = [Sp(n)/En, z+1 ]. Since z+1 is be-

neath every G ∈ D1,p(n)/En ∪ D2,p(n)/En by Proposition 2.2.2, we have D1,p(n)/En ∪

D2,p(n)/En ⊆ F3((Sp(n)/En)+). Note that [E, x2, z+1 ] ∈ F3((Sp(n)/En)+) for every edge

E of (Sp(n)/Gn)+. Hence,

{[x2, Vi, z+1 ] | 3 ≤ i ≤ p− 4} ∪ {[x2,Wi, z+1 ] | p− 4 ≤ i ≤ n− 4} ⊆ F3((Sp(n)/En)+),

[x2, xn−3, xn−1, z+1 ] = [x2,Wn−3, z

+1 ] ∈ F3((Sp(n)/En)+) and [x2, xn−2, xn−1, z

+1 ] =

[x2, Vn−2, z+1 ] ∈ F3((Sp(n)/En)+). For every edge E of Sp(n)/Gn such that z+2 is be-

neath E, we have [E, xn−1, z+1 ] ∈ F3((Sp(n)/En)+), whence

{[Vi, xn−1, z+1 ] | 3 ≤ i ≤ p− 4} ∪ {[Wi, xn−1, z+1 ] | p− 4 ≤ i ≤ n− 4} ⊆ F3((Sp(n)/En)+).

49

Thus,

F3((Sp(n)/En)+) = (R1,p(n)/En)+ ∪ (R2,p(n)/En)+ ∪ (D1,p(n)/En)+ ∪ (D2,p(n)/En)+

with

(R1,p(n)/En)+ := {[x2, Vi, z+1 ] | 3 ≤ i ≤ p− 4},

(R2,p(n)/En)+ := {[x2,Wi, z+1 ] | p− 4 ≤ i ≤ n− 3},

(D1,p(n)/En)+ := {[V2, Vi] | 4 ≤ i ≤ p− 3} ∪ {[Vi, Vj] | 3 ≤ i ≤ p− 4},

(D2,p(n)/En)+ := {[x2, Ti] | p− 3 ≤ i ≤ n− 3} ∪ {[x2, Vn−2, z+1 ]}.

Lemma 3.2.1. Let n ≥ p. Then

U1((Sp(n)/En)+) = {[Vi] | i ∈ {3, 4, . . . , p− 4, n− 2}} ∪ {[x2, z+1 ], [xn−1, z+1 ]}.

Proof. Let n = p. Then Sp(p)/Ep = C(p)/[x1, xp] is a cyclic 4-polytope with vertex array

x2 < x3 < · · · < xp−1. By Theorem 1.1.8,

U1(Sp(p)/Ep) = {[Vi] | 2 ≤ i ≤ p− 2} ∪ {[x2, xp−1]}.

From Step 2(ii) of Algorithm 2, [x2, z+1 ] and [xp−1, z

+1 ] are universal faces of (Sp(p)/Ep)

+

while [x2, xp−1] is not a universal edge of (Sp(p)/Ep)+. For i ∈ {3, 4, . . . , p − 4, n − 2},

[x2, Vi, z+1 ] is a 3-face of (Sp(p)/Ep)

+, and so [Vi] is a universal edge of (Sp(p)/Ep)+. For

p− 3 ≤ i ≤ n− 3, neither [x2, Vi, z+1 ] nor [Vi, xn−1, z

+1 ] is a 3-face of (Sp(p)/Ep)

+, and so

[Vi] is not a universal edge of (Sp(p)/Ep)+.

We now assume that for some n = k ≥ p,

U1((Sp(k)/Ek)+) = {[Vi] | i ∈ {3, 4, . . . , p− 4, k − 2}} ∪ {[x2, z+1 ], [xk−1, z

+1 ]}.

From Theorem 2.2.3, we have (Sp(k)/Ek)+ ∼= [Sp(k), z+1 ]/[x1, z

+1 ], which is combinatori-

ally equivalent to [Sp(k), xk+1]/[x1, xk+1] = Sp(k + 1)/Ek+1. Hence,

U1(Sp(k + 1)/Ek+1) = {[Vi] | i ∈ {3, 4, . . . , p− 4, k − 2} ∪ {[x2, xk], [xk−1, xk]}.

50

Again following Step 2(ii) of Algorithm 2, the proof follows in a similar fashion to the

case n = p, and the details are omitted.

This completes Step 2 of Algorithm 2. Step 3 yields the odd-dimensional universal

faces of Sp(n). We skip Step 3(i) since the facets of Sp(n) were derived in Theorem 3.1.8.

We define the following 3-dimensional subpolytopes of Sp(n):

U1,p(n) = {[V1, Vi] | i ∈ {3, 4, . . . , p− 4, n− 3, n− 2}} and

U2,p(n) = {[x1, Vi, xn] | i ∈ {3, 4, . . . , p− 4, n− 3, n− 2}}⋃

{[Vi, Vn−1] | i ∈ {1, 3, 4, . . . , p− 4, n− 3}}.

Theorem 3.2.2. Let n ≥ p+ 1. Then U3(Sp(n)) = U1,p(n) ∪ U2,p(n).

Proof. Let n = p + 1. Recall that Sp(p) = C(p) is a cyclic 6-polytope with vertex array

x1 < x2 < · · · < xp, and hence by Theorem 1.1.8,

U3(Sp(p)) = {[Vi, Vj]} ∪ {[x1, Vi, xp]}.

Following Step 3(ii) of Algorithm 2, we obtain that Gp = [V1, Vp−1] is a universal 3-face

of Sp(p+1). Since [V1, Vi, xp, xp+1] is a facet of Sp(p+1) for i ∈ {3, 4, . . . , p−4, p−2}, we

have [V1, Vi] is a universal 3-face of Sp(p+1) and U1,p(p+1) ⊂ U3(Sp(p+1)). Since neither

[x1, G, xp+1] nor [G, xp, xp+1] is a facet of Sp(p+ 1) for every G ∈ U3(Sp(p)) \ U1,p(p+ 1),

it follows that G is not a universal 3-face of Sp(p+ 1). From Lemma 3.2.1, we have that

U1((Sp(p))+) = {[Vi] | i ∈ {3, 4, . . . , p− 4, p− 2}} ∪ {[x2, z+1 ], [xp−1, z+1 ]}.

Since [Vi] is a universal edge of Sp(p)+ for i ∈ {3, 4, . . . , p − 4, p − 2}, we have that

[x1, Vi, xp+1] and [Vi, Vp] are universal 3-faces of Sp(p + 1). Also, with the convention

that the vertex z+1 of Sp(p)+ corresponds to the vertex xp of Sp(p + 1), we obtain that

[x1, x2, xp, xp+1] = [V1, Vp] and [x1, xp−1, xp, xp+1] = [x1, Vp−1, xp+1] are universal 3-faces

51

of Sp(p + 1). Thus, U2,p(p + 1) ⊂ U3(Sp(p + 1)). We conclude that U3(Sp(p + 1)) =

U1,p(p+ 1) ∪ U2,p(p+ 1).

The proof that U3(Sp(n)) = U1,p(n) ∪ U2,p(n) for n > p + 1 is proved similarly, and

the details are omitted.

Theorem 3.2.3. Let n ≥ p+ 1. Then

U1(Sp(n)) = {[Vi] | i ∈ {1, 3, 4, . . . , p− 4, n− 3, n− 2, n− 1}} ∪ {[x1, xn]}.

Proof. Let n = p + 1. As in the proof of Theorem 3.2.2, we note that Sp(p) = C(p) is a

cyclic polytope with vertex array x1 < x2 < · · · < xp, and by Theorem 1.1.8,

U1(Sp(p)) = {[Vi]} ∪ {[x1, xp]}.

Following Step 3(iii) of Algorithm 2, we have that Ep = [x1, xp] /∈ U1(Sp(p + 1)) while

[x1, xp+1] and [Vp] are universal edges of Sp(p + 1). Since [x1, Vi, xp+1] ∈ U3(Sp(p + 1))

for i ∈ {3, 4, . . . , p − 4, p − 2, p − 1}, we have that [Vi] ∈ U1(Sp(p + 1)). Since [V1, Vp] ∈

U3(Sp(p + 1), it follows that [V1] ∈ U1(Sp(p + 1)). Since neither [x1, Vi, xp+1] nor [Vi, Vp]

is a universal 3-face of Sp(p + 1) for i = 2 and i = p− 3, we see that [V2] and [Vp−3] are

not universal edges of Sp(p+ 1). The case n > p+ 1 is analogous to the case n = p+ 1,

and is omitted.

3.3 Quotient Polytopes of Sp(n)

In this section, we describe the quotient polytopes that will aid us to verify that a polytope

of type Sp(n) satisfies the Separation Conjecture. We recall that Sp(n) = [x1, . . . , xn]

denotes a semi-cyclic 6-polytope with the vertex array x1 < x2 < · · · < xn. We verify

the Separation Conjecture by determining the location of the vertices of Sp(n) relative

to their labelling; that is, the order of the sewings and the resulting vertex array. For

G ∈ U3(Sp(n)), we have that Sp(n)/G is a polygon with n − 4 vertices, and so we have

52

Figure 3.1: Sp(n)/[x1, Vi, xn] for 3 ≤ i ≤ p− 4.

a particular arrangement of the vertices. Ideally, we would like to consider Sp(n)/G

only for G ∈ U3(Sp(n)), however Sp(n) has few universal 3-faces in general, so we allow

for G /∈ U3(Sp(n)). In all cases, the edges of Sp(n)/G are derived from facets of Sp(n)

that contain G. Note that we also include xn+1 in the quotient polytopes, even though

xn+1 /∈ V(Sp(n)). This is to exhibit which facets of Sp(n) containing G that xn+1 would

be beneath or beyond if it were sewn onto Sp(n) to obtain Sp(n+ 1).

We consider first the quotient polytopes Sp(n)/G for G ∈ U3(Sp(n)). If G = [x1, Vi, xn]

for some 3 ≤ i ≤ p − 4, then a representation of Sp(n)/G is shown in Figure 3.1. The

edge [x∗i−2, x∗i−1] of Sp(n)/G represents the facet F = [G, xi−2, xi−1] = [x1, Vi−2, Vi, xn] of

Sp(n), and the line 〈x∗i−2, x∗i−1〉 represents the hyperplane H1 = 〈F 〉. We see that H1

strictly separates xn+1 from those vertices of Sp(n) that do not lie in F , that is, that

xn+1 is beyond F . The line 〈x∗3, x∗p−2〉 represents the hyperplane H2 = 〈x1, Vi,Wp−4, xn〉,

and we see that H2 separates the vertices x2, xp−3, xp−4, . . . , xn−1 of Sp(n) from xn+1 and

the vertices x4, x5, . . . , xp−3 of Sp(n). Thus, [x1, Vi,Wp−4, xn] is not a facet of Sp(n) for

3 ≤ i ≤ p− 4.

If G = [Vi, Vn−1] for 3 ≤ i ≤ p − 4, then Figure 3.2 shows a representation of

53

Figure 3.2: Sp(n)/[Vi, Vn−1] for 3 ≤ i ≤ p− 4.

Figure 3.3: Sp(n)/[V1, V3].

Sp(n)/G. If G = [V1, V3] then Figure 3.3 shows a representation of Sp(n)/G, and if

G = Gn = [V1, Vn−1] then Figure 3.4 shows two possible representations of Sp(n)/G: one

for n− p even, and one for n− p odd.

We obtained the vertex labelling of Sp(n)/G for G ∈ U3(Sp(n)) directly from the list

of facets of Sp(n) to derive Figures 3.1–3.4. In Figures 3.5–3.20, we consider Sp(n)/G

54

(a) n− p even (b) n− p odd

Figure 3.4: Sp(n)/[V1, Vn−1].

for G /∈ U3(Sp(n)), and use the information regarding the vertices obtained in Figures

3.1–3.4 to locate the vertices of Sp(n) in Sp(n)/G that do not determine a face of Sp(n)

with G.

Let G = [x1,Wi, xn] for some p − 4 ≤ i ≤ n − 4. Recalling that Wp−4 = {x3, xp−2},

a representation of Sp(n)/[x1,Wp−4, xn] is shown in Figure 3.5. Since the line 〈x∗i , x∗i+1〉

represents the hyperplane H2 = 〈x1, Vi,Wp−4, xn〉 for 4 ≤ j ≤ p−4, and H2 is represented

by the line 〈x∗3, x∗p−2〉 in Sp(n)/[x1, Vi, xn], we see from Figure 3.1 thatH2 strictly separates

the vertices x2, xp−1, xp, . . . , xn−1 from xn+1 and the vertices x4, x5, . . . , xi−1, xi+2, . . . , xp−3.

From this, it follows that the vertices x5, x6, . . . , xp−4 of Sp(n) lie above both of the lines

H and H of Sp(n)/[x1,Wp−4, xn]. For i = p− 3, a representation of Sp(n)/[x1,Wp−3, xn]

is shown in Figure 3.6. The line 〈x∗i , x∗i+1〉 represents the hyperplane 〈x1, Vi,Wp−3, xn〉 for

3 ≤ i ≤ p − 5, which in turn is represented by the line 〈x∗p−3, x∗p−1〉 in Sp(n)/[x1, Vi, xn].

It now follows that 〈x1, Vi,Wp−3, xn〉 strictly separates the vertices x2, x3, . . . , xi−1, and

xi+2, . . . , xn−1 of Sp(n) from xn+1 and the vertex xp−2. In addition, the line 〈x∗3, x∗p−2〉

represents the hyperplane 〈x1,Wp−4,Wp−3, xn〉, and is represented by the line 〈x∗p−3, x∗p−1〉

in Sp(n)/[x1,Wp−4, xn]. From this, it follows that 〈x1,Wp−4,Wp−3, xn〉 strictly separates

the vertices x2, xp, xp+1, . . . , xn−1 of Sp(n) from xn+1 and the vertices x4, x5, . . . , xp−4.

55

Figure 3.5: Sp(n)/[x1,Wp−4, xn].

Figure 3.6: Sp(n)/[x1,Wp−3, xn].

For i ≥ p − 2, we follow a similar procedure to find the relative locations of those ver-

tices of Sp(n) in Sp(n)/[x1,Wi, xn] that do not determine a face with [x1,Wi, xn]. A

representation of Sp(n)/[x1,Wp−2, xn] is shown in Figure 3.7, while representations of

Sp(n)/[x1,Wi, xn], p − 1 ≤ i ≤ n − 4, are given in Figure 3.8 for i − p even and Figure

3.9 for i− p odd.

It will not always be important for us to know the exact location of those vertices of

56

Figure 3.7: Sp(n)/[x1,Wp−2, xn].

Figure 3.8: Sp(n)/[x1,Wi, xn] for p− 1 ≤ i ≤ n− 4 and i− p even.

Sp(n) that do not determine a face with [x1,Wi, xn] in Sp(n)/[x1,Wi, xn] for p− 4 ≤ i ≤

57

Figure 3.9: Sp(n)/[x1,Wi, xn] for p− 1 ≤ i ≤ n− 4 and i− p odd.

n− 4. To simplify matters in such an event, we introduce the following notation:

z(p− 4) = 4,

z(p− 3) = p− 4,

z(p− 2) = 3,

z(i) = i− 2,

and

Z(p− 4) = {x5, x6, . . . , xp−4},

Z(p− 3) = {x3, x4, . . . , xp−5},

Z(p− 2) = {x4, x5, . . . , xp−3},

Z(i) = {x3, x4, . . . , xi−3, xi−1},

for i ≥ p− 1. The representation shown in Figure 3.10 is valid for any p− 4 ≤ i ≤ n− 4.

In terms of F(Sp(n)) = Rp(n) ∪ Dp(n) with

Rp(n) = R1,p(n) ∪R2,p(n) ∪R3,p(n) ∪R4,p(n) and

Dp(n) = D1,p(n) ∪D2,p(n) ∪D3,p(n),

Figure 3.1 represents information regarding those facets F ∈ R1,p(n) ∪ D1,p(n) that

contain [x1, Vi, xn] for 3 ≤ i ≤ p − 4, Figure 3.2 contains information regarding those

facets F ∈ R1,p(n) ∪ D1,p(n) that contain [Vi, Vn−1] for some 3 ≤ i ≤ p − 4, Figure 3.3

contains information regarding those facets F ∈ R1,p(n)∪D1,p(n) that contain [V 1, V 3],

58

Figure 3.10: Sp(n)/[x1,Wi, xn] for p− 4 ≤ i ≤ n− 4.

Figure 3.4 represents information regarding those facets F ∈ R1,p(n) ∪R2,p(n) ∪D2,p(n)

that contain [V1, Vn−1], and Figures 3.5–3.10 contain information regarding those facets

F ∈

R2,p(n) ∪D1,p(n), for i = p− 4,

R2,p(n) ∪D1,p(n) ∪D2,p(n) ∪D3,p(n), for i = p− 3,

R2,p(n) ∪D2,p(n) ∪D3,p(n), for i ≥ p− 2,

that contain [x1,Wi, xn]. In a similar fashion, we construct representations of Sp(n)/G

with G = [Wi, Vn−1] for p − 4 ≤ i ≤ n − 4 (see Figures 3.11–3.16), G = [Vi, Vn−1] for

p − 3 ≤ i ≤ n − 3 (see Figures 3.17 and 3.18), G = [x1, V2, xn] (see Figure 3.19) and

G = [V2, Vn−1] (see Figure 3.20). Figures 3.11–3.16 contain information about those

facets

F ∈

R1,p(n) ∪R2,p(n) ∪R4,p(n) ∪D3,p(n), for i = p− 4,

R1,p(n) ∪R2,p(n) ∪R3,p(n) ∪R4,p(n) ∪D3,p(n), for i = p− 3,

R2,p(n) ∪R3,4(n) ∪R4,p(n) ∪D3,p(n), for i ≥ p− 2,

59

Figure 3.11: Sp(n)/[Wp−4, Vn−1].

Figure 3.12: Sp(n)/[Wp−3, Vn−1].

that contain [Wi, Vn−1] for p− 4 ≤ i ≤ n− 4, Figure 3.17 contains information regarding

those facets F ∈ R1,p(n)∪R3,p(n)∪R4,p(n) that contain [Vi, Vn−1] for p− 3 ≤ i ≤ n− 4,

Figure 3.18 contains information regarding those facets F ∈ R1,p(n)∪R3,p(n)∪R4,p(n)∪

D2,p(n) ∪ D3,p(n) that contain [Vn−3, Vn−1], Figure 3.19 contains information regarding

those facets F ∈ R1,p(n) ∪ R2,p(n) ∪ D1,p(n) that contain [x1, V2, xn], and Figure 3.20

contains information about those F ∈ R1,p(n) ∪R2,p(n) that contain [V2, Vn−1].

60

Figure 3.13: Sp(n)/[Wp−2, Vn−1].

Figure 3.14: Sp(n)/[Wi, Vn−1] for p− 1 ≤ i ≤ n− 4 and i− p even.

61

Figure 3.15: Sp(n)/[Wi, Vn−1] for p− 1 ≤ i ≤ n− 4 and i− p odd.

Figure 3.16: Sp(n)/[Wi, Vn−1] for p− 4 ≤ i ≤ n− 4.

62

(a) i− p even (b) i− p odd

Figure 3.17: Sp(n)/[Vi, Vn−1] for p− 3 ≤ i ≤ n− 4.

(a) n− p odd (b) n− p even

Figure 3.18: Sp(n)/[Vn−3, Vn−1].

Figure 3.19: Sp(n)/[x1, V2, xn].

63

Figure 3.20: Sp(n)/[V2, Vn−1].

64

Chapter 4

Separation in Semi-Cyclic 6 -Polytopes

In this chapter, we verify the Separation Conjecture for the semi-cyclic 6-polytopes

Sp(n) = [x1, x2, . . . , xn] with vertex array x1 < x2 < · · · < xn. To simplify matters,

we present only the case p = 9 as the verification for p > 9 follows very closely to that

of p = 9. Indeed, noting that the quotient polytopes used to verify separation below are

obtained from the quotient polytopes derived in Section 3.3 with p = 9, one sees imme-

diately that the results of this section still hold for p > 9. As a further simplification, we

let S(n) = S9(n) and recall that in this construction, n ≥ 9 and S(9) = C(9). Since C(9)

is cyclic with x1 < x2 < · · · < x9, it follows that C(8) is cyclic with x1 < x2 < · · · < x8.

We let S(m) = [x1, x2, . . . , xm] for 7 ≤ m ≤ n.

Let n ≥ 9. From Chapter 3, we have

R1(n) = {[V1, Vi, Vj] | 3 ≤ i ≤ 5} ∪ {[V2, Vi, Vj] | 4 ≤ i ≤ 6} ∪ {[Vi, Vj, Vk] | 3 ≤ i ≤ r},

R2(n) = {[V1,Wi, Vj] | 5 ≤ i ≤ n− 4},

R3(n) = {[x2, Ti, Vj] | 6 ≤ i ≤ n− 4},

R4(n) = {[Wi, Vj, Vk] | 5 ≤ i ≤ n− 6},

D1(n) = {[x1, V2, Vi, xn]} | 4 ≤ i ≤ 6} ∪ {[x1, Vi, Vj, xn] | 3 ≤ i ≤ 5},

D2(n) = {[V1, Ti, xn] | 6 ≤ i ≤ n− 3} and

D3(n) = {[x1,Wi, Vj, xn] | 5 ≤ i ≤ n− 5},

with the restrictions that r = 4 if n = 9, r = 5 if n ≥ 10 (cf. page 36) and that

R3(9) = D3(9) = R4(9) = R4(10) = ∅. Then with

R(n) = R1(n) ∪R2(n) ∪R3(n) ∪R4(n) and

65

D(n) = D1(n) ∪D2(n) ∪D3(n),

Theorem 3.1.8 states that F(S(n)) = R(n) ∪ D(n). We further recall that

U1(n) = {[V1, Vi] | i ∈ {3, 4, 5, n− 3, n− 2}} and

U2(n) = {[x1, Vi, xn] | i ∈ {3, 4, 5, n− 3, n− 2}} ∪ {[Vi, Vn−1] | i ∈ {1, 3, 4, 5, n− 3}}

for n ≥ 10, and that by Theorem 3.2.2,

U3(S(n)) = U1(n) ∪ U2(n).

Finally, Theorem 3.2.3 shows that for n ≥ 10,

U1(S(n)) = {[Vi] | i ∈ {1, 3, 4, 5, n− 3, n− 2, n− 1}} ∪ {[x1, xn]}.

With respect to the quotient polytopes S(n)/[x1,Wi, xn] and S(n)/[Wi, Vn−1], we

recall the notation introduced on page 57, which is presented here again with p = 9.

z(5) = 4,

z(6) = 5,

z(7) = 3,

z(i) = i− 2,

and

Z(5) = {x5},

Z(6) = {x3, x4},

Z(7) = {x4, x5, x6},

Z(i) = {x3, x4, . . . , xi−3, xi−1},

(4.1)

4.1 Overview

Let O ∈ int(S(n)). We call a hyperplane H = H(O) a separating hyperplane for O if

O /∈ H and there is a facet of S(n) that is separated from O by H. A complete collection

of separating hyperplanes for O is a family of hyperplanes H such that any facet of S(n)

is separated from O by some H ∈ H. The goal of this chapter is to prove the following:

Main Theorem 1. Let n ≥ 9, and O ∈ int(S(n)). Then there exists a complete

collection H of separating hyperplanes for O such that |H| ≤ 29 and any facet of S(n) is

strictly separated from O by some H ∈ H.

66

We choose H ∈ H that are spanned by vertices of S(n). Let H be a separating

hyperplane for O, and F be a facet of S(n) that is separated from O by H. If H contains

a vertex of F , then H does not strictly separate F from O. However, since O /∈ H,

we achieve strict separation by translating H towards O (see Figure 4.1). Thus, it is

indeed sufficient to prove Main Theorem 1 by finding a complete collection of separating

hyperplanes for O, which are affinely spanned by vertices of S(n).

Given O ∈ int(S(n)), consider the smallest integer m = m(O) such that

8 ≤ m ≤ n and O ∈ int(S(m)). (4.2)

It follows that either O ∈ int(S(8)) or O ∈ int(S(m)) \ int(S(m − 1)) for some m ≥ 9.

The three sets

A(m) = F(S(n)) ∩ F(S(m− 1)),

B(m) =(F(S(n)) ∩ F(S(m))

)\ F(S(m− 1)) and

C(m) = {F ∈ F(S(n)) |F ∩ {xm+1, xm+2, . . . , xn} 6= ∅}

= B(m+ 1) ∪B(m+ 2) ∪ · · · ∪B(n)

partition F(S(n)) (note that if m = n, then C(m) = ∅). For a given a collection of

hyperplanes H and the smallest m satisfying (4.2), the following Lemma shows that it

Figure 4.1: From separation to strict separation.

67

is sufficient to verify that H is a complete collection of separating hyperplanes for O by

verifying that any facet F ∈ A(m) ∪ B(m) ∪ B(m + 1) is separated from O by some

H ∈ H.

Remark 4.1.1. For n ≥ 13, there is crucial distinction: whether or not O ∈ int(S(12)).

If O /∈ int(S(12)) then the arguments for any location of O are generalizations of the

arguments for O ∈ int(S(13)).

Lemma 4.1.2. Let 8 ≤ m ≤ n− 1, O ∈ int(S(m)), and H be a collection of hyperplanes

spanned by vertices of S(m). If O is separated from any F ∈ B(m+ 1) by some H ∈ H,

then O is separated from any F ∈ C(m) by some H ∈ H.

Proof. Assume H is a collection of hyperplanes spanned by vertices of S(m) such that

any facet F ∈ B(m+ 1) is separated from O ∈ int(S(m)) by some H ∈ H. If m = n− 1,

then we are done, so we assume m < n−1. We show that any F ∈ B(m+2) is separated

from O by some H ∈ H. The proof that any F ∈ B(j) is separated from O by some

H ∈ H for j = m+ 3,m+ 4, . . . , n is similar.

Let F ∈ B(m+2). Then F = [G, xm+2] with G = F ∩F for some F , F ∈ F(S(m+1))

such that xm+2 is beneath F and beyond F . By assumption, O is separated from F by

some H ∈ H, and thus G is separated from O by H. Since xm+2 is beyond F , it follows

that Em+1 = [x1, xm+1] ⊂ F . Since H is spanned by vertices of S(m), it follows that

xm+1 /∈ H and O is strictly separated from xm+1 by H. Since [xm+1, xm+2] is a universal

edge of S(m+ 2), Theorem 1.1.11 shows that xm+1 and xm+2 lie on the same side of H.

Hence, it follows that F and O are separated by H.

Theorem 4.1.3. Let O ∈ int(S(8)). Then there is a complete collection H of separating

hyperplanes with |H| ≤ 16.

Proof. Note that |F(S(8))| = 16. Let H1, H2, . . . , H16 denote the hyperplanes spanned

by F ∈ F(S(8)). Then it is clear that any facet F ∈ A(8) ∪ B(8) is separated from O

68

by some H ∈ H = {H1, H2, . . . , H16}. Let F ∈ B(9). Then x9 ∈ F , and F = [G, x9]

with G = F ∩ F for some F , F ∈ F(S(8)) such that x9 is beneath F and beyond F . By

construction, there is an H ∈ H that separates F from O. In particular, H separates

G from O. Since x9 is beyond F , F = [G, x9] is separated from O by H. Thus, any

F ∈ B(9) is separated from O by some H ∈ H, and the assertion follows from Lemma

4.1.2.

Lemma 4.1.2 shows that it is sufficient to verify that a given collection H of hyper-

planes is a complete collection of separating hyperplanes for O by verifying that any

F ∈ A(m) ∪B(m) ∪B(m+ 1) is separated from O by some H ∈ H. We note that

A(m) ∪B(m) =

R(m), for m < n,

F(S(n)), for m = n

and

B(m+ 1) =

R(m+ 1) \ R(m), for m < n− 1

F(S(n)) \ F(S(n− 1)), for m = n− 1

∅, for m = n.

From this, it follows that

A(m) ∪B(m) ∪ D(m) = F(S(m)) and

B(m+ 1) ∪ D(m+ 1) = F(S(m+ 1)) \ F(S(m))

for any m = 8, 9, . . . , n. We obtain that

A(m) ∪B(m) ∪ D(m) ∪B(m+ 1) ∪ D(m+ 1) = F(S(m)) ∪(F(S(m+ 1)) \ F(S(m))

)= F(S(m+ 1)) ∪ D(m).

69

To avoid handling the cases m = n − 1 and m = n separately, we verify a stronger

condition: given a collection H of hyperplanes, we demonstrate that H a complete col-

lection of separating hyperplanes for O by proving that any F ∈ F(S(m+ 1))∪D(m) is

separated from O by some H ∈ H.

In the following, we assume that 11 ≤ m ≤ n; the cases m = 9, 10 being addressed in

Section 4.5 below. Since O ∈ int(S(m))\ int(S(m−1)), there exists a F ∈ D(m−1) such

that O ∈ [F , xm]. The case F ∈ D2(m− 1) is handled in Section 4.2, F ∈ D3(m− 1) is

dealt with in Section 4.3 and the case F ∈ D1(m−1) is addressed in Section 4.4. In each

case we determine first a complete collection of separating hyperplanes for O under the

assumption that O ∈ int([F , xm]). Lemma 4.1.4 below is then used to handle the case

O ∈ bd([F , xm]).

Let P be a neighbourly 6-polytope, and O ∈ int(P ). Consider a hyperplane H

spanned by the vertices of P such that H ∩ int(P ) 6= ∅. Let H+ be a closed halfspace

determined by H, and Q = P ∩H+. Then Q a neighbourly 6-subpolytope of P .

Lemma 4.1.4. Let P,H and Q be defined as above. If O ∈ H, then any F ∈ F(P ) ∩

F(Q) is separated from O by one of at most 5 hyperplanes.

Proof. Since Q = P ∩ H+ is neighbourly and H is spanned by vertices of P , there is a

facet F of Q such that F = H ∩Q, from which it follows that O ∈ F . We note that F is

not a facet of P because H does not support P . Denote the vertices of F by y1, y2, . . . , y6.

For i = 1, 2, . . . , 6, Gi = [V(F ) \ {yi}] is a 4-face of F . Since dim(F ) = 5, F is not a

missing face of P (cf. [22]), and thus, one of G1, G2, . . . , G6, say G6, is not a face of P .

For j = 1, 2, 3, 4, 5, let Qj = [V(Q)\{yj}]. Note that every F ∈ F(P )∩F(Q) is a facet of

Qj for some j = 1, 2, 3, 4, 5. Let Hj be a hyperplane that separates [O, yj] from Qj. Then

any F ∈ F(P )∩F(Q) is separated from O by one of the hyperplanes H1, H2, H3, H4 and

H5.

70

4.2 The Case F ∈ D2(m− 1)

Let F = [V1, Ti, xm−1] ∈ D2(m − 1) for some 6 ≤ i ≤ m − 4, and O ∈ int(S(m)) \

int(S(m− 1)) such that

O ∈ [F , xm] = [V1, Ti, xm−1, xm]

⊂ [V1, V3, Vi−1, Vi+1, Vm−1]

= [x1, x2, x3, x4, xi−1, xi, xi+1, xi+2, xm−1, xm],

a subpolytope of S(m) with ten vertices. Consider the hyperplanes

H1 = 〈V1, Ti, xm〉, H2 = 〈V1,Wi, Vm−1〉,

H3 = 〈x1, Ti, Vm−1〉, H4 = 〈x2, Ti, Vm−1〉,

H5 = 〈V1, Vi, Vm−1〉, H6 = 〈V1, Vi+1, Vm−1〉,

H7 = 〈V1, V3, xi, xm〉, H8 = 〈V1, Ti−1, xm〉,

H9 = 〈V1, Ti, xm−1〉,

(4.3)

all of which are spanned by the vertices of this subpolytope. They appear in Figures

4.2–4.6 below. We remark that each hyperplane depends on i; as i changes, so too do

the hyperplanes and the sets of facets of S(n) that they each separate from O. In Figure

4.4, the quotient polytope S(m)/[V1, Vm−1] is shown only for the case m odd. For the

case m even, the hyperplanes H2, H5 and H6 are unchanged, as are the sets of facets

of S(n) that they separate from O (see Figure 3.4 with p = 9). We proceed under the

assumption that O ∈ int([F , xm]).

Lemma 4.2.1. Let O ∈ int([F , xm]) and F = [V1, Ti, xm−1] ∈ D2(m−1). Then any facet

of S(n) is separated from O by one of H1, H2, . . . , H9.

Proof. We first handle the facets of S(n) that belong to A(m). Since O ∈ int([F , xm]) ⊂

int(S(m)) \S(m− 1) and F ∈ F(S(m− 1)), it follows that O is beyond F and H9 = 〈F 〉

71

separates O from S(m − 1), and thus, every facet of S(n) belonging to A(m) is strictly

separated from O by H9.

We next deal with those facets F ∈ S(n) such that

F ∈ B(m) ∪ D(m)

= (R(m) \ R(m− 1)) ∪ D(m)

=4⋃i=1

(Ri(m) \Ri(m− 1)

)∪

3⋃j=1

Dj(m).

Let F ∈ R1(m) \ R1(m− 1). If F = [V1, Vα, Vm−1] for α = 3, 4, 5, then F appears as

the edge [Vα] = [xα, xα+1] in the quotient polytope S(m)/[V1, Vm−1], and it follows from

Figure 4.4 that H5 separates F from O. If F = [V2, Vα, Vm−1] for α = 4, 5, 6, then F

appears in S(m)/[V1, Vm−1] as the set [x3, Vβ] = [x3, xβ, xβ+1], and it again follows from

Figure 4.4 that F is separated from O by H5. If F = [Vα, Vβ, Vm−1] for α = 3, 4, 5 and

β = α + 2, α + 3, . . . ,m− 3, then x2 /∈ F , and it follows that H3 separates F from O as

evidenced by Figure 4.3.

Let F ∈ R2(m) \R2(m− 1). Then F = [V1,Wα, Vm−1] for α = 5, 6, . . . ,m− 4, and it

Figure 4.2: S(m)/[x1,Wi, xm] and the hyperplanes H1, H2 and H3.

72

follows from Figure 4.4 that F is separated from O by

H5, for α ≤ i− 1,

H2, for α = i and

H6, for α ≥ i+ 1.

In the case F ∈ R3(m) \ R3(m − 1), we have that F = [x2, Tα, Vm−1] for α =

6, 7, . . . ,m− 4, and F is separated from O by

H5, for α ≤ i− 1,

H4, for α = i and

H6, for α ≥ i+ 1,

as displayed in Figures 4.3 and 4.4.

Let F ∈ R4(m) \ R4(m − 1). Then F = [Wα, Vβ, Vm−1] for α = 5, 6, . . . ,m − 6, and

H3 separates F from O as evidenced in Figure 4.3.

Figure 4.3: S(m)/[Wi, Vm−1] and the hyperplanes H2, H3 and H4.

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Let F ∈ D1(m). If F = [x1, V2, Vα, xm] for α = 4, 5, 6, then F is separated from

O by H5, as displayed in Figure 4.4. If F = [x1, Vα, Vβ, xm] for α = 3, 4, 5 and β =

α + 2, α + 3, . . . ,m− 2, then F is separated from O by H3 as exhibited in Figure 4.2.

In the case that F ∈ D2(m), we have that F = [V1, Tα, xm] for α = 6, 7, . . . ,m − 3,

and F is separated from O by

H5, for α ≤ i− 1,

H1, for α = i and

H6, for α ≥ i+ 1,

as indicated in Figures 4.2 and 4.4.

Lastly, if F ∈ D3(m), then we have that F = [x1,Wα, Vβ, xm] for α = 5, 6, . . . ,m− 5

and β = α+ 3, α+ 4, . . . ,m− 2, and F is separated from O by H3 as displayed in Figure

4.2.

Finally, we handle the case F ∈ B(m+1)∪D(m+1) = (R(m+1)\R(m))∪D(m+1).

Let F ∈ R1(m+ 1) \R1(m). If F = [V1, Vα, Vm] for α = 3, 4, 5, then H7 separates F and

Figure 4.4: S(m)/[V1, Vm−1] and the hyperplanes H2, H5 and H6.

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O as evidenced in Figure 4.5. If F = [V2, Vα, Vm] for α = 4, 5, 6, then F is separated from

O by

H7, for α ≤ i− 1 and

H8, for α = i,

as displayed in Figures 4.5 and 4.6. If F = [Vα, Vβ, Vm] for α = 3, 4, 5 and β = α +

2, α+ 3, . . . ,m− 2, then F is separated from O by H3 as displayed in Figure 4.2. In the

case F ∈ (R2(m + 1) \ R2(m)) ∪ (R3(m + 1) \ R3(m)), we have that F = [V1,Wα, Vm]

for α = 5, 6, . . . ,m − 3 or F = [x2, Tα, Vm] for α = 6, 7, . . . ,m − 3. It then follows from

Figures 4.2, 4.4, 4.5 and 4.6 that F is separated from O by

H7, for α ≤ i− 2,

H8, for α = i− 1,

H1, for α = i and

H6, for α ≥ i+ 1.

Figure 4.5: S(m)/[V1, V3] and the hyperplane H7.

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Let F ∈ R4(m + 1) \ R4(m). Then F = [Wα, Vβ, Vm] for α = 5, 6, . . . ,m − 5 and

β = α+ 3, α+ 4, . . . ,m− 2, and it follows from Figure 4.2 that H3 separates F from O.

Let F ∈ D1(m+ 1). If F = [x1, V2, Vα, xm+1] for α = 4, 5, 6, then F is separated from O

by

H7, for α ≤ i− 1 and

H8, for α = i,

as displayed in Figures 4.5 and 4.6. If F = [x1, Vα, Vβ, xm+1] for α = 3, 4, 5 and β =

α + 3, α + 4, . . . ,m − 1, then it follows from Figure 4.2 that F is separated from O by

H3. In the case F ∈ D2(m+ 1), we have that F = [V1, Tα, xm+1] for α = 6, 7, . . . ,m− 2,

and F is separated from O by

H7, for α ≤ i− 2,

H8, for α = i− 1,

H1, for α = i and

H6, for α ≥ i+ 1,

Figure 4.6: S(m)/[x1,Wi−1, xm] and the hyperplane H8.

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as displayed in Figures 4.2, 4.4, 4.5 and 4.6. Lastly, let F ∈ D3(m + 1). Then F =

[x1,Wα, Vβ, xm+1] for α = 5, 6, . . . ,m − 4 and β = α + 3, α + 4, . . . ,m − 1, and H3

separates F from O, as depicted in Figure 4.2.

We have thus shown that if O ∈ int([F , xm]) with F = [V1, Ti, xm−1] ∈ D3(m − 1),

then any facet F ∈ F(S(m + 1)) ∪ D(m) is separated from O by one of H1, H2, . . . , H9.

Since A(m) ∪ B(m) ∪ B(m + 1) ⊆ F(S(m + 1)) ∪ D(m), Lemma 4.1.2 shows that any

facet F of S(n) is separated from O by one of H1, H2, . . . , H9.

We now remove the restriction that O is in the interior of [F , xm]. This relaxation

allows for O to lie on some of the separating hyperplanes Hi in (4.3).

Lemma 4.2.2. Let O ∈ [F , xm], with F = [V1, Ti, xm−1] ∈ D2(m − 1). Then any facet

of S(n) is separated from O by one of at most twenty-one hyperplanes.

Proof. Let Q = [F , xm]. By Lemma 4.2.1, we may assume that O ∈ bd(Q). Lemma

4.2.1 holds for any O ∈ int(Q), and hence, if a facet F of S(n) is separated from O by

some Hi, it follows that F is separated from Q by Hi, and thus, Hi ∩ int(Q) = ∅. From

this, we obtain that if Hi ∩Q 6= ∅ then Hi supports Q. Thus, we need to determine the

Hi that may contain O. If there are none then O is separated from any facet of S(n) by

one of H1, H2, . . . , H9.

We show that O may be contained in at most three of the Hi, and hence, O is

separated from any facet of S(n) by one of at most 9 − 3 + 3(5) = 21 hyperplanes by

Lemma 4.1.4. From Figures 4.2, 4.3 and 4.6, we note that each of H1, H2, H4 and H8

support S(m). Hence, O /∈ H1 ∪H2 ∪H4 ∪H8.

We note that Hi contains exactly the six vertices of S(m) which span Hi. Also, since

Q is a simplex, any subset of V(Q) determines a face of Q. It follows that if O ∈ Hi then

O ∈ [V(Q) ∩Hi] ∈ F(Q). We find that

Q ∩H3 = [x1, xi, xi+1, xi+2, xm−1, xm]

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Q ∩H5 = [x1, x2, xi, xi+1, xm−1, xm]

Q ∩H6 = [x1, x2, xi+1, xi+2, xm−1, xm]

Q ∩H7 = [x1, x2, xi, xm]

Q ∩H9 = [x1, x2, xi, xi+1, xi+2, xm−1].

We proceed by determining how many of H3, H5, H6, H7 and H9 may simultaneously

contain O. We compute first the intersection of Q with any two of H3, H5, H6, H7 and

H9, and conclude that O does not lie on the two hyperplanes if their intersection with Q

is a face of S(m). We obtain

Q ∩H3 ∩H5 = [x1, xi, xi+1, xm−1, xm],

Q ∩H3 ∩H7 = [x1, xi, xm],

Q ∩H5 ∩H6 = [x1, x2, xi+1, xm−1, xm],

Q ∩H5 ∩H9 = [x1, x2, xi, xi+1, xm−1],

Q ∩H6 ∩H9 = [x1, x2, xi+1, xi+2, xm−1],

Q ∩H3 ∩H6 = [x1, xi+1, xi+2, xm−1, xm],

Q ∩H3 ∩H9 = [x1, xi, xi+1, xi+2, xm−1],

Q ∩H5 ∩H7 = [x1, x2, xi, xm],

Q ∩H6 ∩H7 = [x1, x2, xm],

Q ∩H7 ∩H9 = [x1, x2, xi, xm].

Since S(m) is a neighbourly 6-polytope, [x1, xi, xm] and [x1, x2, xm] are faces of S(m), and

O /∈ H3 ∩H7 and O /∈ H6 ∩H7. Also, [x1, x2, xi+1, xm−1, xm] ⊂ [V1,Wi+1, Vm−1] ∈ R2(m)

shows that O /∈ H5 ∩H6, [x1, x2, xi, xm] ⊂ [V1, Ti, xm] ∈ D2(m) verifies that O /∈ H5 ∩H7

and [x1, x2, xi, xm] ⊂ [V1, Ti, xm] ∈ D2(m) proves that O /∈ H7 ∩H9.

Since the intersection of any four of H3, H5, H6, H7 and H9 is contained in one of

H3 ∩H7, H6 ∩H7, H5 ∩H6, H5 ∩H7 and H7 ∩H9, we obtain that O may not lie on four

of H1, H2, . . . , H9.

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4.3 The Case F ∈ D3(m− 1)

Let F = [x1,Wi, Vj, xm−1] ∈ D3(m − 1) for some 5 ≤ i ≤ m − 6 and O ∈ int(S(m)) \

int(S(m− 1)) such that

O ∈ [F , xm] = [x1,Wi, Vj, xm−1, xm]

⊂ [V1,Wi, Tj−1, Vm−1]

= [x1, x2, xi, xi+2, xj−1, xj, xj+1, xm−1, xm],

a subpolytope of S(m) with nine vertices. As in Section 4.2, we pick separating hyper-

planes spanned by the vertices of this subpolytope. We assume that O ∈ int([F , xm])

and consider two cases depending on the parity of j − i.

4.3.1 The Case j − i Even

Consider the hyperplanes

H1 = 〈V1,Wi, xj, xm〉, H2 = 〈x1,Wi, xj+1, Vm−1〉,

H3 = 〈x1,Wi, xj, Vm−1〉, H4 = 〈x1,Wi, Vj, xm〉,

H5 = 〈Wi, Vj, Vm−1〉, H6 = 〈V1, Vj, Vm−1〉,

H7 = 〈V1, Tj−1, xm〉, H8 =

〈x1, x3,Wj−1, Vm−1〉, for i = 5,

〈x1, xi,Wj−1, Vm−1〉, for i ≥ 6,

H9 = 〈x1, Tj−1, Vm−1〉, H10 =

〈x1, x3, Tj−1, xm〉, for i = 5,

〈x1, xi, Tj−1, xm〉, for i ≥ 6,

H11 =

〈x3, Tj−1, Vm−1〉, for i = 5,

〈xi, Tj−1, Vm−1〉, for i ≥ 6,H12 = 〈x1, xi+2, Vj, Vm−1〉,

H13 = 〈x1,Wi, Vj, xm−1〉.(4.4)

These thirteen hyperplanes appear in Figures 4.7 to 4.10 below. As they depend on

the values of i and j, the facets of S(n) that they separate from O vary with both i and

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j. For simplicity, we consider only the case i ≥ 6, the argument for i = 5 being similar.

Figures 4.8, 4.9 and 4.10 are shown for the case i and j even; the case i and j odd is similar

(see Figures 3.4, 3.9 and 3.17 with n = m and p = 9). For example, from Figure 4.9

we find that H8 strictly separates O from the vertices x3, x4, . . . , xi, xi+1, xi+3, . . . , xj−3

of S(m), and it follows from Figure 3.9 (with p = 9) that H8 also separates O from

x3, x4, . . . , xi, xi+1, xi+3, . . . , xj−3 for i, j odd.

Lemma 4.3.1. Let O ∈ int([F , xm]), F = [x1,Wi, Vj, xm−1] ∈ D3(m − 1) and j − i be

even. Then any facet of S(n) is separated from O by one of H1, H2, . . . , H13.

Proof. Analogous to the proof of Lemma 4.2.1, the hyperplane H13 = 〈F 〉 separates O

from S(m− 1), and thus, every facet of S(n) belonging to A(m) is separated from O by

H13.

We next handle the case F ∈ B(m) ∪ D(m). Let F ∈ R1(m) \ R1(m − 1). If

Figure 4.7: S(m)/[x1,Wi, xm] and hyperplanes H1, H2, H3, and H4.

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F = [V1, Vα, Vm−1] for α = 3, 4, 5, then it follows from Figure 4.7 that H3 separates

F from O. If F = [V2, Vα, Vm−1] for α = 4, 5, 6, then H3 again separates F from O as

displayed in Figure 4.7. If F = [Vα, Vβ, Vm−1] for α = 3, 4, 5 and β = α+2, α+3, . . . ,m−3,

then it follows from Figures 4.7, 4.9 and 4.10 that F is separated from O by

H3, for β ≤ j − 1,

H11, for β = j and

H8, for β ≥ j + 1.

In the event that F ∈ (R2(m) \ R2(m − 1)) ∪ (R3(m) \ R3(m − 1)), we have that F =

[V1,Wα, Vm−1] for α = 5, 6, . . . ,m − 4, or F = [x2, Tα, Vm−1] for α = 6, 7, . . . ,m − 4. In

either case, we see from Figures 4.7, 4.8 and 4.9 that F is separated from O by

H3, for α ≤ j − 2,

H9, for α = j − 1 and

H6, for α ≥ j.

Figure 4.8: S(m)/[V1, Vm−1], j − i even, and the hyperplane H6.

81

Let F ∈ R4(m) \ R4(m − 1). Then F = [Wα, Vβ, Vm−1] for α = 5, 6, . . . ,m − 6 and

β = α + 3, α + 4, . . . ,m− 3. If α ≤ i− 1 then F is separated from O by

H3, for β ≤ j − 1,

H11, for β = j and

H8, for β ≥ j + 1,

as displayed in Figures 4.7, 4.9 and 4.10. If α = i then F and O are separated by

H3, for β ≤ j − 1,

H5, for β = j and

H2, for β ≥ j + 1,

as evidenced in Figures 4.7 and 4.10. In the case i + 1 ≤ α ≤ j − 2, it follows from

Figure 4.9: S(m)/[x1,Wj−1, xm], j − i even, and the hyperplanes H7, H8, H9, and H10.

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Figures 4.7, 4.9 and 4.10 that F and O are separated by

H3, for β ≤ j − 1,

H12, for β = j and j − α even,

H11, for β = j and j − α odd,

H12, for β ≥ j + 1 and j − α even and

H8, for β ≥ j + 1 and j − α odd.

If α = j−1 then, as seen in Figure 4.9, H8 separates F and O. If α ≥ j then F and O are

separated by H6, as evidenced in Figure 4.8. Let F ∈ D1(m). If F = [x1, V2, Vα, xm] for

α = 4, 5, 6, it follows from Figure 4.7 that H3 separates F and O. If F = [x1, Vα, Vβ, xm]

for α = 3, 4, 5 and β = α + 2, α + 3, . . . ,m− 2, then F and O are separated by

H3, for β ≤ j − 1,

H10, for β = j and

H8, for β ≥ j + 1,

Figure 4.10: S(m)/[Vj, Vm−1], j − i even, and the hyperplanes H5, H11 and H12.

83

as demonstrated in Figures 4.7, 4.9 and 4.10. In the case F ∈ D2(m), we have that

F = [V1, Tα, xm] for α = 6, 7, . . . ,m− 3, and F and O are separated by

H3, for α ≤ j − 2,

H9, for α = j − 1 and

H6, for α ≥ j,

as indicated in Figures 4.7, 4.8 and 4.9. Finally, for F ∈ D3(m), we have that F =

[x1,Wα, Vβ, xm] for α = 6, 7, . . . ,m− 3 and β = α+ 3, α+ 4, . . . ,m− 2. If α ≤ i− 1 then

it follows from Figures 4.7, 4.9 and 4.10 that F and O are separated by

H3, for β ≤ j − 1,

H10, for β = j and

H8, for β ≥ j + 1.

If α = i then F and O are separated by

H3, for β ≤ j − 1,

H4, for β = j and

H2, for β ≥ j + 1,

as exhibited in Figure 4.7. In the event that i + 1 ≤ α ≤ j − 2, we find that F and O

are separated by

H3, for β ≤ j − 1,

H12, for β = j and j − α even,

H10, for β = j and j − α odd,

H12, for β ≥ j + 1 and j − α even and

H8, for β ≥ j + 1 and j − α odd,

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as displayed in Figures 4.7, 4.9 and 4.10. If α = j − 1 then it follows from Figure 4.9

that F and O are separated by H8. If α ≥ j then H6 separates F and O as evidenced in

Figure 4.8.

Lastly, we deal with the case F ∈ B(m+ 1)∪D(m+ 1). Let F ∈ R1(m+ 1) \R1(m).

If F = [V1, Vα, Vm] for α = 3, 4, 5, or F = [V2, Vα, Vm] for α = 4, 5, 6, then we find that F

is separated from O by H1, as displayed in Figure 4.7. If F = [Vα, Vβ, Vm] for α = 3, 4, 5

and β = α + 2, α + 3, . . . ,m − 1, it follows from Figures 4.7 and 4.9 that F and O are

separated by

H1, for β ≤ j − 1,

H10, for β = j and

H8, for β ≥ j + 1.

In the case F ∈ (R2(m+1)\R2(m))∪(R3(m+1)\R3(m)), we have that F = [V1,Wα, Vm]

for α = 5, 6, . . . ,m− 3, or F = [x2, Tα, Vm] for α = 6, 7, . . . ,m− 3, and in either case, F

and O are separated by

H1, for α ≤ j − 2,

H7, for α = j − 1 and

H6, for α ≥ j,

as displayed in Figures 4.7, 4.8 and 4.9. In the event that F ∈ R4(m + 1) \ R4(m), it

follows that F = [Wα, Vβ, Vm] for α = 5, 6, . . . ,m− 5 and β = α+ 3, α+ 4, . . . ,m− 2. If

α ≤ i− 1 then F and O are separated by

H1, for β ≤ j − 1,

H10, for β = j and

H8, for β ≥ j + 1,

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as evidenced in Figures 4.7 and 4.9. If α = i then it follows from Figure 4.7 that F and

O are separated by

H1, for β ≤ j − 1,

H4, for β = j and

H2, for β ≥ j + 1.

In the case i+ 1 ≤ α ≤ j − 2, F and O are separated by

H1, for β ≤ j − 1,

H12, for β = j and j − α even,

H10, for β = j and j − α odd,

H12, for β ≥ j + 1 and j − α even and

H8, for β ≥ j + 1 and j − α odd.

as evidenced by Figures 4.7, 4.9 and 4.10. If α = j − 1 then H8 separates F from O,

as displayed by Figure 4.9. If α ≥ j then it follows from Figure 4.8 that F and O are

separated by H6. Let F ∈ D1(m + 1). Then F = [x1, V2, Vα, xm+1] for α = 4, 5, 6, and

it follows from Figure 4.7 that H1 separates F from O. If F = [x1, Vα, Vβ, xm+1] for

α = 3, 4, 5 and β = α + 2, α + 3, . . . ,m− 1, we find that F and O are separated by

H1, for β ≤ j − 1,

H10, for β = j and

H8, for β ≥ j + 1.

as illustrated in Figures 4.7 and 4.9. In the case F ∈ D2(m + 1), we obtain that F =

[V1, Tα, xm+1] for α = 6, 7, . . . ,m− 2, and it follows from Figures 4.7, 4.8 and 4.9 that F

is separated from O by

H1, for α ≤ j − 2,

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H7, for α = j − 1 and

H6, for α ≥ j.

Lastly, for F ∈ D3(m + 1), we have that F = [x1,Wα, Vβ, xm+1] for α = 5, 6, . . . ,m − 4

and β = α + 3, α + 4, . . . ,m− 1. If α ≤ i− 1 then F is separated from O by

H1, for β ≤ j − 1,

H10, for β = j and

H8, for β ≥ j + 1,

as depicted in Figures 4.7 and 4.9. If α = i then F and O are separated by

H1, for β ≤ j − 1,

H4, for β = j and

H2, for β ≥ j + 1,

as supported by Figure 4.7. In the case i + 1 ≤ α ≤ j − 2, it follows from Figures 4.7,

4.9 and 4.10 that F and O are separated by

H1, for β ≤ j − 1,

H12, for β = j and j − α even,

H10, for β = j and j − α odd,

H12, for β ≥ j + 1 and j − α even and

H8, for β ≥ j + 1 and j − α odd.

If α = j − 1 then H8 separates F and O, as demonstrated in Figure 4.9. It follows from

Figure 4.8 that F and O are separated by H6 in the case α ≥ j.

Thus, we have shown that if O ∈ int([F , xm]) with F = [x1,Wi, Vj, xm−1] ∈ D3(m−1)

and j − i even, then any facet F ∈ F(S(m + 1)) ∪ D(m) is separated from O by one of

87

H1, H2, . . . , H13. Since A(m) ∪ B(m) ∪ B(m + 1) ⊆ F(S(m + 1)) ∪ D(m), Lemma 4.1.2

shows that any facet F of S(n) is separated from O by one of H1, H2, . . . , H13.

4.3.2 The Case j − i Odd

We handle first the case j 6= i + 3. Recall the hyperplanes H1, H2, . . . , H13 defined in

(4.4) and consider the following collection of hyperplanes:

H1 = 〈V1,Wi, xj, xm〉, H2 = 〈x1,Wi, xj+1, Vm−1〉,

H3 = 〈x1,Wi, xj, Vm−1〉, H4 = 〈x1,Wi, Vj, xm〉,

H5 = 〈Wi, Vj, Vm−1〉, H6 = 〈V1, Vj, Vm−1〉,

H7 = 〈V1, Tj−1, xm〉, H ′8 = 〈x1, xi+2,Wj−1, Vm−1〉,

H9 = 〈x1, Tj−1, Vm−1〉, H ′10 = 〈x1, xi+2, Tj−1, xm〉,

H ′11 = 〈xi+2, Tj−1, Vm−1〉, H ′12 =

〈x1, x3, Vj, Vm−1〉, for i = 5,

〈x1, xi, Vj, Vm−1〉, for i ≥ 6,

H13 = 〈x1,Wi, Vj, xm−1〉.

(4.5)

We note first that quotient polytope S(m)/[x1,Wi, xm] represented in Figure 4.7 does

not depend on the parity of j − i and thus we refer to Figure 4.7 regarding statemets

involving the hyperplanes H1, H2, H3 and H4 in the proof of Lemma 4.3.2. Second, the

hyperplanes H5, H6, H7 and H9 appearing in Figures 4.11, 4.12 and 4.13 separate the

same sets of vertices as in the case j − i even (compare to Figures 4.8, 4.9 and 4.10).

Finally, Figures 4.11, 4.12 and 4.13 show only the case i odd and j even, the case i even

and j odd being similar (see Figures 3.4, 3.8 and 3.17).

Lemma 4.3.2. Let O ∈ int([F , xm]), F = [x1,Wi, Vj, xm−1] ∈ D3(m − 1), j − i be odd

and j 6= i+ 3. Then any facet of S(n) is separated from O by one of

H1, H2, H3, H4, H5, H6, H7, H′8, H9, H

′10, H

′11, H

′12, H13.

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Proof. We note that in most instances, the proof follows from the proof of Lemma 4.3.1

with Hk replaced by H ′k for k = 8, 10, 11, 12, and we mention here only the cases for

which these four substitutions are not sufficient. If

F = [Vα, Vβ, Vm−1] ∈ R1(m) \R1(m− 1)

for α = 3, 4, 5 and β = α + 2, α + 3, . . . ,m− 3, or

F = [Wα, Vβ, Vm−1] ∈ R4(m) \R4(m− 1)

for α = 5, 6, . . . , i− 1 and β = α + 3, α + 4, . . . ,m− 3, or

F = [x1, Vα, Vβ, xm] ∈ D1(m)

for α = 3, 4, 5 and β = α + 2, α + 3, . . . ,m− 2, or

F = [x1,Wα, Vβ, xm] ∈ D3(m)

for α = 5, 6, . . . , i− 1 and β = α + 3, α + 4, . . . ,m− 2,

then F and O are separated by

H3, for β ≤ j − 1 and

Figure 4.11: S(m)/[V1, Vm−1], j − i odd, and the hyperplane H6.

89

H ′12, for β ≥ j,

as evidenced by Figures 4.7 and 4.13. If

F = [Vα, Vβ, Vm] ∈ R1(m+ 1) \R1(m)

for α = 3, 4, 5 and β = α + 1, α + 3, . . . ,m− 2, or

F = [Wα, Vβ, Vm] ∈ R4(m+ 1) \R4(m)

for α = 5, 6, . . . , i− 1 and β = α + 3, α + 4, . . . ,m− 2, or

F = [x1, Vα, Vβ, xm+1] ∈ D1(m+ 1)

for α = 3, 4, 5 and β = α + 2, α + 3, . . . ,m− 1, or

F = [x1,Wα, Vβ, xm+1] ∈ D3(m+ 1)

for α = 5, 6, . . . , i− 1 and β = α + 3, α + 4, . . . ,m− 1,

Figure 4.12: S(m)/[x1,Wj−1, xm], j − i odd, and the hyperplanes H7, H′8, H9 and H ′10.

90

then it follows from Figures 4.7 and 4.13 that F and O are separated by

H1, for β ≤ j − 1 and

H ′12, for β ≥ j.

Thus, if F = [x1,Wi, Vj, xm−1] ∈ D3(m − 1), j − i odd and j 6= i + 3, then any facet of

S(m+ 1) ∪ D(m) is separated from O by one of

H1, H2, H3, H4, H5, H6, H7, H′8, H9, H

′10, H

′11, H

′12 and H13.

Since A(m) ∪B(m) ∪B(m+ 1) ⊆ F(S(m+ 1)) ∪D(m), the result follows from Lemma

4.1.2.

We now address the case j = i+ 3. Recall the sets defined in (4.5). For j = i+ 3, we

find that

H ′8 = 〈x1, xi+2, Ti+2, xm〉 = 〈x1, xi+2, xi+2, xi+3, xi+4, xm〉 = 〈x1, xi+2, xi+3, xi+4, xm〉

Figure 4.13: S(m)/[Vj, Vm−1], j − i odd, and the hyperplanes H5, H′11 and H ′12.

91

is not a hyperplane of R6. Similarly, neither H ′10 nor H ′11 is a hyperplane of R6. We note

that

H1 = 〈V1,Wi, xi+3, xm〉 H2 = 〈x1,Wi, xi+4, Vm−1〉,

H3 = 〈x1,Wi, xi+3, Vm−1〉, H4 = 〈x1,Wi, Vi+3, xm〉,

H5 = 〈Wi, Vi+3, Vm−1〉, H6 = 〈V1, Vi+3, Vm−1〉,

H7 = 〈V1, Ti+2, xm〉, H9 = 〈x1, Ti+2, Vm−1〉

H ′12 = 〈x1, xi, Vi+3, Vm−1〉, H13 = 〈x1,Wi, Vi+3, xm−1〉,

(4.6)

are hyperplanes.

Lemma 4.3.3. Let O ∈ int([F , xm]), with F = [x1,Wi, Vi+3, xm−1] ∈ D3(m − 1). Then

any facet of S(n) is separated from O by one of

H1, H2, H3, H4, H5, H6, H7, H9, H′12, H13.

Proof. The proof follows that of Lemma 4.3.1 with Hk replaced with H ′k for k = 8, 10,

11, 12 with the exception of those instances addressed in Lemma 4.3.2. It follows from

Lemmata 4.3.1 and 4.3.2 that if a facet F is separated from O by one of H ′8, H′10, H

′11

then

F ∈(R4(m) \R4(m− 1)

)∪D3(m) ∪

(R4(m+ 1) \R4(m)

)∪D3(m+ 1).

Let F ∈ R4(m) \ R4(m − 1). Then F = [Wα, Vβ, Vm−1] for α = 5, 6, . . . ,m − 6 and

β = α + 3, α + 4, . . . ,m− 3. If α ≤ i− 1 then F is separated from O by

H3, for β ≤ i+ 2 and

H ′12, for β ≥ i+ 3

as evidenced in Figures 4.7 and 4.13. If α = i then it again follows from Figures 4.7 and

4.13 that F and O are separated by

H5, for β = i+ 3 and

92

H2, for β ≥ i+ 4.

If α = i+ 1 then F and O are separated by H ′12 as indicated in Figure 4.13. In the case

α = i + 2, it follows from Figure 4.7 that F and O are separated by H2. If α ≥ i + 3

then F and O are separated by H6, as depicted in Figure 4.11.

The remaining cases F ∈ D3(m), F ∈ R4(m + 1) \ R4(m) and F ∈ D3(m + 1) are

handled analogously.

Finally, we remove the restriction that O lies in the interior of [F , xm], allowing for

O to lie on some of the separating hyperplanes.

Lemma 4.3.4. Let O ∈ [F , xm], F = [x1,Wi, Vj, xm−1] ∈ D3(m − 1), 5 ≤ i ≤ m − 6

and i + 3 ≤ j ≤ m − 3. Then any facet of S(n) is separated from O by one of at most

twenty-nine hyperplanes.

We omit the proof that O does not lie on five separating hyperplanes as the details

are similar to the proof of Lemma 4.2.2, and thus, any facet of S(n) may be separated

from O by one of at most 13− 4 + 4(5) = 29 hyperplanes.

4.4 The Case F ∈ D1(m− 1)

Let F ∈ D1(m − 1). Then either F = [x1, V2, Vi, xm−1] for some i = 4, 5, 6, or F =

[x1, Vi, Vj, xm−1] for some i = 3, 4, 5. We handle the latter case first. We assume that

O ∈ int(S(m)) \ int(S(m− 1)) such that

O ∈ [F , xm] = [x1, Vi, Vj, xm−1, xm]

⊂ [V1, Vi, Tj−1, Vm−1]

= [x1, x2, xi, xi+1, xj−1, xj, xj+1, xm−1, xm],

93

a subpolytope of S(m) with nine vertices. As in Section 4.2, we pick separating hyper-

planes spanned by the vertices of this subpolytope. We note that this section closely

follows Section 4.3. We assume that O ∈ int([F , xm]) and consider two cases depending

on the parity of j.

4.4.1 The Case j Even

We handle first the case (i, j) 6= (3, 6). Consider the collection of hyperplanes:

H1 = 〈V1, Vi, xj, xm〉, H2 = 〈x1, Vi, xj+1, Vm−1〉,

H3 = 〈x1, Vi, xj, Vm−1〉, H4 = 〈x1, Vi, Vj, xm〉,

H5 = 〈Vi, Vj, Vm−1〉, H6 = 〈V1, Vj, Vm−1〉,

H7 = 〈V1,Wj−1, xj, xm〉, H8 = 〈x1, xi,Wj−1, Vm−1〉,

H9 = 〈x1,Wj−1, xj, Vm−1〉, H10 = 〈x1, xi,Wj−1, xj, xm〉,

H11 = 〈xi,Wj−1, xj, Vm−1〉, H12 = 〈x1, xi+1, Vj, Vm−1〉,

H13 = 〈x1, Vi, Vj, xm−1〉.

(4.7)

Lemma 4.4.1. Let O ∈ int([F , xm]), F = [x1, Vi, Vj, xm−1] ∈ D1(m − 1), j be even and

(i, j) 6= (3, 6). Then any facet of S(n) is separated from O by one of H1, H2, . . . , H13.

Proof. As in the proofs of Section 4.3, H13 separates all of those facets of S(n) that

belong to A(m), so we turn our attention to the case F ∈ B(m) ∪ D(m). Let F ∈

R1(m) \R1(m− 1). If F = [V1, Vα, Vm−1] for α = 3, 4, 5, then F and O are separated by

H3 as evidenced in Figure 4.14. If F = [V2, Vα, Vm−1] for α = 4, 5, 6, then (since α ≤ j

with equality if, and only if, j = 6) it follows from Figures 4.14 and 4.16 that F and O

are separated by

H3, for α ≤ j − 1 and

H9, for α = j.

94

If F = [Vα, Vβ, Vm−1] for α = 3, 4, 5 and β = α + 2, α + 3, . . . ,m − 3, then we have that

α ≤ j − 1. If α ≤ i − 1, it follows from Figures 4.14, 4.16 and 4.17 that F and O are

separated by

H3, for β ≤ j − 1,

H11, for β = j and

H8, for β ≥ j + 1.

If α = i, then as evidenced in Figures 4.14 and 4.17, F and O are separated by

H3, for β ≤ j − 1,

H5, for β = j and

H2, for β ≥ j + 1.

In the case that i+ 1 ≤ α ≤ j − 1, F and O are separated by

H3, for β ≤ j − 1 and

Figure 4.14: S(m)/[x1, Vi, xm] and the hyperplanes H1, H2, H3 and H4.

95

H12, for β ≥ j,

as displayed in Figures 4.14 and 4.17. Let F ∈ (R2(m)\R2(m−1))∪(R3(m)\R3(m−1)).

Then either F = [V1,Wα, Vm−1] for α = 5, 6, . . . ,m − 4 or F = [x2, Tα, Vm−1] for α =

6, 7, . . . ,m− 4. In either case, F and O are separated by

H3, for α ≤ j − 2,

H9, for α = j − 1 and

H6, for α ≥ j,

as depicted in Figures 4.14, 4.15 and 4.16. With F ∈ R4(m) \ R4(m − 1), we have that

F = [Wα, Vβ, Vm−1] for α = 5, 6, . . . ,m−6 and β = α+3, α+4, . . . ,m−3, and we obtain

that α ≥ i. If α ≤ j − 2 then F and O are separated by

H3, for β ≤ j − 1,

H12, for β = j and j − α even,

Figure 4.15: S(m)/[V1, Vm−1] and the hyperplane H6.

96

H11, for β = j and j − α odd,

H12, for β ≥ j + 1 and j − α even and

H8, for β ≥ j + 1 and j − α odd,

which is indicated in Figures 4.14, 4.16 and 4.17. If α = j−1, then it follows from Figure

4.16 that H8 separates F and O. If α ≥ j then H6 separates F and O, which can be seen

in Figure 4.15. Let F ∈ D1(m). If F = [x1, V2, Vα, xm] for α = 4, 5, 6, then (since α ≤ j

with equality if, and only if, j = 6) we find that F is separated from O by

H3, for α ≤ j − 1 and

H9, for α = j,

as exhibited in Figures 4.14 and 4.16. If F = [x1, Vα, Vβ, xm] for α = 3, 4, 5, then α ≤ j−1.

If α ≤ i− 1, then it follows from Figures 4.14, 4.16 and 4.17 that F and O are separated

Figure 4.16: S(m)/[x1,Wj−1, xm], j even, and the hyperplanes H7, H8, H9 and H10.

97

by

H3, for β ≤ j − 1,

H10, for β = j and

H8, for β ≥ j + 1.

If α = i, then F and O are separated by

H3, for β ≤ j − 1,

H4, for β = j and

H2, for β ≥ j + 1.

as demonstrated in Figure 4.14. If i + 1 ≤ α ≤ j − 1, then it follows from Figures 4.14

and 4.17 that F and O are separated by

H3, for β ≤ j − 1 and

H12, for β ≥ j.

Figure 4.17: S(m)/[Vj, Vm−1], j even, and the hyperplanes H5, H11 and H12.

98

For F ∈ D2(m), we find that F = [V1, Tα, xm] for α = 6, 7, . . . ,m − 3, from which it

follows that α > i. As evidenced in Figures 4.14, 4.15 and 4.16, F and O are separated

by

H3, for α ≤ j − 2,

H9, for α = j − 1 and

H6, for α ≥ j.

Lastly, we let F ∈ D3(m). Then F = [x1,Wα, Vβ, xm] for α = 5, 6, . . . ,m − 5 and

β = α+3, α+4, . . . ,m−2. We note that α ≥ i. If α ≤ j−2 then F and O are separated

by

H3, for β ≤ j − 1,

H12, for β = j and j − α even,

H10, for β = j and j − α odd,

H12, for β ≥ j + 1 and j − α even and

H8, for β ≥ j + 1 and j − α odd,

as displayed in Figures 4.14, 4.16 and 4.17. If α = j − 1 then Figure 4.16 exhibits that

H8 separates F and O. If α ≥ j then H6 separates F and O as depicted in Figure 4.15.

Lastly, we handle the case F ∈ B(m+ 1)∪D(m+ 1). Let F ∈ R1(m+ 1) \R1(m). If

F = [V1, Vα, Vm] for α = 3, 4, 5, then α ≤ j − 1, and H1 separates F and O as depicted

in Figure 4.14. If F = [V2, Vα, Vm] for α = 4, 5, 6, then α ≤ j and it follows from Figures

4.14 and 4.16 that F and O are separated by

H1, for α ≤ j − 1 and

H7, for α = j.

99

If F = [Vα, Vβ, Vm] for α = 3, 4, 5, and β = α + 2, α + 3, . . . ,m − 2, then α ≤ j − 1. If

α ≤ i− 1, then F and O are separated by

H1, for β ≤ j − 1,

H10, for β = j and

H8, for β ≥ j + 1,

which is demonstrated in Figures 4.14 and 4.16. If α = i then it follows from Figure 4.14

that F and O are separated by

H1, for β ≤ j − 1,

H4, for β = j and

H2, for β ≥ j + 1.

If i+ 1 ≤ α ≤ j − 1 then F and O are separated by

H1, for β ≤ j − 1 and

H12, for β ≥ j,

as evidenced in Figures 4.14 and 4.17. Let F ∈ (R2(m+1)\R2(m))∪(R3(m+1)\R3(m)).

Then either F = [V1,Wα, Vm] for α = 5, 6, . . . ,m − 3, or F = [x2, Tα, Vm] for α =

6, 7, . . . ,m− 3. In either case, α ≥ i and as evidenced in Figures 4.14, 4.15 and 4.16, F

and O are separated by

H1, for α ≥ j − 2,

H7, for α = j − 1 and

H6, for α ≥ j.

With F ∈ R4(m+ 1) \R4(m), we find that F = [Wα, Vβ, Vm] for α = 5, 6, . . . ,m− 5 and

β = α + 3, α + 4, . . . ,m − 2, and it follows that α ≥ i. If α ≤ j − 2 then F and O are

100

separated by

H1, for β ≤ j − 1,

H12, for β = j and j − α even,

H10, for β = j and j − α odd,

H12, for β ≥ j + 1 and j − α even and

H8, for β ≥ j + 1 and j − α odd,

as displayed in Figures 4.14, 4.16 and 4.17. If α = j − 1 then F and O are separated

by H8 as depicted in Figure 4.16. If α ≥ j, then it follows from Figure 4.15 that H6

separates F and O. Let F ∈ D1(m + 1). If F = [x1, V2, Vα, xm+1] for α = 4, 5, 6, then

(since α ≤ j with equality if, and only if, j = 6), F and O are separated by

H1, for α ≤ j − 1 and

H7, for α = j,

as evidenced in Figures 4.14 and 4.16. If F = [x1, Vα, Vβ, xm+1] for α = 3, 4, 5 and

β = α + 2, α + 3, . . . ,m − 1, then we have that α ≤ j − 1. If α ≤ i − 1 then it follows

from Figures 4.14 and 4.16 that F and O are separated by

H1, for β ≤ j − 1,

H10, for β = j and

H8, for β ≥ j + 1.

If α = i then, as displayed in Figure 4.14, F and O are separated by

H1, for β ≤ j − 1,

H4, for β = j and

H2, for β ≥ j + 1.

101

If i+ 1 ≤ α ≤ j − 1 then F and O are separated by

H1, for β ≤ j − 1 and

H12, for β ≥ j,

as indicated in Figures 4.14 and 4.17. Let F ∈ D2(m + 1). Then F = [V1, Tα, xm+1] for

α = i+ 1, i+ 2, . . . ,m− 2. It follows from Figures 4.14, 4.15 and 4.16 that F and O are

separated by

H1, for α ≥ j − 2,

H7, for α = j − 1 and

H6, for α ≥ j.

Finally, let F ∈ D3(m + 1). Then F = [x1,Wα, Vβ, xm+1] for α = 5, 6, . . . ,m − 4 and

β = α + 3, α + 4, . . . ,m− 1, from which it follows that α ≥ i. If α ≤ j − 2, then F and

O are separated by

H1, for β ≤ j − 1,

H12, for β = j and j − α even,

H10, for β = j and j − α odd,

H12, for β ≥ j + 1 and j − α even and

H8, for β ≥ j + 1 and j − α odd,

as depicted in Figures 4.14, 4.16 and 4.17. If α = j − 1 then F and O are separated by

H8 as exhibited in Figure 4.16. If α ≥ j then it follows from Figure 4.15 that F and O

are separated by H6.

Hence, any F ∈ F(S(m+ 1))∪D(m) is separated from O by one of H1, H2, . . . , H13.

Since A(m) ∪ B(m) ∪ B(m + 1) ⊆ F(S(m + 1)) ∪ D(m), Lemma 4.1.2 yields that any

facet of S(n) is separated from O by one of H1, H2, . . . , H13.

102

We now deal with the case i = 3 and j = 6. We note that from H1, H2, . . . , H13, only

H8, H10 and H11 are not hyperplanes, and that the remaining hyperplanes are sufficient

to achieve separation. The proof is similar to that of Lemma 4.4.1.

4.4.2 The Case j Odd

We assume first that (i, j) /∈ {(3, 5), (5, 7)}. Recall the hyperplanes H1, H2, . . . , H13

defined in (4.7) and consider the following collection of hyperplanes:

H1 = 〈V1, Vi, xj, xm〉, H2 = 〈x1, Vi, xj+1, Vm−1〉,

H3 = 〈x1, Vi, xj, Vm−1〉, H4 = 〈x1, Vi, Vj, xm〉,

H5 = 〈Vi, Vj, Vm−1〉, H6 = 〈V1, Vj, Vm−1〉,

H7 = 〈V1,Wj−1, xj, xm〉, H ′8 = 〈x1, xi+1,Wj−1, Vm−1〉,

H9 = 〈x1,Wj−1, xj, Vm−1〉, H ′10 = 〈x1, xi+1,Wj−1, xj, xm〉,

H ′11 = 〈xi+1,Wj−1, xj, Vm−1〉, H ′12 = 〈x1, xi, Vj, Vm−1〉,

H13 = 〈x1, Vi, Vj, xm−1〉.

(4.8)

Lemma 4.4.2. Let O ∈ int([F , xm]), F = [x1, Vi, Vj, xm−1] ∈ D1(m − 1), j be odd and

(i, j) /∈ {(3, 5), (5, 7)}. Then any facet of S(n) is separated from O by one of

H1, H2, H3, H4, H5, H6, H7, H′8, H9, H

′10, H

′11, H

′12, H13.

Proof. We note that in most instances, the proof follows from the proof of Lemma 4.4.1

with Hk replaced by H ′k for k = 8, 10, 11, 12, and we mention here only the cases for

which these four substitutions are not sufficient. Let

F = [Vα, Vβ, Vm−1] ∈ R1(m) \R1(m− 1)

for α = 3, 4, 5, and β = α + 2, α + 3, . . . ,m− 3, or

103

F = [x1, Vα, Vβ, xm] ∈ D1(m)

for α = 3, 4, 5, and β = α + 2, α + 3, . . . ,m− 2.

If α ≤ i− 1 then F and O are separated by

H3, for β ≤ j − 1 and

H ′12, for β ≥ j,

as evidenced in Figures 4.14 and 4.19. If i + 1 ≤ α ≤ j − 1 then it follows from Figures

4.14, 4.18 and 4.19 that F and O are separated by

H3, for β ≤ j − 1,

H ′11, for β = j and

H ′8, for β ≥ j + 1.

Figure 4.18: S(m)/[x1,Wj−1, xm], j odd, and the hyperplanes H7, H′8, H9 and H ′10.

104

Let

F = [Vα, Vβ, Vm] ∈ R1(m+ 1) \R1(m)

for α = 3, 4, 5, and β = α + 2, α + 3, . . . ,m− 2, or

F = [x1, Vα, Vβ, xm+1] ∈ D1(m+ 1)

for α = 3, 4, 5, and β = α + 2, α + 3, . . . ,m− 1.

If α ≤ i− 1 then it follows from Figures 4.14 and 4.19 that F and O are separated by

H1, for β ≤ j − 1 and

H ′12, for β ≥ j.

If i+ 1 ≤ α ≤ j − 1 then F and O are separated by

H1, for β ≤ j − 1,

H ′10, for β = j and

H ′8, for β ≥ j + 1,

Figure 4.19: S(m)/[Vj, Vm−1], j odd, and the hyperplanes H5, H′11 and H ′12.

105

as depicted in Figures 4.14 and 4.18.

Let i = 3 and j = 5. We argue as in the proof of Lemma 4.4.2 that O is separated

from any facet of S(n) by one of H1, H2, H3, H4, H5, H6, H7, H9, H′12 with

H1, . . . , H5 as defined in (4.8),

H6 = 〈V1, x3, x6, Vm−1〉,

H7 = 〈V1, x3, V5, xm〉,

H9 = 〈x1, x3, V5, Vm−1〉 and

H ′12 = 〈x1, x4, V5, Vm−1〉.

The hyperplanes H7 and H9 appear in the quotient polytope S(m)/[x1, V5, xm] instead of

S(m)/[x1,Wj−1, xm]. We note that [V5, Vm−1] is a universal face of S(m), and hence, the

quotient polytope S(m)/[V5, Vm−1] used to represent H ′12 appears in the form of Figure

3.2.

In the case i = 5 and j = 7, we find that only H ′8, H′10 and H ′11 fail to be hyperplanes

from (4.8). We let H ′′10 = 〈x1, x5,W6, x7, xm〉 and find that the hyperplanes H1, H2, H3,

H4, H5, H6, H7, H9, H′′10 and H ′12 separate O from any facet of S(n).

4.4.3 The Case F = [x1, V2, Vi, xm−1], i = 4, 5, 6

We briefly address the case O ∈ int([F , xm]) with F = [x1, V2, Vi, xm−1] for some i =

4, 5, 6. We give only the necessary hyperplanes for each i = 4, 5, 6 and do not supply

proofs. For i = 4, the hyperplanes

H1 = 〈x1, V2, x4, Vm−1〉, H2 = 〈x1, V2, x5, Vm−1〉,

H3 = 〈x1, V2, V4, xm〉, H4 = 〈x1, x3, V4, Vm−1〉,

H5 = 〈V1, V4, Vm−1〉, H6 = 〈V2, V4, Vm−1〉,

H7 = 〈x1, V2, V4, xm−1〉,

(4.9)

106

are sufficient to acheive separation while for i = 5, the hyperplanes

H1 = 〈x1, V2, V4, xm〉, H2 = 〈x1, V2, x6, Vm−1〉,

H3 = 〈x1, V2, x5, Vm−1〉, H4 = 〈x1, V2, V5, xm〉,

H5 = 〈V2, V5, Vm−1〉, H6 = 〈x1, x3, V5, Vm−1〉,

H7 = 〈V1, V5, Vm−1〉, H8 = 〈x1, V2, V5, xm−1〉,

(4.10)

do the task. Finally, if i = 6 then the hyperplanes

H1 = 〈V1, V3, x6, xm〉, H2 = 〈x1, V2, x7, Vm−1〉,

H3 = 〈x1, V2, x6, Vm−1〉, H4 = 〈x1, V2, V6, xm〉,

H5 = 〈x1, x3, V6, Vm−1〉, H6 = 〈V2, V6, Vm−1〉,

H7 = 〈V1, V6, Vm−1〉, H8 = 〈x1, V2, V6, xm−1〉,

(4.11)

separate any facet of S(n) from O.

Lastly, we remove again the restriction that O lies in the interior of [F , xm].

Lemma 4.4.3. Let O ∈ [F , xm] and F ∈ D1(m−1). Then any facet of S(n) is separated

from O by one of at most twenty-nine hyperplanes.

4.5 The Case m ∈ {9, 10}

As we remarked earlier, our verification of the Separation Conjecture for S(n) is for

11 ≤ m ≤ n. The case m = 8 is verified in Theorem 4.1.3. We recall that

R(9) = R1(9) ∪R2(9),

R(10) = R1(10) ∪R2(10) ∪R3(10), and

R(n) = R1(n) ∪R2(n) ∪R3(n) ∪R4(n)

for n ≥ 11, and that

D(9) = D1(9) ∪D2(9) and

107

D(n) = D1(n) ∪D2(n) ∪D3(n)

for n ≥ 10. Let

R(8) = R1(8) = {[Vi, Vj, Vk] | 1 ≤ i ≤ 3, i+ 2 ≤ j ≤ 5, j + 2 ≤ k ≤ 7} and

D(8) = D1(8) = {[x1, Vi, Vj, x8] | 2 ≤ i ≤ 4, i+ 2 ≤ j ≤ 6},

and R2(8) = R3(8) = R4(8) = D2(8) = D3(8) = ∅. We note that if O ∈ int(S(m)) \

int(S(m− 1)) for m = 9, 10, then O ∈ [F , xm] with

F ∈

D1(8), for m = 9, and

D1(9) ∪D2(9), for m = 10.

(4.12)

Moreover, we see that for m = 9,

F(S(10)) ∪ D(9) = F(S(9)) ∪(F(S(10)) \ F(S(9))

)= A(9) ∪B(9) ∪ D(9) ∪B(10) ∪ D(10)

= A(9) ∪(R(9) \ R(8)

)∪ D(9) ∪

(R(10) \ R(9)

)∪ D(10)

= A(9) ∪

(2⋃

k=1

Rk(9) \Rk(8)

)∪ D(9) ∪

(3⋃

k=1

Rk(10) \Rk(9)

)∪ D(10),

(4.13)

and similarly for m = 10,

F(S(11)) ∪ D(10) =

A(10) ∪

(3⋃

k=1

Rk(10) \Rk(9)

)∪ D(10) ∪

(4⋃

k=1

Rk(11) \Rk(10)

)∪ D(11).

(4.14)

Together, (4.12), (4.13) and (4.14) show that the case m = 9 follows entirely from Section

4.4, while the case m = 10 follows from Sections 4.2 and 4.4.

108

In summary, Main Theorem 1 now follows from the preceding discussion and Lemmata

4.2.2, 4.3.4 and 4.4.3.

As a result of Lemma 4.1.2, we found that for O ∈ int(S(m)) \ int(S(m− 1)), it was

sufficient to show that any facet of S(n) belonging to A(m)∪B(m)∪B(m+1) is separated

from O by some H ∈ H in order to verify that H is a collection of separating hyperplanes

for O. However, to avoid additional cases for m = n− 1 and m = n, we showed instead

that any F ∈ F(S(m + 1)) ∪ D(m) is separated from O by some H ∈ H. We note only

that the hyperplanes used to separate F ∈ D(m) ∪ D(m + 1) are also used to separate

facets of S(n) contained in A(m) ∪B(m) ∪B(m+ 1). Hence, no additional hyperplanes

were introduced when verifying this stronger condition for m ≤ n − 1. For m = n, we

disregard any hyperplanes introduced to separate only those facets of B(n+1)∪D(n+1).

109

Chapter 5

X-Raying Mirror Symmetric Convex Bodies in R3

In this Chapter we turn out attention to the problems of Illumination and X-Raying. We

write vectors in boldface, and for v = (v1, v2, . . . , vd) ∈ Rd, the norm of v is given by

||v|| =√v21 + v22 + · · ·+ v2d. The d-dimensional open ball of radius ε centered at x ∈ Rd

is denoted by Dd(x, ε) = {y ∈ Rd | ||x− y|| < ε}. A non-zero vector is called a direction.

The open ray emanating from a point x ∈ Rd with direction v is denoted by r xv and the

line through x with direction v is denoted as `xv . The notation `v will be used to denote

an arbitrary line of Rd with direction v. We recall that for sets A,B ⊂ Rd, [A,B] denotes

the convex hull, and for x, y ∈ Rd, [x, y] = [{x}, {y}] is the line segment determined by

x and y. We define (x, y) to be the relative interior of [x, y].

For a d-dimensional convex body K and F ⊂ K, we call F a face of K if for every

x, y ∈ K, (x, y) ∩ F 6= ∅ implies that [x, y] ⊂ F . A face F of K is exposed if it is the

intersection of K and a supporting hyperplane of K. We remark that if K is a convex

polytope then every face is an exposed face.

5.1 Illumination and X-raying

In this section, K denotes a d-dimensional convex body. A direction v is said to illuminate

a point p ∈ bd(K) if r pv intersects the interior of K. We say that K is illuminated by

the family of directions v1,v2, . . . ,vn if every p ∈ bd(K) is illuminated by at least one of

v1,v2, . . . ,vn. This definition of illumination is due to Boltyanski [6] and is equivalent to

Hadwiger’s defininition of illumination which uses “point sources” rather than directions

(see [16]). The Illumination Problem is to find the least number of directions, denoted

110

by I(K), required to illuminate K. We recall that the Illumination Conjecture states

that I(K) ≤ 2d.

A variation of illumination is the following: a line `v is said to X-ray a point

p ∈ bd(K) if ` pv has non-empty intersection with the interior of K. A collection of

lines `v1 , `v2 , . . . , `vn is said to X-ray K if every p ∈ bd(K) is X-rayed along one of

`v1 , `v2 , . . . , `vn . The X-ray Problem is to find the least number of lines, denoted by

X(K), necessary to X-ray K. Formulated by Bezdek and Zamfirescu [4] the X-ray Con-

jecture states X(K) ≤ 3 · 2d−2.

Note that for x ∈ Rd and a direction v, we have `xv = r xv ∪ {x} ∪ r x−v. From this, we

may view X-raying a convex body K as illuminating K by pairs of opposite directions.

It follows that we can use X(K) to obtain bounds on I(K):

Lemma 5.1.1. For any convex body K we have

X(K) ≤ I(K) ≤ 2X(K). (5.1)

Proof. Assume that I(K) = n. Then there are n directions, v1,v2, . . . ,vn, such that for

any x ∈ bd(K) there is a 1 ≤ i ≤ n with r xvi∩ int(K) 6= ∅. From this it follows that

`xvi∩ int(K) 6= ∅, and thus x is X-rayed along `vi

and X(K) ≤ I(K).

Now assume that X(K) = m. Then there are m directions, w1,w2, . . . ,wm, such

that for any x ∈ bd(K) there is a 1 ≤ j ≤ m with `xwj∩ int(K) 6= ∅. Thus, either

r xwj∩ int(K) 6= ∅ or r x−wj

∩ int(K) 6= ∅ and thus x is illuminated by one of wj and −wj

and I(K) ≤ 2X(K).

We say that a convex body K exhibits mirror symmetry, or that K is symmetric

about a hyperplane, if there exists a hyperplane H such that the reflection of K through

H leaves K unchanged. In [10], Dekster outlined a proof that every convex body of R3

symmetric about a plane is illuminated by eight directions. The aim of this chapter is to

prove the following.

111

Main Theorem 2. Let K ⊂ R3 be a convex body symmetric about a plane. Then

X(K) ≤ 6.

In light of the relationship between illumination and X-raying, one would expect the

proof of Main Theorem 2 to follow closely that of [10]. However as we will see below, the

proof of Main Theorem 2 takes a different approach than Dekster’s, resulting in a proof

that is much shorter and more concise.

Since X-raying is defined in terms of illumination, we include here some facts about

illumination that will be useful below.

Lemma 5.1.2. Let F be a face of K and v ∈ Rd be a direction. If v illuminates x ∈ F ,

then v illuminates every y ∈ relint(F ).

Proof. Let x ∈ F and y ∈ relint(F ). Since v illuminates x, there is a z ∈ rxv ∩ int(K).

Since y ∈ relint(F ), there is a p ∈ F with y ∈ [x, p). Note that points x, p and z determine

a triangle and (p, z) ⊂ int(K). Since rxv is parallel to ryv, there is a q ∈ (p, z)∩ryv ⊂ int(K).

Hence, v illuminates y.

Lemma 5.1.3. Let v ∈ Rd be a direction. Then v illuminates a relatively open subset

of bd(K).

Proof. Let v ∈ Rd be a direction and x ∈ bd(K) be illuminated by v. Then there exists

a y ∈ r xv ∩ int(K). Since y ∈ int(K), there is an ε > 0 such that D3(y, ε) ⊂ int(K).

Define

R =⋃

z∈D3(y,ε)

r z−v =⋃λ>0

(D3(y, ε)− λv)

and U = R∩bd(K). We have that R is open in Rd as the union of open sets, from which

it follows that U is an open set of bd(K). By construction, x ∈ U and every p ∈ U is

illuminated by v.

Lemma 5.1.4. For any convex body K, I(K) <∞.

112

Proof. Let v1,v2, . . . be a set of directions that illuminate K. Lemma 5.1.3 shows that

each direction vi illuminates an open subset U(vi) of bd(K), from which it follows that⋃U(vi) forms an open covering of bd(K). Since bd(K) is a compact set, there exists a

finite open subcovering of⋃U(vi). The directions that induce these finitely many open

neighbourhoods of bd(K) serve to illuminate K.

It follows from Lemma 5.1.3 that for any direction v ∈ Rd, a line `v X-rays an open

subset of bd(K), and Lemma 5.1.4 shows that X(K) is finite.

Lemma 5.1.5. For any d-dimensional convex body K, X(K) ≥ d.

Proof. Let n < d, and consider n directions v1,v2, . . . ,vn. Let H be a hyperplane such

that span{v1,v2, . . . ,vn} ⊆ H and H ′ be a hyperplane parallel to H that supports K at

some x ∈ bd(K). Since int(K) ∩ H ′ = ∅, it follows that x is not X-rayed along any of

the lines `v1 , `v2 , . . . , `vn . Hence X(K) ≥ d.

Lemma 5.1.6. For any d-dimensional convex body K with smooth boundary, X(K) = d.

Proof. Let K ⊂ Rd denote a convex body with smooth boundary, x ∈ bd(K) and

v1,v2, . . . ,vd ∈ Rd be linearly independent. Since K has smooth boundary, there exists

a unique hyperplane H that supports K at x. Because at most d − 1 of the directions

v1,v2, . . . ,vd lie on H, we assume without loss of generality that v1 /∈ H. Since x is a

smooth point, it follows that `xv1∩ int(K) and x is X-rayed along `v1 .

5.2 The X-Ray Conjecture in the Plane

Let C ⊂ R2 be a planar convex body. The goal of this section is to verify the X-ray

Conjecture in the plane; that is, that X(C) ≤ 3. The Illumination Conjecture was

verified in the plane by Levi [18], who showed that I(C) = 4 if C is a parallelogram,

and that I(C) = 3 otherwise. From this we deduce that X(C) ≤ 3. Indeed, if C is a

113

parallelogram then the two pairs of non-adjacent vertices of C determine two lines that

serve to X-ray C by Lemma 5.1.2 . If C is not a parallelogram then I(C) = 3 and it

follows from (5.1) that X(C) ≤ 3. However, we require a stronger result: we wish to

show that X(C) ≤ 3 with equality if, and only if, C is a triangle. This does not follow

from the result of Levi.

A point p ∈ R2 is said to be between two distinct parallel lines if p lies in the strip

determined by the two lines.

Lemma 5.2.1. Let C be a planar convex body, v ∈ R2 be a direction and x, y ∈ bd(C)

be such that ` yv 6= `xv . If z ∈ bd(C) is between `xv and ` yv , then z is X-rayed along `v.

Proof. We argue by contradiction. Suppose that C is supported by ` zv . Since z is between

`xv and ` yv , the line ` zv strictly separates x ∈ bd(C) and y ∈ bd(C), a contradiction. Thus,

z is X-rayed along `v.

Let x ∈ bd(C). We say x is a smooth point if there is a unique line that supports C

at x; otherwise we say x is a corner point. If x is a corner point, we say that a line L

supports C at exactly x if L supports C and L ∩ C = {x}. If L is a supporting line of

C at x, we denote by L+ the supporting halfspace of C determined by L. We define the

tangent cone of C at x by

TC(x) =⋂x∈LC⊂L+

L+.

We remark that TC(x) is a convex planar cone and is thus polyhedral. It follows that if

x is a corner point of bd(C), then there are two distinct directions v1 and v2 such that

TC(x) = [x, r xv1, r xv2

], and we see that `xv1and `xv2

support C at x.

Lemma 5.2.2. Let C be a planar convex body that is not a triangle and x ∈ bd(C)

be a corner point. Then there is a face F of C such that F and x lie in distict parallel

supporting lines, one which supports C at exactly x, the other which supports C at exactly

F . Moreover, x and F are X-rayed along one line.

114

Proof. Consider the tangent cone TC(x). From above, there there are two distinct direc-

tions v1 and v2 such that TC(x) = [x, r xv1, r xv2

], and the lines `xv1and `xv2

support C at x.

Since x is a corner point of C, the lines `xv1and `xv2

are not parallel. Let p ∈ bd(C) be

such that ` pv1supports C and is distinct from `xv1

, and q ∈ bd(C) be such that ` qv2sup-

ports C and is distinct from `xv2. Let y ∈ R2 be such that ` pv1

∩ ` qv2= {y}. Furthermore,

define P = ` pv1∩ bd(C) and Q = ` qv2

∩ bd(C) (see Figure 5.1).

We assume first that P ∩ Q = {y} ∈ bd(C). Since y ∈ ` pv1∩ ` qv2

, we have that

y /∈ `xv1∩ `xv2

, and it follows that [x, y] is not an edge of C. From this, we have that x

and y are X-rayed along `y−x. Because x is a corner point, there is a direction w such

that `xw supports C at exactly x, and since TC(y) = [y, r x−v1, r x−v2

] = −TC(x) + (y − x),

` yw supports C at exactly y.

We now assume that P ∩ Q = ∅. Let q0 ∈ Q and p0 ∈ P be such that [x, P,Q] lies

in the closed halfspace determined by ` p0p0−q0, and note that since P ∩Q = ∅, y /∈ C and

p0 − q0 /∈ TC(x) (see Figure 5.1).

Case 1. [p0, q0] is an edge of C. Then both p0 and q0 lie in two distinct supporting

lines of C, and are thus both corner points of C. Since C is not a triangle,

we have that p0 /∈ `xv1or q0 /∈ `xv2

. Without loss of generality, we assume that

p /∈ `xv1, and it follows that [x, p0] is not an edge of C. Thus x and p0 are

X-rayed along `p0−x. Let w be a direction such that ` p0w strictly separates y

and q0. Then ` p0w supports C at exactly p0 and `xw supports C at exactly x.

Case 2. [p0, q0] is not an edge of C. It is clear that `xp0−q0supports C at exactly x. Let

z ∈ bd(C) be such that ` zp0−q0supports C with ` zp0−q0

6= `xp0−q0and define

F = ` zp0−q0∩ bd(C). It follows that F ∩ ` p0p0−q0

= ∅, from which we find

` ry−x ∩ (p0, q0) 6= ∅ for every r ∈ F . Since (p0, q0) ⊂ int(C), we have that x

and F are X-rayed along `y−x.

115

Figure 5.1: The planar convex body C and the supporting lines `xv1, `xv2

, ` pv1and ` qv2

Theorem 5.2.3. Let C be a planar convex body. Then X(C) ≤ 3, with equality if, and

only if, C is a triangle.

Proof. We have already seen that if C is a triangle, then no direction illuminates more

than one vertex of C and X(C) = 3. If C has a smooth boundary then X(C) = 2 by

Lemma 5.1.6. Thus we assume that C is not a triangle and that the boundary of C has

a corner point, denoted by x.

By Lemma 5.2.2, there is a face F of C and two directions v,w such that F and x

are X-rayed along `v, the line `xw supports C at exactly x and for any z ∈ F , the line

` zw supports C at exactly F . By construction, every point p ∈ bd(C) \ ({x} ∪ F ) lies

between `xw and ` zw and is X-rayed along `w by Lemma 5.2.1. Hence X(C) ≤ 2, and we

obtain X(C) = 2 from Lemma 5.1.5.

It is interesting to note for planar convex bodies C, a proof that I(C) ≤ 4 follows

from Theorem 5.2.3. Indeed, if C is a triangle then I(C) = 3 as noted above. Otherwise

C is not a triangle, and we have that X(C) = 2. It then follows from (5.1) that I(C) ≤ 4.

116

5.3 X-Raying Convex Bodies of R3 Symmetric about a Plane

In this section, we prove that if K ⊂ R3 = {(x1, x2, x3) |xi ∈ R} is a convex body

that is symmetric about a plane then there is a collection of six lines that X-ray K. The

notations used closely follow those of Dekster [10]. Let K be symmetric about a plane H,

which we assume satisfies x3 = 0. We denote by e1, e2 and e3 the canonical basis vectors

of R3. For any x = (x1, x2, x3) ∈ R3, let Pr(x) = (x1, x2, 0). Let H+ = {x ∈ R3 |x3 ≥ 0}

and H− = {x ∈ R3 |x3 ≤ 0} be the upper and lower closed halfspaces of R3 determined

by H. Let B = K ∩H and for any X ⊆ B, define X± = Pr−1(X) ∩ bd(K) ∩H±. For a

direction v ∈ R3 and a real number t ≥ 0, we define v+(t) = v− te3 and v−(t) = v+ te3.

Let x ∈ relbd(B). If Pr−1({x}) ∩ bd(K) = {x} then we say that x is a ground point,

and denote the set of ground points of relbd(B) by G. Otherwise there exist distinct

x ∈ B− and x ∈ B+ such that Pr−1({x}) ∩ bd(K) = [x, x]. In this instance, we refer to

x as a cliff point, and call [x, x] the cliff at x.

Lemma 5.3.1. Let v ∈ H be a direction, and x ∈ (relint(B))+ (resp. (relint(B))−).

Then there exists a t ≥ 0 such that x is illuminated by v+(t) (resp. v−(t)).

Proof. We prove the case x ∈ (relint(B))+, and note that the case x ∈ (relint(B))− follows

from the symmetry of K. From x ∈ (relint(B))+, it follows that x′ = Pr(x) ∈ relint(B).

Thus there exists an ε > 0 such that D3(x′, ε) ∩ H ⊂ relint(B), and a δ > 0 such

x′ + δv ∈ D3(x′, ε) ∩H. Let t = ||x− x′||/δ. Then

x+ δv+(t) = x+ δ

(v − ||x− x

′||δ

e3

)= x+ δv − ||x− x′||e3

= x′ + δv.

Thus, x′ + δv ∈ r xv+(t) ∩ int(K) and v+(t) illuminates x.

117

Lemma 5.3.2. Let x ∈ relbd(B) be a cliff point and [x, x] be the cliff at x. If a direction

v ∈ H illuminates x, then there exists a t ≥ 0 such that every y ∈ [x, x] is illuminated

by one of v+(t) and v−(t).

Proof. Since v illuminates x ∈ relbd(B), it follows from Lemma 5.1.2 that v illuminates

every y ∈ (x, x), and by Lemma 5.3.3, we find that y is illuminated by both v+(t)

and v−(t) for any t ≥ 0. Thus, we need only show that there is a t ≥ 0 such that

x is illuminated by v+(t) and x is illuminated by v−(t). We have that there exists a

δ > 0 such that x+ δv ∈ relint(B) ⊂ int(K). Following the proof of Lemma 5.3.1, we let

t = ||x−x||/δ and find that x+δv ∈ rxv+(t). Hence v+(t) illuminates x and the symmetry

of K guarantees that v−(t) illuminates x.

The following Lemma shows that if a point on the boundary of K is illuminated

by some direction then it is illuminated by any “steeper” direction provided it is not a

ground point. This displays the difficulty ground points pose: if x is a ground point of

K and a direction v illuminates x, then v±(t) will not necessarily illuminate x for large

t.

Lemma 5.3.3. Let x ∈ B+ \G (resp. B− \G). If v ∈ H is a direction such that v+(t1)

(resp. v−(t1)) illuminates x for some t1 ≥ 0, then v+(t2) (resp. v−(t2)) for any t2 > t1.

Proof. We prove only the case x ∈ B+ \ G as the case x ∈ B− \ G follows from the

symmetry of K. If Pr(x) = x, then x ∈ relbd(B). It now follows that x is a cliff point

and [x, x] is the cliff at x. Let

x′ =

x, if Pr(x) = x,

Pr(x), otherwise.

In either case, x′ ∈ r x−e3 . Since v+(t1) illuminates x, there exists a y ∈ r xv+(t1)∩ int(K),

from which it follows that the open segment (x′, y) ⊂ int(K). For any t2 > t1, there exists

118

a z ∈ r xv+(t2)∩ r y−e3 . The segments [x, x′] and [y, z] are parallel. Since [y, z]∩ `x−e3 = ∅, we

have that Q = [x, x′, y, z] is a convex quadrilateral with diagonals [x′, y] and [x, z]. These

diagonals intersect at some point q ∈ (x′, y) ∩ (x, z) ⊂ int(K). Since (x, z) ⊂ r xv+(t2),

v+(t2) illuminates x.

Lemma 5.3.4. Let v1,v2, . . . ,vn be a collection of directions in H such that any p ∈

relbd(B) is illuminated by at least one of v1,v2, . . . ,vn. Then there is a finite real number

M > 0 such that any x ∈ bd(K)\G is illuminated by one of v±1 (M),v±2 (M), . . . ,v±n (M).

Proof. Let x ∈ bd(K) and x′ = Pr(x). For i = 1, 2, . . . , n, define

di(x) = max{d ∈ R |x′ + dvi ∈ B}.

Since x′ ∈ B ⊂ K, each di(x) is nonnegative and finite. Since v1,v2, . . . ,vn illuminate

B, there exists an 1 ≤ i ≤ n such that di(x) > 0. It follows that for i = 1, 2, . . . , n,

ti(x) :=

2||x− x′||di(x)

, for di(x) 6= 0,

∞, for di(x) = 0.

is finite and nonnegative. We further define T (x) = min{ti(x) | 1 ≤ i ≤ n}. It follows

from Lemmas 5.3.1 and 5.3.2 that for every x ∈ bd(K), there is an i = 1, 2, . . . , n such

that x is illuminated by v+i (T (x)) or by v−i (T (x)). It is clear that x′ = Pr(x) and

each of di(x) and ti(x) varies continuously with x. Thus, T (x) is continuous on bd(K).

Since bd(K) is a compact set, it follows that T (x) attains its maximum M on bd(K).

It then follows from Lemma 5.3.3 that any x ∈ bd(K) \ G is illuminated by one of

v±1 (M),v±2 (M), . . . ,v±n (M).

We are now in a position to prove Main Theorem 2.

Proof of Main Theorem 2. Our argument depends on the shape of B = K ∩H. If B is

not a triangle, then Theorem 5.2.3 guarantees that there are two directions, v1,v2 ∈ H,

119

such that any x ∈ relbd(B) is X-rayed along one of `v1 and `v2 . It follows that any x ∈

relbd(B) is illuminated by one of ±v1 and ±v2. From Lemma 5.3.4, we see that there is

an M > 0 such that any x ∈ bd(K)\G is illuminated by one of v±1 (M),−v±1 (M),v±2 (M)

and −v±2 (M). Thus, any x ∈ bd(K) \G is X-rayed along one of `v+1 (M), `v−

1 (M), `v+2 (M),

and `v−2 (M). Since G ⊆ relbd(B), the lines

`v1 , `v2 , `v+1 (M), `v−

1 (M), `v+2 (M), `v−

2 (M)

serve to X-ray K, and X(K) ≤ 6.

If B is a triangle, then we label the vertices of B as x1, x2 and x3. For i = 1, 2, 3, let

vi be a direction in H that illuminates xi. From Lemma 5.1.2, we see that v1,v2 and

v3 are sufficient to illuminate relbd(B). We consider whether or not x1, x2 and x3 are

ground points or cliff points.

Let x1, x2 and x3 be ground points. From Lemma 5.1.3, we see that for i = 1, 2, 3,

each vi illuminates an open set N(vi) of bd(K) and −vi illuminates an open set N(−vi)

of bd(K). It follows that

K ′ = bd(K) \

(3⋃i=1

(N(vi) ∪N(−vi)

))

is a closed subset of K, and thus, it is compact. Since any x ∈ relbd(B) \ {x2, x3} is

illuminated by one of v1 and −v1, we argue as in the proof of Lemma 5.3.4, and find

there is an M > 0 such that any x ∈ K ′ is illuminated by one of v±1 (M) and −v±1 (M).

Thus K ′ is X-rayed along one of `v+1 (M) and `v−

1 (M). It follows that any x ∈ bd(K) is

X-rayed along one of

`v1 , `v+1 (M), `v−

1 (M), `v2 , `v3 .

Let x1 be a cliff point, and x2, x3 be ground points. Since x1 is illuminated by v1,

Lemma 5.3.3 guarantees it is illuminated by v±1 (t) for any t ≥ 0. Similar to the above,

120

we define

K ′ = bd(K) \

(3⋃i=2

(N(vi) ∪N(−vi)

))which is a compact subset of bd(K). Since any x ∈ relbd(B) \ {x2, x3} is illuminated by

one of v1 and −v1, we use again Lemma 5.3.4 and see that there is an M > 0 such that

any x ∈ K ′ is illuminated by one of v±1 (M) and −v±1 (M). It follows that any x ∈ bd(K)

is X-rayed along one of

`v+1 (M), `v−

1 (M), `v2 , `v3 .

Let x1, x2 be cliff points, and x3 be a ground point. Proceeding as above, we find that

any x ∈ bd(K) is X-rayed along one of

`v+1 (M), `v−

1 (M), `v+2 (M), `v−

2 (M), `v3 .

Finally, let x1, x2 and x3 be cliff points. Analogous to the above, we see that

`v+1 (M), `v−

1 (M), `v+2 (M), `v−

2 (M), `v+3 (M), `v−

3 (M)

serve to X-ray any x ∈ bd(K).

We have now exhausted all cases, and so, conclude that X(K) ≤ 6 for any convex

body K ⊂ R3 with mirror symmetery.

121

Conclusion

In Chapter 2, we developed an algorithm to generate the list of odd-dimensional univer-

sal faces of a totally-sewn 6-polytope. As a generalization of sewing, Finbow-Singh [11]

introduced tailoring, where the universal tower used for sewing is replaced with any tower

containing an increasing sequence of universal faces of a given polytope. Our algorithm

could be adapted to handle tailoring as well as sewing. In addition, Bisztriczky [5] gen-

eralized Shemer’s Sewing Construction for odd-dimensional neighbourly polytopes. As

stated in Section 2.2, our algorithm works for all even-dimensional neighbourly polytopes,

and it would be useful to extend it to odd-dimensions as well.

In Chapter 4, we argued that for any O ∈ int(Sp(n)), 9 ≤ p ≤ n, at most twenty-

nine hyperplanes were neccessary to separate O from any facet of Sp(n), thus verifying

the Separation Conjecture for the semi-cyclic 6-polytopes Sp(n). This was achieved by

locating O in representations of appropriate 2-dimensional quotient polytopes. Although

successful, this method relied on having a complete facial description of Sp(n), and would

be difficult to generalize to larger classes of neighbourly 6-polytopes where the facial

structure is not explicity known. A first step in the continuation of this work would be to

provide a shorter, more concise proof of Main Theorem 1 in Chapter 4. In light of Section

2.3, any facet F ∈ F(Sp(m)) \F(Sp(m− 1)) satisfies F ∩{[x1, xm], [xm−1, xm]} 6= ∅, with

{[x1, xm], [xm−1, xm]} ⊆ U1(Sp(m)). Therefore, the quotient polytopes Sp(m)/[x1, xm]

and Sp(m)/[xm−1, xm] are neighbourly 4-polytopes with m − 2 vertices. If these two

quotient polytopes can be shown to be semi-cyclic, one can apply the results of Finbow-

Singh and Oliveros [12] to obtain our results with less work.

From Chaper 3, recall the construction of the semi-cyclic 6-polytopes Sp(n), 9 ≤ p ≤

n. We defined Sp(p) = C(p) and and for p ≤ m ≤ n − 1 we sewed xm+1 onto Sp(m)

122

through the universal tower

Tm = {Em, Gm, Fm}

with

Em = [x1, xm] ∈ U1(Sp(m)),

Gm = [Em, x2, xm−1] ∈ U3(Sp(m)) and

Fm = [Gm, xm−3, xm−2] ∈ F(Sp(m)).

As a generalization of our work in Chapters 3 and 4, consider any sequence of neighbourly

6-polytopes P (9), P (10), . . . , P (n), with P (9) = C(9) and P (m + 1) being obtained by

sewing xm+1 onto P (m) through the universal tower T m = {Em, Gm, Fm} for 9 ≤ m ≤

n − 1. Here, Fm is any facet of P (m) containing Gm. We note that this larger class of

totally-sewn polytopes includes both the cyclic and semi-cyclic polytopes. It is natural

to examine the Separation Problem with these polytopes as a next step in verifying the

Separation Conjecture for all neighbourly 6-polytopes.

Finally, we gave a verification of the X-ray Conjecture in the plane that did not depend

on the results of Levi [18] in Chapter 5. We proved that in the plane, at most two lines

were required to X-ray any convex body with the exception of a triangle, which requires

three lines. One possible avenue for future research is to consider which d-dimensional

convex bodies require the conjectured maximum of 3 ·2d−2 lines to be X-rayed. For d ≥ 3,

it is clear that a (d−2)-fold d-cylinder over a triangle requires 3 ·2d−2 lines to be X-rayed,

but it is not known if this is the only instance where the maximum number of lines is

required.

In [1], Bezdek verifies the Illumination Conjecture for 3-dimensional convex poly-

topes with affine symmetry; that is, central symmetry, mirror symmetry and rotational

symmetry. However, the proof does not generalize for an arbitrary 3-dimensional con-

vex bodies C. This was done by Lassak [17] for the case C having central symmetry,

123

and by Dekster [10] for the case C having mirror symmetry. To date, the Illumination

Conjecture has not been verified for C having rotational symmetry. In this thesis, we

have verified the X-ray Conjecture for C having mirror symmetry, and the verification of

the X-ray Conjecture for C having central symmetry follows from [17]. A verification of

the X-ray Conjecture for C having rotational symmetry might be sufficient to verify the

Illumination Conjecture (see (5.1)), or give us some insight as to how one might proceed.

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