Contractions of Polygons in Abstract Polytopes

by Ilya Scheidwasser

B.S. in Computer Science, University of Massachusetts Amherst M.S. in Mathematics, Northeastern University

A dissertation submitted to

The Faculty of the College of Science of Northeastern University in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy

March 31, 2015

Dissertation directed by Egon Schulte Professor of Mathematics Acknowledgements

First, I would like to thank my advisor, Professor Egon Schulte. From the first class I took with him in the first semester of my Masters program, Professor Schulte was an engaging, clear, and kind lecturer, deepening my appreciation for mathematics and always more than happy to provide feedback and answer extra questions about the material. Every class with him was a sincere pleasure, and his classes helped lead me to the study of abstract polytopes. As my advisor, Professor Schulte provided me with invaluable assistance in the creation of this thesis, as well as career advice. For all the time and effort Professor Schulte has put in to my endeavors, I am greatly appreciative. I would also like to thank my dissertation committee for taking time out of their sched- ules to provide me with feedback on this thesis. In addition, I would like to thank the various instructors I’ve had at Northeastern over the years for nurturing my knowledge of and interest in mathematics. I would like to thank my peers and classmates at Northeastern for their company and their assistance with my studies, and the math department’s Teaching Committee for the privilege of lecturing classes these past several years. Finally, I would like to thank my girlfriend Rachel for her support and company during my work on this dissertation. Her presence in my life has been a priceless grounding force. She helps keep me happy and sane.

ii Abstract of Dissertation

There are several well-known constructions of new polytopes from old, such as the pyramid and prism constructions. This thesis defines two new local constructions on abstract polytopes. The first construction, called digonal contraction, allows digonal sections to be removed by merging their two edges into a single edge. The second construction, called polygonal contraction, allows polygonal sections with at least four vertices to be converted to two smaller polygons by merging two non-adjacent vertices. Neither of these contractions can be applied arbitrarily; we present necessary and sufficient conditions for the use of each contraction. In the case of polygonal contraction, we find that the graph-theoretical properties of the polytope affect whether the contraction can be applied. Both of these contractions have interesting interactions with another new construction, the k-bubble construction introduced by Helfand (2013, Section 2.2), which are described. We also investigate how polygonal contraction can be performed somewhat globally, and what the effects are on the automorphism group of the polytope.

iii Table of Contents

Acknowledgements ii

Abstract of Dissertation iii

Table of Contents iv

List of Figures vii

1 Introduction 1

2 Basic Notions 4 2.1 Basics of Convex Polytopes ...... 4 2.2 Posets and Symmetry ...... 6 2.3 Abstract Polytopes ...... 13 2.4 Graph Theory and Polytopes ...... 17

3 Digonal Contraction 21 3.1 Preliminary Results for Digonal Contraction ...... 21 3.2 The Digonal Contraction Theorem ...... 23 3.3 Reversing Digonal Contraction ...... 34

4 The Helfand Construction 38 4.1 The Global and Local Helfand Constructions ...... 38 4.2 Connections to Digonal Contraction ...... 43

5 Polygonal Contraction 51 5.1 Deﬁning Polygonal Contraction ...... 51

iv 5.2 The Polygonal Contraction Theorem ...... 57 5.3 Polygonal Contraction for Polytope Lattices ...... 75 5.4 Connections to the Helfand Construction ...... 76

6 Polygonal Contraction and Symmetry 84 6.1 Multiple Polygonal Contraction ...... 84 6.2 Connections to Symmetry ...... 87

Bibliography 104

v List of Figures

2.1 A ﬂag of a 3-cube...... 6 2.2 The Hasse diagram of a square...... 8 2.3 The pyramid and prism over a pentagon...... 10 2.4 The distinguished generators of a square...... 10 2.5 A triangular 2-section of a cube...... 12 2.6 Examples of abstract polytopes which are not convex polytopes...... 15

3.1 Digonal contraction on a cube-like polytope...... 24 3.2 Four non-contractible digons...... 31 3.3 An example of Condition (4) breaking for digonal contraction...... 33 3.4 Merging two polygons with digonal contraction...... 35

4.1 Two diﬀerent applications of the global Helfand construction to the 3-cube. 40 4.2 Producing a digon via the Helfand construction...... 44 4.3 Producing and contracting digons with the Helfand construction...... 45 4.4 Merging two triangles into a square with digonal contraction...... 48 4.5 Merging two polygons with digonal contraction, in connection with the Helfand construction...... 50

5.1 Polygonal contraction on a hexagon...... 52 5.2 A polytope nontrivially satisfying the conditions in Prop. 5.1.2...... 55 5.3 Edge graphs satisfying and violating the conditions in Prop. 5.1.2...... 56 5.4 Polygonal contraction failing due to apeirogons...... 67 5.5 Examples of connectivity and non-connectivity in polygonal contraction. . . 68 5.6 Two similar polygons with diﬀerent results under polygonal contraction. . . 70

vi 5.7 Five diﬀerent examples of non-contractible polygonal sections F/G in polytopes...... 71 5.8 The edge graphs G for the examples in Figure 5.7...... 72 5.9 Breaking or preserving the lattice condition via polygonal contraction. . . . 77 5.10 Producing and contracting polygons with the Helfand construction...... 80 5.11 The creation of a triangle via the Helfand construction...... 82 5.12 The creation of another triangle via the Helfand construction...... 82

6.1 The truncated square tiling...... 91

6.2 The polyhedron P1 obtained by contracting all octagons in the truncated square tiling...... 92

6.3 The polyhedron P2 obtained by contracting all squares in the truncated square tiling...... 93

6.4 The polyhedron P12 obtained by contracting all squares and octagons all in the truncated square tiling...... 95 6.5 Further contraction of the truncated square tiling...... 96 6.6 The ﬁrst step in a maximal contraction of the truncated square tiling. . . . 97 6.7 The second step in a maximal contraction of the truncated square tiling. . . 98 6.8 The third step in a maximal contraction of the truncated square tiling. . . . 99 6.9 The ﬁnal step in a maximal contraction of the truncated square tiling. . . . 100

vii Chapter 1

Introduction

An abstract polytope is a combinatorial object generalizing the face lattice of a convex polytope (McMullen and Schulte, 2002, Section 2A). Specifically, an abstract polytope is a poset. The elements of the set represent faces of the polytope and one face being less than another in the poset represents two faces being incident, such as a vertex contained in an edge. In order for a poset to be an abstract polytope, it must satisfy four conditions - some are very straightforward, such as the requirement that there is exactly one least face and one greatest face, while others are more complicated, such as the strong connectivity condition. As such, testing whether an arbitrary poset is an abstract polytope is not simple. A construction on abstract polytopes is a way to perturb an abstract polytope, whether locally or globally, in order to get another abstract polytope. Two well-known constructions are the pyramid construction and the prism construction: just as we can take the pyramid over a convex polytope or the prism over a convex polytope to get a new convex polytope of one higher dimension (Grünbaum, 2003, Sections 4.2, 4.4), we can generalize these constructions to apply to abstract polytopes, and performing either of these constructions on an abstract polytope always produces an abstract polytope of one higher rank (dimension). Helfand introduced a new global construction on abstract polytopes called the k-bubble construction (Helfand, 2013), or in this paper, the global Helfand construction (we define a local version later). This construction is a generalization of the operation of truncating a convex polytope. For example, if we truncate a cube, we imagine slicing the cube near each of its vertices, so that there is now a new triangle for every vertex; the resulting convex polytope is the well-known truncated cube. This new polytope is not regular like the original polytope was, but is still highly symmetric. The global Helfand construction allows us to

1 apply a similar operation, in various ways, to an abstract polytope, and always produces a new abstract polytope of the same rank. The two new constructions examined in this thesis are local constructions focused on polygons in abstract polytope. A polygon is any two-dimensional section of the face poset of a polytope: for example, the cube clearly has square faces, but it also has triangular sections (its vertex figures). The first construction examined is the digonal contraction. A digon is an abstract polygon with just two vertices and two edges, and the presence of digonal sections in an abstract polytope is sometimes viewed as a degenerate condition. Digonal contraction is the operation of taking a digonal section, removing the greatest face of that section altogether, and merging the two edges into a single edge. The second construction examined is the polygonal contraction. Unlike digonal contraction, the aim here is not to remove the polygon altogether, but to split it into two smaller polygons. This is done by taking two non-adjacent vertices of the polygon (hence, the polygon must be at least a square) and merging them, thus producing two smaller polygons in a natural way. Neither of these contractions can be applied to arbitrary polygons in arbitrary polytopes; both have various conditions that must be satisfied in order for the contraction to produce an abstract polytope (otherwise, the contraction will just produce a poset which does not qualify as an abstract polytope for one reason or another). However, there are specific cases in which these contractions are very easy to apply (i.e. the conditions required for them to produce abstract polytopes become much simpler). One example is contracting polygons in a polytope lattice: a polytope lattice is a polytope which, when viewed as a poset, satisfies the lattice condition. Polytope lattices cannot have digonal sections, and their (non-triangular) polygonal sections can be contracted if and only if they lie on the lowest level, i.e. their greatest face has rank 2. Contracting a polygon in a polytope lattice in this way does not necessarily produce a polytope lattice, but will always produce a polytope. Both contractions also have interesting interactions with the Helfand construction: after applying the Helfand construction to a polytope (or locally to part of a polytope), we find that certain new faces introduced by the Helfand construction can be easily contracted. While both the digonal and the polygonal contraction are local constructions, it is sometimes possible to repeatedly apply them in a global way. Since the polygonal contraction is more interesting, we examine it in particular. Given a (possibly infinite) set of contractible polygons in a polytope, it is sometimes possible to contract all of them simultaneously. If the initial polytope is highly symmetric and we choose polygons to contract in a symmetric

2 way, we ﬁnd that the resulting polytope will have a similar automorphism group to the original polytope (although it may have new combinatorial symmetries as well). We examine several cases like this when applied to highly symmetric face-to-face tessellations of the plane by regular polygons.

3 Chapter 2

Basic Notions

In this chapter, we will introduce some well-known concepts relating to convex and abstract polytopes. Important, well-established results will be presented without proof. The reader is referred to Gr¨unbaum (2003) for convex polytopes and to McMullen and Schulte (2002) for abstract polytopes.

2.1 Basics of Convex Polytopes

n Deﬁnition 2.1.1. The convex hull of a set of points M ⊆ R is the smallest convex set containing M.

n Proposition 2.1.1. For M = {x1, ..., xm} ⊂ R a finite set, the convex hull of M is the m m P n P set λixi λi ∈ R ∀ i, 0 ≤ λi ≤ 1 ∀ i, λi = 1 . i=1 i=1 For example, the convex hull of two different points is the line segment between the two points. The convex hull of three different points is either a triangle with those points as vertices (specifically, a ‘filled in’ triangle), or a line segment; this latter case occurs if the points are colinear.

n Definition 2.1.2. The affine hull of a set of points M ⊆ R is the smallest affine subspace n of R containing M (an affine subspace is a translation of a linear subspace).

n Definition 2.1.3. A convex polytope is the convex hull of a finite set of points in R . The rank or dimension, m, of a convex polytope P is the dimension of the affine hull of P ; the polytope is then called an m-polytope. A 0-polytope is called a vertex, a 1-polytope is

4 called an edge or line segment, a 2-polytope is called a polygon, and a 3-polytope is called a polyhedron.

For example, the square is a convex polytope. It can be represented as the convex 2 hull of four points in R , or just as easily as the convex hull of four (coplanar) points 10 in R ; regardless, it is a rank-2 polytope. Other well-known convex polytopes include the three-dimensional cube, the icosahedron, the truncated icosahedron (soccer ball), and the tesseract (four-dimensional cube).

n Deﬁnition 2.1.4. A proper face of a convex n-polytope P in R is any nonempty inter- n section of P with a closed half-space H of R such that none of the points of H lie in the interior of P . The two improper faces of P are P itself and the empty set, which is considered a (−1)-face (these two faces are sometimes referred to as Fn and F−1 respectively). Note that the intersection of any two faces of a convex polytope is also a face of the polytope

(if the intersection is empty, this is considered to be F−1).

For example, a 3-cube has eight 0-dimensional faces, or 0-faces (i.e. the vertices), twelve 1-faces (i.e. the edges), and six 2-faces (i.e. the squares). Note that every i-face of a convex polytope P is a convex i-polytope (since it is the convex hull of the vertices of P contained in the half-space deﬁning that i-face).

Definition 2.1.5. A chain of a convex polytope is a set of faces such that every face is incident to every other face. A flag of a convex polytope is a maximal chain - this implies that the flag has exactly one face of each rank.

In Figure 2.1, we see a cube with a flag marked in red: the flag includes a vertex, an edge, and a 2-face (note that the flag must also include the maximal face, i.e. the entire cube, and the minimal face, i.e. the empty set, although these are not marked). The term ‘flag’ is consistent with the fact that such a drawing resembles a physical flag: the marked edge is the flagpole, the marked vertex is the tip of the flagpole, and the marked 2-face is the flag itself. In the Hasse diagram (Definition 2.2.2), a flag of an n-polytope is a strictly rank-ascending path from the (−1)-face to the n-face of the polytope. Convex n-polytopes have the property that given any flag Φ and any proper rank i, there is exactly one flag Φi which is identical to Φ except at the rank-i face; this is called the i-adjacent flag to Φ. We can confirm visually that the flag shown in the figure has exactly one 0-adjacent flag (we would get this by marking the bottom vertex of the marked edge rather than the

5 Figure 2.1: A 3-cube, with a flag marked in red. top vertex), exactly one 1-adjacent flag (we would get this by marking the top edge of the marked square instead of the left edge), and exactly one 2-adjacent flag (we would get this by marking the other square containing the marked vertex and edge; this square is not visible in the diagram).

2.2 Posets and Symmetry

Deﬁnition 2.2.1. A partially ordered set, or poset (Stanley, 1986, Chapter 3), is a set P with relation ‘≤’ satisfying: (1) a ≤ a ∀ a ∈ P (reﬂexivity) (2) a ≤ b ∧ b ≤ a ⇒ a = b (antisymmetry) (3) a ≤ b ∧ b ≤ c ⇒ a ≤ c (transitivity)

This is called a partial ordering since not every two elements in P need to have the relation apply to them. A lattice is a poset in which every two elements have a supremum (least upper bound) and infimum (greatest lower bound). A convex polytope P can be naturally viewed as a poset: P is identified with its face set, and the relation F ≤ G can be read as ‘F is incident to G and has rank less than or equal to that of G’. Moreover, they are ranked posets, meaning that all flags (maximal chains) have the same length. This poset is called the face lattice of the polytope. Note that the face lattice is indeed a lattice: given any two faces F and G of a polytope, we can

6 ﬁnd a face A containing both and contained in every face containing both, and a face B contained in both and containing all faces containing both (this is a nontrivial result).

Deﬁnition 2.2.2. A Hasse diagram of a convex n-polytope is a certain graphical description of the faces of the polytope, their ranks, and their incidences (i.e. which faces are contained in which), which is independent of any geometrical representation of the polytope. More precisely, it is a graph in which the nodes represent the faces and in which edges represent incidences between faces whose rank diﬀers by 1.

In Figure 2.2, we see a square with every face labeled (except for the (−1)-face), and below it we see its Hasse diagram: every vertex in the Hasse diagram is one of the faces of the square, and they are arranged vertically in order of rank. Edges in the Hasse diagram represent incidences between faces whose rank differs by 1. The Hasse diagram is independent of the geometrical representation of the square in the following sense: suppose the square pictured in Figure 2.2 has vertices at coordinates (0, 0), 2 (0, 1), (1, 0), and (1, 1) in R . If we imagine a different square with vertices at other coor- 2 n dinates in R , or indeed in any R with n ≥ 2, it would still have the same Hasse diagram. Indeed, we could even deform the square into a rectangle, rhombus, or any quadrilateral, as long as the vertices were coplanar (and no vertex lay inside the convex hull of the other three, since that would produce a triangle), and the Hasse diagram would be the same. All of these different convex polytopes are called combinatorially equivalent/isomorphic to the square: they correspond to the same combinatorial structure of faces and incidences.

Deﬁnition 2.2.3. A geometrical symmetry of a convex polytope is any isometry (distance- preserving aﬃne transformation) taking the polytope to itself.

Deﬁnition 2.2.4. A combinatorial symmetry of a convex polytope with face set F (this is the set of all the faces of the polytope of every rank) is a bijection σ : F → F that preserves rank and incidences.

Every geometrical symmetry is clearly a combinatorial symmetry, but the converse is not necessarily true. For example, consider a non-square rectangle: geometrically, there is no symmetry that rotates the rectangle by 90 degrees or reﬂects across a diagonal, but combinatorially, there are such symmetries. We note that the set of geometrical symmetries of a polytope P form a group, called G(P ), as do the set of combinatorial symmetries (which we will refer to as ‘automorphisms’). This latter group will be referred to as the

7 Figure 2.2: The Hasse diagram of a square.

8 automorphism group of P , written Γ(P ). In this thesis we are interested in combinatorial rather than geometrical symmetries. Often, when there is little chance of confusion, we use the term ’symmetries’ in place of ’combinatorial symmetries’ or ’automorphisms’.

Definition 2.2.5. A convex polytope P is regular when Γ(P ) is transitive on the flags of P , meaning that there exists an automorphism taking any flag of P to any other flag of P .

This is the notion of being combinatorially regular; there is also the notion of being geometrically regular, which is just the same statement applied to G(P ). By a result of McMullen (1967), each combinatorially regular convex polytope is isomorphic to a geometrically regular convex polytope (see also McMullen and Schulte, 2002, Section 1B). The regular 2-polytopes are exactly the polygons; in fact, the only 2-polytopes are the polygons, so all convex 2-polytopes are regular, although they may not be geometrically regular, depending on how they are represented on the plane (e.g. a square versus a rectangle). The regular convex 3-polytopes are exactly the tetrahedron, the cube, the octahedron, the icosahedron, and the dodecahedron, as well as all 3-polytopes combinatorially equivalent to the aforementioned (e.g. parallelepipeds, which are combinatorially equivalent to cubes). Most convex 3-polytopes are not regular. For example, any pyramid over a non-triangular n polygon is not regular (a pyramid over an (n − 1)-polytope Q in R is the convex hull of Q and a single point, called the apex, outside the affine hull of Q; this is an n-polytope). Similarly, any prism over a non-square polygon is not regular (a prism over an (n − 1)- n polytope Q in R is the convex hull of Q and a translate of Q outside the affine hull of Q; this is an n-polytope). Figure 2.3 shows the prism and pyramid over a pentagon. Note that a non-regular polytope may still be highly symmetric: for example, the soccer ball is vertex-transitive, meaning that there is a symmetry taking any vertex to any other vertex. An important property of regular polytopes is the following (McMullen and Schulte, 2002, Proposition 2B8): for any regular n-polytope P , given any flag Φ, called the base flag, i we have that Γ(P ) = hρ0, ..., ρn−1i, where ρi is the automorphism taking Φ to Φ . The

{ρi} are called the distinguished generators of Γ(P ) relative to the base ﬂag Φ. Note that an automorphism on a regular polytope is uniquely determined by its action on a base ﬂag.

Also, each ρi is given by a reflection if P is a geometrically regular convex polytope. We see a simple example of this in Figure 2.4. In this figure, we have as our polytope P the square and n = 2. The base flag Φ is shown in red: this is represented as a triangle which contains the base vertex v, lies along the base edge e, and moves into the square

9 Figure 2.3: The pyramid and prism over a pentagon.

Figure 2.4: The distinguished generators of a square.

10 itself, P (such a representation of a flag is called a fundamental region). The 0-adjacent flag to Φ, denoted Φ0, is shown in orange to the left of Φ: this flag differs from Φ only in its 0-face, i.e. its vertex. The 1-adjacent flag to Φ, denoted Φ1, is shown in orange below Φ: this flag differs from Φ only in its 1-face, i.e. its edge. We can visually verify that the 0 1 reflection ρ0 (shown in blue) moves Φ to Φ and the reflection ρ1 moves Φ to Φ ; moreover, these two reflections generate the entire symmetry group of the square (composing them gives us a 90-degree rotation; performing this rotation repeatedly on either Φ or Φ0 lets us move to any flag of the square).

Deﬁnition 2.2.6. Given a convex polytope with two faces F ≤ G, the section G/F is the set of faces {H | F ≤ H ≤ G}. This is isomorphic to a convex polytope of dimension rk(G) − rk(F ) − 1.

For example, consider the cube: let F be a vertex of the cube and let G be the cube itself. Then G/F consists of the vertex F , the three edges incident to that vertex, the three squares incident to the vertex, and the cube itself. We see in Figure 2.5 that the Hasse diagram of this section is the same as the Hasse diagram of a triangle; hence, we can view this section as a triangle. In fact, every section of a convex polytope is isomorphic to a convex polytope in this same sense.

Deﬁnition 2.2.7. An (i-face)-ﬁgure of a polytope P is a section P/F where rk(F ) = i.

For example, Figure 2.5 shows a vertex ﬁgure of a cube.

Deﬁnition 2.2.8. A polygon at level i of a polytope P is a rank-two section of P , with greatest face of rank i in P .

For example, Figure 2.5 shows that cubes have triangles at level 3.

Definition 2.2.9. A convex polytope is called equivelar when, for every rank i, every rank-2 section at that rank (i.e. every two-dimensional section where the difference in rank between the least and greatest faces of the section is 3 and the greatest face of the section has rank i) is a polygon with the same number of vertices. Then the Schläflisymbol of such a polytope lists the sizes of these polygons in order of rank.

For example, the cube is equivelar (this is implied by it being regular): all sections G/F with rk(G) = 2 are squares, and all sections G/F with rk(G) = 3 are triangles. Hence, the Schl¨aﬂisymbol of the cube is {4, 3}.

11 Figure 2.5: A triangular 2-section of a cube.

12 2.3 Abstract Polytopes

Deﬁnition 2.3.1. An abstract polytope is a poset P with elements called faces such that it has the following properties:

(P1) P has a least face (a face F−1 such that ∀ F ∈ P,F−1 ≤ F ) and greatest face (a face Fn such that ∀ F ∈ P,F ≤ Fn). (P2) All ﬂags have the same number of faces, i.e. P is ranked. (P3) P is strongly connected. In particular, P is connected, meaning that we can get from any proper face to any proper face via a ﬁnite sequence of successive incidences between proper faces, or P is rank 1. But in fact, P is even strongly connected, meaning that every section of P is connected. (P4) For every pair of faces F ≤ G with rk(G) = rk(F ) + 2, the section G/F has exactly two proper faces. This is called the diamond condition.

A non-trivial, but very important fact, is that every convex polytope, if viewed as a poset as described above, is an abstract polytope (this helps justify the term ‘polytope’ for the above posets). From here on out, if we simply refer to a ‘polytope’, we mean an abstract polytope. Another important fact (this one is fairly easy to show) is that every section of a polytope is itself a polytope. Just as with convex polytopes, we can define a Hasse diagram for any abstract polytope, in the same way: the nodes of the diagram represent the faces and the edges of the diagram represent incidences between faces whose rank differs by 1. Then every abstract polytope is completely described by its Hasse diagram. A useful note about connectivity: the condition of a poset being connected is equivalent to it being flag-connected: P is flag-connected if and only if any flag can be joined to any other flag via a finite sequence of successively adjacent flags. Similarly, the condition (P3) of strong connectivity is equivalent to the condition of strong flag-connectivity: P is strongly flag-connected if and only if every section is flag-connected, or equivalently, any flag Φ can be joined to any other flag Ψ via a finite sequence of successively adjacent flags such that every flag in the sequence contains Φ ∩ Ψ. There are many abstract polytopes which are not convex polytopes, and they can often (though not always) be hard to visualize. In rank 1, there is exactly one abstract polytope, the line segment (which is also convex). In rank 2, the abstract polytopes consist of all

13 convex polygons as well as the digon and the apeirogon: the digon is a polygon with two vertices and two edges, and the apeirogon is a polygon with infinitely many vertices and edges. In rank 3, there are too many different kinds of polytopes to list, but well- known examples include the planar tessellation by squares, and the hemicube. All of the aforementioned polytopes are also regular (an abstract polytope P is regular if Γ(P ) is flag-transitive). Note that the hemicube, shown in Figure 2.6, has only four vertices, six edges, and three faces (e.g. the two nodes of the diagram labeled b represent the same vertex); in this way it can be thought of as half of a cube. Unlike convex polytopes, abstract polytopes need not be a lattice: for example, the two edges of a digon do not have an infimum (since both vertices of the digon are contained in both edges, but neither vertex contains the other), nor do any two 2-faces of a hemicube (since every pair of 2-faces shares two edges). If an abstract polytope is a lattice, we call it a polytope lattice. Again, every convex polytope is a polytope lattice, but there are polytope lattices which are not convex polytopes (e.g. {4, 4}, shown in Figure 2.6).

Note: the pyramid and prism constructions on convex polytopes, described earlier, can be extended to abstract polytopes as follows:

Deﬁnition 2.3.2. Given an n-polytope P , the pyramid over P is the poset P 0 obtained by taking P as a subposet and creating a new (i + 1)-face F 0 for every i-face F in P (with i ≥ −1) so that every F 0 contains its corresponding old face F as well as every face G0 such that F > G in P . Then P 0 is in fact an (n + 1)-polytope.

Deﬁnition 2.3.3. Given an n-polytope P , the prism over P is the poset P 0 obtained by taking P as a subposet and creating a new i-face F 0 for every i-face F in P with i ≥ 0 so that every new face F 0 contains the new face G0 if F > G in P ; additionally, for every i-face F in P with i ≥ 0 a new (i + 1)-face Fˆ is created such that Fˆ > F, F 0 and every Fˆ contains the face Gˆ if F > G in P . Then P 0 is in fact an (n + 1)-polytope.

Deﬁnition 2.3.4. In a polytope, if G is a j-face of an i-face F then the corank of G in F is given by i − j, and G is called a corank i − j face of F .

Deﬁnition 2.3.5. The 1-skeleton of an n-polytope P is the poset composed of its vertices and edges.

14 Figure 2.6: Examples of abstract polytopes which are not convex polytopes.

15 For example, the 1-skeleton of a cube consists of all eight vertices and twelve edges of the cube, but not the (−1)-face, the six squares, or the 3-face.

Lemma 2.3.1. For any polytope P , a sequence of successively incident proper faces can be found from any vertex or edge F , to any vertex or edge G, such that each member of the sequence is itself a vertex or edge (or, abusing notation, we might say that the skeleton of P is connected).

Proof. Since P is connected, we have a sequence of successively incident proper faces {F =

P0, ..., Pm = G} taking F to G. If no member of the sequence has rank at least 2, we are done. Suppose otherwise: choose a member Pi of the sequence with greatest rank among all members of the sequence (there may be several such members). Since the section Pi/F−1 is a polytope, it is connected, and since it has rank at least 2, we can ﬁnd a sequence of successively incident proper faces in Pi/F−1 taking Pi−1 to Pi+1; note that each member of this new sequence has rank less than that of Pi. Then we can take our original sequence of faces and remove Pi by replacing it with the new sequence we just found. We can continue doing this until every member of the sequence has rank at most 1, and thus produce the desired sequence.

Deﬁnition 2.3.6. The dual of an n-polytope P is the poset P ∗ with the same face set but reversed incidence structure. Thus each i-face of P = Fn/F−1 is an (n − i − 1)-face of ∗ P = F−1/Fn.

It is clear that the dual of a polytope is also a polytope (it is easy to see that the conditions for being a polytope still hold on the dual given that they hold for the original polytope), and the dual of a polytope lattice is also a polytope lattice. It is less clear, but still true, that the dual of a convex polytope is also a convex polytope. For example, the dual of the 3-cube is the octahedron: the eight vertices of the cube, which have triangular vertex ﬁgures, become the triangular faces of the octahedron; the twelve edges of the cube become the twelve edges of the octahedron; and the six square faces of the cube become the six vertices of the octahedron, which have square vertex ﬁgures. An easy way to visualize the dual of a relatively simple polyhedron is to put a new vertex in the middle of each of the old polytope’s facets, and connect two of these vertices with an edge whenever the two corresponding facets share an edge. Throughout this thesis, we will implicitly refer to dual polytopes whenever we reverse the order of two faces: i.e., if we have a section G/F of a polytope, then F/G refers to the dual of that section.

16 2.4 Graph Theory and Polytopes

Here, we present some basic graph theory that will be used later. The reader is referred to West (2001) for this material.

Deﬁnition 2.4.1. A graph is a pair (V,E) consisting of a set V whose members are called vertices, and a multiset E of unordered pairs of distinct vertices whose members are called edges. (Hence, multiple edges may contain the same two vertices.)

For example, the Hasse diagram of a square, shown in Figure 2.2, can be viewed as a graph: the vertices are the nodes corresponding to each face of the polytope, and the edges are the incidences between faces which diﬀer in rank by 1. Of course, viewing the Hasse diagram as a graph only, means that we no longer know about the rank of each vertex; graphs are independent of their graphical representation. Hence, the same graph corresponds to multiple Hasse diagrams; for example, dual polytopes have the same Hasse diagram.

Remark 2.4.1. Although this is not important for this thesis, an interesting point is the following: we can define a digraph, or directed graph, similarly to a graph, but instead of the edges being unordered pairs, they are ordered pairs; we can view each edge as ‘pointing’ out of one vertex and into another. Then we can convert a Hasse diagram to a digraph by setting the edges so that the first vertex in the ordered pair has lower rank than the second vertex. If we do this, we find that the resulting digraph has exactly one vertex with the property that all edges containing that vertex are pointing out of the vertex (i.e. the vertex corresponding to the (−1)-face), and one vertex with the property that all edges containing the vertex are pointing into the vertex (i.e. the vertex corresponding to the n-face); using this, it becomes clear that we can actually reconstruct the rank of each vertex from the digraph, and so the digraph completely represents the Hasse diagram and therefore the polytope itself.

Deﬁnition 2.4.2. A subgraph of a graph G = (V,E) is a graph G0 = (V 0,E0) so that V 0 ⊆ V and E0 ⊆ E (since G0 is a graph as well, it must be the case that every vertex in any edge of E0 is a member of V 0).

For example, if we take G as the graph corresponding to the Hasse diagram of the square, one possible subgraph G0 consists of all vertices of G corresponding to faces of rank at least 1, and all edges whose least-rank vertex has rank at least 1. This means that G0 is basically the top third of the Hasse diagram.

17 Deﬁnition 2.4.3. Given a graph G = (V,E) and a set V 0 ⊆ V , the induced subgraph of G (induced by V 0) is the subgraph consisting of all vertices of V 0 and all edges in E which join two vertices of V 0.

For example, the subgraph G0 described above is induced by the set of the ﬁve vertices of G corresponding to the faces of the square with ranks 1 and 2.

Deﬁnition 2.4.4. A path of a graph G is a sequence of vertices and edges of G of the form v1, e1, v2, e2, ..., ek−1, vk (k ≥ 1), such that all vertices (and therefore, edges) in the sequence are distinct and every edge in the sequence contains the vertices directly before and after it.

For example, if G is again the graph corresponding to the Hasse diagram of the square, each ﬂag of the square corresponds to two paths: one path moves from F−1 up through the

ﬂag to F2, and the other path moves from F2 down through the ﬂag to F−1.

Deﬁnition 2.4.5. A graph G = (V,E) is connected if for all pairs of vertices v1, v2 ∈ V , the vertices v1 and v2 are joined by a path in G; otherwise, it is disconnected.

Deﬁnition 2.4.6. A connected component of a graph G is a maximal, connected, induced subgraph of G.

Remark 2.4.2. Every graph can be partitioned into its connected components, i.e. no two connected components have a vertex in common and the union of (the vertex and edge sets, respectively, of) all connected components is (the vertex and edge set, respectively, of) the graph itself.

So far, all graphs we have looked at have been connected; however, if we take the graph corresponding to the Hasse diagram of the square and remove all edges between vertices corresponding to 0-faces and vertices corresponding to 1-faces, then the result is a disconnected graph with two connected components (one of which is exactly the graph G0 described above).

Deﬁnition 2.4.7. An edge cut of a connected graph G = (V,E) is a set E0 ⊆ E such that the subgraph G0 = (V,E \ E0) is disconnected.

For example, if G is the graph corresponding to the Hasse diagram of the square, then the set of edges joining a vertex corresponding to a 0-face to a vertex corresponding to a 1-face is an edge cut of G.

18 Deﬁnition 2.4.8. The edge graph of an abstract polytope P is the graph (possibly with multiple edges) whose vertices are the vertices of P and whose edges are the edges of P (note that this is indeed a graph).

For example, the edge graph of the 3-cube has eight vertices (i.e. the eight vertices of the cube) and twelve edges (i.e. the twelve edges of the cube).

Deﬁnition 2.4.9. A cycle (of length k) of a graph G is a sequence of vertices and edges of G of the form v1, e1, v2, e2, ..., ek, v1 (k ≥ 2), such that all vertices (and therefore, edges) in the sequence are distinct (except for the ﬁrst and last vertices, which are identical) and every edge in the sequence contains the vertices directly before and after it.

Proposition 2.4.3. Let G be a connected graph with a cycle. Let e be an edge of the cycle. Then {e} is not an edge cut of G.

Proof. Let G0 be the graph obtained by removing e from G. We show that given any two 0 0 vertices v1, v2 of G , we can ﬁnd a path from v1 to v2 in G , or equivalently, a path from v1 to v2 in G not passing through e. Since G is connected, we have a path from v1 to v2 in G. If this path does not pass through e, we are done. Suppose the path passes through e - let v3 be one vertex of e and v4 be the other vertex of e, so that the path starts at v1, eventually hits v3, then passes through e to v4, and eventually hits v2. Then we produce a new sequence of vertices and edges as follows: we start at v1 and move to v3 as in the old path. Then we move along the cycle to pass from v3 to v4 without passing through e (this is possible since there are always two exactly distinct paths from one vertex to another vertex in a cycle, one ‘clockwise’ and one ‘counter-clockwise’). Then we move from v4 to v2 as in the previous path. It is possible that this new sequence is not a path because vertices are repeated, e.g. some vertex v5 appears at one point in the sequence and then appears again later. If this happens, we shorten the sequence by removing everything after the ﬁrst occurrence of v5 but before the last occurrence of v5; we perform this process until there are no repeated vertices, and as a result, we have a path from v1 to v2 in G not passing 0 0 through e, i.e. a path from v1 to v2 in G . Hence, G is connected, and so {e} is not an edge cut of G.

Proposition 2.4.4. Let G be a connected graph and let e be an edge of G so that {e} is an edge cut of G. Then G0, the graph obtained by removing e from G, has exactly two connected components.

19 Proof. Suppose that G0 has more than two connected components (clearly it must have at 0 least two). Choose three vertices v1, v2, v3 of G so that they lie in three diﬀerent connected components. Then there is no path between any two of these vertices in G0. Since G is connected, any two of these vertices have a path between them in G. Since these paths are 0 not paths in G , each of these paths must pass through e in G. Let v4 and v5 be the vertices of e in G. Denote the path from v1 to v2 in G by P12, the path from v1 to v3 by P13, and the path from v2 to v3 by P23. Say P12 passes through v4 before v5.

Suppose P23 passes through v4 before v5. Then we can produce a new path by taking P12 from v1 to v4, and then taking P23 backwards from v4 to v2 (as in the previous proposition, we can truncate as needed to avoid repeated vertices); this path moves from v1 to v2 0 without passing through e, so v1 and v2 lie in the same connected component of G , giving us a contradiction.

Hence, P23 must pass through v5 before v4. Then we can produce a new path by taking

P12 from v1 to v4, and then taking P23 from v4 to v3; this path moves from v1 to v3 without 0 passing through e, so v1 and v3 lie in the same connected component of G , again giving us a contradiction. Therefore, we have a contradiction in every case, and so G0 must have exactly two connected components.

Corollary 2.4.5. Let G be a connected graph with a cycle, and let e1 and e2 be two distinct 0 edges of the cycle. Then if {e1, e2} forms an edge cut of G, the graph G = G\{e1, e2} :=

(V,E \{e1, e2}) has exactly two connected components.

Proof. Let H be the graph obtained by removing e1 from G. By Proposition 2.4.3, H is 0 0 connected. Note that G is the result of removing e2 from H. Then by Proposition 2.4.4, G has exactly two connected components.

20 Chapter 3

Digonal Contraction

In this chapter, we will introduce a new construction on abstract polytopes which lets us remove digons under certain conditions. We ﬁrst present some useful lemmas.

3.1 Preliminary Results for Digonal Contraction

Lemma 3.1.1. Suppose P is an n-polytope and E a non-empty set of facets of P such that, for every (n − 2)-face of P , the facets containing the (n − 2)-face either both belong to E or both do not belong to E. Then E is the set of facets of P .

Proof. Suppose there is a facet J not contained in E. Since P is flag-connected, we must have a sequence of pairwise-adjacent flags taking any flag containing J to any flag containing any E ∈ E. This means that we must at some point move from a flag with its facet not in E, to an (n − 1)-adjacent flag with its facet in E. Then those two flags share an (n − 2)-face, which (by the diamond condition) is contained in exactly one facet from E. However, this contradicts our assumption on E, so no such J can exist.

Lemma 3.1.2. Suppose we have a polytope P with an i-face F containing (i − 1)-faces H and H0 and a common corank-2 face of H and H0. If all common corank-2 faces R of H and H0 are such that the 1-sections H/R and H0/R of P have the same proper faces, then F contains no (i − 1)-faces other than H and H0.

Proof. Let R denote the set of all common corank-2 faces of H and H0. Suppose all faces R in R are such that H/R and H0/R have the same proper faces. Suppose that F contains an (i − 1)-face I distinct from H and H0.

21 Suppose there is an (i − 2)-face A contained in both I and either H or H0: WLOG let A < H. Then we cannot have A < H0, since this would violate the diamond condition between F and A. Examine the section H/F−1. Let E be the set of (i − 2)-faces in H/F−1 containing a face in R: clearly this is a nonempty set since R is nonempty and every face in R is contained in H. Thus every face E in E contains a face R in R. Since H > R and H0 > R and the two sections H/R and H0/R have the same proper faces, we have E < H0.

Now let B be any (i − 3)-face in H/F−1 with B < E for some E in E. Then we have B < H and B < H0, so B ∈ R, so both proper faces of H/B are in E. Hence, E is a nonempty set of (i − 2)-faces of the (i − 1)-polytope H/F−1 such that every (i − 3)-face B containing any face in E must contain another face in E, so by Lemma 3.1.1, A must also be in E and hence contain a face in R; however, this is impossible since we established that we do not have A < H0, so we have a contradiction. Hence, in P , there can be no (i − 2)-face A contained in both I and either H or H0. Then, since I here is just any (i − 1)-face distinct from H and H0, we see that in fact any (i − 2)-face A contained in one of H or H0 must also be contained in the other: by (P4) we know that F/A must have two proper faces, but if one proper face is H then the other proper face cannot be any face other than H0, and similarly with H and H0 interchanged. 0 Now consider the section F/F−1. Let H be the set of (i − 1)-faces {H,H }. We know that any (i − 2)-face contained in any face in H must be contained in another face in H, so by Lemma 3.1.1 applied to H, I must be in H; however, this is impossible since by assumption I is distinct from H and H0, so we have a contradiction. Hence, F contains no (i − 1)-faces other than H and H0.

Lemma 3.1.3. Suppose we have a polytope P with an i-face F containing (i − 1)-faces H and H0 and a common corank-2 face of H and H0. If all common corank-2 faces R of H and H0 are such that the 1-sections H/R and H0/R of P have the same proper faces, then each face contained in both H and H0 is also contained in a common corank-1 face of H and H0.

Proof. Suppose all common corank-2 faces R of H and H0 are such that the 1-sections H/R and H0/R have the same proper faces. Then by Lemma 3.1.2, F contains no (i − 1)-faces other than H and H0. Let A be any face such that A < H, H0. Then A < F and A has rank at most i − 2. Find an (i − 2)-face B (possibly equal to A) such that A ≤ B < F . By the diamond condition between F and B, there must be two (i − 1)-faces containing B and

22 contained in F ; by Lemma 3.1.2, these faces must be exactly H and H0. Hence, B < H, H0, so A is contained in B, which is a common corank-1 face of H and H0.

Proposition 3.1.4. Given a polytope P with a 1-section L whose greatest face, of rank i ≥ 2 (in P ), is called F and whose intermediate faces are called H and H0, we can produce a poset P 0 from P by removing F and every other i-face containing both H and H0 (we call the set of all such faces F, with F one of them), and combining H and H0 into a single face. In P 0, the combined face inherits all incidences of H and H0 in P (except for those involving the removed faces in F) while all other incidences are unchanged. In particular, P 0 can only be a polytope if for every corank-1 face G of the least face of L, the 2-section F/G in P is a digon (equivalently, every F 0/G is a digon for every F 0 in F). If P 0 is indeed a polytope, we call the operation of moving from P to P 0 a digonal contraction.

Proof. Suppose we have some corank-1 face G of the least face A of L such that F/G is not a digon. Then there exist faces B and C of P such that A and B lie between H and G, and A and C lie between H0 and G. We must have B 6= C since we would otherwise have a digon F/G. Then in P 0, we have three distinct faces A, B, and C between H and G, violating (P4), so P 0 is not a polytope.

Note that removing only some of the faces in F and not others will never result in P 0 being a polytope since (P4) will be violated between every non-removed face in F and the merged face H (there will be exactly one face between them in P 0).

3.2 The Digonal Contraction Theorem

Theorem 3.1 (Digonal Contraction Theorem). Suppose P is an n-polytope with a digon D whose greatest face is at rank i and is called F , and whose (i − 1)-faces are called H and H0, with the set of all i-faces containing both H and H0 called F (so F ∈ F). Suppose that we have produced a poset P 0 by performing digon contraction on D as in Proposition 3.1.4. Then P 0 is a polytope if and only if the following conditions hold in P : (1) No two of the i-faces in F are contained in a common (i + 1)-face. (2) All faces containing both H and H0 also contain some face in F. (3) All common corank-2 faces R of H and H0 are such that F 0/R is a digon for every F 0 in F. (4) For every (i + 2)-face I incident to both H and H0 in P , the rank-2 section I/H is

23 Figure 3.1: Digonal contraction on a cube-like polytope. connected in P 0. In particular, if P 0 is a polytope, it is rank-(n − 1) or rank-n, according as F is the greatest face of P or not.

Note that Condition (4) is a condition on P 0 and not P : this is for simplicity, as the corresponding equivalent condition on P is nontrivial to state. We examine Condition (4) in more detail after the proof of this theorem. An example of digonal contraction can be seen in Figure 3.1: on the left, we have a polyhedron P which is identical to a cube, except that it has a digon added in where one of the edges of the cube would be; on the right, we have the cube P 0, the result of the digonal contraction.

Proof. In the first part of this proof, we will show that for P 0 to be a polytope, all four conditions must hold. We show this by proving first that P 0 is not a polytope if (1) does not hold, then that P 0 is not a polytope if (1) holds but (2) does not hold, then that P 0 is not a polytope if (1) and (2) hold but (3) does not hold, and finally that P 0 is not a polytope if all of the conditions hold except (4). In the second part of the proof, we will show that the conditions are sufficient for P 0 to be a polytope, i.e. if all four of the conditions hold then P 0 is a polytope. First, suppose that (1) does not hold, i.e. some (i + 1)-face G of P contains more than 0 00 one face in F; say it contains both F and F . Examine the section G/F−1 of P , and

24 0 0 compare it to the section G/F−1 of P (if G does not contain F−1 in P then of course P 0 is not a polytope, so we assume that it does). We note that it is impossible for G in P 0 to contain the merged face H, since this would mean we have some i-face I in P 0 such that H < I < G, but then in P , we have I < G and either H < I or H0 < I (but not both, because otherwise I would have been removed), which violates the diamond condition between H and G or H0 and G in P 0, respectively, since we already have F 0 and 00 0 0 F between H and G and between H and G. Hence, in G/F−1 in P , the face H does not 0 occur. Therefore, we see that G/F−1 in P consists of the section G/F−1 of P with some 0 0 i-faces removed. However, this means that G/F−1 is not a polytope in P , so P itself is not a polytope: given any polytope, removing faces from it while preserving the rank will necessarily produce a poset that is not a polytope - by removing faces we remove flags, and by removing flags we necessarily deprive some (not removed) flag of its m-adjacent flag for some m, which breaks connectivity. Hence, (1) is necessary for P 0 to be a polytope. Now suppose that (1) holds but (2) does not hold, so in particular H and H0 are both contained in some face G (necessarily of rank at least i + 1) which does not contain any face in F. Let A be a common corank-1 face of H and H0. We may assume that every proper face of G/A in P contains at most one of H or H0 (otherwise, we can replace G by a proper face of G containing both H and H0; since none of these proper faces contain any of the faces from F, which are the only i-faces containing both H and H0 by definition, we will eventually find a face G of rank at least i + 1 containing both H and H0, with each of its proper faces containing at most one of them). Suppose G has rank i + 1: then the proper faces of G/H in P are both distinct from the proper faces of G/H0 since G contains no face from F, but then in P 0 there are four proper faces of the 1-section G/H, violating (P4). Hence, G must have rank at least i + 2. If P 0 is a polytope, then in the section G/H of P 0 we can find a sequence of proper incident faces taking us from any proper face to any other proper face (note that G/H is at least of rank 2) - in particular, we can take a corresponding sequence from a face of P containing H and not H0 to a face of P containing H0 and not H (in P 0, the faces H and H0 are identified, so all faces containing the merged face H in P 0 contain H or H0, or both, in P ). Since G/H is a section of G/A in P 0, every proper face of G/H in P 0 must contain exactly one of H or H0 in P (we established earlier that every proper face of G/A in P contains at most one of H or H0). But if we have a sequence of incident proper faces of G/H in P 0 taking us from a face containing only H to a face containing only H0, then at some point we must pass through a face containing both:

25 if some face B contains H and some face C contains H0, and the two faces are incident (say B contains C), then one of them must contain both H and H0 (so B would contain both). This gives us a contradiction. Hence, P 0 is not a polytope. Therefore, (2) is necessary for P 0 to be a polytope. Now suppose that (1) and (2) hold but (3) does not hold, i.e. we have a common corank- 2 face R of H and H0 such that F 0/R is not a digon for some F 0 in F. Then in the 2-section F 0/R of P , the subsections H/R and H0/R do not share both of their proper faces with each other, so we have at least three distinct faces between H and R in P 0, violating the diamond condition. Hence, (3) is necessary for P 0 to be a polytope. Finally, (4) is clearly necessary for P 0 to be a polytope: if we have a disconnected section I/H in P 0, then (P3) does not hold for P 0, and so P 0 is not a polytope. Thus, all four conditions are necessary for P 0 to be a polytope; it remains to show that the conditions are suﬃcient. Also note that (3) implies, by Lemma 3.1.2, that each face in F contains only two (i − 1)-faces, namely H and H0. We now check the four deﬁning conditions of a polytope on P 0 given the four conditions of this theorem. (P1): We see that P 0 has exactly one least face, i.e. the least face of P . This is because we can still trace a path of incident faces descending in rank from any face of P 0 to the least face of P since any (i + 1)-face of P containing any face in F has at least one other corank-1 face by the diamond condition, and this face cannot be in F by (1). Suppose F is the greatest face of P : then P 0 has exactly one greatest face, i.e. the merged face H, since now P only has two rank-(n − 1) faces. Suppose otherwise: then by the diamond condition at rank i and by (1), any (i − 1)-face contained in a face in F is also contained in at least one face not in F, so since (P1) holds for P , it also holds for P 0, with the same greatest face. (P2): Suppose F is the greatest face of P : then since P 0 has exactly one greatest face, i.e. the merged face H, it has property (P2) and is of rank n − 1 since P has property (P2) and is of rank n. Suppose otherwise: then every (i + 1)-face of P containing a face from F has at least one corank-1 face not in F by the diamond condition and (1), and in P every corank-1 face of every face in F in P is contained in at least one i-face not in F, again by the diamond condition and (1), so since P has property (P2) and is of rank n, P 0 also has property (P2) and is of rank n. (P4): Consider a pair of faces A and B in P 0 such that A < B and rank(A) + 2 =

26 rank(B). The arguments of the proof depend on rank(A). Suppose rank(A) > i: clearly A and B have exactly two faces between them in P 0, since no faces of rank greater than i have been changed via our operation. Then suppose rank(A) = i: then A and B still have exactly two faces between them, since the only way the i-faces changed via our operation was in the removal of the faces in F, so A must correspond to some face in P which is not in F, so we have the diamond condition for this case as well. Then suppose rank(A) = i − 1. If A does not lie in a face of F in P , then A is not in D and the set of i-faces containing A in P 0 is the same as the set of i-faces containing A in P , so we again have the diamond condition. If A < F 0 in P for some F 0 in F, then (3) with Lemma 3.1.2 implies that A = H or A = H0. Suppose in P the (i + 1)-face B contains some face of F: then by (1) we have a unique i-face FH ∈/ F such that H < FH < B, 0 as well as a unique face FH0 ∈/ F such that H < FH0 < B; then, FH 6= FH0 , since the only faces having both H and H0 as corank-1 faces are in F. Then in P 0, the faces H and 0 0 H are identified, and so we find that FH and FH0 are exactly the faces of P between A and B, so the diamond condition is satisfied in this case. Now if B does not contain any face from F in P , and A = H (say), then by the diamond condition we have exactly two i-faces between H and B in P , and neither can contain H0 since the faces in F are the only i-faces containing both H and H0. Moreover, by (2), B cannot contain H0, so in P 0 there are exactly two faces between A and B, giving us the diamond condition. The argument for A = H0 is similar. This covers all cases for rank(A) = i − 1. Then suppose rank(A) = i − 2: first suppose A is in D. If both faces between A and B in P are outside of D, then clearly there are still exactly two faces between A and B in P 0. If exactly one face between A and B in P is in D, then we clearly still have exactly two faces between A and B in P 0 (say we have A < H < B in P ; we also know A < H0, so we cannot have H0 < B in P , so in P 0 the section B/A is unchanged). Finally, it is impossible that both faces between A and B in P are in D, since the only i-faces containing both H and H0 are in F and so we would have B ∈ F, but this is impossible since the faces in F are not in P 0. Now suppose A is not in D. Again, by similar arguments, we see that there are exactly two faces between A and B in P 0, and so the diamond condition is satisfied in this case. Next, suppose rank(A) = i − 3: the only case we need consider is when B is in D (i.e. B = H or B = H0), since otherwise the relevant faces and incidences are identical for P

27 and P 0. WLOG say B = H. If A is not incident to H0, then clearly the faces in B/A in P 0 are exactly the faces in B/A in P , so the diamond condition holds. If A is incident to H0, it is contained in both H and H0, and so by (3) we find that the two faces between A and H in P are the same as the two faces between A and H0 in P : therefore, we have exactly two faces between A and B in P 0, so again the diamond conditon holds. Finally, if rank(A) < i − 3, then clearly the diamond condition holds in P 0 since it holds in P and no relevant faces or incidences are modified. Therefore, the diamond condition holds for all cases. 0 (P3): We wish to show that P is strongly connected. We will first show that H/F−1 is 0 0 strongly flag-connected in P . Consider a section of H/F−1 in P : if this section does not contain H, it is identical to a section of P , and is therefore a polytope and hence strongly connected. If the section does contain H, call it H/A and say we want to get from some proper face X to some proper face Y of H/A in P 0 via a sequence of proper successively incident faces. If in P both X and Y are contained in H (or both X and Y are contained in H0, respectively), then since H/A (or H0/A, respectively) is a polytope in P , we can get a sequence of proper successively incident faces of H/A in P 0 taking X to Y . If this is not the case, say in P , the face X is only contained in H and Y is only contained in H0. Then by (3) and Lemma 3.1.3, we have some common corank-1 face B of H and H0 such that A < B. Then we can find a sequence of proper successively incident faces taking X to B in H/A in P , and we can then find a sequence of proper successively incident faces taking B to Y in H0/A in P ; concatenating these two sequences, we get a sequence of proper successively 0 0 incident faces taking X to Y in H/A in P . Hence, H/F−1 is strongly connected in P , and is therefore strongly flag-connected. Now, we will show that given any face C > H in P 0, the section C/H is connected. Suppose in P the face C contains only one of the faces H or H0, say H (it must contain at least one of the faces H or H0 in P since C > H in P 0). Then C/H in P 0 is identical to C/H in P , so clearly it is connected; we can make a similar argument if C > H0 in P . Now suppose otherwise, i.e. that C is incident to both H and H0 in P 0. If rk(C) < i + 2, then C/H is trivially connected in P 0, and if rk(C) = i + 2, then C/H is connected in P 0 by (4). Then suppose rk(C) > i + 2; let rk(C) = k. It is sufficient to show that the 1-skeleton of H/C is connected in (P 0)∗, the dual of P 0: if this is the case, then we can move between any two proper faces X and Y of C/H in P 0 by moving from X up to a (k−1)-face, then moving along the 1-skeleton of H/C (i.e. along the (k − 1)-faces and (k − 2)-faces of C/H in P 0) to

28 a (k − 1)-face containing Y , and then moving down to Y . To this end, we determine which faces are in the 1-skeleton of H/C in (P 0)∗. Let E be a face in the 1-skeleton of H/C in P ∗, i.e. H < E < C in P and rk(E) = k − 2 or rk(E) = k − 1. Since k > i + 2, we know that rk(E) > i. Therefore, E/∈ F, and so E occurs in P 0. Moreover, by (2), E contains some face F 0 ∈ F. Let J be an (i + 1)-face in P such that F 0 < J ≤ E. By (P4) on the 1-section J/H, we have an (i + 1)-face F¯ 6= F such that H < F¯ < J, and by (1), F/¯ ∈ F (if we had F¯ ∈ F, then J would contain two i-faces in F, violating (1)). Therefore, in P 0, E ≥ J > F¯ > H, and so in particular, E is in the 1-skeleton of H/C in (P 0)∗. Hence, every face in the 1-skeleton of H/C in P ∗ also occurs in the 1-skeleton of H/C in (P 0)∗, and we see by similar reasoning that every face in the 1-skeleton of H0/C in P ∗ also occurs in the 1-skeleton of H/C in (P 0)∗. Since the 1-skeletons of H0/C and H/C in P ∗ share faces (since C is incident to both H and H0 in P , it contains faces in F, and hence contains (k − 1)-faces and (k − 2)-faces incident to faces in F, which are incident to both H and H0) and digonal contraction does not affect the incidences between (k−1)-faces and (k−2)-faces for k > i + 2, we find that the 1-skeleton of H/C in (P 0)∗ is connected, and hence, C/H is connected in P 0 (and therefore flag-connected). We will now proceed by showing that P 0 is strongly flag-connected (and therefore strongly connected). In order to do this, we choose an arbitrary section S0 = B/A of P 0 and demonstrate that it is flag-connected. Since every face of P 0 corresponds to at least one face of P , we can examine the corresponding section S = B/A of P . Examine two flags Φ and Ψ 0 of S : they correspond to two flags Φ and Ψ of S, so we have a sequence Φ = Φ0, ..., Φr = Ψ of successively adjacent flags of S taking Φ to Ψ in S. If this sequence does not contain a flag that passes through any of the faces in F, we can produce a corresponding sequence of successively adjacent flags in S0, as desired (since all the relevant flags will still exist in P 0, up to the renaming of the face H0 as H, if it occurs in any of the flags). Suppose the sequence contains a flag that passes through a face in F. If the maximum face of P is an 0 element of F (and hence F = {F }), then we are done, since we then have P = H/F−1, which we already showed is strongly flag-connected. Suppose that this is not the case: then 0 let Φj be the first flag containing any member of F in the sequence - call this member F . 0 Then a number of subsequent flags will all contain F until we reach some flag Φk again not 0 containing F - note that Φk cannot contain any of the faces in F because no two of these faces are contained in a common (i + 1)-face, by (1). 0 We now find a sequence of successively adjacent flags taking Φj−1 to Φk in S . Any flag

29 of S containing F 0 must also contain H or H0: Condition (3) with Lemma 3.1.2 tells us that F 0 has only H and H0 as corank-1 faces, and of course we cannot have A = F 0 since 0 0 0 A is a face of P , whereas F does not occur in P . Therefore, Φj−1 and Φk must each also contain H or H0, since they are i-adjacent to flags containing F 0. Hence, in S0, the 0 flags Φj−1 and Φk both contain H (the combination of H and H from P ). Since H/F−1 is strongly flag-connected in P 0 (as established earlier), we know that H/A is flag-connected 0 0 in P . Hence, we can move Φj−1 to a flag Φ identical to Φj−1 for all faces containing H, and identical to Φk for all faces contained in H. Since we know that B/H is flag-connected 0 0 in P (as established earlier), we can move Φ to Φk, since these two flags coincide in all 0 faces contained in H. Hence, we can move Φj−1 to Φk in S . This means that we can take 0 Φ = Φ0 to Φk in S . Now, we can continue taking the original sequence of successively adjacent flags in S from Φk onwards to get from Φ to Ψ; if we again pass through a face in F, we can again find an alternative sequence of subsequently adjacent flags, as we just did from Φj−1 to Φk. Hence, we can modify the sequence of subsequently adjacent flags from Φ to Ψ in S to get a sequence of subsequently adjacent flags from Φ to Ψ in S0, and so S0 is flag-connected. Since S0 is an arbitrary section of P 0, we see that every section of P 0 is flag-connected, and so P 0 is strongly flag-connected, and therefore strongly connected. Therefore, since we have demonstrated all four conditions (P1) through (P4) on P 0, we see that P 0 is a polytope.

In Figure 3.2, we see four polytopes with digons that cannot be contracted. In particular, the first polytope violates Condition (1) of Theorem 3.1 while satisfying (2) and (3), the second polytope violates Condition (2) of Theorem 3.1 while satisfying (1) and (3), and the third polytope violates Condition (3) of Theorem 3.1 while satisfying (1) and (2). The combinatorial structure of these examples is too tight to allow the contraction of the specified digons. The fourth polytope violates Condition (4) of Theorem 3.1 while satisfying the other three conditions. We do not examine Condition (4) on the first three polytopes: we do not know whether (4) actually implies some of the other conditions, but even if it were the case that (4) implied another condition, we would want to include all four conditions in the theorem, since (4) is a condition on P 0 with a complicated corresponding condition on P which we would like to keep as simple as possible. The first polytope is a “dihedron over a digon”. This is a polyhedron consisting of two digonal faces sharing both edges and both vertices of the polyhedron. Both digonal faces are in F (so i = 2), and they are both contained in the (i + 1)-face P , violating Condition (1)

30 Figure 3.2: Four non-contractible digons.

31 of Theorem 3.1. The diagram showing the second polytope does not actually show the polytope itself, but a related polytope. To get the polytope that we are actually interested in, we increase the rank of every face of the displayed polytope (which is a polyhedron made up of two triangles and a square) by 1, and then have two vertices contained in every higher-rank face. The resulting polytope is then rank-4 and consists of two vertices, four edges (corresponding to the four vertices of the displayed polytope), two of which are H and H0, five digons (corresponding to the five edges of the displayed polytope), one of which is F , and three polyhedra (corresponding to the triangles and square of the displayed polytope). We see that the polyhedron corresponding to the square of the displayed polytope contains both H and H0 but does not contain F , violating Condition (2) of Theorem 3.1. The third polytope is a “pyramid over a digon”. The apex of the pyramid, shown on the left, is contained in two triangles, which are H and H0. Then F is the polytope itself. We see that Condition (3) of Theorem 3.1 is violated since, taking R as either of the vertices of the polyhedron other than the apex, the vertex figure of R is a triangle rather than a digon. For the fourth polytope, as with the second, the diagram does not actually show the polytope itself. Again, to get the polytope P that we are actually interested in, we increase the rank of every face of the displayed polytope (which is a polyhedron made up of four triangles, two in the front containing F and two in the back containing F 0) by 1, and then have two vertices contained in every higher-rank face. The resulting polytope is then rank-4. Figure 3.3 demonstrates that Condition (4) of Theorem 3.1 is in fact violated by this polytope, with I = P . We denote the four facets of P corresponding to the four triangles in the diagram as follows: I corresponds to the triangle XWF , II to YZF , III to XWF 0, and IV to YZF 0. The first two diagrams in Figure 3.3 show the sections P/H and P/H0. The result of digonal contraction is shown in the third diagram in Figure 3.3: F and F 0 have been removed, and the remaining faces and their incidences have been adjusted accordingly. As we can see, P 0/H is not connected, hence violating (4). If the original polytope was ‘twisted’ in such a way that the order of the faces in P/H was the same but the faces I and II were swapped in P/H0 (so that I contained Z and II contained W ), then P 0/H would be a square, and hence connected. Hence, we see the complexity of Condition (4) in P : in order to see whether Condition (4) is satisfied in P , we need to examine the incidences between the faces in F

32 Figure 3.3: An example of Condition (4) breaking for digonal contraction.

33 and the faces in I/H and I/H0. However, there exist simple conditions on P which are suﬃcient for Condition (4) to be satisﬁed.

Remark 3.2.1. Suppose that for a polytope P as in Theorem 3.1, we have F = {F } and for every (i + 2)-face I in P , one of the polygonal sections I/H and I/H0 is ﬁnite. Then Condition (4) is satisﬁed.

We see why this is true in Figure 3.4. Here, we note that the (i + 1)-faces A and B coincide in I/H and in I/H0, since both faces A and B contain F , which contains H and H0. It is possible that x and y coincide, or that z and w coincide, but this does not meaningfully affect the result. If both polygonal sections I/H and I/H0 in P are infinite, then we see that the resulting section I/H in P 0 cannot be connected, since there is no way to get from A to B. Similarly, we see that the following condition is also sufficient for Condition (4) to be satisfied:

Remark 3.2.2. Suppose that for a polytope P as in Theroem 3.1, every (i + 2)-face I in P incident to both H and H0 is such that I contains exactly one face from F, and one of the polygonal sections I/H and I/H0 is ﬁnite. Then Condition (4) is satisﬁed.

Another important note about Theorem 3.1 is that it extends naturally to the question of contracting an edge. Suppose P is an n-polytope with two vertices H and H0 joined by an edge F , and denote by F the set of all edges containing both H and H0. Suppose that we have produced a poset Pˆ by combining H and H0 and removing the faces in F, as in digonal contraction, although of course F is not a digon but an edge. Then Pˆ is a polytope if and only if the conditions of Theorem 3.1 hold for P (note that Condition (3) holds trivially since H and H0 have no corank-2 faces); this follows from the proof of Theorem 3.1. In particular, if P is a polytope lattice, then the edge F is contractible (i.e., Pˆ is a polytope) if and only if for every facet I in P containing both vertices H and H0, one of the polygonal sections I/H and I/H0 is ﬁnite. Note that this is always the case if P is in fact a convex polytope.

3.3 Reversing Digonal Contraction

A natural question is if, and how, we may perform the opposite operation to digonal contraction, which we might call digonal insertion. Whereas digonal contraction replaces a

34 Figure 3.4: Merging two polygons with digonal contraction.

35 digon with an edge, digonal insertion would replace an edge with a digon. To formalize this, let P 0 be an n-polytope with an (i − 1)-face H (for i ≤ n). We wish to produce a poset P by replacing H by two copies of it, H and H0, which are both contained in a set of new i-faces F. Faces of P 0 other than H and H0 are copied over to P , and their incidences between each other are preserved, but we must determine the number of faces in F, their incidences, and the incidences of the faces H and H0, with the goal of P being a polytope so that performing digonal contraction on P produces again P 0. It is not clear if there is any simple way of deciding how to set the number of faces in F and the incidences of the faces in F, H, and H0 so that P is a polytope and performing digonal contraction on P produces P 0, or moreover if there is any simple condition on P 0 which tells us whether it is even possible to produce such a polytope P . The following considerations highlight some of the diﬃculties. First, we see that the new i-faces in F each should only have two corank-1 faces (i.e. H and H0): since our goal is to reverse digonal contraction, the poset P should have the properties required for digonal contraction, and in particular each face in F can only have two corank-1 faces, by Lemma 3.1.2 with Condition (3) for digonal contraction. Moreover, the faces H and H0 must have identical lower-rank incidences: given any corank-1 face I of H, (P4) for F/I (for any F ∈ F) tells us that I must be incident to another (i − 1)-face of P , but we know that only one other (i − 1)-face is incident to F , i.e. H0, so I is also a corank-1 face of H0; we can make a similar argument assuming ﬁrst that I is instead a corank-1 face of H0. Hence, for the new faces we have produced (i.e. H, H0, and the faces in F), it is clear how to set their lower-rank incidences. The problem is in setting the higher-rank incidences so that P is a polytope. In P 0, we have a set of i-faces containing H, and in P , for each such i-face we need to decide whether it is incident to H or to H0 (it cannot be incident to both: any i-face containing both H and H0 in P is a member of the set F and is removed in P 0). Clearly, we cannot make this decision completely arbitrarily: if we view the 1-skeleton of P 0/H as a graph, we need to assign each vertex to H or to H0 in P , i.e. to partition the vertices of the graph into two sets, so that each corresponding subgraph is nonempty and connected. Then the edges of the graph, corresponding to (i + 1)-faces in P 0 containing H, have obvious lower-rank incidences in P : if such an edge contains two vertices both assigned to H, then the corresponding (i + 1)-face in P must be incident to H; if such an edge contains two vertices both assigned to H0, then the corresponding (i + 1)-face in P must be incident

36 to H0; and if such an edge contains one vertex assigned to H and one vertex assigned to H0, then the corresponding (i + 1)-face in P must be incident to both H and H0, and therefore must be incident to exactly one of the faces in F. However, this still leaves us, in general, many diﬀerent ways to assign the vertices of the graph to H or H0 in P , and also many ways to decide on how many faces are in F and which edges of the graph they are incident to. It is not clear if there is any simple condition telling us which choices give us an actual polytope for P . Hence, it seems as though reversing digonal contraction is more complicated than performing digonal contraction.

37 Chapter 4

The Helfand Construction

Helfand (2013, Section 2.2) gives us a rank-preserving construction that we can apply to any (abstract) polytope of rank n ≥ 3 to get another n-polytope. This construction has useful connections to the digonal contraction and to the (not yet deﬁned) polygonal contraction, and so we present it here.

4.1 The Global and Local Helfand Constructions

Let P be an n-polytope. For Helfand’s construction, we choose a rank k with 0 ≤ k ≤ n − 1 and replace (combinatorially) every k-face F of P with a number of new k-faces, one for each (k +1)-face containing F , and add an i-face for i > k for every incident pair of a k-face F of P and an (i + 1)-face G of P . All faces of P with rank other than k are retained, as well as all incidences between faces with rank other than k. The new k-faces each contain the same proper faces as their corresponding original faces did, and each is contained in the original (k + 1)-face it corresponds to. Every new i-face for i > k is contained in its corresponding original (i + 1)-face G, and contains the new faces corresponding to the original faces above F contained in G. We call this process the global Helfand construction

(the resulting polytope is called [P ]k, the k-bubble of P ).

In Figure 4.1, we have an example of what happens when the global Helfand construction is applied to the cube. The first figure is the cube itself. If k = 0, we get the second figure, known as the truncated cube (in general, applying the Helfand construction with k = 0 to any convex polyhedron produces a convex polyhedron

38 obtained by truncating the vertices of the original polyhedron). Here, we see that each vertex (i.e. each k-face) is replaced with three new vertices, corresponding to the three edges of the original cube containing the vertex. Additionally, we have three new edges for each vertex, corresponding to the three squares of the original cube containing the vertex. Finally, we have a new triangle for each vertex, corresponding to the incidence between the vertex and the cube itself. If k = 1, we get the third ﬁgure, in which every edge of the cube is replaced by a digon. This occurs because every edge is contained in exactly two squares in the cube, and so each edge of the cube is replaced with two new edges. Additionally, we have a new 2-face for each edge (the digon), corresponding to the incidence between the edge and the cube itself. If k = 2, the construction can be applied, but we just get the same cube back. We produce a local version of this construction, again combinatorially: instead of removing every k-face of P , we only remove one k-face F of P , and replace it with a collection of new k-faces, one for each (k + 1)-face of P containing F . We also add an i-face for i > k for every incident pair (F,G) of P with G an (i + 1)-face of P . All faces other than F are retained, as are all incidences between faces other than F . The new k-faces each contain the same proper faces as F , and each is contained in the (k +1)-face containing F from which it was derived. Every new i-face for i > k is contained in its corresponding original (i+1)-face G, and contains the new faces corresponding to the original faces above F contained in G. We call this operation the local Helfand construction (at level k), or for short, the Helfand construction. It is easy to see the eﬀect of the local Helfand construction on the cube: looking at Figure 4.1, the case k = 0 would have us truncate a single vertex of the cube instead of every vertex, and the case k = 1 would have us replacing a single edge by a digon instead of every edge. Note that the global Helfand construction is the same as applying the local Helfand construction in turn to every (original) k-face of a polytope.

Theorem 4.1. The (local) Helfand construction always produces a polytope.

Proof. Say we apply the local Helfand construction to a k-face F of an n-polytope P to produce an object P 0. In P 0, we classify the faces as ‘old’ or ‘new’ depending on whether those faces exist in P or were produced by the construction, respectively. Each new i-face corresponds to an old (i + 1)-face containing F ; for each old (i + 1)-face G, we call the new i-face Gˆ. We now check whether P 0 satisﬁes each of the four deﬁning conditions for being a polytope.

39 Figure 4.1: Two diﬀerent applications of the global Helfand construction to the 3-cube.

40 We first see that (P1) and (P2) hold in P 0: each new i-face Gˆ is incident to the (i + 1)- face G above, and to at least one (i − 1)-face (for i > k, the (i − 1)-faces of Gˆ are incident to the new faces corresponding to the corank-1 faces of G above F ; for i = k, the (i − 1)-faces of Gˆ are incident to the corank-1 faces of F in P ). Hence, we can see that the ranks still behave as expected and the conditions (P1) and (P2) hold. We now examine (P4). Consider two faces B > A in P 0 with rk(B) = rk(A)+2. Several cases need to be considered. First suppose A and B are both old faces. Then B/A in P 0 is identical to B/A in P unless rk(B) = k + 1 and F lies in B/A: in this case, B > F > A in P and we see that B/A in P 0 has the same structure as in P , but F is replaced by Bˆ. Hence, (P4) holds in this case. Now suppose B is new, B = Cˆ (say), and A is old. The only situation in which a new i-face contains an old (i − 1)-face occurs when i = k, so we must have either rk(B) = k or rk(B) = k + 1. In the first case, B/A is identical in P 0 to F/A in P (except for the name of the k-face); in the second case, we know that in P we have exactly two faces H and H0 between C and F ; then we have exactly two faces Hˆ and Hˆ0 between Cˆ = B and A in P 0 (since F < B in P ). Hence, (P4) also holds in this case. Now suppose B is old and A = Dˆ is new. We then see that we must have exactly the faces Bˆ and D between B and A in P 0, so (P4) holds in this case. Finally, if B = Cˆ and A = Dˆ are both new, then (P4) must hold because the section B/A is identical in structure to the section C/D. Hence, in all cases, (P4) holds. It remains to show (P3). We consider a section B/A in P 0 of rank at least 2, and wish to show that it is connected; it suffices to show that we can move from any proper face G to any proper face H. Suppose G and H are both old faces in B/A. If B is a new face, then G and H must both be contained in F in P (since the only old faces contained in new faces of P 0 are faces that were contained in F in P ): in this case, we can take the same sequence of proper faces to get from G to H in B/A in P 0 as we would take in F/A in P . If B is an old face and A = Cˆ is a new face, all old faces in B/A = B/Cˆ are also in B/C since C is the only old face having Cˆ as a corank-1 face. Then we can take the same sequence of (old) faces as in B/C in P to get from G to H in B/A. If A and B are both old, either there are no new faces in B/A (in which case it is identical to B/A in P and is hence connected) or we have B > F > A in P : in this case, we take the sequence of (old) faces from G to H

41 in B/A in P , and if it passes through F , we need to instead use the new k-faces that have replaced F . Denote by I the face directly before F in the sequence in P , and denote by J the face directly after F in the sequence. If both I and J are (k − 1)-faces, then we can easily get from I to J in B/A in P 0 by using any arbitrary k-face that replaced F instead of F itself, which has been removed: every new k-face in P 0 has identical lower-rank incidences to F . If one of the faces I or J is a (k + 1)-face and the other is a (k − 1)-face - say I is a (k + 1)-face and J is a (k − 1)-face - then we simply replace F with Iˆ in the sequence in P 0, since I > Iˆ > J in P 0. If both I and J are (k + 1)-faces, then in B/A in P 0 we can move from I to Iˆ and from Jˆ to J, so it suffices to move from Iˆ to Jˆ. If rk(A) < k − 1, then we can do this by moving to any (k − 1)-face of Iˆ, since Iˆ and Jˆ have identical lower-rank incidences. Otherwise, we must have rk(A) = k − 1. If rk(B) > k + 2, then B/F has rank at least 2 in P , and so we have a sequence of proper faces taking I to J in B/F in P : replacing each face in the sequence by its corresponding new face in P 0, we have a sequence of proper faces taking Iˆ to Jˆ in B/A in P 0, as desired. Otherwise, rk(B) ≤ k + 2: if rk(B) < k + 2 then B/A is trivially connected, so consider rk(B) = k + 2, so B/A is a 2-section. Then in B/A in P 0, we can move from Iˆ to Jˆ by passing through Bˆ: since the (k + 2)-face B is incident to the (k + 1)-faces I and J in P , the (k + 1)-face Bˆ is incident to the k-faces I and J in P 0. Hence, in all cases, we can move from Iˆ to Jˆ in B/A in Pˆ, so we are done. Thus, in all cases, we can move from any old face to any old face in B/A in P 0. Now suppose G = Cˆ and H = Dˆ are both new faces. We first find an appropriate corresponding section in P to B/A in P 0: if A and/or B are new faces, we take their corresponding old faces in P (if either face is old, we use it in P 0). Then in this section in P , the faces C and D are proper and we have a sequence of proper faces moving from C to D. Then in B/A in P 0, we can take a similar sequence from G to H, replacing every old face with its corresponding new face if it corresponds to a new face. Hence, in B/A in P 0 we can move from any new face G to any new face H. It remains to show that in B/A in P 0 we can move from some old face G to some new face H = Dˆ. Suppose we cannot: then the previous discussion implies that the set of proper faces of B/A is composed of exactly two connected components, one entirely of old faces and one entirely of new faces. This means that for every proper new face in B/A, its corresponding old face is not in B/A.

42 Suppose A = Eˆ is a new face. Let rk(A) = j; then there must be a rank-(j + 1) old face in B/A (otherwise any ﬂag containing a proper old face in B/A would also contain a proper new (j + 1)-face, thus giving one connected component), and this face can only be E. Then take a proper (j + 2)-face K > E in B/A and examine the section K/A; by (P4), we must have a second (j + 1)-face in this section. This must be a new face (the only old (j + 1)-face containing A is E), and hence can only be Kˆ ; hence, we have two incident proper faces in B/A, one new, Kˆ , and one old, K, giving us a contradiction. Then suppose A is an old face. Then we must have rk(A) = k − 1, since this is the only rank j in P 0 with old j-faces contained in new (j + 1)-faces. Suppose B is also an old face. Then let L be a rank-(k + 1) face of B/A in P containing F : then L must also be in B/A in P 0, as well as Lˆ, so we again have two incident proper faces, one new and one old, giving us a contradiction. Now suppose B is a new face. Let rk(B) = i. Since B/A contains old proper faces, it contains old faces of rank at least k. Every face containing an old face of rank at least k must also be old, so B must in face be an old face, and hence we again have a contradiction. Hence, in all cases, we can move from some old proper face in B/A to some new proper face. This means that, with the previous results, we can move from any proper face in B/A to any proper face, so all sections B/A in P 0 are connected, so we have (P3). Therefore, P 0 is a polytope.

4.2 Connections to Digonal Contraction

We now note that the (local) Helfand construction, when applied to a k-face F of an n- polytope P for 1 ≤ k ≤ n − 2, produces digons (as rank-2 sections) with greatest faces of rank k + 1. We see this illustrated in the Hasse diagrams in Figure 4.2. A digon I/Gˆ is produced for every incident pair of faces (G, I) of P such that G is a corank-2 face of F and F is a corank-2 face of I. This follows from the fact that there are two (k + 1)-faces J and J 0 between F and I. After the construction, we get a new k-face Jˆ corresponding to J and a new k-face Jˆ0 corresponding to J 0. Both of these faces contain the (k − 2)-face G and the two (k − 1)-faces between F and G in the original polytope, and both are contained in the new (k + 1)-face Iˆ corresponding to I. Hence, we get a digon I/Gˆ in Pˆ. We can now ask when a single such digon can be removed via digon contraction.

Theorem 4.2. Let P be an n-polytope with a k-face F , let 1 ≤ k ≤ n − 2, and let Pˆ

43 Figure 4.2: Producing a digon via the Helfand construction. be the n-polytope obtained from P by applying the Helfand construction at F . Let G be a (k − 2)-face of P and I be a (k + 2)-face of P with G < F < I. Let J and J 0 be the two proper (k+1)-faces of the section I/F in P . Let I be the set of (k+2)-faces in P containing both J and J 0 (so I ∈ I). Then the digon I/Gˆ in Pˆ is contractible (in the sense that the resulting structure is an n-polytope) if and only if the following conditions hold in P : (1) No two faces in I are contained in a common (k + 3)-face. (2) Every face containing both J and J 0 also contains a face in I. (3) Denote by P 0 the poset obtained by combining the faces J and J 0 in P as in digonal contraction, removing every face in I, and leaving all other faces and incidences of P unchanged. Then for every (k + 4)-face L containing both J and J 0 in P , the rank-2 section L/J is connected in P 0.

In Figure 4.3, we see an example of such a process: we start with the cubic tessellation of 3-space (this is a 4-polytope; in the diagram, only the neighborhood of one edge is shown - the edge is the intersection of four cubes), apply the Helfand construction to an edge (replacing the edge with a 3-polytope consisting of four digons sharing the two vertices and joined edge-to-edge to form the tessellation {2, 4} of the 2-sphere), then contract one of the digons in the result (the new 3-polytope now has three digons sharing the two vertices joined edge-to-edge to form the tessellation {2, 3} of the sphere).

Proof. We ﬁrst examine the incidence structure of Pˆ. An old face C in P/F (a face that

44 Figure 4.3: Producing and contracting digons with the Helfand construction.

45 exists in P as well) contains exactly (1) the old faces in P that C contains, and (2) the new faces (i.e. those newly-introduced in Pˆ, each corresponding to an old face in P/F ) corresponding to either C itself or a face in P/F contained in C. A new face C = Dˆ contains exactly the new faces corresponding to the faces in P/F contained in D, as well as all faces properly contained in F in P . We note that in Pˆ, the digon we are considering contracting is I/Gˆ , which means that the faces Jˆ and Jˆ0 correspond to the faces H and H0 in the digonal contraction theorem, Theorem 3.1, (since they are the edges of the digon) and the set Iˆ := {Iˆ0|I0 ∈ I} corresponds to the set F in the digonal contraction theorem (since they are exactly the faces having Jˆ and Jˆ0 as common corank-1 faces). We ﬁrst show that if any of the conditions for the present theorem do not hold, then the digon I/Gˆ is not contractible in Pˆ. First, suppose that (1) of the present theorem does not hold, i.e. some (k + 2)-faces

I1,I2 ∈ I are contained in a common (k + 3)-face C in P . Then in Pˆ, we have Iˆ1, Iˆ2 ≤ Cˆ, so we violate condition (1) of Theorem 3.1 (with i = k + 1). Hence, condition (1) of the present theorem is necessary for contractibility. Next, suppose that (2) of the present theorem does not hold, i.e. we have some face K in P containing J and J 0 that does not contain any face in I. Then in Pˆ, we have J,ˆ Jˆ0 ≤ Kˆ , but none of the faces in Iˆ are contained in Kˆ . Hence, we violate condition (2) of Theorem 3.1, so condition (2) of the present theorem is necessary for contractibility. Finally, suppose that (3) of the present theorem does not hold, i.e. we have some (k+4)- face L in P containing both J and J 0, with the section L/J disconnected in P 0. Denote by Pˆ0 the poset obtained by contracting the digon I/Gˆ in Pˆ. Condition (4) for Theorem 3.1 tells us that the 2-section L/ˆ Jˆ of Pˆ0 must be connected in order for the digon I/Gˆ in Pˆ to be contractible in Pˆ. However, the 2-section L/ˆ Jˆ of Pˆ0 corresponds to the 2-section L/J of P 0, by deﬁnition of the Helfand construction: every face of L/ˆ Jˆ corresponds to a face of L/J, and the incidence structure is the same. Therefore, L/ˆ Jˆ is disconnected in Pˆ0, and so the digon I/Gˆ in Pˆ is not contractible in Pˆ. Hence, condition (3) of the present theorem is necessary for contractibility. Now suppose that conditions (1) through (3) of the present theorem hold. We show that then conditions (1) through (4) of Theorem 3.1 must hold in Pˆ, so the suﬃciency (of (1) through (3)) follows from Theorem 3.1. To prove (1) of Theorem 3.1 (with i = k + 1) for Pˆ, observe that each Iˆ0 ∈ Iˆ is a

46 corank-1 face of only one old face, i.e. I0, and I0 cannot contain any of the other faces in Iˆ since each old face contains only one new corank-1 face. Hence, no two of the (k + 1)-faces in Iˆ can be contained in a common old (k + 2)-face. Furthermore, no two faces Iˆ1, Iˆ2 in Iˆ can be contained in a common new (k + 2)-face C = Dˆ of Pˆ: if this were the case, then we would have D > I1,I2 in P , violating (1) of the present theorem. Hence, condition (1) of Theorem 3.1 must hold for Pˆ. To prove (2) of Theorem 3.1 for Pˆ, suppose some face C in Pˆ contains Jˆ and Jˆ0. First, suppose C is an old face: then it must contain J and J 0 (every face contained in C is either an old face or a new face corresponding to an old face above F contained in C). Then by (2) of the present theorem, C contains some I0 in I, and therefore it contains Iˆ0 in Iˆ, so (2) of Theorem 3.1 holds in this case. Second, suppose C = Dˆ is a new face: then D must contain J and J 0, so by the same reasoning as before, D contains a face from Iˆ, and therefore Dˆ = C does as well. Hence, in both cases, (2) of Theorem 3.1 holds for Pˆ. To prove (3) of Theorem 3.1 for Pˆ, let R be a common corank-2 face of Jˆ and Jˆ0 in Pˆ. We note that Jˆ and Jˆ0 have the same set of incident faces of lower rank (i.e. the set of faces of F in P , by definition of the Helfand construction). Then the two faces between Jˆ and R are identical to the two faces between Jˆ0 and R, so Iˆ0/R is a digon for every Iˆ0 in Iˆ. Hence, (3) of Theorem 3.1 holds for Pˆ. To prove (4) of Theorem 3.1 for Pˆ, let E be a (k + 4)-face of Pˆ incident to both Jˆ and Jˆ0. We examine the rank-2 section E/Jˆ in Pˆ0. First, suppose that E is a new face in Pˆ, so E = Lˆ for a (k +5)-face L in P . Then by (3) of the present theorem, the 2-section L/J is connected in P 0, and hence the corresponding rank-2 section L/ˆ Jˆ = E/Jˆ is connected in Pˆ0. Now suppose that E is an old face in Pˆ. Then consider the 2-section E/Jˆ of Pˆ. Clearly, this rank-2 section contains Eˆ as an edge. Also, since E is an old face containing Jˆ, the 2-section E/Jˆ must contain J as a vertex. Furthermore, since E is incident to both Jˆ and Jˆ0, it is incident to both J and J 0, and therefore to some I0 in I; hence, E/Jˆ must contain I0 as an edge and Iˆ0 as a vertex. Finally, by (P4) on the section E/J in Pˆ, we see that E/Jˆ must contain an additional edge K (this must be an old face since it lies in E/J) and a corresponding vertex Kˆ . Examining the incidence structure of these three vertices and three edges, we see that E/Jˆ is in fact a triangle in Pˆ, as shown in Figure 4.4; similarly, as shown in the same figure, E/Jˆ0 is a triange in Pˆ as well (here, we have an analogous face K0 to K, from (P4) on the section E/J 0). Examining the effect of digonal contraction

47 Figure 4.4: Merging two triangles into a square with digonal contraction. on these triangles, we see, again in Figure 4.4, that E/Jˆ is a square in Pˆ0, and is hence connected. Hence, in both cases, the rank-2 section E/Jˆ in Pˆ0 is connected in Pˆ0, satisfying (4) of Theorem 3.1 for Pˆ. Thus, conditions (1) through (4) of Theorem 3.1 must hold for Pˆ, and so the digon I/Gˆ of Pˆ is contractible.

Corollary 4.2.1. Suppose P is a polytope lattice and we apply the (local) Helfand construction to a k-face F of P to obtain a polytope Pˆ as in the previous theorem. Let I/Gˆ be an arbitrary digon of Pˆ obtained as a result of this construction, with G < F < I in P , rk(I) = k + 2, and rk(G) = k − 2. Denote by J and J 0 the two proper (k + 1)-faces of the section I/F in P . Then I/Gˆ is contractible in Pˆ if and only if for every (k + 4)-face L of P containing both J and J 0, one of the polygonal sections L/J and L/J 0 in P is ﬁnite.

48 Proof. We will proceed with this proof with only the assumption that P is a polytope lattice. We first note that we must have I = {I}, since I is the supremum of J and J 0 in P . Hence condition (1) of the previous theorem must hold trivially. Similarly, condition (2) of the previous theorem must hold, again since I is the supremum of J and J 0 in P . We now consider condition (3) of the previous theorem. Denote by P 0 the poset obtained by combining the faces J and J 0 in P , removing I, and leaving all other faces and incidences of P unchanged. Let L be a (k + 4)-face of P containing both J and J 0. We see the effect of contracting J and J 0 on the polygons L/J and L/J 0 in Figure 4.5: here, A and B are the (k + 3)-faces containing I in L/J (A and B must also contain I in L/J 0 since I contains both J and J 0). Since I = {I}, the rank-2 section L/J in P 0 is connected if and only if at least one of the paths shown from x to y or from z to w is finite; hence, the digon I/Gˆ is contractible in Pˆ if and only if one of the polygonal sections L/J and L/J 0 is finite for every such choice of L.

Note 4.2.2. Suppose we apply the (local or global) Helfand construction with 1 ≤ k ≤ n−2 to an n-polytope P , and then contract one of the new digons of Pˆ to obtain a polytope Pˆ0. We can then contract another of the new digons of Pˆ in Pˆ0, as long as it was unaffected by our previous contraction (i.e. neither of the new faces H and H0 of this second digon is the merged face of the first digon), by the same reasoning. We can then continue contracting digons produced by the Helfand construction, again under the assumption that all contracted digons are unaffected by previous contractions. In general, however, we cannot repeat this process to get the original polytope P back: for instance, in Figure 4.3, after contracting one digon (the digon in front), there is only one other digon that can be chosen that was not affected by the first contraction (the digon in back); after contracting both, we cannot contract further, since Condition (1) of Theorem 3.1 is violated.

49 Figure 4.5: Merging two polygons with digonal contraction, in connection with the Helfand construction.

50 Chapter 5

Polygonal Contraction

In this chapter, we introduce a second new construction on abstract polytopes called polygonal contraction, letting us combine non-adjacent vertices of polygons under certain conditions. These polygons can be 2-faces of the polytope or sections of rank 2.

5.1 Deﬁning Polygonal Contraction

We will first define what we mean by polygonal contraction. As we will see, it is not immediately obvious exactly how this operation should be defined.

Note 5.1.1. Let P be an n-polytope with a 2-section D = F/G at level rk(F ) = i < n (so rk(G) = i − 3), and let v1 and v2 be two vertices of D (faces in D of rank i − 2 in P ) which are non-adjacent in D. See Figure 5.1 for an illustration. We wish to produce a poset P 0 by merging v1 and v2, splitting F into two new faces F1 and F2, and removing every (i−1)-face of P containing both v1 and v2 (the set of all such faces in P is denoted E). We carry over all incidences between faces other than v1, v2, F , and those from E. The incidences of the 0 0 merged vertex v1 in P carry over from P (v1 in P inherits all incidences of v1 and v2 0 in P ); what is not immediately clear is how the incidences of F1 and F2 in P should be assigned in order for P 0 to be a polytope. In particular, we wish to copy every incidence 0 F < H in P to get new incidences F1 < H and F2 < H in P ; however, we then also need to determine how to assign incidences H < F in P (for rk(H) = i − 1) to incidences 0 H < F1 and/or incidences H < F2 in P .

51 Figure 5.1: Polygonal contraction on a hexagon.

In Figure 5.1, we have a (local) example of what should occur during polygonal contraction. On the left side, in P , we see a hexagon (this hexagon is not P itself but some polygonal section in P at some level i), with distinguished vertices v1 and v2 shown. On 0 the right side, in P , the vertices v1 and v2 have been merged, and F has been split. The labeling of F1 versus F2 is arbitrary, but it is clear that F should split into two triangles, and it is clear which incidences of F in P should go to which triangle (note: all incidences of F in P to faces of greater rank are copied to both F1 and F2). Hence, it is clear how we should assign the incidences H < F with rk(H) = i − 1 to incidences in P 0: if H lies in F/G, then we have a natural incidence between H and exactly 0 one of the two faces F1 or F2 in P . However, we still need to determine how to assign incidences H < F (with rk(H) = i − 1) to incidences in P 0 if H does not lie in F/G, i.e. H is not incident to G. Let H, with H < F , be an (i − 1)-face of P outside of D, and B > F be an (i + 1)-face of P (note that i < n). Then by (P4), we have another face besides F between H and 0 B in P . In P , in order to preserve (P4) for B/H, exactly one of the new faces F1 or F2 needs to be incident to H. In fact, upon further investigation, we can see that there is at most one way to assign each of the corank-1 faces H of F in P to be incident to (exactly 0 one of) the faces F1 or F2 in P in order to produce a polytope. We present the following

52 proposition, allowing us to deﬁne polygonal contraction:

Proposition 5.1.2. Let P be a polytope with a polygonal section F/G. Let rk(F ) = i < n and let v1 and v2 be non-adjacent vertices of F/G. Let G denote the edge graph of the dual polytope of the i-polytope F/F−1 (so v1 and v2 are edges of G). We consider creating a new poset P 0 by contracting F/G as described above, and arbitrarily assigning each corank-1 face 0 0 of F (the vertices of G) to exactly one of the faces F1 or F2 in P . If P is a polytope, then the subset of edges {v1, v2} of G is an edge cut of G and the incidences of the corank-1 faces of F are assigned to F1 or F2 according to the edge cut (i.e., the edge cut splits G into several connected components; every vertex of G in a given connected component must be 0 assigned to the same face F1 or F2 in P ). We note that this condition only depends on the structure of F/F−1 in P .

Proof. Let P 0 be a polytope. In order to prove this statement, we ﬁrst prove the following: for any edge v 6= v1, v2 of G (note that v, viewed as a face of P , has rank i − 2), both vertices E1 and E2 of G contained in v (which, viewed as faces of P , have rank i − 1) must 0 0 be assigned to the same face F1 or F2 in P since P is a polytope. Suppose they are not 0 0 assigned to the same face, say E1 < F1 in P and E2 < F2 in P . In P , the section F/v has 0 exactly two proper faces (by (P4)), namely E1 and E2. However, in P , the sections F1/v and F2/v then each have exactly one proper face, namely E1 and E2 respectively. Hence, 0 0 (P4) fails in P , and so P would not be a polytope. Therefore, each edge v 6= v1, v2 of G 0 has both of its vertices in G assigned to the same face F1 or F2 in P .

Suppose that {v1, v2} is not an edge cut of G. We know that the two vertices of v1 in G 0 are naturally assigned to different faces F1 and F2 of P (as discussed above). Note that G is connected in P by Lemma 2.3.1. Since {v1, v2} is not an edge cut of G, the graph obtained by removing these two edges from G is still connected, and so we have a path in G from one vertex of v1 to the other using neither v1 nor v2 as an edge. By the above conditions, each vertex in the path must be assigned to the same face F1 or F2; however, we already saw that the first and last vertices of the path are assigned to different faces, which gives a contradiction. Hence, {v1, v2} must be an edge cut of G.

Moreover, since {v1, v2} is an edge cut of G and every edge v 6= v1, v2 of G has both of 0 its vertices assigned to the same face F1 or F2 in P , we similarly see that each connected component in G − {v1, v2} must have all of its vertices assigned to the same face F1 or F2: if not, we have a path between two vertices in G − {v1, v2} assigned to diﬀerent faces F1,

53 F2, and so again we get a contradiction, since every vertex in the path must be assigned to the same face F1 or F2. Thus the vertices in a connected component of G − {v1, v2} all are assigned to the same face F1 or F2 in P .

Note 5.1.3. Suppose that F/G is not an apeirogon. Then the vertices and edges of G corresponding to the proper faces of F/G form a cycle, and so by Corollary 2.4.5 applied to the subgraph of G determined by F/G, the graph G − {v1, v2} has exactly two connected components. However, if F/G is an apeirogon, then F/G does not correspond to a cycle in G, and so the graph G −{v1, v2} has either two or three connected components (regardless, Proposition 5.1.2 holds).

Note 5.1.4. Proposition 5.1.2 finally allows us to define polygonal contraction. If the necessary conditions of Proposition 5.1.2 for P 0 to be a polytope are satisfied, then there is 0 exactly one way to assign (i − 1)-faces of P (incident to F in P ) to F1 or F2 that could 0 result in P being a polytope (up to renaming F1 and F2). This is because every connected component of G − {v1, v2} contains at least one of the vertices of v1 or v2. These vertices correspond to the proper faces of F/v1 and F/v2 in P , for which the assignments to F1 0 or F2 in P are known - hence, their assignments give us the assignments of all the other (i − 1)-faces contained in F .

Figure 5.3 shows us a graph G for which the conditions in Proposition 5.1.2 are satisfied and a graph G for which they fail. The first graph of Figure 5.3 corresponds to the polyhedron shown in Figure 5.2. This figure shows a polyhedron obtained by the following process: First, two vertices (the top and bottom ones in the figure) are given four edges between them, creating a polyhedron with four digons tessellating the 2-sphere (the edge in grey is visualized as being in the back). Then, one of the edges is split by adding a square as shown in the figure; call this square A. This polyhedron then has six vertices (the top vertex is G), nine edges (the left and right edges are v1 and v2), and five 2-faces (two digons in the back, the square in the front, and two pentagons next to the square). We then take this polyhedron and produce a prism over it, which is the entire 4-polytope P we are interested in. For F we take the base of P , that is the polyhedron over which P is the prism. We then see the edge graph G of the polyhedron dual to F in the first example of Figure 5.3. The five vertices of the graph correspond to the 2-faces contained in F , and the nine edges correspond to the edges contained in F . In this graph, v1 and v2 are the left

54 Figure 5.2: A polytope nontrivially satisfying the conditions in Prop. 5.1.2. and right edges of the graph, and the square consisting of the four edges drawn as straight lines represents the proper faces of F/G. The inner vertex of the graph corresponds to the 0 square A. We then get a natural assignment of faces to F1 or F2 in P : the red part of the graph is assigned to F1 (in Figure 5.2, this is the part of F facing us) and the green part of the graph is assigned to F2 (in Figure 5.2, this is the part of F facing away from us). The second graph G is the dual of the octahedral base facet F in a pyramid P over an octahedron, and hence is the edge graph of a 3-cube. Again, P has rank 4. The standard 3 octahedron in R has each vertex at a point ±ei (1 ≤ i ≤ 3). Then the vertices of an abstract octahedron can be labeled e1, e¯1, e2, e¯2, e3, e¯3; the edges can be labeled e1e2, e1e¯2, ..., e¯2e¯3; and the 2-faces can be labeled e1e2e3, ..., e¯1e¯2e¯3. Then the polygon being considered for contraction (in P ) in this example is F/G where F is the octahedron and G is the vertex e1; we set v1 = e1e2 and v2 = e1e¯2. The vertices of the graph G correspond to the 2-faces contained in F , i.e. triangles in the octahedron, and the edges of the graph correspond to the edges in the octahedron. Again, v1 and v2 are the left and right edges of the graph; the square consisting of the four vertices whose label includes e1 (rather thane ¯1) represents the proper faces of F/G. Here, we see that there is no way to assign faces to F1 or F2 to 0 produce a polytope P , according to Proposition 5.1.2, since the removal of v1 and v2 does not disconnect the edge graph.

55 Figure 5.3: Edge graphs satisfying and violating the conditions in Prop. 5.1.2.

56 5.2 The Polygonal Contraction Theorem

Now that we have deﬁned polygonal contraction, we categorize when the contracted poset is in fact a polytope. To help do so, we present the following lemma, which is a consequence of Proposition 5.1.2.

Lemma 5.2.1. Let P be an n-polytope with a polygon F/G at rank i < n as in Propo- sition 5.1.2 with distinguished vertices v1 and v2. Suppose that the condition in Proposi- tion 5.1.2 holds, i.e. the set {v1, v2} is an edge cut of the edge graph G the dual polytope 0 of F/F−1, and we deﬁne the poset P as the result of contracting F/G via v1 and v2 as described in Proposition 5.1.2. 0 Let E1 be the set of corank-1 faces of F in P which are incident to F1 in P , and let E2 0 be the set of corank-1 faces of F in P which are incident to F2 in P . Then for every face

K < F with K incident to neither v1 nor v2, each facet of F/K lies in of E1 or each facet S lies in E2 (note that each facet of F/K must lie in E1 E2 since every corank-1 face of F lies in exactly one of these two sets).

Proof. The proof is by downward induction of the rank of K. Clearly the condition holds if rk(K) = i − 1 since then K is a corank-1 face of F and F/K has rank 0 (with K the only facet). The condition must also hold if rk(K) = i − 2: such faces K correspond to edges of G, and from the proof of Proposition 5.1.2 we know that every edge of G other than v1 and v2 has both of its vertices assigned to the same face F1 or F2. Now suppose that rk(K) < i − 2. W.L.O.G. suppose F/K has a facet that belongs to

E1 (as opposed to E2). Assuming that our statement holds for faces of rank i − 2, then by

(P4) we know that every (i − 2)-face of F/K contained in an (i − 1)-face in E1 must be contained in another (i − 1)-face in E1. Then by Lemma 3.1.1, all facets of F/K are in E1, so the statement holds for rk(K) < i − 2 as well. Hence, the statement holds for faces K of any rank.

Theorem 5.1 (Polygonal Contraction Theorem). Let P be an n-polytope with a polygon 2-section D at level i < n whose i-face is called F and whose (i − 3)-face is called G, and let v1 and v2 be two non-adjacent vertices of D (faces of rank i − 2 in P ). Let G denote ∗ the edge graph of the dual polytope (F/F−1) of the i-polytope F/F−1. Suppose we produce 0 a poset P by contracting D, i.e. merging v1 and v2, splitting F into two new faces F1 and

F2, and removing (if need be) every (i − 1)-face of P containing both v1 and v2 (the set

57 of all such faces in P is denoted E). Then P 0 is a polytope (of rank n) if and only if the following conditions hold in P : 0 (0) The set {v1, v2} is an edge cut of G (and the incidences of faces to F1 or F2 in P are assigned according to this edge cut). (Note that this condition is numbered (0) since, as discussed earlier, it is needed to deﬁne polygonal contraction in the ﬁrst place.) (1) No two faces in E are contained in a common i-face of P .

(2) For every common corank-2 face H of v1 and v2, the proper faces of v1/H are the same as the proper faces of v2/H.

(3) For every (i + 1)-face I of P containing v1 and v2, the rank-2 section I/v1 is connected in P 0.

(4) For every face J of P containing both v1 and v2, we have J ≥ F or J ≥ E for some E in E.

Note 5.2.2. We will see later when Condition (3) for P 0 does or does not hold; Condi- tion (3) could be described in terms of P , but the corresponding condition on P is somewhat complicated to state. However, there are simple conditions on P which are suﬃcient for Condition (3) to hold.

Proof. In the first part of this proof, we will show that for P 0 to be a polytope, Conditions (0) through (4) must hold. We show this by proving first that P 0 is not a polytope if (0) does not hold, then that P 0 is not a polytope if (0) holds but (1) does not hold, then that P 0 is not a polytope if (0) and (1) hold but (2) does not hold, and so on. In the second part of the proof, we will show that the conditions are sufficient for P 0 to be a polytope, i.e. if all of the conditions hold then P 0 is a polytope. First, we know by Proposition 5.1.2 that Condition (0) is necessary for P 0 to be a polytope. Next, assume that (0) holds but (1) does not hold, i.e. we have some i-face K in P containing two (i−1)-faces E1 and E2 from E. Then K 6= F . Suppose K has all its corank-1 faces belonging to E: then K has no corank-1 faces in P 0 (since all of the faces in E are removed), and so P 0 is not a polytope. Suppose otherwise, so K has a corank-1 face L 0 outside of E. Examine the section K/F−1 in P . Suppose K/F−1 contains the merged face v1. Then in P , we have more than two faces between K and v1: we have E1, E2, and L, so 0 (P4) does not hold in P , which is a contradiction. Hence, K/F−1 does not contain v1 in P . 0 This means that K/F−1 in P is a subset of K/F−1 in P , but with some faces removed.

58 Therefore, as we discussed in the proof of Theorem 3.1 on digonal contraction, K/F−1 is not strongly connected, and so P 0 is not a polytope. Hence, (1) is necessary for P 0 to be a polytope. Next, assume that (0) and (1) hold but (2) does not hold, i.e. we have some common corank-2 face H of v1 and v2 such that the proper faces of v1/H are not both the same 0 as the proper faces of v2/H. We then see that in P , the section v1/H has more than two proper faces (since it includes all proper faces from v1/H and v2/H in P ), violating (P4). Hence, (2) is necessary for P 0 to be a polytope. Note that by Lemma 3.1.2 (applied to the faces in E, and with i replaced by i − 1), Condition (2) implies that every face of E has only v1 and v2 as corank-1 faces. Next, assume that (0) through (2) hold, but (3) does not hold. Clearly (3) is necessary for (P3), so if (3) does not hold then P 0 is not a polytope. Next, assume that (0) through (3) hold, but (4) does not hold, i.e. we have some face

J > v1, v2 of P not containing F nor any of the faces in E. We note that J cannot be of rank i − 1 since this would mean J is among the faces in E. Suppose J is of rank i. Then

J cannot be F . If J/v1 and J/v2 have any proper faces in common, those faces must lie in E, meaning J does not violate (4). Then suppose J/v1 and J/v2 have no proper faces in 0 0 common: in this case, J/v1 in P has all four of these faces, violating (P4), and so P is not a polytope. Now suppose rk(J) > i. Examine the proper faces of J/v1 in P : if any of these faces contains v2, take any such face with minimal rank and replace J by it (again, if the new face has rank i, we already see that P 0 is not a polytope). Thus, we can assume that

J/v1 in P has no proper faces containing v2, and therefore J/v2 in P also has no proper faces containing v1. Let A be a proper face of J/v1 in P and let B be a proper face of J/v2 0 in P : note that neither A nor B is F nor any of the faces in E, and so J/v1 in P contains both A and B. However, there cannot be any sequence of proper successively incident faces 0 in J/v1 in P taking A to B: if there was such a sequence, it would start with A, a face which contains v1 but not v2 in P , and end with B, a face which contains v2 but not v1 in P ; at some point, the sequence would need to pass through a face containing both v1 and v2 in P (since in P every face in the sequence would be in J/v1 or J/v2), but no such 0 0 face exists. Hence, the section J/v1 is not connected in P , and so P is not a polytope. Therefore, (4) is necessary for P 0 to be a polytope. We have shown that Conditions (0) through (4) are necessary for P 0 to be a polytope. We will now show suﬃciency: if Conditions (0) through (4) hold, then P 0 is a polytope. We

59 check the four defining conditions of a polytope on P 0 given our conditions. (P1): We easily find that there is still exactly one greatest face and one least face (and they are the same faces as before). (P2): We also easily find that P 0 is a ranked poset with rank n. We will examine (P4) next and (P3) at the end of the proof. (P4): Consider a 1-section B/A of P 0. We will check (P4) on B/A for various choices of rk(A). If rk(A) ≥ i, then this section is identical in structure to the corresponding section in P , so (P4) holds in this case. This is still true if A = F1 or A = F2 since the faces F1 and F2 each have the same higher-degree incidences in P 0 as F does in P . 0 Now suppose rk(A) = i − 1. If we do not have A < F1 or A < F2 in P , then we do 0 not have A < F in P (since the new faces F1 and F2 in P only contain (i − 1)-faces that 0 F contains in P ), so B/A in P is identical to B/A in P , so (P4) holds. If A < F1 (say), then we note that we do not have A < F2 (since F1 and F2 share no (i − 1)-faces). Then we have A < F < B in P . By (P4) in P , we have exactly one face F 0 6= F in P such that A < F 0 < B. Then in P 0, we must have A < F 0 < B (since these incidences are not affected 0 by the construction), so B/A in P has at least two proper faces, F1 and F . Suppose it has 0 0 a third proper face K, i.e. A < K < B in P with K 6= F1, K 6= F . Then in P , we must have A < K < B with K 6= F , K 6= F 0, so (P4) is violated in P , giving us a contradiction. Hence, B/A in P 0 must have exactly two proper faces, so (P4) is satisfied. We can make a similar argument if A < F2, so if rk(A) = i − 1 then we always have (P4) for B/A. Now suppose rk(A) = i − 2. If in P , neither A nor B are in D, then clearly B/A in P 0 is identical to B/A in P , so (P4) holds in this case. 0 If in P , the face A is in D but B is not, then A is a vertex of F1/G (say) in P . If A is not the merged vertex, then clearly B/A is the same in P 0 as in P . Suppose A is the 0 merged vertex. If in P , the face B only contains one of v1 or v2, then in P there will be exactly two proper faces between A, the merged vertex, and B (i.e. the same ones as in P ).

If in P , the i-face B contains both v1 and v2, then Conditions (1) and (4) tell us that in P , 0 the 1-sections B/v1 and B/v2 share exactly one proper face, i.e. some E from E; then in P , the new 1-section B/v1 has exactly two proper faces, since the common face E is removed by the contraction. Hence, (P4) holds when, in P , the face A is in D but the face B is not. 0 If in P , the face B is in D but A is not, then B is F in P and B is F1 or F2 in P .

60 Then in P , the (i − 2)-face A is incident to F but not incident to G. By Condition (0), this means that the (i − 1)-faces in B/A = F/A in P are both incident to the same face F1 0 or F2 in P , because the corresponding vertices of G lie in the same connected component 0 of G − {v1, v2}. Hence, B/A is identical in P and in P , except that the greatest face F is renamed F1 or F2. 0 Finally, if in P , both A and B are in D (so B = F ), then in P , we have B = F1 or 0 B = F2 and A is a vertex of F1/G or F2/G in P ; since these 2-sections are in fact polygons, (P4) holds in this case. This covers all cases for rk(A) = i − 2. Now suppose rk(A) = i − 3. If none of the proper faces of B/A is the merged vertex 0 0 v1 in P , then B/A in P is identical to B/A in P and so (P4) holds in this case. Suppose 0 one of the intermediate faces is v1. Consider B/A in P : we note that since B exists in P , in P it cannot be one of the faces in E, and so B does not contain both v1 and v2 in P , so w.l.o.g. say B/A contains v1 but not v2. Then B/A must have two proper faces in P : one 0 is v1, and we call the other K. Then both v1 and K are also proper faces of B/A in P (the 0 merged face v1 in P inherits all the incidences of v1 in P , and the face K in P is unaffected by the construction), and we cannot have any additional such faces (the construction does not introduce new incidences between (i − 3)-faces and (i − 2)-faces, so there cannot be any new incidence between A and some (i − 2)-face in P 0). Hence, (P4) holds in this case as well. 0 Now let rk(A) = i − 4. If B is not v1 in P , clearly (P4) holds since the section B/A of 0 0 P is identical to the corresponding section in P . If B is v1 in P , then in P , A may or may not be a common corank-2 face of v1 and v2. If A is not a common corank-2 face, then B/A in P 0 corresponds to B/A in P . If A is a common corank-2 face, then by Condition (2) the 0 section B/A in P has exactly two proper faces (since the proper faces in P of v1/A and v2/A are identical by assumption). Hence, (P4) also holds in this case. Finally, if rk(A) < i − 4, then (P4) holds since no relevant faces or incidences of P are modified by the contraction. Hence, (P4) holds in all cases. It remains to show that (P3) holds. 0 (P3) We will first show that every section of the form F1/K or F2/K is connected in P . 0 W.L.O.G. let K < F1 in P and consider the section F1/K. Note that F1/K is trivially connected if rk(K) = i − 1 or rk(K) = i − 2, so we only need to examine the case where rk(K) ≤ i − 3. 0 First, suppose that F1/K does not contain the merged face v1 in P : then F/K in P

61 does not contain v1 or v2, so by Lemma 5.2.1 (and its notation) we know that every facet of F/K in P is in E1 (it is not possible that every facet of F/K in P is in E2 since we would 0 0 then not have K < F1 in P ). In that case, F1/K in P is identical to F/K in P , and is therefore connected. 0 Now suppose that F1/K contains v1 in P . We label the vertices of F/K in P as follows, again bearing in mind Lemma 5.2.1: a vertex is labeled “1” if in P it is only contained in a facet of F/K in E1, it is labeled “2” if in P it is only contained in a facet of F/K in E2, and it is labeled “3” if in P it is contained in v1 or v2. Clearly no vertex is labeled both “1” and “2” since E1 and E2 are disjoint; moreover, no vertex is labeled both “1” and “3” since v1 and v2 are each contained in exactly one element of E1 and exactly one element of E2 (similarly, no vertex is labeled both “2” and “3”). Hence, each vertex of F/K is assigned at most one label. Suppose that some vertex v is not assigned a label: this means that it is contained in a facet in E1 as well as a facet in E2 in P , but not contained in v1 or v2; however, this then violates Lemma 5.2.1 applied to F/v. Therefore, every vertex must in fact be assigned exactly one of these three labels. Also note that no edge of F/K in P can contain a vertex of F/P labeled “1” as well as a vertex labeled “2” (the edge would not lie in any facets in E2 since it contains a vertex labeled “1” and it would not lie in any facets in E1 since it contains a vertex labeled “2”, so it would not lie in any facets at all, since every facet is contained in one of the two sets). 0 Finally, note that the vertices of F1/K in P are exactly the vertices of F/K in P that 0 were labeled “1” or “3”, and the edges of F1/K in P are exactly the edges of F/K in P joining two vertices each labeled “1” or “3”; any edge containing a vertex labeled “2” does not exist in F1/K since such edges are only incident to facets in E2. 0 We will now show that we can move from any vertex X of F1/K in P to any other vertex Y . First, suppose that X and Y are both labeled “3”: then clearly we can move from X to v1 to Y , since X and Y in P lie in v1 or v2. Then suppose that one of the vertices is labeled “1” (say X) and the other is labeled “3” (say Y ). In F/K in P , we can move along the 1-skeleton to get from X to Y . At some point in this sequence, we pass from a vertex labeled “1” to a vertex labeled “3” (via an edge); until the ﬁrst occurrence of this, every vertex in the sequence is labeled “1”. So in P 0, we can take the same sequence from X onwards until we hit the ﬁrst vertex labeled

“3”, and then move from this vertex, which lies in v1 or v2, directly to Y .

62 Finally, suppose that X and Y are both labeled “1”. In F/K in P , we can move along the 1-skeleton to get from X to Y . If this sequence only passes through vertices labeled “1”, we can take the same sequence in P 0. If it passes through vertices labeled “3”, we can denote by W and Z the first and last such vertices in the sequence, respectively; then 0 in F1/K in P we can move from X to W by the same sequence as in F/K in P , then we can move from W to Z as described above, and finally we can move from Z to Y by the same sequence as in F/K in P . 0 Therefore, in all cases, we can move from any vertex of F1/K in P to any other vertex. Hence, we can move from any proper face X to any other proper face Y by moving from X down to a vertex, then moving over to a vertex contained in Y , then moving up to Y . 0 Hence, every section F1/K is connected in P if F/K contains v1 or v2; similarly, every 0 section F2/K is connected in P if F/K contains v1 or v2. We have now examined every 0 possible case, so we see that every section F1/K or F2/K is connected in P . We will now proceed to show that P 0 is strongly connected, i.e. every section of P 0 is flag-connected, or equivalently, connected. Consider a section B/A of P 0: we want to move from any flag Φ of B/A to any other flag Ψ of B/A via a sequence of successively adjacent flags of B/A. We note that B/A corresponds to at least one section of P (it could 0 0 correspond to two if B or A is v1 in P ), and every flag of B/A in P corresponds to at least one flag of B/A in P . So we start by examining the sequence of flags of B/A in P taking Φ to Ψ. Suppose we have two adjacent flags in this sequence neither of which contains F nor a face in E: then the two corresponding flags in P 0 are also adjacent. So when moving from Φ to Ψ in B/A in P , it is trivial to translate to P 0 moves between flags not involving F or a face in E. Now consider a sequence of successively adjacent flags in B/A in P that contains a flag containing F : translating the flags in the sequence to P 0, we may need to move a flag of

B/A containing F1 to a flag of B/A, adjacent in P , containing F2. Note that we have 0 established that every section in P having F1 or F2 as its greatest face is connected, so we deal here with the case that B 6= F1 and B 6= F2 in P . We would only need to do this move if the two flags in P 0 were (i − 1)-adjacent in P ; if two flags in the sequence are k-adjacent 0 in P for k > i, we can use F1 or F2 for both flags in P since F1 and F2 are contained in the same set of faces; if they were k-adjacent in P for k < i − 1, then both k-faces would be contained in the same (i − 1)-face and hence in the same face F1 or F2. Moreover, we would only need to do this move if the common (i − 2)-face of the two (i − 1)-adjacent flags

63 was v1 or v2 in P ; otherwise, again, we see that both flags would have the same i-face, F1 0 or F2, in P . In that case, calling the common (i + 1)-face of the two flags I, we can move 0 one flag to the other in B/A in P since the rank-2 section I/v1 of B/A is connected (hence 0 flag-connected) in P by (3) (we can apply (3) since we know I contains both v1 and v2 in P since I > F ). Hence, when translating the sequence of successively adjacent flags in B/A in P to a sequence of successively adjacent flags in B/A in P 0, we can move through F without incident. Now say we have a sequence of successively adjacent flags in B/A in P that contains a flag containing some face E in E. Listing this sequence of successively adjacent flags from Φ to Ψ as Φ0 = Φ, Φ1, ..., Φm = Ψ, denote the first flag containing a face E from E by Φr and examine all subsequent flags still containing E; let Φs be the last such flag (so that between 0 Φr and Φs, every flag in the sequence contains E). Then in B/A in P , there are no flags corresponding to Φr or Φs and any of the flags between them (so in particular, we note that 0 we cannot have r = 0 or s = m, since Φ0 = Φ and Φm = Ψ are flags in B/A in P ), but there are flags corresponding to Φr−1 and Φs+1 (note that the two corresponding flags in P cannot contain F or any of the faces in E, since F does not contain E, the (i − 1)-face of

Φr, and since no two of the faces in E are contained in a common i-face by (1)). We then 0 want to move in B/A in P from the flag Φr−1 to the flag Φs+1. Since Φr−1 and Φr are (i−1)-adjacent (one contains E, the other does not), they both include the same (i−2)-face, 0 and by (2) with Lemma 3.1.2 (applied with F,H,H replaced by E, v1, v2 respectively) this face is either v1 or v2; by the same reasoning, Φs+1 includes v1 or v2. So in the section B/A 0 of P , the flags Φr−1 and Φs+1 both include the merged face v1; so in particular, A ≤ v1 ≤ B 0 in P . We will see that to move from Φr−1 to Φs+1, we can first move the part of Φr−1 below v1 to the part of Φs+1 below v1, and then move the part of Φr−1 above v1 to the part of Φs+1 above v1. 0 We will first show that v1/A is connected in P and hence that we can move the lower parts of one flag to the other. By (2) with Lemma 3.1.2 (applied as above), we know that in P , the section E/A is a polytope with exactly two facets (namely v1 and v2); hence, by

(P4), every ridge of this polytope is contained in both v1 and v2, and so v1/A has the same 0 set of proper faces (and incidences) as v2/A in P , and so v1/A in P also has the same set 0 of proper faces (and incidences). Hence, v1/A is connected in P . 0 Next, we will show that B/v1 is connected in P and hence that we can move the upper 0 parts of one ﬂag to the other. Note that B/v1 in P consists of the faces and incidences

64 of B/v1 and B/v2 in P except that all faces in E are removed and F is split (if F < B); so 0 in particular, given any proper face of B/v1 in P , there is a corresponding face of B/v1 or

B/v2 in P . 0 Suppose we want to move from one proper face in B/v1 in P to another, and moreover that in P both of the corresponding faces contain v1 or both of the corresponding faces contain v2. Then since every section of P is connected, we have a sequence of proper successively incident faces in B/v1 or B/v2 in P taking the ﬁrst face to the second. If this sequence does not pass through any face in E, we can use it in P 0 as well and so we are done. Otherwise, it does pass through some face E of E: then denote the face directly before E in the sequence by C, and the face directly after E in the sequence by D; WLOG, we may 0 take C and D to have rank i. Then C and D must both exist in B/v1 in P ; we wish to ﬁnd a sequence of proper successively incident faces from one to the other. Note that since E is a vertex of B/v1 or B/v2 in P , the faces C and D must both be edges of B/v1 or B/v2, i.e. rk(C) = rk(D) = i. Suppose rk(B) ≥ i + 2. Then since B/E is a polytope (of rank at least 2) in P , we have a sequence of proper successively incident faces in B/E taking C to D; we can take this identical sequence in P 0 to move from C to D without passing through E (the sequence in P never passes through F since F is not incident to E). This argument does not apply if rk(B) = i + 1: in this case, B/E in P is a line segment, so we cannot pass from C to D in B/E in P . However, in this case B is an (i + 1)-face containing both 0 v1 and v2 in P , and so by (3) the rank-2 section B/v1 is connected in P , and so again we 0 see that we can move from C to D in B/v1 in P . Hence, we can move from any proper face in B/v1 to any other proper face provided that they both contain v1 or both contain v2 in P . 0 Now suppose that in P we want to move from a proper face X1 of B/v1 to a proper face X2 of B/v1, where X1 contains v1 and not v2 in P , and X2 contains v2 and not v1 in P .

Since B then contains v1 and v2 in P , the face B must contain F or some face from E by 0 (4). Suppose B contains F in P . Then in P , the face B contains F1 or F2 (B may coincide with F1 or F2, but in that case, B/v1 is a line segment and hence is trivially connected). 0 Hence we may assume that F < B in P , and that F1,F2 < B in P . We know that the two 0 proper faces of F1/v1 in P are such that, in P , one contains v1 and the other v2; the same 0 holds for the proper faces of F2/v1 in P . We established above that we can move from any 0 proper face in B/v1 to any other proper face in B/v1 in P given that they both contain v1 or they both contain v2 in P , so we can move from X1 to a proper face H1 of F1/v1 which

65 contains v1 in P , and we can move from X2 to a proper face H2 of F2/v1 which contains v2 0 in P . It remains to show that we can move from H1 to H2 in B/v1 in P . If rk(B) ≥ i + 2, then we know that every (i + 1)-face I containing F1 also contains F2 and vice versa, and so we can move from H1 up to I, and move from I down to H2, hence connecting X1 and

X2 in B/v1. If rk(B) = i + 1, then B is a rank-(i + 1) face in P containing both v1 and v2, 0 and so by (3) we know that B/v1 in P is connected. So in any case, we can move from X1 0 to X2 in B/v1 in P if X1 contains v1 and not v2 in P , and X2 contains v2 and not v1 in P . 0 We established earlier that we can move from any proper face in B/v1 in P to any other proper face if they both contain v1 or both contain v2 in P . Hence, we now see that 0 we can move from any proper face at all in B/v1 in P to any other proper face, and so 0 0 B/v1 is connected in P . Hence, B/v1 is flag-connected in P , so we can move the part of the previously defined flag Φr−1 of B/A above v1 to the part of Φs+1 above v1. We already know that we can move the part of Φr−1 below v1 to the part of Φs+1 below v1, and so we can move the full flag Φr−1 to the full flag Φs+1. Hence, given any two flags Φ and Ψ in B/A in P 0, we can take the corresponding sequence of flags in B/A in P taking Φ to Ψ, and get a (possibly longer) sequence of flags in B/A in P 0 taking Φ to Ψ, and so P 0 is strongly flag-connected, therefore strongly connected, and so we have (P3), and so P 0 is a polytope. This completes the proof.

Remark 5.2.3. Unlike the other conditions, Condition (3) of the previous theorem is a condition on the contracted poset P 0, not the original polytope P . There is an equivalent condition on P which, however, is nontrivial to state. Condition (3) says that for every 0 rank-(i + 1) face I of P containing v1 and v2, the rank-2 section I/v1 of P is connected.

A necessary condition for (3) to hold is that I/v1 and I/v2 in P are not both apeirogons: 0 if they are, we see that I/v1 in P cannot be connected, as illustrated in Figure 5.4, since there is no sequence of successively incident proper faces connecting the face A to the face B 0 within the section I/v1 in P . Note that this ﬁgure shows part of the polyhedron I/G in P , where G is the least face of the polygonal section being contracted, and the corresponding 0 part of the poset I/G in P ; the rank-2 sections I/v1 and I/v2 in P are apeirogonal vertex ﬁgures.

Using the notation of Remark 5.2.3 and supposing that I/v1 and I/v2 are not both apeirogons in P , we examine the sections I/v1 and I/v2 in P and see what the section I/v1 then looks like in P 0. In Figure 5.5, we see a few simple situations that may occur. In

66 Figure 5.4: When contracting a polygon with two apeirogonal vertex ﬁgures as shown, we get a disconnected vertex ﬁgure.

general, I/v1 and I/v2 in P may possibly contain some faces from E and may or may not contain F ; for each of these faces, if it appears in one section, it must also appear in the other (since all of these faces contain both v1 and v2), so we say the face is shared. The examples in Figure 5.5 show the polygons representing the sections I/v1 and I/v2 of P and 0 the section I/v1 in P . 0 We see that in the ﬁrst two examples, I/v1 in P must be connected (they are clearly polygons). However, in both of the latter two examples, I/v1 and I/v2 both contain exactly 0 two of the faces in E and not F , and yet I/v1 in P may or may not be connected, depending on how the faces A, B, C, and D connect to each other in I/v1 and I/v2 in P : as we can see, in Case 1, the order of the four edges A, B, C, D is the same in I/v1 and in I/v2 in P , 0 and the section I/v1 in P is diconnected as a result; on the other hand, in Case, 2, the 0 order is A, B, D, C in I/v1 and A, B, C, D in I/v2, and the section I/v1 in P is connected as a result. (We see in Figure 5.6 polyhedra in which these types of sections actually occur; we will discuss these examples in more detail later.) This is why Condition (3) is somewhat complex; we cannot simply say that certain faces need to be in I/v1 and I/v2 in P and certain other faces must not be in those sections. In general, as we consider putative sections possibly sharing faces from E and/or the face F , for a given set-up (e.g. I/v1 and I/v2 do not contain F but contain exactly two faces from E) there is almost always a way to produce 0 both a putative connected section I/v1 in P as well as a putative disconnected section I/v1 0 in P by changing how faces are connected in I/v1 and I/v2 in P (as demonstrated in the last two examples in the above diagram). The only deﬁnite cases are when I/v1 and I/v2 in P contain only F but no faces in E (in which case, we see by the corresponding diagram

67 Figure 5.5: Examples of connectivity and non-connectivity in polygonal contraction.

68 0 in Figure 5.5 that the rank 2-section I/v1 in P is connected); when the sections contain only one face from E and not F (in which case, we again see by the diagram that the section is connected); and when the sections contain none of the faces in E, nor the face F (in which case it is easy to see that the section is disconnected).

In Figure 5.6, we see two examples of polyhedra I/G in which the scenarios for I/v1 and I/v2 described in the last two examples in Figure 5.5 actually occur; as before, I is an

(i + 1)-face of P containing v1 and v2, and G is the least face of the polygonal section of P 0 to be contracted. Note that there are edges E, E connecting v1 and v2 in these examples (so E 6= ∅), but this does not mean that the two vertices are adjacent in F/G, since in these examples we do not have I > F . Returning to the general discussion, we note that each of the following simple conditions on P (alone) is suﬃcient for Condition (3) of Theorem 5.1 to hold.

• For all rank-(i + 1) faces I > v1, v2, the rank-2 sections I/v1 and I/v2 are not both apeirogons, I contains no faces from E, and I > F .

• For all rank-(i + 1) faces I > v1, v2, the rank-2 sections I/v1 and I/v2 are not both apeirogons, and I contains exactly one of the faces in E but not F , or I contains F but none of the faces in E.

In order to show that none of the five conditions in the Polygonal Contraction Theorem (Theorem 5.1) is redundant, we require examples of polytopes with polygonal sections satisfying any four of the conditions for polygonal contraction but not the fifth. Figure 5.7 has diagrams representing five polytopes on which we will examine the conditions. Figure 5.8 has the corresponding edge graphs G, the edge graphs of the dual polytopes of F/F−1. Recall that Condition (3) is the most complicated one, and it may actually imply some of the other conditions; however, if we omit other conditions which are implied by (3), then (3) becomes more complicated still. In order to keep (3) relatively easy to understand, we present the polygonal contraction theorem with five separate conditions. As such, we are not as interested in examining Condition (3) for these polytopes, and do not examine it in the following examples (except for Example 3). In Example 0 in Figure 5.7, we have a polyhedron embedded in a torus: the top and bottom of the figure are identified as usual, as well as the left and right, as indicated by the arrows. The polyhedron has four vertices, marked in red: the middle vertex is G. The edges v1 and v2 are incident to G on the left and right; there are a total of eight edges

69 Figure 5.6: Two similar polygons with diﬀerent results under polygonal contraction.

70 Figure 5.7: Five diﬀerent examples of non-contractible polygonal sections F/G in polytopes.

71 Figure 5.8: The edge graphs G for the examples in Figure 5.7.

72 in the polyhedron (four horizontal and four vertical), and four 2-faces (squares). The face F is the polyhedron itself; we take a prism over F in order to get the entire polytope P we are interested in, then consider contracting the square F/G in P . Thus, n = 4 and i = 3. Condition (0) fails for P , as we can see from the edge graph in Figure 5.8: the removal of the edges v1 and v2 from the edge graph still leaves us with a connected graph.

Condition (1) holds trivially for P since there are no 2-faces containing both v1 and v2, so

E = ∅. Condition (2) also holds: the only common corank-2 face of v1 and v2 is F−1, and we see that the sections v1/F−1 and v2/F−1 share both of their proper faces (G and another vertex). Condition (4) holds since the only faces containing both v1 and v2 are F and P . As noted above, we do not examine Condition (3). In Example 1 in Figure 5.7, we have a polyhedron P with four vertices (two of which are 0 v1 and v2), a square face F in the back, a digon with the edges E and E in front, and two triangles above and below this digon. We consider contracting the square F/F−1. Thus, n = 3 and i = 2. Condition (0) holds for P , as we can see from the edge graph (of course,

Condition (0) holds trivially if rk(F ) = 2, since the graph G is a polygon with v1 and v2 as non-adjacent edges). Condition (1) fails for P since E and E0 are contained in a common 2-face, i.e. the digon. Condition (2) holds trivially since there are no common corank-2 faces of v1 and v2 (again, this condition holds trivially if rk(F ) = 2). Condition (4) holds: every 2-face of the polyhedron contains v1 and v2, and is either F itself or contains E or E0. As noted above, we do not examine Condition (3). In Example 2 of Figure 5.7, we have a polyhedron F with an infinite number of faces. The polyhedron F has only five vertices: A, B, C, D, and G. It has an infinite number of edges between each pair of vertices other than G, shown receding into the corners of the figure (but there are no vertices at the corners). It has exactly one edge between G and each other vertex (so F/G is a square). The polyhedron has four triangular faces in the center and infinitely many digonal faces. Then we get a 4-polytope P by taking a prism over F . Thus, n = 4 and i = 3. In the graph G, we see that G has become a square and the other vertices of F have become apeirogons (the vertices in the graph go on forever in all four directions). Removing v1 and v2 clearly disconnects the graph into a left part and a right part, so Condition (0) holds for P . No 2-faces exist containing both v1 and v2, so

Condition (1) holds trivially since E = ∅. Condition (2) fails with H = F−1 since v1 and v2 do not have both of their vertices in common. Condition (4) succeeds for P since the only faces containing both v1 and v2 are F and P . As noted above, we do not examine

73 Condition (3). In Example 3 of Figure 5.7, we have a simple polyhedron P with four vertices (two of which are v1 and v2), ﬁve edges (one of which is E), and three 2-faces (two triangles in front and one square, F , in the back). We consider contracting F/F−1 in P . Thus, n = 3 and i = 2. Condition (0) holds trivially for P since rk(F ) = 2. Condition (1) also holds trivially since there is only one edge E containing both v1 and v2. Condition (2) holds trivially since 0 0 rk(F ) = 2. Condition (3) fails for P : we see that in the poset P , the vertex ﬁgure P /v1 is not connected since there is no way to move from the faces in the top half of the diagram to the faces in the bottom half of the diagram. Condition (4) holds: all three 2-faces contain both v1 and v2, and they are either the square F or the two triangles containing E. In Example 4 of Figure 5.7, we have a simple polyhedron P with four vertices (two of which are v1 and v2), four edges, and two squares (these squares contain the same vertices and edges; we can visualize one, F , being in front, and the other, F 0, in the back). We consider contracting the square F/F−1 in P . So again, n = 3 and i = 2. Condition (0) holds trivially for P since rk(F ) = 2. Condition (1) holds trivially since there are no edges containing both v1 and v2, so E = ∅. Condition (2) also holds trivially since rk(F ) = 2. 0 Condition (4) fails for P since F contains v1 and v2 but does not contain F . As noted above, we do not examine Condition (3). Condition (0) has some interesting consequences on a polytope P . For example, let A be an (i − 3)-face of P such that A is incident to exactly one of the two (i − 2)-faces v1 or v2. Then if the rank-2 section F/A is not an apeirogon, it corresponds to a cycle in G which contains exactly one of the edges v1 or v2. But every edge of G other than v1 and v2 0 must have both of its vertices assigned to the same face F1 or F2 of P , so since the edges v1 and v2 each have one vertex assigned to F1 and the other vertex to F2, having such a cycle gives us a contradiction (there is no way to assign each vertex of the cycle to a face

F1 or F2 so that every edge of the cycle other than v1 or v2 has its vertices assigned to the same face). Therefore, every such rank-2 section F/A must be an apeirogon. In fact, if no such faces A exist, then Condition (2) immediately follows - i.e., in a polytope without apeirogons, Condition (0) implies Condition (2).

74 5.3 Polygonal Contraction for Polytope Lattices

In this section, we examine the contractability of polygonal sections in polytope lattices and in particular, convex polytopes.

Corollary 5.3.1. Suppose P is a polytope lattice. Then a rank-2 section D = F/G of P with rk(F ) = i and distinguished non-adjacent vertices v1 and v2 is contractible if and only if i = 2 and there is no (i + 1)-face I of P such that I/v1 and I/v2 are both apeirogons.

Proof. We ﬁrst show that the conditions are necessary. So let D be contractible in P . Suppose rk(F ) > 2. Then we have a corank-1 face H of G in P . Then H is a common corank-2 face of v1 and v2 in P . Since P is a polytope lattice and G is a common corank-1 face of v1 and v2, the (i − 3)-face G must be the inﬁmum of v1 and v2 in P . Hence, it 0 0 is not possible that we have another (i − 3)-face G such that v1, v2 > G > H, and so Condition (2) for polygonal contraction fails. Therefore, D cannot be contractible in P , which is a contradiction.

Clearly if there exists an (i + 1)-face I such that I/v1 and I/v2 are both apeirogons, 0 then I/v1 in P is disconnected, as we saw earlier, and so D cannot be contractible, again a contradiction. Now suppose the conditions of the corollary are satisﬁed, that is, rk(F ) = 2 and no face I as described above exists. We verify that the conditions for the polygon contraction hold with i = 2. Then Theorem 5.1 applies, and completes the proof of the corollary. Condition (0) holds trivially since rk(F ) = 2. Condition (1) also holds trivially since E = ∅

(suppose there was some edge E in E; it would be the supremum of the two vertices v1 and v2 of D, and therefore would be contained in F , which clearly cannot be the case in a polytope lattice). Condition (2) holds trivially since there are no corank-2 faces of any vertices. Condition (3) holds because every 3-face containing v1 and v2 must contain F , the supremum of v1 and v2 in P , and so we always are in the case represented by the ﬁrst diagram of Figure 5.5. Condition (4) holds because every face containing v1 and v2 must contain F by the same reasoning as for (3).

Note 5.3.2. If P is a convex polytope, then a 2-section F/G of P with distinguished non- adjacent vertices is contractible if and only if rk(F ) = 2 (since convex polytopes cannot contain apeirogonal sections).

75 Clearly if we perform the polygon contraction on a 2-face of a polytope lattice and the distinguished vertices are two steps apart in the edge graph of the 2-face, we will produce digons, and therefore not get a polytope lattice from the contraction. Another way we can fail to produce a polytope lattice is seen in the example obtained by merging the opposite vertices of the hexagon in a hexagonal pyramid: call these vertices v1 and v2, and call the 0 apex z. Then in P , we have two distinct edges (v1, z) and (v2, z) with identical vertex 0 sets (since v1 and v2 are identified), so clearly P is not a polytope lattice. However, some polytopes produced via polygon contraction are in fact polytope lattices (and even convex polytopes): an example is the resulting polytope when merging the opposite vertices of one hexagon in a hexagonal prism. We see these two examples in Figure 5.9. Note that both examples are looking at a top-down view of the pyramid or prism respectively; in the first example, v1 and z are not identified, but rather z lies above v1 (then after contraction, we have two edges between v1 and z). The result of contraction in the second example may not appear to be quite convex, but the vertices can in fact be arranged in 3-space in such a way that the polytope is convex.

5.4 Connections to the Helfand Construction

As we have seen, Condition (0) of Theorem 5.1 for polygonal contraction is quite strong. It has major consequences on the polytope, and it never holds for polytope lattices unless rk(F ) = 2. We can then ask whether there are ways to ensure Condition (0) if F is of higher rank. In fact, the Helfand construction produces examples with polygonal sections F/G at higher ranks for which Condition (0) holds, as well as Condition (2), as we will show in Theorem 5.2. The following propositions show this:

Proposition 5.4.1. Suppose we apply the Helfand construction to a k-face A of an n- polytope P to get an n-polytope Pˆ. In P , let F be a (k + 3)-face and G a (k − 1)-face with G < A < F . For any face X of P with X > A, we denote by Xˆ the corresponding new face of Pˆ. Then in Pˆ, if the polygonal section Fˆ /G has two non-adjacent vertices vˆ1 and vˆ2, the set {vˆ1, vˆ2} is an edge cut of the graph G, the edge graph of the dual polytope of Fˆ /F−1. In particular, Condition (0) of Theorem 5.1 holds.

Proof. Recall that Fˆ has rank k +2 in Pˆ, so Fˆ /G is a rank-2 section of Pˆ. We ﬁrst examine the corank-1 faces of Fˆ in Pˆ. These are exactly the new faces corresponding to the corank-1

76 Figure 5.9: One example of polygonal contraction in convex polyhedra breaking the lattice condition, and one example (at the bottom) preserving it.

77 faces of F in P which contain A in P , and so they are exactly the edges of the polygon Fˆ /G in Pˆ (since G < A in P ). We next examine the corank-2 faces of Fˆ in Pˆ. These are exactly the new faces corresponding to the corank-2 faces of F in P which contain A in P , and so they are exactly the vertices of the polygon Fˆ /G in Pˆ (since G < A in P ). Hence, G consists entirely of the single cycle corresponding (dually) to the rank-2 sub- section of Fˆ /F−1, so clearly {vˆ1, vˆ2} is an edge cut of G.

Proposition 5.4.2. Suppose we apply the Helfand construction as described in Proposi- tion 5.4.1 to obtain a polytope Pˆ from P . Then for every common corank-2 face H of vˆ1 and vˆ2 in Pˆ, the proper faces of vˆ1/H are the same as the proper faces of vˆ2/H, so in particular, Condition (2) of Theorem 5.1 holds.

Proof. We note thatv ˆ1 andv ˆ2 have rank k in Pˆ. When the Helfand construction is applied to a k-face A of a polytope P , that face is removed and replaced by new k-faces, one for each (k + 1)-face containing A in P . In particular,v ˆ1 andv ˆ2 are new k-faces of this kind. Each of these new k-faces have the same lower-rank incidences as A in P ; they only diﬀer in their higher-rank incidences. Therefore,v ˆ1 andv ˆ2 are incident to the exact same set of faces of lower rank in Pˆ, and so our condition clearly holds.

We can now examine the conditions on P which are necessary and suﬃcient for Fˆ /G to be contractible in Pˆ:

Theorem 5.2. Let P be an n-polytope with a polygonal section D = F/A with rk(F ) = i such that i ≥ 3 (so rk(A) = i−3), and such that we have distinguished non-adjacent vertices v1 and v2 of F/A. Let G be an (i−4)-face of P with G < A. Let Pˆ be the n-polytope achieved from P by applying the Helfand construction (at level i − 3) to A, with polygonal section Dˆ = Fˆ /G determined by the (i − 1)-face Fˆ of Pˆ corresponding to F . Let E be the set of all

(i − 1)-faces in P containing both v1 and v2. Then Dˆ (with distinguished vertices vˆ1 and vˆ2, of ranks i − 3 in Pˆ) is contractible in Pˆ if and only if the following conditions hold in the original polytope P : (1) No two faces in E are contained in a common i-face of P .

(2) For every (i + 1)-face I > v1, v2, the section I/v1 is connected in the poset P¯ obtained by performing polygonal contraction on F/A in P itself (even if P¯ is not a polytope).

(3) For every face J in P containing both v1 and v2, we have J ≥ F or J ≥ E for some E in E.

78 In Figure 5.10, we see an example of the entire process for n = 4 and i = 4. We start with the cubic tessellation P of 3-space and apply the Helfand construction to an edge A (shown at the top of Figure 5.10). Since A is a 1-face, the 4-face F can only be the entire polytope itself, P . The 0-face G can be either of the vertices of this edge; the result is the same. In the 4-polytope Pˆ, the face Fˆ is a 3-face (the middle diagram of Figure 5.10) consisting of four digons tessellating a 2-sphere in an edge-to-edge manner, and then the polygon Fˆ /G is a square (so, the ‘vertices’ of Fˆ /G are the edges of the digons, and its ‘edges’ are the digons themselves). (As a word of caution, although F coincides with the maximal face of P , the corresponding face Fˆ of Pˆ does not coincide with the maximal face of Pˆ; in short, F = P does not imply Fˆ = Pˆ.) Finally, we contract two opposite vertices of Fˆ /G, i.e. two opposite edges in our four-digon polyhedron, which results in the bottom polytope of Figure 5.10: we now have two adjacent polyhedra as facets where A used to be, each consisting of two digons joined edge-to-edge.

Proof of Theorem 5.2. We ﬁrst show that if any of the conditions of this theorem are violated, then the polygon Dˆ is not contractible; that is, the conditions are necessary. Suppose (1) of the present theorem does not hold in P , i.e. we have two faces E0 and E00 in E contained in a common i-face L. In Pˆ, we have a set of new faces Eˆ := {Eˆ|E ∈ E}, with each new face in Eˆ having rank i − 2, one lower than the rank of each face in E. This set of faces Eˆ is exactly the set of (i − 2)-faces in Pˆ containing bothv ˆ1 andv ˆ2. Then in Pˆ, we violate Condition (1) of Theorem 5.1 for polygonal contraction in Pˆ, since Eˆ0 and Eˆ00 are (i − 2)-faces containing bothv ˆ1 andv ˆ2, and are contained in a common (i − 1)-face Lˆ. Suppose (2) of the present theorem does not hold. Then in Pˆ, we violate Condition (3) 0 of Theorem 5.1, since I/ˆ vˆ1 in the poset Pˆ obtained by contracting Fˆ /G in Pˆ is naturally isomorphic to I/v1 in P¯, but is moved down in rank by 1.

Suppose (3) of the present theorem does not hold, i.e. we have some face J > v1, v2 in P such that J does not contain F nor any face in E. Then in Pˆ, we would like to conﬁrm that indeed, Jˆ (which clearly containsv ˆ1 andv ˆ2) does not contain Fˆ nor any Eˆ in Eˆ. This is clear since incidences between new faces in Helfand’s construction are exactly identical to the incidences between their corresponding old faces, i.e. two new faces Bˆ and Cˆ are incident in Pˆ if and only if B and C are incident in P . Hence, Condition (4) of Theorem 5.1 is also violated in Pˆ. Therefore, the three conditions in the present theorem are necessary for Dˆ to be contractible. It remains to show that they are suﬃcient.

79 Figure 5.10: Producing and contracting polygons with the Helfand construction.

80 Suppose the three conditions in the present theorem are satisﬁed for P and F/G. We ﬁrst note that then Conditions (0) and (2) of Theorem 5.1 hold for Pˆ and Fˆ /G, by Propo- sitions 5.4.1 and 5.4.2. We examine Condition (1) of Theorem 5.1 for Pˆ: the (i − 2)-faces of Pˆ containing both distinguished verticesv ˆ1 andv ˆ2 of Fˆ /G are exactly the faces in Eˆ. Clearly no old (i − 1)-face of Pˆ contains two faces from Eˆ since every new face is contained in exactly one old face of one rank higher than itself, and no new (i − 1)-face contains two faces from Eˆ by Condition (1) of the present theorem (i.e. if some new (i − 1)-face Hˆ of Pˆ contained two faces Eˆ1 and Eˆ2, then we would have H > E1,E2 in P , violating Condition (1) of the present theorem), so Condition (1) of Theorem 5.1 holds.

Next, we examine Condition (4) of Theorem 5.1: let J ≥ vˆ1, vˆ2 in Pˆ. Suppose J = Bˆ is a new face. Then by Condition (3) of the present theorem, B ≥ vˆ1, vˆ2, so B ≥ F or B ≥ E for some E in E, so J = Bˆ ≥ Fˆ or J = Bˆ ≥ Eˆ, satisfying Condition (4) of Theorem 5.1 for Pˆ. Now suppose J is an old face. Since J > vˆ1, vˆ2 in Pˆ, we must have J > v1, v2 in P . Then by Condition (3) of the present theorem, J ≥ F or J ≥ E for some E in E, and therefore J ≥ Fˆ or J ≥ Eˆ. Hence, we have Condition (4) of Theorem 5.1 for Pˆ in this case as well, and so Condition (4) is satisﬁed in all cases. Finally, we examine Condition (3) of Theorem 5.1 for Pˆ: let I be an i-face of Pˆ containing vˆ1 andv ˆ2. Suppose I is a new face with I = Cˆ for some old face C. Then as mentioned earlier, the section I/vˆ1 in the contracted version of Pˆ is identical to the section C/v1 in P¯ (up to a shift in rank), and so Condition (3) of Theorem 5.1 for Pˆ holds by Condition (2) of the present theorem for P . Now suppose I is an old face. Then if I > Fˆ, we must have I = F , in which case I in P contains none of the faces from E and hence, as a face of

Pˆ, none of the faces from Eˆ; moreover, we note that I/vˆ1 (and likewise I/vˆ2) must be a triangle, and hence not an apeirogon, as we can see in Figure 5.11. This gives us Condition (3) of Theorem 5.1 for Pˆ in this case. Now suppose that the old face I does not contain Fˆ. Then since I containsv ˆ1 andv ˆ2, I must contain v1 and v2. Then by Condition (4) of Theorem 5.1 for Pˆ (which we already know must hold), I > Eˆ for some Eˆ in Eˆ (since we already know that I does not contain Fˆ and hence does not contain F ). Moreover, I, as a face of P , cannot contain more than one face in E by Condition (1) of the present theorem. We then see, again, that I/vˆ1 must be a triangle (this is illustrated in Figure 5.12), and so Condition (3) of Theorem 5.1 is again satisﬁed for Pˆ. Hence, we have Condition (3) of Theorem 5.1 for Pˆ in all cases.

81 Figure 5.11: The Helfand construction, in this case, produces a triangle from a line segment.

Figure 5.12: The Helfand construction produces a triangle here as well.

82 Therefore, since all the conditions for polygonal contraction hold for Pˆ, the section Fˆ /G in Pˆ is contractible.

Note 5.4.3. With some basic investigation, we can see that Theorem 5.2 still holds if we apply the global Helfand construction instead of the local Helfand construction.

Corollary 5.4.4. Let P be a polytope lattice of rank n, and let Pˆ be an n-polytope obtained from P by applying the Helfand construction to a k-face A with 0 ≤ k ≤ n − 4. Let F and G be faces of P with rk(F ) = k + 3, rk(G) = k − 1, and G < A < F . Then the polygonal section Fˆ /G in Pˆ is contractible if and only if Fˆ /G has at least four vertices and for every

(i + 1)-face I of P , at least one of the sections I/v1 and I/v2 of P is a ﬁnite polygon. In particular, if P is a convex polytope then Fˆ /G in Pˆ is contractible if and only if Fˆ /G has at least four vertices.

Proof. We first show that Fˆ /G is not contractible in Pˆ if Fˆ /G has at most three vertices or there exists an (i + 1)-face I of P such that I/v1 and I/v2 are both apeirogons in P . Clearly if Fˆ /G is a digon or triangle then there are no two non-adjacent vertices in Fˆ /G, so Fˆ /G is not contractible. If I/v1 and I/v2 are both apeirogons in P , then the resulting rank-2 section I/v1 in P¯ is not connected, so Condition (3) of Theorem 5.2 fails. We now show that if Fˆ /G has at least four vertices and for every (i + 1)-face I in P at least one of the sections I/v1 and I/v2 is a finite polygon, then Fˆ /G is contractible in Pˆ. We do this by examining the three conditions of Theorem 5.2. Since P is a polytope lattice, we immediately know that E = ∅, and so Condition (1) of Theorem 5.2 holds trivially. As noted in the earlier investigation of polygonal contraction in polytope lattices, Condition (2) of Theorem 5.2 holds if and only if there are no (i + 1)- faces I such that I/v1 and I/v2 are both apeirogons. We can also see that Condition (3) of Theorem 5.2 must hold since F must be the supremum of v1 and v2. Therefore, since all three conditions of Theorem 5.2 hold, Fˆ /G is contractible in Pˆ. Finally, if P is a convex polytope, then every section of P is finite, so only the condition on the size of Fˆ /G remains.

83 Chapter 6

Polygonal Contraction and Symmetry

In this chapter, we discuss possible global versions of polygonal contraction: instead of contracting a single polygonal section, we contract a set of polygonal sections successively. If the contraction is performed in some way symmetrically, then we may know something about the automorphism group of the resulting polytope.

6.1 Multiple Polygonal Contraction

We first consider the question: suppose we have two polygonal sections F/G and F 0/G0 in a polytope P , both of which are contractible; can they both be contracted successively? In other words, if we first contract F/G to get a polytope P 0, then is F 0/G0 still contractible in this new polytope? In general, this question is not clear (and we can find various examples in which both polygonal sections cannot be contracted), but we can produce some sufficient conditions on the two polygonal sections in P that will allow both to be contracted:

Proposition 6.1.1. Let P be an n-polytope with two contractible polygonal sections F/G and F 0/G0, where F and F 0 both have rank i in P . If F 6= F 0 and the two polygonal 0 0 sections do not share distinguished vertices (meaning that the sets {v1, v2} and {v1, v2} of distinguished vertices for F/G and F 0/G0, respectively, are disjoint), then both polygonal sections can be contracted (meaning that in P 0, the polytope produced by contracting F/G, the section F 0/G0 still exists and is contractible).

84 Proof. We confirm that each of the five conditions of Theorem 5.1 holds for F 0/G0 in P 0. First, we examine Condition (0) for P 0. We know that in P , Condition (0) holds for 0 0 0 0 F /G . The contraction of F/G in P has an effect on F /F−1 (and hence the edge graph G 0 ∗ 0 of its dual (F /F−1) ) only if F contains the distinguished vertices v1 or v2 of F/G in P . 0 0 0 Hence, if F does not contain v1 or v2, then F /F−1 survives in P and Condition (0) holds for F 0/G0 in P 0 since the graphs G0 are identical in P 0 and in P . 0 Then suppose that F contains exactly one of the faces v1 or v2 in P (although not as a distinguished vertex of F 0/G0). Clearly in P 0, the graph G0 will be identical to G0 in P , up to the renaming of the edge v1 or v2, respectively, as the merged “edge” v1: the edge will 0 still be present, and it must still contain the same vertices, since the merged face v1 in P has all of the incidences that v1 and v2 do in P . 0 Finally, suppose that F contains both of the faces v1 and v2 in P (again, neither as a distinguished vertex of F 0/G0). Then by Condition (4) of Theorem 5.1 for F/G and P , the face F 0 must contain some (i − 1)-face E from E, the set of (i − 1)-faces of P containing 0 0 v1 and v2. Suppose F contains a second such face E : we know from Condition (2) of

Theorem 5.1 for F/G in P along with Lemma 3.1.2 (applied with v1 and v2 in place of H 0 0 0 and H ) that E and E each contain no (i − 2)-faces other than v1 and v2. Then in G in P , 0 each of the vertices E and E can only be contained in the two edges v1 and v2 joining E and 0 0 0 0 E . However, G in P must have other edges (e.g. v1 and v2), so the graph is disconnected, which gives us a contradiction to the assumption that there are two faces of E containing 0 v1 and v2. So F must contain a face E from E, but cannot contain any other face from E. 0 Then in G in P , E is a vertex contained only in the two edges v1 and v2; this follows as 0 before. Let A be the second vertex of v1 and B be the second vertex of v2. Then in G 0 in P , the edges v1 and v2 have been merged and E has been removed, so that the merged 0 0 0 edge v1 now contains the two vertices A and B. Now since {v1, v2} is an edge cut of G in P 0 0 0 0 and v1 and v2 are both distinct from v1 and v2, we can see that {v1, v2} is also an edge cut 0 0 of G in P (the only change to the graph was the replacement of the two edges v1 and v2 adjacent at the vertex E of degree 2, with one edge). Hence, in all cases, Condition (0) of Theorem 5.1 holds for F 0/G0 and P 0. We see that Condition (1) of Theorem 5.1 holds since no (i − 1)-faces of P acquire any additional incidences with i-faces of P (the only change is that F has been split into two new faces F1 and F2, but each (i−1)-face incident to F in P is incident to exactly one of F1 0 and F2 in P ). We see that Condition (2) of Theorem 5.1 holds since all (i − 2)-faces have

85 0 the same lower-rank incidences in P as they did in P , except for v1 and v2, which are by assumption not the distinguished vertices of F 0/G0, and hence irrelevant to Condition (2) on F 0/G0. Now consider Condition (3) of Theorem 5.1 for F 0/G0 and P 0: Let I be an (i + 1)-face 0 0 0 0 0 with I > v1, v2 in P . We note that I must then contain v1 and v2 in P . Examining the 0 0 0 polygonal sections I/v1 and I/v2 in P , we see in the following argument that they must in 0 0 fact be identical to I/v1 and I/v2, respectively, in P . First of all, the only i-face affected by 0 0 0 the construction of P is F , but since v1 and v2 are not the distinguished vertices of F/G, 0 0 Condition (0) on F/G and P tells us that, if F > v1 or F > v2 in P , then both proper faces 0 0 0 of F/v1 or F/v2, respectively, in P will lie in the same face F1 or F2 in P , meaning that 0 0 the relevant incidences in I/v1 and I/v2 are unaffected; F is merely renamed to F1 or F2. Second, the only (i − 1)-faces affected by the construction of P 0 are the (i − 1)-faces (in E) containing both v1 and v2 in P , which by Condition (2) with Lemma 3.1.2 cannot contain 0 0 any (i − 2)-faces other than v1 and v2, which are distinct from v1 and v2 by assumption; 0 0 0 hence, none of these faces from E appear in I/v1 or I/v2 in P . Therefore, we see that I/v1 0 0 0 0 and I/v2 are identical in P and in P , and so Condition (3) of Theorem 5.1 holds for F /G and P 0 since it holds for F 0/G0 and P . Finally, consider Condition (4) of Theorem 5.1 for P 0: let J be a face in P 0 containing 0 0 both v1 and v2. If a face K in P of rank greater than i is incident to F , then as above, we 0 0 0 see that its incidences (or lack thereof) to v1 and v2 are the same in P as in P . If K is not incident to F , then clearly the incidences are again the same, since any sections of the 0 0 0 form K/v1 or K/v2 are entirely unaffected by the construction of P . Hence, the sections 0 0 0 0 0 J/v1 and J/v2 are identical in P and in P , so by Condition (4) on F /G and P , we know 0 0 0 that J is incident to F or some face in E , the set of (i − 1)-faces of P containing v1 and 0 0 0 0 0 v2. In P this is still the case, so Condition (4) must also hold for F /G and P . Therefore, all five conditions of Theorem 5.1 hold for P 0, and so F 0/G0 is contractible in P 0.

Note 6.1.2. The above proposition extends naturally to the following statement: given a ﬁnite set of contractible polygonal sections {F/G}, all with top faces of the same rank, if no two such polygons share greatest faces and no two share distinguished vertices, then all of them can be contracted successively. For example, given the truncated square tiling (the semiregular tiling of the plane with squares and octagons), each square is contractible and no two squares share vertices, so we can contract all squares successively into pairs of digons

86 (choosing pairs of opposite vertices as distinguished vertices arbitrarily for each square).

Note 6.1.3. When contracting multiple polygonal sections as above (i.e. under the conditions of the proposition), the order in which the polygonal sections are contracted is irrelevant to the structure of the resulting polytope, so we can think of these contractions occurring simultaneously rather than in sequence. Similarly, we can think of contracting a countably infinite number of polygonal sections simultaneously, so we need not restrict ourselves to finite numbers of polygons (it follows from the defining condition (P3) of abstract polytopes that an abstract polytope must have a countable number of faces, and hence a countable number of polygonal sections).

6.2 Connections to Symmetry

Proposition 6.2.1. Let P be a polytope with a (countable) set of distinguished simultaneously contractible polygonal sections (as in Proposition 6.1.1) {D = F/G}. Let Λ denote the subgroup of the automorphism group Γ(P ) consisting of all automorphisms that permute the polygonal sections D as well as the corresponding sets of distinguished vertices (i.e., we are considering sections D along with their distinguished vertices for contraction, and we consider automorphisms of P that take a distinguished polygonal section and its set of distinguished vertices to a distinguished polygonal section and its set of distinguished vertices). Let P 0 be the polytope achieved after contracting every section D. Then Λ induces a subgroup Λ0 of Γ(P 0) isomorphic to Λ, or more informally, Λ ≤ Γ(P 0).

Note 6.2.2. There is some abuse of notation in Proposition 6.2.1, as we are using the same symbol Λ for a subgroup of Γ(P ) and of Γ(P 0). For this to make sense, for every automorphism σ ∈ Λ we need a well-defined induced face map of P 0 which we can also call σ (the proposition then says that this induced map σ is in fact an automorphism of P 0, and that the set of all such induced maps is a subgroup of Γ(P 0) isomorphic to Λ). We define the induced map σ on P 0 as follows. First, consider the effect of the original σ in P on a face other than a greatest face of a distinguished polygonal section or a distinguished vertex. Since such faces survive in P 0 under the contraction and no automorphism in Λ can send such a face to a greatest face of a distinguished section or a distinguished vertex, we can define σ in P 0 to act identically on such faces to σ in P . Next, consider the action of σ in P on a distinguished vertex: it must send any given distinguished vertex v to some distinguished vertex v0. In P 0, we have corresponding vertices to v and v0, since the contraction has merged

87 v with some other vertex and v0 with some other vertex, so we can have σ in P 0 send the ﬁrst merged vertex to the second. Finally, consider the action of σ in P on a distinguished greatest polygon face D: it must send it to a distinguished greatest polygon face D0. 0 0 0 0 In P , the section D is split into polygons D1 and D2, and D is split into D1 and D2; 0 0 we need to send each of D1 and D2 to one of D1 and D2 in a natural way. To do this, we examine the automorphism σ on P : it sends each proper face of D to a proper face of D0, and sends the sets of distinguished vertices of D to that of D0. This means that there is a 0 0 natural correspondence between each of D1 and D2 and one of D1 and D2, which we can

ﬁnd by examining their proper faces (so given a proper face of D1, we see what σ in P 0 0 sends that face to, and the result will be some proper face of either D1 or D2; we use this 0 0 to determine which of D1 or D2 the face D1 should be sent to). Hence, we see that each σ ∈ Λ in P has a natural corresponding face map in P 0, which we can also call σ, and so we have a well-deﬁned set of face maps in P 0 induced by Λ; it remains to show that they are all automorphisms, and that they form a group isomorphic to Λ.

Proof. Let σ be the face map on P 0 induced by σ ∈ Λ, so σ is a bijection of P 0. In order for σ to be an automorphism, it needs to preserve incidences. We consider incidences between faces with difference 1 in rank (clearly if σ preserves these incidence, it preserves all incidences). We first note that σ must send greatest faces of contracted polygonal sections to greatest faces of contracted polygonal sections, merged vertices to merged vertices, and all faces not directly affected by the contraction (i.e. all other faces) to faces not directly affected by the contraction; this is clear from the definition of Λ in P . So consider the incidence I < J in P 0, where rk(J) = rk(I) + 1. If both I and J were not directly affected by the contraction, then σ(I) and σ(J) in P 0 are identical to the original σ(I) and σ(J), respectively, in P , and so σ(I) < σ(J) in P 0. Suppose I is the greatest face of a contracted polygonal section, so rk(I) = i; then J is some (i + 1)-face incident to I. In P , the original σ is an automorphism, so it sends this incidence (replacing I with the greatest face of the original, not contracted polygonal section of P ) to another incidence; in P 0, since the greatest faces of contracted polygonal sections inherit all incidences to higher-rank faces, we see that σ also preserves this incidence, that is, σ(I) < σ(J) in P 0. Now suppose J is the greatest face of a contracted polygonal section. In P , the face I is also incident to J (replacing J with the greatest face of the original, not contracted polygonal section of P ); the automorphism σ in P sends this incidence to another incidence of the greatest face of a distinguished polygonal section with one of its

88 corank-1 faces. In P 0, the face σ(J) is the greatest face of a contracted polygonal section, and σ(I) is incident to σ(J); we are assured that σ(I) in P 0 is incident to σ(J) in P 0 and not to the other contracted polygonal section corresponding to σ(J) in P , by the definition of the action of σ in P 0 as described in Note 6.2.2. Hence, σ(I) < σ(J) in P 0. Now suppose I is a merged vertex. In P , the face J contains exactly one of the distinguished vertices corresponding to I, and σ sends this incidence in P to another incidence of a rank-(i − 1) face containing a distinguished vertex. Hence, in P 0, the face map σ sends the corresponding incidence to another incidence of a rank-(i − 1) face containing a merged vertex, and again σ(I) < σ(J). Now suppose J is a merged vertex: then I is an (i − 3)-face incident to J in P 0. In P , the corresponding I is an (i − 3)-face incident to one or both of a pair of distinguished vertices, and σ sends this incidence to another incidence of an (i − 3)-face with one or both of a pair of distinguished vertices. Hence, in P 0, the induced map σ sends this incidence to another incidence, and so σ(I) < σ(J) as before. Therefore, σ preserves all incidences in P 0, and so is an automorphism. Let Λ0 be the set of all face maps of P 0 induced by automorphisms of P in Λ. Then it is 0 0 clear that Λ is a group. In fact, given two automorphisms σ1 and σ2 in Λ , the composition 0 σ1 · σ2 is a well-defined automorphism in P , and moreover naturally corresponds to the composition σ1 · σ2 of the original automorphisms σ1 and σ2 of P . Then all other defining properties for groups follow immediately. Thus, Λ0 ≤ Γ(P 0). It remains to show that Λ0 is isomorphic to Λ. To do this, we show that an automorphism σ in P 0 can only be the identity map if the original σ in P is the identity map. Suppose we indeed have an automorphism σ in P 0 which is the identity map. Then, as is evident from the discussion in Note 6.2.2, the original σ in P must be the identity map on all faces of P other than, possibly, some distinguished vertices: it may be that σ in P sends a pair 0 of distinguished vertices v1 and v2 to v2 and v1, respectively. In P , denote by {L1,L1} the 0 set of edges containing v1 in the polygonal section D, and denote by {L2,L2} the set of edges containing v2 in the polygonal section D. Since v1 and v2 are non-adjacent vertices 0 0 in D, the two sets {L1,L1} and {L2,L2} are disjoint; moreover, the original σ in P must send each of these four edges to themselves. However, we then have a contradiction, since

σ in P does not preserve the incidence v1 < L1. Hence, σ in P must send v1 to v1 and v2 0 to v2, and is hence the identity map. Therefore, Λ is isomorphic to Λ.

Note 6.2.3. Although we must have Λ0 ≤ Γ(P 0), it is not always the case that Λ0 = Γ(P 0); this is sometimes true and sometimes false. We see some examples below.

89 In the first example, we start with the Archimedean tiling of the plane by squares and octagons, known as the truncated square tiling - call this polyhedron P (see Grünbaum and Shephard (1987) for information on Archimedean tilings). We show part of P in Figure 6.1, with three reflections marked in blue so that P has the following automorphism group:

2 4 2 4 Γ(P ) = hr0, r1, r2 | ri , (r0r1) , (r0r2) , (r1r2) i

1 This is the string Coxeter group [4, 4] and a wallpaper group of type W4 (see Martin (1982, Chapter 11) and Grünbaum and Shephard (1987) for information on wallpaper groups). We consider either contracting all octagons (distinguished vertices marked in green, path of contraction shown as a green line) or all squares (distinguished vertices marked in red, path of contraction shown as a red line). Note that each individual square is indeed contractible, and so is each individual octagon; we can in fact simultaneously contract all octagons and/or all squares according to the distinguished vertices shown, as noted earlier. Also note that there are infinitely many ways to choose which squares to contract (in this case, we contract every square) and/or which octagons to contract (in this case, we contract every octagon), and how to choose their distinguished vertices; in this case, we choose to contract all squares and/or all octagons with distinguished vertices chosen in a highly symmetric way in order to produce highly symmetric polyhedra via contraction. In fact, in this example, the sets of distinguished squares and distinguished vertices can be expressed as the orbits of a single distinguished square and pair of distinguished vertices under a subgroup of Γ(P ); this of course is not true in general for an arbitrary choice of distinguished squares and distinguished vertices (the same applies to the octagons in this example). We first consider contracting (only) all octagons. This gives us a tiling of the plane by

(combinatorial) squares, shown in Figure 6.2. For this polyhedron, which we will call P1, the group Γ(P1) consists precisely of the automorphisms of P1 induced by automorphisms of P which permute the sets of distinguished vertices (clearly they also permute the octagons).

The automorphism group Γ(P1), in terms of r0, r1, and r2, is hr2r1r0r1, r0r1r2r1, r0r2i (the generators here are a vertical translation moving an octagon to an adjacent octagon, a horizontal translation moving an octagon to an adjacent octagon, and a 180-degree rotation, respectively), which is a wallpaper group of type W2. We next consider contracting all squares in P and not the octagons. This gives us a

90 Figure 6.1: The truncated square tiling, with some reﬂections and paths of contraction shown.

91 Figure 6.2: The polyhedron P1 obtained by contracting all octagons in the truncated square tiling.

92 Figure 6.3: The polyhedron P2 obtained by contracting all squares in the truncated square tiling. tiling of the plane by octagons and digons, shown in Figure 6.3. For this polyhedron, which we will call P2, the group Γ(P2) consists precisely of the automorphisms of P2 induced by automorphisms of P which permute the set of distinguished vertices (of course they also permute the squares). The automorphism group is hr2r1r0r1, r0r1r2r1, r0, r2i, which is a 2 wallpaper group of type W2 . Finally, we consider contracting all squares and all octagons in P (we can equivalently think of this as contracting all squares in P1 or all octagons in P2) to get the polyhedron P12 shown in Figure 6.4. The polyhedron P12 is a tiling of the plane by squares and digons, and we can compare Γ(P12) to Γ(P ), Γ(P1), and Γ(P2). The automorphisms of P permuting sets of distinguished vertices in octagons of P induce automorphisms of P1, and those which also permute the sets of distinguished vertices in squares of P induce automorphisms of P1 which permute the sets of induced distinguished vertices in squares of P1, and in turn the latter induce automorphisms of P12. The same holds with “squares” and “octagons” and

93 P1 and P2 interchanged, and yields the same set of automorphisms of P12, a subgroup of

Γ(P12). Moreover, this subgroup also coincides with the set of all automorphisms of P12 induced by all automorphisms of P which permute the sets of distinguished vertices in squares and the sets of distinguished vertices in octagons. However, this subgroup of Γ12 is proper; in fact, P12 has additional automorphisms, for example the reﬂection r3 depicted in

Figure 6.4 which does not exist in any of the polyhedra P , P1, or P2. The automorphism 2 group Γ(P12) is again a wallpaper group of type W2 , the same type as Γ(P2). The group is diﬀerent from Γ(P2) and is given by the following:

2 4 4 Γ(P12) = hr2r1r3r2, r3r1, r2r3r2, r3 | ..., r3, (r0r3) , (r2r3) , r3r1r3r0r2r1r2r0i

Here, the four listed generators are two translations along the main diagonals of Figure 6.4 and two reﬂections in lines parallel to the main diagonals of Figure 6.4. The same relations hold for Γ(P12) as for Γ(P ) (we have replaced those with ellipses in the list of relations), but we have several new relations as well, which concern the new reﬂection r3.

A natural question now is whether we can continue contracting polygons in P12 in such a way that we get some minimal polyhedron, i.e. a polyhedron with no contractible polygons.

In the case of P12, it seems as though no matter how many contractions we perform (of course, it only makes sense to talk about performing a finite number of global contractions of this type), we will still have a polyhedron with contractible squares. For example, one natural way to contract P12 is to contract the squares in the orbit of a single square and a pair of opposite vertices under the subgroup of Γ(P12) generated by a single translation ‘over’ (i.e. to an adjacent square) and a double translation ‘down’ (i.e. across a digon, a square, and another digon). This produces a similar polyhedron (with an isomorphic automorphism group), which we can again contract in a similar way, and we can repeat this process indefinitely; the first step is shown in Figure 6.5 (note that we start with a combinatorially identical but geometrically ‘straightened out’ version of P12, and that the resulting polytope as drawn is also ‘straightened out’). If the number of digons between a pair of horizontally-adjacent vertices is n on a given step, then it will be 2n + 2 at the next step. Note that we can imagine this sequence of contractions tending towards a certain polyhedron, i.e. one with a single (horizontal) line of adjacent vertices, each of which has a (countably) infinite number of digons to the next vertex; however, this is not actually a polytope, as the vertex figures are not connected.

94 Figure 6.4: The polyhedron P12 obtained by contracting all squares and octagons all in the truncated square tiling.

95 Figure 6.5: Further contraction of the truncated square tiling.

96 Figure 6.6: The truncated square tiling, with a diﬀerent plan to contract the octagons.

While we may not be able to get a maximally-contracted polytope from P12, we can actually get one from the original tessellation P via the sequence of contractions shown in Figures 6.6 through 6.9 (note that the marked vertices in each figure refer to the vertices that will be contracted in the next step, not the ones that were contracted in the previous step). The final polyhedron that we get in Figure 6.9 is a tessellation of the plane by triangles and digons; since there are no polygons with at least four vertices, clearly we cannot contract polygon faces any further, although in principle, contractions of vertex-figures might still be possible. A natural question to ask is: given some polyhedron with an infinite number of 2-faces, does it have a maximal contraction? In other words, is it possible to contract polygon faces in the polyhedron to get a polyhedron with no contractible polygon faces? A related

97 Figure 6.7: The resulting polyhedron from the previous contraction deﬁned by Figure 6.6, with a plan to contract the pentagons.

98 Figure 6.8: The resulting polyhedron from the previous contraction deﬁned by Figure 6.7, with a plan to contract the squares.

99 Figure 6.9: The ﬁnal, maximally-contracted polyhedron, with only digonal and triangular 2-faces.

100 question is: does every maximally-contracted polyhedron have only digons and triangles as faces, or are there maximally-contracted polyhedra with larger faces?

Definition 6.2.1. An abstract (n + 1)-polytope P (with finite vertex-figures) is a (locally n finite) face-to-face tessellation of R (Schulte, 1993, Chapter 3.5) if P can be embedded n in R so that each of its facets is a geometrically convex n-polytope and the intersection of any two faces is a (possibly empty) face of P .

Note 6.2.4. The speciﬁcation of geometric convexity in the above deﬁnition is important n since it may be the case that an (n + 1)-polytope P is embedded in R and has facets which are isomorphic to convex n-polytopes without actually being geometrically convex; such an example is discussed below.

2 For example, the truncated square tiling is a face-to-face tessellation of R by convex tiles, as are the polyhedra in Figures 6.2 and 6.7. The polyhedra in the other related figures are not, since they have digonal facets, which are not convex 2-polytopes. However, if we were to contract all of the digons in those figures, we would indeed get face-to-face 2 tessellations of R , except in Figure 6.3: in Figure 6.3, there is no way to embed the resulting 2 polyhedron in R while making all of the octagonal facets geometrically convex octagons. An important fact, which follows quickly from the definition of a face-to-face tessellation, is that every face-to-face tessellation is a polytope lattice. n By Corollary 5.3.1, we see that a polygonal section of a face-to-face tessellation P of R is contractible if and only if the polygonal section is at level 2 (that is, it is actually a 2-face of P ) and has at least four vertices (face-to-face tessellations cannot contain apeirogonal faces since every facet of a face-to-face tessellation is a convex polytope). In other words, n the only way to have a face-to-face tessellation of R which is maximally contracted is to have every 2-face be a triangle. Also, with Proposition 6.1.1, we see that we can contract a n set of 2-faces in a face-to-face tessellation of R if no two of the 2-faces share either of their distinguished vertices. We would like to extend this result to the other polytopes we have seen in this section which arise from face-to-face tessellations via polygonal contraction; however, as we will see, we run into some issues.

Lemma 6.2.5. Let P be an abstract polytope with the property that (*) no non-adjacent vertices of a 2-face of P are joined by an edge of P . Let P 0 be a polytope obtained from P by contracting certain 2-faces of P once: this means that we may perform a multiple polygonal

101 contraction as in Prop. 6.1.1 on P to contract a ﬁnite or inﬁnite number of 2-faces of P at once. Then P 0 has the property (*).

0 Proof. Let F be a 2-face of P with non-adjacent vertices v1 and v2. Then F corresponds to exactly one 2-face of P , since the contraction of 2-faces in P aﬀects each 2-face of P by either leaving it unchanged or splitting it into two 2-faces. 0 Suppose F in P is identical to its corresponding 2-face in P , and an edge E joins v1 and 0 v2 in P . Then E must join v1 and v2 in P since polygonal contraction does not produce new edges, but this violates our assumption about P , since then F in P has two non-adjacent vertices joined by an edge in P . Therefore, we have a contradiction, and so no edge can 0 join v1 and v2 in P . 0 Now suppose F in P is the result of contracting a 2-face G in P , and an edge E joins v1 0 and v2 in P . Then v1 and v2 are also non-adjacent vertices of G in P and are also joined by E in P , so again our assumption about P is violated and we have a contradiction. Again, 0 we see that no edge can join v1 and v2 in P . 0 Hence, in all cases, v1 and v2 cannot be joined by any edges in P , and so the property (*) holds for P 0.

By applying Lemma 6.2.5 repeatedly, we see that if we start with any abstract polytope P with the property (*) (in particular, a polytope lattice has this property) and perform multiple polygonal contraction (as in Prop. 6.1.1) any number of times, we end up with a polytope with the property (*). Using this result, we can see that a polytope achieved from a lattice polytope by polygonal contractions satisfies certain conditions for polygonal contraction. Given a face F at rank 2 of the polytope with two non-adjacent vertices, we see that the set E as defined in Theorem 5.1 is empty. Since F has rank 2 and E = ∅, several conditions for polygonal contraction are satisfied, but we run into issues with Condition (4) of Theorem 5.1. Condition (4) holds for any distinguished 2-face (with distinguished non-adjacent vertices) of a polytope lattice, but may not hold for all distinguished 2-faces of a polytope achieved from a polytope lattice via polygonal contractions. We see such an example in

Figure 6.3: let F be any octagon of the polyhedron shown, and set v1 and v2 to be non- adjacent red and green (respectively) vertices at the top of the octagon. Then let J be the octagon adjacent to F and directly above it: we see that J contains v1 and v2 but not F (and E = ∅), so Condition (4) of Theorem 5.1 is violated and we cannot contract the

102 octagon F . Of course, as we have already seen, we can contract octagons in this polyhedron; we just need to pick the appropriate distinguished vertices. Hence, it is possible that given a polytope arising from a polytope lattice via polygonal contraction, we can contract any 2-face with at least four vertices as long as we choose the correct vertices to merge. Whether this is in fact the case is not clear.

103 Bibliography

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