From polytopes to enumeration Ed Swartz - Cornell Universitypi.math.cornell.edu/~karola/Ed.Swartz.pdf · 2013. 1. 21. · 1. V-polytopes 11 2. Examples of Polytopes 13 3. Faces of

From polytopes to enumeration

Ed Swartz

c� Draft date July 22, 2012

Contents

Contents i

List of Figures vii

List of Tables ix

Preface 1

Preface 1

Chapter 1. Introduction 3

Chapter 2. A�ne and convex geometry 51. A�ne subspaces 52. Convexity 73. Simplices 9

Chapter 3. What is a polytope? 111. V-polytopes 112. Examples of Polytopes 133. Faces of polytopes 16

Chapter 4. Convexity 191. Projection 192. Extreme points 203. V and H-polytopes 21

Chapter 5. Shelling 251. Shellings 252. Geometric simplicial complexes 293. Line shellings 31

Chapter 6. Duality 371. Duality for convex sets 372. Polytope duality 40

Chapter 7. The Mobius function 431. The Mobius function 43

iii

iv CONTENTS

2. Mobius inversion 44

Chapter 8. Hyperplane arrangements and Zonotopes 511. Hyperplane arrangements 512. Zonotopes 543. The intersection poset 564. Acyclic orientations 57

Chapter 9. Matroids 611. Flats 622. Broken circuits 633. Matroid duality 65

Chapter 10. Graphs of polytopes 691. Acyclic orientations of G(P ) 692. Simple polytopes 703. Balinkski’s theorem 714. f -vectors of 3-polytopes 73

Chapter 11. Hints 751. A�ne and convex geometry 752. What is a polytope? 753. Convexity 764. Shelling 765. Duality 766. Mobius inversion 777. Hyperplane arrangements and zonotopes 778. Matroids 779. Graphs of polytopes 78

Chapter 12. Solutions 791. A�ne and convex geometry 792. What is a polytope? 813. Convexity 834. Shelling 855. Duality 896. The Mobius function 917. Hyperplane arrangements and zonotopes 938. Matroids 979. Graphs of polytopes 100

Appendix A. Analysis 1051. Compactness 105

Appendix. Bibliography 107

CONTENTS v

Bibliography 107

List of Figures

1 Examples of convex subsets of R2 8

1 An H-polytope 13

2 Prism (P ) 14

1 ch(ext K) = K 21

2 P = \Hi 23

1 A shelling of the boundary of the octahedron 25

2 The beginning of a non-shelling of the boundary of theoctohedron 26

3 Minimal new faces of the shelling 26

4 G and � 28

5 A geometric simplicial complex in R2 30

6 Two examples which are not geometric simplicial complexes inR2 30

7 Line shelling of a polygon 33

1 A planar graph and its dual 39

1 Computing µ(x, y). 44

2 A simple graph 47

3 LG 48

1 Several sign vectors 52

2 Face fan of a polygon 53

3 Cube, translation and prisms as Minkowski sums 55

4 Two orientations of a graph, one acyclic 58

1 Two di↵erent edge orderings for the same graph 64

1 Before and after chopping o↵ a vertex 102

vii

List of Tables

ix

Preface

1

CHAPTER 1

Introduction

The main goal of the text is give students an opportunity to discoversome the remarkable connections between geometry and combinatorics.In contrast to many branches of mathematics, the ideas here are onesthat beginning students, those who have finished a calculus sequence,linear algebra and one rigorous course, can appreciate.

A significant portion of the development of the material is left tothe student in the form of problems. The hope is that the presentation,first as a problem, then a brief hint and finally a full solution, allowsstudents at di↵erent parts of their education to get as much out of thetext as possible.

None of the material here is new. The polytope material is basedon the two well known texts by Grunbaum [11] and Ziegler [29]. Forinstance, the proof of the equivalence of H-polytopes and V-polytopesis based on the convexity presentation in [11] while the zonotope sectionclosely follows [29]. However, we do believe that bringing together inone place all of these closely related ideas and notions is new.

3

CHAPTER 2

A�ne and convex geometry

1. A�ne subspaces

In order to prepare for our journey from polytopes to enumerationwe begin with two concepts crucial to the geometry of polytopes, a�nesubspaces and convexity. An a�ne subspace of Euclidean space is anytranslation of a linear subspace. The translation of a subset A of Rd

by v 2 Rd is defined by

A + v = {x + v : x 2 A}.An a�ne subspace of Rd is a subset of Rd which is either empty orof the form W + v, where W is a linear subspace. Familiar examplesinclude a point, a line or a plane. Since most geometric concepts suchas dimension, distance, and angle are invariant under translation, weshould expect them to behave reasonably in a�ne spaces.

Problem 1. Suppose A = W + v = W 0 + v0 is an a�ne subspace ofRd. Prove that W = W 0. Also, A + (�y) = W for every y 2 A.

As is the case with linear subspaces, the notions of dimension, spanand independence are critical to understanding a�ne subspaces. Thedimension of the empty set is defined to be �1, otherwise the dimensionof an a�ne subspace W +v is defined to be the dimension of the linearsubspace W. The previous problem shows that this is well defined andthat the dimensions of points, lines and planes are what we expect: 0, 1and 2 respectively.

Problem 2. The intersection of a set of a�ne subspaces of Rd is ana�ne subspace of Rd.

Let A ✓ Rd. Consider the intersection of all a�ne subspaces whichcontain A. This is a nonempty intersection since Rd is such a set. ByProblem 2 this intersection is the smallest a�ne subspace which con-tains A. It is called the a�ne span of A and we denote it by aspan(A).

The a�ne analogues of linear combinations are called a�ne combi-nations. A linear combination of the form

a1x1 + · · · + anxn

5

6 2. AFFINE AND CONVEX GEOMETRY

is an a�ne combination if a1 + · · · + an = 1.

Problem 3. The a�ne span of A equals the set of all a�ne combina-tions of elements of A.

What is the a�ne notion corresponding to linear independence? If{x1, . . . ,xn} is a finite set of distinct point of Rd, then we say they area�nely independent if their a�ne span has dimension n� 1.

Problem 4. A subset A of Rd is a�nely independent if and only ifno element of A can be written as an a�ne combination of the otherelements of A.

The definition implies that the empty set is a�nely independent asis any set of cardinality one.

Just as linear maps are the natural morphisms for linear spaces,a�ne maps are the natural morphsims for a�ne spaces. A functionf : Rd ! Re is an a�ne map if f(x) = T (x) + v for some linear mapT : Rd ! Re and fixed vector v 2 Re.

Problem 5. If f : Rd ! Re is an a�ne map and A is an a�nesubspace of Rd, then f(A) is an a�ne subspace of Re. Suppose B is ana�ne subspace of Re. Is f�1(B) an a�ne subspace of Rd?

Exercise 1.1.

(1) Prove that A ✓ Rd is an a�ne subspace if and only if A is thesolution to a set of simultaneous linear equations in d vari-ables.

(2) Recall that for two distinct elements x and y the line deter-mined by them consists of all points of the form tx + (1 �t)y, t 2 R. Prove that A is an a�ne subspace if and only iffor all x 6= y in A, the line determined by x and y is containedin A.

(3) Let A and B be two a�nely independent subsets of Rd. Provethat if |B| > |A|, then there exists x 2 B�A such that A[{x}is a�nely independent.

(4) An a�ne dependence relation is a dependence relation

a1x1 + · · · + anxn = 0

such that a1 + · · · + an = 0. Prove that {x1, . . . ,xn} ✓ Rd

is a�nely independent if and only if there are no nontriviala�ne dependence relations involving the xi.

(5) An a�ne hyperplane of Rd is an a�ne subspace of dimen-sion d� 1. If H is an a�ne hyperplane, then there exists con-stants a1, . . . ad, and b such that H = {(x1, . . . , xd) 2 Rd :

2. CONVEXITY 7

a1x1 + · · · + anxn = b}. Let A be an a�ne subspace and Han a�ne hyerplane which does not contain A. Prove that ifA \H 6= ;, then dim(A \H) = dim A� 1.

(6) Prove that if A is an i-dimensional a�ne subspace of Rd, thenA is the intersection of d� i a�ne hyperplanes. (See Exercise5 for the definition of a�ne hyperplane.)

(7) Let A be an a�ne subspace of Rd and x,y two points not in A.Then y 2 aspan(A [ {x}) if and only if x 2 aspan(A [ {y}).

(8) Let

x1 =

2

4x1,1...

xd,1

3

5 , . . . ,xn =

2

4x1,n...

xd,n

3

5

be n elements of Rd written as column vectors. Prove thatx1, . . . ,xn are a�nely independent if and only if

x1 =

2

664

1x1,1...

xd,1

3

775 , . . . , xn =

2

664

1x1,n...

xd,n

3

775

are linearly independent in Rd+1.(9) Let T : Rd ! Re be a linear map and v a fixed vector in

Rd. Then S : Rd ! Re defined by S(x) = T (x + v) is ana�ne map. Show that not all a�ne maps are of this form.Characterize which ones are.

2. Convexity

One way to view a�ne subspaces is as those sets which are closedunder forming lines (Exerc. 1.1.2). Convex sets are subsets that areclosed under forming line segments. For two points x,y 2 Rd, theline segment between them, which we denote by [x,y], is the set of allpoints of the form t x + (1� t)y, 0 t 1.

Definition 2.1. A subset A of Rd is convex if for all x,y 2 A theline segment [x,y] ✓ A.

Problem 6. The intersection of a set of convex sets is convex.

As was the case with a�ne spaces, the above problem implies thatfor any subset A of Rd there is a smallest convex set which contains A.It is called the convex hull of A and we denote it by ch(A).

The examples in Figure 1 demonstrate that convex sets can beempty, open, closed, neither open nor closed, bounded or unbounded.

8 2. AFFINE AND CONVEX GEOMETRY

Figure 1. Examples of convex subsets of R2

In a fashion similar to using a�ne combinations to form the a�ne span,convex combinations can be used to form the convex hull of a set. Aconvex combination is a linear combination

a1x1 + · · · + anxn

such that 0 ai 1 and a1 + · · · + an = 1. Equivalently, a convexcombination is an a�ne combination with all of the scalars nonnegative.

Problem 7. The convex hull of A consists of all convex combinationsof elements of A.

At this point we know that if y 2 ch(A), then y is a convex com-bination of points of A. However, we do not yet have a bound on thenumber of elements needed.

Problem 8. (Caratheodory) Let y 2 ch(A), where A ✓ Rd. Then thereexists x1, . . . ,xd+1 in A such that y is a convex combination of the xi.

Exercise 2.2.

(1) If K is convex and T is an a�ne map, then T (K) is convex.(2) Let K be a convex set and x,y be distinct points not in K. If

x 2 ch(K[{y}), then y /2 ch(K[{x}). (Compare to Exercise1.1. 7.)

(3) Let f : R ! R be a twice di↵erentiable function with f 00(x) 0for all x 2 R. Prove that for any a < b the points under thecurve y = f(x), i.e.

{(x, y) : a x b, y f(x)}

is a convex subset of R2.

3. SIMPLICES 9

3. Simplices

Simplices are one of the fundamental examples and building blocksof polytopes. They are the higher dimensional analogues of line seg-ments, triangles and tetrahedra. The d-simplex, which we denoteby �d, is the convex hull of the standard unit coordinate vectorse1, . . . , ed+1 in Rd+1. Equivalently,

�d = {(x1, . . . , xd+1) : 0 xi 1, and x1 + · · · + xd+1 = 1.}The dimension of a convex set is the dimension of its a�ne span.

Since the a�ne span of �d is the hyperplane x1 + · · · + xd+1 = 1, it isd-dimensional.

A d-simplex (as opposed to the d-simplex) is the convex hull ofd + 1 a�nely independent points. A d-simplex can live in any Re aslong as e � d. For instance, any triangle is a 2-simplex.

Exercise 3.1.

(1) A d-simplex is d-dimensional.(2) Some texts define the d-simplex as the convex hull of {~0, e1, . . . , ed

in Rd.} Denote this version of the simplex by �d. Prove thatif S is a d-simplex in Re, then there exists an injective a�nemap T : Rd ! Re such that T (�) = S. To what extent is Tuniquely defined?

(3) Let S = ch(x0, . . . ,xd} be a d-simplex. A face of S is theconvex hull of any subset of {x0, . . . ,xd}. Prove that if F isa face of S other than S itself, then there exists an a�nehyperplane H such that F = S \H.

(4) Is the closure of a convex set convex?(5) Is the convex hull of a closed set closed?

CHAPTER 3

What is a polytope?

Some of the first geometric objects that a child encounters are con-vex polygons and polyhedra. Squares, hexagons, cubes and pyramidsare familiar to almost anyone over the age of five. What are the higherdimensional analogues of these shapes? There are several approachesto this question. Here we will consider two apparently di↵erent pos-sibilities. The first idea is that a polyhedron is the smallest convexsubset which contains its vertices. The second approach is to examinethe problem from the point of view of linear optimization.

1. V-polytopes

Definition 1.1. A subset P of Rd is a V-polytope if it is the convexhull of a finite set of points.

By definition the empty set is a V-polytope. What are some ex-amples of V-polytopes? Perhaps the easiest to write down are thed-simplex, the d-cube and the d-crosspolytope. We have already seenthe d-simplex. The d-cube is defined by

⇤d = {x = (x1, . . . , xd) : |xi| 1, i = 1, . . . , d}.So, the 2-cube is a square and the 3-cube is a cube. The d-cube is ad-dimensional V-polytope. It is the convex hull of all points in Rd ofthe form "1e1 + · · · + "ded, with "i = ±1. The d-crosspolytope is theconvex hull of {±ei : 1 i d} in Rd. By definition it is a V-polytopeand it is easy to see that it is d-dimensional. We denote it by ⇧d. The3-crosspolytope is the octahedron.

Problem 9. Let f : Rd ! Re be an a�ne map and P a V-polytope. Isf(P ) necessarily a V-polytope? What about f�1(P )?

Problem 10. Let P be a V-polytope in Rd. If A is an a�ne subspaceof Rd, is P \ A a V-polytope?

An alternative way to generalize convex polygons is from the pointof view of linear optimization. The general idea of linear optimizationis that you are asked to find the maximum (or minimum) of a linearobjective function when restricted to a number of linear inequalities.

11

12 3. WHAT IS A POLYTOPE?

Here is an example to illustrate the point. Suppose that you are re-sponsible for purchasing the gold and platinum your company needsduring the coming year. The company has contracts with two mines,one for each metal. In order to insure continued business with themines you must purchase at least 2 tons of platinum and 3 tons ofgold. Your budget, plus the fact that platinum is approximately 50percent more expensive than gold, implies that if you purchase x tonsof platinum and y tons of gold, then 2x+3y 24. Your last constraintis that due to political pressure from the owner of the platinum mineyou must be sure to purchase at least 2 more tons of platinum thangold. As usual, your objective is to maximize the company’s profit.Currently the items with more platinum have a greater profit margin,so you can estimate the profit by the expression 6x + y. We can sumup your problem as trying to maximize 6x + y given that

x � 2y � 3

2x + 3y 24x � y � 2.

The situation is depicted in Figure 1. The shaded quadrilateral is theintersection of the four inequalities above and represents the feasibleregion, those points (x, y) which satisfy all the constraints. The dottedlines represent the objective function. From the diagram the optimalanswer is 6 tons of platinum and 4 tons of gold. Notice that no matterwhat the objective function is, as long as it is linear and your goal isto maximize (or minimize) it, then an optimal solution is at one ofthe vertices of the polygon. (Can you explain why?) In practice thenumber of variables and constraints can easily be several thousand.

With the above model in mind we now introduce H-polytopes. Thedual of Rd, which we denote by (Rd)?, is the real vector space of alllinear functions a : Rd ! R. We will denote the value of a at x 2 Rd

by a · x. This notation suggests the usual dot product on Rd, and ifwe coordinatise (Rd)? in the usual way it is the ‘usual’ dot product. Aclosed half-space of is any subset of the form

H�a,b = {x 2 Rd : a · x � b}

for some fixed a 6= 0 in (Rd)? and b 2 R. While H�0,b is not a closed

half-space, we will use this notation where appropriate. Obviously H�0,b

is either all of Rd or empty depending on the sign of b.

Definition 1.2. An H-polytope is a bounded set which is the intersec-tion of finitely many closed half-spaces.

2. EXAMPLES OF POLYTOPES 13

(5,3)

X=2

Y=3

2X + 3Y = 24

(2,3)

(2, 20/3)

(6,4)

6 X + Y = B

X − Y = 2

Figure 1. An H-polytope

Since closed half-spaces are convex, any H-polytope is convex.

Problem 11. ⇤d, ⇧d, and any simplex are H-polytopes.

As a hyperplane is the intersection of two closed half-spaces we cansee that if P is an H-polytope and H a hyperplane, then P \ H isan H-polytope. By (Exerc. 1.1.6) any a�ne subspace can be writtenas the intersection of hyperplanes. So, Problem 10 is very easy whenapplied to H-polytopes.

Problem 12. Is the a�ne image of an H-polytope an H-polytope?

2. Examples of Polytopes

Here we give some examples of polytopes. In each case it will beobvious from the definition that it is either an H or V-polytope, butnot so obvious if it is the other.


Figure 2. Prism (P )

The permutahedron in Rd is the convex hull of all possible permu-tations of the coordinates of the vector

2

664

12...d

3

775 .

For instance, the permutahedron in R2 is the line segment between

12

�

and

21

�. The permutahedron in R3 is a familiar polygon. What is it?

Let P be a convex subset of Rd. The prism of P is defined by

prism(P ) = P⇥[0, 1] = {(x1, . . . , xd, xd+1) 2 Rd+1 : (x1, . . . , xd) 2 P and 0 xd+1 1.}Figure 2 is a simple example. As we will see in Exercise 3, if P is anH or V-polytope, then so is prism (P ).

In order to define our next example, order polytopes, we recall thedefinition of a partially ordered set. A partially ordered set, which isusually called a poset, is a pair (⇧,) such that for all x, y, z 2 ⇧

• x x. (reflexivity)• If x y and y x, then x = y (antisymmetry).• If x y and y z, then x z (transitivity).

When the binary relation is clear we will frequently suppress . Ifx y or y x, then x and y are comparable. Otherwise they areincomparable. It may well happen that the binary relation is empty,

2. EXAMPLES OF POLYTOPES 15

in which case all pairs of distinct elements of the poset are incomparableand ⇧ is called an antichain.

Examples of posets are everywhere in mathematics. Familiar onesinclude the rationals or reals with the usual ordering. For any set X,we use BX to represent the poset of all subsets of X with ✓ as thebinary relation. In particular, the poset of all subsets of [n] is denotedBn. Another interesting poset is (Z+, |) where the binary relation is“divides”. For instance, in this poset 3|6 but 3 6 | 5. We will use Dn todenote (Z+, |) restricted to the finite poset consisting of those elementswhich divide n. For instance, D4 has three elements, 1 < 2 < 4.

An isomorphism between posets (⇧,) and (⇧0,0) is a bijection� : ⇧ ! ⇧0 such that x y if and only if �(x) 0 �(y). The face posetof a convex set K, denoted F(K), is the partially ordered set of facesof K with the subset relation. (Faces of convex sets are defined below,Definition 3.1)

Let [n] = {1, . . . , n}. A natural poset on [n] is a poset structure = ([n], ) such that if i j, then i j (the usual on theintegers). The order polytope for is

{(x1, . . . , xn) 2 [0, 1]n : If i j, then xi xj.

By definition, order polytopes are H-polytopes.

Exercise 2.1.

(1) What is the dimension of the permutahedron in Rd?(2) What is the dimension of prism (P )?(3) Show that if P is an H or V-polytope, then so is prism (P ).(4) Let (L,) be a finite poset with |L| = n. Show that there exists

a natural poset on [n] isomorphic to (L,).(5) Let (L,) be a poset with L countably infinite. Is there a

natural poset on Z isomorphic to (L,)?(6) Let be a natural poset on [n]. What is the dimension of its

order polytope?(7) Show that the order polytope of the usual linear order on [n] is

a simplex. What are its vertices?

As you can see, sometimes V-polytopes appear to be better whileother times H-polytopes look more useful. Our analysis of the exactnature of the relationship between these two definitions will occupy thenext two chapters. In the mean time we will use ‘polytope’ to refer toboth types. Central to the study of any polytope are its faces.


3. Faces of polytopes

What are the higher dimensional generalizations of the vertices,edges and faces of a three-dimensional polyhedron? A brief examina-tion of the cube shows that each of these is the intersection of the cubeand an a�ne hyperplane where the cube is completely contained onone side of the hyperplane. (See Exerc. 1.1.5 for the definition of a�nehyperplane.)

An inequality of the form a ·x b, a 2 (Rd)?,x 2 Rd, b 2 R is validfor a convex set K if it holds for all x 2 K. An important special caseis ~0 · x 0.

Definition 3.1. A face of a convex set K in Rd is any subset of theform H�

a,b \K, where a · x b is a valid inequality for K.

Since ~0 · x 0 is always valid, K is always a face of K. It is calledthe improper face of K. The maximal proper faces of K are called thefacets of K. If K is bounded, then it is easy to see that the empty set isa face of K. The vertices of K are its zero-dimensional faces. Similarly,the edges of K are its one-dimensional faces. Since the intersection ofconvex sets is convex, faces of convex sets are convex.

Problem 13. Let P = ch(x1, . . . ,xn) be a V-polytope. Then the ver-tices of P are a subset of {x1, . . . ,xn}.

We can easily see that if F is a face of a 3-dimensional polyhedronand G is a face of F, then G is also a face of the polyheron. We willsee later that this holds for any polytope. However, it is not true forarbitrary convex sets.

Problem 14. Construct an example in R2 of a convex set K with aface F such that F contains a face G which is not a face of K.

When are two polytopes ‘the same’? Two possible approaches tothis question are a�ne and combinatorial equivalence.

Definition 3.2. Two convex subsets K ✓ Rd and K 0 ✓ Re are a�nelyequivalent if there exists an a�ne map f : Rd ! Re such that frestricted to K is a bijection onto K 0.

A�ne equivalence allows us to assume that whatever convex set weare investigating has full dimension.

Problem 15. Let K be a convex set. Prove that K is a�nely equivalentto a convex set K 0 ✓ Rd where d is the dimension of K.

3. FACES OF POLYTOPES 17

The notion of a�ne equivalence is very restrictive. For instance,there are convex quadrilaterals in the plane that are not a�nely equiv-alent (Exercise 5). In order to make precise the idea that any twopolygons with the same number of sides should be considered equiva-lent we introduce the face poset of a convex set.

Definition 3.3. Two convex sets K and K 0 are combinatoriallyequivalent if their face posets F(K) and F(K 0) are isomorphic.

It is now easy to see that any two polygons with the same number ofsides are combinatorially isomorphic.

Problem 16. Prove that if P and Q are two a�nely equivalent convexsets, then P and Q are combinatorially equivalent.

One class of polytopes which have received a great deal of atten-tion are simplicial polytopes. A polytope P is simplicial if all of itsproper faces are simplices. For instance, any polygon is simplicial asis the octahedron and the icosahedron. However, the cube and thedodecahedron are not simplicial.

One of the most fundamental algorithms in linear optimization isDantzig’s simplex algorithm [7]. The idea is that the maximum (orminimum) of an objective function will always be at the vertices of thefeasible region. Combined with an intelligent method for searching thevertices yields an algorithm which is very e↵ective in practice. Thisleads to the natural question of “What is the maximum number of ver-tices the simplex algorithm will have to examine given a fixed numberof constraints?” In Chapter 6 we will see that this is equivalent to ask-ing what is the maximal number of facets in a simplicial polytope witha given number of vertices. It turns out that this is answered by thecyclic polytopes.

The moment curve in Rd is given by � : R ! Rd, �(t) = (t, t2, . . . , td).For n � d + 1 the cyclic polytope C(n, d) is defined to be the convexhull of {�(t1), . . . , �(tn)}, where t1 < · · · < tn. Evidently C(n, d) de-pends on the choices for ti. However, as we are about to see, the faceposet, and in particular the number of facets, does not depend on thesechoices!

Problem 17. C(n, d) is a simplicial d-dimensional V-polytope.

We will eventually show that the vertices of a face F of a polytope Pare the vertices of P contained in F. Furthermore, any face (includingP ) is the convex hull of its vertices. Hence, we can determine all of thefaces of a simplicial polytope P by specifying which vertices are thevertices of facets. By Problem 13 this amounts to figuring out when


the hyperplane determined by a d-subset of {�(t1), . . . , �(tn)} gives avalid inequality.

Problem 18. (Gale evenness condition) A subset V = {�(ti1), . . . , �(tid)}, ti1 <· · · < tid is a set of vertices for a facet of C(n, d) if and only if forall tj < tm with tj and tm not in V, the cardinality of {l : j < l <m and tl 2 V } is even.

Exercise 3.4.

(1) Suppose K is a convex proper subset of Rd. Is the empty facenecessarily a face of K?

(2) Let K be a convex set. If F is a proper face of K, then dim F <dim K. If the dimension of a face of K is exactly one dimensionless than the dimension of K, then F is a facet of K. Is theconverse true?

(3) What are the faces of ⇤d?(4) What are the faces of ⇧d?(5) Give an example of two convex quadrilaterals in R2 which are

not a�nely equivalent.(6) Give an example of two compact convex sets of di↵erent di-

mensions whose face posets are isomorphic. Can this happenif the two convex sets are polytopes?

(7) Assuming that any face of a face of C(n, d) is a face of C(n, d),show that cyclic polytopes of dimension four or higher arebd/2c-neighborly. A polytope is k-neighborly if every subsetof k-vertices are the vertices of a face of the polytope.

(8) Let fi(P ) be the number of i-dimensional faces a d-polytope P.Compute

dX

i=0

(�1)ifi

for the d-cube, d-simplex and d-crosspolytope.

CHAPTER 4

Convexity

In this chapter we develop the tools necessary to understand therelationship between V and H-polytopes. Our presentation is based onGrunbaum’s classic text [11].

1. Projection

Our initial investigation of the relationship between H and V-polytopesrevolves around convexity. Here we establish a number of importantproperties of convex sets. Perhaps the single most useful property isthe existence of projections for closed convex sets.

Problem 19. Let K ✓ Rd, K 6= ;, be closed and y 2 Rd. Prove thereexists x 2 K which minimizes the distance to y. Specifically, thereexists x 2 K such that for all z 2 K, ||x� y|| ||z� y||. Prove that ifK is also convex, then there exists a unique such x.

The point x whose existence and uniqueness is given by the aboveproblem is called the projection of y onto K and is denoted proj K(y).

Any hyperplane {x 2 Rd : a · x = b} partitions Rd into three sets,H+

a,b = {x 2 Rd : a · x > b}, H�a,b = {x 2 Rd : a · x < b}, and H=

a,b

the hyperplane itself. The two half-spaces H+a,b and H�

a,b are calledthe sides of the hyperplane and when no confusion is possible we willsimply write H+ and H�. The hyperplane separates two sets A and Bif A ✓ H+ and B ✓ H� or vice versa.

Problem 20. If K is a closed convex set and x /2 K, then there existsa hyperplane which separates K and {x}.

Looking at polygons and three-dimensional polyhedra it is fairlyobvious that any point on the boundary is contained in a proper face,and the only face a point in the relative interior is contained in is theimproper face. This holds for any closed convex set.

Theorem 1.1. Let K be a closed d-dimensional convex subset of Rd.If x is on the boundary of K, then x is contained in a proper face ofK.

19

20 4. CONVEXITY

Proof: Without loss of generality we can assume that 0 is in the in-terior of K. (Why?) Let K✏ = {✏x : x 2 K}. Since x 2 @K,x /2 K(1�✏)

for 0 < ✏ < 1. (Why?) By the previous proposition, for each such ✏there exists a hyperplane H✏ which separates x and K(1�✏). The collec-tion of all the H✏ must have at least one limit hyperplane H which willsatisfy the proposition. (What is a limit hyperplane? Why does oneexist? Why does it work?) ⇤

Problem 21. Fill in the details left out in the above proof.

Exercise 1.2.

(1) Every closed convex set is the intersection of the closed half-spaces which contain it.

(2) Is the interior of a convex set convex?(3) For any closed convex K, proj K is a continuous idempotent.

A function f is an idempotent if f � f = f.(4) If x 2 H+

a,b and y 2 H�a, b, then any path from x to y passesthrough H=

a,b.

(5) Let K and K 0 be closed convex sets in Rd. Prove that if one ofthem is bounded, then there exists a hyperplane which separatesthem. What if neither is bounded?

2. Extreme points

Our intuition from two and three dimensional polyhedra suggeststo us that a compact convex set is the convex hull of its vertices. ButProbem 14 warns us that this is not the case. The next best thinginvolves the extreme points of a convex set.

Definition 2.1. x is an extreme point of a convex set K if x is neverin the relative interior of a line segment contained in K. We denotethe extreme points of K by ext K.

If you look at any compact solution to Problem 14, you will noticethat the convex set is the convex hull of its extreme points. Before wecan prove this in general we establish some elementary properties ofextreme points.

Lemma 2.2. Let K be convex.

• Every vertex of K is an extreme point of K.• If K = ch(A), then ext K ✓ A.• If F is a face of K, then ext F = F \ ext K.

Problem 22. Prove the above lemma.

3. V AND H-POLYTOPES 21

Theorem 2.3. If K is a compact convex set, then ch(ext K) = K.

Proof: Evidently, ch(ext K) ✓ K. We prove the reverse inclusionby induction on the dimension of K. Dimension one is obvious.

Suppose that x 2 K. If x is an extreme point then obviouslyx 2 ch(ext K). So, let [y0, z0] be a line segment in K containing xin its relative interior. Extend this segment in both directions. It willintersect the boundary of K at two points, y and z. By the previousproposition there exist faces F

y

and Fz

which contain y and z respec-tively. See Figure 1. Each of these faces has dimension less that K, soby the induction hypothesis, F

y

= ch(ext Fy

) and Fz

= ch(ext Fz

). Asx 2 ch(ext F

y

[ ext Fz

) and the lemma tells us that ext Fy

[ ext Fz

✓ext K, we are done. ⇤

K

y=Fy

Fz z

z

x

y

0

0

Figure 1. ch(ext K) = K

3. V and H-polytopes

As the reader may (or may not) of guessed by now, the main the-orem of polytopes is that V and H-polytopes are the same. We arefinally ready to prove this.

Theorem 3.1. P ✓ Rd is a V-polytope if and only if P is an H-polytope.

Proof: Using an a�ne equivalent polytope if necessary, we can as-sume that P is d-dimensional. Let P be an H-polytope. We prove Pis a V-polytope by induction on dimension. As usual, dimension one istrivial.

Write P as a minimal intersection of closed half-spaces P = \mi=1Hi.

By minimal we mean that if we set Pi = \j 6=iHj, then Pi 6= P for any

22 4. CONVEXITY

i. Minimality guarantees that the boundary of P is contained in theunion of the (d� 1)-dimensional faces of P and there are m of these.

Problem 23. Why?

By induction, each such face has a finite number of extreme points.All of the extreme points of P are on the boundary. Hence, the previ-ous lemma and the last theorem imply that P has a finite number ofextreme points and must be a V-polytope.

Now suppose that P is a d-dimensional V-polytope in Rd. WriteP = ch(V ), |V | = n. By the previous theorem, we might as well assumethat V = ext P. Any k-face of P is determined by (k + 1) a�nelyindependent points in the face. Hence, for every k the number of k-dimensional faces is less than or equal to

� |V |k+1

�. In particular it is finite.

Let Fi enumerate the facets of P and let Hi be the corresponding closedhalf-spaces. The proof is complete once we show the following.

Claim: P = \Hi.Proof: (of claim) Certainly P ✓ \Hi. Suppose x /2 P. For each

face Gj of P of dimension d � 2 or less, let Aj be a�ne span of Gj

and x. The Aj form a finite collection of a�ne subspaces each of whichhas dimension at most d� 1. Therefore their union does not cover theinterior of P.

Problem 24. Why?

So, there exists y in the interior of P such that y is not in theunion of the Aj. Consider the line segment from y to x. Since y 2 Pand x /2 P there exists z on the boundary of P and the line segmentfrom y to x. By Theorem 1.1 there is a proper face of P which containsz. As y is not in any of the Aj, it must be the case that z is on theboundary of one of the Hi. Hence, x /2 \Hi. ⇤

Theorem 3.1 makes it clear the V and H-polytopes are the same.As we have already seen, each point of view has its advantages anddisadvantages. In chapter 6 we will see that this is only the beginningof the story.

Exercise 3.2.

(1) Open convex sets have no extreme points.(2) A convex set K is centrally symmetric if whenever x 2 K,

then �x 2 K. Prove that if P ✓ Re is a nonempty centrallysymmetric polytope, then there exists d > 0 and an a�ne mapf : Rd ! Re such that f(⇧d) = P.

(3) Every proper face of a polytope is a face of a facet.

3. V AND H-POLYTOPES 23

H

PA

A

A

G

G

G

y

z

x 3

2

1

i

1

2

3

Figure 2. P = \Hi

(4) Let F be a face of a polytope P , and let G be a face of F. Provethat G is a face of P.

(5) Let F be a k-dimensional face of a d-dimensional polytope P.Show that F is the intersection of d� k facets of P.

(6) (The basis of the simplex algorithm) Let {v1, . . . , vn} be thevertices of a polytope P ✓ Rd. Prove that for any a 2 (Rd)?

supx2P

a · x = max{a · v1, . . . , a · vn}.

CHAPTER 5

Shelling

The boundary of a d-polytope is homeomorphic to the (d � 1)-sphere. It turns out that this boundary can be built up by gluing thefacets together in a ‘nice’ way. This will lead us to some remarkableenumerative results for the f -vector of the polytope, especially when Pis simplicial. The f -vector of a d-polytope is (f�1, f0, f1, . . . , fd), wherefi is the number of i-dimensional faces of P. Until further notice, P isa simplicial d-polytope in Rd.

1. Shellings

A shelling of P is an ordering F1, . . . , Fm of the facets of P suchthat for all j � 2, Fj \ ([j�1

i=1Fi) is a nonempty union of facets of Fj.If P has a shelling order, then we say P is shellable. Figure 1 showsa shelling of the boundary of the octrahedron, while figure 2 shows anordering of the four facets of the boundary of the octahedron which isnot the beginning of any shelling.

Now view the ordering of the facets as a way of building up @P.Each facet adds new faces of @P as we take the union of the facetwith the previous ones. The shelling condition insures that there is aunique new minimal face added at each step. Indeed, at the jth stepFj \ ([j�1

i=1Fi) is a union of facets of Fj. The minimal face is Mj =

Figure 1. A shelling of the boundary of the octahedron

25

26 5. SHELLING

Figure 2. The beginning of a non-shelling of the bound-ary of the octohedron

(empty set)

Figure 3. Minimal new faces of the shelling

ch({vj1 , . . . , vjm}), where the vjkare the vertices opposite the facets of

Fj in the intersection. Figure 3 shows the minimal new faces of theshelling in Figure 1.

Problem 25. Prove that Mj is the minimal new face at the jth stepof the shelling.

Exercise 1.1.

(1) Any ordering of the facets of �d is a shelling of its boundary.(2) Are there any other simplicial polytopes such that any ordering

of their facets is a shelling?(3) Is @ ⇧d shellable?(4) Can a simplicial polytope with f -vector (1, 7, 19, 20, 8, 1) be

shellable?

The concept of shellability is very important and extends to a num-ber of situations including abstract simplicial complexes. An abstractsimplicial complex consists of a set V, the vertices of the complex,and a set of faces � ✓ 2V The faces must be closed under subsets.

1. SHELLINGS 27

If F 2 � and G ✓ F, then G 2 �. The maximal faces (with respectto inclusion) of � are the facets of the complex. The dimension ofa face G 2 � is |G| � 1. The dimension of � is the maximum of thedimensions of its faces.

Example 1.2. Some examples of abstract simplicial complexes.

(1) Boundaries of simplicial polytopes. The faces are the subsetsof vertices whose convex hull is a proper face of the polytope.

(2) Simple graphs. The faces are the pairs of vertices which havean edge between them.

(3) Let V be a subset of vectors in a vector space and let � be thesubsets of V which form independent subsets of vectors. Whatare the facets of this complex?

The definition of shellable for a pure abstract simplicial complexis almost exactly the same as for simplicial polytopes; for j � 2, Fj \([j�1

i=1Fi) must be a union of facets of the boundary of Fj. The boundaryof a face of � is all subsets of Fj of cardinality less than or equal to|Fj| � 1. An abstract simplicial complex is pure if all facets have thesame dimension. For instance, if V = [4] and the facets of � are{1, 2, 3}, {1, 4}, {2, 4}, then � is not pure, but does satisfy the aboveconditions. There is a notion of nonpure shellability, but we will notdiscuss that here. Whenever we are considering shellability the complexwill be pure.

Problem 26. Show that connected graphs and the complexes describedin Example 1.2 (3) are shellable.

Example 1.3. Let G be the graph in Figure 4. Let the vertices of � bethe edges of the graph. The faces of � are those subsets of edges whoseremoval does not disconnect the graph. Specifically, ;, all singleton, andall doubletons except {a, b} and {c, d} are the faces of �. We represent� in Figure 4. Singletons are represented by vertices, doubletons byedges. Ordering the edges (which are the facets of �) as shown, we seethat ; is the minimal new face for 1, and the minimal new face is asingle vertex for faces 2, 3, 4 and an edge for faces 5, 6, 7, 8.

Suppose that G above represents a network and each edge has equaland independent probability of failing p, 0 < p < 1. What is the proba-bility that that network will remain connected? Directly checking eachpossible subset of edges which keep the graph connected we see thatthis probability is

(1� p)5 + 5p(1� p)4 + 8p2(1� p)3 = (1� p)3[1 + 3p + 4p2].

28 5. SHELLING

a

c de1

2

3

4

5 6

78

|�|

a

b

c

d

e

G b

Figure 4. G and �

Notice that the coe�cient in each degree i of the last factor is thenumber of steps in the shelling in which we added a minimal new faceof cardinality i.

Suppose � is a shellable abstract simplicial complex and we aregiven a shelling. For each i, let hi be the number of facets whose uniquenew minimal face has cardinality i. The h-vector of � is (h0, . . . , hd).

Example 1.4. The shelling in Figure 1 gives h-vector (1, 3, 3, 1).

The shelling polynomial of the shelling F1, . . . , Fm is

(1) h�(x) = h0xd + h1x

d�1 + · · · + hd�1x + hd.

The shelling polynomial appears to depend on the shelling. We can alsoencode the f -vector of � in a polynomial. The f -vector of an abstractsimplicial complex � is defined like the f -vector of a polytope, fi isthe number of i-dimensional faces and the f(�) = (f�1, f0, . . . , fd�1),where (d � 1) is the dimension of �. Define the face polynomial of �to be

f�(x) = f�1xd + f0x

d�1 + f1xd�2 + · · · + fd�1.

Theorem 1.5. h�(x + 1) = f�(x).

Problem 27. Prove this theorem.

Corollary 1.6. The h-vector of � does not depend on the shellingorder.

Problem 28. Write down formulas for hi in terms of fj and viceversa. Which fj does hi depend on (and vice versa)? Show that if

2. GEOMETRIC SIMPLICIAL COMPLEXES 29

D is a collection of abstract simplicial complexes and � 2 D has theproperty that hi(�) � hi(�0) for all 0 i d and �0 2 D, then fi(�)also maximizes all fi(�0) in D. Is this still true if we reverse the roleof the f - and h-vector?

Aside from the formulas established in Problem 28 there is anothermethod for computing h-vectors known as ‘Stanley’s trick’. We writethe f -vector (including f�1 = 1) along the right-hand side of a trianglethat looks like Pascal’s triangle. Then put ones along the left-hand side.Fill in the rest of the triangle as you would Pascal’s triangle, exceptsubtract instead of add. Compute one extra row of subtractions andyou get the h-vector. Here is an example which shows that the f -vector(1, 7, 21, 14) becomes the h-vector (1, 4, 10,�1).

11 7

1 6 211 5 15 14

1 4 10 �1

Problem 29. Prove that Stanley’s trick works.

2. Geometric simplicial complexes

We saw in Example 1.2 (1) a geometric object, the boundary ofa simplicial polytope, represent the combinatorial data of an abstractsimplicial complex. Is this possible for all abstract simplicial com-plexes? The answer lies with geometric simplicial complexes.

Definition 2.1. A geometric simplicial complex � in Rd is a finiteset of simplices in Rd such that

• � 6= ;.• If F 2 � and G is a face of F , then G 2 �.• If F1 and F2 are in �, then F1 \ F2 is a face of both F1 and

F2. (Recall that the empty set is always a face of both.)

Examples illustrating the above definition are in Figures 5 and 6

Suppose that � is a geometric simplicial complex. Let V be the 0-dimensional faces in �. The abstract simplicial complex associated to� has as its vertices V , and as its faces those subsets of vertices whoseconvex hull is a simplex in �. For instance, the figure on the right

30 5. SHELLING

Figure 5. A geometric simplicial complex in R2

Figure 6. Two examples which are not geometric sim-plicial complexes in R2

hand side of Figure 4 is a geometric simplicial complex in R2 whoseassociated abstract simplicial complex is the one described in Example1.3. Does every abstract simplicial complex come from a geometricsimplicial complex?

Let � be an abstract simplicial complex with vertex set [n]. Thegeometric realization of � is the subset of Rn given by

|�| =[

F2�

ch([i2F ei).

It is immediate from the definition that the abstract simplicial com-plex associated to |�| is �. A geometric realization of an abstractsimplicial complex � (as opposed to the geometric realization) is anygeometric simplicial complex whose associated abstract simplicial com-plex is �. Any two geometric realizations of the same abstract simplicialcomplex are homeomorphic. The idea is simple. Map correspondingvertices of the two complexes to each other and ‘extend linearly’. Be-cause it does not matter what geometric realization is used, we willsimply say that � is homeomorphic to a given space if any of its geo-metric realizations is.

3. LINE SHELLINGS 31

There are many geometric realizations for a given abstract simplicialcomplex �. One of the big disadvantages of |�| is that it lies in Rn for nmuch bigger than necessary. The geometric realization of the complexin Example 4 would be in R5, but obviously only R2 is needed. Ingeneral, what is the smallest Rd needed to realize a given complex?

Problem 30. Let � be an abstract simplicial complex of dimension d.Prove that there is a geometric realization of � in R2d+1.

Note that the bound in the above problem is optimal. Indeed, thereare one-dimensional complexes, such as K5, which can not be realizedin R2. Because of the close connection between abstract and geometricsimplicial complexes we will simply say simplicial complex from hereon and depend on the context to make it clear which type we mean.

3. Line shellings

A little experimentation shows that not all simplicial complexesare shellable. Disconnected graphs and two triangles with exactly onevertex in common are two examples. What about the boundary of asimplicial polytope? For a long time it was simply assumed that anysimplicial complex homeomorphic to a ball or sphere was shellable.However, this turned out not to be true! In 1958 Rudin demonstratedhow to construct a simplicial complex homeomorphic to a ball whichwas not shellable [18]. Finally, in 1971 Brugessor and Mani proved thatall polytopes were shellable [6]. What does it mean for a nonsimplicialpolytope to be shellable? To make sense of this we need the notion ofa polytopal complex.

Definition 3.1. A polytopal complex in Rd is a nonempty finite setof polytopes � in Rd such that

• If P is in � and F is a face of P , then F is in �.• If P1 and P2 are in �, then P1 \ P2 is a face of P1 and P2.

The prototypical example of a polytopal complex is the boundary of apolytope. Of course, any geometric simplicial complex is a polytopalcomplex. Just as in the case of simplicial complexes a polytopal com-plex is pure if all of its maximal polytopes (in the sense of inclusion)have the same dimension. As before, the maximal polytopes of � arecalled facets. The definition of a shelling of a polytopal complex isalmost the same as for a simplicial complex.

Definition 3.2. A shelling of a polytopal complex � is an orderingP1, . . . , Pt of the facets of � such that for all j � 2,

32 5. SHELLING

Pj

\

j�1[

i=1

Pi

!

is a union of a subset of the facets of Pj which form an initial segmentof a shelling order for @Pj.

In view of Exercise 1.1 (1) the above definition reduces to our pre-vious definition if � is a geometric simplicial complex. Note that thedefinition is very inductive!

The shellings that Bruggessor and Mani constructed for polytopesare called line shellings. Let x be a point in the interior of P, a d-polytope in Rd, not necessarily simplicial. Now choose a line l throughx with the following two properties:

(1) l intersects all the hyperplanes of the facets of P. (These arethe a�ne spans of the facets.)

(2) l does not intersect any nontrivial intersection of the hyper-planes of the facets of P .

Any line through x satisfying the above properties is called a shellingline of P.

Problem 31. Why is it obvious we can always choose such a line?

Now imagine you are in a rocket ship inside the planet P headingalong the line l in a chosen positive direction. Initially you can notsee anything as you are inside the planet. By (2) you will emerge ata facet, F1. As you travel away from P you will see one new faceteach time you pass through its corresponding hyperplane. The shellingorder for these facets is the same as the order you see them. By (2) youwill never see two (or more) new facets simultaneously. Eventually youwill go far enough towards “+1” so that you can see as many facetsas possible in that direction. Now begin your return to P coming fromthe “�1” direction. At this point you can see all the facets you couldnot see from the “+1” direction. As you pass through each of theircorresponding hyperplanes the corresponding facet will disappear fromyour vision. The shelling order continues from before in the order inwhich the facets disappear from your vision. By (1) each facet of Poccurs exactly once in your shelling order.

How can we make the above description precise? We start with thenotion of being able to ‘see’ a face.

Definition 3.3. Let y /2 P and F a face of a convex set K. Then ycan see F if for all z 2 F the intersection of the line segment [z,y]and P equals z.


1

2

3

4

5

Figure 7. Line shelling of a polygon

Let l be a shelling line and F a facet of P. Since F is a face thereis a valid inequality a · x b such that F = P \ H�

a,b. The definitionof shelling line implies that H=

a,b \ l is a single point. Therefore thereare points z on l such that a · z > b. Linearity implies that z cansee F. So every facet is eventually seen by l. Conversely, suppose zis on l and a · z < b. Then z can not see F. If dim F is zero, this isobvious. Otherwise, let y be a point in the relative interior of F. Anyline segment ending y such that a · x < b for all the points of thesegment other than y must intersect the interior of P.

The above discussion has given us a concrete realization of the facetorder described by Brugessor and Mani. There only remains one issue.

Problem 32. Show that this is a shelling of P .

The fact that every simplicial polytope is shellable already showsthere are strong restrictions on their f -vectors. For instance, (1, 6, 15, 18, 7, 1)is not the f -vector of any simplicial 4-polytope. Indeed, this would give(1, 2, 3, 2,�1) as the h-vector of its boundary, and this is impossible.In fact, for the same reason, (1, 6, 15, 18, 7) is not the f -vector of anyshellable abstract simplicial complex.

What happens if we travel along l in the opposite direction? Thissimply reverses the shelling order. But now, if P is a simplicial poly-tope, each facet which originally contributed to hi, contributes to hd�i.Since the h-vector is independent of the shelling order we obtain theDehn-Sommerville equations.

Theorem 3.4. [8], [20] If P is a simplicial d-polytope, then hi = hd�i.

34 5. SHELLING

This is an even stronger restriction on the h-vector of P. Combiningthis theorem with Problem 28, we can see now that once we knowf0, . . . , fbd/2c, we know the entire f -vector of P. Are there any otherrestrictions on the h-vectors of simplicial polytopes? In order to discussthis we must introduce some notation.

Let a, i be natural numbers. Then there is a unique way of writing

a =

✓ai

i

◆+

✓ai�1

i� 1

◆+ · · · +

✓aj

j

◆, ai > ai�1 > · · · > aj � j.

Problem 33. Prove the above statement.

Example 3.5. a = 14, i = 3. Then 14 =�53

�+�32

�+�11

�.

With a decomposed as above define

a<i> =

✓ai + 1

i + 1

◆+

✓ai�1 + 1

i

◆+ · · · +

✓aj + 1

j + 1

◆.

Example 3.6. So, 14<3> =�64

�+�43

�+�22

�= 15 + 4 + 2 = 21.

In 1971 P. McMullen conjectured that necessary and su�cient con-ditions for (h0, . . . , hd) to be the h-vector of the boundary of a simpliciald-polytope are

• hi � 0.• hi = hd�i.• h0 h1 · · · hbd/2c.• For i d/2, define gi = hi � hi�1. Then for i d/2, gi+1

g<i>i .

Stanley [22] proved necessity, while Billera and Lee [2] proved suf-ficiency. These results give a complete characterization of h-vectors(and hence f -vectors) of the boundaries of simplicial polytopes. Forinstance, here are two problems that are now fairly easy.

Problem 34. Show that Cd(n) maximizes every fi among all possiblesimplicial d-polytopes with n vertices. What is fewest number faces indimension i that a simplicial d-polytope with n vertices can have?

Looking at many examples, it appears that the f -vector of a sim-plicial polytope is unimodal. A sequence (f0, f1, . . . , fd) is unimodal ifthere exists i such that f0 f1 · · · fi � fi+1 � · · · � fd). ByStanley’s proof of the necessity of McMullen’s conditions, the h-vectorof a simplicial polytope is symmetric and unimodal. It can be shownthat the f -vector of a simplicial d-polytope is unimodal when d 19[9]. Shortly after Billera and Lee proved the su�ciency of McMullen’s


conditions, Bjorner [4] and Lee [14] gave examples of 20-dimensionalsimplicial polytopes with trillions of vertices with f11 > f12 < f13.

What can we say about f -vectors of nonsimplicial polytopes? Youmay have guessed from Exercise 3.4 (8) that the alternating sum of theface numbers of a polytope only depended on the dimension. Analo-gously to our definition for simplicial complexes, for a polytopal com-plex � we define fi(�) to be the number of i-dimensional polytopes in�.

Definition 3.7. Let � be a d-dimensional polytopal complex. TheEuler characteristic of � is

�(�) =dX

i=0

(�1)ifi(P ).

Theorem 3.8. Let P be a d-polytope. Then �(P ) = 1. Equivalently,�(@P ) is zero if dim P is even (dim @P is odd) and two if dim P is odd(dim @P is even).

Proof: The proof is by induction on d. In fact, we will prove some-thing stronger. Let F1, . . . , Ft be a line shelling of the facets of P.Define �i = [i

j=1Fj. So �1 = F1 and �t = @P. We will show that�(�i) = 1 for i < t and �(�t) = �(@P ) is as claimed in the theorem.Our induction hypothesis is this stronger statement. It is easy to checkthat it holds for d equal to zero, one or two.

Now assume the induction hypothesis holds for (d � 1)-polytopes.We proceed by induction on i. For i = 1, �1 = F1, a (d� 1)-polytope.So �(�1) = 1. What is �i�1 \ Fi? When i < t the intersection is theinitial segment of a partial shelling of the boundary of Fi. When i = tthe intersection is the entire boundary of Fi. In either case

�(�i) = �(�i�1 [ Fi) = �(�i�1) + �(Fi)� �(�i�1 \ Fi).

The theorem now follows by all of the inductive hypotheses. ⇤

The fact that the Euler characteristic of a polytope is always oneis just the tip of a remarkable iceberg. For instance, the Euler charac-teristic of any polytopal complex homeomorphic to a ball is one, andif �1 and �2 are two polytopal complexes which are homeomorphic,then �(�1) = �(�2). The interested reader is invited to check out anyof the many texts on algebraic topology where vast generalizations ofthese results are derived.

Exercise 3.9.

36 5. SHELLING

(1) Let F be a face of a convex set K and y /2 K. Prove that ycan see F if and only if for at least one point z in the relativeinterior of F, K \ [z,y] = z.

(2) Let � be an abstract simplicial complex with vertex set V .Define Cone(�) to be the abstract simplicial complex whosevertex set is V [ v, where v /2 V, and whose faces are

{F 2 �} [ {F [ v : F 2 �}.What is the f -vector of Cone(�) in terms of the f -vector of�? What is the Euler characteristic of Cone(�)?

(3) Let K be a compact convex subset of Rd with the origin inits interior. Embed K in Rd+1 be setting the last coordinateto zero. The pyramid(K) is the convex hull of K [ ed+1.What are the faces of pyramid(K)? Suppose K is a geometricsimplicial complex. What relationship, if any, is there betweenpyramid(K) and Cone(K)?

(4) Let K be a compact convex subset of Rd with the origin in itsinterior. Embed K in Rd+1 be setting the last coordinate tozero. The bipyramid(K) is the convex hull of K [ ed+1 [�ed+1. What are the faces of bipyramid(K)? What relation-ship, if any, is there between bipyramid(K) and prism(K)?

(5) Let � be an abstract simplicial complex. Prove that � isshellable if and only if Cone(�) is shellable.

CHAPTER 6

Duality

Duality takes many shapes and sizes in mathematics. We havealready met (Rd)?, the linear dual of Rd. In this chapter we take acloser look at polytope duality, and more generally, duality for closedconvex sets. Anyone who has done Exercise 3.4.8 will (hopefully!) havenoticed that fi(⇤d) = fd�i(⇧d). This is one manifestation of the factthat the d-cube and the d-cross polytope are dual to one another. Fora myriad of reasons, some of which we will see, duality in this contextis an important idea with many applications.

1. Duality for convex sets

In general we have written a · x for the value of a(x) whenevera 2 (Rd)? and x 2 Rd. However, the possible confusion when thedouble dual of Rd is being discussed leads us to use the functionalnotation a(x) in this chapter.

Definition 1.1. Let K be a closed convex subset of Rd. Then the dualof K is

K? = {a 2 (Rd)? : a(x) 1 for all x 2 K.}

Example 1.2.

• K = Br(~0) ✓ Rd. K? = B1/r(~0) (in (Rd)?).• K = [1, 2] ✓ R. K? = (�1, 1/2] ✓ R?.• K = {(x, 0) 2 R2 : �1 x 1}. K? = {(x, y) 2 (R2)? : �1

x 1.}• (⇤d)? = ⇧d and (⇧d)? = ⇤d.

Without further restrictions the dual of a closed convex set maybe of a di↵erent dimension, be unbounded when the set is bounded, orvice versa. However, with very mild restrictions it is very well behaved.

Lemma 1.3. Let K be a compact convex subset of Rd with the originin its interior. Then K? is a also a compact convex subset with theorigin in its interior.

Problem 35. Prove the above lemma.

37

38 6. DUALITY

One of the most useful properties of duality for convex sets is thatunder mild conditions K = K??. In order to even make sense of thisstatement we need a method of identifying Rd and (Rd)?? = ((Rd)?)?.So far we have frequently identified Rd with its dual (Rd)? using thedot product. This isomorphism between Rd and its dual depends onthe choice of the unit coordinate vectors as a basis for Rd. In a certainsense (which we will not follow up on here) this is unavoidable. Soperhaps it is a little surprising to discover that there is a natural choiceof isomorphism between Rd and its double dual which does not dependon any choices.

Let x 2 Rd. Define an element of (Rd)?? by

x??(a) = a(x)

for a 2 (Rd)?. While it is immediate that x?? is a function from (Rd)? !R, the real value of this definition is the following.

Problem 36. Prove that x ! x?? is a linear isomorphism between Rd

and (Rd)??.

Now we can make precise one of the most useful properties of K?.

Proposition 1.4. If K is a closed convex subset which contains theorigin, then under the above identification K = K??.

Before we begin the proof we observe that the hypothesis that K con-tains the origin is unavoidable since K?, and hence K??, contains theorigin.

Proof: Using the isomorphism from Problem 36 and Definition 1.1

K?? = {x 2 Rd : a(x) 1 for all a 2 K?}= {x 2 Rd : a(x) 1 for all a 2 (Rd)? such that a(y) 1 for all y 2 K.}

Hence any x 2 K is in K??. To see that K?? ✓ K we show that ifx /2 K, then x /2 K??. So suppose that x is not in K. Then, since theorigin is in K, there exists a hyperplane H=

a,1 that separates K and x

with K ✓ H�a,1. In particular, a 2 K?. However, a ·x = a(x) > 1. Thus

x /2 K??. ⇤

In addition to linear duality, the reader has probably seen graphduality. For any (embedded) planar graph G its dual is defined byassigning a vertex to each region of the plane that G divides R2 into,and an edge between any two regions separated by a single edge of G.See Figure 1 for a typical example.

One way to view this process is as a way of changing the dissectionof R2 induced by the graph G. Zero-dimensional faces (the vertices of

1. DUALITY FOR CONVEX SETS 39

Figure 1. A planar graph and its dual

G) are replaced by two-dimensional faces (regions) of its dual. Simi-larly, edges are replaced by other edges and two-dimensional faces arereplaced by vertices. Under mild conditions this is another kind ofduality encoded by K?.

Let K be a compact convex subset of Rd with the origin in itsinterior. For each subset F of K define (F ) ✓ K? by

(F ) = {a 2 K? : a(x) = 1 for all x 2 F.}

Theorem 1.5. Let K be as above.

(1) If G ✓ F, then (G) ◆ (F ).(2) If F is a face of K, then (F ) is a face of K?.(3) If F is a face of K, then ( (F )) = F.

Proof: The first property follows easily from the definition. For thesecond, we first consider two extreme cases. If F = K, then, since Kcontains the origin, (F ) = (K) = ;. On the other hand, if F = ;,then it is immediate that (F ) = (;) = K?. So from here on weassume that F is neither of these extremes. Let x be in the relativeinterior of F (equal to F when F is just a point).

Problem 37. Hx

??,1 is a valid inequality for K? and (F ) = K? \H

x

??,1. Hence (F ) is a face of K?.

40 6. DUALITY

The last property we leave as a problem. ⇤

Problem 38. Prove that if F is a face of K, then ( (F )) = F.

2. Polytope duality

One way to sum up the result of the last problem would be toobserve that if K is a compact subset of Rd with the origin in itsinterior, then induces an inclusion reversing bijection between F(K)and F(K?). The order dual of a poset (⇧,), which we denote by(⇧?,?), is the poset whose ground set is also ⇧ and whose orderrelation is given by x ? y if and only if y x. Hence, F(K) isisomorphic to F(K?)?.

What consequences can we derive for face posets of polytopes? Tobegin with, we need the following.

Problem 39. If P is a polytope, F a face of P and G a face of F, thenG is a face of P.

Problem 40. If P is a d-polytope in Rd, then P ? is also a d-polytope.

A chain in a poset is a sequence

x0 < x1 < · · · < xm

of distinct elements of the poset. The length of the chain is one lessthan the number of elements in the chain. The above chain has lengthm. A chain is maximal if it is not contained in any strictly longerchain. For instance, in D30, 1 < 3 < 6 < 30 is a maximal chain as is1 < 5 < 10 < 30. (Recall the definition of Dn in Chapter 3 However, inB3, ; < {2} < {1, 2, 3} is not a maximal chain. A poset is graded ifthe length of every maximal chain is the same. If ⇧ is a graded poset,then the rank of an element x 2 ⇧ is the maximum length among allchains which end with x. The rank of a graded poset is the maximumrank obtained by its elements. Equivalently, the common length of allof its maximal chains. For examples of graded posets see Exercises 2and 3

Problem 41. If P is a d-polytope, then F(P ) is a graded poset of rankd+1. More generally, the rank of any face F of P in F(P ) is 1+dim F.

The dual of a simplicial polytope is called a simple polytope. Asimple combinatorial characterization of simple polytopes is containedin the exercises below. By Problem 41, interchanges i-dimensionalfaces of a d-polytope with (d � i)-dimensional faces of P ?. Thus thecomplete description of possible f -vectors of simplicial polytopes with

2. POLYTOPE DUALITY 41

n vertices given in Chapter 5 also gives a complete description of thepossible f -vectors of simple polytopes with n facets. In addition, Prob-lem 34 gives an exact formula for the maximum number of vertices forsuch a polytope.

Exercise 2.1.

(1) Let K be a closed convex subset of Rd which contains the ori-gin. Then K? contains an a�ne subspace of dimension i ifand only if dim K d� i.

(2) Bn is a graded poset and the rank of A ✓ [n] is equal to |A|.(3) Dn is a graded poset. If a 2 n and the prime factorization of

a isa = pe1

1 · pe22 · · · · · pem

m ,

then the rank of a is e1 + · · · + em.(4) Let P be a d-polytope. If F is a (d� 2)-dimensional face, then

F is contained in exactly two facets.(5) P is a simple d-polytope if and only if every vertex is incident

to d edges.(6) Any simplex is both simple and simplicial. Are there any other

polytopes with this property?(7) Are there any polytopes such that P = P ?? Polytopes with the

property that F(P ) ' F(P ?) are called self-dual. Simplicesare self dual. Are there other self dual polytopes?

(8) What can you say about self dual convex subsets?

CHAPTER 7

The Mobius function

In order to further advance our study of face posets of polytopes,their connections to hyperplane arrangements and graph coloring, wenow turn to Mobius inversion. This is now one of the standard tech-niques of modern combinatorics. Introduced by Weisner [25], Mobiusinversion is a poset based generalization of inclusion-exclusion. Its realvalue as a powerful approach to many enumerative problems was pi-oneered by Rota [17]. Throughout this chapter ⇧ is a finite poset.While the theory applies to locally finite posets, those such that everyinterval [x, y] is finite, we will have no need of this generality.

1. The Mobius function

The Mobius function is a simultaneous generalization of inclusion-exclusion and its more famous number theory namesake. It is thefunction µ : ⇧ ⇥ ⇧ ! Z defined inductively using the following prop-erties:

(1) If x 6 y, then µ(x, y) = 0.(2) µ(x, x) = 1.(3) If x < y, then

X

xzy

µ(x, z) = 0.

As long as [x, y] is finite, µ(x, y) can be computed using the lasttwo properties inductively. See Figure 1 for a simple example whereµ(x, y) is computed for a specific poset and a fixed x in the poset.

While the above inductive procedure gives a finite algorithm tocompute the Mobius function for any two elements in ⇧, it is knownthat for general posets and elements it is not computationally fast.However, in many specific cases it can be computed easily. Here aretwo examples. (Recall the definitions of Bn and Dn in Chapter 3,)

Problem 42. Compute µ(S, T ) for any S, T 2 Bn.

Problem 43. Compute µ(s, t) for any s, t 2 Dn.

43

44 7. THE MOBIUS FUNCTION

0

x 1

�1 �1

2

�1

�1

0

0

1

Figure 1. Computing µ(x, y).

Another class of posets for which the Mobius function is (very)easy to compute are Eulerian posets. A poset is called Eulerian if it isgraded, contains a unique minimal element, a unique maximal element,and its Mobius function satisfies

µ(x, y) = (�1)⇢(y)�⇢(x)

for all x y. Here, ⇢(x) is the rank of x. The easiest example of anEulerian poset is Bn. Perhaps the most interesting class of Eulerianposets are face posets of polytopes.

Problem 44. Let P be a (finite) ranked poset with minimum 0 andmaximum 1. Then P is Eulerian if and only if for every interval [x, y] ✓P the number of elements of odd rank equals the number of elementsof even rank.

Problem 45. If P is a polytope, then F(P ) is an Eulerian poset.

Another large class of Eulerian posets are the face posets of simplicialcomplexes homeomorphic to spheres or odd-dimensional manifolds.

2. Mobius inversion

The Mobius inversion formula is probably best understood by in-troducing the incidence algebra of a poset. The incidence algebraof ⇧, denoted by I(⇧), is the space of all functions ⌘ : ⇧ ⇥ ⇧ ! Rsuch that ⌘(x, y) = 0 whenever x 6 y. It comes with the usual realvector space structure associated to spaces of functions to the reals. Italso has a multiplicative structure which looks suspiciously like matrixmultiplication (if you are an algebraist) or convolution (if you are an

2. MOBIUS INVERSION 45

analyst). For ⌘, 2 I(⇧) define ⌘ � in I(⇧) by

(⌘ � )(x, y) =

(0 if x 6 yP

xzy ⌘(x, z) · (z, y) if x y.

This multiplicative structure is easily deciphered from the point ofview of matrix multiplication. Let n = |⇧| and let � : [n] ! ⇧ be abijection so that the induced order relation [n], given by i [n] j ifand only if �(i) �(j), is natural. (See Section 2.1.)

Problem 46. The incidence algebra I(⇧) is isomorphic to the matrixalgebra of n⇥ n real matrices M such that Mi,j = 0 if i 6[n] j.

All of the properties of I(⇧) below follow easily from the aboveproblem.

Corollary 2.1.

(1) (⌘ � ) � ⌫ = ⌘ � ( � ⌫) (associativity).

(2) Define �(x, y) =

(1 x = y

0 x 6= y.

Then ⌘ � � = � � ⌘ (existence of identity)

(3) Define ⇣(x, y) =

(1 x y

0 x 6 y.

Then µ � ⇣ = ⇣ � µ = � (inverse of the Mobius function).

Let f and g be functions from ⇧ to Z. Mobius inversion is a methodfor analyzing the situation when

g(y) =X

xy

f(x)

or its order dual relation

g(x) =X

xy

f(y).

It might seem that such a relationship would be quite rare. As we willsee, it occurs quite frequently. Here is a simple example. Let �(n) bethe Euler totient - �(n) is the number of positive integers less than orequal to n which are relatively prime to n. (Have no common factorsother than one.) For instance, �(12) = 4 since 1, 5, 7, 11 are the onlysuch integers.

Proposition 2.2. X

d|n

�(d) = n.


Proof: Write down the fractions 1n, 2

n, . . . , n

n. Then put them in low-

est terms. For instance, for n = 12 we end with

1

12,1

6,1

4,1

3,

5

12,1

2,

7

12,2

3,3

4,5

6,11

12,1

1.

Notice that only denominators d which divide n occur, and each suchdenominator occurs exactly �(d) times. ⇤

Theorem 2.3. (Mobius inversion) Let f and g be functions from ⇧ toZ.

g(y) =X

xy

f(x)

if and only if

f(y) =X

xy

g(x)µ(x, y).

Dually,

g(x) =X

xy

f(y)

if and only if

f(x) =X

xy

µ(x, y) g(y).

The above formulas look like multiplication of vectors and matricesand that is the main idea behind the proof of the theorem.

Problem 47. Prove the Mobius inversion formulas.

The number theoretic Mobius function, for which the poset versiongets its name is defined for n 2 Z+ by

µ(n) =

8><

>:

1 n = 1

(�1)k n = p1 · · · pk, the pi distinct primes

0 otherwise.

Corollary 2.4.

�(n) =X

d|n

µ(n/d) · d.

Proof: Problem 43 and Mobius inversion. ⇤

Our first combinatorial application of Mobius inversion involvesgraph coloring. Let G be a graph without loops. Denote the ver-tices of G by V (G). A �-coloring of G is a function f : V (G) ! [�].The coloring f is proper if no vertices with the same color (i.e. image)


D

A

B C

Figure 2. A simple graph

share an edge. Define �G(�) to be the number of proper �-colorings ofG. Evidently �G is only defined for � � 1.

What kind of function is �G? It is called the chromatic polyno-mial, so it is a good guess that �G is a polynomial! Any �-coloring ofG induces a partition of the vertices which has at least � many blocks.The partition-induced subgraph of G corresponding to the coloringf is the graph whose vertices are the same as G, but whose edges arethose whose end points have the same color. Thus a proper coloringcorresponds to the graph consisting of the vertices of G and no edges.For another example, consider the graph in Figure 2. If f mappedvertices A and D to color 1, and vertices B and C to color 2, then thepartition-induced subgraph would consist of the vertices and the singleedge between B and C. Notice that the same thing would happen if fmapped vertex A to 1, vertex D to 2 and vertices B and C to 3.

The poset of vertex-induced subgraphs of G, which we denote byLG, consists of all possible partition-induced subgraphs of G (over allpossible colorings, not just those of a fixed �) ordered by inclusion. Itsleast element is the graph with no edges and its maximal element isG itself. The Hasse diagram for the graph in Figure 2 can be seen inFigure 3. What else can we say about LG?

Problem 48. LG is graded. The rank of a partition-induced subgraphH of G is |V |�# components of H.

In order to connect LG with the chromatic polynomial, for H 2 LG

define f�(H) to be the number of �-colorings of G such that H is the


Figure 3. LG

associated partition-induced subgraph. By definition, �G(�) is f�(0),where 0 is the least element of LG. Can we apply Mobius inversion?For any graph H let c(H) be the number of components of H.

Problem 49.

�G(�) =X

H2LG

µ(0, H)�c(H) = �c(G)X

H2L(G)

µ(0, H)�⇢(G)�⇢(H).

From this we easily see that �G(�) is a polynomial of degree |V |, con-tains �c(G) as a factor and the coe�cient of �c(G) is µ(0, G). The lastterm of the right-hand side of the above equality makes sense for any(finite) graded poset with a unique minimal element 0.


Definition 2.5. Let ⇧ be a (finite) graded poset with unique minimalelement 0 and let r be the rank of ⇧. The chracteristic polynomialof ⇧ is

�⇧(�) =X

x2⇧

µ(0, x)�r�⇢(x).

We will meet the characteristic polynomial again in the next chapteron hyperplane arrangements.

Exercise 2.6.

(1) An element ⌫ 2 I(⇧) is multiplicatively invertible if and onlyif ⌫(x, x) 6= 0 for all x 2 ⇧.

(2) Inclusion-exclusion is the following combinatorial principle.Let A1, . . . , Am be finite sets. For each nonempty subset Iof [m] define AI = \i2IAi. Then

|A1 [ · · · [ Am| =mX

i=1

X

|I|=i

(�1)i+1|AI |.

Prove this by showing it is a consequence of Mobius inversionon Bm.

(3) Let G be a connected graph. Prove that if � does not divideµLG(0, G), then G has a proper �-coloring. Prove the converseholds if � = 2.

(4) Fix n � 1. The partition lattice, ⇡n, is the poset whoseelements are partitions of [n] and whose order relation is givenby refinement. For instance, in ⇡3 the least element is thepartition with three blocks, {1}, {2}, {3}, the greatest elementis the partition with one block, {1, 2, 3}, and there are a totalof five elements. In general ⇡n has a unique minimum, 0 ={1}, . . . , {n} and a unique maximum 1 = {1, . . . , n}. Computeµ(0, 1) in ⇡n.

(5) A di↵erent approach to the chromatic polynomial is throughdeletion-contraction. Let G be a graph. It may have paralleledges (more than one edge between a pair of vertices) and/orloops. Define �G(�) as before. If G has a loop, then �(G) ⌘ 0.Let e be an edge of G. The deletion of e, denoted G � e, isthe graph obtained by removing the edge e. The contraction ofG along e, denoted G/e is the graph obtained by contractingthe edge down to a vertex and identifying the two vertices ofthe edge down to one vertex. This may introduce loops and/orparallel edges. For instance, if G is a triangle, then the con-traction of G along any of its edges is a graph with two vertices


and two parallel edges. Prove that for any edge e of a looplessgraph,

�G(�) = �G�e(�)� �G/e(�).

(6) Prove that the coe�cients of �G(�) alternate in sign.(7) Compute �G when G is a tree, a connected graph with no cir-

cuits.

CHAPTER 8

Hyperplane arrangements and Zonotopes

On the surface, hyperplane arrangements are easy to describe. Theyare simply a finite collection of hyperplanes in Rd. As noted earlier,any hyperplane H=

a,b divides space into three disjoint convex sets, thehyperplane itself and two open half spaces, H+

a,b and H�a,b. The two

full-dimensional sets are called regions. As we add hyperplanes therewill be more regions. The main question we will address is, “Howmany regions are there?” Along the way we will discover a connectionbetween this question, “Does �(�1) mean anything”, and how manypieces a d-dimensional cake can be cut into using n slices.

1. Hyperplane arrangements

A hyperplane arrangement in Rd is a finite collection A ={H1, . . . , Hn} of hyperplanes in Rd. The arrangement is called centralif the hyperplanes are all linear, i.e. all contain the origin. OtherwiseA is called a�ne. If the intersection of all the hyperplanes in a centralarrangement is exactly the origin, then it is essential.

A central arrangement divides space into various subsets which arecalled cones.

Definition 1.1. C is a cone in Rd if for all x 2 C and t � 0, tx 2 C.

By definition, the empty set and Rd are cones in Rd. Warning:while related, the notion of cone here is di↵erent from that in Exercise5.3.9 #2. Examples of cones include closed half-spaces of the formH

a,0, a 6= ~0 and their intersections. A cone is polyhedral if it isthe intersection of a finite number of such half-spaces. In general,the intersection of cones is a cone. As usual, this means that for anysubset Y of Rd there is a smallest cone containing Y which we denote byco(Y ). It consists of all of the rays starting at the origin and containinga nonzero point of Y. Hence nonempty cones in Rd are in one-to-onecorrespondence with subsets of the unit sphere in Rd.

Let A = {H=a1

, . . . , H=an} be a central arrangement in Rd. For x 2

Rd we can record using {+, 0,�}n whether ai · x is positive, zero ornegative. The resulting sign vector sA(x) is in {+, 0,�}n. As long as

51

52 8. HYPERPLANE ARRANGEMENTS AND ZONOTOPES

1

+ �

+�+�

H

H

H

2

3

(+��) (�++)

(��)

(0��)

(+++)(++0)

Figure 1. Several sign vectors

there is no confusion we will suppress the subscript A. Note that sA(x)depends on the choice of direction of ai. For a potential sign vectors 2 {+, 0,�}n we want to describe the points of Rd with the given signvector. For this purpose, define

c(s) =

(x 2 Rd :

8><

>:

ai · x = 0, s(i) = 0

ai · x 0, s(i) = �ai · x � 0, s(i) = +

).

Figure 1 shows a typical example in R2. The covectors of A are thosesign vectors s 2 {+, 0,�}n such that

(x 2 Rd :

8><

>:

ai · x = 0, s(i) = 0

ai · x < 0, s(i) = �ai · x > 0, s(i) = +

)

is not empty. For instance, for the arrangement in Figure 1, c(+,�, +) =c(+,�, 0) = c(�, 0, 0) = c(0, 0, 0) = {~0}, but only (0, 0, 0) is a covectorof the arrangement. With this notation, part 3 of the next proposi-tion shows us that our original question can be rephrased, “How manycovectors are there with no zeros in their sign vector?”

Proposition 1.2.

(1) c(s) is a closed convex cone.(2) Suppose that s1 and s2 are covectors of A. Then c(s1) ✓ c(s2)

if and only if s1 can be obtained from s2 by changing some +’s

1. HYPERPLANE ARRANGEMENTS 53

Figure 2. Face fan of a polygon

and/or �’s to zeros and this holds if and only if c(s1) is a faceof c(s2).

(3) If s is a covector of A, then dim c(s) = dim\i:s(i)=0Hi.

Problem 50. Prove the above proposition.

Looking at Figure 1 and Proposition 1.2 suggests that the variouscones c(s) are arranged in a way somewhat similar to the way polytopalcomplexes are configured. To make this precise requires the notion ofa fan.

Definition 1.3. A finite set of cones C = {C1, . . . , Cn} in Rd is a fanif

(1) Any nonempty face of a cone in the fan is in the fan.(2) For all Ci and Cj in the fan, Ci \ Cj is a face of both Ci and

Cj.

The fan is called complete if its union is all of Rd.

Problem 51. If A is a central hyperplane arrangement, then the col-lection of the cones c(s), s a covector of A, is a complete fan.

A second example of a complete fan is the face fan of a polytope.Let P be a d-polytope in Rd with the origin in its interior. We form afan by collecting all subsets of the form co(F ), F a proper face of Pand then including the origin. See Figure 2 for an example of the facefan of a polygon. Since the dimension of co(F ) is the dimension of Fplus one, the number of cones of dimension i, 1 i d in the face fanof P is just fi�1 of P .

Problem 52. Let C be the face fan of a d-polytope P with the originin its interior and let fi(C) be the number of i-dimensional cones in C.


Prove thatdX

i=0

(�1)ifi(C) = (�1)d.

The key to relating face fans to the fans coming from hyperplanearrangements are zonotopes.

2. Zonotopes

An polytope Z in Re is a zonotope if it is the a�ne projectionof a cube. Specifically, there exists an a�ne map f : Rd ! Re suchthat f(⇤d) = Z. Since faces of cubes are translates of cubes, it is notunreasonable to hope that faces of zonotopes are zonotopes.

Problem 53. Let f be a projection of a polytope P onto a polytope Q.If F is a face of Q, then f�1(F ) \ P is a face of P.

Corollary 2.1. Faces of zonotopes are zonotopes.

At first sight it is not apparent how zonotopes might be relatedto hyperplane arrangements. A hint that there might be a connectioncomes from an alternative characterization of zonotopes via Minkowskisums.

Definition 2.2. Let X and Y be nonempty subsets of Rd. The Minkowskisum of X and Y is

X + Y = {x + y : x 2 X,y 2 Y.}Since addition is commutative and associative Minkowski sums are

commutative and associative. Furthermore, it is easy to see that {~0}acts as the identity with respect to Minkowski sums. Several operationswe have seen can be described in terms of Minkowski sums. Figure 3suggests how to construct cubes, translate subsets, and form prismsusing Minkowski sums.

Problem 54. Z is a zonotope in Rd if and only if there exists x1, . . . ,xn

in Rd such that Z can be written as a translation of a Minkowsi sum

Z = [�x1,x1] + · · · + [�xn,xn].

(Recall that [�xi,xi] is the line segment from �xi to xi.)

Now suppose A = {H=a1,0, . . . , H

=an,0} is an essential central hyper-

plane arrangement. Consider the zonotope in ZA in (Rd)? given by

[�a1, a1] + · · · + [�an, an].

Since A is essential, ZA is a d-polytope with the origin in its interior.What does the face fan of Z?

A in (Rd)?? = Rd look like?

2. ZONOTOPES 55

+ = + =

+=

Figure 3. Cube, translation and prisms as Minkowski sums

Theorem 2.3. The face fan of Z?A equals the fan of the hyperplane

arrangement A. Specifically, F is a proper face of ZA if and only ifthe smallest cone containing (F ) equals c(s) for some covector s 6=(0, . . . , 0) of A.

Proof: We begin by considering the faces of ZA. By Problems 53 and54 we know that a proper face F of Z is the image of a face of the cube⇤n. So there is a partition of [n] into three blocks [n] = Pl [Mi [ Zesuch that

(2) F =

(a : a =

nX

i=0

�iai,

8><

>:

if i 2 Pl, then �i = 1

if i 2 Mi, then �i = �1

if i 2 Ze, then � 1 �i 1

).

Note that while every face is of this form, it is not true that everypartition corresponds to a face.

When is x = (x1, . . . , xd) 2 Rd in (F ) (the face in Z?A dual to F )?

When a · x = 1 for all a 2 F and a · x 1 for all a in Z. From thiswe can see that tx, t > 0 is in co( (F )) if and only if x · a = a · x ismaximized over all a 2 ZA on F . Looking at F we first notice thatthis means that ai ·x must be zero for all i 2 Ze. Otherwise, for a 2 Fof the form specified in (2), a · x would not be constant. Similarly,ai · x � 0 for i 2 Pl, and ai · x 0 for i 2 Mi. Otherwise a · x is notmaximized over ZA on the face F. Thus, each co(F ) is equal to c(s) forthe covector s which is + for all i 2 Pl,� for all i 2 Mi and 0 for alli 2 Ze. Conversely, any nonzero covector s determines F by reversingthe above reasoning. ⇤


Corollary 2.4. Let A be a central arrangement. Let C be the fan ofthe arrangement. Then

dX

i=0

(�1)ifi(C) = (�1)d.

Proof: When A is essential this follows immediately from Theorem2.3 and Corollary 52.

Problem 55. Prove the above formula when A is not essential.

3. The intersection poset

In order to use the above result on the number of cones in thefan of a central hyperplane arrangement we introduce the intersectionposet of A. It consists of all intersections of hyperplanes ordered byreverse inclusion. This includes the intersection of no hyerplanes, Rd.The intersection poset is denoted LA and always has a minimal element0 = Rd, and a maximal element 1 equal to the intersection of all of thehyperplanes. One way to get a feeling for LA is to look at graphicarrangements.

Let G be a graph without loops or parallel edges, whose vertexset is [d] and edge set is E. For each edge e = {i, j} 2 E define thehyperplane He in Rd by xi � xj = 0. The set AG = {He}e2E of all theHe is the graphic arrangement associated to G. Evidently AG is acentral arrangement. However, it is never essential (see Exerc. 5).

Problem 56. LAG ' LG.

The most well known graphic arrangements are the braid arrangements- the graph arrangements associated to complete graphs, Kn. There isa close connection between the fan of the arrangement AKn and thepermutahedron (see Exerc. 6).

This suggests several properties of intersection posets of all centralhyperplane arrangements. A poset is a lattice if every pair of elementsof the poset have a least upper bound and a greatest lower bound.Suppose L is a lattice. The least upper bound of x and y in L is calledtheir join and is denoted by x _ y. The greatest lower bound of x andy is called their meet and is denoted x ^ y. The operations of meetand join are commutative and associative. When L is finite we cantake the join of all of the elements to see that L has a unique maximalelement 1. Similarly, the meet of all the elements of L shows that thereis a unique minimal element 0. In any poset which contains a uniqueminimal element 0 the elements which cover 0 are called the atoms ofthe poset.

4. ACYCLIC ORIENTATIONS 57

Problem 57. Let A be a central arrangement in Rd.

(1) LA is graded. The rank of A 2 LA is d� dim A.(2) LA is a lattice.(3) Let A, B 2 LA. Then ⇢(A)+⇢(B) � ⇢(A_B)+⇢(A^B). (LA

is semi-modular.)(4) Every A > 0 in LA is the join of atoms of LA. (LA is atomic.)

Any finite lattice which satisfies the last two properties of the aboveproblem is called a geometric lattice.

Problem 58. Are there any geometric lattices which are not isomor-phic to LA for some central hyperplane arrangement?

We are finally ready to prove the promised formula for the numberof regions of a central hyperplane arrangement. For A 2 LA definef(A) to be the number of cones in the fan of A whose a�ne spanequals A. Thus the number of regions of the hyperplane arrangementequals f(0).

Theorem 3.1. [28], [13] Let A be an essential central hyperplane ar-rangement in Rd. Then

f(0) = (�1)d�LA(�1).

Problem 59. Prove the above formula.

What if A is not essential?

Problem 60. Let A be a nonessential central arrangement, dim\Hi =m > 0. Then there exists an essential central arrangement A in Rd�m

such that LA ' LA and the covectors of the two arrangements areidentical.

4. Acyclic orientations

At the beginning of the chapter we asked whether or not �G(�1)meant anything. Since �G(�) = �c(G)�LG(�), �LG = �LAG

and The-orem 3.1 tells us that the number of regions of the hyperplane ar-rangement AG is (�1)|V (G)|�LAG

(�1), we could say that �G(�1) is

(�1)|V (G)|�c(G) times the number of regions of the hyperplane arrange-ment AG. This does not seem like a very satisfactory answer. At thevery least an interpretation of �G(�1) should involve objects which are‘obviously’ graph invariants. In 1973 Richard Stanley showed how todo this via acyclic orientations.

An orientation of a graph is a choice of direction for every edgeof the graph. If G has n edges, then there are 2n possible orientations


Figure 4. Two orientations of a graph, one acyclic

for G. An orientation of G is acyclic if there are no directed circuits.See Figure 4 for examples.

Let O be an orientation of G. Define a binary relation O on thevertices of G by v1 v2 if there is a directed path from v1 to v2. Theempty path is permitted (as a directed path from v1 to v1).

Problem 61. Suppose G has no loops. Then an orientation O isacyclic if and only if O is a partial order on the vertices of G. Con-versely, if (⇧,) is a poset, then there exists a graph G and an acyclicorientation O on the graph so that (⇧,) is isomorphic to (V (G),O).

Stanley’s original observation was the following.

Problem 62. [21] If G is a graph, then the number of acyclic orien-tations of G is (�1)|V |�c(G)�G(�1).

Acyclic orientations occur in a variety of contexts. We will seea simple application of them in the next chapter as we examine thegraphs of polytopes.

Exercise 4.1.

(1) Classify all open cones in Rd.(2) Classify all convex cones in R2.(3) An n-gon is a zonotope if and only if n is even.(4) The Minkowski sum of two polytopes is a polytope.(5) If G is a simple graph, then dim\e2EHe = c(G).(6) Let A be the arrangement guaranteed by Problem 60 for AKn .

Show that ZA is combinatorially equivalent to a permutahe-dron.

(7) Give an example of a lattice L with no maximal element.(8) Let L be a finite poset such that x ^ y exists for all x, y 2 L.

Show that L, the poset obtained by adding a unique maximalelement to L, is a lattice. Does this work if L is infinite?

4. ACYCLIC ORIENTATIONS 59

(9) Let E be a finite set of vectors in a vector space V. Define LE

to be poset of all subsets of the form E \ W, where W is alinear subspace of V, ordered by inclusion. Prove that LE is ageometric lattice.

CHAPTER 9

Matroids

Many of the ideas of the last chapter, in particular the characteristicpolynomial of a hyperplane arrangement, should be approached fromthe point of view of matroids. Introduced by Whitney in his 1935 paper[26], matroids are a combinatorial abstraction of linear independenceand planar graph duality (see Section 3).

A matroid M is a pair (E, I) such that E is a finite set, called theground set of M , and I is a subset of the power set of M, called theindependent subsets of M, which satisfy the following axioms.

(1) ; 2 I.(2) Subsets of independent subsets of M are independent.(3) If A and B are independent subsets of M and |A| < |B|, then

there exists e 2 B � A such that A [ {e} is independent.

The second axiom says that the independent subsets of M are an ab-stract simplicial complex on E. The third axiom is sometimes calledthe Steinitz exchange axiom.

In truth, there are a large number of equivalent definitions of ma-troids. In the exercises there are two examples, one in terms of minimaldependent sets and another in the language of simplicial complexes.

The canonical example of a matroid is a finite set of vectors. Let Ube a vector space and E a finite subset of U . Then (E, I), where I arethe linearly independent subsets of E, form a matroid. Similarly, if Uis an a�ne subspace and E is a finite subset, the a�nely independentsubsets of E form a matroid (Chap. 2 Exerc. 1.1 # 3). A slightgeneralization of the first example, which allows repeated vectors to betreated as distinct elements of the matroid, is to let E be the set ofcolumns of a matrix and I the subsets which are independent in thecolumn space of the matrix. In this case we say M is represented bythe matrix.

Problem 63. Let G be a finite graph and E the edges of G. Set I to bethe subsets of edges which do not contain a circuit. Prove that (E, I)is a matroid.

61

62 9. MATROIDS

The matroid in the above problem is denoted MG and is called thecycle matroid of the graph.

Many concepts from linear algebra and graph theory have matroidanalogs. One of the most useful is the notion of rank.

Problem 64. Let M be a matroid and A a subset of the ground set.Then all maximal (in the sense of containment) independent subsets ofA have the same cardinality.

The common cardinality referred to in the previous problem iscalled the rank of A. We use ⇢(A) for the rank of A. It is easy tosee that for any e 2 E, ⇢(A) ⇢(A [ {e}) ⇢(A) + 1. For a rep-resented matroid the rank of a subset of columns is the dimension oftheir linear span as column vectors. The rank of a matroid is the rankof its ground set. As in linear algebra, a maximal independent subsetis call a basis and every independent set can be extended to a basis.

Problem 65. Let A be an independent subset of a matroid M . Provethat there exists a basis B of M such that A ✓ B.

1. Flats

The flats of a matroid approximate the subsets of a vector spaceclosed under linear combinations. For this reason they are frequentlycalled the closed subsets of M. A subset A of the ground set is a flatof a matroid if for all e /2 A the rank of A [ {e} is strictly larger thanthe rank of A. For represented matroids the flats are the subsets ofcolumns which are of the form E \W, where W is a linear subspace ofthe column space of the matrix. The set of flats of M is denoted LM .One key to understanding the flats of a matroid is determining when⇢(A) = ⇢(A [ {e}).Lemma 1.1. Suppose A ✓ E and e /2 A. Then the following areequivalent.

(1) ⇢(A) = ⇢(A [ {e}).(2) For every basis B of A, the union B [ {e} is dependent.(3) For one basis B of A, the union B [ {e} is dependent.

Problem 66. Prove the lemma.

Problem 67. If A and B are flats of M , then A \B is a flat of M.

As we have seen before, a result like Problem 67 implies that forany A ✓ E there is a (unique) minimal flat, called the closure of A,

A =\

F2LMA✓F

F

2. BROKEN CIRCUITS 63

containing A.The collection of flats of a matroid has a natural poset structure

given by subset. The poset (LM ,✓) is in fact a geometric lattice withgrading given by the rank function. See Exerc. 5. Obviously E is themaximal element of LM . The minimal element is a little more interest-ing.

Problem 68. Let M be a matroid and let Z be the minimal flat of LM .Show that for all independent subsets I of M the intersection I \ Z isempty.

The elements of Z in the above problem are called the loops of thematroid.

As in chapter 7 we can define the Mobius function for LM and itscharacteristic polynomial.

Problem 69. Let M be a matroid without loops. Then

µLM (;, F ) =X

A✓E:A=F

(�1)|A|.

What happens if we allow loops?

2. Broken circuits

Whitney introduced the notion of a broken circuit in order to studythe chromatic polynomial of a graph [27]. Later, Rota [17] realized thathis ideas could be extended to matroids.

To even define broken circuits we have to linearly order the elementsof the matroid. For simplicity we will assume that E = [n] with itsusual ordering. A circuit of a matroid is a minimal dependent set. Asyou might guess, the term comes from the fact that for a graph G thecircuits of MG are the edges which form a circuit in G. A broken circuitof M is a circuit with its least element removed. The broken circuitcomplex of M is the abstract simplicial complex whose faces are thosesubsets of E which do not contain a broken circuit.

If M has a loop, say i, then ; is a broken circuit. Since every subsetof E contains the empty set as a subset, the broken circuit complexwould be empty! For this reason we will assume that M contains noloops for the remainder of this section.

Example 2.1. Consider two orderings of the same graph in Figure 1.The ordering depicted on the left-hand side has circuits {1, 2, 3}, {3, 4, 5}, {1, 2, 4, 5}so the broken circuits are {2, 3}, {4, 5}, {2, 4, 5}. The ordering depictedon the right-hand side has circuits {1, 2, 5}, {3, 4, 5}, {1, 2, 3, 4}, so the

64 9. MATROIDS

4

1

3

2

45

5

1 2

3

Figure 1. Two di↵erent edge orderings for the same graph

broken circuits for this ordering are {2, 5}, {4, 5}, {2, 3, 4}. The two bro-ken circuit complexes are not isomorphic as the nonedges in the firsthave distinct vertices while in the second complex the vertex 5 is inboth. However, the f -vector of both complexes is (1, 5, 8, 4).

As can be seen from the above example, di↵erent orderings of thematroid may lead to di↵erent broken circuit complexes. The remark-able fact is that the f -vector does not depend on the ordering.

Theorem 2.2. Let M be a rank r matroid without loops. The charac-teristic polynomial of LM is

�LM (�) = f�1�r � f0�

r�1 + · · · + (�1)rfr�1,

where (f�1, f0, . . . , fr�1) is the f -vector of the broken circuit complexof M with respect to any ordering of the ground set.

The proof of this theorem follows easily after solving the followingproblem and applying the result to each element of LM .

Problem 70. Let M be a rank r matroid with without loops, r � 1.Then (�1)rµLM (;, E) equals the number of bases of M which do notcontain any broken circuit.

Theorem 2.2 has several immediate consequences for the character-istic polynomial of a matroid. Evidently the coe�cients alternate insign and are constrained by the fact that their signless counterpartsmust form the f -vector of a simplicial complex. In fact, the brokencircuit complex is always shellable (Exerc. 8) - so we have even morerestrictions from this fact.

3. MATROID DUALITY 65

3. Matroid duality

One of Whitney’s original motivations for introducing matroids wasto answer the question, “What is the dual of a nonplanar graph?”Before we explain how matroids answer this question, we review thegeometric duals of planar graphs.

If G is the graph of a 3-polytope P , then the geometric dual of G isthe graph of P ?. This idea can be generalized to an arbitrary graph Gembedded in the plane as follows. The geometric dual of G, which wedenote by G?, is the graph whose vertices are the regions in which thegraph divides the plane and whose edges correspond to regions whichlie on opposite sides of an edge of G. See Figure ?? for an example.

Di↵erent embeddings of the same graphs can give geometric dualswhich are not isomorphic as graphs. See Exerc. 9. Nonetheless, planargraph duality has many uses and several properties of dual graphs donot depend on the embedding. The edges of a circuit in G correspondto a minimal set of edges of G? whose removal increases the numberof components of G?. A di↵erent connection between G and G? pointsto Whitney’s original definition of matroid duality. Suppose G is con-nected. Let T be the edges of a spanning tree of G. Then the edgesof G? corresponding to the complement of T is a spanning tree of G?.Spanning trees of a connected graph G are the bases of MG. In otherwords, the bases of MG are the complements of bases of M(G?).

The same phenomenon can be observed in matroids represented bymatrices. Let N be an r ⇥ n matrix, r < n. For notational simplicitywe assume that N has rank r and that the first r columns of N arelinearly independent. The matroid represented by N is not changed byelementary row operations, so let us assume that the r ⇥ r submatrixconsisting of the first r columns of N is the identity matrix. So N lookslike this.

2

66664

1 0 0 . . . 0 c1,r+1 c1,r+2 . . . c1,n

0 1 0 . . . 0 c2,r+1 c2,r+2 . . . c2,n

0 0 1 . . . 0 c3,r+1 c3,r+2 . . . c3,n...

......

0 0 0 . . . 1 cr,r+1 cr,r+2 . . . cr,n

3

77775

Now define N? to be2

664

�c1,r+1 �c2,r+1 �c3,r+1 . . . �cn,r+1 1 0 . . . 0�c1,r+2 �c2,r+2 �c3,r+2 . . . �cn,r+2 0 1 . . . 0

......

...�c1,n �c2,n �c3,n . . . �cr,n 0 0 . . . 1

3

775

66 9. MATROIDS

The inner product of row i of N and row j of N? is �ci,r+j +ci,r+j = 0.Hence the two row spaces are orthogonal, even for fields of positivecharacteristic. For this reason, in older literature the two associatedmatroids were called orthogonal. It is the following relationship be-tween the matroids represented by N and N? that interests us.

Problem 71. Show that a subset of the columns of N is a basis of thecolumn space of N if and only if the complement of those columns is abasis of the column space of N?.

Problem 72. Let M = (E, I) be a matroid with bases B. Then M? =(E, I?), with I? the subsets of E whose complements contain a basis inB, is a matroid.

The matroid M? is called the dual matroid to M. Another way todescribe M? is as the matroid whose bases are the complements of thebases of M. From our previous discussions we can see that for a planargraph M?(G) = M(G?). It is in this sense that we can define the dualof a nonplanar graph G as the matroid M?

G.There are many applications of matroid duality. We end with one

originally discovered by Tutte [24]. Let � be a finite abelian group and~G a directed graph with edge set E which we think of as ordered pairsof vertices. A �-flow on ~G is a function � : E ! � such that at eachvertex v of ~G

X

(w,v)2G

�((w, v))�X

(v,w)2G

�((v, w)) = 0.

As with most flows on graphs this condition can be imagined as sayingthe the in-flow at each vertex equals the out-flow. A nowhere zero �-flow on ~G is a �-flow where �(e) 6= 0 for all e 2 E. The question wewant to answer is, “How many nowhere zero �-flows are there on ~G?”The first surprise is that the answer does not depend on the orientationgiven ~G.

Problem 73. Prove the the number of nowhere zero �-flows on ~G onlydepends on the underlying undirected graph G.

Denote the number of nowhere zero � flows on G by F (G, �). If eis a loop, then by definition e is in every basis of M?. For this reasonany element that is in every basis of a matroid is called a coloop.

Problem 74. If e is a coloop of G and � is a �-flow on G, then�(e) = 0. In particular, if G has a coloop, then F (G, �) = 0.

3. MATROID DUALITY 67

Theorem 3.1. Suppose MG has no coloops. Then

F (G, �) = �LM?G(|�|).

For once we do not list the proof as a Problem. However, the interestedreader will learn quite a bit by finding a Mobius inversion proof in thestyle of Probelm 49.

One of the most well known conjectures in graph theory, which hasbeen open for over fifty years, is Tutte’s 5-flow conjecture.

Conjecture 3.2. [24] Let G be a graph such that MG has no coloops.Then F (G, 5) > 0.

Seymour was able to show that if MG has no coloops, then F (G, 6) > 0[19].

Exercise 3.3.

(1) Two distinct elements {e1, e2} of a matroid are parallel if {e1, e2}are a dependent set. Prove that ‘is parallel to’ is an equivalencerelation.

(2) Let C be the circuits of a matroid M .(a) Prove that if C1 ✓ C2 are in C then C1 = C2. This prop-

erty is sometimes referred to as C is a clutter.(b) (Circuit elimination) Prove that if e 2 C1\C2 and C1, C2 2

C, then there exists a circuit C3 such that C3 ✓ (C1[C2)�{e}.

(c) (Strong circuit elimination) Better yet, prove that if e 2C1 \ C2, C1, C2 2 C, and e0 2 C1 � C2, then there existsC3 2 C such that e0 2 C3 and e /2 C3.

(d) Prove that if C is a clutter of subsets of E which satisfycircuit elimination, then C is the set of circuits of thematroid whose independent subsets are the subsets of Ewhich do not contain any member of C.

(3) Let � be a simplicial complex with vertex set E, a finite set.For any subset A of E let �A be the simplicial complex whosefaces are the faces of E contained in A. Prove that the faces of� are the independent subsets of a matroid if and only if �A

is a pure simplicial complex for all A.(4) Prove that LG = LMG , where LG is the poset from Chapter 7.(5) Show that for any matroid LM is a geometric lattice whose

rank function is the rank function of the matroid.(6) Let L be a geometric lattice and E be the set of atoms of L.

Show that if I = {A ✓ E : ⇢(W

A) = |A|}, then ML = (E, I)is a matroid. Show that not all matroids are of the form ML.

68 9. MATROIDS

(7) Prove that M is a matroid without parallel elements or loops,then M = ML for some geometric lattice.

(8) Order the facets of the broken circuit complex lexicographically.Prove that this is a shelling.

(9) Show by example that di↵erent embeddings of a graph withone vertex and 4 loops can have geometric duals that are notisomorphic as graphs.

CHAPTER 10

Graphs of polytopes

The simplex algorithm optimizes a linear function over a polytopeby examining a vertex and then considering the vertices adjacent tothat vertex to see if a better choice is available. This requires anunderstanding of the graph of the polytope - the graph given by thevertices and edges of the polytope. For a polytope P we denote thisgraph by G(P ).

How much information does G(P ) tell us about P? A naturalquestion to ask is whether or not G(P ) determines the face poset ofP. Can we determine which subsets of vertices of the graph correspondto faces of the polytope? Looking at Exercise 1 we can see that G(P )does not even determine the dimension of P ! There are at least twosituations where G(P ) does determine F(P ).

One is when P is three-dimensional. In this case we already knowthe vertices and edges, so all that remains is to determine which circuitsof the graph are the boundaries of facets. If we remove the boundary ofa facet from the boundary of the polytope, then the graph will still beconnected. However, if we remove a circuit that is not the boundary ofa 2-face then the graph will be disconnected. This gives us an algorithmfor figuring out which circuits correspond to facets.

Another situation when G(P ) determines F(P ) is when P is a sim-ple polytope. A remarkably simple argument due to Kalai shows thatG(P ) does indeed determine the face poset of P. The acyclic orienta-tions we met at the end of the last chapter play a crucial role in Kalai’sproof.

1. Acyclic orientations of G(P )

The most common method for obtaining an acyclic orientation ofthe graph of a polytope is by using a generic element of (Rd)?.

Definition 1.1. Let P be a polytope in Rd. An element a 2 (Rd)? isgeneric for P if for all distinct vertices v1 and v2 of P, a · v1 6= a · v2.

Let a be generic for the polytope P . The orientation of the graphof the polytope induced by a is given by orienting an edge {v1, v2} so

69

70 10. GRAPHS OF POLYTOPES

that it points toward v1 if a ·v1 > a ·v2 and points toward v2 otherwise.Since all of the arrows point toward vertices whose dot product with ais greater, there cannot be any directed circuits and the orientation isacyclic. For this to be useful we need to know that there are generic a.

Problem 75. Let P be a d-polytope and let GP be the linear maps in(Rd)? which are generic for P . Then GP is an open subset and hasnonvoid intersection with every nonempty open ball.

Corollary 1.2. Let F be a face of a polytope P and let V (F ) be thevertices of F. Then there exists an acyclic orientation O of G(P ) suchthat the vertices of F form an initial segment of O .

An initial segment of a poset with minimal element 0 is a subset ofthe poset of the form [0, x] for some x in the poset.

Proof: Let Ha,b be a closed half-space which shows that F is a face

of P . By Problem 75 there exist generic a0 arbitrarily close to a. Inthe partial order associated to �a0 the vertices of F must come first. ⇤

2. Simple polytopes

Most properties of simple polytopes are either self-evident or followeasily by looking at their simplicial duals.

Problem 76. Let P be a simple d-polytope.

• Any k-dimensional face of P is the intersection of exactly d�kfacets.

• Let v be a vertex of P and {e1, . . . , em} edges incident to v.There exists a unique m-dimensional face of P which contains{e1, . . . , em} and v.

How can we identify the faces of a simple polytope knowing onlythe graph of the polytope? The key to Kalai’s algorithm are specialclass of acyclic orientations of G(P ). An acyclic orientation of G(P )is good if the graph of every nonempty face of P has a unique sink.The prototypical example of a good orientation is one associated to ageneric linear map. Is it possible to know which acyclic orientations Oare good without knowing the faces of P? For an acyclic orientation Odefine hO

i to be the number of vertices with indegree i. Now set

(3) fO = hO0 + 2hO

1 + · · · + 2dhOd .

Problem 77. Prove that fO is greater than or equal to the number offaces of P . Furthermore, fO equals the number of nonempty faces ofP if and only if O is a good acyclic orientation.

3. BALINKSKI’S THEOREM 71

Notice that this allows us to determine which acyclic orientationsof G(P ) are good without knowing the faces of P . We compute fO forall of them and O is good if it minimizes fO.

Are there any other properties of the vertices of a face of P thatcould help us identify them? Every face of a simple polytope is a simplepolytope and the graph of any simple polytope is regular. A graph isregular if the degree of every vertex is the same. Of course, the graphof a face is connected.

Theorem 2.1. [12] Let P be a simple polytope. Then a vertex inducedsubgraph H of G(P ) equals G(F ) for some face of P if and only if

• H is regular.• The vertices of H form an initial segment of O for some good

acyclic orientation of G(P ).• H is connected.

Problem 78. Prove the above theorem.

Theorem 2.1 gives us an algorithm for determining the face posetof a simple polytope from its graph. This is not the end of the story.The first complete proof that F(P ) was determined by G(P ) for simplepolytopes was due to Blind and Mani [5]. However, it was not construc-tive. Kalai’s algorithm, while elegant and transparent, is exponentialin time and it was an open question whether or not a polynomial timealgorithm for determining the faces of a simple polytope from its graphexists. In [10] Friedman provided such an algorithm.

3. Balinkski’s theorem

Another property of polytope graphs is that they are connected. Infact, they are highly connected. A simple graph G is k-connected ifit has at least k vertices and is connected whenever (k � 1) or fewervertices (and their incident edges) are removed. The following is usuallycalled Balinski’s theorem.

Theorem 3.1. [1] Let P be a d-polytope. Then G(P ) is d-connected.

Before we begin the proof we state a lemma that we will need.

Lemma 3.2. Let P be a polytope in Rd, a 2 (Rd)? and v a vertex ofP . Then there exists a path in G(P )

v = v0 ! v1 ! · · ·! vm

such that for all 0 i m � 1, a · vi > a · vi+1 and a · vm minimizesa · x for all x 2 P.

72 10. GRAPHS OF POLYTOPES

Problem 79. Prove this lemma.

By applying the lemma to �a, the same result holds for paths withincreasing a · vi instead of decreasing.

Proof (Balinski’s theorem): Let S be a subset of the vertices of Pwith |S| d� 1. We consider two cases.

Case one: The vertices in S all lie on a proper face F of P. ThenF = P \ H

a,b for some hyperplane H=a,b. Let c be the minimum of

a · x,x 2 P. By Exercise 3.2 (6) Fmin = P \ H=a,c is a face of P

containing at least one vertex. By Lemma 3.2 any vertex not in S ispart of a path which ends in Fmin. Since Fmin is connected, all verticesnot in S are connected by paths which do not go through S.

Case two: The vertices in S do not lie in a proper face of P. Let vbe any vertex of P not in S. Consider S[{v}. Since |S[{v}| d, thisset lies in at least one hyperplane H=

a,b. (More than one if S [ {v} isa�nely dependent.) Furthermore, there are points in P on both sidesof H=

a,b. Otherwise, all of the vertices of S would lie in the proper faceP \H=

a,b. Define

H+ = {x 2 P : a · x � b}, H� = {x 2 P : a · x b.}Now apply Lemma 3.2 to both H+ and H� in the appropriate directionto see that all the vertices of P not in S are connected to v by pathswhich do not contain any vertices of S. ⇤

An immediate consequence of Balinski’s theorem is that the graphsof 3-polytopes are simple, planar and 3-connected. The remarkablefact is that this is a characterization of such graphs.

Theorem 3.3. (Steinitz’ theorem) [16], [23] A graph is the graph of a3-polytope if and only if it is simple, planar and 3-connected.

The proof of this theorem is not easy. We will not try to prove thishere, except to observe that one direction is above and refer the readerto one of the extant proofs, [11, Sect. 13.1], [29, Chapter 4] for theother direction. In the next section we will see how Steinitz used thisstructural result to get a complete description of all possible f -vectorsof 3-polytopes.

Exercise 3.4.

(1) Give examples of two polytopes P and Q such that G(P ) isisomorphic to G(Q), but dim P 6= dim Q.

(2) Prove that hOi from Equation (3) equals hi(P ?).

(3) Does every good acyclic orientation of a G(P ) for a simplepolytope come from a generic a?

4. f -VECTORS OF 3-POLYTOPES 73

(4) Let P be a 3-polytope with n facets. Then for any two verticesof P there exists a path of length at most n� 3 between themin G(P ). (The length of a path is the number of edges in thepath.)

4. f-vectors of 3-polytopes

Using the Dehn-Sommerville equations it is easy to show that thef -vector of a simplicial 3-polytope is always of the form (1, n, 3n �6, 2n� 4, 1), n � 4. By duality, the f -vector of any simple polytope isof the form (1, 2n� 4, 3n� 6, n, 1), n � 4. Do all n � 4 actually occur?

Problem 80. For every n � 4 show how to construct a simple polytopewith f -vector (1, 2n� 4, 3n� 6, n, 1).

What about other 3-polytopes?

Problem 81. Prove that if (1, f0, f1, f2, 1) is the f -vector of a 3-polytope, then there exist uniquely determined nonnegative integers a, bsuch that

(1, f0, f1, f2, 1) = (1, 4, 6, 4, 1) + a(0, 1, 1, 0, 0) + b(0, 0, 1, 1, 0).

If we can determine all possible pairs (a, b) that occur in the aboveproblem, then we have a complete description of all possible f -vectorsof three polytopes.

Problem 82. Let (1, f0, f1, f2, 1) be the f -vector of a 3-polytope P anddefine a, b as in Problem 81. Prove that a 2b and the a = 2b if andonly if P is simple. Show that b 2a and b = 2a if and only if P issimple.

The main point of this section is that the above conditions on(1, f0, f1, f2, 1) are also su�cient.

Problem 83. Prove that for any (1, f0, f1, f2, 1) that satisfies the condi-tions of Problem 82 there is a 3-polytope whose f -vector is (1, f0, f1, f2, 1).

Steinitz’ description of all possible f -vectors of 3-polytopes datesback to 1906. Over one-hundred years later there is not even a con-jecture as to what all possible f -vectors of 4-polytopes look like!

CHAPTER 11

Hints

1. A�ne and convex geometry

1. If W is a linear subspace, then W + v is a linear subspace if andonly if v 2 W and (hence) W + v = W.2. Translate.3. Show by induction on the number of terms in the a�ne combinationthat every a�ne combination is in the a�ne span.4. Equate to linear independence.5. Write in terms of linear maps.6. Just like Problem 2.7. The set of all convex combinations of elements of A is convex.8. Write y as a convex combination of elements of A with as few termsas possible. If there are more than d + 1 terms then there must be ana�ne dependence among the elements of A in the expression.

2. What is a polytope?

9. Compare to Problem 5.10. This will be easier in the future.11. (x1, . . . , xd) 2 ⇧d if and only if |x1| + · · · + |xd| 1.12. See 10.13. If y 2 P is not one of the xi, then it must lie on a line segmentcontained in P.14. You have seen an example.15. K lies in an a�ne space of dimension d.16. Use the previous problem.17. Use Exerc. 1.1.8 to show that �(ti1), . . . , �(tid+1

) are a�nely inde-pendent.18. For a potential set {�(ti1), . . . , �(tid)} of vertices for a facet ofC(n, d) use determinants to find a formula for the hyperplane contain-ing those vertices.

75

76 11. HINTS

3. Convexity

19. Compactness for existence. If K is convex and there are twopoints x1,x2 which minimize the distance to y, consider the triangle�(x1,x2,y).20. Consider the line segment [y,x] where y is the projection of x ontoK.21. For a fixed a�ne hyperplane H✏ there are many possible ways towrite H✏ = H=

a✏,b✏Try to normalize them so that a✏ and b✏ converge.

22. How do hyperplanes intersect line segments?23. If x is in the interior of each closed half space then x is in theinterior of P.24. Induction on dimension.

4. Shelling

25. When is a face new?26. For connected graphs, start with a spanning tree. For the last, orderthe vertices any way you want, and list the facets lexicographically.27. Induction on the number of facets.28. Use the previous problem.29. What is the e↵ect on the coe�cients of a polynomial in t when youmultiply by t� 1?30. Put the vertices in general position.31. Which lines through x are not allowed?32. Induction on dimension. For the induction step, observe thatthrough any y /2 P, there is a shelling line through y.33. Choose the biggest ai possible.34. Use the g-theorem.

5. Duality

35. Mostly straightforward. For instance, if Br(~0) ✓ K, then K? iscontained in B1/r(~0).36. dim Rd = dim(Rd)? = dim(Rd)?? = d.37. Suppose that a is in K? \H

x

??,1 but not in (F ). Then a · x = 1,and a · y < 1 for some y 2 F.38. There is an a in K? such that F = H

a,1 \K.39. Let F = (H=

a,b \ P ) where a · x b is a valid inequality for F.Similarly, let G = (H=

a

0,b0 \F ), where a0 ·x b0 is a valid inequality forF, but perhaps not P. If t > 0, then (a + t a0) · x b + t b0 is a validinequality for F.40. What are the hyperplanes which define P ??

8. MATROIDS 77

41. Problem 39 and Exercise 3.2.3

6. Mobius inversion

42. First compute µ(;, [n]) in Bn.43. First compute µ(1, n) in Dn. For this, consider two cases. Whenn is the product of distinct primes and when n has a factor which is asquare of a prime.44. Induction on rank.45. Euler characteristic and Problem 44.46. For ⌫ 2 I(⇧) define �(⌫) to be the matrix whose i, j entry is⌫(�(i), �(j)).47. For f : ⇧ ! R define �(f) to be the vector in Rn whose ith

coordinate is f(�(i)). What is �(⇣) ·�(f)? (Notation as in the previoushint.)48. If H ( G can they have the same number of components?49. Consider f : LG ! R, where f(H) is the number of �-colorings ofG such that H is the associated partition-induced subgraph.

7. Hyperplane arrangements and zonotopes

50. If s is a covector, then the a�ne span of c(s) = \s(i)=0Hi. To showthat c(s1) ✓ c(s2) if and only if c(s1) is a face of c(s2) observe that ifa =

Ps1(i)=0 � ai with the sign of � equal to �s1(i) for all i such that

s1(i) = 0 and s2(i) 6= 0, then a · x 0 is valid on c(s2).51. Proposition 1.2.52. Polytope Euler characteristic.53. Use the projection f : P ! Q and the a 2 (Rd)? which defines F.54. A linear map is determined by the image of the unit coordinatevectors.55. Let H = \Hi. Consider A \H.56. x is in He, where e = {i, j} if and only if the ith and the jth

coordinates are equal.57. Translate to linear algebra.58. The Fano plane.59. Corollary 2.4.60. See the hint for Problem 55.61. Use ⇧ for the vertices of your graph.62. How many regions does LAG have?

8. Matroids

63. A cycle matroid can be represented by a matrix.64. Steinitz exchange axiom.

78 11. HINTS

65. Steinitz exchange axiom.66. Steinitz exchange, again!67. Any basis of A \B can be extended to bases of A and B.68. Lemma 1.1.69. For any nonempty finite set E,

PA✓E(�1)|A| = 0.

70. Use Problem 69 and try to cancel all those A which contain abroken circuit. The closure of a set with a given circuit and with oneelement of that circuit removed is the same.71. The columns are independent if and only if the induced squarematrices are nonsingular.72. Let A, B be in I? with |A| < |B|. So there are basis B1 and B2

of M with A ✓ Bc1 and B ✓ Bc

2. If there is an e 2 B2 which is not inB1 [ A, then obviously e satisfies the Steinitz exchange axiom. Fromthe cardinalities of the sets |B1 \B| > 0. Can you arrange to move oneof these elements down to A and replace it with something appropriatefrom B2?73. Move from one orientation to another one edge at a time.74. If e is a coloop of MG, then removing e increases the number ofcomponents of G.

9. Graphs of polytopes

75. Problem 24.76. Translate to the simplicial setting.77. The second item of Problem 76.78. Suppose H satisfies the conditions of the theorem and O is thegood orientation guaranteed by these conditions. Look at the minimumvertex in H with respect to O. Its incident edges correspond to a faceF (Problem 76). Show that H = G(F ).79. What does it mean if for all vertices w which share an edge withv, a · w � a · v?80. Chop o↵ a vertex.81. Euler characteristic.82. What is the minimum number of edges incident to each vertex?83. What e↵ect does chopping a vertex and its dual operation have ona and b?

APPENDIX A

Analysis

For the sake of completeness we include the basic facts from analysisthat we need. More detailed information can be found in any numberof textbooks.

1. Compactness

The notions of open, closed, bounded and compact sets are someof the most fundamental ideas behind understanding the geometry ofRd.

Definition 1.1. A ✓ Rd is open if for all x 2 A, there exists an openball centered at x, Br(x) = {y 2 Rd : ||x � y|| < r}, r > 0, which iscontained in A.

Examples of open subsets of R2 are open balls, the empty set, allof AR2 and the interior of any polygon. A line segment and a polygonare not open. The union of open sets is always open.

Definition 1.2. A subset A of Rd is closed if it is the complement ofan open set.

Polygons, a�ne subspaces, any finite number of points and {(0, 1/n) :n 2 Z+} [ {(0, 0)} are examples of closed subsets of R2. The intersec-tion of closed sets is closed. As usual, this means there is a smallestclosed set containing a given set A. It is called the closure of A. Forinstance, the closure of the rationals in R is all of R.

A bounded subset of Rd is any subset which is contained in someball Br(0). Compact sets form one of the most well-behaved classes ofsubsets of Rd.

Definition 1.3. A ✓ Rd is compact if it is closed and bounded.

Two of the most useful properties of compact sets are described bythe following theorem.

Theorem 1.4. Let C be a nonempty compact subset of Rd.

• Let f : Rd ! R be a continuous function. Then there existsx 2 C which maximizes f on C. Similarly, there exists y 2 Cwhich minimizes f on C.

105

106 A. ANALYSIS

• Let (x1,x2, . . . ) be an infinite sequence in C. Then there existsa subsequence (xi1 ,xi2 , . . . ) which converges to some x 2 C.

The hypotheses that C be closed and bounded are both necessary.

Exercise 1.5.

(1) Polytopes are compact.(2) Construct a nonempty subset Ac ✓ Rd which is closed and a

continuous function f : Rd ! R such that Ac does not containa point which maximizes f on Ac.

(3) Construct a nonempty subset Ab ✓ Rd which is bounded and acontinuous function f : Rd ! R such that Ab does not containa point which maximizes f on Ab.

(4) Construct a nonempty subset Ac ✓ Rd which is closed and aninfinite sequence of elements of Ac which does not contain aconvergent subsequence.

(5) Construct a nonempty subset Ab ✓ Rd which is bounded andan infinite sequence of elements of Ab which does not containa convergent subsequence.

Intuitively it seems obvious that the proper faces of a convex setlie on its ‘boundary’. To make this precise requires we define what theboundary is!

Definition 1.6. The boundary of a subset A of Rd consist of allpoints x 2 Rd such that any nonempty ball Br(x) contains at least onepoint of A and at least one point not in A.

The boundary of A is denoted by @A. Note that it is not requiredthat the points in the boundary of a set are actually in the set. Forinstance, the boundary of the half-open interval (0, 1] in R is {0, 1}.One of many properties of the boundary of a set is the following.

The complement of the boundary of a set A is its interior. A pointx in A is an interior point of A if there exists a nonempty ball Br(x)contained in A. Any closed set is the (disjoint) union of its interior andboundary.

Proposition 1.7. x is in the boundary of A if and only if there exists asequence of points in A which converge to x, and there exists a sequenceof points in the complement of A which converge to x.

Note that the sequence may repeat the same point as often as needed.

Bibliography

[1] M. Balinski, On the graph structure of a convex polythedra in n-space, PacificJ. Math, 11 (1961), 431–434.

[2] L. Billera and C. Lee, A proof of the su�ciency of McMullen’s conditions forf-vectors of simplicial polytopes, J. Comb. Theory, Ser. A, 31 (1981), 227–255.

[3] A. Bjorner, The homology and shellability of matroids and geometric latticesin N. White editor Matroid Applications, Cambridge Univ. Press, 1992.

[4] A. Bjorner, The unimodality conjecture for convex polytopes, Bull. Amer. Math.Soc., 4 (1981), 187–188.

[5] R. Blind and P. Mani-Levitska, On puzzles and polytope isomorphism, Aeq.Math., 34 (1987), 287–297.

[6] H. Bruggesser and P. Mani, Shellable decompositions of cells and spheres,Math. Scan., 29 (1971), 197–205.

[7] G. Dantzig, Linear Programming and Extensions, Princeton University Press,Princeton 1963.

[8] M. Dehn, Die Euersche Formel im Zusammenhang mit dem Inhalt in der nicht-Euklidischen Geometrie, Math. Ann., 61 (1905), 561–568.

[9] J. Eckho↵, Normat, 54 (2006), 146–159.[10] E. Friedman, Finding a simple polytope from its graph in polynomial time,

Disc. Comp. Geom., 41 (2009), 249–256.[11] B. Grunbaum, Convex Polytopes 2nd. ed., Springer-Verlag, New York, 2003.[12] G. Kalai, A simple way to tell a simple polytope from its graph, J. Comb.

Theory Ser. A, 49 (1988), 381–383.[13] M. Las Vergnas, Convexity in oriented matroids, J. Comb. Theory, Ser. A, 29

(1980), 231–243.[14] C. Lee, Counting the faces of simplicial polytopes, Ph.D. thesis, Cornell Uni-

versity, 1981.[15] J. Oxley, Matroid theory, 2nd ed., Oxford Univ. Press, 2011.[16] H. Rademacher and E. Steinitz, Vorlesungen uber die Theorie der Polyeder,

Springer-Verlag, Berlin 1934.[17] G.C. Rota, On the foundations of combinatorial theory I. Theory of Mobius

functions, Z. Wahrscheinlichkeitstheorie 2 (1964), 340–368.[18] M. Rudin, An shellable triangulation of a tetrahedron, Bull. Amer. Math. Soc.,

64 (1958), 90–91.[19] P. Seymour, Nowhere zero 6-flows, J. Comb. Theory. Ser. B, 30, 1981, 130–135.[20] D. Sommerville, The relations connecting the angle-sums and volume of a poly-

tope in space of n dimensions. Proc. Royal Soc. London, Ser. A, 115 (1927),103–119.

[21] R. Stanley, Acyclic orientations of graphs, Disc. Math., 5 (1973), 171–178.

107

108 BIBLIOGRAPHY

[22] R. Stanley, The number of faces of a simplicial convex polytope, Adv. Math.,35 (1980), 236–238.

[23] E. Steinitz, Polyeder und Raumeinteilungen, Encyklopaedia der mathematis-chen Wissenshaften, Band 3 (Geomtrie), Teil 3AB12, 1922.

[24] W. Tutte, A contribution to the theory of chromatic polynomials, Can. J.Math., 6 (1954), 80–91.

[25] L. Weisner, Abstract theory of inversion of finite series, Trans. Amer. Math.Soc., 38 (1935), 474–484.

[26] H. Whitney, On the abstract properties of linear dependence, Amer. J. Math.,57 (1935), 509–533.

[27] H. Whitney, A logical expansion in mathematics, Bull. Amer. Math. Soc., 38(1932), 572 – 579.

[28] T. Zaslavsky, facing up to arrangements: face-count formulas for partitions ofspace by hyperplanes, Mem. Amer. Math. Soc., 1 (1) no. 154 (1975).

[29] G. Ziegler, Lectures on polytopes, Springer-Verlag, 1995.

Documents

From polytopes to enumeration Ed Swartz - Cornell Universitypi.math.cornell.edu/~karola/Ed.Swartz.pdf · 2013. 1. 21. · 1. V-polytopes 11 2. Examples of Polytopes 13 3. Faces of