Interval Vector Polytopes

By Gabriel Dorfsman-Hopkins

SENIOR HONORS THESIS ROSA C. ORELLANA, ADVISOR DEPARTMENT OF MATHEMATICS DARTMOUTH COLLEGE JUNE, 2013 Contents

Abstract iv

1 Introduction 1

2 Preliminaries 4 2.1 Convex Polytopes ...... 4 2.2 Lattice Polytopes ...... 8 2.2.1 Ehrhart Theory ...... 10 2.3 Faces, Face Lattices and the Dual ...... 12 2.3.1 Faces of a Polytope ...... 12 2.3.2 The Face Lattice of a Polytope ...... 15 2.3.3 Duality ...... 18 2.4 Basic Graph Theory ...... 22

3IntervalVectorPolytopes 23 3.1 The Complete Interval Vector Polytope ...... 25 3.2 TheFixedIntervalVectorPolytope ...... 30 3.2.1 FlowDimensionGraphs ...... 31

4TheIntervalPyramid 38

ii 4.1 f-vector of the Interval Pyramid ...... 40 4.2 Volume of the Interval Pyramid ...... 45 4.3 Duality of the Interval Pyramid ...... 50

5 Open Questions 57 5.1 Generalized Interval Pyramid ...... 58

Bibliography 60

iii Abstract An interval vector is a 0, 1 -vector where all the ones appear consecutively. An { } interval vector polytope is the convex hull of a set of interval vectors in Rn.Westudy several classes of interval vector polytopes which exhibit interesting combinatorial- geometric properties. In particular, we study a class whose volumes are equal to the catalan numbers, another class whose legs always form a lattice basis for their ane space, and a third whose face numbers are given by the pascal 3-triangle.

iv Chapter 1

Introduction

An interval vector is a 0, 1 -vector in Rn such that the ones (if any) occur consecu- { } tively. These vectors can nicely and discretely model scheduling problems where they represent the length of an uninterrupted activity, and so understanding the combinatorics of these vectors is useful in optimizing scheduling problems of continuous activities of certain lengths. Much of this work is ﬁrst done in a paper on which I was a cowriter of at the MSRI-UP in 2012.[BDDPR]

In this thesis we consider interval vectors as geometric points in Rn and our goal is to understand their geometric relationship to eachother To do this we take sets of interval vectors as vertices of objects called polytopes, which are n-dimensional generalizations of 2d polygons or 3d polyhedra. We call a polytope constructed from interval vectors an interval vector polytope,introducedin[Da].Inessence,thisproject catalogues di↵erent interval vector polytopes and proves some elegant and interesting properties about them. In chapter 3, we ﬁrst consider the polytope formed by taking the convex hull of

n every interval vector in R (section 3.1), and notice that it has volume Cn the n-th

1 catalan number. The surprising fact that this polytope, named the complete interval vector polytope is a catalanotope is interesting on its own, but what is more fascinating is that many of its subpolytopes (generated by taking smaller sets of interval vectors in Rn)alsohaveveryinterestingstructuretothem. For example, in section 3.2 we consider the fixed interval vector polytope which is the polytope whose vertices are all the interval vectors in a dimension with a fixed interval length (i.e. the same number of ones). This polytope turns out to be a unimodular simplex (which is the generalization of a unit triangle). This is interesting because a unimodular simplex is the ’smallest’ polytope possible whose vertices are all integer valued, and it can tessellate to fill the entire plane hitting each lattice point. Essentially its legs form a basis for the lattice points of the plane. Finally, we consider the interval pyramid,whichisthepolytopewhosevertices are the standard unit vectors in Rn,plustheintervalvectorswithone0.Though this may seem like an arbitrary selection of interval vectors as vertices, we still see the interval pyramid exhibiting fascinating properties. It’s volume is very easily calculated to be 2(n 2) and its face numbers (the number of faces the polytope has in each dimension) reflect a combinatorial sequence called the pascal 3-triangle, which is like Pascal’s triangle summed with a shifted Pascal’s triangle, and thus is a sum of binomial coecients. 3 While each of these results are at least mildly interesting, together, they seem to say something quite deep. Essentially, the geometric locations of the interval vectors on the plane are such that, even seemingly arbitrary sets of interval vectors create beautiful combinatorial-geometric objects. Somehow a number of famous combinatorial sequences are contained within the geometric relationships between interval vectors, making them geometric objects of note. The second chapter of this thesis introduces the necessary terminology and basic theorems necessary to study convex polytopes and their combinatorial structure.

2 Much of my preliminary research in the field of convex polytopes and their combinatorics in general can be found in [Gr] and [Zi], and a for lattice polytopes and Ehrhart theory, in [Be]. Many definitions are pulled from these sources, which would be good to consult for a more complete introduction to the field. In this second chapter we define the convex hull, the ane subspace, and introduce lattice polytopes and the basics of Ehrhart theory. We also introduce posets and explain how they represent the face structure of a polytope, so that we can define duality of a polytope. Finally we introduce basic graph theory for use in certain proofs later in the paper. Those already familiar with this material may skip to chapter 3, though a brief skim of the first chapter may familiarize the reader with the notation we use. The third chapter introduces interval vector polytopes and proves basic results about the entire class of polytopes. We then present the results on the complete and fixed interval vector polytopes. The entire fourth chapter is dedicated to studying the face structure, volume and duality of the interval pyramid. The final chapter presents some open questions and suggests future work for further study.

3 Chapter 2

Preliminaries

2.1 Convex Polytopes

The definitions in this section are adapted from [Gr], [Zi], and [Be]. Intuitively, given asetofverticesinR2 we can ’connect the dots’ to form a polygon. This idea is precisified and generalized to Rn with the following definition.

n Deﬁnition 1. If A = v1,...,vk R , we deﬁne the convex hull of A to be the set { }⇢ of nonnegative linear combinations of elements of A whose coecients sum to 1.

k conv(A):= 1v1 + 2v2 + + kvk 1,2,...,k R 0 and i =1 . ··· | 2 ( i=1 ) X

Deﬁnition 2. A convex polytope Rn is the convex hull of ﬁnitely many points P✓ in Rn.

Example 1. Let V = (0, 0), (0, 1), (1, 0), (1, 1) R2. Then conv(V )=I I is the { }⇢ ⇥ unit square in R2.

We can envision a polytope in Rn might have less than n dimensions (for example,

4 (0, 1) (1, 1)

(0, 0) (1, 0)

Figure 2.1: conv(V ). asquareembeddedin3space).Suchanobjectstillspansasubspaceofitsentire ambient space which we call the polytope’s ane space.

n Deﬁnition 3. Let A = v1,v2 ...,vk R . The ane hull of A is deﬁned as the { }⇢ set of linear combinations of elements of A whose coecients sum to 1.

k a↵(A):= v + v + + v =1 . 1 1 2 2 ··· k k | i ( i=1 ) X This is similar to the deﬁnition of the convex hull, without the restriction that all the coecients are nonnegative. The ane hull of a set corresponds to the ane subspace of Rn spanned by the convex polytope generated by that set.

Example 2. Let U = (0, 0, 0), (0, 1, 0), (1, 0, 0), (1, 1, 0) R3. Then conv(U) R3 { }⇢ ⇢ is the unit square lying ﬂat on the xy plane and a↵(U) is the xy plane itself, x { 2 3 R x3 =0 . Notice a↵(U) is a 2 dimensional vector space. | }

Example 3. Let W = (0, 0, 1), (0, 1, 1), (1, 0, 1), (1, 1, 1) R3. Then conv(W ) is { }⇢ 3 the unit square hovering at the plane x3 =1 R . Then a↵(W ) is exactly the plane { }⇢ z =1 R3. We notice that this ane subspace is not a linear subspace, but can still { }⇢ be viewed as a 2 dimensional vector space with addition deﬁned as (a, b, 1)+(c, d, 1) =

5 z

conv(W )

conv(U) y

Figure 2.2: Non full dimensional polytopes.

(a + c, b + d, 1).

We would like a polytope to have a unique vertex set, but notice that various sets could have the same convex hull, so the vertices of a polytope =conv( v ,...,v ) P { 1 k} ⇢ n R is not well deﬁned as just v1,...,vk .Insteadwemustﬁndaminimalsetwhich { } generates . P

Deﬁnition 4. We call a set of points convexly independent if each point is not in the convex hull of the rest. That is, A is convexly independent if for all v A, 2 v/(conv(A v ). 2 \{ }

The following proposition allows us to uniquely deﬁne the vertex set of a polytope.

Proposition 1. [Gr] Let A, B Rn be two convexly independent sets. If conv(A)= ✓ conv(B), then A = B.

This justiﬁes us in deﬁning the vertex set as follows.

Deﬁnition 5. Let Rn be a convex polytope. The vertex set of , denoted vert( ) P✓ P P is deﬁned as the set of convexly independent points whose convex hull is . P

6 (0, 1) (1, 1)

1 1 ( 2 , 2 )

(0, 0) (1, 0)

Figure 2.3: Convexly dependent point in B.

Example 4. The set B = (0, 0), (0, 1), (1, 0), (1, 1), (1/2, 1/2) is not convexly inde- { } pendent because (1/2, 1/2) = 1 (0, 1)+ 1 (1, 0) so that (1/2, 1/2) conv(A (1/2, 1/2) ). 2 2 2 \{ } The set A = (0, 0), (0, 1), (1, 0), (1, 1) is convexly independant, and conv(A)= { } conv(B). The vertex set of the polytope vert(conv(B)) = A.

We can similarly deﬁne ane independence of a set.

Deﬁnition 6. A set of points is called anely independent if each point is not in the ane hull of the rest. That is, A is anely independent if for all v A, v/a↵(A). 2 2

Now that we have uniquely deﬁned vertices, we can begin to discuss the dimension of a convex polytope . P

Definition 7. We define the ane space of a convex polytope to be the ane hull of its vertices. We can always view the ane space of dimension d of a polytope as a vector space, and we define the dimension of Rn to be the dimension of its ane P⇢ space. We denote this dim( ), and we call a d polytope. If d = n, then is full P P P dimensional.

Notice that the polytopes described in examples 2 and 3 are 2 dimensional polytopes embedded in 3 dimensional space, so they are not full dimensional. On the other

7 Figure 2.4: 1, 2, and 3 dimensional simplices hand, the polytope described in example 1 is full dimensional. A very important class of d-dimensional polytopes are those with exactly d +1vertices.

Deﬁnition 8. A d dimensional polytope with exactly d +1 vertices is called a d- simplex.

Simplexes are very important, because they have a very uniform structure, and their volumes are very easy to compute (in fact, I will deﬁne volume using simplexes). Most importantly, any polytope can be triangulated,thatiscanbeexpressedas the union of simplexes. Since the volume of each simplex is easy to compute, a triangulation of a convex polytope makes the volume easy to compute.

Example 5. A line segment is a 1-simplex. A triangle is a 2-simplex. A tetrahedron is a 3-simplex.

2.2 Lattice Polytopes

Most of the polytopes we consider are those with integral vertices, called lattice polytopes. All examples considered so far are lattice polytopes.

Deﬁnition 9. A point p Zn is called a lattice point of Rn.Alattice polytope is a 2 polytope whose vertices are all lattice points.

8 n Deﬁnition 10. [Be] Let R be a d-dimensional ane subspace. A set v1,...,vd A✓ { } of d integer valued vectors in Rn is said to be a lattice basis for if, ﬁxing a lattice A point p , any other lattice point q can be expressed as an integral linear 2A 2A combination of the lattice basis, plus p. That is any lattice point q can be written 2A as

q = p + 1v1 + ...+ dvd with each i Z. The lattice basis of an ane subspace is essentially a basis for 2 A the lattice points of viewed as a vector space with origin p. A

Deﬁnition 11. We call a d-simplex with vertex set v ,...,v to be a uni- P { 1 d+1} modular d-simplex, if the legs v v ,v v ,...,v v form a lattice basis for { 2 1 3 1 d+1 1} a↵( ). P

Example 6. Let =conv( (0, 0), (0, 1), (1, 0) ). Then the legs of this polytope are P { } (1, 0) and (0, 1) which form a lattice basis for R2. So is a unimodular simplex. P

We can now assign a normalized volume of 1 to any unimodular simplex, so that for any arbitrary d-polytope ,wedefineitsnormalized volume with respect to P unimodular d-simplexes having volume 1. Thus, we can triangulate into unimodular P d-simplexes, and the number of simplexes in such a triangulation is the normalized volume of ,denotedvol( ). It is known that this definition of volume is well- P P defined.[Be]

Example 7. If =conv( (0, 0), (0, 1), (1, 0), (1, 1) ), then we can triangulate it via P { } =conv( (0, 0), (0, 1), (1, 0) ) and =conv( (0, 1), (1, 0), (1, 1) ). Both and P1 { } P2 { } P1 are unimodular, and = so that the vol( )=2. P2 P P1 [P2 P

9 (0, 1) (1, 1)

P2 P1 (0, 0) (1, 0)

Figure 2.5: Triangulation of the unit square.

Proposition 2. [Da] Let Rn be a full dimensional n-simplex with vertex set P2 v ,...,v . Then: { 0 n}

vol( )=det(v v ,v v ,...,v v ). P 1 0 2 0 n n

This determinant is called the Cayley-Menger determinant.

Notice that we would expect the volume of the 2-cube to be 1. It turns out that for a d-polytope the the normalized volume of is d!timesthevolumeof . P P P

2.2.1 Ehrhart Theory

How a polytope behaves when it is dilated or contracted can tell us important things about its combinatorial structure. For t Z 0,thet-dilation of a polytope is 2 P

t := tv v . P { | 2P}

Each lattice d-polytope has associated with it an Ehrhart polynomial,denoted P th L (t). When t Z 0, the polynomial yields the number of lattice points in the t P 2 dilate of the polytope. It is known that the constant term of any Ehrhart polynomial

10 L (t) 1 4 9 16 25 L (t) 1 3 6 10 15 P Q

Figure 2.6: The Ehrhart polynomials of the unit triangle and unit square produce the triangle numbers and square numbers respectively. is 1, and that the degree of this polynomial is the dimension d of .Importantly,the P leading coecient of the Ehrhart polynomial is its volume, i.e. 1 vol( ).[Be] d! P

Example 8. Let =conv( (0, 0), (0, 1), (1, 0), (1, 1) ). Then L (t)=(t+1)2, which P { } P are the square numbers as we would expect. Notice that the leading coecient is 1, therefore the volume of the square is 1, and the normalized volume is 2.

Example 9. Let =conv((0, 0), (0, 1), (1, 0) ). Then L (t)= (t)(t+1) which are Q { } Q 2 the triangular numbers as we would expect. We notice that the leading coecient is

1 1 2 , therefore the volume of the triangle is 2 and the normalized volume is 1 as desired (since the triangle is unimodular).

Call a transformation lattice preserving if it takes a lattice basis to a lattice basis. Linear, lattice preserving bijections between polytopes preserve combinatorial structure in the following way.

Proposition 3. [Be] A linear, lattice preserving bijection between polynomials preserves the Ehrhart polytnomial. Thus if and are polytopes, and T : P Q is a P Q ! linear lattice preserving bijection, then L (t)=L (t). P Q

11 2.3 Faces, Face Lattices and the Dual

Alotoftheinterestingcombinatorialdatainaconvexpolytopeliveswithinitsface structure. That is, the way faces of di↵erent dimensions intersect can often have very interesting combinatorial properties. A question we hope to answer in chapter 4, is whether the interval pyramid is self-dual. In this section we define the necessary terminology to begin to try and solve this problem. We first rigorously define the inequality description of a polytope and use that to define the faces of a polytope, stating basic but important theorems. We then briefly review posets and define the face lattice of a polytope as a poset. Next we define the dual form of a polytope and define what it would mean for a polytope to be self-dual.

2.3.1 Faces of a Polytope

The sequence of deﬁnitions and results in this section are adapted from [Zi]. Proofs of all these results can be found there.

n n Definition 12. Let R .Ifc R , we define the linear equality c x c0 to be P✓ 2 ·  valid for if it is satisfied for all points x P . P 2

Deﬁnition 13. A face F of a d-polytope Rn is any set of the form: P⇢

n F = x R : c x = c0 P\{ 2 · } where c x c is a valid inequality. A face of a polytope is a polytope, and thus ·  0 has dimension. We call a face F a k face of if it has dimension k. A 0-face is P a vertex, a 1-face is called an edge, and a (d 1)-face is called a facet.If has f P k

12 k faces, we deﬁne the f-vector of to be P

f( ):=(f0,f1,...,fd 1) P

It turns out than any polytope deﬁned by vertices can be equivalently deﬁned by it’s n 1 dimensional faces, in what is called a facet description or inequality description of the polytope. This is summarized in the following theorem.

Theorem 1. (See [Zi] theorem 1.1). P Rn is a convex polytope if and only if it ⇢ can be described as

n n P = x R : ci x di for some c1,...,cm R and d1,...,dm R { 2 ·  2 2 }

This is the unique facet description exactly when each c x d is a valid inequality i ·  i describing an n 1 dimensional face.

2 Example 10. The triangle =conv( (0, 0), (0, 1), (1, 0) ) R has 3 vertices v1 = P { } ⇢ (0, 0),v =(0, 1), and v =(1, 0). It also has three faces f = x =0 ,f = y = 2 3 1 { }\P 2 { 0 , and f = x + y =1 . Thus is a 2-dimensional polytope with 3 vertices }\P 3 { }\P P and 3 edges has f-vector f( )=(3, 3), and its facet description is P

= c R2 : c ( 1, 0) 0,c (0, 1) 0,c (1, 1) 1 . P { 2 ·  ·  ·  }

Example 11. Consider the unit 2-cube =conv( (0, 0), (0, 1), (1, 0), (1, 1) ). Name P { } the vertices u ,u ,u ,u respectively. We see that the linear inequality x 1 is valid 1 2 3 4 1  2 for . So g1 = x R x1 =1 is a 1 dimensional face of the 2-cube. Similarly, P P\{ 2 | } 2 (x1 +x2) 0 is a valid inequality, and g2 = P x R x1 +x2 =0 = (0, 0) =  \{ 2 | } { } u1 is a zero dimensional face (or vertex) of the 2-cube. The 2-cubes facet description

13 v3

f3 f2

v1 v2 f1

Figure 2.7: The faces of the unit triangle.

u4 u3

g2 = u1 u2

Figure 2.8: Some faces of the unit square. is

= c R2 : c ( 1, 0) 0,c (0, 1) 0,c (1, 0) 1,c (0, 1) 1 . P { 2 ·  ·  ·  ·  }

Notice that for any d polytope ,theinequality0 x 0isvalid,and0 x =0 P ·  · for all x ,sothat is itself a d face of .Similarly,theinequality0 x 1 2P P P ·  is valid, but 0 x =1foranyx .Thus is also always a face of (deﬁned to · 6 2P ; P have dimension 1). What follows is an important fact about the face structure of a

14 polytope.

Proposition 4. (See [Zi] proposition 2.3). (See for example: Ziegler Lectures on

Polytopes Proposition 2.3). Let Rn be a polytope, and V its vertex set. Let F be P✓ a face of . P 1. F is a polytope, with vertex set F V . \ 2. Every intersection of faces of is a face of . P P 3. The faces of F are exactly the faces of that are contained in F . P 4. F = P a↵(F ). \ Example 12. Consider the 3-cube C =conv(V ) where V = (0, 0, 0), (0, 0, 1), (0, 1, 0), 3 { (1, 0, 0), (0, 1, 1), (1, 1, 0), (1, 0, 1), (1, 1, 1) . We can consider the inequality x 1 } 1  which is valid for C and defines the face F = C x C : x =1.We 3 3 \{ 2 3 1 } know that this must be a polytope, and we know its vertex set is V 0 = F V = \ (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1) . Thus F is the unit square lying flat on the plane { } x =1 . { 1 } Next consider the inequality x 1 which is valid for C and the face F˜ = C x 2  3 3\{ 2 C : x =1 , with vertex set V˜ = F˜ V = (0, 1, 0), (0, 1, 1), (1, 1, 0), (1, 1, 1) so that 3 2 } \ { } F˜ is the unit square lying flat on the plane x =1 . { 2 } We know that F F˜ must be a face of C as well, and its vertex set is (1, 1, 0), (1, 1, 1) , \ 3 { } so that F F˜ is the line connecting those two points. It is also a face of C defined \ 3 by the valid inequality x + x 2. 1 2 

2.3.2 The Face Lattice of a Polytope

We earlier introduced the concept of an f-vector of a polytope, which tells us the number of faces of each dimension that a convex polytope may have. In fact, we can

15 x3

F F˜ x2 F F˜ \

Figure 2.9: Some faces of the unit 3-cube. consider a more structured way to consider the faces of a polytope, using a poset called a face lattice. Some terminology must ﬁrst be introduced.

Definition 14. [Zi] A poset (S, ) is a finite set S equipped with a relation ‘ ’ which   satisfies the following properties.

1. Reﬂexivity: for all x S, x x. 2 

2. Transitivity: if x y and y z then x z.   

3. Antisymmetry: if x y and y x then x = y.   For two elements x, y S,ifx y or y x then we say x and y are comparable. 2   A maximal element x S is an element such that there is no y S such that x y. 2 2  A minimal element y S is an element such that there is not x S such that x y. 2 2  Aposetisbounded if there is a unique maximal element and unique minimal element. Achain(oflengthn)isalistofelementsx ,...,x S such that x ... x . 1 n 2 1   n Aboundedposetisgraded if every maximal chain has the same length. For a poset S,wedeﬁnetheopposite poset Sop to have the same underlying set as S,withthe relation x y being in Sop if and only if y x holds in S.Wedeﬁnetwoposets op  16 Figure 2.10: The poset (P ( x, y, z ), ).[Ks] { } ✓

(S, ), (T, ) to be isomorphic, denoted (S, ) = (T, )ifthereisabijection S T S ⇠ T f : S T such that for all x, y S,itholdsthatx y if and only if f(x) f(y). ! 2 S T Example 13. The power set of x, y, z ordered by inclusion is the graded poset { } (P ( x, y, z , )), of length 4. An example of a maximal chain is z y, z { } ✓ ;⇢{ }⇢{ }⇢ x, y, z . Notice also that the opposite poset P x, y, z op is isomorphic to P x, y, z . { } { } { }

The face structure of a polytope can be nicely coded into posets in the following way.

Deﬁnition 15. Let be a d polytope whose faces are the set . Then we deﬁne P F the face lattice of a convex polytope to be ( , ), all the faces of ordered by P F ✓ P inclusion, and call this poset L( ). The length of L( ) is d +2. P P

Deﬁnition 16. Two polytopes , Rn are deﬁned to be combinatorially isomor- P Q⇢ phic if their face lattices are isomorphic as posets, i.e. L( ) = L( ). P ⇠ Q Example 14. Two n simplices are always combinatorially isomorphic, with face lattices equivalent to the poset of the power set n +1elements, ordered by inclusion.

17 Figure 2.11: The face lattice of a square pyramid.[Ep]

2.3.3 Duality

A natural construction to take for a polytope is to consider its dual. Intuitively, this is turning a polytope inside-out, replacing each facet with a vertex and reconstructing the polytope from there. An interesting class of polytopes are those which are self- dual, meaning that the dual has the same combinatorial structure as the original polytope. We make these notions more precise here.

Definition 17. For a d polytope , we define an interior point y to be a point P 2P which is not contained in a face of of dimension smaller than d. The collection of P interior points of is called int . P P We hope to define the dual of a polytope ,butfortheconstructionweuse,we P must have that 0 int .Thiscangenerallybeachievedforanypolytope(aslong 2 P as the polytope has nonempty interior) by a simple ane translation of the polytope

18 Figure 2.12: Taking the dual.[Zi] so that 0 is in the interior.[Zi]

Lemma 1. ([Zi] Lemma 2.8). Let be a polytope in Rn.Ifp can be represented P 2P as p = 1 n x for n +1 anely independant points x ,...,x , then p is an n+1 i=0 i 0 n 2P interior pointP of . P The following construction is from [Zi].

Deﬁnition 18. Let Rn be a polytope with 0 in the interior. The dual of is P✓ P deﬁned by

:= y Rn : y x 1 for all x , P { 2 ·  2P} where y x = y x + ...+ y x is the dot product. · 1 1 n n Figure 2.3.3 shows a convex pentagon on a plane given by its ﬁve vertices, and it’s dual given by ﬁve inequalities.

Deﬁnition 19. A polytope Rn with 0 in the interior is said to be self-dual if P⇢ L( ) = L( ). P ⇠ P Proposition 5. ([Zi] Corollary 2.14)Let Rn be a polytope with 0 in the interior. P⇢ The face lattice of is the the opposite poset of . That is L( )=(L( ))op. P P P P 19 Corollary 1. A polytope Rn with 0 in the interior is self-dual if and only if P⇢ L( ) = L(( ))op. P ⇠ P

The following proposition conﬁrms the idea that our deﬁned notion of taking the dual of a polytope in fact it replaces facets with vertices and so on.

Proposition 6. Let Rn be a polytope with 0 in the interior, and vertex set P✓ n v1,...,vm .Ifc R , then c if and only if c vi 1 for i =1,...,m. { } 2 2P · 

Proof. First if c ,thenclearlyc v 1foralli =1,...,m since c x 1must 2P · i  ·  be a valid inequality for all x (by how we deﬁned the dual construction). 2P Conversely, assume c v 1holdsforeachi.Toshowc we must show that · i  2P c x 1forallx . Fix some x .Weknowthatx = v + ...+ v with ·  2P 2P 1 1 m m =1andeach 0. But then since c v 1foreachi,then (c v ) for i i i · i  i · i  i eachP i.Letussum (c v ) for each i so that i · i  i

(c v )+...+ (c v ) + ...+ =1 1 · 1 m · m  1 m

Clearly (c v )=c ( v )sothat i · i · i i

c ( v )+...+ c ( v ) 1 · 1 1 · m m  c ( v + ...+ v ) 1 · 1 1 m m  c x 1 · 

Since x was arbitrary, c x 1forallx ,sothatc . ·  2P 2P

Deﬁnition 20. We deﬁne an ane translation of a polytope Rn to be just a P⇢ translation of all the points in the polytope by a single vector v Rn. We denote the 2

20 Pv

Figure 2.13: Ane translation of the unit triangle. translated polytope = + v Pv P

Example 15. Let =conv((0, 0), (1, 0), (0, 1) ) and v =(1, 1). Then = P { } Pv conv( (1, 1), (2, 1), (1, 2) ). { }

Notice that for any polytope Rn with nonempty interior, if v int( ), then P⇢ 2 P

0 int( v). Thus it is not dicult to have an ane translation of with 0 on the 2 P P interior. It is important to notice that ane translation of a polytope does not a↵ect the polytopes face lattice poset[Zi]. In fact, it does not a↵ect any of the combinatorial

data of the polytope, and so we are justiﬁed in calling v the dual of .Therefore, P P we can restate Proposition 19 for the more general case of polytopes with nonempty interiors.

Corollary 2. [Zi] Let Rn be a polytope with nonempty interior, and let v P⇢ 2 int( ). Then, P L(P )=(L( ))op v P

Thus any is self dual if any only if L( )=(L( ))op. P P P

21 Figure 2.14: An example of a graph.

2.4 Basic Graph Theory

Relating to any interval vector polytope, there is associated a graph called the flow dimension graph which tells us the ane-dimension of the polytope.[Da] I will introduce the necessary basics of graph theory and wait until later to define the flow dimension graph.

Deﬁnition 21. A graph is an ordered pair G =(V,E) where V is a set of vertices, and E V V is a set of edges between pairs of vertices. ✓ ⇥

Example 16. Consider the graph G =(V,E) with six nodes, where V = 1, 2, 3, 4, 5, 6 , { } and the edge set E = (1, 2), (1, 5), (2, 3), (3, 4), (4, 5), (4, 6) . Each element of E can { } be visualized as an edge connecting the two vertices.

Deﬁnition 22. Two nodes a, b in a graph G =(V,E) are said to be connected if there exists a path from a to b, that is there exist q ,...,q V such that (a, q ), (q ,q ),...,(q ,b) 0 s 2 0 0 1 s 2 E.Aconnected component of a graph is a maximal set of vertices which are all connected.

Notice that the poset in example 16 has one connected component.

22 Chapter 3

Interval Vector Polytopes

We begin by deﬁning interval vectors and interval vector polytopes in general. We then consider several classes of interval vector polytopes and the interesting combinatorial properties that arise within them.

Deﬁnition 23. [Da] A 0, 1 vector is a vector whose entries are all in the set { } 0, 1 .Aninterval vector is a 0, 1 vector in Rn such that the ones (if any) occur { } { } n consecutively. More precicely, a vector x =(x1,...,xn) R for i

Example 17. Each standard unit vector is an interval vector with interval length 1.

6 (0, 1, 1, 1, 0, 0) = ↵2,4 R is an interval vector with interval length 3. (1, 1, 1, 1) = 2 4 ↵1,4 R is an interval vector with interval length 4. The 0 vector is always trivially 2 an interval vector with interval length 0. (1, 0, 0, 1) is not an interval vector, since there are 0s between the the ﬁrst and fourth coordinates (both 1s). (0, 2, 2, 0) is not an interval vector because it is not a 0,1 -vector. { }

23 Deﬁnition 24. A lattice polytope whose vertices consist entirely of interval vectors is called an interval vector polytope.IfI Rn is a set of interval vectors, then ✓ (I):=conv(I) is the corresponding interval vector polytope. Pn

Example 18. ( (0, 0), (0, 1), (1, 0) ), which is the triangle we have been considering P2 { } in many examples, is an interval vector polytope. Similarly, ( (0, 0), (0, 1), (1, 0), (1, 1) ) P2 { } which is the unit 2-cube is also an interval vector polytope.

Let [n]:= 0, 1,...,n .ThenforanysetS [n], we define I = interval vectors { } ✓ S { with interval length t : t S .Define (I ), as the interval vector with vertices of 2 } Pn S interval length in S.Thisisausefulnotationaldevicetoallowustoeasilydefine interval vector polytopes coming from interval vectors of specific lengths.

Example 19. Let S = 2 [4]. Then (I )=conv( (1, 1, 0, 0), (0, 1, 1, 0), (0, 0, 1, 1) ). { }✓ P4 S { } If T = 1, 2 [3] then P (I )=conv( (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (0, 1, 1) ). { }✓ 3 T { } We will look further at polytopes which are the convex hull of all interval vectors in

n R : n(I[n])=conv 0,e1,e2,...,↵1,n . P { }

Notice that the vertex set of an interval vector polytope is not dicult to describe.

Lemma 2. Let I Rn be a set of interval vectors. Then I is convexly independent. ⇢

Proof. Since a subset of a convexly independent set is convexly independent, it suf-

n ﬁces to show that the set of all interval vectors in R , I[n],isconvexlyindependent.

Consider ↵i,j where i can but does not need to equal j.Eachotherintervalvector either has a 0 in the entry where ↵i,j has a 1, or a 1 where ↵i,j has a 0.

If a vector of the ﬁrst type (with a 0 in the kth entry that ↵i,j has a 1) were to have nonzero coecient, then the sum of the rest of the coecients would be less than 1. But then, since no interval vector has a value greater than 1 in the kth entry,

24 the kth entry of any convex combination would be less than 1, and thus certainly the sum could not be ↵i,j.

If a vector of the second type (with a 1 in the kth entry where ↵i,j has a 0) were to have nonzero coeent, then the kth entry of any convex combination would be greater than 0, and thus the sum could certainly not be ↵i,j.Thus↵i,j / conv(I [n] ↵i,j ), 2 { } \{ } and I [n] is convexly independent, along with all of its subsets. { }

n Corollary 3. Let I R be a set of interval vectors. Then the vert( n(I)) = I. ⇢ P

3.1 The Complete Interval Vector Polytope

We call Pn(I[n])theComplete Interval Vector Polytope. We compute the normalized volumes of the ﬁrst few of these polytopes using polymake [Ga] and noticed a pattern. Let us look at a few examples.

Example 20. 1(I[1])=conv( 0, 1 )=[0, 1] R which is the unit interval. P { } ⇢ vol( (I )) = 1. P1 [1]

2 Example 21. 2(I[2])=conv((0, 0), (0, 1), (1, 0), (1, 1) ) R which is the unit P { } ⇢ 2-cube. vol( (I )) = 2. P2 [2]

3 Example 22. 3(I[3])=conv(e1,e2,e3,↵1,2,↵2,3,↵1,3 ) R . This looks like the P { } ⇢ unit 3-cube with a corner cut out and vol( (I )) = 5. P3 [3]

4 Example 23. 4(I[4])=conv( e1,e2,e3,e4,↵1,2,↵2,3,↵3,4,↵1,3,↵2,4,↵1,4 ) R . P { } ⇢ vol( (I )) = 14. P4 [4] An important sequence of numbers that arises often in combinatorics is called the

Catalan numbers, C 1 ,deﬁnedby { n}n=1 25 x3

Figure 3.1: The third complete interval vector polytope, P3(I[3]).

1 2n Cn = 0 1 (3.1) n +1 n B C @ A These numbers are often the solutions to counting problems and represent, among other things, the number of monotonic paths between opposite corners in an n n ⇥ grid not crossing above the line y = x.Wefindthisexamplein[St],alongwithover 200 occurrences of the catalan numbers in an appendix. The catalan numbers are fascinating because of their frequent appearances in di↵erent combinatorial problems. We notice that the first four numbers in the sequence are 1, 2, 5, 14, and some computation shows the first 10 interval vector polytopes have Catalan volume.

n In his paper [Po], Postnikov deﬁnes the complete root polytope Qn R as the ✓ convex hull of 0 and ei ej for all i

As it turns out, we can provide a lattice preserving linear bijection between Qn and

n 1(I[n 1]). By proposition 3, this implies the two polytopes have the same Erhart P polynomial, and since the ﬁrst term of the Erhart polynomial is the volume of the polytope, the two must share volumes.

26 Figure 3.2: The fourth catalan number represented as monotonic paths in a 4 4 ⇥ grid.[Dm]

Theorem 2. LQn (t)=L n 1(I[n 1]). P

Iwillﬁrstpresenttheproofbyprovidingalatticepreservinglinearbijection between the two polytopes. To illustrate how it works, I then compute the transformation on a series of examples.

Proof. Each of the vertices of Qn are vectors with entries that sum to zero, so any linear combination (and speciﬁcally any convex combination) of these vertices also has entries who sum to zero:

x = y =0= ax + by = a x + b y =0. i j ) i j i j i j i j i j X X X X X X

n n Deﬁne B := x R xi =0 ;thenQn B. B is an (n 1)-dimensional ane { 2 | i=1 } ⇢ subspace of Rn. P Consider the linear transformation T given by the n n lower triangular (0, 1)- ⇥ matrix where t =1ifi j and t =0otherwise.Thentheimage ij ij

n T (B) x R xn =0 =: A. ✓{ 2 | } 27 A is also an n 1dimensionalsubspaceofRn. Since T has determinant 1, it is injective when restricting the domain to B.Forthesamereason,weknowthatfor

n n any y A,thereexistsx R such that y = T (x). But since yn = xi =0,then 2 2 i=1 x B,sothatT : B A is surjective, and therefore a linear bijection.P 2 |B ! n 1 Now consider the projection ⇡ : A R given by !

⇡ ((x1,...,xn 1, 0)) = (x1,...,xn 1).

The transformation is clearly linear, and has the inverse

1 ⇡ ((x1,...,xn 1)) = (x1,...,xn 1, 0), so that ⇡ is a bijection.

n 1 Now we show that the linear bijection ⇡ T B : B R is a lattice-preserving | ! map. First we ﬁnd a lattice basis for B.Considertheset

S := e = e e i

We notice that any integer point of B can be represented as

n 1 n 1 a1,...,an 1, ai = aiei,n. i=1 ! i=1 X X Any integer point of B is an integer combination elements of S,soS is a lattice basis of B.

Note that ⇡ T (ei,n)=ei + + en 1 =: ui.Therefore ···

⇡ T (S)= u i n 1 =: U. { i|  }

28 We notice that en 1 = un 1 and ei = ui ui+1,sothateachofthestandardunit n 1 vectors ei of R is an integral combination of the vectors in U.Sincethestandard

n 1 basis is a lattice basis of R ,soisU,thus⇡ T B is a lattice-preserving map. | Since our bijection is linear and lattice-preserving, all we have left to show is that the vertices of Qn map to those of n 1(I[n 1]). By linearity, ⇡ T (0) = 0, and given P any vertex ↵i,j for n 1(I[n 1]), we know that ⇡ T (ei,j+1)=↵i,j where i

I’ll illustrate how this works in the following examples.

Example 24. Consider the third complete root polytope,

Q =conv( 0,e e ,e e ,e e )=conv( (0, 0, 0), (1, 1, 0), (1, 0, 1), (0, 1, 1) ). 3 { 1 2 2 3 1 3} { }

Since the transformation is linear, it preserves convex combinations (since they are just linear combinations), so we need only notice that it takes vertices to vertices.

⇡(T (0, 0, 0)) = ⇡(0, 0, 0) = (0, 0) = 0,

⇡(T (1, 1, 0)) = ⇡(1, 0, 0) = (1, 0) = e , 1

⇡(T (1, 0, 1)) = ⇡(1, 1, 0) = (1, 1) = ↵ , 1,2

⇡(T (0, 1, 1)) = ⇡(0, 1, 0) = (0, 1) = e . 2

These are exactly the vertices P2(I[2]).

Example 25. Consider the third complete interval vector polytope

P (I )=conv( (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0), (0, 1, 1), (1, 1, 0), (1, 1, 1) ). 3 [3] { }

29 Transforming the vertices we see

1 1 1 T (⇡ (0, 0, 0) = T (0, 0, 0, 0) = (0, 0, 0, 0) = 0,

1 1 1 T (⇡ (0, 0, 1) = T (0, 0, 1, 0) = (0, 0, 1, 1) = e e , 3 4

1 1 1 T (⇡ (0, 1, 0) = T (0, 1, 0, 0) = (0, 1, 1, 0) = e e , 2 3

1 1 1 T (⇡ (1, 0, 0) = T (1, 0, 0, 0) = (1, 1, 0, 0) = e e , 1 2

1 1 1 T (⇡ (0, 1, 1) = T (0, 1, 1, 0) = (0, 1, 0, 1) = e e , 2 4

1 1 1 T (⇡ (1, 1, 0) = T (1, 1, 0, 0) = (1, 0, 1, 0) = e e , 1 3

1 1 1 T (⇡ (1, 1, 1) = T (1, 1, 1, 0) = (1, 0, 0, 1) = e e . 1 4

These are exactly the vertices of Q4.

Acorollarytothistheoremdescribesthenormalizedvolumeofthecomplete interval vector polytope.

2n 1 Corollary 4. vol(Pn(I[n])) = Cn = n+1 0 1 . n B C @ A 3.2 The Fixed Interval Vector Polytope

Given an interval length i and and a dimension n, we deﬁne the ﬁxed interval vector

n n polytope n(I i ) R as the convex hull of all interval vectors in R with interval P { } ⇢ length i.

Example 26. The ﬁxed interval vector polytope with n =5,i=3is

5(I 3 )=conv( (1, 1, 1, 0, 0), (0, 1, 1, 1, 0), (0, 0, 1, 1, 1) )=conv( ↵1,3,↵2,4,↵3,5 ). P { } { } { }

30 x3

Figure 3.3: 3(I 1 ). P { }

Example 27. The ﬁxed interval vector polytope with n =3,i=1is

3(I 1 )=conv( (1, 0, 0), (0, 1, 0), (0, 0, 1) )=conv( e1,e2,e3 ). P { } { } { }

This is a unit triangle in the ane subspace x + y + z =1 of R3. { }

The last example is not an exception. In fact, the main theorem of this section shows us that that each ﬁxed interval vector polytope is a unimodular simplex in its ane subspace. The proof requires some graph theory.

3.2.1 Flow Dimension Graphs

This entire construction is due to [Da]. Related to any interval vector polytope, there is associated a graph can be shown to tell us the ane-dimension of the polytope. Here I define the flow dimension graph, which will allows us to prove the main theorem of this section. For more information on flow dimension graphs and their construction and uses see [Da]. Denote e := e e for i

31 Figure 3.4: The flow dimension graph G5(I 3 ). { } containing all such ei,j,eachunitvectorei,andthezerovector.LetT be the lower triangular (0, 1)-matrix, as in the proof of Theorem 2. We notice that T (ei)=↵i,n and T (ei,j)=↵i,j 1.Sotheimageofanelementaryvectorisanintervalvector.Since T is invertible, for any set of interval vectors I,thereisauniqueset of elementary E 1 vectors such that T ( )=I,namelyT (I)= . E E n Thus for any interval vector polytope n(I) R ,wecanconstructthecorre- P ⇢ 1 sponding flow-dimension graph G (I)=(V,E)asfollows.Let = T ( ). We let n E I the vertex set V =[n], specify a subset V = j V e ,anddefinetheedgeset 1 { 2 | j 2E} E = (i, j) e . Also we let k denote the number of connected components { | i,j 2E} 0 C of the graph G (ignoring direction) so that V = . C\ 1 ;

Example 28. Recall that the ﬁxed interval-vector polytope with n =5, i =3is

5(I 3 )=conv (1, 1, 1, 0, 0), (0, 1, 1, 1, 0), (0, 0, 1, 1, 1) P { } { }

The corresponding ﬂow dimension graph of G5(I 3 )=(V,E) has vertex set V = { } 1, 2, 3, 4, 5 . The corresponding elementary vectors are = e ,e ,e . Thus the { } E { 1,4 2,5 3} edge set E = (1, 4), (2, 5) and the speciﬁed subset V = 3 V . The constant k { } 1 { }⇢ 0 representing the number of connected components of the graph not intersecting V1 is k0 =2.

32 Theorem 3. [Da] If 0 a↵(I), then the dimension of (I) is n k . Else, if 2 Pn 0 0 / a↵(I) then the dimension of (I) is n k 1. 2 Pn 0

For n(I i ), we have I = ↵j,j+i 1 j n i +1 which translates to the P { } { |  } elementary vector set = ek,k+i k n i en i+1 .Wecandeﬁnethe E { |  }[{ } corresponding ﬂow-dimension graph Gn(I i )=(V,E)whereV = 1,...,n and { } { } E = (k, k + i) k [n i] corresponding to each e . Also V := n i +1 { | 2 } i,j 2E 1 { } corresponding to en i+1 . 2E

Lemma 3. Let a, b be nodes in the ﬂow-dimension graph Gn(I i )=(V,E). Then a { } and b are connected i↵ a b mod i. ⌘ Proof. Assume without loss of generality a b.Supposea and b are connected by  the path q ,...,q V .ThereforebydeﬁnitionofE,wehave 0 s 2

q0 = a + i

q1 = q0 + i = a +2i . .

qs = qs 1 + i = a +(s +1)i

b = qs + i = a +(s +2)i

Thus a b mod i by deﬁnition. ⌘

Now suppose that a b mod i where a b,thenthereexistsm N such that ⌘  2

b = a + mi

= a +(m 1)i + i.

33 Since b and a +(m 1)i di↵er by i, then by deﬁnition of E,thereisanedgebetween these nodes. Call this edge (q ,b) E.Similarly,wehave t 2

a +(m 1)i = a +(m 2)i + i (qt,qt 1) E ) 2

a +(m 2)i = a +(m 3)i + i (qt 1,qt 2) E ) 2

. .

a +2i =(a + i)+i (q ,q ) E ) 1 0 2

a + i = a + i (q ,a) E. ) 0 2

Hence q ,q ,...,q V ,deﬁneapathfroma to b,soa and b are connected. 0 1 t 2

Proposition 7. n(I i ) is an (n i)-dimensional simplex. P { }

Proof. By Lemma 3 we know there are i connected components in the ﬂow-dimension graph Gn(I i )andsinceV1 has only one element, k0 = i 1. Thus by Theorem 3 { } the dimension of n(I i )isn i.Since n(I i )hasn i +1 vertices,itisan P { } P { } (n i)-dimensional simplex.

Theorem 4. n(I i ) is an (n i)-dimensional unimodular simplex. P { }

Proof. Consider the ane space where the sum over every ith coordinate is 1,

n A = x R xj =1, k [i] . ( 2 8 2 ) j k mod i ⌘X 34 Since the vertices of n(I i ) have interval length i,theyareinA.Thus n(I i ) A. P { } P { } ⇢

We want to show that the w1,w2,...,wn i of n(I i )formalatticebasisforA P { } where

w1 = ↵1,i ↵n i+1,n

w2 = ↵2,i+1 ↵n i+1,n . . .

wn i = ↵n i,n 1 ↵n i+1,n We will do this by showing that any integer point p A can be expressed as a integral 2 linear combination of the proposed lattice basis, that is, there exist integer coecients

Y1,...,Yn i so that Y1w1 + ...+ Yn iwn i + ↵n i+1,n = p. We ﬁrst notice that p can be expressed as

p1,p2,...,pn i, ( pj)+1, ( pj)+1,..., ( pj)+1 0 1 j n i j n i j n i j t Xi+1 mod i j t Xi+2 mod i j nX=mod i B ⌘ ⌘ ⌘ C @ A by solving for the last term in each of the equations deﬁning A.Let

p1 t =1 8 Yt = > pt pt 1 1  <> pt Yt i i > :> Then each Yt is an integer. We claim that

Y1w1 + + Yn iwn i + ↵n i+1,n = p. ···

Clearly the ﬁrst coordinate is p1 since w1 is the only vector with an element in the ﬁrst coordinate. Next consider the tth coordinate of this linear combination for 1

35 by summing the coecients of all the vectors who have a 1 in the tth position:

Yt + Yt 1 + Yt 2 + + Y1 = pt pt 1 + pt 1 pt 2 + + p2 p1 + p1 = pt ··· ···

We next consider the tth coordinate of the combination for i

Yt + Yt 1 + + Yt i+1 =(pt Yt 1 Yt i+1)+Yt 1 + + Yt i+1 = pt ··· ··· ···

Finally, we consider the tth coordinate of the combination for n i

(Y1 + Y2 + + Yt i)+1. ···

Applying the two relations we have deﬁned between coordinates, and calling t the h i least residue of t mod i,wesee:

(Y1 + Y2 + + Yt i)+1 = (Y1 + Y2 + + Yt 2i + pt i)+1 ··· ···

= (Y1 + Y2 + + Yt 3i + pt 2i + pt i)+1 ···

= Y1 + Y2 + + Y t + pj +1 0 ··· h i 1 i

Thus p = Y1w1 + Y2w2 + + Yn iwn i + ↵n i+1,n and so w1,...,wn i form a lattice ··· basis of A.So n(I i )andisaunimodularsimplex. P { }

36 Corollary 5. vol n(I i )=1. P { }

To help with understanding this proof, I will walk through an example.

Example 29. Consider again P5(I 3 ) as in example 28. We saw that the number of { } connected components not intersecting V1 is k0 =2. We also know that 0 / a↵(I 3 ) 2 { } so that by theorem 3, dim( 5(I 3 )) = 5 2 1=2. So 5(I 3 ) is a 2 dimensional P { } P { } simplex. Since each vertex is, we also know that 5(I 3 ) is a subset of the ane P { } space

5 A = x R x1 + x4 =1,x2 + x5 =1,x3 =1 . 2 ⇢ But since this is 2 dimensional, we know that A is in fact the ane hull of 5(I 3 ). P { } So to show that this polytope is a unimodular simplex, we must notice that its legs w ,w = (1, 1, 1, 0, 0) (0, 0, 1, 1, 1), (0, 1, 1, 1, 0) (0, 0, 1, 1, 1) form a lattice basis { 1 2} { } for A. Take any lattice point p A, it can be written as (x ,x , 1, 1 x , 1 x ). If we 2 1 2 1 2 let Y = x and Y = x x , then 1 1 2 2 1

Y w + Y w + ↵ = x (1, 1, 0, 1, 1) + (x x )(0, 1, 0, 0, 1) + (0, 0, 1, 1, 1) 1 1 2 2 3,5 1 2 1 =(x ,x + x x , 1, x 1, x x + x +1) 1 1 2 1 1 1 2 1 =(x ,x , 1, 1 x , 1 x )=p. 1 2 1 2

Since, ↵3,5 is a lattice point of A, p is an arbitrary lattice point of A and each Yi is integral, then w1,w2 is a lattice basis of A and P5(I 3 ) is unimodular. { } { }

37 Chapter 4

The Interval Pyramid

n Given a dimension n,wedeﬁnetheinterval pyramid n(I 1,n 1 ) R to be the P { } ✓ convex hull of all the standard unit vectors of Rn along with the two interval vectors with interval length n 1: ↵1,n 1 and ↵2,n

Example 30. For n =3and n =4and in general,

3(I 1,2 )=conv( (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (0, 1, 1) ), P { } { }

4(I 1,3 )=conv( (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (1, 1, 1, 0), (0, 1, 1, 1) ), P { } { }

n(I 1,n 1 )=conv( e1,e2,...,en,↵1,n 1,↵2,n) . P { } { }

We can use the ﬂow dimension graph to show that dim(Pn(I 1,n 1 )) = n,sothat { } its ane space is all of Rn.

Proposition 8. The dimension of n(I 1,n 1 ) is n. P { }

38 Figure 4.1: The ﬂow dimension graph of the interval pyramid: G(I 1,n 1 )) { }

Proof. For n 3, the vertices of n(I 1,n 1 )formtheset P { }

e1 =(1, 0,...,0, 0) 8 9 > e2 =(0, 1,...,0, 0)> > > > . > > . > > > = > > . I > > <> en =(0, 0,...,0, 1)=>

>↵1,n 1 =(1, 1,...,1, 0)> > > > > > ↵ =(0, 1,...,1, 1)> > 2,n > > > > > :> ;> We convert the interval vectors to the corresponding elementary vector set

= e1,2,e2,3,...,en 1,n,e1,n,e2,en . E { }

From this we construct the ﬂow-dimension graph G( n(I 1,n 1 )) = (V,E)asseenin P { } Figure 4.1, where V =[n]and

E = (k, k +1)k [n 1] (1,n) { | 2 }[{ } corresponding to each e in . The subset of vertices V = 2,n (circled in Figure i,j E 1 { }

39 4.1) corresponds to each e in .Sincetheunderlyinggraphisconnected,weknow i E

k0 =#connected components C in G( n(I 1,n 1 )) such that C V1 = =0. { P { } \ ;}

Next we notice that

1 1 1 1 e1 + e2 + + en 1 ↵1,n 1 = 0 n 2 n 2 ··· n 2 n 2 where the sum of the coecients is

n 1 1 n 2 = =1 n 2 n 1 n 2

So 0 a↵( )andbyTheorem3,dim( n(I 1,n 1 )) = n k0 = n. 2 I P { }

4.1 f-vector of the Interval Pyramid

Recall that the f-vector of a polytope tells us the number of faces a polytope has of each dimension. We will see that the f-vector of Pn(I 1,n 1 )withn 3, is precisely { } the nth row of the Pascal 3-triangle without 1’s. The Pascal 3-triangle [Az] is an analogue of Pascal’s Triangle, where the third row, instead of being 1 2 1, is replaced

40 Interval-Vector Polytopes Jessica De Silva, Gabriel Dorfsman-Hopkins, Joseph Pruitt Advisors: Prof. Matthias Beck and Amanda Ruiz 2012 Mathematical Sciences Research Institute - Undergraduate Program

Background Abstract: An interval vector is a (0, 1)-vector where all the ones appear consecutively. Polytopes whose vertices 4. Generalized Interval Pyramid are among these vectors have some astonishing properties. We present a number of interval-vector polytopes, A convex polytope is formed including one class whose volumes are the Catalan numbers and another class whose volumes are the even Due to the interesting properties of , we Pn,1 by taking the convex hull of numbers and face numbers mirror Pascal’s triangle. studied a related class of polytopes d a set A = v1,v2,...,vn R , { } ⇢ 1. Complete Interval-Vector Polytope 3. Interval Pyramid n,i := conv(e1,...,en, ↵1,n i, ↵2,n i+1,...,↵i+1,n) conv(A), which is defined as P Given a dimension n, define to be the n Let n = ↵i,j i, j [n],i j . n,1 Pn where n>2 and i 2. n I { | 2  } convex hull of all vectors in R with interval  1v1 + 2v2 + + nvn 1, 2,...,n R 0 and i =1 . 8 ··· 2 9 length 1 or n 1. Example 6. For n =5and i =2 < i=1 = The complete interval-vector polytope is X : ; defined as n := conv( n). Example 5. For n =4, 4,2 = conv (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), A simplex is an n-dimensional polytope with PI I P n +1 4,1 = conv (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (1 , 1, 0, 0), (0, 1, 1, 0), (0, 0, 1, 1) vertices. We form a lattice-preserving bijection between P the complete interval-vector polytope and (0, 0, 0, 1), (1, 1, 1, 0), (0, 1, 1, 1) . Given an n-dimensional polytope with fk P Postnikov’s complete root polytope in [2]. Proposition 1. The dimension of is n. k-dimensional faces, the f-vector of is Pn,i P Theorem 1. The volume of the n-dimensional Theorem 3. The f-vector for n,1 for n 3 is the written as th P interval-vector polytope is the nth Catalan n row of the Pascal 3-triangle. Proposition 2. Let f( ):=(f ,f ,...,f ). number. 0 1 n 1 = conv e1,e2,...,ei,en i+1,...,en, P 1 2n B { vol( n)= . PI n +1 n ↵1,ni, ↵2,n i 1 ...,↵i+1,n . ✓ ◆ n =1:3 } Example 1. Tetrahedron Then adding each vector in ei+1,ei+2,...,en i n =2:44 { } Simplex: 3-dimensional polytope, 4 vertices 2. Fixed Interval-Vector Polytope sequentially pyramids over the previous base. Given an interval length i and a dimension n n =3:585 f-vector: (4,6,4) we define the fixed interval-vector polytope n =4: 6 13 13 6 Finally, we have conjectured the volume of n,i n P 4 vertices, 6 edges, 4 planes n,i as the convex hull of all vectors in R and plan to prove it by proving a triangulation Q n =5 i with interval length i : 7 19 26 19 7 of the base of n,i contains 2 simplices whose P n =6: 8 26 45 45 26 8 volume as one pyramids over them is Denote the volume of a polytope as vol( ). n,i := conv( ↵j,j+i 1 j n i +1 ). n (i +1). P P Q { |  } We project down to its ambient Triangulation of the base of n,1: n n,i Conjecture 1. An interval vector [1] is a (0, 1)-vector x R Q P 2 dimension and prove using Dahl’s such that, if x = x =1for i

n =1: 3

n =2: 4 4

n =3: 5 8 5 (4.1) n =4: 6 13 13 6

n =5: 7 19 26 19 7

n =6: 8 26 45 45 26 8

The proof of this correspondence requires a few preliminary results. First we deﬁne

the face =conv(e1,en,↵1,n 1,↵2,n ) n(I 1,n 1 )asthebase of the interval B { } ✓P { } pyramid.

Lemma 4. The base of the interval pyramid is 2 dimensional.

Proof. We ﬁrst consider =conv(en,↵1,n 1,↵2,n). The corresponding elementary A vectors of the vertex set are e ,e ,e .Sowebuildtheﬂow-dimensiongraphas { 1,n 2 n} seen in Figure 4.1, G( )) = (V,E)whereV =[n], E = (1,n) corresponding to A { } e .ThesubsetV = 2,n (circled in Figure 2) corresponds to e and e .This 1,n 1 { } 2 n

41 graph has n 1connectedcomponents,twoofwhichcontainelementsofV so that 1 k = n 3. 0

If we let 1en + 2↵1,n 1 + 3↵2,n = 0,weﬁrstnoticethat2 =0since↵1,n 1 is the only vector with a nonzero ﬁrst coordinate. But this implies that 1 = 3 =0.

Since the coecients cannot sum to 1, we conclude that 0 / a↵(en,↵1,n 1,↵2,n). 2 So now by Theorem 3

dim(conv(en,↵1,n 1,↵2,n)) = n k0 1=n (n 3) 1=2.

Finally e1 =(1)↵1,n 1 +( 1)↵2,n +(1)en is in the ane hull of and does not add A a dimension. Thus we conclude that dim( )=2. B

Corollary 6. Each ei for 2 i n 1 adds a dimension to n(I 1,n 1 ), that is   P { } ei / a↵ 1,n 1 ) ei . 2 I{ } \{ }

Proof. This follows from Theorem 4.1 and Lemma 4. Since the base of the interval pyramid has dimension 2 and n(I 1,n 1 )hasdimensionn,thenthen 2 remaining B P { } vertices must add the remaining n 2dimensions.Clearlynonecanaddmorethan one, so each must add precisely one dimension.

Lemma 5. The base of the interval pyramid, has f-vector (4, 4). B

Figure 4.3: G( ): The ﬂow dimension graph of part of the base of the interval A pyramid.

42 Proof. Since has dimension 2, f1 = f0.Weknowthat en,↵1,n 1,↵2,n are three B { } vertices of since they form a 2-dimensional object. If e1 conv(en,↵1,n 1,↵2,n) B 2 then

e1 = 1en + 2↵1,n 1 + 3↵2,n (4.2) where the coecients sum to 1. Since ↵1,n 1 is the only vector with a nonzero coordinate in the ﬁrst position, that implies 2 =1. This in turn implies that 1 = 3 =0, contradicting (4.2). So e1 / conv(en,↵1,n 1,↵2,n)andthereforeformsafourthvertex. 2

Thus f0 =4=f1 completing the proof.

We can tie all this together with the following theorem. First we deﬁne (as in [Gr])

d d 1 a d-pyramid P as the convex hull of the union of a (d 1)-dimensional polytope K d d 1 d (the basis of P )andapointA/a↵(K )) (the apex of P ). 2

d d 1 Theorem 5. [Gr] If P is a d-pyramid with (d 1)-dimensional basis K then

d d 1 f0(P )=f0(K )+1

d d 1 d 1 fk(P )=fk(K )+fk 1(K ) for 1 k d 2   d d 1 fd 1(P )=1+fd 2(K ).

Example 31. Consider 3(I 1,2 ). The vertices of the base are e1,e3,↵1,2,↵2,3 , P { } { } which has f vector (4, 4). The vertex (0, 1, 0) serves as an apex over the base, and completes the pyramid, which now by Theorem 5 has f vector (5, 8, 5).

Since each vertex of the interval pyramid which is not a vertex of is anely B independent, we can imagine each one as the apex of a pyramid, and imagine the interval pyramid as being formed adding each apex one by one as a succession of pyramids over pyramidal bases all the way down to . B

43 e3 ↵2,3

↵1,2

Figure 4.4: The 3 dimensional interval pyramid, with the base shaded in dark gray.

We notice that the rows of Pascal’s 3-triangle act in the same manner as face numbers for pyramids, and we claim the face numbers for n(I 1,n 1 )canbederived P { } from Pascal’s 3-triangle.

th Theorem 6. The f-vector for n(I 1,n 1 ) for n 3 is the n row of the Pascal P { } 3-triangle.

Proof. Let = e1,e2,...,en,↵1,n 1,↵2,n be the vertex set for n(I 1,n 1 )with I { } P { } n 3, and call k =conv( ek,ek+1,...,en 1 )for1 k

n 1 and en 1 / a↵( n 1)(byCorollary6),andthusisapyramidanditsface R 2 R numbers can be computed as in Theorem 5 from the face numbers of n 1. R

Notice next that n 1 is the convex hull of the union of the (n 2)-dimensional R polytope n 2 and en 2 / a↵( n 2)(againbyCorollary6),sowecancomputethe R 2 R face numbers of n 1 from those of n 2 as in Theorem 5. R R We can continue this process until we get that is the convex hull of and R3 R2 e / a↵( ). But we notice that = is the base of the interval pyramid, so 2 2 R2 R2 B by Lemma 5, f ( )=f ( )=4.Fromherewecanbuildusingf-vectors of 0 R2 1 R2

n(I 1,n 1 ) from Theorem 5 which are exactly those of the Pascal 3-triangle. We P { } do this n 1timestoreach n(I 1,n 1 ), and since (4,4) is the second row of the P { }

44 th triangle, then the f-vector of n(I 1,n 1 )isthen row of the Pascal 3-triangle, as P { } desired.

We can rewrite (4.1) as

2+1

3+1 3+1

4+1 6+2 4+1

5+1 10 + 3 10 + 3 5+1

6+1 15 + 4 20 +6 15 + 4 6+1

7+1 21 + 5 35 + 10 35 + 10 21 + 5 7+1

(4.3) which is Pascal’s triangle added to a shifted Pascal’s triangle. Thus we can derive a formula for the number of k-faces for n(I 1,n 1 )intermsofbinomialcoecients. P { }

n 1 n +1 Corollary 7. For n 3, fk( n(I 1,n 1 )) = 0 1 + 0 1. P { } k k +1 B C B C @ A @ A 4.2 Volume of the Interval Pyramid

The base of the interval pyramid can be easily triangulated into B

1 =conv(e1,en,↵1,n 1) and 2 =conv(en,↵1,n 1,↵2,n). 4 4

Since the remaining vertices are anely independent and triangulations of a base extend to a pyramid, this extends to a triangulation of n(I1,n 1)into2n-simplexes P

S1 =conv(e1,e2,...,en 1,en,↵1,n 1) and S2 =conv(e2,...,en 1,en,↵1,n 1,↵2,n).

45 Interval-Vector Polytopes Jessica De Silva, Gabriel Dorfsman-Hopkins, Joseph Pruitt Advisors: Prof. Matthias Beck and Amanda Ruiz 2012 Mathematical Sciences Research Institute - Undergraduate Program

n that for R ,afulldimensionaln simplex with vertex set v0,...,vn , P2 { }

vol( )=det(v v ,v v ,...,v v ). P 1 0 2 0 n n

So this triangulation allows us to easily calculate the volume of the interval pyramid.

Lemma 6. The determinant of the n n-matrix ⇥

01 1 1 ··· 210 1 13 ··· 6 7 6 ... 7 6 7 6 7 6 7 61 1017 6 ··· 7 6 7 611 107 6 ··· 7 4 5

n 1 is ( 1) (n 1).

Proof. Let A be the n n matrix whose diagonal entries are 0, and all entries o↵ n ⇥ the diagonal are 1. E.g.,

46 01 A2 = 2 3 10 6 7 4 5 k 1 and so det(A )= 1. Assume det(A )=( 1) (k 1). A is the (k+1) (k+1)- 2 k k+1 ⇥ matrix of the form 01 11 ··· 210 1 13 ··· 6 . 7 Ak+1 = 6 .. 7 . 6 7 6 7 6 7 611 017 6 ··· 7 6 7 611 107 6 ··· 7 4 5 Subtracting the second row from the ﬁrst, which does not change the value of the determinant, will give us the matrix

11 0 0 ··· 2 101 13 ··· 6 7 6 ... 7 . 6 7 6 7 6 7 6 11 017 6 ··· 7 6 7 6 11 107 6 ··· 7 4 5

Now the determinant of Ak+1 is the sum of two determinants by cofactor expansion. Speciﬁcally it is ( 1) det(A )minusthedeterminantofthematrixobtainedbytaking k out the ﬁrst row and second column. We know that ( 1) det(A )=( 1)k(k 1) by k the inductive hypothesis. So what we have left to compute is the determinant of the

47 (k k)-matrix ⇥ 11 11 ··· 210 1 13 ··· 6 . 7 6 .. 7 . 6 7 6 7 6 7 611 017 6 ··· 7 6 7 611 107 6 ··· 7 4 5 We will subtract the ﬁrst row from each of the rows below it, also not changing the determinant, to give us the upper triangular matrix

11 111 ··· 2 3 0 10 00 6 ··· 7 6 7 600 10 0 7 6 ··· 7 6 . 7 6 .. 7 6 7 6 7 600 0 107 6 7 6 ··· 7 6 7 600 00 17 6 ··· 7 4 5 k 1 whose determinant is ( 1) . Furthermore,

k 1 det(A )=( 1) det(A ) ( 1) k+1 k =( 1)k(k 1) + ( 1)k =( 1)kk.

n 1 Therefore, by induction, det(An)=( 1) (n 1), for all n Z 2. 2

Theorem 7. vol n(I 1,n 1 ) =2(n 2) for n 3 P { } Proof. In order to calculate the volume of n(I 1,n 1 )wewillﬁrsttriangulatethe P { }

48 2-dimensional base of the pyramid from Lemma 4

1 =conv(e1,en,↵1,n 1) and 2 =conv(en,↵1,n 1,↵2,n). 4 4

4 Let x be a point in the base, then for some 0, where =1, i i i=1 X

x = 1e1 + 2en + 3↵1,n 1 + 4↵2,n =( + , + , , + , + ) 1 3 3 4 ··· 3 4 2 4

=(1 4)e1 +(2 + 4)en +(3 + 4)↵1,n 1

=(1 + 2)en +(1 + 3)↵1,n 1 +(4 1)↵2,n.

So x is a point in if and x is a point in if .Thus and 41 1 4 42 4 1 41 42 is a triangulation of the 2-dimensional base of the pyramid.

By Corollary 6, each e2, ,en 1 adds a dimension so that the convex hull of these ··· points and is an n-dimensional simplex. The same can be said of .Callthese 41 42 simplices S1 and S2 respectively. Thus S1 and S2 triangulate n(I 1,n 1 ). Therefore P { } the sum of their volumes is equal to the volume of n(I 1,n 1 ). In order to calculate P { } the volume of S1 and S2, we will use the Cayley Menger determinant [Da] once again.

Consider S1,whosevolumeisthedeterminantofthematrix

0 1 1 1 1 ··· 2 3 10 1 1 1 6 ··· 7 6 7 6 1 10 1 17 e ↵ e ↵ ... e ↵ = 6 ··· 7 . 1 1,n 1 2 1,n 1 n 1,n 1 6 . 7  6 .. 7 6 7 6 7 6 1 1 10 17 6 7 6 ··· 7 6 7 6 000 017 6 ··· 7 4 5 49 Cofactor expansion on the last row will leave us with the determinant, up to a sign, of the (n 1) (n 1) matrix ⇥

0 1 1 1 ··· 2 10 1 13 ··· 6 7 6 ... 7 , (4.4) 6 7 6 7 6 7 6 1 10 17 6 ··· 7 6 7 6 1 1 107 6 ··· 7 4 5 which when ignoring sign by Lemma 6 is n 2. Therefore the volume of S is n 2. 1

Now consider the Cayley Menger determinant of S2, the determinant of

1000 0 ··· 2 3 00 1 1 1 6 ··· 7 6 7 6 0 10 1 ... 17 ↵ ↵ e ↵ e ↵ e ↵ = 6 7 . 1,n 1 2,n 2 2,n 3 2,n n 2,n 6 . 7  ··· 6 .. 7 6 7 6 7 6 0 1 1 0 17 6 7 6 ··· 7 6 7 6 1 1 1 107 6 ··· 7 4 5 By cofactor expansion on the ﬁrst row we are left with the positive determinant of the matrix (4.4) which is n 2. Therefore the volume of S is n 2andsothevolume 2 of n(I 1,n 1 )is2(n 2), as desired. P { }

4.3 Duality of the Interval Pyramid

Since the interval pyramid has a symmetric f-vector, it is natural to wonder about its duality. That is, if we replace each k face with an n k face (take the dual of the

50 polytope), are you left with an identical polytope? How does the volume and face lattice structure change? Clearly the f-vector will remain the same, but I wonder whether the interval pyramid is actually self-dual. Recall that in Section 2.3.3, we o↵ered the following construction to deﬁne the dual of a polytope Rn for the cases where 0 int( ).[Zi] P⇢ 2 P

:= y Rn : y x 1forallx , P { 2 ·  2P}

where y x = y x + ... + y x is the dot product. But we know that 0 / · 1 1 n n 2 int( n(I 1,n 1 )) for any interval pyramid. Still, we noted in the previous section that P { } we can anely translate a polytope so that 0 is in the interior without losing any information about the face structure or volume. Let us take several examples of translating the interval pyramid so that it contains zero and taking the dual.

Example 32. Consider 3(I 1,2 )=conv((1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (0, 1, 1)). P { } 1 1 1 Call the vertex set V . Let v = 2 , 2 , 4 The ane translation 1 1 1 ( 3(I 1,2 ))v =conv V , , P { } 2 2 4 ✓ ✓ ◆◆ leaves the structure of the polytope intact and has zero in its interior as we will show.

First let’s call the translated vertices of ( 3(I 1,2 ))v (in order) v1,...,v5. Then notice P { } that

1/2 1/2 1/2 1/2 0 0 1 0 1 0 1 0 1 0 1 v1 + v2 + v3 + v4 = 1/2 + 1/2 + 1/2 + 1/2 = 0 B C B C B C B C B C B C B C B C B C B C B 1/4C B 1/4C B 1/4 C B 1/4C B0C B C B C B C B C B C @ A @ A @ A @ A @ A 51 Thus if we let 1 = ...= 4 =1/4 and 5 =0. Then

1 v + ...+ v = (v + ...+ v )=0 1 1 5 5 4 1 4

With + ...+ =1and each 0. 1 5 i Applying to dual construction to this translated polytope gives us a half space

description for ( 3(I 1,2 ))v given by the following inequalities: P { }

1 1 1 x x x 1 2 1 2 2 4 3  1 1 1 x + x x 1 2 1 2 2 4 3  1 1 3 x x + x 1 2 1 2 2 4 3  1 1 1 x + x x 1 2 1 2 2 4 3  1 1 3 x + x + x 1. 2 1 2 2 4 3 

Example 33. Consider 4(I 1,3 )=conv( (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1) P { } { (1, 1, 1, 0), (0, 1, 1, 1) ). Call the vertex set V . Letting u = 2 , 2 , 2 , 1 , we can trans- } 5 5 5 5 late this polytope so that 0 is an interior point as follows:

2 2 2 1 ( 4(I 1,3 ))u =conv V , , , . P { } 5 5 5 5 ✓ ✓ ◆◆

Now we can apply the dual construction and have (( 4(I 1,3 )))u given by the following P { }

52 ineqalities:

3 2 2 1 x x x x 1 5 1 5 2 5 3 5 4  2 3 2 1 x + x x x 1 5 1 5 2 5 3 5 4  2 2 3 1 x x + x x 1 5 1 5 2 5 3 5 4  2 2 2 4 x x x + x 1 5 1 5 2 5 3 5 4  3 3 3 1 x + x + x x 1 5 1 5 2 5 3 5 4  2 3 3 4 x + x + x + x 1. 5 1 5 2 5 3 5 4 

We want to prove that 0 is an interior point of the translated interval pyramid. We ﬁrst recall lemma 1[Zi].

Lemma 7. [Zi] Let be a polytope in Rn.Ifp can be represented as p = P 2P 1 n x for n+1 anely independant points x ,...,x , then y is an interior n+1 i=0 i 0 n 2P pointsP of . P

Proposition 9. Let n(I 1,n 1 ) be the n dimensional interval pyramid. Let u = P { } 2 2 2 1 , , , , . Then 0 is an interior point of ( n(I 1,n 1 ))u. This ane n+1 n+1 ··· n+1 n+1 P { } translation doesn’t disturb any combinatorial structure (i.e. it is a purely linear translation leaving all structure of the polytope intact, essentially just repositioning the origin).

Proof. First we show that 0 ( n(I 1,n 1 ))u.Thevertexsetof( n(I 1,n 1 )))u is 2 P { } P { }

v1,...,vn+2 = e1 u, e2 u, . . . , en u, ↵1,n 1 u, ↵2,n u . { } { }

53 Recall that ↵1,n 1 = e1 + ...+ en 1 and notice that:

v1 + ...+ vn+1 =(e1 u)+...+(en u)+(↵1,n 1 u)

= e1 + ...+ en +(e1 + ...+ en 1) (n +1)u

=2e1 + ...+2en 1 + en (2,...,2, 1) = 0

1 So if we let 1 = ...= n+1 = n+1 and n+2 =0,then

1 v + ...+ v = (v + ...+ v )=0 1 1 n+2 n+2 n +1 1 n+1

Notice that v1,...,vn+1 are n +1anelyindependentpoints. Thisiseasierto show using the untranslated vertices e1,...,en,↵1,n 1 (which is equivalent). Clearly n the unit vectors are each anely independent, and a↵(e1,...,en)= x R : xi = { 2 i

1 ↵1,n 1.Sosincev1,...,vn+1 are n+1 anely independent points in n(I 1P,n 1 ) }63 P { } 1 n+1 u,wecanrefertothepreviouslemmatonoticethatsince0=n+1 i=1 vi then 0 is in fact an interior point of ( n(I 1,n 1 ))u. P P { }

We recall Proposition 6 to take the dual construction and calculate the half space description of the interval pyramid in general (albeit translated).

Proposition 10. Let Rn be a polytope with 0 on its interior, and vertex set P✓ n v1,...,vm .Ifc R , then c if and only if c vi 1 for i =1,...,m. { } 2 2P · 

This is enough to give us the facet description we need.

2 2 2 1 Corollary 8. Let u = , , , , and assume that n(I 1,n 1 )u has n+1 n+1 ··· n+1 n+1 P { } vertices v ,...,v . Then we deﬁne the dual by: { 1 n+2}

n n(I 1,n 1 ) = c R : c vi 1 for i =1,...,n+2 P { } u { 2 ·  } 54 Thus, n(I 1,n 1 ) has the following inequality description. P { } u 2 2 1 x + 1 x x 1 for all j [n 1] 0 n +1 i1 n +1 j n +1 n  2 i [n 1],i=j ✓ ◆ 2 X 6 @ An 1 2 n x + x 1 n +1 i n +1 n  i=1 ✓ ◆! n 1X 2 1 1 x x 1 n +1 i n +1 n  i=1 ✓ ◆ ! Xn 1 2 2 n x + 1 + x 1. n +1 1 n +1 n +1 n  i=2 ! X ✓ ◆ Proof. Because we know that the f-vector of the interval pyramid is symmetric, we know that there are exactly n +2facetsofthedualoftheintervalpyramid. Because of the Proposition 10, we know that a point is in the dual of the interval pyramid if and only if it satisﬁes the n +2facetequationsgiven,sothesefacetsaresucientto completely describe the dual of the interval pyramid. But because we know we need at least n +2 facets,weknowaswellthattheyarenecessary.Thusthisisa complete

facet description of ( n(I 1,n 1 ))u . P { }

Recall that two polytopes are called combinatorially isomorphic if their face lattices are isomorphic as posets. If a polytope is combinatorially isomorphic to its dual then we call the polytope self dual. Let u be as deﬁned in the previous corollary, to prove that the interval pyramid is self dual we must prove that

L((( n(I 1,n 1 )u ) = L( n(I 1,n 1 )). P { } ⇠ P { }

But by proposition 2 we know that

op L(( n(I 1,n 1 ))u ) = L(( n(I 1,n 1 )) . P { } ⇠ P { } 55 And because the ane translation preserved all of the combinatorial structure of the polytope we know that:

op op L(( n(I 1,n 1 )u) = L(( n(I 1,n 1 ))) . P { } ⇠ P { }

So in fact, to show that the interval pyramid is self dual, we need only to show that:

op (L( n(I 1,n 1 ))) = L( n(I 1,n 1 )) P { } ⇠ P { }

To fully understand the combinatorial self-duality of the interval pyramid, we must fully describe its face lattice in general.

56 Chapter 5

Open Questions

First we note that we have not yet answered the question on the self-duality of the Interval Pyramid. To do so, we must get a general look at its face lattice poset and see if we can prove an isomorphism between that poset and its opposite. Computing examples in Sage and Polymake for the ﬁrst 10 Interval Pyramids leads me to this ﬁrst conjecture.

n Conjecture 1. Let n(I 1,n 1 ) R be the interval pyramid. P { } ⇢

op (L( n(I 1,n 1 ))) = L( n(I 1,n 1 )) P { } ⇠ P { }

That is, the interval pyramid is self dual.

As the nature of this project was cataloguing an interesting class of combinatorial polytopes, and we only touched on a few initially, there is plenty of interesting work to be done further investigating polytopes with di↵erent sets of interval vectors as vertices. Since all of the collections of interval vectors we’ve considered thus far have yielded very interesting combinatorial properties when considered as vertices of lattice polytopes, there is a good chance that there are many interesting interval

57 vector polytopes left to study. Some possibilities are n(I i)or n(I i), where the P P  interval vectors all have length greater than or less than some ﬁxed constant. Here is one more interval vector polytope we have begun doing work on.

5.1 Generalized Interval Pyramid

We can make a generalization of the previous polytope by deﬁning the generalized interval pyramid n(I 1,n i )tobetheconvexhullofallthestandardunitvectorsin P { } Rn and all the interval vectors with interval length n i.Werestrictourselvestothe cases where n 2andi n .  2

Example 34. For n =6and i =2,

6(I 6,2 )=conv(e1,e2,e3,e4,e5,e6,↵1,3,↵2,4,↵3,5,↵4,6). P { }

We are able to generalize many of the results from the interval pyramid. The important similarity is that the generalized interval pyramid is an n dimensional polytope, which is formed as a series of pyramids over a 2i dimensional base. I hope to focus on triangulating this base in general. Repeated computations in of examples in polymake [Ga] have led to the following conjecture.

i Conjecture 2. vol( n(I 1,n i )) = 2 (n (i +1)) P { }

In all the examples I considered, the f-vector was a combination of pascals triangle rows, and so the face numbers should all be easily expressible in terms of binomial coecients. This is an area for further study.

58 Bibliography

[Az] Azarian, M. Pascal 3-Triangle, OEISA028263.

[Dm] Dmharvey Catalan number 4x4 grid example http://upload.wikimedia.org/wikipedia/commons/f/f4/ Catalan_number_4x4_grid_example.svg.

[Ep] Eppstein D. The face lattice of a square pyramid, https://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/ Pyramid_face_lattice.svg/500px-Pyramid_face_lattice.svg.png.

[Ks] KSmrq Hasse Diagram of x, y, z , http://upload.wikimedia.org/wikipedia/commons/e/ea/{ } Hasse_diagram_of_powerset_of_3.svg.

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