ON �-POLYNOMIALS FOR LATTICEPARALLELEPIPEDS

Emily McCullough

Version of July 20, 2016

ON �-POLYNOMIALS FOR LATTICE PARALLELEPIPEDS

Emily McCulloughSan Francisco State University

2016

We introduce the (B, `)-Eulerian numbers, a refined descent statistic on the set of signed

permutations on {1, 2, . . . , d}, and show that these numbers appear as the coe�cients of

the �-polynomial (also known as the Ehrhart h-polynomial or the h⇤-polynomial) for the

d-dimensional half-open ±1-cube with ` non-translate facets removed. Using an interplay of

the geometry and combinatorics, we characterize the �-polynomials for specific families of

half-open parallelepipeds and improve upon known inequality relations on their coe�cients.

In particular, we prove that the coe�cients of the �-polynomial for half-open parallelepipeds

with lattice centrally symmetric edges are alternatingly increasing. Our results extend nat-

urally to the �-polynomials for closed lattice zonotopes.

ACKNOWLEDGMENTS

First and foremost I want to thank my advisor, Matt Beck. Thank you for intro-

ducing me to the art of proof, for honing my mathematical senses of smell and

taste, and for sharing with me the worlds of algebraic and geometric combina-

torics. From my first day as an undergraduate at SFSU you have shown me what

it looks like to be a great teacher, mentor and mathematician. You have given

me much to aspire to and for that I am deeply grateful.

Additionally, I want to express my appreciation for Professors Federico Ardila

and Serkan Hosten. Your teaching and mentorship have been integral to my

development as a mathematician. Thank you.

My deepest thanks go toKatharina Jochemko for her incredible generosity, friend-

ship and mentorship. I am also grateful for my friend and colleague Hannah

Winkler — it was a true pleasure studying with you.

I express my gratitude to the Math Department at SFSU and the NSF GK-12

2013-14 Fellowship (Grant DGE-0841164).

I extend special thanks to Felix Breuer for fruitful discussions and great fun at

the Villa.

Finally, I want to thank my family for teaching me tenacity and courage and for

supporting me every step of the way.

3

Contents

1 Introduction 5

2 Basics 9

2.1 An Introduction to Lattice Polytopes and Ehrhart Theory . . . . . . . . . . 9

2.2 Parallelepipeds, Unit Cubes and Zonotopes . . . . . . . . . . . . . . . . . . . 14

3 Half-Open Parallelepipeds and the (A, j)-Eulerian Numbers 19

3.1 Descent Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2 Half-Open Unit Cubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.3 Half-Open Parallelepipeds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4 Half-Open ±1-Cubes and the (B, `)-Eulerian Numbers 44

4.1 Signed Permutation Descent Statistics . . . . . . . . . . . . . . . . . . . . . 44

4.2 Half-Open ±1-Cubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

5 Lattice Centrally Symmetric Parallelepipeds 68

5.1 Parallelepipeds with Lattice Centrally Symmetric Edges . . . . . . . . . . . . 68

5.2 Extensions and Open Questions . . . . . . . . . . . . . . . . . . . . . . . . . 71

Bibliography 73

4

Chapter 1

Introduction

This thesis deals with results in the fields of algebraic and geometric combinatorics, specif-ically in the area of Ehrhart theory. Our main objects of study are half-open lattice d-parallelepipeds — higher-dimensional analogues to 2-dimensional parallelograms with integervertices and j d non-translate facets removed — and the associated �-polynomials (alsoknown as h⇤-polynomials or Ehrhart h-polynomials).

Using the interplay of the geometry provided by disjoint decompositions of the paral-lelepipeds and the combinatorics provided by relevant permutation descent statistics, weimprove upon known inequality relations on the coe�cients of the �-polynomials for specificfamilies of half-open parallelepipeds. In particular, we improve these inequalities for certainfamilies of half-open parallelepipeds with an interior lattice point. Our results for half-openlattice parallelepipeds extend naturally to the �-polynomials for closed lattice zonotopes.

Chapter 2 is a review of the basics. The first section is an introduction to Ehrharttheory. We start from the beginning with formal definitions for lattice polytope and relatedterminology such as dimension, hyperplane, face, boundary, relative interior, and variousnotions of volume. We introduce the lattice point enumerator, or Ehrhart polynomial, of alattice polytope P , and the related Ehrhart series of P . The numerator of the Ehrhart seriesin its rational form is the �-polynomial of P . In this section we also define what it meansfor a �-polynomial to be unimodal or alternatingly increasing. These inequality relations onthe coe�cients of the �-polynomial are the focus of this thesis.

In the second section of Chapter 2 we introduce the main geometric objects of our study:lattice parallelepipeds, in particular unit cubes, and lattice zonotopes — all integrally closedlattice polytopes. We consider a theorem due to Shephard which depicts the close relation-ship between zonotopes and parallelepipeds [11]. We also note the close relationship betweensubdivisions and dissections of a polytope. In this way we set up and motivate our laterresults on the �-polynomials for zonotopes.

This thesis is part of a joint project with Matthias Beck and Katharina Jochemko. Chap-ter 3, with the exception of Lemma 3.4, is a reformulation of Jochemko’s work on zonotopesin [8]. It is the starting point for our extensions and contributions, which appear in Chapters4 and 5.

A conjecture by Stanley [13] infers that the coe�cients of the �-polynomial for every inte-grally closed lattice polytope are unimodal. Schepers and Van Langenhoven [10] prove that

5

this weakening of Stanley’s conjecture holds for (closed) lattice parallelepipeds1, the simplestexample of an integrally closed lattice polytope after unimodular simplices. Jochemko [8] ex-tends their work to half-open lattice parallelepipeds by first interpreting the A-polynomialsof [10] in terms of refined descent statistics on Sd — statistics we call the (A, j)-Euleriannumbers. Using the symmetric and recursive properties of the (A, j)-Eulerian numbers pro-vided by Brenti and Welker [5, Lemma 2.5], Jochemko proves the (A, j)-Eulerian numbersare unimodal and specifies the peak according to d and j. Jochemko further provides a char-acterization of the �-polynomial for half-open parallelepipeds as a nonnegative linear com-bination of (A, j)-Eulerian polynomials. Together these results imply that the coe�cientsof the �-polynomials for half-open parallelepipeds are unimodal. Using the aforementionedtheorem of Shephard, combined with the Beneath and Beyond construction of Koppe andVerdoolaege [9], Jochemko further extends the unimodality results to the �-polynomials forlattice zonotopes — a (large) family of integrally closed lattice polytopes. In Chapter 3,we carefully build up to these results. Additionally, through an extension of Jochemko’sunimodality proof, we show that the (A, j)-Eulerian numbers are alternatingly increasingfor su�ciently large j.

Chapter 4 deals with the relationship between the (B, `)-Eulerian numbers, a refineddescent statistic on Bd, the set of signed permutations on [d], and the �-polynomial for[�1, 1]d` , the d-dimensional half-open ±1-cube with ` non-translate facets removed. Ourwork is motivated by a result due to Brenti [4, Theorem 3.4] as seen from the geometricperspective presented by Beck and Braun in [1]. The result (for q = 1) says that thetype-B Eulerian polynomial B(d, t) is the �-polynomial for the (closed) ±1-cube [�1, 1]d.In Section 4.1 we characterize B(d, t) as a positive linear combination of (A, j)-Eulerianpolynomials of the same degree. We do this via a disjoint decomposition of [�1, 1]d into unitcells Ud

I where I ✓ [d], geometric objects which are congruent to unit d-cubes. Using thesymmetric properties of the (A, j)-Eulerian numbers [5, Lemma 2.5], we give a geometricproof that the coe�cients of B(d, t) are symmetric and unimodal, and thus alternatinglyincreasing. This is not a new result. It follows from [4, Theorem 2.4] and can also bededuced from [10, Proposition 2.17].

In Section 4.2 we extend the methods of [1] to half-open ±1-cubes. We consider a disjointdecomposition of [�1, 1]d` into half-open unimodular simplices indexed by signed permuta-tions (⇡, ✏) 2 Bd and show that the removed facets of the simplex 4d,`

(⇡,✏) are enumeratedby our refined descent statistic. In this way we prove that the (B, `)-Eulerian polynomialB`+1

(d+ 1, t) is the �-polynomial for [�1, 1]d` .We further prove that the (B, `)-Eulerian polynomial is alternatingly increasing. As

before this is seen via a disjoint decomposition of [�1, 1]d` into unit cells. From the decompo-sition we obtain a characterization of the �-polynomial for [�1, 1]d` as a linear combinationof (A, j)-Eulerian polynomials of the same degree. The nonnegative integral coe�cients ofthis linear combination have symmetric, unimodal and recursive properties which we use toshow that B`+1

(d + 1, t) is alternatingly increasing. Our geometric perspective results inan alternative geometric proof for the palindromic and unimodal properties of the binomial

1Stanley’s original conjecture has been disproved in its general form.

6

coe�cients as well as a geometric interpretation of the well-known recursive formula✓

d

j

◆

=

✓

d� 1

j

◆

+

✓

d� 1

j � 1

◆

.

In Chapter 5 we consider the �-polynomials for lattice parallelepipeds and zonotopeswith lattice centrally symmetric edges. An edge of a polytope is lattice centrally symmetricif the midpoint of that edge is a lattice point. Using the symmetry of the edges as well asresults of Jochemko and Schepers and Van Langenhoven seen in Chapter 3, we express the�-polynomial for half-open parallelepipeds with lattice centrally symmetric edges in terms ofthe �-polynomials for half-open ±1-cubes of the same dimension. In particular, we provide acharacterization for the �-polynomial as a nonnegative linear combination of (B, `)-Eulerianpolynomials of the same degree. Applying our results about the inequality relations on thesepolynomials from Chapter 4, we see that the �-polynomial for half-open parallelepipedswith lattice centrally symmetric edges is alternatingly increasing. We further extend thealternatingly increasing results to the �-polynomials for closed lattice zonotopes with latticecentrally symmetric edges using [11, Theorems 54 and 56] and [8, Corollary 3.5.3].

We conclude with a short discussion of extensions and open questions.

7

Chapter 2

Basics

2.1 An Introduction to Lattice Polytopes and EhrhartTheory

Consider a finite set of points in Rm: v1

,v2

, . . . ,vn. Imagine shrink-wrapping these pointsin m-space. The object you obtain, called a polytope, is a d-dimensional analogue of a2-dimensional convex polygon. Like a polygon, a polytope is a closed and convex geometricobject with flat faces and extreme points called vertices. The vertices are necessarily somesubset of the original set of points. See Figure 2.1.

P = conv {(�2, 1), (�2,�1), (�1, 0), (0,�1), (1, 0), (1,�1)}

Figure 2.1: A lattice polytope P . The vertices appear as solid points.

Formally we define a polytope P as the convex hull of v1

,v2

, . . . ,vn:

P = conv {v1

,v2

, . . . ,vn} :=

8

<

:

X

i2[n]

�ivi : �i � 0 andX

i2[n]

�i = 1

9

=

;

.

A lattice polytope1 is the convex hull in Rm of finitely many lattice points — points inthe integer lattice Zm. We are interested in lattice polytopes exclusively, so from here onwe shall take the term polytope to mean lattice polytope while omitting the descriptor. Weare also interested in positive integer dilates of P , denoted

nP := {nx : x 2 P} .

1A lattice polytope is also known as an integral polytope.

8

H1

H2

Figure 2.2: Supporting hyperplanes H1

and H2

in R2 define a vertex and a facet (also anedge) of polytope P = conv {(0, 0), (2, 0), (2, 2), (0, 2)} .

The (a�ne) span of polytope P ,

span(P) := {µx+ �y : x,y 2 P and µ+ � = 1} ,

is the translate of a vector space, called an a�ne space. The dimension of an a�ne space isequal to the dimension of the vector space of which it is a translate. We define the dimensionof P as the dimension of span(P) [2]. If P has dimension d, we call P a d-polytope andwrite dim(P) = d.

A hyperplane in Rm is a set of the form

H = {x 2 Rm : a · x = b} for some a 2 Rm and b 2 R .

When a is the zero vector and b = 0, H = Rm. When this is not the case, the dimension ofH is one less than the dimension of the ambient space. Just as a line (a hyperplane in R2)divides R2 into two spaces, the hyperplane H divides Rm into two spaces, called halfspaces.We denote the halfspaces by

H : = {x 2 Rm : a · x b} and

H� : = {x 2 Rm : a · x � b} .

We call H a supporting hyperplane of the polytope P if P lies entirely in one halfspaceof H [2]. See Figure 2.2.

A face F of a polytope P is a subset of the polytope defined by the intersection of P withone of its supporting hyperplanes. That is, F = P \H, where H is a supporting hyperplaneof P . When H = Rm we have

F = P \H = P \ Rm = P ,

meaning P is a face of P . When H is a supporting hyperplane that does not meet P , thenP \ H is empty, meaning F = ; is a face of P . The faces of a polytope are themselvespolytopes [7]. P and ; are the trivial faces of P of dimension d and �1, respectively. Thefaces of P of dimension 0 are called vertices, the faces of P of dimension 1 are called edges,and the faces of P of dimension d� 1 are called facets. A face of P of dimension r is calledan r-face. When r < d, we say r-face F is a proper face of P .

9

11

1

Figure 2.3: The relative volume of the blue facet of the 3-polytope is equal to 1.

We distinguish between the boundary and the relative interior of a polytope P inthe following manner. The boundary of P consists of those points contained in some properface of P and the relative interior of P (denoted by P�) consists of those points in P thatare not contained in the boundary.

The relative volume of polytope P is the Euclidean volume of P relative to its a�nespan [2]. This definition of volume allows us to assign a non-zero measure of volume toobjects that are not full-dimensional, such as the proper faces of a polytope. Considerthe blue facet of the 3-polytope in Figure 2.3 for example. The span of the blue facet isrepresented in yellow; it is a 2-dimensional hyperplane. We see that the blue facet is a unitsquare relative to its span. Therefore, the blue facet has relative volume 1.

We further define the normalized volume of a d-polytope P to be d! times the relativevolume of P and write vol(P) to denote this value [10]. Normalized volume changes the unitof volume measure from the d-dimensional unit cube to a d-dimensional unimodular simplex.We describe these objects in detail below.

A fundamental combinatorial object in Ehrhart theory is the lattice point enumeratorof a polytope. The lattice point enumerator of P counts the number of lattice points in thenth positive integer dilate of P and is denoted by

ehr(P , n) := # (nP \ Zm) .

This counting function for lattice polytopes is a polynomial in n of degree d [6]. We observethis in a simple example in Figure 2.4. The result is due to Eugene Ehrhart and accordingly,ehr(P , n) is called the Ehrhart polynomial of P . When embedded in a generating function,we obtain the Ehrhart series of P :

Ehr(P , t) := 1 +X

n�1

ehr(P , n)tn .

The Ehrhart series of every lattice polytope can be written as a rational function in thefollowing form:

�(P , t)

(1� t)d+1

=�0

+ �1

t+ · · ·+ �dtd

(1� t)d+1

, (2.1)

where the �i are nonnegative integers and �0

= 1 [2,12]. The numerator �(P , t) is called the

10

n = 1 n = 2 n = 3

ehr(P , 1) = 4

ehr(P , 2) = 9

ehr(P , 3) = 16

ehr(P , n) = (n+ 1)2

Figure 2.4: Observing the Ehrhart polynomial of the 2-cube.

�-polynomial of P and the sequence of coe�cients (�0

, �1

, . . . , �d) is called the �-vector2 ofP . A property of this polynomial (vector) with particular significance to Ehrhart theory isthe fact that the normalized volume of a polytope P is equal to the sum of the �i [2]; thatis,

vol(P) = �0

+ �1

+ · · ·+ �d . (2.2)

Specific relationships among these coe�cients for particular families of polytopes arethe focus of this thesis. One property we are interested in is unimodality. A �-vector isunimodal if it has a single peak; that is, if its entries increase up to some point, thendecrease. Symbolically, (�

0

, �1

, . . . , �d) is unimodal if there exists a k 2 {0, 1, . . . , d} suchthat

�0

�1

· · · �k � · · · � �d .

We are also interested in whether a �-vector is alternatingly increasing; that is, if

�0

�d �1

· · · �b d+12 c .

This is a stronger inequality property than unimodality: If (�0

, �1

, . . . , �d) is alternatinglyincreasing, then (�

0

, �1

, . . . , �d) is unimodal with peak at⌅

d+1

2

⇧

. We say a �-polynomial isunimodal or alternatingly increasing if the corresponding �-vector is.

2.2 Parallelepipeds, Unit Cubes and Zonotopes

Given d linearly independent generating vectors v1

, . . . ,vd 2 Zm, we define the d-dimensional(lattice) parallelepiped

⌃(v1

, . . . ,vd) :=

8

<

:

X

i2[d]

�ivi : 0 �i 1

9

=

;

.

A zero-dimensional parallelepiped is a point. A one-dimensional parallelepiped is a linesegment. A two-dimensional parallelepiped is a parallelogram. A three-dimensional paral-lelepiped (known by this name in layperson’s terms as well) is a 3-polytope with three pairsof parallel faces. See Figure 2.5.

2Other names for the �-polynomial (vector) are Ehrhart h-polynomial (vector) and h⇤-polynomial (vector).

11

⌃(;) ⌃(v1

)

v1

⌃(v1

,v2

)

v1

v2

⌃(v1

,v2

,v3

)

v1

v2

v3

Figure 2.5: Parallelepipeds living in R3 with dimension d = 0, 1, 2, 3 respectively.

The unit d-cube Cd := [0, 1]d forms the simplest example of a d-dimensional paral-lelepiped. Its generating vectors are the standard unit basis vectors e

1

, . . . , ed:

Cd = ⌃(e1

, . . . , ed) .

A related object is the (lattice) zonotope

Z(u1

, . . . ,ur) :=

8

<

:

X

i2[r]

�iui : 0 �i 1

9

=

;

,

where u1

, . . . ,ur are vectors in Zm. We call u1

, . . . ,ur the generators of Z(u1

, . . . ,ur).When the generators are clear, we will drop the argument and simply write Z. If the gen-erators of Z are linearly independent, then Z(u

1

, . . . ,ur) is an r-dimensional parallelepipedand

Z(u1

, . . . ,ur) = ⌃(u1

, . . . ,ur) .

A subdivision C of a polytope P is a non-empty, finite collection of polytopes such that

(1) Q 2 C implies all faces of Q are also in C ,

(2) Q1

,Q2

2 C implies Q1

\Q2

is a common face of both Q1

and Q2

,

(3)S

Q2C Q = P .

The polytopes that comprise the subdivision C are called cells of C . Furthermore, thosecells Q 2 C with dim(Q) = dim(P) are called maximal [3].

We further note that the collection of maximal cells Q1

,Q2

, . . .Qs in a subdivision of apolytope P forms a dissection of P . That is, [3]

P = Q1

[Q2

[ · · · [Qs

andQ�

i \Q�j = ; whenever i 6= j .

The following theorem due to Shephard makes explicit the close relationship betweenzonotopes and parallelepipeds.

12

u1

= (0, 1)

u2

= (0, 2)u3

= (2, 2)

u4

= (3, 1)

⌃ (u2

,u4

)

⌃ (u2

,u3

)

⌃ (u1

,u4

)

⌃ (u1

,u3

)

⌃ (u3

,u4

)

(a) generators in Z2 (b) Z(u1

,u2

,u3

,u4

) (c) a subdivision of Z

Figure 2.6: Generators in Z2, the 2-dimensional zonotope Z(u1

,u2

,u3

,u4

), and a subdivisionof Z into parallelepipeds.

Theorem 2.1 [11, Theorems 54 and 56]. Every zonotope admits a subdivision into paral-lelepipeds. In particular, every d-dimensional zonotope Z(u

1

, . . . ,ur) admits a subdivision Cinto parallelepipeds where the maximal cells in C are generated by the linearly independentsubsets of {u

1

, . . . ,ur} of size d.

Figure 2.6 shows a 2-dimensional zonotope Z generated by vectors u1

,u2

,u3

,u4

2 Z2

and a subdivision C of Z into parallelepipeds. The maximal cells in C are labeled3. Observethat the generators of these maximal cells are exactly those linearly independent subsets of{u

1

, . . . ,u4

} of size two. Further observe how the maximal cells of C form a dissection of Z.A lattice polytope P is integrally closed if for all integers n � 1 and for all lattice

points p 2 nP , there exist lattice points p1

,p2

, . . . ,pn 2 P such that

p = p1

+ p2

+ · · ·+ pn .

Parallelepipeds and zonotopes are both examples of integrally closed polytopes. We useinduction and a simple tiling argument to show that parallelepipeds are integrally closed. Asimilar argument holds for zonotopes as a result of Theorem 2.1.

Let P = ⌃(v1

, . . . ,vd) ⇢ Rm be a lattice d-parallelepiped. When n = 1, the desiredresult is trivially true. Let n � 2. We can tile nP with ndim(P) = nd copies of P :

nP =[

(P + v) ,

where the union runs over all linear combinations of v = k1

v1

+ · · · + kdvd with ki 2{0, 1, . . . , n� 1}. See Figure 2.7 for an example with n = 3 and d = 2. Choose p 2 nP \Zm

such that p /2 (n� 1)P \ Zm. Then p 2 (P \ Zm) + v for some v = k1

v1

+ · · ·+ kdvd withat least one ki equal to n� 1. If this is not the case, then we have ki 2 {0, 1, . . . , n� 2} forall i 2 [d]. But this implies p 2 (n� 1)P \ Zm, a contradiction.

3We abuse notation here: each maximal cell in C is in fact a translate of the parallelepiped by which itis labeled.

13

v1

v2

P

P + v1

P + 2v1

P + v2

P + v1

+ v2

P + 2v1

+ v2

P + 2v2

P + v1

+ 2v2

P + 2v1

+ 2v2

Figure 2.7: We can tile the third dilate of 2-parallelepiped P with 32 copies of P .

We know v 2 (n� 1)P \ Zm by definition. Therefore,

v = p1

+ · · ·+ pn�1

for some p1

, . . . ,pn�1

2 P\Zm by the induction hypothesis. Furthermore, p 2 (P \ Zm)+vimplies p = p

0

+ v for some p0

2 P \ Zm. Thus we have

p = p0

+ p1

+ · · ·+ pn�1

,

concluding our argument that parallelepipeds are integrally closed.

14

Chapter 3

Half-Open Parallelepipeds and the(A, j)-Eulerian Numbers

3.1 Descent Statistics

Let Sd denote the set of all permutations on [d] := {1, . . . , d}. We use permutation words� = �

1

�2

. . . �d in single line notation to denote permutations in Sd. For example, the per-mutation word � = 4213 in S

4

is equivalent to the permutation seen in two-line notationhere:

1 2 3 44 2 1 3

�

.

We call i a descent of � if �i > �i+1

. For example, the descents of � = 4213 are 1 and 2.Observe these descents in a permutation diagram of � = 4213 in Figure 3.5. The collectionof descents of a permutation � is called the descent set of �. The cardinality of the descentset is called the descent number of �. We write

Des(�) := {i 2 [d� 1] : �i > �i+1

} and

des(�) := |Des(�)|

for the descent set and the descent number of �, respectively. We similarly define the ascentset and the ascent number of a permutation, respectively, as follows:

Asc(�) := {i 2 [d� 1] : �i < �i+1

} and

asc(�) := |Asc(�)| .

We note that d is neither a descent nor an ascent of � 2 Sd. It follows that 0 des(�), asc(�) d� 1.

We define the (type-A) Eulerian number a(d, k) to be the number of permutations inSd with exactly k descents. That is,

a(d, k) = | {� 2 Sd : des(�) = k} | .

See Figure 3.1 for the Eulerian numbers when d = 3. We further define the (type-A)

15

123

a(3, 0) = 1

132 213 231 312

a(3, 1) = 4

321

a(3, 2) = 1

Figure 3.1: Permutation diagrams of S3

with descents in red, grouped by descent number.

Eulerian polynomial A(d, t) to be the degree-(d � 1) polynomial whose kth term hascoe�cient a(d, k):

A(d, t) :=d�1

X

k=0

a(d, k)tk .

From Figure 3.1 we see that when d = 3 we have the Eulerian polynomial A(3, t) = 1+4t+t2.Eulerian numbers play an important role in the Ehrhart theory of parallelepipeds. For

example, the Eulerian polynomial is the �-polynomial of the unit cube. That is,

�(Cd, t) = A(d, t) . (3.1)

See [2] for example, where Eulerian numbers are in fact defined via this equality. As wewill see, variants of the Eulerian numbers also appear in the �-vector for other families ofparallelepipeds. Let us motivate the results described in later sections and the methods usedto prove them by sketching a proof of (3.1).

The main idea behind our proof of (3.1) is the decomposition of Cd into disjoint half-openlattice polytopes, P

1

, . . . ,Pr, whose �-polynomials we already know. If

Cd =G

i2[r]

Pi ,

where the Pi are lattice d-polytopes, then

Ehr(Cd, t) =X

i2[r]

Ehr(Pi, t) =X

i2[r]

�(Pi, t)

(1� t)d+1

=

P

i2[r] �(Pi, t)

(1� t)d+1

,

which implies�(Cd, t) =

X

i2[r]

�(Pi, t) . (3.2)

The �-vector of a polytope encodes information about the lattice point count of thatpolytope. Thus it is necessary that the Pi in the decomposition are disjoint. If the Pi arenot disjoint and they share any lattice points, then the sum of the lattice points in the unionwill not reflect the lattice point count of Cd.

To obtain the disjoint union we desire, we decompose Cd into half-open, unimodu-

16

4a = conv {(0, 0), (1, 1)} 4b = conv {(0, 0), (1, 1), (1, 0)} 4c = 4b \4a

Figure 3.2: A unimodular 1-simplex, a unimodular 2-simplex and a half-open unimodular2-simplex.

lar simplices. A simplex 4 is a d-dimensional polytope with d + 1 vertices. As thenotation suggests, simplices are higher-dimensional analogues of triangles. A d-simplex4 = conv {v

0

,v1

, . . . ,vd} is unimodular if

v1

� v0

,v2

� v0

, . . . ,vd � v0

form a basis for the sub-lattice Zm \ span(4) [14]. In Figure 3.2, 4a and 4b are unimod-ular simplices of dimension 1 and 2 respectively. This is best seen by computing the set{vj � v

0

: j 2 [d]} for each simplex, where v0

is the vertex of your choice. The span of eachsimplex is represented in yellow. For 4a, the intersection span(4a)\Z2 is isomorphic to Z,whereas span(4b) \ Z2 is all of Z2. We say a polytope P is half-open if one or more of itsfacets have been removed. In Figure 3.2, 4c is a half-open simplex with exactly one facetremoved.

A lattice simplex 4 is unimodular if and only if it has normalized volume 1 [14]. Com-bining this with (2.1) and (2.2), we learn that �(4, t) = 1 if and only if 4 is unimodular.Therefore, the Ehrhart series of the unimodular simplices 4a and 4b are

1

(1� t)2and

1

(1� t)3,

respectively, and thus the Ehrhart series of the half-open unimodular simplex 4c is

1

(1� t)3� 1

(1� t)2=

t

(1� t)3.

More generally, the Ehrhart series of a d-dimensional half-open unimodular simplex with kfacets removed is

tk

(1� t)d+1

.

See [14], for example.

17

x2

x1

42

� = {x 2 C2 : x1

x2

}

x2

x1

42

µ = {x 2 C2 : x2

< x1

}

42

� t42

µ = C2

x1

x2

Figure 3.3: The decomposition of C2 into disjoint half-open unimodular simplices.

Let � 2 Sd and define

4d� :=

�

x 2 Cd : x�1 x�2 · · · x�dwith x�i < x�i+1 when i 2 Des(�)

.

A strict inequality in 4d� corresponds to the removal of one facet. So 4d

� is a half-open,unimodular simplex with exactly des(�) facets removed. Every point x in Cd is containedin 4d

� for some permutation �. Furthermore, the strict inequalities at the descents ensurethat the union is disjoint [1, 14]. Thus

Cd =G

�2Sd

4d� .

We see this in Figure 3.3 for the 2-cube. We can now conclude with (3.2) that

�(Cd, t) =X

�2Sd

�(4d�, t) =

X

�2Sd

tdes(�) =d�1

X

k=0

a(d, k)tk = A(d, t) .

That is, the �-polynomial of Cd is the Eulerian polynomial, and this proves (3.1).It is well known that the Eulerian polynomial is palindromic. Equivalently we say that

A(d, t) is symmetric with center of symmetry at d�1

2

and write

a(d, k) = a(d, d� 1� k)

for all d � 1 and 0 k d � 1. This symmetry is easily seen via a bijection betweenpermutations in Sd with k descents and permutations in Sd with d� 1� k descents:

: Sd ! Sd

� 7! �rev := �d�d�1

· · · �1

.

Observe the symmetry via permutation diagrams of � = 4213 and �rev = 3124 in Figure 3.5.In addition to being palindromic, the Eulerian polynomial is also unimodal (see Theorem

3.3 below). Together these imply that A(d, t) is alternatingly increasing. More generally,the symmetry and unimodality of a polynomial together imply that it must peak at the

18

123 213 132 312 231 321

A1

(3, t) = 1 + t A2

(3, t) = 2t A3

(3, t) = t+ t2

Figure 3.4: The permutation diagrams of S3

grouped by last letter reveal the (A, j)-Eulerianpolynomials for d = 3.

coe�cient(s) closest to the center of symmetry. Therefore, when d is odd, A(d, t) has asingle peak at k = d�1

2

, and when d is even, A(d, t) has a double peak at k =⌅

d�1

2

⇧

andk =

⌅

d�1

2

⇧

+ 1. (This is confirmed by Theorem 3.3.)

3.2 Half-Open Unit Cubes

We begin this section by introducing another descent statistic and a refinement of the Eule-rian numbers. For d � 1, 1 j d, and k 2 Z we define the (A, j)-Eulerian number

aj(d, k) := |{� 2 Sd : �d = d+ 1� j and des(�) = k}|

and the (A, j)-Eulerian polynomial

Aj(d, t) :=d�1

X

k=0

aj(d, k)tk .

For all k < 0 and for all k � d we have aj(d, k) = 0. We treat Aj(d, t) as having degree d�1,though its leading term may be zero. See Figure 3.4 for the (A, j)-Eulerian polynomials ford = 3.

The following lemma is due to [5]. We elaborate on the details of their argument in theproof below which gives a bijection between permutations � 2 Sd with k descents and lastletter d + 1 � j and permutations �flip 2 Sd with d � 1 � k descents and last letter j. Theresult appears later in [10, Lemma 2.3], though in this case the coe�cients of Aj(d, t) arenot defined in terms of descent statistics on Sd.

Lemma 3.1 [5, Lemma 2.5]. For all d � 1, 1 j d and k 2 Z,

aj(d, k) = ad+1�j(d, d� 1� k) , equivalently

Aj(d, t) = td�1Ad+1�j

✓

d,1

t

◆

.

19

Proof. Fix d � 1. Consider the bijection

: Sd ! Sd

� 7! �flip ,

where �flip

i := d+ 1� �i (for a diagram see Figure 3.5). Then

i 2 Des(�) () �i > �i+1

() d+ 1� �i < d+ 1� �i+1

() �flip

i < �flip

i+1

() i 2 Asc�

�flip

�

.

Therefore,

des(�) = k () asc�

�flip

�

= k () des�

�flip

�

= d� 1� k .

Also observe �d = d+ 1� j if and only if �flip

d = j. We conclude

aj(d, k) = ad+1�j(d, d� 1� k) .

This gives us the following:

td�1Ad+1�j

✓

d,1

t

◆

= td�1

d�1

X

k=0

ad+1�j(d, k)t�k

=d�1

X

k=0

ad+1�j(d, k)td�1�k

=d�1

X

k=0

ad+1�j(d, d� 1� k)tk

=d�1

X

k=0

aj(d, k)tk

= Aj(d, t) .

The following lemma gives a recursive formula for the (A, j)-Eulerian numbers and poly-nomials. The result is due to [5]. Once again, we provide an elaboration of their argument.

Lemma 3.2 [5, Lemma 2.5].

aj(d+ 1, k) =j�1

X

l=1

al(d, k � 1) +dX

l=j

al(d, k) , equivalently

Aj(d+ 1, t) = t

j�1

X

l=1

Al(d, t) +dX

l=j

Al(d, t) .

20

� = 4213

1

2

3

4

1 2 3 4

�rev = 3124

1

2

3

4

1 2 3 4

�flip = 1342

1

2

3

4

1 2 3 4

Figure 3.5: Permutation diagrams of �, �rev and �flip respectively.

Proof. Fix d � 1, 1 j d and k 2 Z. Let � 2 Sd+1

with k descents and last letterd+ 2� j. Furthermore, let µ = �

1

�2

. . . �d and K = [d+ 1] \ {d+ 2� j}. Then µ 2 SK , theset of all permutations on the set K. We note that SK is isomorphic to S|K| or Sd.

Case 1: d 2 Des(�).

d 2 Des(�) () �d > �d+1

() �d > d+ 2� j () d+ 3� j �d d+ 1 .

Because � has k descents and d 2 Des(�), µ must have exactly k � 1 descents. Then

|{� 2 Sd+1

: d 2 Des(�) and �d = d+ 2� j and des(�) = k}| =|{µ 2 SK : d+ 3� j µd d+ 1 and des(µ) = k � 1}| =

|{µ0 2 Sd : d+ 2� j µ0d d and des(µ0) = k � 1}| =

j�1

X

l=1

al(d, k � 1) .

Case 2: d /2 Des(�).

d /2 Des(�) () �d < �d+1

() �d < d+ 2� j () 1 �d d+ 1� j .

Furthermore, des(�) = k and d /2 Des(�) implies µ has k descents. This implies

|{� 2 Sd+1

: d /2 Des(�) and �d = d+ 2� j and des(�) = k � 1}| =|{µ 2 SK : 1 µd d+ 1� j and des(µ) = k}| =

dX

l=j

al(d, k) .

We conclude

aj(d+ 1, k) =j�1

X

l=1

al(d, k � 1) +dX

l=j

al(d, k) ,

21

from which it follows that

Aj(d+ 1, t) =dX

k=0

aj(d+ 1, k)tk

=dX

k=0

"

j�1

X

l=1

al(d, k � 1) +dX

l=j

al(d, k)

#

tk

=dX

k=0

j�1

X

l=1

al(d, k � 1)tk +dX

k=0

dX

l=j

al(d, k)tk

=j�1

X

l=1

"

dX

k=0

al(d, k � 1)tk#

+dX

l=j

"

dX

k=0

al(d, k)tk

#

=j�1

X

l=1

"

d�1

X

k=0

al(d, k)tk+1

#

+dX

l=j

"

dX

k=0

al(d, k)tk

#

= t

j�1

X

l=1

"

d�1

X

k=0

al(d, k)tk

#

+dX

l=j

"

dX

k=0

al(d, k)tk

#

= t

j�1

X

l=1

Al(d, t) +dX

l=j

Al(d, t) .

The following theorem is due to Jochemko. It can be seen as a reincarnation of theunimodality results for closed parallelepipeds by Schepers and Van Langenhoven in [10].The proof is taken from [8].

Theorem 3.3 [8, Theorem 3.2.2]. For all d � 1, 1 j d, the coe�cients of Aj(d, t) areunimodal. More specifically, in the case that d is even we have

aj(d, 0) . . . aj(d,d2

� 1) � . . . � aj(d, d� 1) if 1 j d2

,aj(d, 0) . . . aj(d,

d2

) � . . . � aj(d, d� 1) if d2

< j d,

and if d � 3 is odd we have

a1

(d, 0) . . . a1

(d, bd2

c � 1) = a1

(d, bd2

c) � . . . � a1

(d, d� 1, )ad(d, 0) . . . ad(d, bd

2

c) = ad(d, bd2

c+ 1) � . . . � ad(d, d� 1),

aj(d, 0) . . . aj(d, bd2

c) � . . . � aj(d, d� 1) if 2 j d� 1 .

Proof. When d = 1, the result is trivially true. For d � 2, we argue by induction on d. Thecase d = 2 is easily checked. For d = 3, we see from Figure 3.4 that the result holds.

Let d+ 1 be even. We then distinguish two cases:Case: 1 j < d+1

2

. Then

Aj(d+ 1, t) = t

j�1

X

l=1

Al(d, t) +d+1�jX

l=j

Al(d, t) +dX

l=d+2�j

Al(d, t)

22

by Lemma 3.2. The first and the third summand added give by Lemma 3.1 a palindromicpolynomial with center of symmetry at d

2

which, by induction, has unimodal coe�cients withpeaks at

⌅

d2

⇧

and⌅

d2

⇧

+1. The second summand has by induction unimodal coe�cients withpeak at

⌅

d2

⇧

= d+1

2

� 1.Case: d+1

2

< j d+ 1 . Then

Aj(d+ 1, t) = t

d+1�jX

l=1

Al(d, t) + t

j�1

X

l=d+2�j

Al(d, t) +dX

l=j

Al(d, t) .

The first and the third summand added give a palindromic polynomial with center of sym-metry at d

2

, which has unimodal coe�cients with peaks at⌅

d2

⇧

and⌅

d2

⇧

+1. In this case, thecoe�cients of the second summand form a unimodal sequence with peak at

⌅

d2

⇧

+ 1 = d+1

2

.If d+ 1 is odd, we distinguish again two cases:Case: 1 j d+1

2

. By Lemma 3.2 we have

Aj(d+ 1, t) = t

j�1

X

l=1

Al(d, t) +d+1�jX

l=j

Al(d, t) +dX

l=d+2�j

Al(d, t) .

The second summand is by induction and Lemma 3.1 a palindromic polynomial with uni-modal coe�cients and peaks at d

2

� 1 and d2

. The coe�cients of the first and third summandare unimodal with peak at d

2

=⌅

d+1

2

⇧

.Case: d+1

2

< j d+ 1 . We have

Aj(d+ 1, t) = t

d+1�jX

l=1

Al(d, t) + t

j�1

X

l=d+2�j

Al(d, t) +dX

l=j

Al(d, t) .

As in the previous case, the coe�cients of the summand in the middle are unimodal andpalindromic, this time with peaks at d

2

and d2

+ 1. The coe�cients of the first and thirdsummand form again a unimodal sequence with peak at d

2

=⌅

d+1

2

⇧

.

Theorem 3.3 proves that the coe�cients of the (A, j)-Eulerian polynomials are unimodaland specifies the peak(s). From the proof of Proposition 2.17 in [10], we also know thatthe coe�cients of these polynomials are alternatingly increasing for su�ciently large j. Weformally record this result and provide a proof, an extension of the proof of Theorem 3.3,below.

Lemma 3.4. For all d � 0 and⌅

d+1

2

⇧

< j d + 1, the coe�cients of Aj(d + 1, t) arealternatingly increasing.

Proof. Fix d � 0 and⌅

d+1

2

⇧

< j d+ 1. Let

b(t) = t

d+1�jX

l=1

Al(d, t) +dX

l=j

Al(d, t)

23

and

c(t) =j�1

X

l=d+2�j

Al(d, t) .

Then by Lemma 3.2

Aj(d+ 1, t) = b(t) + t c(t) and aj(d+ 1, k) = bk + ck�1

,

where bi and ci are the coe�cient of ti in b(t) and c(t) respectively, and c�1

:= 0. We wishto show that the coe�cients of Aj(d+1, t) are alternatingly increasing (for our given valuesof j). Equivalently, we will show that

aj(d+ 1, k) aj(d+ 1, d� k) for 0 k ⌅

d2

⇧

, (3.3)

andaj(d+ 1, d� k) aj(d+ 1, k + 1) for 0 k

⌅

d�1

2

⇧

. (3.4)

By Lemma 3.1 and Lemma 3.3 we know b(t) is symmetric and unimodal with peak(s) ofunimodality in agreement with center of symmetry at d

2

; and c(t) is symmetric and unimodalwith peak(s) of unimodality in agreement with center of symmetry at d�1

2

.Fix 0 k

⌅

d2

⇧

. Then by the symmetry and unimodality of b(t) and c(t),

aj(d+ 1, k) = bk + ck�1

= bd�k + cd�k bd�k + cd�k�1

= aj(d+ 1, d� k) .

This establishes (3.3).Now fix 0 k

⌅

d�1

2

⇧

. Then

aj(d+ 1, d� k) = bd�k + cd�k�1

= bk + ck bk+1

+ ck = aj(d+ 1, k + 1) ,

also by the symmetry and unimodality of b(t) and c(t). This establishes (3.4). We concludethat Aj(d+ 1, t) is alternatingly increasing for

⌅

d+1

2

⇧

< j d+ 1.

For j 2 {0, . . . , d} we define the half-open unit cube

Cdj := [0, 1]d \ {xd = 1, xd�1

= 1, . . . , xd+1�j = 1} .

The subscript j indicates the number of facets removed from Cd to obtain Cdj . We note that

for j > 0, the intersection of the removed facets of Cdj is non-empty.

We define the j-descent set Desj(�) ✓ {1, . . . , d} of a permutation � 2 Sd by

Desj(�) :=

(

Des(�) [ {d} if d+ 1� j �d d ,

Des(�) otherwise .

Further, we define the j-descent number desj(�) to be the cardinality of the j-descent set.We can now describe the (A, j)-Eulerian numbers in terms of j-descents, as the following

lemma shows us. The result is due to Jochemko and appears in [8].

24

C3

0

x1

x2

x3

C3

1

C3

2

C3

3

Figure 3.6: Half-open unit cubes C3

j for j = 0, 1, 2, 3.

Lemma 3.5 [8, Lemma 3.3.1]. Let � 2 Sd. Then for all 0 j d and 0 k d

|{� 2 Sd : desj(�) = k}| = aj+1

(d+ 1, k) .

Proof. Fix 0 j d and 0 k d. Consider � 2 Sd with desj(�) = k. By definition

desj(�) =

(

des(�) + 1 if d+ 1� j �d d ,

des(�) otherwise .

If d+1�j �d d, then desj(�) = k implies that des(�) = k�1. If 1 �d d�j, thendesj(�) = k implies that des(�) = k. So the number of permutations in Sd with j-descentnumber k is equal to the number of permutations in Sd with descent number k � 1 and lastletter between d+1�j and d (inclusive) plus the number of permutations in Sd with descentnumber k and last letter strictly less than d+ 1� j. Equivalently,

|{� 2 Sd : desj(�) = k}| =jX

l=1

al(d, k � 1) +dX

l=j+1

al(d, k)

= aj+1

(d+ 1, k) ,

where the last equality follows from Lemma 3.2.

Remark. Lemma 3.5 also holds as a result of the proof of equations (4.3) and (4.4). Thebijection we use in the proof is a modified version of that used by Jochemko in [8].

The following result comes from [8]; it is a generalization of property (1) of Lemma 2.5in [10]. We provide an altered version of Jochemko’s proof below.

Proposition 3.6 [8, Theorem 3.3.2]. Let d � 1 and 0 j d. Then

Ehr�

Cdj , t

�

=Aj+1

(d+ 1, t)

(1� t)d+1

.

Proof. Let d � 1, 0 j d and � 2 Sd. Define the half-open unimodular simplex

4d,j� :=

(

x 2 Cdj : x�1 x�2 · · · x�d

with x�i < x�i+1 when i 2 Des(�)

)

.

25

This gives us the disjoint union,Cd

j =G

�2Sd

4d,j� .

We also have the following equivalent definitions

4d,j� =

8

>

<

>

:

x 2 Rd : 0 x�1 x�2 · · · x�d 1

with x�i < x�i+1 when i 2 Des(�)

and x�d< 1 when d+ 1� j �d d

9

>

=

>

;

=

8

>

<

>

:

x 2 Rd : 0 x�1 x�2 · · · x�d 1

with x�i < x�i+1 when i 2 Desj(�)

and x�d< 1 when d 2 Desj(�)

9

>

=

>

;

.

Each strict inequality corresponds bijectively to a missing facet in the simplex. We seethat 4d,j

� has des(�) + 1 missing facets when d + 1 � j �d and des(�) missing facetsotherwise. Therefore, the half-open unimodular simplex 4d,j

� has exactly desj(�) missingfacets. Together with the disjoint union, this implies

Ehr�

Cdj , t

�

=X

�2Sd

Ehr�

4d,j� , t

�

=

P

�2Sd��

4d,j� , t

�

(1� t)d+1

=

P

�2Sdtdesj(�)

(1� t)d+1

=

Pdk=0

aj+1

(d+ 1, k)tk

(1� t)d+1

=Aj+1

(d+ 1, t)

(1� t)d+1

,

where the second to last inequality follows from Lemma 3.5.

This proposition tells us that the �-polynomial for the half-open unit cube Cdj is the

(A, j)-Eulerian polynomial Aj+1

(d + 1, t). Together with Theorem 3.3 and Lemma 3.4, wesee that the �-polynomial for Cd

j is unimodal for all 0 j d and alternatingly increasingfor

⌅

d+1

2

⇧

j d.

3.3 Half-Open Parallelepipeds

Let v1

, . . . ,vd be linearly independent vectors in Zd and let J ✓ [d]. Then we define respec-tively the closed parallelepiped ⌃(J), the standard half-open parallelepiped ⇧(J),

26

v1

= (1,�1)

v2

= (2, 1)

⌃([2]) {1}([2]) {2}([2]) {1,2}([2]) = ⇧([2])

Figure 3.7: A closed 2-dimensional parallelepiped and its half-open variants.

and the open parallelepiped 2(J) generated by J [10]:

⌃(J) :=(

X

j2J

�jvj : 0 �j 1

)

,

⇧(J) :=

(

X

j2J

�jvj : 0 �j < 1

)

,

2(J) :=

(

X

j2J

�jvj : 0 < �j < 1

)

.

Let Fi be the facet of ⌃(J) generated by J \ {i} and let F 0i be the translate of Fi in ⌃(J):

Fi := ⌃ (J \ {i}) and F 0i := vi + Fi .

Note that⇧(J) = ⌃(J) \

[

i2J

F 0i

and2(J) = ⌃(J) \

[

i2J

(Fi [ F 0i ) .

Let I ✓ J ✓ [d]. Then we define

I(J) :=

(

X

j2J

�jvj : 0 �j 1 with �j < 1 for all j 2 I

)

to be the half-open parallelepiped generated by J with omitted facets F 0i , for all i 2 I.

In Figure 3.7 we see a closed 2-dimensional parallelepiped and its half-open variants.The lemma below is a modification of a well-known result which states that a polytope

is the disjoint union of the relative interiors of its faces. That is,

P =G

F✓P

F� .

27

The lemma also appears in a more general form in [8]. The stronger result seen there appliesmore generally to Zd-valuations1.

Lemma 3.7 [8, Lemma 3.4.3]. Let J ✓ [d]. Then

#�

⇧(J) \ Zd�

=X

;✓J 0✓J

#�

2(J 0) \ Zd�

.

Lemma 3.8 below is a generalization of [10, Lemma 2.1]. The original result is for closed(lattice) parallelepipeds ⌃ ([d]) = ; ([d]). Jochemko extends it to half-open (lattice) par-allepipeds in [8, Lemma 3.4.1]. In fact, she proves a stronger statement which applies toarbitrary Zd-valuations. We follow the proof from [10] for closed parallelepipeds, adjustingas necessary for the half-open condition.

Lemma 3.8 [10, Lemma 2.1]. Let I ✓ [d]. Then

#�

n I ([d]) \ Zd�

=X

J 0: I✓J 0✓[d]

n|J 0| #�

⇧(J 0) \ Zd�

.

Proof. Let I ✓ [d], 2(nJ) := n2(J), and k0 := |{[d] \ (J [ I)}|. Then

#�

n I ([d]) \ Zd�

=X

J✓[d]

2k0#�

2(nJ) \ Zd�

=X

J✓[d]

X

J✓J 0✓[d] and I✓J 0

#�

2(nJ) \ Zd�

=X

J,J 0✓[d]: J✓J 0and I✓J 0

#�

2(nJ) \ Zd�

=X

J 0✓[d]: I✓J 0

#�

⇧(nJ 0) \ Zd�

=X

J 0: I✓J 0

n|J 0| #�

⇧(J 0) \ Zd�

.

The second to last equality follows from Lemma 3.7. The last equality follows becausen⇧(J 0) is covered by ndim(⇧(J 0

)) translates of ⇧(J 0) and the dimension of ⇧(J 0) is preciselythe order of the generating set J 0.

Corollary 3.9 [8, Corollary 3.4.2]. Let d � 1, 0 j d and n 2 Z>0

. Then

ehr�

Cdj , n

�

=X

[j]✓J✓[d]

n|J | ,

where we define [0] := ;.1A Zd-valuation is a map ' from the set of all lattice polytopes in Rd to an abelian group such that

' (;) = 0 and ' (P \Q) = ' (P) + ' (Q) � ' (P \Q) for all lattice polytopes P,Q such that P [ Q isalso a lattice polytope; additionally, a Zd-valuation ' is invariant under translation by vectors in Zd, so' (P + v) = ' (P) for all lattice polytopes P and all v 2 Zd [8].

28

Proof. Fix d � 1 and 0 j d. Let n 2 Z>0

and vi = ei for all i 2 [d]. Then Cdj is

congruent to[j]([d]). By Lemma 3.8

ehr�

Cdj , n

�

=X

[j]✓J✓[d]

n|J | #�

⇧(J) \ Zd�

=X

[j]✓J✓[d]

n|J | ,

where the last equality holds because the origin is the only lattice point in the half-open unitcube ⇧(J).

The following theorem and corollary are due to [8], as are their proofs. The theorem isa generalization of [10, Proposition 2.2].

Theorem 3.10 [8, Theorem 3.4.4]. Let I ✓ [d]. Then the Ehrhart series of the half-openparallelepiped I([d]) is

Ehr ( I ([d]) , t) =

P

K✓[d] #�

2(K) \ Zd�

A|I[K|+1

(d+ 1, t)

(1� t)d+1

.

Proof.

Ehr ( I ([d]) , t) = 1 +X

n�1

ehr ( I ([d]) , n) tn by definition

=X

n�0

tnX

I✓J

n|J | #�

⇧(J) \ Zd�

by Lemma 3.8

=X

n�0

tnX

I✓J

n|J |X

K✓J

#�

2(K) \ Zd�

by Lemma 3.7

=X

K✓[d]

#�

2(K) \ Zd�

X

n�0

tnX

(I[K)✓J

n|J |

=X

K✓[d]

#�

2(K) \ Zd�

X

n�0

tn ehr�

Cd|I[K|, n

�

by Corollary 3.9

=

P

K✓[d] #�

2(K) \ Zd�

A|I[K|+1

(d+ 1, t)

(1� t)d+1

by Proposition 3.6.

The third to last equality holds because for a fixed subset K and positive integer n, thenumber of times that #

�

2(K) \ Zd�

is counted in both expressions is equal to the numberof supersets J of I that also contain K.

Corollary 3.11 [8, Corollary 3.4.5]. The �-polynomial of the d-dimensional half-open par-allelepiped I([d]) is unimodal with peak at d

2

if d is even and with peak at d�1

2

or d+1

2

if d isodd.

Proof. From Theorem 3.10 we have

� ( I([d]), t) =X

K✓[d]

#�

2(K) \ Zd�

A|I[K|+1

(d+ 1, t) .

29

Furthermore #�

2(K) \ Zd�

� 0 for all K ✓ [d]. By Theorem 3.3, the coe�cients ofA|I[K|+1

(d + 1, t) form a unimodal sequence with peak at⌅

d+1

2

⇧

= d2

if d is even, and peakat d+1

2

� 1 = d�1

2

or d+1

2

if d is odd. The same is true for the coe�cients of any nonnegativelinear combination of the A|I[K|+1

(d+ 1, t) where K ✓ [d].

Theorem 3.10 also implies the unimodality of the �-vector for lattice zonotopes. This isrecorded formally in the following corollary, a weaker version of Jochemko’s result from [8].

Corollary 3.12 [8, Theorem 3.5.4]. The �-polynomial of a d-dimensional zonotope is uni-modal with peak at d

2

if d is even and with peak at d�1

2

or d+1

2

if d is odd.

In order to prove this result we require a corollary to Theorem 2.1. The corollary isdue to Jochemko and can be found in [8]. Her proof relies on the Beneath and Beyondconstruction of Koppe and Verdoolaege [9] (stated here in terms of visible faces) which givesrise to disjoint half-open decompositions of polytopes for an appropriately chosen referencepoint p 2 Rd. We motivate and outline Jochemko’s proof below.

Corollary 3.13 [8, Corollary 3.5.3]. Every d-zonotope admits a disjoint decomposition intohalf-open d-parallelepipeds of the -type.

We say a face F of a polytope P is visible2 from p 2 Rd if

[p,x] \ P = {x}

for all x 2 F . In part (a) of Figure 3.8, all faces of the 2-parallelepiped P that are visiblefrom p

1

/2 P appear in blue. We note that identifying all visible facets of P is su�cient inidentifying all visible faces of P . This is true in general, provided the reference point lies onno facet-defining hyperplane of the polytope [3]. In part (b), the facets of P that are notvisible from p

1

appear in red. Observe that if facet F is visible from p1

, then facet F 0, thetranslate of F in P , is not visible from p

1

. This also holds in general [3]. In part (c), we seethat no faces of P are visible from p

2

2 P�.From a subdivision C into parallelepipeds of a d-zonotope Z ⇢ Rd we obtain the dissec-

tionZ = ⌃

1

[ ⌃2

[ · · · [ ⌃s ,

where ⌃1

,⌃2

, . . . ,⌃s are the maximal cells of C . This dissection, in utilizing our notion ofvisible faces from a fixed reference point, gives rise to a disjoint decomposition of Z intohalf-open variants of the ⌃i.

Fix p 2 Z� such that p lies on no facet-defining hyperplane of any of the maximal cellsin C . For each i 2 [s], the half-open d-parallelepiped ⌃0

i is obtained by removing those facesof ⌃i that are visible from p:

⌃0i := ⌃i \

[

F is visible from p

F .

By construction, no two of the removed facets in ⌃0i are translates of one another. Thus

the intersection of the removed facets is empty and we rewrite ⌃0i as i, as ⌃0

i is indeed a

2Alternatively, we say point p 2 Rd is beyond F .

30

p1

P

(a)

⇥⇥

⇥⇥

p1

(b)

p2

(c)

Figure 3.8: (a) All faces of P that are visible from p1

/2 P appear in blue; (b) The two facetsof P that are not visible from p

1

appear in red; (c) No faces of P are visible from p2

2 P�.

p

Figure 3.9: A 2-zonotope Z, a subdivision of Z into parallelepipeds, and a disjoint decom-position of Z into half-open 2-parallelepipeds.

half-open parallelepiped of the -type. This gives us the following disjoint decomposition ofd-zonotope Z into d-parallelepipeds of the -type:

Z =1

t2

t · · · t s .

See Figure 3.9 for an example in dimension 2 of how a subdivision into parallelepipeds anda reference point p give rise to such a decomposition.

Proof of Corollary 3.12. Let Z be a d-dimensional zonotope. By Corollary 3.13 there existsa disjoint decomposition of Z into half-open d-dimensional parallelepipeds of the -type, say

Z =G

i2[s]i ,

which implies�(Z, t) =

X

i2[s]

�( i, t) .

31

By Corollary 3.11, we know �( i, t) is unimodal for all i 2 [s] with peak depending on thedimension of i. But all i have the same dimension, so all �( i, t) have the same peak. Itfollows that any sum of the �-polynomials is unimodal with peak equal to that of each ofthe �( i, t), and so �(Z, t) is unimodal with peak at d

2

if d is even and with peak at d�1

2

ord+1

2

if d is odd.

32

Chapter 4

Half-Open ±1-Cubes and the(B, `)-Eulerian Numbers

4.1 Signed Permutation Descent Statistics

A signed permutation on [d] is a pair (⇡, ✏) with ⇡ 2 Sd and ✏ 2 {±1}d. To each letter⇡i in the permutation word ⇡ we assign the sign of ✏i, the ith entry of ✏. For a given d, theset of signed permutations is denoted by Bd and has 2d · d! elements. We will use one-linenotation to denote signed permutation words with the following convention: those lettersassociated with a negative sign will be followed by an accent mark. So for d = 5, ⇡ = 42135and ✏ = (�1,�1, 1,�1, 1) we write (⇡, ✏) = 40201305.

Let ⇡0

:= 0 and ✏0

:= 1 for all (⇡, ✏) 2 Bd and all d � 1. Then i 2 [d � 1] [ {0}is a (natural) descent of (⇡, ✏) 2 Bd if ✏i⇡i > ✏i+1

⇡i+1

. So 0 and 3 are the descents ofsigned permutation word 40201305. We easily observe the descents, seen here in red, from thetwo-line notation for 40201305:

0 1 2 3 4 50 �4 �2 1 �3 5

�

.

We define the (natural) descent set and the (natural) descent number of (⇡, ✏) 2Bd, respectively, as follows:

NatDes(⇡, ✏) := {i 2 [d� 1] [ {0} : ✏i⇡i > ✏i+1

⇡i+1

} and

natdes(⇡, ✏) := |NatDes(⇡, ✏)| .

Notice that with 0 as a possible descent we have 0 natdes(⇡, ✏) d for all (⇡, ✏) 2 Bd.This di↵ers from the descent statistic des(�) for permutations in Sd where 0 des(�) d�1for all � 2 Sd.

The number of signed permutations on [d] with exactly k descents is a descent statisticon Bd. We call these descent statistics the type-B Eulerian numbers and write

b(d, k) := | {(⇡, ✏) 2 Bd : natdes(⇡, ✏) = k} | .

33

The type-B Eulerian polynomial is

B(d, t) :=dX

k=0

b(d, k)tk .

Remark. The descent statistic natdes on signed permutations in Bd agrees with the descentstatistic des on permutations in Sd when we fix the sign vector ✏ = (1, 1, . . . , 1).

Consider the d-dimensional ±1�cube,

[�1, 1]d :=�

x 2 Rd : �1 xi 1 for all i

,

and the half-open unimodular simplex indexed by the signed permutation (⇡, ✏) 2 Bd,

4d(⇡,✏) :=

(

x 2 [�1, 1]d : 0 ✏1

x⇡1 · · · ✏dx⇡d 1

with ✏ix⇡i < ✏i+1

x⇡i+1 when i 2 NatDes(⇡, ✏)

)

.

The definitions imply the following disjoint union

[�1, 1]d =G

(⇡,✏)2Bd

4d(⇡,✏) ,

see [1]. We also observe 4d(⇡,✏) has k missing facets if and only if natdes(⇡, ✏) = k. Therefore,

Ehr�

[�1, 1]d, t�

=X

(⇡,✏)2Bd

Ehr�

4d(⇡,✏), t

�

=X

(⇡,✏)2Bd

�(4d(⇡,✏), t)

(1� t)d+1

=

P

(⇡,✏)2Bdtnatdes(⇡,✏)

(1� t)d+1

=B(d, t)

(1� t)d+1

.

In this way we see that the Ehrhart �-polynomial for [�1, 1]d is the type-B Eulerianpolynomial B(d, t) [4, Theorem 3.4]; that is,

��

[�1, 1]d, t�

= B(d, t) . (4.1)

The coe�cient of tk counts the half-open unimodular simplices with exactly k facets removedin the Bd-induced decomposition of [�1, 1]d, or equivalently, the signed permutations in Bd

with exactly k descents. See Figure 4.1 for this decomposition with d = 2.We wish to show that B(d, t) is alternatingly increasing using the interplay of the ge-

ometry provided by the ±1-cube and the combinatorics provided by relevant permutation

34

412

421

42

01

0

41

02

0

421

0

412

0

42

01

41

02

Figure 4.1: Decomposition of [�1, 1]2 into disjoint half-open unimodular simplices indexedby signed permutations in B

2

.

descent statistics. To do this we introduce the unit cell indexed by I ✓ [d],

UdI :=

�

x 2 [�1, 1]d : xi � 0 for all i 2 I and xi < 0 for all i /2 I

.

Notice that UdI is a half-open unit d-cube with d�|I| facets removed and that the intersection

of the removed facets is non-empty, so no two of the removed facets are translates of oneother. Therefore, Ud

I is congruent to Cdj where j = d� |I|. We write Ud

I⇠= Cd

j . There are 2d

unit cells indexed by the 2d subsets of [d], exactly�

dj

�

of which are congruent to Cdj .

The union of the unit cells is the d-dimensional ±1-cube. Furthermore, the union isdisjoint. That is,

[�1, 1]d =G

I✓[d]

UdI .

See Figure 4.2 for this decomposition with d = 2. Therefore,

Ehr�

[�1, 1d], t�

=X

I✓[d]

Ehr�

UdI , t

�

=X

I✓[d]

Ehr�

Cdd�|I|, t

�

=X

I✓[d]

�⇣

Cdd�|I|, t

⌘

(1� t)d+1

=

Pdj=0

�

dj

�

��

Cdj , t

�

(1� t)d+1

=

Pdj=0

�

dj

�

Aj+1

(d+ 1, t)

(1� t)d+1

,

where the last equality follows from Proposition 3.6. From this we arrive at the followingtheorem and proof for the corollary.

35

U2

{1,2}⇠= C2

0

U2

{1}⇠= C2

1

U2

{2}⇠= C2

1

U2

;⇠= C2

2

Figure 4.2: Decomposition of [�1, 1]2 into unit cells indexed by subsets of {1, 2}.

Theorem 4.1.

B(d, t) =dX

j=0

✓

d

j

◆

Aj+1

(d+ 1, t) and

b(d, k) =dX

j=0

✓

d

j

◆

aj+1

(d+ 1, k) .

Corollary 4.2 [4, Theorem 2.4]. The type-B Eulerian numbers are symmetric and unimodal.In particular, they are alternatingly increasing.

Proof. In order to show that the coe�cients of B(d, t) are alternatingly increasing, we willconsider B(d, t), as in Theorem 4.1, as a positive linear combination of the (A, j)-Eulerianpolynomials. We will show that this linear combination is both symmetric and unimodal,thus implying that the center of symmetry is in agreement with the peak(s) of unimodality.We will also see that the peaks are in the middle, further implying that the coe�cients ofB(d, t) are alternatingly increasing. We proceed by cases.

Case 1: d is odd.We rearrange B(d, t) into a linear combination of polynomial pairs:

dX

j=0

✓

d

j

◆

Aj+1

(d+ 1, t) =

d+12 �1

X

j=0

✓

d

j

◆

Aj+1

(d+ 1, t) +dX

j= d+12

✓

d

d� j

◆

Ad+1�j(d+ 1, t)

=

d+12 �1

X

j=0

✓

d

j

◆

[Aj+1

(d+ 1, t) + Ad+1�j(d+ 1, t)] . (4.2)

From Lemma 3.1 we know the polynomial Aj+1

(d+1, t)+Ad+1�j(d+1, t) is symmetric withcenter of symmetry at d

2

=⌅

d+1

2

⇧

. Furthermore, from Theorem 3.3 we know that for alld � 1 odd, the polynomial Aj+1

(d+ 1, t) is unimodal with peak at

k =

(

d+1

2

� 1 if 1 j + 1 d+1

2

() 0 j d+1

2

� 1 ,d+1

2

if d+1

2

< j + 1 d+ 1 () d+1

2

j d .

36

Therefore, the first polynomial in a pair is unimodal with peak at d+1

2

� 1 and the secondpolynomial in a pair is unimodal with peak at d+1

2

. After adding the two polynomials in a

pair, the coe�cient of td+12 �1 will be equal to the coe�cient of t

d+12 and the resulting sum

will be unimodal with double peak at d+1

2

� 1 and d+1

2

.

The coe�cient�

dj

�

is nonnegative for all d and j and thus will not a↵ect the symmetryor the unimodality of the sum of each polynomial pair. It follows that the entire sum in(4.2) is symmetric and unimodal with centers of symmetry in agreement with the peaks ofunimodality. Because the peaks are in the middle, the entire sum is alternatingly increasingwhen d is odd.

Case 2: d is even.Once again we rearrange B(d, t):

dX

j=0

✓

d

j

◆

Aj+1

(d+ 1, t) =

d2�1

X

j=0

✓

d

j

◆

[Aj+1

(d+ 1, t) + Ad+1�j(d+ 1, t)]

+

✓

dd2

◆

A d2+1

(d+ 1, t) .

As before, the sum of each polynomial pair is symmetric with center of symmetry at d2

.By Theorem 3.3 we know that Aj+1

(d + 1, t) is unimodal with peak at⌅

d+1

2

⇧

= d2

for allvalues of j with d � 2 and even. By the same reasoning, the polynomial A d

2+1

(d + 1, t) is

also symmetric and unimodal with center of symmetry and peak in agreement at d2

. The

nonnegative coe�cient�

dj

�

will not a↵ect the symmetry or the unimodality of A d2+1

(d+1, t)or of the sum of the polynomial pairs. Therefore, the entire sum is symmetric and unimodalwith center of symmetry equal to the peak of unimodality. Because the peak is in the middlewe conclude that the entire sum is alternatingly increasing.

Remark. Corollary 4.2 can also be deduced from Proposition 2.17 in [10] and equation (4.1).The proposition states that the �-vector for a lattice parallelepiped with at least one interiorpoint is alternatingly increasing. Our UI-induced decomposition of the ±1-cube provides ageometric proof of the result for this specific class of parallelepipeds.

4.2 Half-Open ±1-Cubes

We begin this section by introducing the (B, `)-Eulerian numbers, a refinement of the type-BEulerian numbers defined by

b`(d, k) := {(⇡, ✏) 2 Bd : ✏d⇡d = d+ 1� ` and natdes(⇡, ✏) = k} ,

where 1 ` d. We further introduce the (B, `)-Eulerian polynomial,

B`(d, t) =dX

k=0

b`(d, k)tk .

37

[�1, 1]20

[�1, 1]21

[�1, 1]22

Figure 4.3: Half-open ±1-cubes [�1, 1]2` for ` = 0, 1, 2, respectively.

What else do the (B, `)-Eulerian numbers count? To answer this question we introducethe half-open ±1-cube [�1, 1]d` , where d � 1 and 0 ` d. This object is the ±1-cubewith ` non-translate facets removed:

[�1, 1]d` : = [�1, 1]d \ {xd = 1, xd�1

= 1, . . . , xd+1�` = 1}=�

x 2 Rd : �1 xi 1 with xi < 1 when d+ 1� ` i d

.

See Figure 4.3. Additionally we define the (natural) `-descent set, NatDes`(⇡, ✏) ✓{0, 1, . . . , d}, of signed permutation (⇡, ✏) 2 Bd by

NatDes`(⇡, ✏) :=

(

NatDes(⇡, ✏) [ {d} if d+ 1� ` ✏d⇡d d ,

NatDes(⇡, ✏) otherwise .

The cardinality of this set is the (natural) `-descent number of (⇡, ✏), denoted by

natdes`(⇡, ✏) := |NatDes`(⇡, ✏)| .

We observe that

natdes`(⇡, ✏) =

(

natdes(⇡, ✏) + 1 if d+ 1� ` ✏d⇡d d ,

natdes(⇡, ✏) otherwise .

As with [�1, 1]d, we can decompose [�1, 1]d` into half-open unimodular simplices indexedby signed permutations in Bd. However, we must first adjust our definition of the simplicesto account for the changes in the ±1-cube. Let

4d,`(⇡,✏) :=

(

x 2 [�1, 1]d` : 0 ✏1

x⇡1 · · · ✏dx⇡d 1

with ✏ix⇡i < ✏i+1

x⇡i+1 when i 2 NatDes(⇡, ✏)

)

.

These simplices are disjoint, and their union is [�1, 1]d` :

[�1, 1]d` =G

(⇡,✏)2Bd

4d,`(⇡,✏) .

38

Unlike the half-open simplices 4d(⇡,✏) we introduced earlier, the number of missing facets

of 4d,`(⇡,✏) is not always enumerated by the number of descents of the indexing signed per-

mutation. The number of missing facets of a simplex 4d,`(⇡,✏) that lies on a removed facet of

[�1, 1]d` is one more than its descent number. This follows by construction: 4d,`(⇡,✏) lies on the

hyperplane {xi = 1} if and only if the last signed letter of the permutation is +i.The removed facets of [�1, 1]d` are defined by supporting hyperplanes {xd+1�` = 1}, . . . , {xd = 1}.

Therefore, the simplex4d,`(⇡,✏) has natdes(⇡, ✏)+1 open facets if and only if ✏d⇡d 2 {d+ 1� `, . . . , d}.

It follows that the number of open facets of 4d,`(⇡,✏) is natdes`(⇡, ✏). The `-descent set of (⇡, ✏)

also determines the location of the strict inequalities in the construction of 4d,`(⇡,✏) and thus

the location of the open facets in the simplex itself. We see this in the following equivalentdefinition:

4d,`(⇡,✏) =

8

>

<

>

:

x 2 Rd : 0 ✏1

x⇡1 · · · ✏dx⇡d 1

with ✏ix⇡i < ✏i+1

x⇡i+1 when i 2 NatDes`(⇡, ✏)

and ✏dx⇡d< 1 when d 2 NatDes`(⇡, ✏)

9

>

=

>

;

.

Consider the following map:

' : Bd ! {(⇡, ✏) 2 Bd+1

: ✏d+1

⇡d+1

= d+ 1� `}

⇡ 7!

0

B

@

i 7!

8

>

<

>

:

⇡i if 0 ⇡i d� ` ,

⇡i + 1 if d+ 1� ` ⇡i d ,

d+ 1� ` if i = d+ 1 ,

1

C

A

✏ 7! (✏1

, . . . , ✏d, 1) .

The map ' defines a bijection between signed permutations of order d and signed permuta-tions of order d + 1 with last signed letter ✏d+1

⇡d+1

= d + 1 � `. Furthermore, the descentnumber of the image is equal to the `-descent number of the pre-image. That is,

natdes ('(⇡, ✏)) = natdes`(⇡, ✏) . (4.3)

In fact,NatDes`(⇡, ✏) = NatDes ('(⇡, ✏)) . (4.4)

We prove these equalities in the following argument.Let (⇡, ✏) 2 Bd and i 2 {0, 1, . . . , d� 1}.Case 1. Suppose the signs of the two adjacent signed letters ✏i⇡i and ✏i+1

⇡i+1

aredi↵erent. Then the inequality relation � between ✏i⇡i and ✏i+1

⇡i+1

is determined by thesigns of ✏i and ✏i+1

. These signs are preserved under the map '. So

✏i⇡i � ✏i+1

⇡i+1

() '(✏i⇡i) � '(✏i+1

⇡i+1

) .

Case 2. Suppose the signs of the two adjacent signed letters ✏i⇡i and ✏i+1

⇡i+1

are thesame. Then the inequality relation between ✏i⇡i and ✏i+1

⇡i+1

is preserved if and only if theinequality relation between ⇡i and ⇡i+1

is. In this case we let � represent the inequalityrelation between ⇡i and ⇡i+1

.

39

• Let ⇡i, ⇡i+1

< d+ 1� `. Then '(⇡i) = ⇡i and '(⇡i+1

) = ⇡i+1

, so

⇡i � ⇡i+1

() '(⇡i) � '(⇡i+1

) .

• Let ⇡i, ⇡i+1

� d+ 1� `. Then

⇡i � ⇡i+1

() '(⇡i) = ⇡i + 1 � ⇡i+1

+ 1 = '(⇡i+1

) .

• Let ⇡i < d+ 1� ` and ⇡i+1

� d+ 1� `. Then ⇡i < ⇡i+1

and

'(⇡i) = ⇡i < ⇡i+1

< ⇡i+1

+ 1 = '(⇡i+1

) .

• Let ⇡i � d+ 1� ` and ⇡i+1

< d+ 1� `. Then ⇡i > ⇡i+1

and

'(⇡i) = ⇡i + 1 > ⇡i > ⇡i+1

= '(⇡i+1

) .

We see that the inequality relation � between ⇡i and ⇡i+1

is preserved in all instancesand so is the inequality relation between ✏i⇡i and ✏i+1

⇡i+1

. From Cases 1 and 2 we concludethat for all i 2 {0, 1, . . . , d� 1}

i 2 NatDes(⇡, ✏) () i 2 NatDes('(⇡, ✏)) .

Now consider the dth signed letter ✏d⇡d of (⇡, ✏) 2 Bd. Suppose ✏d = 1 and d + 1 � ` ⇡d d. This implies d 2 NatDes`(⇡, ✏). The dth signed letter of '(⇡, ✏) is

'(✏d⇡d) = ✏d(⇡d + 1) = ⇡d + 1 .

The (d+1)st signed letter of '(⇡, ✏) is d+1�` by definition. We assumed d+1�` ⇡d d,so d+ 1� ` < ⇡d + 1. This implies d 2 NatDes('(⇡, ✏)).

Now suppose ✏d = 1 and d + 1 � ` ⇡d d do not both hold. This implies d /2NatDes`(⇡, ✏). We consider two cases.

a) Suppose ✏d = �1. Then

'(✏d⇡d) = �'(⇡d) < 0 < d+ 1� ` .

From this we see that d /2 NatDes('(⇡, ✏)).

b) Suppose ✏d = 1 and ⇡d < d+ 1� `. It follows that '(✏d⇡d) = ⇡d. The (d+ 1)st signedletter of '(⇡, ✏) is d+ 1� `, so d /2 NatDes('(⇡, ✏)).

All implications are bi-directional. Therefore,

d 2 NatDes`(⇡, ✏) () d 2 NatDes('(⇡, ✏)) .

Together with Cases 1 and 2 we now conclude that i is an `-descent of (⇡, ✏) if and only

40

if i is a descent of '(⇡, ✏). Therefore,

NatDes`(⇡, ✏) = NatDes ('(⇡, ✏)) and

natdes`(⇡, ✏) = natdes ('(⇡, ✏)) .

This gives us the following lemma.

Lemma 4.3. Let d � 1 and 0 ` d. Then

b`+1

(d+ 1, k) = |{(⇡, ✏) 2 Bd : natdes`(⇡, ✏) = k}| .

Theorem 4.4.

Ehr�

[�1, 1]d` , t�

=B`+1

(d+ 1, t)

(1� t)d+1

.

Proof. We saw above that the number of open facets of 4d,`(⇡,✏) is the `-descent number of

(⇡, ✏). We also have the disjoint decomposition

[�1, 1]d` =G

(⇡,✏)2Bd

4d,`(⇡,✏) .

Therefore,

Ehr�

[�1, 1]d` , t�

=X

(⇡,✏)2Bd

Ehr⇣

4d,`(⇡,✏), t

⌘

=X

(⇡,✏)2Bd

�⇣

4d,`(⇡,✏), t

⌘

(1� t)d+1

=

P

(⇡,✏)2Bdtnatdes`(⇡,✏)

(1� t)d+1

=

Pdk=0

b`+1

(d+ 1, k)tk

(1� t)d+1

=B`+1

(d+ 1, t)

(1� t)d+1

.

Theorem 4.5. For d � 1 and 0 ` d, the coe�cients of ��

[�1, 1]d` , t�

are alternatinglyincreasing.

In order to prove Theorem 4.5, we first introduce another disjoint decomposition of[�1, 1]d` . From the resulting geometry we obtain a characterization for �

�

[�1, 1]d` , t�

as alinear combination of (A, j)-Eulerian polynomials. We then use the symmetric, unimodaland recursive properties of the nonnegative integral coe�cients of the linear combination toprove that the coe�cients of �

�

[�1, 1]d` , t�

are alternatingly increasing.Consider the UI-induced decomposition of [�1, 1]d` into half-open unit cells:

[�1, 1]d` =G

I✓[d]

Ud,`I ,

41

where

Ud,`I : =

�

x 2 [�1, 1]d` : xi � 0 if i 2 I and xi < 0 if i /2 I

=

8

>

<

>

:

x 2 Rd : 0 xi 1 for all i 2 I with

xi < 1 when d+ 1� l i d,

and � 1 xi < 0 for all i /2 I

9

>

=

>

;

.

By construction, the decomposition is disjoint so the �-polynomial of the half-open ±1-cube is equal to the sum of the �-polynomials of the half-open unit cells. Furthermore, Ud,`

I

is congruent to Cdj , where

j = |[d] \ I|+ |{i 2 I : d+ 1� ` i d}| .

Let I ✓ [d], then the unit cell Ud,`I has |[d] \ I| = d� |I| missing facets before the ` facets of

the ±1-cube are removed. This corresponds to the strict inequalities at 0 in the definitionof Ud,`

I . After the ` facets of the ±1-cube are removed, {xd = 1}, . . . , {xd+1�` = 1}, we seestrict inequalities at 1 for all xi with i 2 I and d + 1� ` i d. These correspond to onemissing facet each, giving us a total of j = |[d] \ I| + |{i 2 I : d+ 1� ` i d}| missingfacets.

It follows that the �-polynomial of Ud,`I is the �-polynomial of Cd

j and

��

[�1, 1]d` , t�

=X

I✓[d]

�⇣

Ud,`I , t

⌘

=dX

j=0

cd,`j · ��

Cdj , t

�

=dX

j=0

cd,`j · Aj+1

(d+ 1, t) ,

wherecd,`j :=

�

�

�

n

I ✓ [d] : Ud,`I

⇠= Cdj

o

�

�

�

.

By Lemma 3.4 we know that the coe�cients of Aj+1

(d+1, t) are alternatingly increasingfor all j �

⌅

d+1

2

⇧

. Additionally, from the proof of Corollary 4.2 we know that for all 0 j d,the sum of the polynomial pair Aj+1

(d+ 1, t), Ad�j+1

(d+ 1, t) is alternatingly increasing.In order to prove that �

�

[�1, 1]d` , t�

is alternatingly increasing we will show that for

each I ✓ [d] with Ud,`I

⇠= Cdj and j <

⌅

d+1

2

⇧

, there exists I 0 ✓ [d] with Ud,`I0

⇠= Cdd�j. This

will allow us to pair each (A, j)-polynomial where j <⌅

d+1

2

⇧

with a second polynomialsuch that the polynomial sum is alternatingly increasing. The index sets not paired in thisprocess correspond to j-values greater than or equal to

⌅

d+1

2

⇧

and thus the associated (A, j)-polynomials are alternatingly increasing on their own. In this way, we will see that thesum of the (A, j)-polynomials,

Pdj=0

cd,`j · Aj+1

(d + 1, t), is itself an alternatingly increasingpolynomial.

It is important to note that we consider all (A, j)-polynomials Aj+1

(d + 1, t) as degree-d polynomials, although the coe�cient of td may in fact be zero. Furthermore, we usethe term “alternatingly increasing” in its strictest sense. That is, when we say that adegree-d polynomial is alternatingly increasing, we mean that the polynomial is alternatinglyincreasing in degree d (whether or not s < d, where s is the largest index with a non-zero

42

cd,`0 cd,`1 cd,`2 cd,`3 cd,`4

d = 1 ` = 0 1 1d = 1 ` = 1 0 2

d = 2 ` = 0 1 2 1d = 2 ` = 1 0 2 2d = 2 ` = 2 0 0 4

d = 3 ` = 0 1 3 3 1d = 3 ` = 1 0 2 4 2d = 3 ` = 2 0 0 4 4d = 3 ` = 3 0 0 0 0 8

d = 4 ` = 0 1 4 6 4 1d = 4 ` = 1 0 2 6 6 2d = 4 ` = 2 0 0 4 8 4d = 4 ` = 3 0 0 0 8 8d = 4 ` = 4 0 0 0 0 16

Table 4.1: The Cd,`j -vector for d = 1, 2, 3, 4.

coe�cient). Symbolically, if a0

+ a1

t+ · · ·+ asts + · · ·+ adt

d is alternatingly increasing, then

a0

ad a1

· · · as · · · ab d+12 c .

For fixed d � 1 and fixed ` such that 0 ` d, let⇣

cd,`0

, cd,`1

, . . . , cd,`d

⌘

be called the Cd,`j -vector. From the definition of cd,`k above, we see that the kth entry of this

vector is equal to the number of unit cells congruent to Cdk in the UI-induced decomposition

of [�1, 1]d` . For j < 0 and j > d, we define cd,`j := 0. This aligns with the combinatorial

and geometric interpretations of cd,`j . To get a sense of what these Cd,`j -vectors look like, see

Table 4.1.In our proof of Theorem 4.5 we will need the following lemma regarding the Cd,`

j -vector.

Lemma 4.6. Let d � 1, j 2 Z and 0 ` d. Then the Cd,`j -vector is symmetric and

unimodal with center of symmetry at⌅

d+`2

⇧

and peak(s) of unimodality at:

(

d+`2

when d+ ` is even ,⌅

d+`2

⇧

and⌅

d+`2

⇧

+ 1 when d+ ` is odd .

In particular, the center of symmetry is in agreement with the peak(s) of unimodality.

Proof. First we will prove that the Cd,`j -vector is symmetric about

⌅

d+`2

⇧

by showing a bi-jection between unit cells in [�1, 1]d` congruent to Cd

j and unit cells in [�1, 1]d` congruent to

Cdd+`�j. We will then use induction to prove that the Cd,`

j -vector is unimodal with peak(s)of unimodality in agreement with the center of symmetry.

43

Fix d � 1 and ` such that 0 ` d. In order to prove symmetry of the Cd,`j -vector we

will show that Ud,`I

⇠= Cdj if and only if Ud,`

ˆI⇠= Cd

d+`�j, where I = [d] \ I.Suppose the index set I ✓ [d] is such that Ud,`

I⇠= Cd

j . We know j = |[d] \ I| +|{i 2 I : d+ 1� ` i d}|. Let F := {i 2 I : d+ 1� ` i d}. Then

j = d� |I|+ |F | .

Now consider the index set I = [d] \ I. Notice that |I| = d� |I|. Thus the unit cell Ud,`ˆI

hasj = d� |I|+ |F | = d� (d� |I|) + |F | = |I|+ |F |

removed facets.We note that F and F are disjoint because I and I are disjoint. We further note that

the union of F and F is the set of indices corresponding to all removed facets of [�1, 1]d` . So

F t F = {d+ 1� `, . . . , d� 1, d} =) |F |+ |F | = ` .

Therefore,

j = d� |I|+ |F |�j = �d+ |I|� |F |

d+ `� j = |I|+ (`� |F |)d+ `� j = |I|+ |F | ,

from which it followsj = d+ `� j and Ud,`

ˆI⇠= Cd

d+`�j .

We conclude Ud,`I

⇠= Cdj if and only if Ud,`

I\[d]⇠= Cd

d+`�j. This means cd,`j = cd,`d+`�j for all

j 2 Z. It further implies the Cd,`j -vector is symmetric about

j + (d+ `� j)

2=

d+ `

2.

We will prove the unimodality of the Cd,`j -vector for all 0 ` d by induction on d.

When d = 1, we have C1,`j -vectors (1, 1) and (0, 2) for ` = 0 and ` = 1, respectively. In

this case the peaks of unimodality are trivial. When d = 2, we have C2,`j -vectors (1, 2, 1),

(0, 2, 2) and (0, 0, 4) for ` = 0, 1, 2, respectively. See Figure 4.4. We observe a single peak ofunimodality at index 2+`

2

when 2 + ` is even, and a double peak of unimodality at indices⌅

2+`2

⇧

and⌅

2+`2

⇧

+ 1 when 2 + ` is odd.

Let d � 2. Suppose the Cd�1,`j -vectors are unimodal for all ` satisfying 0 ` d � 1

with the desired peak(s) of unimodality. We will proceed by cases on `.Case 1: ` = 0 .Let I 0 ✓ [d� 1]. In dimension d, Ud,0

I0⇠= Cd

j where j = |[d] \ I 0| = d� |I 0|. In dimension

d � 1, Ud�1,0I0

⇠= Cd�1

j�1

because |[d� 1] \ I 0| = d � 1 � |I 0| = j � 1. Therefore, when K ✓ [d]

44

C2,0j -vector: (1, 2, 1)

[�1, 1]20

C2

0

{1, 2}

C2

1

{1}

C2

1

{2}

C2

2

;

C2,1j -vector: (0, 2, 2)

[�1, 1]21

C2

1

{1, 2}

C2

1

{1}

C2

2

{2}

C2

2

;

C2,2j -vector: (0, 0, 4)

[�1, 1]22

C2

2

{1, 2}

C2

2

{1}

C2

2

{2}

C2

2

;

Figure 4.4: Decomposition of [�1, 1]2` into unit cells indexed by subsets of {1, 2} for ` = 0, 1, 2.

such that d /2 K,Ud,0K

⇠= Cdj () Ud�1,0

K⇠= Cd�1

j�1

.

Let I = I 0 [ {d}. In dimension d, Ud,0I

⇠= Cdj where j = |[d] \ I| = d � |I|. In dimension

d � 1, the index i = d does not contribute to the geometry of the unit cell so Ud�1,0I

⇠=Ud�1,0I0

⇠= Cd�1

j because

|[d� 1] \ I| = d� 1� (|I|� 1) = d� |I| = j .

Therefore, when K ✓ [d] such that d 2 K,

Ud,0K

⇠= Cdj () Ud�1,0

K⇠= Cd�1

j .

It follows that

cd,0j =�

�

�

n

I ✓ [d] : Ud,0I

⇠= Cdj

o

�

�

�

=�

�

�

n

I ✓ [d] : d /2 I and Ud,0I

⇠= Cdj

o

�

�

�

+�

�

�

n

I ✓ [d] : d 2 I and Ud,0I

⇠= Cdj

o

�

�

�

=�

�

�

n

I 0 ✓ [d� 1] : Ud�1,0I0

⇠= Cd�1

j�1

o

�

�

�

+�

�

�

n

I 0 ✓ [d� 1] : Ud�1,0I0

⇠= Cd�1

j

o

�

�

�

= cd�1,0j�1

+ cd�1,0j .

By symmetry, to prove the unimodality of the Cd,0j -vector, it is su�cient to show

cd,0j cd,0j+1

(4.5)

holds for all j <⌅

d2

⇧

.Let j <

⌅

d2

⇧

. We note that inequality (4.5) is equivalent to

cd�1,0j�1

+ cd�1,0j cd�1,0

j + cd�1,0j+1

45

and thus tocd�1,0j�1

cd�1,0j+1

. (4.6)

By the induction hypothesis, if d is odd, then the (single) peak of unimodality of theCd�1,0

j -vector is d�1

2

which implies cd�1,0k cd�1,0

k+1

for all k < d�1

2

. We note j <⌅

d2

⇧

isequivalent to j + 1 d�1

2

when d is odd. Therefore, (4.6) holds by the induction hypothesisand (4.5) holds for all j <

⌅

d2

⇧

when d is odd.

If d is even, then the (double) peaks of unimodality of the Cd�1,0j -vector are

⌅

d�1

2

⇧

and⌅

d�1

2

⇧

+ 1 =⌅

d+1

2

⇧

= d2

. It follows that cd�1,0k cd�1,0

k+1

for all k < d2

, or equivalently, for allk+1 d

2

. We still have j <⌅

d2

⇧

which is equivalent to j+1 d2

when d is even. Therefore,(4.6) holds by the induction hypothesis, implying that (4.5) holds for all j <

⌅

d2

⇧

when d iseven.

By the symmetry of the Cd,0j -vector we have

cd,0j � cd,0j+1

for all j �⌅

d2

⇧

. We conclude that the Cd,0j -vector is unimodal for d � 1 and 0 j d.

Remark. From Theorem 4.1 we know cd,0j =�

dj

�

for all d � 1 and j 2 Z. This gives us the

Cd,0j -vector

✓✓

d

0

◆

,

✓

d

1

◆

, . . . ,

✓

d

d

◆◆

.

Common knowledge about the binomial coe�cients tells us that this vector is palindromicand unimodal and thus alternatingly increasing. Case 1 gives a geometric interpretation ofthis result in addition to a geometric interpretation of the well-known recursive formula forbinomial coe�cients:

cd,0j = cd�1,0j�1

+ cd�1,0j ()

✓

d

j

◆

=

✓

d� 1

j

◆

+

✓

d� 1

j � 1

◆

.

Case 2: ` > 0 .Let I 0 ✓ [d� 1]. In dimension d, Ud,`

I0⇠= Cd

j where

j = |[d] \ I 0|+ |{i 2 I 0 : d+ 1� ` i d}| .

By definition, this unit cell is a subset of [�1, 1]d` , a half-open ±1-cube with ` facets missingcorresponding to the ` supporting hyperplanes xd = 1, xd�1

= 1, . . . , xd+1�` = 1.If we consider this same ±1-cube in one dimension lower, then we observe `� 1 missing

facets corresponding to supporting hyperplanes xd�1

= 1, . . . , xd+1�` = 1. Therefore, whenconsidered as a (d�1)-dimensional object, the unit cell indexed by I 0 is a subset of [�1, 1]d�1

`�1

46

and is congruent to Cd�1

k where

k = |[d� 1] \ I 0|+ |{i 2 I 0 : (d� 1) + 1� (`� 1) i d� 1}|= (d� 1)� |I 0|+ |{i 2 I 0 : d+ 1� ` i d� 1}|= |[d] \ I 0|� 1 + |{i 2 I 0 : d+ 1� ` i d}|= j � 1 .

Therefore, when K ✓ [d] such that d /2 K,

Ud,`K

⇠= Cdj () Ud�1,`�1

K⇠= Cd�1

j�1

.

Let I = I 0 [ {d}. In dimension d, Ud,`I

⇠= Cdj where

j = |[d] \ I|+ |{i 2 I : d+ 1� ` i d}| .

In dimension d � 1, we consider the unit cell indexed by I as a subset of [�1, 1]d�1

`�1

just as

before. Therefore, Ud,`I

⇠= Ud,`�1

I0⇠= Cd�1

j�1

in dimension d� 1 because

j � 1 = d� |I|+ |{i 2 I : d+ 1� ` i d}|� 1

= |[d� 1] \ I 0|+ |{i 2 I 0 : d+ 1� ` i d� 1}| .

Therefore, when K ✓ [d] such that d 2 K,

Ud,`K

⇠= Cdj () Ud�1,`�1

K⇠= Cd�1

j�1

.

It follows that

cd,`j =�

�

�

n

I ✓ [d] : Ud,`I

⇠= Cdj

o

�

�

�

=�

�

�

n

I ✓ [d] : d /2 I and Ud,`I

⇠= Cdj

o

�

�

�

+�

�

�

n

I ✓ [d] : d 2 I and Ud,`I

⇠= Cdj

o

�

�

�

=�

�

�

n

I 0 ✓ [d� 1] : Ud�1,`�1

I0⇠= Cd�1

j�1

o

�

�

�

+�

�

�

n

I 0 ✓ [d� 1] : Ud�1,`�1

I0⇠= Cd�1

j�1

o

�

�

�

= 2 · cd�1,`�1

j�1

.

Suppose j <⌅

d+`2

⇧

. We wish to show

cd,`j cd,`j+1

, (4.7)

or equivalently,cd�1,`�1

j�1

cd�1,`�1

j . (4.8)

By the induction hypothesis, inequality (4.8) holds for all ` > 0 when j � 1 <⌅

d�1+`2

⇧

.Equivalently, inequality (4.8) holds for all ` > 0 when j <

⌅

d+1+`2

⇧

. However,⌅

d+`2

⇧

⌅

d+1+`2

⇧

. So j <⌅

d+`2

⇧

implies (4.8) which further implies (4.7). From the symmetry of the

Cd,`j -vector,

cd,`j � cd,`j+1

47

for all j �⌅

d+1+`2

⇧

. Therefore, the Cd,`j -vector is unimodal for d � 1, 0 j d and ` > 0.

This proves the unimodality of the Cd,`j -vector.

We are now equipped to prove that the coe�cients of ��

[�1, 1]d` , t�

are alternatinglyincreasing.

Proof of Theorem 4.5. Fix d � 1 and ` such that 0 ` d. Suppose j <⌅

d+1

2

⇧

. We wishto show that

cd,`j cd,`d�j . (4.9)

First note j <⌅

d+1

2

⇧

implies j < d � j. If j and d � j are both less than or equal to

p1

:=⌅

d+`2

⇧

, then inequality (4.9) holds by unimodality of the Cd,`j -vector: cd,`k cdk0,` for all

k k0 p1

.If j p

1

< d� j, then we need to show that the index d� j is closer to p1

+ 1 than theindex j is to p

1

. That is, we need to show

p1

� j � (d� j)� (p1

+ 1) ,

or equivalently,2p

1

� d� 1 .

Inequality (4.9) will follow from the symmetry of the Cd,`j -vector.

When d+ ` is even we have

2p1

= 2

✓

d+ `

2

◆

= d+ ` > d� 1 .

When d+ ` is odd we have

2p1

= 2

✓

d+ `

2� 1

2

◆

= d+ `� 1 � d� 1 .

Therefore,2p

1

� d� 1

holds for all d � 1 and 0 ` d.We conclude that the number of unit cells congruent to Cd

j in the UI-induced decompo-sition of [�1, 1]d` is less than or equal to the number of unit cells congruent to Cd

d�j in thedecomposition. Therefore we can pair each of the (A, j)-polynomials with j-value strictlyless than

⌅

d+1

2

⇧

with one of the (A, d � j)-polynomials such that no (A, d � j)-polynomialis paired twice. The sum of each polynomial pair is alternatingly increasing. Furthermore,any (A, j)-polynomial not paired in this process is itself alternatingly increasing. Thus, forfixed d � 1 and 0 ` d, the coe�cients of

��

[�1, 1]d` , t�

=dX

j=0

cd,`j · Aj+1

(d+ 1, t)

are alternatingly increasing.

48

We arrive at the following corollary immediately.

Corollary 4.7. B`+1

(d+ 1, t) is alternatingly increasing for d � 1 and 0 ` d.

49

Chapter 5

Lattice Centrally SymmetricParallelepipeds

5.1 Parallelepipeds with Lattice Centrally SymmetricEdges

A polytope P is centrally symmetric about the origin if p 2 P implies �p 2 P . Apolytope P is lattice centrally symmetric if there exists v 2 Zd such that v + P iscentrally symmetric about the origin. Moreover, we say the half-open lattice parallelepiped

I(J) is lattice centrally symmetric if its closure ⌃(J) is.The ±1-cube is a simple example of a lattice centrally symmetric parallelepiped. We

have already seen that the �-vector for the ±1-cube (closed and half-open) is alternatinglyincreasing.

Theorem 5.1. Let I(v1

, . . . ,vd) be a half-open parallelepiped with lattice centrally symmet-ric edges. Then

Ehr ( I (v1

, . . . ,vd) , t) =

P

K✓[d] #⇣

2(wi1 , . . . ,wi|K|) \ Zd⌘

B|I[K|+1

(d+ 1, t)

(1� t)d+1

,

where wi =1

2

vi 2 Zd for all i 2 [d].

Proof. Let I(v1

, . . . ,vd) be a half-open parallelepiped with lattice centrally symmetricedges. Then there exists wi 2 Zd such that wi =

1

2

vi. In this way we see that

I(v1

, . . . ,vd) = I(2w1

, . . . , 2wd) = 2 I(w1

, . . . ,wd) .

We also note that [�1, 1]d`⇠= 2Cd

` . This implies

ehr�

[�1, 1]d` , n�

= ehr�

2Cd` , n

�

= ehr�

Cd` , 2n

�

=X

[`]✓J✓[d]

(2n)|J | ,

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where the last equality follows from Corollary 3.9. This allows us to express the Ehrhartseries of I(v1

, . . . ,vd) in terms of the Ehrhart series of half-open ±1-cubes:

Ehr ( I(v1

, . . . ,vd), t)

= Ehr (2 I(w1

, . . . ,wd), t)

= 1 +X

n�1

#�

2n I(w1

, . . . ,wd) \ Zd�

tn

=X

n�0

tnX

I✓J✓[d]

(2n)|J | #⇣

⇧(wi1 , . . . ,wi|J|) \ Zd⌘

by Lemma 3.8

=X

n�0

tnX

I✓J✓[d]

(2n)|J |X

K✓J

#⇣

2(wi1 , . . . ,wi|K|) \ Zd⌘

by Lemma 3.7

=X

K✓[d]

#⇣

2(wi1 , . . . ,wi|K|) \ Zd⌘

X

n�0

tnX

I[K✓J

(2n)|J |

=X

K✓[d]

#⇣

2(wi1 , . . . ,wi|K|) \ Zd⌘

X

n�0

ehr�

[�1, 1]d|I[K|, n�

tn

=X

K✓[d]

#⇣

2(wi1 , . . . ,wi|K|) \ Zd⌘

Ehr�

[�1, 1]d|I[K|, t�

=

P

K✓[d] #⇣

2(wi1 , . . . ,wi|K|) \ Zd⌘

B|I[K|+1

(d+ 1, t)

(1� t)d+1

,

where the last equality follows from Theorem 4.4.

Corollary 5.2. The �-polynomial of a half-open parallelepiped with lattice centrally symmet-ric edges has alternatingly increasing coe�cients.

Proof. From Theorem 5.1 we have

� ( I(v1

, . . . ,vd), t) =X

K✓[d]

#⇣

2(wi1 , . . . ,wi|K|) \ Zd⌘

B|I[K|+1

(d+ 1, t) ,

where I(v1

, . . . ,vd) is a half-open parallelepiped with lattice centrally symmetric edges and

wi =1

2

vi 2 Zd. Furthermore, #⇣

2(wi1 , . . . ,wi|K|) \ Zd⌘

� 0 for all K ✓ [d]. By Corollary

4.7, the coe�cients of B|I[K|+1

(d+1, t) are alternatingly increasing. The same is true for thecoe�cients of any nonnegative linear combination of the B|I[K|+1

(d+1, t) where K ✓ [d].

Corollary 5.3. The �-polynomial of a zonotope with lattice centrally symmetric edges hasalternatingly increasing coe�cients.

Proof. Let Z (u1

, . . . ,ur) be a d-dimensional zonotope with lattice centrally symmetric edges.Then there exists wi 2 Zd such that wi =

1

2

ui for all i 2 [r] and

Z (u1

, . . . ,ur) = Z (2w1

, . . . , 2wr) .

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Let C be a subdivision of Z into parallelepipeds whose maximal cells ⌃1

,⌃2

, . . . ,⌃s aregenerated by the linearly independent subsets of {2w

1

, . . . , 2wr}. The existence of such asubdivision exists by Theorem 2.1. By Corollary 3.13, subdivision C gives rise to a disjointdecomposition of Z into half-open parallelepipeds of the -type. In particular, the half-openparallelepipeds of the decomposition are half-open variants of the maximal cells in C . Thuswe have

Z =G

i2[r]i

and� (Z, t) =

X

i2[r]

� ( i, t) ,

where i is the appropriate half-open variant of ⌃i in C .For all i 2 [r], the edges of i are lattice centrally symmetric. Therefore, by Corollary 5.2

the coe�cients of � ( i, t) are alternatingly increasing for all i 2 [r] and so are the coe�cientsof � (Z, t).

5.2 Extensions and Open Questions

We have shown that the coe�cients of the �-polynomial for the following families of latticepolytopes are alternatingly increasing:

• Half-open lattice d-parallelepipeds with j �⌅

d+1

2

⇧

non-translate facets removed;

• Half-open lattice parallelepipeds with lattice centrally symmetric edges;

• Closed lattice zonotopes with lattice centrally symmetric edges.

We are interested in what more we can say about the inequality relations on the coe�-cients of the �-polynomial for the following more general families of lattice polytopes:

• Half-open lattice centrally symmetric lattice parallelepipeds;

• Half-open lattice parallelepipeds with an interior lattice point;

• Lattice centrally symmetric lattice zonotopes;

• Lattice zonotopes with an interior lattice point;

• Half-open lattice zonotopes (i.e., lattice zonotopes with non-translate facets removed).

A further extension of our work which remains open is whether we can define appropriatebivariate permutation statistics which generalize the (A, j)- and (B, `)-Eulerian numbersparallel to Brenti’s q-Eulerian numbers in [4]. In particular, can we define bivariate statisticsand polynomials which will further generalize the classical theory of Eulerian numbers andpolynomials?

52

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