Box Polynomials of Lattice Simplices

DEGREE PROJECT IN MATHEMATICS, SECOND CYCLE, 30 CREDITS STOCKHOLM, SWEDEN 2018

NILS GUSTAFSSON

KTH ROYAL INSTITUTE OF TECHNOLOGY SCHOOL OF ENGINEERING SCIENCES

Box Polynomials of Lattice Simplices

NILS GUSTAFSSON

Degree Projects in Mathematics (30 ECTS credits) Degree Programme in Mathematics (120 credits) KTH Royal Institute of Technology year 2018 Supervisor at KTH: Liam Solus Examiner at KTH: Svante Linusson

TRITA-SCI-GRU 2018:160 MAT-E 2018:24

Royal Institute of Technology School of Engineering Sciences KTH SCI SE-100 44 Stockholm, Sweden URL: www.kth.se/sci

Abstract

The box polynomial of a lattice simplex is a variant of the more well-known h∗-polynomial, where the open fundamental parallelepiped is considered instead of the half-open. Box polynomials are connected to h∗- polynomials by a theorem of Betke and McMullen from 1985. This theorem can be used to prove certain properties of h∗-polynomials, such as unimodality and symmetry. In this thesis, we investigate box polynomials of a certain family of simplices, called s-lecture hall simplices. The h∗-polynomials of these simplices are a generalization of Eulerian polynomials, and were proven to be real-rooted by Savage and Visontai in 2015. We use a modified version of their proof to prove that the box polynomials are also real-rooted, and show that they are a generalization of derangement polynomials. We then use these results to partially answer a conjecture by Brändénand Leander regarding unimodality of h∗-polynomials of s-lecture hall order polytopes.

Sammanfattning

Boxpolynomet av ett gittersimplex ären variant av det mer kända h∗-polynomet, därden öppnafunda- mentala parallelepipeden användsiställetförden halvöppna.Boxpolynom ärkopplade till h∗-polynom tack vare en sats av Betke och McMullen fr˚an1985. Denna sats kan användasföratt bevisa vissa egenskaper hos h∗-polynom, som t.ex. unimodalitet och symmetri. I denna uppsats undersöker vi boxpolynomen hos en särskildfamilj av simplex, de s˚akallade s-hörssalssim- plexen. Fördessa simplex är h∗-polynomen en generalisering av de Eulerska polynomen, och visades ha endast reella rötterav Savage och Visontai 2015. Vi använderen modifierad version av deras bevis föratt bevisa att även boxpolynomen bara har reella rötter,och att de ären generalisering av derangemangpoly- nom. Vi användersedan dessa resultat föratt delvis besvara en förmodan av Brändénoch Leander ang˚aende unimodaliteten hos h∗-polynomen av s-hörsalsordningspolytoper.

Contents

1 Introduction 4

2 Ehrhart theory 6 2.1 Polytopes ...... 6 2.2 Ehrhart’s theorem ...... 8 2.3 Distributional properties of h∗-polynomials ...... 12 2.4 Triangulations and Box Polynomials ...... 13

3 s-Eulerian Polynomials 16 3.1 Lehmer codes ...... 16 3.2 s-inversion sequences ...... 17 3.3 s-lecture hall simplices ...... 19

4 s-Derangement Polynomials 22 4.1 The box polynomial of an s-lecture hall polytope ...... 22 4.2 Derangements and the sequence s = (n, n − 1, n − 2, ..., 2) ...... 23 4.3 Colored permutations ...... 25 4.4 Real-rootedness ...... 30 4.5 Faces of s-lecture hall simplices ...... 34

5 s-Lecture Hall Order Polytopes 37 5.1 Order polytopes ...... 37 5.2 s-lecture hall order polytopes ...... 38

Chapter 1

Introduction

Polytopes are fundamental geometric objects that have been studied since ancient times. Informally, a polytope is an object with “ﬂat sides” and “sharp edges”. Examples include cubes, triangles and intervals. A circle is not a polytope. Polytopes are useful in a wide variety of situations. For example, they can be used to approximate things that are not polytopes, like in computer graphics, or to describe the set of possible solutions to an optimization problem, like in linear programming.

Most people would agree that a polytope is an object of continuous nature, as opposed to say a graph or a list of positive integers. However, polytopes can also be useful in more discrete settings in combinatorics and number theory. One such link between polytopes (in the case of convex lattice polytopes) and combinatorics is Ehrhart theory. Ehrhart theory was developed by EugèneEhrhart in the 1960:s, and concerns the so called discrete volume of polytopes. The discrete volume of a polytope is the number of points with integer coordinates contained in it. When the polytope is scaled up by integer factors and the discrete volume is computed for each of scaled up versions, a sequence of numbers is obtained. Ehrhart showed in [14] that this sequence can be described by a polynomial, called the Ehrhart polynomial. Not only that, but this polynomial contains the continuous volume as one of its coefficients, as well as a lot of other information. The Ehrhart polynomial can also be written in terms of another polynomial, called the h∗-polynomial. In the case of a simplex, the h∗-polynomial has a simple geometric interpretation, which makes it relatively easy to compute. For other polytopes, it is harder to interpret what the coefficients of the h∗-polynomial mean, but a theorem due to Stanley [26] says that the coefficients are always, at least, nonnegative integers.

Sometimes it is discovered that a sequence of integers from combinatorics is the h∗-polynomial of some polytope. When that happens, Ehrhart theory acts as a kind of bridge between the continuous world of polytopes and the discrete world of combinatorics, which can be of great beneﬁt for both sides. One such example are the Eulerian numbers. The Eulerian number An,k is the number of permutations of [n] with k descents, i.e. k places where πi > πi+1. If we let the sequence {An,0,An,1, ··· An,n−1} be the coeﬃcients of a polynomial, then it is also the h∗-polynomial of a simplex called the lecture hall simplex. This example can be generalized into a rather broad family of polynomials called s-Eulerian polynomials, which are the h∗-polynomials of s-lecture hall simplices.

When we have a sequence of positive integers that describe a distribution such as “the number of permutations with a certain number of descents”, it is often interesting to know whether the sequence is unimodal or not. Unimodal means that the sequence only has one “peak” (or local maximum). It turns out that unimodality often follows if the polynomial corresponding to the sequence is real-rooted. In [23], Savage and Visontai used tools from algebra to show that the s-Eulerian polynomials are real-rooted, and therefore unimodal.

4 Figure 1.1: A dodecahedron is a polytope.

In this thesis, we will investigate a variant of the h∗-polynomials of lattice simplices, called box polynomials. Like h∗-polynomials, we will see in this thesis that box polynomials can also correspond to interesting combinatorial sequences. Due to a theorem of Betke and McMullen [8], the h∗-polynomial of a polytope can be written as a kind of weighted sum of these box polynomials. Sometimes, we can then use unimodality of the box polynomials to recover unimodality of the h∗-polynomial.

In Chapter 2, Ehrhart’s theory will be introduced in more detail, as well as some basics on polytopes and unimodality. Most of the Ehrhart theory is based on [7], and we will use the same notation. In Chapter 3, we will introduce s-inversion sequences, s-lecture hall polytopes, and s-Eulerian polynomials. Most definitions and results are collected from various articles, such as [23], [22] and [18]. Our definitions and notation may differ slightly as compared to them. For example, we use a different definition of descents compared to the one in [22]. In Chapter 4, we study the properties of box polynomials of s-lecture hall simplices. First, the theory from Chapter 3 is used to describe them in terms of s-inversion sequences. Then, we turn our attention to permutations and colored permutations, where the box polynomials are shown to be a well- known family of polynomials called derangement polynomials. Then, we use a modified version of the proof in [23] to show that they are real-rooted. Finally, we show that the faces of s-lecture hall simplices also have real-rooted box polynomials. In Chapter 5, we apply the real-rootedness results from Chapter 4 combined with the theorem of Betke-McMullen to s-lecture hall order polytopes, a very broad family of polytopes introduced in [12]. We show that the h∗-polynomials for some of these polytopes are unimodal.

5 Chapter 2

Ehrhart theory

Ehrhart theory deals with the problem of counting lattice points in convex polytopes. It provides a kind of bridge between geometry and combinatorics. This chapter contains some Ehrhart theory that will be used throughout the article. For a comprehensive discussion on the topic we refer the reader to [7].

2.1 Polytopes

A convex polytope P in Rn is the convex hull of ﬁnitely many points:

n P = {λ1v1 + ... + λmvm ∈ R : λi ≥ 0 and λ1 + ... + λm = 1}.

The points v1, ..., vm are called the vertices of P. If the vertices have integer coordinates, then we say that P is a lattice polytope (or integral polytope). The dimension of P is the dimension of the aﬃne space

span(P) = {x + λ(y − x): x, y ∈ P, λ ∈ R}. If P has dimension d, then P is called a d-polytope. The definition above of a polytope as the convex hull of finitely many points is called the vertex description of P. Every polytope also has an equivalent description as the bounded intersection of finitely many half-spaces and hyperplanes. This is called the hyperplane description. These descriptions are useful in different situations, and we will often use both of them. The interior of a polytope P, denoted P◦, is the set of points of P such that all points within distance from them also belong to P, for some > 0. The interior can be obtained by taking the hyperplane description of P and turning all inequalities strict. A face of a convex polytope is a set on the form P ∩H, where H is a hyperplane such that P lies entirely on one side of it. The (d − 1)-dimensional faces are called facets, and the 2-dimensional faces are called edges.A d-simplex is a d-polytope with exactly d + 1 vertices. This is the smallest number of vertices a d-polytope can have, because otherwise it can’t have dimension d. A 1-simplex is an interval, a 2-simplex is a triangle, and a 3-simplex is a tetrahedron. Simplices are often easier to work with than general polytopes. Therefore, it can be useful to decompose a polytope into simplices.

Deﬁnition 2.1.1. Let S ⊆ Rn.A triangulation of S is a ﬁnite set of simplices T such that S 1. S = ∆∈T ∆, 2. If ∆ ∈ T and ∆0 is a face of ∆, then ∆0 ∈ T , and 3. If ∆ ∈ T and ∆0 ∈ T , then ∆ ∩ ∆0 ∈ T . If the vertices of the simplices in T are the same as the vertices of a polytope P, then we say that P is triangulated using no new vertices. All convex polytopes can be triangulated using no new vertices, because

6 Figure 2.1: A convex 2-polytope along with its second and third dilates. The polytope is the convex hull of (1, 1), (0, −1) and (−1, 0).

a triangulation can be constructed in the following way: Take the convex d-polytope P = conv(v1, ..., vn), and lift it to (d+1)-dimensional space by adding an extra coordinate to every vertex. Call this new polytope P0. This lifting can be done in such a way that the vertices of P0 are in general position. Now, take the convex hull of P0. Since the vertices are in general position, the facets of the convex hull will be simplices. If we take the lower hull of this (the facets such that all points “below” them in the xd+1-direction are outside of the convex hull), and move it back to Rd by deleting the last coordinate, we get a triangulation of P.A triangulation obtained in this way is called a regular triangulation, and they will be of special interest later.

One main problem of Ehrhart theory is to count the lattice points in a polytope when it is scaled up by diﬀerent factors. For a convex polytope P and a positive integer t, the t:th dilate of P is deﬁned as:

n tP := {tx ∈ R : x ∈ P}. For positive integers t, deﬁne the lattice-point enumerator of P to be the number of lattice points in the t:th dilate of P: n LP (t) = #(tP ∩ Z ).

The generating function containing LP (t) as coeﬃcients is called the Ehrhart series of P:

X t EhrP (z) = 1 + LP (t)z . t≥1

Example 2.1.2. One of the most important examples of a polytope is the unit cube d:

d d d := [0, 1] = conv (x1, ..., xd) ∈ R : all xi are 0 or 1 . d can be written as the intersection of 2d half-spaces. d \ d = {xi ≥ 0} ∩ {xi ≤ 1}. i=1 d Here {xi ≥ 0} is short for {(x1, ..., xd) ∈ R : xi ≥ 0}. This is the hyperplane description of d. The t:th dilate of d is d td = [0, t] , which implies that L (t) = (t + 1)d, d and X Ehr (z) = (t + 1)dzt. d t≥0

7 Figure 2.2: The unit cube 2 along with its second, third, and fourth dilates.

2.2 Ehrhart’s theorem

In this section, some motivation behind the deﬁnition of LP (t) and EhrP (z) will be provided. Both functions turn out to be very useful, as they contain a lot of information about the polytope. For example, Ehrhart’s theorem says that LP (t) is a polynomial:

Theorem 2.2.1 ([14]). If P is a convex lattice polytope of dimension d, then LP (t) is a polynomial in t of degree d.

Due to Theorem 2.2.1, LP (t) is called the Ehrhart polynomial (not to be confused with the Ehrhart series EhrP (z)). Ehrhart’s theorem implies that LP (t) can be written as

d d−1 LP (t) = cdt + cd−1t + ... + c1t + c0,

for some rational coefficients c0, c1, ··· , cd. Now we could try to make sense of these coefficients. The first one, cd, actually turns out to be a very familiar number: the volume of P. To see why, consider the following quantity: L (t) #tP ∩ n lim P = lim Z . t→∞ td t→∞ td The numerator on the right-hand-side is the number of lattice points inside P when P is scaled up by a factor of t. Each such lattice point contributes t−d to the expression on the right. So it is a bit like we covered P with hypercubes of side length t−1 and added up their volumes. For large t, this should give a good approximation of the P:s volume (in fact, the volume can be defined as the limit above). Finally we note that the expression on the left is cd, so cd really is the volume of P.

This means that once we have found LP (t), we also know the volume of P. However, there is a lot more information in LP (t) and EhrP (z). To find some of it, we will need some definitions that naturally turn up in the proof of Theorem 2.2.1. It turns out that it suffices to prove Theorem 2.2.1 in the case when P is a simplex. To see this, note that if P is not a simplex, then it can be triangulated using no new vertices, and we can express LP (t) with the inclusion-exclusion formula

X d−dim(∆) LP (t) = (−1) L∆(t). ∆∈T

Here T is a triangulation using no new vertices. So if Theorem 2.2.1 is true for simplices, then LP (t) for a general lattice polytope is a sum of polynomials of degree d, and is therefore a polynomial of degree at most d. If the degree was smaller than d, then the volume of P would be 0 using the argument above. However, this would imply that the dimension of P is smaller than d. So the degree of LP (t) must be exactly d.

8 Figure 2.3: The cone over the interval P = [−1, 2]. Note that the intersection with the hyperplane {y = 1} is the polytope itself.

To prove Theorem 2.2.1 for simplices, we first need some definitions. For a set S ⊆ Rn, the integer-point transform σS(z) is defined as X m σS(z) := z , n m∈S∩Z

m m1 m2 mn where z := z1 z2 ...zn . So it is a kind of generating function that lists the lattice points of S. For a convex lattice polytope P with vertices v1, ..., vk, deﬁne the cone over P as

cone(P) = {λ1w1 + ... + λkwk : all λi ≥ 0}, where wi := (vi, 1). Geometrically, we added an extra dimension, lifted P into the hyperplane {xn+1 = 1}, made a cone with the origin as the “tip” and made it go through the lifted version of P. Note that cone(P) ∩ {xn+1 = 1} is just P itself in disguise. This is because if

w = λ1w1 + ... + λkwk ∈ cone(P), then the last coordinate of w is λ1 + ··· + λk. So cone(P) ∩ {xn+1 = 1} is the collection of points of cone(P) where λ1 + ... + λk = 1. This is the convex hull of w1, ..., wk, also known as P lifted to {xn+1 = 1}. Generalizing this argument a little bit gives the following: Proposition 2.2.2. For a convex lattice polytope P and a positive integer t,

cone(P) ∩ {xn+1 = t} = tP.

If we take the integer-point transform of the cone, then interesting things start to happen: Proposition 2.2.3. For a convex lattice polytope P,

σcone(P)(1, 1, ..., 1, z) = EhrP (z).

Proof. Recall that the integer-point transform listed all the lattice points of a set S. Take one such lattice point w ∈ cone(P) and see how it contributes to σcone(P)(1, 1, ..., 1, z). The monomial associated to w is w1 wn+1 wn+1 z1 ...zn+1 . When evaluated at (1, 1, ..., 1, z), it becomes z , so it contributes 1 to the coeﬃcient in wn+1 front of z in the series σcone(P)(1, 1, ..., 1, z). However, thanks to Proposition 2.2.2, we know that w corresponds to a point in wn+1P. So σcone(P)(1, 1, ..., 1, z) enumerates points in tP. In other words,

X t σcone(P)(1, 1, ..., 1, z) = 1 + LP (t)z = EhrP (z). t≥1

9 Now we will express σcone(P)(1, 1, ..., 1, z) in a different way, by tiling it with parallelepipeds. Definition 2.2.4. For a d-dimensional lattice simplex P, the fundamental parallelepiped Π is defined as Π := {λ1w1 + ... + λd+1wd+1 : 0 ≤ λi < 1 for all i = 1, 2, ..., d + 1}. Note that the fundamental parallelepiped is half-open. This is important because it allows us to tile the cone using translated copies of it. d+1 Proposition 2.2.5. Let P be a d-dimensional lattice simplex. For some m ∈ Z≥0 , let

Πm := m1w1 + ... + md+1wd+1 + Π be Π translated by some non-negative integer linear combination of w1, ..., wd+1. Then it holds that [ cone(P) = Πm. d+1 m∈Z≥0

Moreover, the sets Πm are disjoint. Proof. Let w ∈ cone(P). If we could prove that w belongs to exactly one of the translated parallelepipeds, then the proof would be done. Here, the fact that P is a simplex is crucial for the uniqueness part, because it implies that w1, ..., wd+1 are linearly independent. Therefore, w can be uniquely written as

w = λ1w1 + ... + λd+1wd+1.

If we let m = (bλ1c, ..., bλd+1c), then w ∈ Πm, and there is no other such parallelepiped. Corollary 2.2.6. For a d-dimensional lattice simplex P, X σcone(P)(z) = σΠm (z). d+1 m∈Z≥0 Now we are ready to prove the following: Proposition 2.2.7. For a d-dimensional lattice simplex P,

σΠ(z) σcone(P)(z) = . (1 − zw1 )...(1 − zwd+1 )

Proof. First, note that it follows from the deﬁnition of σ and Πm that

m σΠm (z) = z σΠ(z). Combining this with Corollary 2.2.6 yields     σ (z) X m X iw1 X iwd+1 Π σcone(P)(z) = σΠ(z) z = σΠ(z)  z  ...  z  = (1 − zw1 )...(1 − zwd+1 ) d+1 i≥0 i≥0 m∈Z≥0

P i 1 In the last step we used the fact that geometric series satisﬁes i≥0 z = 1−z . This result can in turn be combined with Proposition 2.2.3: Proposition 2.2.8. For a d-dimensional lattice simplex P, h∗zd + h∗ zd−1 + ... + h∗ Ehr (z) = d d−1 0 P (1 − z)d+1

∗ where hk is the number of integer points in Π whose last coordinate is k.

10 Proof. In Proposition 2.2.7, evaluate at (1, 1, 1, ..., 1, z), to recover

σ (1, 1, ..., 1, z) σ (1, 1, ..., 1, z) = Π . cone(P) (1 − z)d+1

In the proof of Proposition 2.2.3, we noted that evaluating the integer-point transform in (1, 1, ..., 1, z) gives a series that counts the number of integer points with diﬀerent last coordinates, which is exactly what the h∗-coeﬃcients do. The reason why the series only goes from 0 to d is that there are no integer points in Π whose last coordinate is outside of this interval. Also, by Proposition 2.2.3, we have that the series above is actually the Ehrhart series. Therefore,

h∗zd + h∗ zd−1 + ... + h∗ Ehr (z) = d d−1 0 . P (1 − z)d+1

∗ d ∗ d−1 ∗ ∗ The polynomial hdz + hd−1z + ... + h0 is called the h -polynomial of the simplex P, and it will become very important later. Now we have everything we need to prove Ehrhart’s theorem:

Proof of Theorem 2.2.1. Rewrite the Ehrhart series using the h∗-polynomial to see that   ∗ d ∗ d−1 ∗ d ! d ! h z + h z + ... + h0 X d + t X X X d + t − i Ehr (z) = d d−1 = zt h∗zi = h∗ zt. P (1 − z)d+1  d  i d i t≥0 i=0 t≥0 i=0

By the definition of the Ehrhart series, the expression within the parentheses in the last expression is LP (t). Since the expression is a polynomial in t of degree at most d, the proof is almost done. The only thing remaining is to show that the degree is exactly d. This follows from the fact that the volume of P is positive. To conclude this section, a few more theorems (that will not be proved) and definitions will be presented. Proofs for each of these results can be found in [14]. These results and definitions will play a key role in the main results of this thesis.

From a combinatorial perspective, one very exciting thing about the h∗-polynomial of a lattice simplex is that all of its coeﬃcients are nonnegative integers. In fact, the h∗-polynomial can be deﬁned for general lattice polytopes and not just simplices. This is illustrated by Stanley’s non-negativity theorem: Theorem 2.2.9 ([26]). For a lattice polytope P,

h∗zd + ... + h∗ Ehr (z) = d 0 , P (1 − z)d+1

∗ ∗ where h0, ..., hd are non-negative integers.

Now that we know that LP (t) is a polynomial, we can evaluate it at other points than positive integers, and try to interpret what that means. For negative integers, such an interpretation is given by the following reciprocity theorem: Theorem 2.2.10. If P is a convex lattice polytope with dimension d, then

d LP (−t) = (−1) LP◦ (t).

11 A special kind of polytope that will be useful in Chapter 5 are the reflexive polytopes. Informally, a reflexive polytope is a convex polytope whose vertices have integer coordinates, and whose hyperplane description also only contains integers. More precisely, a lattice d-polytope P is reflexive if it contains the origin in its interior, and can be written as

n P = {x ∈ R : Ax ≤ 1}, for some matrix A only containing integers. In this thesis, we will say that a polytope is reflexive when it really is reflexive up to translation according to the above definition. This means that it can be translated into a reflexive polytope. The reason for this is that most properties of interest to us, like h∗- polynomials and regularity of triangulations, are unaffected by translation. Polytopes can be shown to be reflexive by looking at their h∗-polynomials, using a theorem of Hibi. Theorem 2.2.11 (Theorem 2.1 in [16]). Let P be a lattice d-polytope. Then P is reflexive if and only if the ∗ ∗ ∗ h -polynomial satisfies hi = hd−i for i ∈ {0, 1, 2, ..., d}.

Example 2.2.12. Going back to Example 2.1.2, we proved that for the unit cube d, X Ehr (z) = (t + 1)dzt. d t≥0 This can be rewritten as d X P A(d, k)zk Ehr (z) = (t + 1)dzt = k=0 , d (1 − z)d+1 t≥0 where A(d, k) is the Eulerian number (they are often deﬁned in terms of the identity above). So the ∗ h -polynomial of d is d ∗ X k h (z) = Ad(z) := A(d, k)z . d k=0 Eulerian numbers have many other properties. Most importantly, they count certain statistics over permutations. For a permutation π = π1π2...πd, a descent is an index i where 1 ≤ i < d such that πi > πi+1. Now, it turns out that A(d, k) is the number of permutations of length d with exactly k descents. In other words, d X k X des(π) Ad(z) = A(d, k)z = z ,

k=0 π∈Sd where des(π) is the number of descents in π.

2.3 Distributional properties of h∗-polynomials

In the previous section, we saw that h∗-polynomials for lattice simplices count the number of lattice points on diﬀerent heights of the fundamental parallelepiped. For general lattice polytopes, it is a bit more un- clear what the h∗-polynomials count, but Stanley’s non-negativity theorem seems to imply that they count something. In the case of the unit cube, the h∗-polynomial turned out to count the number of permutations with a certain number of descents. There are many other such distributions that can also be described as an h∗-polynomial of some polytope. Therefore, we are interested in their distributional properties, such as unimodality and symmetry.

A polynomial 2 n p(x) = p0 + p1x + p2x + ... + pnx is unimodal if p0 ≤ p1 ≤ ... ≤ pm ≥ pm+1 ≥ ... ≥ pn for some m ∈ {0, 1, 2, ..., n}. It is log-concave if 2 pi ≥ pi−1pi+1 for every i ∈ {1, 2, ..., n − 1}. It is symmetric with respect to d if pi = pd−i for every

12 i ∈ {0, 1, ..., d}. A polynomial that is symmetric with respect to its degree is sometimes just said to be symmetric. Note that by Hibi’s theorem (Theorem 2.2.11), the h∗-polynomial of a reﬂexive polytope is always symmetric. A polynomial with real coeﬃcients is said to be real-rooted if all its roots are real, or if it is identically zero.

n For a polynomial p(x) = p0 + p1x + ··· + pnx , an internal zero is an index i such that pi = 0 and there exists numbers j < i and k > i where pj 6= 0 and pk 6= 0. Unimodality and log-concavity are related to real-rootedness in the following way:

2 n Theorem 2.3.1 (Theorem 1.2.1 in [9]). If p(x) = p0 +p1x+p2x +···+pnx is real-rooted with nonnegative coeﬃcients, then it is log-concave with no internal zeros which implies that it is unimodal. The following are some other ﬁrst results that are useful in determining when a given polynomial is symmetric and/or unimodal.

n The mode of a unimodal polynomial p(x) = p0 + p1x + ... + pnx is the index i such that pi is maximal. If there are multiple such indices, then the mode is deﬁned as the median of them. Note that the mode of a symmetric and unimodal polynomial is always d/2 where d is the degree. Proposition 2.3.2. The sum of two unimodal polynomials p(x) and q(x) with the same mode is unimodal.

Proof. Let m be the mode. Note that p0 ≤ p1 ≤ ... ≤ pbmc ≥ pbmc+1 ≥ ... ≥ pn and the same is true for q(x), so p0 + q0 ≤ p1 + q1 ≤ ... ≤ pbmc + qbmc ≥ pbmc+1 + qbmc+1 ≥ ... ≥ pn + qn. This proves that p(x) + q(x) is unimodal. Corollary 2.3.3. The sum of a ﬁnite number of unimodal polynomials that are symmetric with respect to d, is also unimodal and symmetric with respect to d. Proposition 2.3.4 (Theorem 1 in [1]). Let

n p(x) = p0 + p1x + ... + pnx

and m q(x) = q0 + q1x + ... + qmx be two unimodal and symmetric polynomials. Then p(x)q(x) is also unimodal and symmetric. In Chapter 5, we will use these various results in order to prove that a certain family of h∗-polynomials are always symmetric and unimodal. To do so we will prove ﬁrst in Chapter 4 that a family of box polynomials are always real-rooted. Box polynomials and their connection to unimodality of h∗-polynomials is the topic of the next section.

2.4 Triangulations and Box Polynomials

In this section, box polynomials and some applications of them will be presented. Box polynomials will be featured heavily in the following chapters, especially Chapter 4. So let us start by defining them: Definition 2.4.1. Let P be a d-dimensional lattice simplex, and Π◦ the interior of its fundamental parallelepiped. The box polynomial BP (z) is defined as

d X k BP (z) := σΠ◦ (1, 1, 1, ..., 1, z) = bkz , k=0

◦ where bk is the number of lattice points of Π with last coordinate k.

13 One useful property of box polynomials is that they are always symmetric.

Proposition 2.4.2. For any d-dimensional lattice simplex P, BP (z) is symmetric with respect to d + 1. Proof. Let d X i BP (z) = biz . i=0 ◦ d+1 To prove that bk = bd+1−k, we will ﬁnd a bijection between the subset of ΠP ∩ Z with last coordinate k, and the subset with last coordinate d + 1 − k. Pick a point

d+1 X w = λiwi i=1

◦ d+1 in ΠP ∩ Z with last coordinate k, and map it to

d+1 X f(w) := (1 − λi)wi. i=1

Since x was in the interior of ΠP , the weights λi are strictly between 0 and 1, so f(w) will also be in ◦ d+1 ΠP ∩ Z , but it will have last coordinate d + 1 − k. Also, f is a bijection since we can invert it by applying it a second time.

The main application of box polynomials is a theorem by Betke and McMullen that gives a way to express h∗-polynomials as a kind of weighted sum of certain box polynomials. Before stating that theorem, we will need some more deﬁnitions.

A simplicial polytope is a convex polytope whose proper faces (faces other than the polytope itself) are all simplices. The faces of a simplicial polytope can be thought of as a triangulation of its boundary (recall from Deﬁnition 2.1.1 that triangulations are deﬁned for any set S ⊆ Rn).

Let S ⊆ Rn, and T a triangulation of S. Deﬁne the f-polynomial f(T ; z) of T as

d X i f(T ; z) := fi−1z , i=0 where fi is the number of i-dimensional simplices of T if i ≥ 0, and f−1 := 1. Similarly, let the h-polynomial 2 d of T , h(T ; z) = h0 + h1z + h2z + ... + hdz , be deﬁned by

d d X d−i X d−i fi−1(z − 1) = hiz . i=0 i=0

n Let S ⊆ R , and T a triangulation of S. For a simplex ∆ ∈ T , deﬁne linkT (∆) to be the set of simplices in T that are disjoint from ∆, but are contained in a simplex of T that also contains ∆. In other words,

0 0 linkT (∆) := {Ω ∈ T such that Ω ∩ ∆ = ∅ and there exists ∆ ∈ T such that ∆, Ω ⊂ ∆ }

Note that linkT (∆) is itself a triangulation, and so it has an f-polynomial and an h-polynomial.

14 Theorem 2.4.3 (Betke-McMullen). Let P be a lattice d-polytope, and T a triangulation of P. Then

∗ X hP (z) = h(linkT (∆); z)B∆(z). ∆∈T

If we have a reﬂexive polytope, then the following corollary to Theorem 2.4.3 (which follows directly from Theorem 10.5 in [7]) can be used instead.

Corollary 2.4.4 (Theorem 10.5 in [7]). Let P be a reﬂexive lattice d-polytope and T a triangulation of its boundary. Then ∗ X hP (z) = h(linkT (∆); z)B∆(z). ∆∈T This corollary will be especially useful in Chapter 5. It can be used to prove unimodality and symmetry ∗ for h -polynomials under certain conditions. First, we need the polynomials h(linkT (∆); z) to be unimodal and symmetric, and have degree d − dim(∆) − 1. Since the box polynomials are symmetric with respect to dim(∆) + 1, this would imply that the product h(linkT (∆); z)B∆(z) is symmetric with respect to d. Therefore, the sum of them would be symmetric with respect to d as well. If, in addition to this, the box polynomials are unimodal, then by Proposition 2.3.4 this would imply that h(linkT (∆); z)B∆(z) is unimodal and symmetric. This would in turn mean that their mode is d/2, and since the sum of unimodal polynomials ∗ with the same mode is unimodal (see Corollary 2.3.3), this would mean that hP (z) is unimodal. This is why we are interested in proving unimodality for box polynomials. For boundaries of simplicial polytopes, the conditions needed above on the h-polynomials are satisﬁed. This is true thanks to the following lemma, which follows from Lemma 2.9 in [27].

Lemma 2.4.5 (Lemma 2.9 in [27]). Let T be the boundary of a simplicial polytope and let ∆ ∈ T . Then h(linkT (∆); z) has degree d − dim(∆) − 1, is symmetric, and is unimodal. On the other hand, the following lemma says that if the triangulation is regular, then the polytope does not have to be simplicial. Lemma 2.4.6 (Lemma 9 in [10]). Let P be a polytope with a regular triangulation that induces a triangulation T on the boundary. Then there exists a simplicial polytope Q such that the boundary triangulation T 0 of Q is combinatorially equivalent to T . In particular, for each ∆ ∈ T ,

0 h(linkT (∆); z) = h(linkT 0 (∆ ); z) for some ∆0 ∈ T 0.

These applications of box polynomials to unimodality of h∗-polynomials were first stated in [24], where the authors collected these various necessary properties into a definition. Definition 2.4.7. A triangulation T of a lattice polytope P is called box unimodal if 1. T is regular, and

2. The box polynomials of all ∆ ∈ T are all unimodal. Note that thanks to Theorem 2.4.3 and the arguments above, a reﬂexive polytope that admits a box unimodal triangulation has a unimodal and symmetric h∗-polynomial.

15 Chapter 3

s-Eulerian Polynomials

In this chapter, a generalization of Eulerian polynomials, called the s-Eulerian polynomials, will be presented. In recent years, the s-Eulerian polynomials have been the focus of numerous research projects in combinatorics (see for example [5, 6, 12, 17, 21, 22, 23]). They turn out to be the h∗-polynomial of a certain kind of lattice simplex, and it is the box polynomials of these polytopes that are studied in this thesis. As a special case of these polynomials, we will recover the regular Eulerian polynomials. Recall that the Eulerian polynomials count descents in permutations. In order to relate these to the lattice points counted by the h∗-polynomial, we will need a diﬀerent representation of permutations, called Lehmer codes.

3.1 Lehmer codes

There are n! permutations of length n. How can we represent each permutation uniquely as a number from 0 to n! − 1 in a good way? Lehmer codes provide an answer to that question.

Deﬁnition 3.1.1. For a permutation π = π1π2...πn, deﬁne the Lehmer code for π as

L(π) := (l1, l2, ..., ln), where li = #j ∈ {i + 1, i + 2, ..., n} such that πi > πj. Let hni denote the set {0, 1, ··· , n − 1}. L(π) is an element in hni × hn − 1i × ... × h1i. We can also map from Lehmer codes to permutations, thereby proving that the map L is a bijection.

Proposition 3.1.2. The map L : Sn → hni × hn − 1i × ... × h1i is a bijection.

Proof. In order to invert the map L, take a Lehmer code l = (l1, l2, ..., ln). First, create a set of numbers A = {1, 2, ..., n} and a (currently empty) permutation p = ∅. Go through l from left to right. For each li, take the li + 1:th smallest number from A, remove it from A, and append it to p. After this is done, p will be the unique permutation such that L(p) = l. To answer the question posed at the beginning of the section, we can get a number from a Lehmer code like this: f(l) = (n − 1)!l1 + (n − 2)!l2 + ... + 0!ln. This number also has the nice property that f(L(π)) < f(L(σ)) if and only if π is lexicographically smaller than σ. More important to us is the following property. Recall that in Example 2.2.12 we deﬁned the descents of a permutation. There is an easy way to spot descents by looking at the Lehmer code:

Lemma 3.1.3. Let π ∈ Sn and L(π) = l1l2...ln. Then i is a descent of π if and only if li > li+1.

16 Proof. Assume that i is a descent. Since πi > πi+1, for any index j > i + 1 such that πi+1 > πj we will also have that πi > πj, so li ≥ li+1. However, since πi > πi+1, li will increase by one, so li > li+1. For the other direction, assume that li > li+1. If i was not a descent, then for any j > i such that πi > πj we would also have πi+1 > πj. This means that li ≤ li+1, which is a contradiction. Lehmer codes give an alternative way of describing permutations, and this will help us in connecting permutations to the objects appearing in the following sections.

3.2 s-inversion sequences

In this section, a kind of generalization of Lehmer codes, called s-inversion sequences, will be introduced.

Deﬁnition 3.2.1. Let s = (s1, s2, ..., sn) be a sequence of positive integers. An s-inversion sequence is a sequence of non-negative integers e = (e1, e2, ..., en) such that 0 ≤ ei < si for all i = 1, 2, ..., n. The set of s-inversion sequences is denoted Is. Just like for permutations, we can deﬁne ascents and descents of s-inversion sequences.

Deﬁnition 3.2.2. For an s-inversion sequence e, the number of ascents is deﬁned as e e asc(e) := #i ∈ {0, 1, 2, ..., n − 1} such that i < i+1 , si si+1

where s0 := 1. Similarly, the number of descents is deﬁned as e e des(e) := #i ∈ {1, 2, ..., n} such that i > i+1 , si si+1

where sn+1 := 1. In other words, the number of ascents (descents) of an s-inversion sequence is the number of pairs of adjacent elements where ei < ei+1 ei > ei+1 , except that you get one extra ascent (descent) if the first si si+1 si si+1 (last) element is non-zero. For reasons that will become more clear later, this turns out to be the most “natural” definition of ascents (descents). We can think of the “extra” (i.e. s0 = 1 and sn+1 = 1) part as adding a 1 to the beginning and end of s. This leads to the following alternative definition. Definition 3.2.3 (Alternative to Definition 3.2.2). For an s-inversion sequence e, let s0 = (1, s, 1), and let e0 be the s0-inversion sequence (0, e, 0). The number of ascents of e is defined as the number of adjacent elements in e0 where 0 0 ei ei+1 0 < 0 . si si+1 The number of descents is defined similarly. To recover the regular Eulerian polynomials, the following choice of s is used: Proposition 3.2.4. Let s = (1, 2, 3, ..., n). Then X X zasc(e) = zdes(π).

e∈Is π∈Sn

Proof. We will ﬁnd a bijection f : Is → Sn such that des(f(e)) = asc(e). For some e ∈ Is, let f(e) be the permutation corresponding to the Lehmer code

le := enen−1...e1.

17 Figure 3.1: An s-inversion sequence with two ascents and two descents. The bars represent s and the dots represent e.

Thanks to Proposition 3.1.2, this is a bijection. Also, since e1 is always 0, we will not have any extra ascents in e. An ascent in e is a pair of adjacent elements such that e e e i < i+1 = i+1 . si si+1 si + 1 This simpliﬁes to siei + ei < siei+1 ⇐⇒ ei < ei+1.

However, in le, ei < ei+1 corresponds to a pair of adjacent elements where lj > lj+1. Due to Lemma 3.1.3, this corresponds to a descent in f(e). So the proof is done. Recall from Example 2.2.12 that P zdes(π) is the Eulerian polynomial A (z). For this reason, the π∈Sn n polynomial X asc(e) Es(z) := z

e∈Is is called the s-Eulerian polynomial. For the regular Eulerian polynomials, it does not matter whether we use des(π) or asc(π) because they are equidistributed:

X des(π) X asc(π) An(z) = z = z .

π∈Sn π∈Sn The reason for this is that each permutation with k descents and l ascents can be mapped bijectively to a permutation with k ascents and l descents by swapping each πi with n − πi + 1 (another way is to reverse π). This is actually true for s-Eulerian polynomials as well: Theorem 3.2.5. For any sequence of positive integers s,

X asc(e) X des(e) Es(z) = z = z .

e∈Is e∈Is

18 Proof. Just like with the regular Eulerian polynomials, the idea is to find a bijection from Is that maps elements with k ascents and l descents to elements with k descents and l ascents. The bijection looks like this: f(e) := −e mod s, meaning that the i:th element of f(e) is −ei modulo si. So, for example, if s = (5, 4, 5) and e = (0, 3, 1) then f(e) = (0, 1, 4). This is a bijection, because we can invert it by just applying f again. Replacing ei with −ei mod si will “turn around” the numbers, meaning that ascents should turn into descents and vice versa. Unfortunately, that is not always the case, because zeros get mapped to themselves. However, it will work out anyway since we fixed two zeros at either end of e0 in Definition 3.2.3. To be a bit more precise: if f(e)i ei ei > 0, then f(e)i = si − ei, and so = 1 − . This implies that if ei > 0 and ei+1 > 0, then si si

1. If ei < ei+1 then f(e)i > f(e)i+1 , si si+1 si si+1

2. If ei > ei+1 then f(e)i < f(e)i+1 , and si si+1 si si+1

3. If ei = ei+1 then f(e)i = f(e)i+1 . si si+1 si si+1 0 Let ae be the number of ascents between non-zero numbers of e = (0, e, 0), and de is the number of descents between non-zero numbers. What we proved above is that ae = df(e) and de = af(e). There is one more 0 0 way an ascent can occur: if ei = 0 and ei+1 > 0. Let αe be the number of such ascents, and let δe be the 0 0 0 number of descents where ei > 0 and ei+1 = 0. Now since e has a zero on both endpoints, both αe and δe 0 are equal to the number of contiguous segments of non-zero numbers in e , and in particular, αe = δe. Also, since f maps zeros to zeros and non-zeros to non-zeros, it holds that αe = αf(e) and δe = δf(e). Combining all this, we get des(e) = de + δe = af(e) + αf(e) = asc(f(e)), and asc(e) = ae + αe = df(e) + δf(e) = des(f(e)). This proves the statement. There are a lot of well-studied polynomials in combinatorics that can be written as s-Eulerian polynomials, so proving things about them is important. As noted in Chapter 2, there is an avid interest in the distributional properties of polynomials whose coeﬃcients form a combinatorial sequence. Thus, Savage and Visontai [23] proved that they are real-rooted, and therefore unimodal. Theorem 3.2.6 (Theorem 1.1 in [23]). For any sequence of positive integers s, the s-Eulerian polynomial Es(z) is real-rooted.

3.3 s-lecture hall simplices

As mentioned at the start of Chapter 3, the s-Eulerian polynomials are actually the h∗-polynomials of certain lattice simplices.

Deﬁnition 3.3.1. For a sequence s of positive integers, the s-lecture hall simplex Ps is deﬁned as n x1 xn Ps := (x1, ..., xn) ∈ R | 0 ≤ ≤ ... ≤ ≤ 1 . s1 sn

This can also be written in terms of the vertices of Ps:         0 0 0 s1 0  0   .  s2  P = conv   ,   ,  .  , ...,   . s .  .     .  .  .  sn−1  .  0 sn sn sn

19 Figure 3.2: The s-lecture hall simplex Ps, where s = (2, 3).

The s-lecture hall simplices are connected to the s-inversion sequences in the following way: Theorem 3.3.2 (Theorem 3.9 in [18]). Let s be a sequence of positive integers. Then the s-Eulerian ∗ polynomial is the h -polynomial of Ps. This was ﬁrst proved in [22]. In [18], there is a proof that uses a certain bijection between s-inversion sequences and the fundamental parallelepiped of Ps, which we will denote by Πs. This bijection will be useful later, so we state it here along with a somewhat sketchy version of the proof:

n Proof of Theorem 3.3.2. The idea is to find a bijection between Πs ∩ Z and Is, such that elements with last coordinate k get mapped to elements with k descents. Recall from Proposition 2.2.8 that the coefficients of ∗ n the h -polynomial count the number of lattice points of Πs ∩ Z with different last coordinates, so this kind n of bijection would prove the theorem. First, let’s examine what elements in Πs ∩ Z look like. Recall that

Πs = {λ1w1 + ... + λnwn such that all 0 < λi ≤ 1}.

Let x ∈ Π ∩ n. Then s Z     x1 λns1  x2  (λn−1 + λn)s2 x =   =   .  ...   ...  xn+1 (λ0 + ... + λn)

x1 Note that λn = s . Thus, 1 x 0 ≤ 1 < 1, s1

and xi+1/si+1 − xi/si = λn−i for i ∈ {1, 2, ..., n − 1}. Therefore, x x 0 ≤ i+1 − i < 1. si+1 si

xn Finally, λ0 = xn+1 − , and so sn xn 0 ≤ xn+1 − < 1. sn

xn However, this just means that xn+1 is always equal to d s e. In fact, any x satisfying the three constraints n n above will be in Πs ∩Z because we can recover the λi:s by taking the diﬀerences xi+1/si+1 −xi/si (they are

20 Figure 3.3: A graphical representation of the REM-function. Notice that the last coordinate of x is 3, which is also the number of descents of REM(x).

the hyperplane description of Πs). If we ignore the last element of x, it looks a little bit like an s-inversion sequence already, but there are two main diﬀerences. First, xi ≤ xi+1 , so it is “weakly increasing” and has si si+1 xi no descents. Also, there is no guarantee that < 1. So xi can go “outside of si”. Therefore, the way si to get a bijection to Is is to ignore the last coordinate of x, and take the other elements modulo si. Let n REM: Πs ∩ Z → Is be the function

n REM(x) := (xi mod si)i=1.

Here is some intuition as to why the last coordinate of x ends up being the number of descents of REM(x): For each i ∈ {1, 2, 3, ..., n}, we can consider the number b xi c as some kind of “winding number”. If b xi c = b xi+1 c si si si+1 then since xi ≤ xi+1 , it will not give a descent in REM(e). If, on the other hand, b xi c = b xi+1 c − 1, then si si+1 si si+1 since xi+1 − xi < 1, this will always give a descent in REM(e). Therefore, the number of descents should si+1 si xn be equal to the “total winding number” b c = bλ1 + ... + λnc. This is equal to xn+1 if and only if λ0 = 0, sn which happens if and only if REM(x)n = 0. Otherwise, xn+1 = bλ1 + ... + λnc + 1 . However, according to the deﬁnition of descents, we should get one extra descent when en > 0. So this works out nicely.

Since Ps is a simplex, it also has an associated box polynomial. These box polynomials turn out to generalize interesting combinatorial polynomials. This is the topic of the next chapter.

21 Chapter 4

s-Derangement Polynomials

In this chapter, we will investigate the box polynomials (see Deﬁnition 2.4.1) of s-lecture hall simplices. In particular, we would like to prove that they generalize the well-studied derangement polynomials, and that ◦ they are real-rooted. Before we can do that, we must translate the open parallelepiped Πs to s-inversion sequences using the “REM”-function from the proof of Theorem 3.3.2.

4.1 The box polynomial of an s-lecture hall polytope

For a sequence of positive integers s, let Bs(z) denote the box polynomial of Ps. Proposition 4.1.1. For a sequence of positive integers s,

X asc(e) X des(e) Bs(z) = z = z ,

e∈Is e∈Is

ei ei+1 where Is is the subset of Is where e1 6= 0, en 6= 0, and 6= for i ∈ {1, 2, ..., n − 1}. si si+1

As we go through the proof, the definition of Is will make more sense, and an alternative definition, similar to Definition 3.2.3, will be presented.

◦ ◦ Proof of Proposition 4.1.1. The open parallelepiped Πs is a subset of Πs. So if we just map Πs to s-inversion sequences using the REM-function, we will get an identity like the one above for some subset Is ⊆ Is. What remains to check is that this subset really is the one described above, and that ascents and descents are equidistributed on this set. ◦ Recall that we found the hyperplane description of Πs in the proof of Theorem 3.3.2. Since Πs is the interior of Πs, it has the same hyperplane description except that all inequalities have turned strict. So ◦ n n Πs ∩ Z is given by the vectors x ∈ Πs ∩ Z that also satisfy: 1. 0 < x1 , s1 2. 0 < xi+1 − xi , for i ∈ {1, 2, ..., n − 1}, and si+1 si

xn 3. 0 < xn+1 − . sn

x1 In terms of the REM-function, > 0 means that x1 ∈ {1, 2, ..., s1 − 1}, and so REM(x)1 6= 0. The second s1 xi+1 xi REM(x)i REM(x)i+1 xn constraint means that > , and so 6= . Recall that xn+1 is always given by d e. si+1 si si si+1 sn Thus, the third constraint is equivalent to xn < d xn e. Equivalently, sn sn

sn 6 | xn,

22 ◦ which translates to REM(x)n 6= 0. Therefore, the image REM(Πs ) is the set Is from the statement of the proposition. What remains is to show that ascents and descents are equidistributed, which can be done by using the bijection from the proof of Theorem 3.2.5, and that this bijection maps Is to itself. This is left as an exercise.

We can now present our alternative deﬁnition of the set Is in analogy to Deﬁnition 3.2.3.

Definition 4.1.2 (Alternative definition of Is). We will call the set Is from Proposition 4.1.1 the strict 0 s-inversion sequences. Let s := (1, s, 1). The set Is is now defined as the set of s-inversion sequences where

0 0 ei ei+1 0 6= 0 si si+1 for any pair of adjacent elements of e0. We will call a pair of indices i and i + 1 such that

0 0 ei ei+1 0 = 0 si si+1 a collision. One useful property of both h∗-polynomials and box polynomials of s-lecture hall simplices is that they are unchanged after reversing s.

∗ Proposition 4.1.3. For a sequence of positive integers s, let s = (sn, sn−1, ..., s1) be the reverse of s. Then

Es(z) = Es∗ (z),

and Bs(z) = Bs∗ (z).

Proof. We need to ﬁnd a bijection between Is and Is∗ so that elements with k descents and l ascents get mapped to elements with k ascents and l descents, and so that Is gets mapped to Is∗ . Then the proposition would follow thanks to the fact that ascents and descents are equidistributed in Is and Is∗ . Such a bijection is given by reversing each element e ∈ Is.

4.2 Derangements and the sequence s = (n, n − 1, n − 2, ..., 2)

The s-Eulerian polynomial corresponding to s = (1, 2, 3, ..., n) are the regular Eulerian polynomials, so this is the most “basic” example in some sense. Therefore, it seems like a good idea to see what the box polynomial corresponding to s = (1, 2, 3, ..., n) looks like. First, let us instead look at s = (n, n − 1, ..., 1). Thanks to Proposition 4.1.3, this makes no diﬀerence, but we do not have to reverse the Lehmer codes all the time (like we did in the proof of Proposition 3.2.4). However, there is a problem with this choice of s: the last element is 1. This means that any element e ∈ Is will have en = 0, and so it will not be an element of Is. Consequently, Bs(z) will be identically zero. Instead, we will consider s = (n, n − 1, ..., 2). Ignoring the last 1 changes nothing in Es(z): the number of ascents and descents is unchanged in Is, and the interpretation of e ∈ Is as a Lehmer code still works because the last element of a Lehmer code is always zero. The box polynomial however, gets much more interesting.

Deﬁnition 4.2.1. A permutation π ∈ Sn is called a derangement if πi 6= i for all i = {1, 2, 3, ..., n}. The set of derangements of size n is denoted Dn.

Deﬁnition 4.2.2. For a permutation π ∈ Sn, the number of of excedances, exc(π), is deﬁned as the number of indices where πi > i.

23 Proposition 4.2.3. Let s = (n, n − 1, ..., 2). Then

X exc π Bs(z) = z .

π∈Dn

Proof. The goal is to ﬁnd a bijection between Is and Sn that maps elements with k ascents to elements with k excedances, and maps Is to Dn. The idea is to use Is as some kind of Lehmer codes, but instead of placing the numbers πi one at a time in order, the bijection will build the cycles of π one element at the time:

For some e ∈ Is, go through each element in (0, e, 0) in order. First, create a set A = {1, 2, 3, ..., n}, the set of available elements, and keep track of the current cycle we are working on (at first, this cycle is 0 empty). If ei = 0, close off the current cycle (if it is non-empty), start a new cycle at the smallest element of 0 0 A, and remove it from A. If ei 6= 0, add the ei:th smallest number of A to the current cycle and remove it from A. The endpoints of (0, e, 0) are a bit special, because we do not have to start a new cycle at the last element, and there is no cycle to close off at the first. The permutation obtained from this process is f(e). It is a bijection. To invert it, take the smallest element from each cycle, sort the cycles according to these elements, and reverse the operations above.

0 0 ei ei+1 An ascent in e ∈ Is is a pair of adjacent elements in (0, e, 0) where 0 < 0 . There are two ways this si si+1 0 0 0 0 0 can happen. First, either both ei and ei+1 are non-zero and ei ≤ ei+1. This corresponds to adding the ei:th 0 0 0 smallest number to a cycle and connecting it to the ei+1:th smallest number. However, since ei ≤ ei+1, the ﬁrst number is smaller than the second. So this edge in the cycle decomposition in f(e) will be an excedance. 0 0 The second way an ascent can happen is if ei is zero and ei+1 is non-zero. This corresponds to starting a new cycle at the smallest number, and then connecting it to some bigger number. Again, this edge in the cycle decomposition is an excedance. Also, all excedances correspond to these ascents, so exc(f(e)) = asc(e). 0 0 ei ei+1 0 0 Finally, the only way a collision 0 = 0 can happen is if ei = ei+1 = 0. So Is is the set of s-inversion si si+1 sequences where no two adjacent elements of e0 are zero. In f(e), two adjacent zeros corresponds to starting a new cycle and then immediately closing it oﬀ. Therefore, Is corresponds to the set of permutations with no cycles of length 1, which is the set of derangements.

24 Algorithm 1 The following Python code represents the bijection described in the proof of Proposition 4.2.3. The input is a sequence (0, e, 0) and the output is a permutation. e = list(map(int,input().split())) n = len(e)-1 unvisited = list(range(0,n)) result = list(range(0,n)) current = -1 cycle = -1 for i in range(0,n+1): if e[i] == 0: if current != -1: result[current] = cycle + 1 if i == n: break current = min(unvisited) cycle = current unvisited.remove(current) else: result[current] = unvisited[e[i]-1] + 1 current = unvisited[e[i]-1] unvisited.remove(current) print(result)

The polynomial d (z) = P zexc π is called the derangement polynomial. For that reason, we n π∈Dn will call Bs(z) for general s the s-derangement polynomial ds(z). The derangement polynomials dn(z) have a long history in enumerative and algebraic combinatorics (see for instance [3], [4], [25]). In fact, a number of their well-studied generalizations are also s-derangement polynomials.

4.3 Colored permutations

An r-colored permutation is a permutation π ∈ Sn together with a list

n c = (c1, c2, ..., cn) ∈ {0, 1, ..., r − 1} .

The number ci is the color of πi. The set of r-colored permutations is denoted Sn,r. A colored permutation c1 c2 c3 cn will be written σ = (π, c) = π1 π2 π3 ...πn .

Deﬁnition 4.3.1. The number of descents of a colored permutation σ = (π, c) ∈ Sn,r, denoted des(σ), is the number of indices i ∈ {1, 2, 3, ..., n − 1} where either ci > ci+1, or ci = ci+1 and πi > πi+1, plus one if cn 6= 0. The polynomial X des(σ) An,r(z) = z

σ∈Sn,r is called the Eulerian polynomial of Sn,r.

Proposition 4.3.2 (Theorem 3 in [19]). For positive integers n and r, An,r(z) is equal to the s-Eulerian polynomial for s = (rn, rn − r, ..., 2r, r).

Proof. There is a bijection between Is and Sn,r that preserves the number of descents. Since it is very similar to the one for s = (n, n − 1, ..., 1), some details will be omitted. The bijection is to take an element n ei n e ∈ Is, let c := (ei mod r)i=1 be the colors, and let L := {b n−i+1 c}i=1 be the Lehmer code that gives the permutation.

25 In the case when s = (n, n−1, ..., 2), the box polynomials turned out to count excedances of derangements in Sn. In this section, we will look at r-colored permutations of size n, for positive integers r and n. Before, when r was 1, we were forced to remove the last element of s in order to get a non-zero box polynomial. When r > 1 we can keep the last element and still get a non-zero box polynomial. However, if we want to generalize the result from r = 1 it seems like a good idea to throw away the last element anyway. So we will look at the box polynomials of both s = (rn, rn − r, rn − 2r, ..., r) and s = (rn, rn − r, rn − 2r, ..., 2r). It turns out that they are both related to derangements, and connected to each other in an interesting way.

1 0 Deﬁnition 4.3.3. For positive integers n and r, denote sn,r := (rn, rn−r, ..., r) and sn,r := (rn, rn−r, ..., 2r). 1 1 0 0 Also, let Bn,r(z) be the box polynomial of sn,r, and let Bn,r(z) be the box polynomial of sn,r. 1 Deﬁnition 4.3.4. Let Sn,r be the set of r-colored permutations of size n whose last color is non-zero, and 0 let Sn,r be the set of r-colored permutations of size n whose last color is zero. Let

b X des(σ) An,r(z) = z b σ∈Sn,r for b ∈ {0, 1}.

0 0 1 Note that An,r(z) is the s-Eulerian polynomial of sn,r, but this is not true for An,r(z). However, it is the s-Eulerian polynomial if we demand the last entry to be non-zero, in the following sense:

1 X des(e) An,r(z) = z .

e∈I 1 where en6=0 sn,r The main result in this section is the following theorem. Theorem 4.3.5. For positive integers n and r, and an integer b ∈ {0, 1},

X n Bb (z) = (−1)n−k Ab (z). n,r k k,r k=0 To prove it, we will ﬁrst need some more deﬁnitions and lemmas.

Deﬁnition 4.3.6. For a colored permutation σ = (π, c) ∈ Sn,r, we will say that a number i ∈ {1, 2, ..., n} is bad with respect to σ if for πj = i it holds that

1. πj < πk for every k > j,

2. πj−1 < πk for every k > j − 1, and

3. πj and πj−1 have the same color.

Here the convention π0 = 0 and c0 = 0 is used. Let Sσ be the set of bad numbers in σ. Note that a number i is bad if and only if ej−1 = ej for the corresponding index j in the corresponding sj−1 sj s-inversion sequence.

b,T b Deﬁnition 4.3.7. For a subset T ⊆ {1, 2, 3, ..., n}, let An,r (z) be An,r(z), but restricted to the permutations where all numbers in T are bad. More precisely,

b,T X des(σ) An,r (z) := z . b σ∈Sn,r and T ⊆Sσ Lemma 4.3.8. For positive integers n and r and an integer b ∈ {0, 1},

b X |T | b,T Bn,r(z) = (−1) An,r (z) . T ⊆{1,2,...,n}

26 b Proof. First, notice that the box polynomial lists the elements of Sn,r whose set of bad numbers is empty. In other words, b X des(σ) Bn,r(z) = z . b σ∈Sn,r and Sσ =∅ 0 1 This is because, in sn,r, collisions correspond bijectively to bad numbers. In sn,r, there is one more way a 1 collision could happen: if the last element of e is zero. However, in Sn,r, the last color is always non-zero, 1 b,T so the last element of the corresponding sn,r-inversion sequence is never zero. Now we note that An,r (z) in a similar way lists the permutations where T is a subset of the bad numbers. Using this, we can prove the lemma with an inclusion-exclusion argument. Note that in the expression on the right-hand-side in the b lemma, every σ ∈ Sn,r contributes X zdes(σ) (−1)|T |.

T ⊆Sσ des(σ) This is 0 if Sσ 6= ∅, and otherwise it is z , so the only elements that “survive” are the ones with Sσ = ∅. As noted earlier, this gives the box polynomial.

b,T Finally, to prove Theorem 4.3.5, we need a nice expression for An,r (z): Lemma 4.3.9. For positive integers n and r, an integer b ∈ {0, 1}, and a subset T ⊆ {1, 2, 3, ..., n},

b,T b An,r (z) = An−|T |,r(z).

b b Proof. The idea is to ﬁnd a bijection between Sn−|T |,r and the subset of Sn,r where all numbers in T are bad, such that it preserves the number of descents. The bijection looks like this:

c1 c2 cn−|T | b 1. Take an element σ = π1 π2 ...πn−|T | ∈ Sn−|T |,r.

2. Replace each element πi = j with the j:th smallest element of {1, 2, 3, ..., n}\ T . (Note that this does not change the number of descents.) 3. We will now insert the numbers in T into our half-ﬁnished permutation, in such a way that they become bad. Pick each i ∈ T in order of size, starting with the smallest. If i = 1, insert i at the front of σ and give it color 0. Otherwise, ﬁnd the rightmost element πj such that πj < i and πj < πk for every k > j. Give i the same color as πj and insert it right after πj.

There are a couple of things we need to note about this function: First, it is a well-defined function that b maps to Sn,r. This follows if we can always find an index j for every number i 6= 1, so that we can insert i after πj. Since we are inserting the numbers in order of size, when we insert i 6= 1, 1 will already be in our permutation. Since 1 satisfies the two conditions above, the set of indices j where πj < i and πj < πk for every k > j is non-empty. Thus, the rightmost one exists. Finally, the number b must be preserved, i.e., if b = 0, then the last number of the result must have color zero, and otherwise, it must have nonzero color. This is true because the color of the last element is unchanged after the bijection. Second, the number of descents does not change. When we insert an element i 6= 1 after πj, then no new descents are created since πj < i. Also, if (πj, πj+1) was a descent earlier, then since i gets the same color as πj and is greater, the pair (i, πj+1) will still be a descent. If i = 1, then inserting it at the start and giving it color 0 does not affect the number of descents. Third, all the numbers in T are bad in the resulting permutation. If 1 ∈ T , then we will insert it at the start and give it color 0, which makes it bad. After that, we will never insert anything in front of it, so it will not stop being bad. For i 6= 1 in T , i will be bad right after inserting it. The reason for this is that it will have the same color as the preceding element πj, it will be greater than πj, and πj will be smaller than all elements to the right of it. The only thing remaining to show is that i is also smaller than everything to the right of it. Assuming that this is not true, pick the rightmost element πk such that πk < i. This element must be smaller than everything to the right of it, but πj was the rightmost element satisfying this, which

27 is a contradiction. Furthermore, the number i will stay bad after the bijection is ﬁnished. This follows from the fact that we will never insert anything between πj and i since all the elements inserted after i will be greater than i. Also, inserting larger numbers into our permutation does not change the fact that πj or i is smaller than everything to the right of it. Finally, we can invert the function. To invert it, remove all numbers in T from the permutation and undo step 2 from the description of the bijection. This completes the proof.

Example 4.3.10. As an example of the bijection in the proof of Lemma 4.3.9, suppose that n = 6, r = 3, 2 1 0 0 b = 0, and T = {1, 3, 4}. Now the bijection will look as follows: First, take an element σ = 2 1 3 ∈ S3,3. Applying step 2 in the description will transform it to 522160. Now we will insert the bad numbers one at a time as in step 3. First, the 1 will end up at the front and get color zero, so the permutation becomes 10522160. Next, the 3 will be inserted after 21 and get color 1, so the permutation becomes 1052213160. Finally, the 4 will be inserted after 31 and get color 1, so the ﬁnal permutation is 105221314160. We are now ready to prove Theorem 4.3.5. Proof of Theorem 4.3.5. Rewrite the inclusion-exclusion expression from Lemma 4.3.8 using Lemma 4.3.9:

n X X X n Bb (z) = (−1)|T |Ab,T (z) = (−1)|T |Ab (z) = (−1)n−k Ab (z) n,r n,r n−|T |,r k k,r T ⊆{1,2,...,n} T ⊆{1,2,...,n} k=0

In order to relate the box polynomials to the Eulerian polynomials Ak,r(z), we will use the following corollary. Corollary 4.3.11. For positive integers n and r,

n X n B0 (z) + B1 (z) = (−1)n−k A (z) n,r n,r k k,r k=0

0 1 Proof. This follows from Theorem 4.3.5 and the fact that Sk,r is the disjoint union of Sk,r and Sk,r, so

0 1 Ak,r(z) + Ak,r(z) = Ak,r(z).

In order to relate these box polynomials to excedances and derangements, we must ﬁrst deﬁne these concepts for colored permutations.

Definition 4.3.12. For a colored permutation σ ∈ Sn,r, the number of excedances, exc(σ), is defined as the number of indices i ∈ {1, 2, 3, ..., n} where either πi > i or πi = i and ci 6= 0. The definition above is due to Steingr´ımsson[28], who also proved that excedances and descents in colored permutations are equidistributed: Theorem 4.3.13 (Theorem 3.15 in [28]). For positive integers n and r,

X exc(σ) An,r(z) = z .

σ∈Sn,r

Deﬁnition 4.3.14. A colored permutation σ ∈ Sn,r is a derangement if for every i ∈ {1, 2, 3, ..., n}, it holds that either πi 6= i, or ci 6= 0. The set of derangements is denoted Dn,r. The derangement polynomial is deﬁned as X exc(σ) dn,r(z) := z .

σ∈Dn,r

28 Using another inclusion-exclusion argument, we can actually show that these derangement polynomials are equal to the right-hand-side of Corollary 4.3.11. Lemma 4.3.15. For positive integers n and r,

n X n d (z) = (−1)n−k A (z) . n,r k k,r k=0 Proof. The idea is to use inclusion-exclusion and the fact that excedances and descents are equidistributed. For a subset T ⊆ {1, 2, 3, ..., n}, let Sn,r,T be the set of colored permutations such that for all i ∈ T , ci = 0 and πi = i (in other words, all indices in T are ﬁxed points). Now it follows from inclusion-exclusion that

X |T | X exc(σ) dn,r(z) = (−1) z .

T ⊆{1,2,3,...,n} σ∈Sn,r,T

Note that Sn,r,T is in bijection with Sn−|T |,r in a way that preserves the number of excedances. The bijection is to just remove the ﬁxed points. Hence, X X zexc(σ) = zexc(σ) .

σ∈Sn,r,T σ∈Sn−|T |,r

Now, combining this with Theorem 4.3.13 gives

n X n d (z) = (−1)n−k A (z) . n,r k k,r k=0

Athanasiadis [2] proved that the polynomials dn,r(z) can be expressed as the sum of two polynomials

+ − dn,r(z) = dn,r(z) + dn,r(z),

+ − n n+1 where dn,r(z) and dn,r(z) are symmetric with centers of symmetry b 2 c and b 2 c, respectively (as well as having a lot of other properties). Expressing a polynomial as this kind of sum is unique:

Pn+1 i Lemma 4.3.16. Any polynomial p(z) = i=0 piz can be written as p(z) = p+(z) + p−(z),

n n+1 where the terms are symmetric with centers of symmetry b 2 c and b 2 c, respectively. Moreover, this sum is unique. Proof. Let i − X pi = pk − pn+1−k, k=0 and i−1 + X pi = pn+1 + pn−k − pk. k=0 Now we can check that

− + 1. pi + pi = pi, − − 2. pi = pn+1−i, and

29 + + 3. pi = pn−i. For uniqueness, note that we can relate the coeﬃcients with a linear system of equations:

Ad∗ = d, where

∗ + + − − T d = (d0 , ··· , dn , d0 , ··· , dn+1) , and T d = (d0, d1, ··· , dn+1) . The matrix A is n × n and its columns are linearly independent, so the solution is unique.

We will call the decomposition of a polynomial p(z) into p+(z) and p−(z) as in Lemma 4.3.16 its symmetric decomposition. We now collect all of this in a theorem. Theorem 4.3.17. For positive integers n and r,

0 + 1 − Bn,r(z) = dn,r(z) and Bn,r(z) = dn,r(z).

That is, the unique symmetric decomposition of the derangement polynomial dn,r(z) is given by the pair of 0 1 s-derangement polynomials Bn,r(z) and Bn,r(z). Proof. Combine Corollary 4.3.11 and Lemma 4.3.15 to get

n X n B0 (z) + B1 (z) = (−1)n−k A (z) = d (z) = d+ (z) + d− (z). n,r n,r k k,r n,r n,r n,r k=0

0 n Note that it follows from Proposition 2.4.2 that Bn,r(z) is symmetric with center of symmetry b 2 c and 1 n+1 Bn,r(z) is symmetric with center of symmetry b 2 c, so they are the symmetric decomposition of dn,r(z). + − But this is true of dn,r(z) and dn,r(z) as well, so now the theorem follows from Lemma 4.3.16. Theorem 4.3.17 provides a new combinatorial and geometric interpretation of the symmetric decomposition of the derangement polynomials dn,r(z). In the next section, we will further show that they are real-rooted, thereby answering Conjecture 2.30 in [4].

4.4 Real-rootedness

The goal of this section is to prove the following theorem.

Theorem 4.4.1. For any sequence of positive integers s, the s-derangement polynomial ds(z) is real-rooted. To prove it, we will use a modiﬁed version of the proof from [23] of the real-rootedness of s-Eulerian polynomials. The main tool is something called compatible polynomials.

Deﬁnition 4.4.2. A set of polynomials {f1(x), f2(x), ..., fm(x)} with real coeﬃcients is called compatible if m X cifi(x) i=1

is real-rooted for every c1, ..., cm ≥ 0. A weighted sum of the form above is called a conic combination of the polynomials. The set is called pairwise compatible if {fi(x), fj(x)} is compatible for every pair of indices i, j.

30 If a set of polynomials is compatible, then, in particular, the sum of them is real-rooted. It is often easier to prove that polynomials are pairwise compatible, but this does not seem to imply that the sum is real-rooted. However, thanks to a lemma by Chudnovsky and Seymour [13], pairwise compatibility often implies compatibility.

Lemma 4.4.3 (Lemma 2.2 in [13]). If f1(x), f2(x), ..., fm(x) are pairwise compatible with positive leading coeﬃcients, then they are compatible. One useful way to show that a set of polynomials is compatible is to use recurrence relations. Thus, our next goal is to prove a recurrence relation similar to Lemma 2.1 in [23].

Definition 4.4.4. For a sequence of positive integers s and a non-negative integer k, let sk = (s0, s1, ··· , sk) be the first k + 1 elements of (1, s, 1). Define Is,k to be the set of sk-inversion sequences where e e i 6= i+1 si si+1 for any pair of adjacent elements. For example, Is,n+1 is just Is.

Deﬁnition 4.4.5. For a sequence of positive integers s, and non-negative integers k, i where i < sk, let

X asc e ds,k,i(z) := χ(ek = i)z ,

e∈Is,k

where χ(ek = i) is 0 if ek 6= i, and 1 otherwise. Notice that

ds,n+1,0(z) = ds(z).

Now we are ready to set up our recurrence relation.

Lemma 4.4.6. For a sequence of positive integers s, and integers k > 0 and i < sk,

t −1 sk−1−1 Xi X ds,k,i(z) = zds,k−1,h(z) + χ(sk 6 | isk−1)ds,k−1,ti (z) + ds,k−1,h(z),

h=0 h=ti+1

isk−1 where ti = d e. The initial condition is ds,0,0(z) = 1. sk

Proof. Let e = (e1, e2, ..., ek−2, h, i). To get ds,k,i(z) we will try all possible values for h. For a ﬁxed h, the box polynomial restricted to the inversion sequences that end with the ordered pair (h, i) will be ds,k−1,h(z), h i multiplied with z if (h, i) is an ascent. An ascent < happens exactly when h < ti. When h ≥ ti the sk−1 sk isk−1 ordered pair (h, i) will not be an ascent, but if is an integer and h = ti, then we will have a collision: sk h = i . So we must exclude this term when that happens. This gives the recurrence above. The initial sk−1 sk condition is true because Is,0 only contains the element (0), which has zero ascents. Now we will prove a modiﬁed version of Theorem 2.3 from [23].

Theorem 4.4.7. For a set of polynomials f0, f1, ..., fm−1 with real coeﬃcients and positive leading coeﬃcient, such that for 0 ≤ i ≤ j < m

(a) fi and fj are compatible, and

(b) zfi and fj are compatible,

31 deﬁne another set of polynomials g0, g1, ..., gm0−1 by

t −1 m−1 Xk X gk(z) = xfi(z) + akftk (z) + fi(z).

i=0 i=tk+1

Here, ak is 0 or 1, and 0 ≤ t0 ≤ t1 ≤ ... ≤ tm0−1 ≤ m. Also, if ti = tj for i < j, then we demand that ai ≥ aj. Then for i ≤ j

(a’) gi and gj are compatible, and

(b’) zgi and gj are compatible. Before we prove Theorem 4.4.7, we will collect some ideas from the proof of Theorem 2.3 [23] in a lemma.

Lemma 4.4.8. Let f0, f1, ..., fm−1 be a set of polynomials with real coeﬃcients and positive leading co- 0 0 eﬃcients, i, j two positive integers such that 0 ≤ i ≤ j < m, ti and tj are positive integers such that 0 0 0 ≤ ti ≤ tj < m, and ci and cj are any non-negative real numbers. Moreover, assume that for 0 ≤ α ≤ β < m

(a) fα and fβ are compatible, and

(b) zfα and fβ are compatible. Then the following two sets of polynomials are compatible:

0 t0 −1 ti−1 S j S m−1 (1) {xfα(x)}α=0 {(ci + xcj)fβ(x)} 0 {fγ (x)} 0 , and β=ti γ=tj

0 t0 −1 ti−1 S j S m−1 (2) {x(cix + cj)fα(x)}α=0 {xfβ(x)} 0 {(cix + cj)fγ (x)} 0 . β=ti γ=tj For a proof of this lemma, see the proof of Theorem 2.3 in [23]. We are now ready to prove Theorem 4.4.7.

Proof of Theorem 4.4.7. For i ≤ j, pick gi and gj as deﬁned in the theorem. Now we want to prove (a’) and (b’).

To prove that gi and gj are compatible, we need to prove that cigi+cjgj is real-rooted for any non-negative real numbers ci, cj. Now we get a few cases:

(i) ti < tj and ai = 0. Expanding cigi(z) + cjgj(z) gives

t −1 tj −1 m−1 Xi X X z(ci + cj)fα(z) + zcjfti (z) + (ci + zcj)fβ(z) + (ci + ajcj)ftj (z) + (ci + cj)fγ (z).

α=0 β=ti+1 γ=tj +1

0 0 This is a conic combination as in Lemma 4.4.8(1), if we let ti := ti + 1 and tj := tj.

(ii) ti < tj and ai = 1. Expanding cigi(z) + cjgj(z) gives

t −1 tj −1 m−1 Xi X X z(ci + cj)fα(z) + (ci + zcj)fti (z) + (ci + zcj)fβ(z) + (ci + ajcj)ftj (z) + (ci + cj)fγ (z).

α=0 β=ti+1 γ=tj +1

0 0 Now we again get a conic combination as in Lemma 4.4.8(1), but with ti := ti and tj := tj.

(iii) ti = tj. Expanding cigi(z) + cjgj(z) gives

t −1 m−1 Xi X z(ci + cj)fα(z) + (ciai + cjaj)fti (z) + (ci + cj)fγ (z).

α=0 γ=ti+1 This is a conic combination as in Lemma 4.4.8(1).

32 So in all these cases, we get that cigi + cjgj is real-rooted. Next, to prove (b’), we need to prove that zcigi + cjgj is real-rooted. Again, a few cases appear:

(iv) ti < tj and aj = 0. Expanding zcigi(z) + cjgj(z) gives

t −1 tj −1 m−1 Xi X X z(zci + cj)fα(z) + z(aici + cj)fti (z) + z(ci + cj)fβ(z) + zciftj (z) + (zci + cj)fγ (z).

α=0 β=ti+1 γ=tj +1

0 0 This is a conic combination as in Lemma 4.4.8(2), with ti := ti and tj := tj + 1.

(v) ti < tj and aj = 1. Expanding zcigi(z) + cjgj(z) gives

t −1 tj −1 m−1 Xi X X z(zci +cj)fα(z)+z(aici +cj)fti (z)+ z(ci +cj)fβ(z)+(zci +cj)ftj (z)+ (zci +cj)fγ (z).

α=0 β=ti+1 γ=tj +1

0 0 This is a conic combination as in Lemma 4.4.8(2) with ti := ti and tj := tj.

(vi) ti = tj. Expanding zcigi(z) + cjgj(z) gives

t −1 m−1 Xi X z(zci + cj)fα(z) + (zciai + cjaj)fti (z) + (zci + cj)fγ (z).

α=0 γ=ti+1

All the polynomials in the two sums are a conic combination of polynomials from Lemma 4.4.8(2). What remains is to show that the middle term is one of the polynomials in Lemma 4.4.8(2), up to a non-zero constant factor. Note that this is true in all cases unless ai = 0 and aj = 1. But this last case is guaranteed not to happen, because in Theorem 4.4.7 we ensured that ai ≥ aj whenever ti = tj.

Again, in all cases the polynomial zcigi + cjgj is real rooted, which is what we wanted to prove. This completes the proof.

Now we are ﬁnally ready to prove that ds(z) is real-rooted for any sequence of positive integers s (Theorem 4.4.1). Proof of Theorem 4.4.1. The idea is to use Theorem 4.4.7 inductively on the recurrence relation stated in Lemma 4.4.6. For the base case, note that the set {ds,0,0(z)} satisﬁes the requirements for f0, f1, ..., fm−1 in Theorem 4.4.7 because it is identically 1 which is real-rooted and therefore compatible with itself. Now

assume that ds,k−1,0(z), ds,k−1,1(z), ..., ds,k−1,sk−1−1(z) satisfy the requirements, when 0 ≤ k − 1 ≤ n. We want to show that this holds for the set corresponding to k as well. The reason for this is that the recurrence relation in Lemma 4.4.6 is a recurrence relation satisfying the requirements for g0, g1, ..., gm0−1 in Theorem isk−1 4.4.7. First, note that since ti = d e, we have that 0 ≤ t0 ≤ t1 ≤ ... ≤ tm0 ≤ m. Furthermore, sk the numbers ai := χ(sk 6 | isk−1) are either 0 or 1, and if ti = tj for i < j, then it cannot happen that ai = 0 while aj = 1, so ai ≥ aj. So applying Theorem 4.4.7 and induction, we conclude that {ds,n+1,0(z)} satisﬁes the requirements for f0, f1, ..., fm−1 in Theorem 4.4.7, and in particular, it is real-rooted. Since ds(z) = ds,n+1,0(z), the s-derangement polynomial ds(z) is real-rooted. Remark: Here, we used the theory of compatible polynomials to prove real-rootedness following the same approach taken by Savage and Visontai [23] for proving real-rootedness of the s-Eulerian polynomials. However, we note that combining Proposition 4.1.1 with Theorem 5.2 in [12] yields a second proof of 4.4.1.

+ − As an immediate corollary to Theorem 4.4.1, we get that the polynomials dn,k(z) and dn,k(z) in the symmetric decomposition of dn,k(z) are real-rooted. This answers Conjecture 2.30 in [4].

+ − Corollary 4.4.9. The polynomials dn,k(z) and dn,k(z) from Section 4.3 are real-rooted.

33 4.5 Faces of s-lecture hall simplices

Recall from Theorem 2.4.3 that in order to apply box polynomials to the study of h∗-polynomials we need to know the box polynomial not only of a simplex itself, but also that of each of its faces. In this section we will investigate the face structure of s-lecture hall simplices and prove that they have real-rooted box ∗ ∗ polynomials as well. Informally, a facet of Ps corresponds to s -inversion sequences, where s is obtained by taking two adjacent elements si, si+1 from s and replacing them with their greatest common divisor. However, it requires some care to make this precise, because a face F is not literally the polytope Ps∗ . We could try to prove that the lattice point enumerators of F and Ps∗ are the same, but unfortunately this does not imply that their box polynomials are the same:

Proposition 4.5.1. If P and Q are two lattice d-simplices such that LP (t) = LQ(t), then it is not always true that BP (z) = BQ(z).

Proof. For a counterexample, consider the s-lecture hall simplices corresponding to the sequences s1 = ∗ 2 (1, 2, 4) and s2 = (2, 2, 2). Their h -polynomials are both 1 + 6z + z , so by Ehrharts theorem (Theorem 2.2.1) their lattice point enumerators are the same. However, the ﬁrst polytope has box polynomial 0, while the second has box polynomial z2. Instead, the goal will be to ﬁnd a bijection between the lattice points of the open parallelepipeds. First, the following lemma will turn out to be useful later:

Lemma 4.5.2. Let s = (s1, s2, ..., sn) be a sequence of positive integers, and let (x1, x2, ..., xn) be some non-negative integers such that x x x 1 = 2 = ... = n . s1 s2 sn

gxi Then is an integer for all i = 1, 2, ..., n, where g = gcd(s1, s2, ..., sn). si Proof. We will prove that gx1 is an integer. By symmetry the same arguments work for the other indices s1 too. The idea is to prove the statement for n = 2, and use induction. Assume that n = 2. Now

s2 s x x1 x = 2 1 = g 2 s1 s1 g

s2 s1 s1 is an integer. But g and g are co-prime, so x1 must be divisible by g , which implies that x gx 1 = 1 s1 g s1

0 is an integer. For n > 2, let g = gcd(s1, ..., sn−1), and assume that the statement is true for s1, ..., sn−1. 0 This implies that g x1 is an integer. Thus, s1 0 g x1 s1 xn 0 = , g sn and we can use the result for n = 2 on this new sequence to conclude that

0 0 g x1 gcd(g , sn) s1 gx1 0 = g s1 is an integer. We are now ready to prove our main theorem of this section.

Theorem 4.5.3. For a sequence of positive integers s, let F be a face of Ps. Then either F is a point, or ∗ BF (z) = Bs∗ (z), for some sequence of positive integers s .

34 Proof. Recall that         0 0 0 s1 0  0   .  s2  P = conv   ,   ,  .  , ··· ,   . s .  .     .  .  .  sn−1  .  0 sn sn sn

We will call these vertices vn, vn−1, ..., v0. Note that this ordering is reversed compared to the one in Section 3.3. Let F be a face of Ps. A face of a simplex is the convex hull of a non-empty subset of the vertices, so we will write

F = conv(vi0 , vi1 , ..., vim ), ∗ ∗ ∗ ∗ where 0 ≤ i0 < i1 < ... < im ≤ n. If m = 0, then F is a point, so assume that m > 0. Let s = (s1, s2, ..., sm) be the sequence of positive integers deﬁned by

∗ sj = gj := gcd(sij−1+1, sij−1+2, ..., sij ),

◦ n+1 ◦ m+1 for j = 1, 2, ..., m. Now the goal is to ﬁnd a bijection between ΠF ∩ Z and Πs∗ ∩ Z such that points with last coordinate k get mapped to points with last coordinate k. The bijection looks like this:

◦ n+1 Take a point x ∈ ΠF ∩ Z . Note that it can be written as     s1l1 x1 m  s2l2   x2  X   x =   = λjwij =  ...  ,  ...    j=0  snln  xn+1 Pm j=0 λj

where wk := (vk, 1), λk ∈ (0, 1), and X lk := λj.

Let f(x) be the point in Πs∗ deﬁned by m X ∗ f(x) := λjwj , j=0 ∗ ∗ ∗ where wj = (vj , 1) and vj is the j:th vertex of Ps∗ . This is an injective function, and since f(x) can be written as   λ0g1  (λ0 + λ1)g2    f(x) =  ...  ,   (λ0 + ... + λm−1)gm Pm j=0 λj its last coordinate is the same as the last coordinate of x. What remains to show is that f is well-deﬁned ◦ m+1 (maps to Πs∗ ∩ Z ) and surjective. For j ∈ {1, 2, 3, ..., m}, note that

lij−1+1 = lij−1+2 = ... = lij .

n+1 We will call this number lj. Also, since x ∈ Z ,

ljsij−1+k = xij−1+k ∈ Z,

35 for k ∈ {1, 2, 3, ..., ij − ij−1}. Thus,

xij−1+1 xij−1+2 xij lj = = = ... = . sij−1+1 sij−1+2 sij

Using Lemma 4.5.2 on this we see that gjlj is an integer. However, note that f(x)j = gjlj for j ∈ {1, 2, ..., m}, m+1 ◦ so f(x) is in Z . Also, since x ∈ ΠF , the numbers λ0, ..., λm are contained in (0, 1). Therefore, f(x) ◦ m+1 is in the interior of Πs∗ , which means that it maps to Πs∗ ∩ Z . Furthermore, for a particular integer f(x)j = gjlj, we can let xij−1+k be sij−1+k xij−1+k = gjlj , gj for k ∈ {1, 2, 3, ..., ij − ij−1}, in order to map to f(x)j. The numbers xij−1+k will then be integers thanks to the fact that gj is a divisor of sij−1+k. This proves that f is surjective. Notice that by Theorem 4.4.1, Theorem 4.5.3 has an immediate corollary. Corollary 4.5.4. Any face of an s-lecture hall simplex has a real-rooted box polynomial. Corollary 4.5.4 tells us that if a lattice polytope P has a regular triangulation all of whose maximal faces are s-lecture hall simplices, then the triangulation is box unimodal. In particular, we have the following immediate corollary.

Corollary 4.5.5. The trivial triangulation of an s-lecture hall simplex Ps into its faces is a box unimodal triangulation. Moreover, X h∗ (z) = B (z). Ps F Fa face of Ps Recent questions have been posed asking which lattice polytopes admit box unimodal triangulations [24]. In the next chapter we will use Corollary 4.5.4 to identify box unimodal triangulations for a large family of lattice polytopes, and use them to answer, in part, some open questions on unimodality of h∗-polynomials.

36 Figure 5.1: A naturally-labeled poset. An edge that goes upwards from a to b represents a relation a ≺ b such that there is no c with a ≺ c and c ≺ b.

Chapter 5 s-Lecture Hall Order Polytopes

In this chapter, we apply the real-rootedness results from Chapter 4 and the theory of box unimodal triangulations described in Section 2.4 to prove unimodality for a family of h∗-polynomials.

5.1 Order polytopes

Order polytopes are associated with partially ordered sets. A partially ordered set (or poset for short) is a pair P = ([n], P ) where P is a partial order. A partial order is a binary relation such that

1. a P b and b P a if and only if a = b, and

2. if a P b and b P c, then a P c.

The poset P is naturally-labeled if whenever i P j in the poset, then i ≤ j as natural numbers. The linear extensions L(P ) of a poset is the set of permutations that respect the partial order:

−1 −1 L(P ) := {π ∈ Sn : i P j implies πi < πj }.

37 For a naturally-labeled poset P , deﬁne the order polytope O(P ) to be

n O(P ) := {(x1, ..., xn) ∈ R : 0 ≤ xi ≤ 1 for i ∈ [n] and xi ≤ xj whenever i P j}. The set of linear extensions of P turn out to describe a triangulation of O(P ). For some π ∈ L(P ), let ∆(π) be the simplex n ∆(π) := {(x1, ..., xn) ∈ R : 0 ≤ xπ1 ≤ xπ2 ≤ · · · ≤ xπn ≤ 1}. Now the collection of n-simplices {∆(π): π ∈ L(P )} induces a triangulation T of O(P ). To see why, note that O(P ) is the union of all ∆(π) since π respects the partial order. Also, given a point x ∈ O(P ), we can recover the simplex it belongs to by sorting its entries in non-decreasing order and taking ∆(π) of the resulting permutation. When x is in the interior of the simplex, then

0 < xπ1 < ··· < xπn < 1, so the permutation π will be unique which implies that x is not contained in any other simplex. The triangulation T of O(P ) is sometimes called the canonical triangulation [20]. Finally, it turns out that the canonical triangulation of O(P ) triangulation is regular (which follows from Theorem 10.1.3 in [15]), a fact that we will use later.

5.2 s-lecture hall order polytopes

In this section, we are interested in a generalization of order polytopes called s-lecture hall order polytopes. They are also a generalization of s-lecture hall simplices.

Deﬁnition 5.2.1. Let s = (s1, s2, ..., sn) be a sequence of positive integers, and P = ([n], P ) a naturally labeled poset. Deﬁne the s-lecture hall order polytope O(P, s) to be

n xi xi xj O(P, s) := {(x1, ..., xn) ∈ R : 0 ≤ ≤ 1 for i ∈ [n] and ≤ whenever i P j}. si si sj Just like with order polytopes, the s-lecture hall order polytopes have a regular triangulation that is described by the linear extensions. For π ∈ L(P ), let ∆(π) be the simplex

n xπ1 xπn ∆(π) := {(x1, ··· , xn) ∈ R : 0 ≤ ≤ · · · ≤ ≤ 1}. sπ1 sπn Note that this is an s-lecture hall simplex. The collection {∆(π): π ∈ L(P )} of n-simplices induces a triangulation T of O(P, s). This is called the s-canonical triangulation of O(P, s). The reason why it is regular is that it is the canonical triangulation of O(P ) from the previous section, but with the i:th coordinate scaled by a factor si, and this preserves regularity. Recall from Section 2.4 that one of the key reasons we were interested in studying real-rootedness and unimodality of box polynomials is because we can use these properties to prove that reflexive polytopes with box unimodal triangulations have unimodal h∗-polynomials. In this regard, the regularity of the s-canonical triangulation is important to us, because it needs it be regular in order to be box unimodal. Using these facts we can now prove that all s-lecture hall order polytopes have a box unimodal triangulation. It follows that any reflexive s-lecture hall order polytope has a unimodal h∗-polynomial. Theorem 5.2.2. The s-canonical triangulation of an s-lecture hall order polytope is box unimodal. Moreover, a reflexive s-lecture hall order polytope has symmetric and unimodal h∗-polynomial. Proof. Let P = O(P, s) be an s-lecture hall order polytope. Let T be the triangulation of the boundary that is induced by the s-canonical triangulation of P. Due to Corollary 2.4.4, the h∗-polynomial can be written as ∗ X hP (z) = h(linkT (∆); z)B∆(z). ∆∈T

38 We know from Corollary 4.5.4 that, since all simplices in T are faces of s-lecture hall simplices, all the box polynomials B∆(z) are real-rooted, and therefore unimodal. Also, Lemma 2.4.6 implies that the h- polynomials in the sum are all h-polynomials of some triangulation of a simplicial polytope, so by Lemma 2.4.5 they are unimodal, symmetric, and have degree d−dim(∆) − 1. This means that P admits a box ∗ unimodal triangulation, and when P is reflexive hP (z) is unimodal and symmetric. Now our goal is to find some s-lecture hall order polytopes that are reflexive, so that we can non-trivially apply Theorem 5.2.2. To do that, we need some more definitions. A chain C in a poset P is a subposet {x1, x2, ..., xm} ⊆ P such that x1 ≺ x2 ≺ · · · ≺ xm. The rank of an element xi in a chain is i − 1. A chain C is maximal if it is not contained in any other chain. The length of a chain is the number of elements in it minus one. A poset P is graded if every maximal chain has the same length. A poset P is ranked if for every maximal element x ∈ P , the subposet {y ∈ P : y P x} is graded. The rank of an element of a ranked poset is its rank in any maximal chain that contains it. Let

ρ :[n] → Z≥0 be the function that sends an element of a ranked poset to its rank.

Theorem 5.2.3 (Theorem 4.2 in [12]). Let P = ([n], P ) be a naturally-labeled ranked poset, and s = ∗ (s1, ..., sn) be the sequence where si := ρ(i) + 1. Then hO(P,s)(z) is symmetric with respect to its degree, which is n − 1. If a d-polytope has an h∗-polynomial with degree d, and is symmetric with respect to d, then it is reﬂexive. Theorem 5.2.3 nearly implies this, except that the degree is one less than the dimension. In order to get a reﬂexive polytope, we would like to raise the dimension by one while keeping the h∗-polynomial constant. This can be achieved with the following lemma:

∗ Lemma 5.2.4. Let P = ([n], P ) be a naturally-labeled poset. Let P = ({0}∪[n], P ∗ ) be the poset obtained by adding a unique minimal element 0 to P . Then O(P, s) has the same h∗-polynomial as O(P ∗, (1, s)). Proof. This follows from Corollary 3.7 in [12].

Now we can prove the following two corollaries:

Corollary 5.2.5. Let P = ([n], P ) be a naturally-labeled ranked poset, and s = (s1, ..., sn) where si := ∗ ρ(i) + 2. Then hO(P,s)(z) is symmetric and unimodal.

∗ ∗ ∗ ∗ Proof. Let P be the poset obtained in the same way as in Lemma 5.2.4. If s := (1, s), then si = ρ (i). Therefore, P ∗ satisﬁes the conditions of Theorem 5.2.3, and the h∗-polynomial of O(P ∗, s∗) has degree n and is symmetric. However, from Lemma 5.2.4 we know that O(P ∗, s∗) and O(P, s) have the same h∗-polynomials. ∗ Thus, O(P, s) is reﬂexive. Now Theorem 5.2.2 implies that hO(P,s)(z) is symmetric and unimodal.

Corollary 5.2.6. Let P = ({0} ∪ [n], P ) be a naturally-labeled graded poset with a unique minimal element ∗ 0, and let s = (s0, ..., sn), where si := ρ(i) + 1. Then hO(P,s)(z) is symmetric and unimodal. Proof. Note that O(P, s) is the polytope O(P ∗, s∗) from the proof of the previous corollary, so its h∗- polynomial is symmetric and unimodal.

∗ In [12], it is conjectured that hO(P,s)(z) is unimodal for any naturally-labeled ranked poset P and si := ρ(i) + 1 (this is the motivating question behind Conjecture 5.4 in [12]). Therefore, Corollary 5.2.6 provides a partial answer to this conjecture, speciﬁcally in the case when P has a unique minimal element.

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