Master of Science in Advanced Mathematics and Mathematical Engineering

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Title: An Introduction to Polytope Theory through Ehrhart's Theorem

Author: Filip Cano Córdoba

Advisor: Julian Pfeifle

Department: Mathematics

Academic year: 2018-2019 Universitat Polit`ecnicade Catalunya Facultat de Matem`atiquesi Estad´ıstica

Master in Advanced Mathematics and Mathematical Engineering Master’s Degree Thesis

An Introduction to Polytope Theory through Ehrhart’s Theorem With applications to Erhart positivity Filip Cano C´ordoba

Supervised by Julian Pfeiﬂe September 2019

I want to thank my colleagues from MAMME, specially Ander, Armando, Marc, Alberto and Marta for their support and for all the time spent together learning mathematics. I also want to thank another mathematician colleague, Zaira, who has been there in important times.

I want to give special thanks to my family, to my mother and father for being always there, to my grandmothers for the prayers, the food and specially the love. Us estimo.

I could not forget my non-mathematics related friends, that have been a huge support in the intellectual distance, and have helped me to view things in perspective, which is very important in the corner of mathematics related to geometry. For the conversations and the random walks. Thank you Sa¨ıd, Alberto, Maribel and Cristina.

I am grateful also to non-vital identities, specially my bike, companion and friend in this journey, and to the Serra de Tramuntana, that always provides a lot of inspiration.

I want to thank also to all my teachers in UPC, specially to Oriol Serra, Sonia Fern´andez,Eva Miranda, Cedric Oms and Juanjo Ru´efor their passion and inspiration, and to Rodrigo and Vera for their patience and support. I would also like to thank to the CFIS community.

Finally, the most important person (apart from me) without whom this project would have never given birth, I want to give a very special thank you to Julian. For the enthusiasm of their classes, for the always challenging problems, for introducing me to the fascinating world of polytopes. For all the hours spent thinking and discussing the topics in this project. And for all the support, patience, understanding and empathy in the days that have been most diﬃcult for me. Abstract

A classic introduction to polytope theory is presented, serving as the foundation to develop more advanced theoretical tools, namely the algebra of polyhedra and the use of valuations. The main theoretical objective is the construction of the so called Berline-Vergne valuation. Most of the theoretical development is aimed towards this goal. A little survey on Ehrhart positivity is presented, as well as some calculations that lead to conjecture that generalized permutohedra have positive coeﬃcients in their Ehrhart polynomials. Throughout the thesis three diﬀerent proofs of Ehrhart’s theorem are presented, as an application of the new techniques developed.

Keywords

Ehrhart polynomials, Generalized permutohedron, Ehrhart positivity, Polytope theory, Algebra of Polyhedra, Brion’s theorem

1 of 62 Contents

1 Polytopes, polyhedra, lattices 5

1.1 Basic concepts ...... 5

1.2 Ehrhart’s theorem ...... 23

1.3 Algebra of polyhedra ...... 26

2 Local formula 35

2.1 The exponential valuation ...... 35

2.2 The counting valuation ...... 40

2.3 Generalizing Ehrhart’s theorem ...... 43

2.4 Berline-Vergne valuation ...... 45

2.5 Ehrhart’s theorem: third proof ...... 50

3 Ehrhart positivity 51

3.1 Known results ...... 51

3.2 Computing Ψ ...... 52

3.3 Generalized permutohedron ...... 58

2 CONTENTS CONTENTS

Introduction

Polytopes and polyhedra are fundamental geometric objects that appear in diﬀerent areas of mathematics, from algebraic geometry to operative research.

The study of planar and 3-dimensional polytopes has been a fascinating area of study since ancient Greek mathematics, as they are appealing geometrical objects that can be represented in the real world and gave rise to a rich theory.

Higher dimensional polytopes have a much more recent history. Swiss mathematician Ludwig Schl¨aﬂi was a pioneer in the study of higher dimensional polytopes in the 18th century, and opened a topic that is today an active area of research in mathematics.

We refer the reader to [7, 11] for a comprehensive text on polytopes.

Polytopes whose vertices lie in integer coordinates are of special interest for the discrete mathematician’s eye. In particular, we are interested in the following problem: Given an integral polytope, how much the number of integer points in it increases when we dilate it by an integer factor? Georg Pick in [9] opened a crack proving the well known Pick’s formula for the area of an integral polygon. But it was Eug`eneEhrhart in [6] who proved in that the behavior is polynomial in the general case. The polynomials arising from the solution to this problem are thus known as Ehrhart polynomials.

The study of Ehrhart polynomials is a useful tool to gain intuition and understanding of how the discrete volume (number of integer points) and the continuous volume of a polytope are related through dilations of the polytope.

In this thesis, the reader starts a journey through the basics of polytope/polyhedra theory that ends in a conjecture about the coefficients of Ehrhart polynomials in a certain family of polytopes, namely generalized permutohedra. This conjecture was first stated in [5] by Castillo and Liu, and has remained open since then. In this journey we will introduce different technical tools, mainly valuations in the algebra of polyhedra, that are used to pose the aforementioned conjecture. Furthermore, throughout the thesis three different proofs of the mentioned Ehrhart’s theorem are presented.

Most of the thesis is self-contained. Nevertheless, some results have only sketched proves, and a few of them, specially in the most advanced part, are presented with no proof at all, because doing so would have supposed a thesis extension way larger than the recommended maximum size.

This thesis has three objectives:

(i) Make a sound introduction to polytope/polyhedra theory, focusing in lattice polytopes. Furhter the introduction to the tools of valuations in the algebra of polyhedra.

(ii) Present three diﬀerent proofs of one of the most relevant theorems in lattice polytope theory, namely Ehrhart’s theorem.

(iii) Present recent developments that lead to active research topics in the area of the study of Ehrhart’s polynomials.

3 of 62 CONTENTS CONTENTS

The original objective of this master’s project was to obtain some new knowledge about the conjecture that is presented in the last section, but unfortunately all attempts to further advance current knowledge resulted in failure.

4 of 62 Chapter 1

Polytopes, polyhedra and lattices

1.1 Basic concepts

1.1.1 Polytopes and polyhedra: Deﬁnitions

In this section we will introduce the main objects of study of this thesis: polytopes and polyhedra. Classically there are two equivalent deﬁnitions of polytopes, one in terms of convex hull and another in terms of half-spaces. We will give both and outline the reasons of why they are equivalent.

1.1.1.1 Polyhedra

N Deﬁnition 1.1.1 (Convexity). A set C ∈ R is convex if for all u, v ∈ C and all 0 ≤ τ ≤ 1, the point τu + (1 − τ)v lies in C. The convex hull of C is the set

conv(C) = \ A. A⊇C,A convex

N Given u1,..., un ∈ R , a convex combination of them is any point of the form n X λiui, with λi ≥ 0 for all i and λ1 + ··· + λn = 1. i=1

Remark 1. The convex hull of a set is the set of the convex combinations of its points.

N Deﬁnition 1.1.2 (Polytope 1). A set C ⊆ R is a polytope if it is the convex hull of ﬁnitely many points.

5 1.1. BASIC CONCEPTS CHAPTER 1. POLYTOPES, POLYHEDRA, LATTICES

N Deﬁnition 1.1.3 (Half-space). A set L ⊆ R is a half-space if it is of the form

N L = {x ∈ R : a(x) ≤ α},

N for some α ∈ R and a : R → R non-zero linear function. The hyperplane deﬁned by {a(x) = α} is called its supporting hyperplane.

N Deﬁnition 1.1.4 (Polytope 2). A set H ⊆ R is a polytope if it is the bounded intersection of ﬁnitely many half-spaces.

Note that in both deﬁnitions we are considering the empty set to be a polytope.

While polytopes are our main interest, we will be working with ”unbounded polytopes”, that we call polyhedra.

N Deﬁnition 1.1.5 (Polyhedron). A set P ∈ R is a polyhedron if it is the intersection of ﬁnitely many half-spaces. In other words, is a set of the form

N P = {x ∈ R : ai(x) ≤ αi, i ∈ I}

for some ﬁnite set I, where ai are linear functions and αi ∈ R. The dimension of a polyhedron is the dimension of its aﬃne span.

A polyhedron is called rational if it can be written as

 d   d X  P = x ∈ R : ai,jxj ≤ αi, i ∈ I  j=1  with ai,j, αi ∈ Z for all i and j.

Note that a polyhedron is always a convex set. Given two points x, y ∈ P satisfying some inequalities of the form ai(x) ≤ αi, then any convex combination also satisﬁes it because of linearity:

ai(γx + (1 − γy)) = γai(x) + (1 − γ)ai(y) ≤ γαi + (1 − γ)αi = αi. Example 1. There are three families of polytopes that are widely known and serve as a good intro- d duction to the concept. We will use {e1,..., ed} to denote the canonical basis in R .

1. The standard simplex in d-dimensions ∆d. Its deﬁnitions in terms of half-spaces and of convex hull are, respectively:

d d d ∆ = {x ∈ R : xi ≥ 0 ∀i, x1 + ··· + xd = 1}, ∆ = conv(0, e1,..., en). The standard simplex is the generalization of the triangle and the tetrahedron.

d 2. The standard d-dimensional (±1)-cube . Its deﬁnitions are: d d d = {x ∈ R : −1 ≤ xi ≤ +1 ∀i}, = conv {±e1 ± · · · ± en} . It is the generalization of the square and the cube.

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3. The standard d-dimensional cross polytope d. Its deﬁnitions are:

d d d ♦ = {x ∈ R : ±x1 + ··· + ±xn ≤ 1}, ♦ = conv {±ei ± ej, ∀i, j} . The cross polytope is the generalization of the square and the octahedron.

3 3 3 (a) Simplex ∆ (b) Cube (c) Cross polytope ♦ Figure 1.1: Three-dimensional version of the three basic families of polytopes.

Figure 1.2: Two examples of 3-dimensional unbounded polyhedra.

1.1.1.2 Faces

Definition 1.1.6 (Face). Given a polyhedron P , a linear inequality of the type {a(x) ≤ α} is a valid inequality if it is satisfied for all x ∈ P . Given a valid inequality, a face F of P is the subset of P where it is satisfied as an equality, F = {x ∈ P : a(x) = α}. The dimension of a face is the dimension of its affine span.

Some faces are specially relevant and have their own names. For a polyhedron of dimension d:

• Faces of dimension 0 are called vertices.

• Faces of dimension 1 are called edges.

• Faces of dimension d − 2 are called ridges.

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• Faces of dimension d − 1 are called facets. Remark 2. Some interesting properties of faces are:

• By Deﬁnition 1.1.5, any face of a polyhedron is in itself a polyhedron.

• The empty set and the polyhedron itself are always faces. A face is called proper if it is neither of them.

• All proper faces can be described with an inequality that is part of the defining inequalities of P . Since a polyhedron can be described with finitely many defining inequalities, the number of faces is finite.

A polyhedron is called integral if its vertices have all integer coordinates.

Deﬁnition 1.1.7 (Relative interior). Let P be a polyhedron and F a face of P . The relative interior of F is the interior of F as a subset of its aﬃne span, denoted by relint(F ).

Note that a point v is in the relative interior of F if and only if non of the deﬁning inequalities of F are active on v. Note also that a point v ∈ P lies in no proper face if and only it lies in the relative interior.

Lemma 1.1.1. A point v ∈ P is a vertex if and only if the only pair v1, v2 ∈ P such that v = (v1 + v2)/2 is v1 = v2 = v.

PROOF. ( =⇒ ) Let a(x) = α be the defining inequality of v. Since dim(span(v)) = 0, v is the only point in P satisfying its defining inequality. Now consider a pair v1, v2 as in the statement. If v1 6= v, then a(v1) < α, and thus a(v) < α. Contradiction. Analogous for v2. ( ⇐= ). We know that v lies in some faces. Let F be a face of minimum dimension where v lies. Let d = dim F . If d = 0, we are done. If d > 0, since v lies in no proper faces of F , it lies in its relative interior, which is not empty because d > 0. Thus we can find v1, v2 6= v satisfying v = (v1 + v2)/2. Contradiction.

1.1.1.3 Cones

Deﬁnition 1.1.8 (Cone, conic combination). A polyhedron K is called a cone or polyhedral cone if for any x ∈ P and λ > 0, we have λx ∈ K. N Given u1,..., un ∈ R , a conic combination of them is any point of the form n X λiui, with λi ≥ 0. i=1 Given a set A, the set of all its conic combinations is called its conic hull and denoted as co(A).

Remark 3. A cone is deﬁned by a system of homogeneous linear inequalities.

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This is true because if a(x) ≤ α is a valid inequality for K, then

(i) α ≥ 0 and

(ii) a(x) ≤ 0 is also a valid inequality.

For the ﬁrst property, take λ = 0, by linearity, 0 ≤ α. For the second one, assume there exists x ∈ K such that a(x) = β > 0. Take λ > α/β. Applying the cone property we have a(λx) = λβ > α. Contradiction.

A cone always contains the origin. Also, for any point u in a cone K, the ray through the origin in the direction of u is contained in K. Therefore, only the origin may be a vertex of a cone. We call them pointed cones.

Sometimes we will want to consider the translation of a cone. Deﬁnition 1.1.9 (Shifted cone). Given a cone K and a point v, a set of the form K +v will be called a shifted cone. If K is pointed, it will be called a shifted pointed cone. In a shifted pointed cone, the vertex is called its apex.

Given a polyhedron, there is a cone that plays a fundamental role in its structural decomposition, see Proposition 1.1.14.

Deﬁnition 1.1.10 (Recession cone). Let P be a polyhedron. Its recession cone is the set

KP = {u : x + τu ∈ P, for some x ∈ P and all τ ≥ 0}.

Alternatively, for a polytope deﬁned by inequalities

P = {x : ai(x) ≤ αi, i ∈ I}, its recession cone can be deﬁned as

KP = {x : ai(x) ≤ 0, i ∈ I}.

This is true by the same argument we made for Remark 3.

Many geometrically meaningful cones can be constructed from lower dimensional polytope.

d Deﬁnition 1.1.11 (Cone over a polytope). Let P be a d-dimensional polytope in R with vertices v1,..., vn, we deﬁne the cone over P as the conic hull of

w1 = (v1, 1),..., wn = (vn, 1).

We will denote it as cone(P ). When a cone can be constructed as a cone over a polytope, the vectors wi are called generators. A cone with exactly d linearly independent generators is called simplicial.

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1.1.1.4 Minkowski sum

N Deﬁnition 1.1.12 (Minkowski sum). Let P,Q ∈ R be sets. Their Minkowski sum is the set

P + Q = {p + q : p ∈ P, q ∈ Q}.

d Example 2. The standard cube can be regarded as a Minkowski sum of the intervals [−ei, ei]: d = [−e1, e1] + ··· + [−ed, ed].

(0,1,0) (0,0,1) (1, 0, 0) (0, 0, 0) + + =

(0, 0, 0) (0, 0, 0)

Figure 1.3: The cube is an example of Minkowski sum.

1.1.1.5 Polarity

N Deﬁnition 1.1.13 (Polar). Let A ⊆ R be a set. The polar of A is the set

◦ N A = {x ∈ R : hx, yi ≤ 1 for all y ∈ A}.

Remark 4. If K is a cone, by the same argument as in Remark 3, the polar K◦ is deﬁned as

K◦ = {x : hx, yi ≤ 0 for all y ∈ K}.

If P = conv(v1,..., vn), then the polar is

◦ P = {x : hx, vii ≤ 1 for all i = 1, . . . , n}.

1.1.2 Polytopes and polyhedra: Basic properties

Now we will deﬁne some important concepts and give fundamental properties of both polytopes and polyhedra. Since the former are a special case of the latter, many deﬁnitions will be stated in terms of polyhedra.

1.1.2.1 Rays and lines

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N Deﬁnition 1.1.14 (Line, ray, interval). Let u, v ∈ R , with u 6= 0, the line through v in the direction of u is the set L = {v + τu : τ ∈ R}. The ray emanating from v in the direction of u is the set R = {v + τu : τ > 0}. The interval with endpoints u, v is the set I = {τv + (1 − τ)u : 0 ≤ τ ≤ 1}.

Now that we know the basic language, we will prove some results that appear very basic and straight- forward. They serve a double purpose. First, they are a sanity check, as we see that definitions make sense, and the properties we expected from them are indeed satisfied. And secondly, they give a taste on what working with polyhedra and polytopes look like, the basic techniques and difficulties.

They are also fundamental results that we will be using through the thesis, so we will need to mention them one way or the other.

Proposition 1.1.2. A polyhedron is bounded if and only if it does not contain any ray.

PROOF. In both implications we will prove the contrapositive statement. ( =⇒ ) Clearly a ray is unbounded, so if a polyhedron contains a ray, it cannot be bounded.

( ⇐= ) Let P be an unbounded polyhedron. Then there exists a sequence of points {un}n∈N in P such that kunk → ∞. Consider the normalized sequence {wn}n∈N, deﬁned by wn = un/kunk. The normalized sequence is bounded, so it has partial limits. Let w one of its limit points. Since the original sequence {un}n∈N is in P , for any valid inequality in P of the form a(x) ≤ α, we have un 1 1 a(wn) = a = a(un) ≤ α −−−→ 0. kunk kunk kunk n→∞ And therefore, a(w) ≤ 0, so for any point x ∈ P , the ray emanating from x in the direction of w is contained in P .

Corollary 1.1.3. A polyhedron is bounded if and only if its recession cone is {0}.

1.1.2.2 Linear transformations

N N−1 Lemma 1.1.4. Let π : R → R be the projection deﬁned by π(x1, . . . , xN ) = (x1, . . . , xN−1). N N−1 Let P ∈ R be a polyhedron. Then π(P ) ∈ R is a polyhedron.

PROOF. Assume that P is deﬁned as a system of linear inequalities,

 N   N X  P = x ∈ R : ai,jxj ≤ αi, for i ∈ I .  j=1 

Classify the deﬁning inequalities with respect to the last coeﬃcient

I+ = {i ∈ I : ai,N > 0},I− = {i ∈ I : ai,N < 0} ,I0 = {i ∈ I : ai,N = 0}.

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N−1 Given y = (y1, . . . , yN−1) ∈ R , we know that y ∈ π(P ), if there exists xN ∈ R such that x = (y, xN ) ∈ P . For this to happen, the following |I| inequalities must be satisﬁed

N−1 X ai,jyj ≤ αi, for all i ∈ I0. (1.1.1) j=1 α N−1 a i − X i,j y ≥ x , for all i ∈ I . (1.1.2) a a j N + i,N j=1 i,N α N−1 a i − X i,j y ≤ x , for all i ∈ I . (1.1.3) a a j N − i,N j=1 i,N (1.1.4)

The inequalities with I0 are already linear inequalities involving y. For ones involving I+ and I−, note that the existence of such xN is equivalent to the fact that every lower bound for xN (the I−- type inequalities) does not exceed every upper bound (the I+-type inequalities). Therefore, they are equivalent to

α N−1 a α N−1 a i+ − X i+,j y ≥ i− − X i−,j y , for all pair (i , i ) ∈ I × I . (1.1.5) a a j a a j + − + − i+,N j=1 i+,N i−,N j=1 i−,N

Now π(P ) is deﬁned by the linear system of inequalities in Eqs. (1.1.1) and (1.1.5).

The process we followed in this proof is called Fourier-Motzkin elimintation. It plays a key role in the algorithm to obtain a polytope in half-space presentation (Deﬁnition 1.1.4) to its convex-hull presentation (Deﬁnition 1.1.2) that we will see in the proof of Theorem 1.1.8.

This lemma was in fact the diﬃcult case in the following result.

N M N Theorem 1.1.5. Let T : R → R be a linear map, and P ⊆ R be a polyhedron. Then the image T (P ) is also a polyhedron.

PROOF. First note that if T is an isomorphism, the result is trivially true because the inequalities are linear. N Also, if T is injective, it is also true because we have an isomorphism in the restriction T : R → Im(T ). Moreover, if T is a surjective map, by applying the previous lemma dim(Ker(T )) times the result holds. Finally, any linear map can be decomposed into an injective and a surjective map, by

N N M M R −→ R × R −→ R x 7−→ (x,T (x)) 7−→ T (x).

This concludes the proof.

1.1.2.3 Weyl-Minkowski theorem

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d Lemma 1.1.6. Let P ∈ R be a d-dimensional polyhedron deﬁned by the inequalities {ai(x) ≤ αi}i∈I , let v ∈ P and let Iv be the set of indices for which the inequalities are active in v. Then v is a vertex of P if and only if

d ∗ span {ai : i ∈ Iv} = (R ) ,

d ∗ d where (R ) is the dual space (in the sense of linear algebra) of R .

PROOF. d ∗ ( ⇐= ) Assume the set {ai : i ∈ Iv} spans (R ) . Let v1, v2 ∈ P be such that v = (v1 + v2)/2. Since v1, v2 ∈ P , we have ai(v1), ai(v2) ≤ αi, for all i ∈ Iv.

And since v = (v1 + v2)/2, we have a(v1) + a(v2) /2 = αi, for all i ∈ Iv.

Therefore we have ai(v1) = ai(v2) = αi, for all i ∈ Iv. d ∗ Since the ai’s span all (R ) , the only solution to the system is v1 = v2 = v. By Lemma 1.1.1 v is a vertex. d ∗ d ( =⇒ ) Assume now that the set {ai : i ∈ Iv} does not span (R ) . Then there exists u ∈ R , u 6= 0 such that ai(u) = 0, for all i ∈ Iv.

Then for suﬃciently small ε > 0, we can set v1 = v + εu and v2 = v − εu with v1, v2 ∈ P , so v is not a vertex.

Lemma 1.1.7. Let P be a polyhedron, F a face of P and v a vertex of F . Then v is a vertex of P .

PROOF. Suppose a(x) ≤ α is a valid inequality for P and F = {x ∈ P : a(x) = α}. Again, let us write v as v = (v1 +v2)/2 with v1, v2 ∈ P . Since a(v) = α and a(v1), a(v2) ≤ α, we have that a(v1) = a(v2) = α. But then v1, v2 ∈ F , and since v is a vertex of F , v1 = v2 = v, so v is a vertex of P .

Theorem 1.1.8 (Weyl-Minkowski). Both deﬁnitions of polytope are equivalent. Concretely:

(i) The convex hull of ﬁnitely many points is a bounded polyhedron.

(ii) A bounded polyhedron is the convex hull of its vertices.

PROOF. N (i) Let P = conv(v1,..., vn), where v1,..., vn ∈ R . Let us consider the standard n-dimensional

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n n simplex ∆ = conv(e1,..., en), where ei’s are the vectors of the canonical basis in R . Consider the linear map n N T : R −→ R ei 7−→ vi. By construction P = T (∆n). Then P is a polyhedron by Theorem 1.1.5. Finally, by compacity of ∆n and continuity of T , we have that P is compact, and thus bounded. (ii) We proceed by induction on d = dim P . If d = 0, the result is clear. Assume d > 0. Assume P is deﬁned by the inequalities P = {x : ai(x) ≤ αi, for i ∈ I}. Take an arbitrary point y ∈ P . Either y lies in the relative interior of P or it lies in a proper face. If y lies in a proper face F of P , by the induction hypothesis it is in the convex hull of the vertices of F , which are vertices of P by Lemma 1.1.7. If y lies in the relative interior of P , there is a line through y in the span of P . Since P is bounded, it cannot contain any ray, so the line intersects with the boundary of P in two points, y1 and y2. This way, y is a convex combination of y1 and y2. By the induction hypothesis, y1 and y2 are convex combination of the vertices of the proper faces they lie in, that again by Lemma 1.1.7, are also vertices of P .

1.1.2.4 Cones

Proposition 1.1.9. A non-empty polyhedron contains a vertex if and only if it contains no lines.

PROOF. ( =⇒ ) We will prove the contrapositive statement. Suppose that P contains a line in the direction of a vector u 6= 0. Then for any valid inequality of P of the form a(x) ≤ α, we must have a(u) = 0. This way, for any v ∈ P , we can write it in the form of v = (v1 + v2)/2 with v1 = v + u and v2 = v − u, with v1, v2 ∈ P . ( ⇐= ) We will proceed by induction on d = dim P . If d = 0, the result is clear. Suppose d > 0. Let y ∈ relint(P ) and consider a line through y that lies in the span of P . Since P contains no lines, it must intersect with the boundary of P in at least one point. Call such a point z. Since it is in the boundary, z lies in a proper face of P , call it F . This proper face contains no lines, so by the induction hypothesis it contains a vertex, which is also a vertex of P by Lemma 1.1.7.

Corollary 1.1.10. A non-empty cone K is pointed if and only if it contains no lines.

Corollary 1.1.11. Let P be an unbounded polyhedron. Then its recession cone KP is a pointed cone if and only if P contains no lines.

PROOF. ( ⇐= ) Recall that by the characterization we give of the recession cone (Deﬁnition 1.1.10), KP ⊆ P . In particular, if P contains no lines, then K contains no lines and therefore it is a pointed cone. ( =⇒ ) We will prove the contrapositive. Assume P contains a line {x + τu : τ ∈ R}. Then KP contains not only u, but {τu : τ ∈ R}, which is a line. Thus KP is not pointed.

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By the pictures we have seen and our general intuition, we expect a pointed cone to be the result of coning over some polytope. This is indeed the case.

d Proposition 1.1.12. Let K ⊂ R be a d-dimensional pointed cone of positive dimension. Then there exists an aﬃne hyperplane H ⊂ H and a polytope P ⊂ H such that K = co(P ).

PROOF. Let K be deﬁned as a system of inequalities by

K = {x : ai(x) ≤ 0, for i ∈ I}.

d P By Lemma 1.1.6 the linear functions {ai}i∈I span (R )∗. Let a = i∈I ai the hyperplane be H = {x : a(x) = −1} and the polytope be P = H ∩ K (we know it is a polyhedron, we have yet to show that it is bounded). First we claim that a 6= 0. If it was, by the construction of K, for all i ∈ I and all x ∈ K, we would have ai(x) = 0. Since K is full dimensional, this would contradict Lemma 1.1.6. For any x ∈ K \{0} we know that a(x) < 0 because K has only one vertex. Therefore there exists λ ≥ 0 such that λx ∈ P . With this we have that K = co(P ) with P a polyhedron. To prove it is bounded, suppose it is not. Then it contains a ray in a direction u parallel to H, so with a(u) = 0. Since the ray is contained in K, we have ai(u) ≤ 0, and thus ai(u) = 0 for all i ∈ I. But then the line {τu : τ ∈ R} is contained in K, contradicting K being a pointed cone. Remark 5. A direct consequence of this result is that any pointed cone can be constructed as the cone over a polytope.

1.1.2.5 Minkowski sum

N Lemma 1.1.13. Let P1,P2 ⊆ R be two polyhedra. Then their Minkowski sum P1 + P2 is a polyhedron.

PROOF. Consider the Cartesian product

P = P1 × P2 = {(x1, x2): x1 ∈ P1, x2 ∈ P2}.

N N We can describe P ⊆ R × R joining the deﬁning inequalities of P1 and P2. Now consider the linear transform N N N T : R × R −→ R (x, y) 7−→ x + y.

Clearly, P1 + P2 = T (P ), so by Theorem 1.1.5 it is a polyhedron.

We are now prepared for two fundamental results on the structure of polyhedra as Minkowski sums.

Proposition 1.1.14. Let P be a non-empty polyhedron without lines. Let K be its recession cone and M be the convex hull of the set of vertices of P . Then

P = K + M.

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PROOF. Note that by Corollary 1.1.3 and Theorem 1.1.8, we have already proved the result for polytopes, so we may assume that P is unbounded, and thus K 6= {0}.

(⊇) Suppose P is defined by the inequalities P = {x : ai(x) ≤ α, i ∈ I}. Recall that the recession cone (see Definition 1.1.10) can be equivalently defined as

K = {x : ai(x) ≤ 0}.

Then clearly P ⊆ K + M. (⊆) We will argue by induction on d = dim P . For d = 0 the result is clear. Consider d > 0. Let y ∈ P , we want to write it as y = u + z, where u ∈ K and z ∈ M. Either y lies in a proper face or it lies in the relative interior of P .

If it lies in a face, by the induction hypothesis it can be written as y = uF + zF , where uF ∈ KF and zF ∈ MF . By Lemma 1.1.7 MF ⊆ MP , and by deﬁnition KF ⊆ KP , so we are done.

If y lies in the relative interior, consider u ∈ K \{0} (exists because K is unbounded). Consider now the line {y + τu : τ ∈ R}. Since P contains no lines, it will intersect a proper face F of P at a point 0 y . This point is of the form y + τ0u, and since u ∈ KP , we have τ < 0 and thus −τ0u ∈ KP . so we 0 0 0 can write y = y + u , where u ∈ KP and, by the induction hypothesis, y ∈ KF + MF . We conclude the proof applying the same argument as in the previous case.

N Proposition 1.1.15. Let P ⊆ R be a non-empty polyhedron. Then it can be decomposed as

P = L + K + M,

N where L is a linear subspace of R , K is a pointed cone and M is a polytope.

PROOF. N Suppose P is deﬁned by the inequalities P = {x ∈ R : ai(x) ≤ α, i ∈ I}. If P contains a line in N the direction u ∈ R , then ai(u) = 0 for all i ∈ I. This justiﬁes the following construction.

Let L = {u : li(u) = 0, i ∈ I}. It is clearly a linear subspace, so we can ﬁnd a decomposition N N R = L ⊕ W , where W ⊆ R is a linear subspace. Then consider the projection map

π : L ⊕ W −→ W u + w 7−→ w, and the image of P by π, Q = π(P ). We have P = L + Q. Since ker π = L, Q contains no lines, and by Proposition 1.1.14 it can be decomposed as Q = K + M.

Note that in the characterization of polyhedra with lines we lost most of the information of what the polytope M and the pointed cone K mean.

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1.1.2.6 Polarity

N Lemma 1.1.16 (Separation lemma). Let A ⊆ R be a closed convex set and a ∈/ A be a point. N Then there exists a non-zero vector u ∈ R and α ∈ R such that

hu, ai < 0 and hu, xi > α for all x ∈ A.

PROOF. If a 6= 0, we can consider the shifted set A − a and ﬁnd u and α there, so without loss of generality we may assume a = 0.

Since A is closed and 0 ∈/ A, there exists (maybe more than one) a point y ∈ A that is the closest point of A to 0. Consider u = y, α = kyk2/2 and the half-space

N H = {x ∈ R : hu, xi ≤ α}. We claim that H contains no points in A. To prove it, suppose there exists z ∈ A ∩ H. Since A is convex, for all 0 ≤ ε ≤ 1, the point yε = (1 − ε)y + εz lies in A. We will see that, for some ε, the point yε is closer to 0 that y: 2 2 2 2 kyεk = (1 − ε) kyk + ε kzk + 2ε(1 − ε)hy, zi ≤ (1 − ε)2kyk2 + ε2kzk2 + 2ε(1 − ε)α (by z ∈ H) h i = (1 − ε)2 kyk2 + ε kyk2 − 2α + ε2kzk2 (rearranging terms) = (1 − ε)kyk2 + ε2kzk2. (by α = kyk2/2)

2 Choosing ε < (kzk/kyk) we have kykε < kyk, which contradicts y being the closest point on A to 0.

N Theorem 1.1.17. Let A ⊆ R be a non-empty set. Then

(i) The polar set A◦ is closed, convex and contains the origin.

(ii) If A is closed and convex, then A◦◦ = A.

(iii) If A is a polyhedron, then A◦ is a polyhedron.

PROOF. (i) By definition A◦ is the intersection of (possibly infinitely many) closed half-spaces, so it is closed and convex. Also by definition 0 ∈ A◦.

(ii)(⊇) This inclusion is always true (even if A is not closed or convex). Points x ∈ A◦ are deﬁned as hx, ai ≤ 1 for all a ∈ A. Thus, given a ∈ A, clearly ha, xi ≤ 1 for all u ∈ A◦, which is the deﬁnition of a ∈ (A◦)◦. (⊆) We will prove this inclusion by contradiction. Assume there exists a ∈ (A◦)◦, a ∈/ A. Since A is closed and convex, by the separation lemma there exists u 6= 0 and α ∈ R such that hu, ai > α and hu, xi < α, for all x ∈ A.

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Since 0 ∈ A, we have α > 0. Consider the vector u0 = α−1u. Now hu0, xi < 1 for all x ∈ A, so u0 ∈ A◦. Since a ∈ (A◦)◦, we must have hu0, ai ≤ 1, which contradicts hu, ai > α.

(iii) Recall that by Proposition 1.1.15 we can decompose any polyhedron as P = L + K + M, where ◦ M is a polytope, K a pointed cone and L a subspace. If M = conv(v1,..., vn), then M is deﬁned as intersection of half-spaces:

◦ M = {x : hx, vii ≤ 1, for i = 1, . . . , n}.

◦ By Proposition 1.1.12, we can write K as K = co(u1, . . . , um), thus K can be deﬁned as intersection of half-spaces: ◦ K = {x : hx, uii ≤ 1, for i = 1, . . . , m}. We will prove that P ◦ = M ◦ ∩ K◦ ∩ L⊥. (⊆) Take an arbitrary x ∈ P ◦. Since M ⊆ P , we have x ∈ M ◦. Choose u ∈ K and y ∈ P . Then y + λu ∈ P for all λ > 0, so we must have hu, xi ≤ 0, and thus x ∈ K◦. Finally, for u ∈ L and y ∈ P , ⊥ we have y + λu ∈ P for all λ ∈ R, so we must have hu, xi = 0, and thus x ∈ L . (⊇) Let x ∈ M ◦ ∩ K◦ ∩ L⊥. Any point in P can be written as

p = l + k + m, where l ∈ L, k ∈ K, m ∈ M.

Thus hp, xi = hl, xi + hk, xi + hm, xi ≤ 1.

1.1.2.7 Triangulations

Deﬁnition 1.1.15 (Simplex). A polytope P is called a simplex if it is the convex hull of d + 1 points and has dimension d. Namely P = conv(v1,..., vd+1) and dim P = d.

Deﬁnition 1.1.16 (Triangulation of a polytope). Given a polytope P, a triangulation of P is a set T of simplices such that

S 1. The simplices cover the polytope, that is ∆∈T ∆ = P. 2. For any pair of simplices ∆, ∆0 ∈ T , the intersection ∆ ∩ ∆0 is a face of both ∆ and ∆0.

We say that a triangulation adds no new vertices if for all ∆ ∈ T , the vertices of ∆ are vertices of P.

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Deﬁnition 1.1.17 (Triangulation of a cone). Given a cone K, a triangulation is a set T of simplicial cones such that

S 1. The simplicial cones cover the original cone, that is S∈T S = K. 2. For any pair of cones S, S0 ∈ T , the intersection S ∩ S0 is a face of both S and S0.

We say that a triangulation adds no new generators if for all S ∈ T , the generators of S are also generators of K.

(a) S = 4, E = 7, V = 0 (b) S = 4, E = 6, V = 0 (c) S = 12, E = 10, V = 1

3 Figure 1.4: Three diﬀerent triangulations of the cube . The ﬁrst number (S) indicates the number of simplices in which the cube is divided, the second one (E) is the number of edges used in the triangulation, and the third one (V ) indicates the number of new vertices added.

Theorem 1.1.18 (Existence of triangulations). Every polytope admits a triangulation with no new vertices.

We refer to [2, Appendix] for a complete proof.

Corollary 1.1.19. Every shifted pointed cone admits a triangulation with no new generators.

PROOF. Just observe that a triangulation of a polytope P with no new vertices corresponds to a triangulation of cone(P ) with no new generators.

1.1.3 Lattices and lattice polyhedra

d Deﬁnition 1.1.18 (Lattice). A set Λ ∈ R is a d-dimensional lattice if it satisﬁes three condi- tions:

d (i) Subgroup: Λ is a subgroup of (R , +). d (ii) Discrete: for any bounded set B ⊂ R , the intersection Λ ∩ B is ﬁnite. d (iii) Spans: the linear span of Λ is R .

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Example 3. The ﬁrst examples of lattices that we have:

d d • The typical example of a lattice in R is Z . d • We know that as an additive group, Z is generated by the basis vectors e1,..., ed. For any d vector λ ∈ R , the group generated by λ1e1, . . . , λded is also a lattice.

• To make things more interesting, we can take any invertible matrix in the integers, A ∈ GLd(Z) and take the group generated by λ1A · e1, . . . , λdA · ed.

We will see in Theorem 1.1.22 that these are essentially the only possible lattices. Notation. We will be using bαc and {α} to denote the ﬂoor integer part and the fractional part of the real number α respectively. So we have α = bαc + {α}, with bαc ∈ Z and 0 ≤ {α} < 1.

d d Lemma 1.1.20. Let Λ ⊂ R be a lattice and let L ⊆ R be a proper linear subspace spanned by points in Λ. Then there exists v ∈ Λ \ L such that

dist(v,L) ≤ dist(w,L) for all w ∈ Λ \ L. (1.1.6)

In other words, there exists a closest point of Λ to L.

PROOF. Let k = dim L and {u1,..., uk} a basis of L consisting of lattice points ui ∈ Λ. Consider the parallelepiped ( k ) X Π = λiui, : 0 ≤ λi ≤ 1, for i = 1, . . . , k , i=1 d and its neighborhoods Πρ = {x ∈ R : dist(x, Π) ≤ ρ}. For sufficiently large ρ, the set Πρ contains points of Λ \ L. Since Λ is discrete, the set Πρ ∩ Λ is finite, so there exists v ∈ Λ \ L that is the closest to Π, i.e. dist(v, Π) ≤ dist(a, Π) for all a ∈ Λ \ L. (1.1.7) We will prove that in fact v satisfies Eq. (1.1.6). Suppose that there exists w ∈ Λ \ L such that dist(w,L) < dist(v,L). Let x ∈ L be a point that satisfies dist(w,L) = dist(w, x). Since x ∈ L, it can be written as k X x = αkuk. i=1 We can decompose it further into x = u + y, where

k k X X u = bαkcuk ∈ Λ ∩ L, and y = {αk}uk ∈ Π. i=1 i=1 Consider w − u ∈ Λ \ L. By construction of x, we have

dist(w − u, Π) ≤ dist(w − u, x − u) = dist(w, x) < dist(v,L) ≤ dist(v, Π).

This contradicts Eq. (1.1.7) by taking a = w − u.

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d Lemma 1.1.21. Let {u1,..., ud} be a basis of R . The additive group generated by them is a lattice.

PROOF. d d Consider the linear transformation T : R → R deﬁned by T (α1, . . . , αd) = α1u1 + . . . αdud. Then d d d T (Z ) is the additive group generated, is discrete because it is the image of Z and spans R because T is a bijective map.

d d Theorem 1.1.22. Let Λ ∈ R be a lattice. Then there exists a basis {u1,..., ud} ⊂ Λ of R such that for every point u ∈ Λ, there is a unique decomposition of the form

d X u = miui, where mi ∈ Z. i=1

We call {u1,..., ud} a basis of Λ.

PROOF. We will proceed by induction on d. For d = 1, since Λ is discrete, there exists (maybe more than one) the smallest (in absolute value) non-zero element of Λ. We call it a. Then Λ = {ma : m ∈ Z}.

For d > 1, choose d − 1 linearly independent points of Λ and let L be the (d − 1)-dimensional subspace they span. Consider the induced lattice Λ1 = Λ ∩ L. By the induction hypothesis, there exists a basis {u1,..., ud−1} of Λ1. By Lemma 1.1.20 there exists ud ∈ Λ \ L such that

dist(ud,L) ≤ dist(w,L) for all w ∈ Λ \ L. (1.1.8)

We will prove that {u1,..., ud−1, ud} is a basis of Λ. We already know that it constitutes a basis of d R , so for a given u ∈ Λ, there exists a unique decomposition

d X u = αiui, αi ∈ R. i=1

Consider as usual the decomposition of the real number αd into its integer and fractional part, αd = bαdc + {αd}. Suppose that {αd} > 0. Then the point

d−1 X v = u − bαdcud = {αd}ud + αiui i=1 lies in Λ \ L and satisﬁes

dist(v,L) = dist ({αd}ud,L) = {αd} dist(ud,L) < dist(ud,L).

This contradicts Eq. (1.1.8). We have just proved αd ∈ Z. To prove that the rest of coeﬃcients are also integers, we just consider w = u−αdud. We have w ∈ Λ∩L = Λ1, so by the induction hypothesis α1, . . . , αd−1 ∈ Z.

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Deﬁnition 1.1.19 (Fundamental parallelepiped). Let Λ be a lattice and {u1,..., ud} a basis of Λ. The fundamental parallelepiped of of Λ associated to {u1,..., ud} is the set

( d ) X Π(u1,..., ud) = αiui : 0 ≤ αi < 1 . i=1

d Lemma 1.1.23. Let Λ ∈ R be a lattice and Π a fundamental parallelepiped. Then every point d x ∈ R can be uniquely written as

x = y + v, where y ∈ Π and v ∈ Λ.

PROOF. d Let {u1,..., ud} be the basis of Λ associated to Π. We know that for all x ∈ R , there is a unique decomposition d X x = αiui, αi ∈ R. i=1 As in the proof of Lemma 1.1.20 we can further decompose it into x = v + y where

k k X X v = bαkcuk ∈ Λ ∩ L, and y = {αk}uk ∈ Π. i=1 i=1 To prove that this decomposition is unique, suppose we have

x = v1 + y1 = v2 + y2, where v1, v2 ∈ Λ and y1, y2 ∈ Π.

Since y1, y2 ∈ Π, they are expressed as

d d X X y1 = βiui, y2 = γiui, where 0 ≤ βi, γi < 1, for all i = 1, . . . , d. i=1 i=1

On the other hand, y2 − y1 = v1 − v2 ∈ Λ, so there exists a unique decomposition

d X y2 − y1 = δiui, where δi ∈ Z. i=1

Since the decomposition is unique, for all i = 1, . . . , d we have δi = γi − βi. The only way that 0 ≤ βi, γi < 1 and βi − γi ∈ Z can be satisﬁed is if γi = βi.

d Theorem 1.1.24. Let Λ be a lattice and Π a fundamental parallelepiped. Let Br ⊂ R be the ball of radius r centered at the origin. Then

|B ∩ Λ| 1 lim r = . (1.1.9) r→∞ vol Br vol Π In particular, the volume of Π is an invariant of Λ. We call it the determinant of Λ, denoted det Λ.

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PROOF. (Sketch only) Choose a particular fundamental parallelepiped Π of Λ. By Lemma 1.1.23, its lattice translates cover d all R . Suppose Π ⊂ Bα for some α > 0. Then for r > α consider [ Xr = (Π + u). u∈Br

We can check that Br−α ⊂ Xr ⊂ Br+α. Therefore

vol Br−α ≤ vol Xr = |Br ∩ Λ| vol Π ≤ vol Br+α. Applying the limit r → ∞ in the previous equation we get Eq. (1.1.9).

Deﬁnition 1.1.20 (Lattice polyhedron). Given a lattice Λ and a polyhedron P , we will say that P is a Λ-lattice polyhedron (or simply lattice polyhedron when the lattice is understood) if the vertices of P are in Λ.

The said concept is just a generalization of integral polyhedron, but we will see in Section 2.4 that it will be useful to think of general lattice polytopes, and not only integral ones.

d d Deﬁnition 1.1.21 (Normalized volume). Given a lattice Λ ⊂ R and a measurable set S ⊂ R , we deﬁne the normalized volume of S as

nvol(S) = vol(S)/ det Λ.

1.2 Ehrhart’s original proof

In this section we will give the ﬁrst proof of Ehrhart’s theorem, using the same method as the original proof in [6].

d Theorem 1.2.1 (Ehrhart, 1962). Let P ⊆ R be a d-dimensional integral polytope, and let tP be its dilation by an integer factor t. Then

d tP ∩ Z = p(t), where p(t) is a polynomial of degree d.

To do this, we ﬁrst need to reformulate the problem with generating functions and cones.

Recall that by Proposition 1.1.12, any pointed cone can be constructed as a cone over a polytope. We will be using shifted pointed cones in this section, so they will look like

K = {v + λ1w1 + . . . , λnwn : λ1, . . . , λn ≥ 0}, where v is the apex and w1,..., wn are the generators.

As we will see, cones are convenient to us because we can have a richer theory that allows us to perform computations that we cannot do otherwise. Recall that given a polytope, there is a natural construction of a shifted pointed cone (Deﬁnition 1.1.11).

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Remark 6. If K is the cone over P , we can recover the original polytope when we intersect with the hyperplane {xd+1 = 1}. Also in general, the intersection

cone(P ) ∩ {xd+1 = t} gives us the dilation of P by a factor t. See Figure 1.5: Cone over a pentagon. Fig. 1.5 for an example. We will prove Ehrhart’s theorem for simplices, and then we will extend the result for general polytopes by triangulations. We have introduced the concept of simplicial cone because the cone over a simplex is always simplicial.

With the spirit of generating functions in mind, we deﬁne the integer-point enumerator of a set.

d Deﬁnition 1.2.1 (Integer-point enumerator). Given a set S ⊆ R , we deﬁne the integer-point enumerator of S as the function X m σS(x) = x , m∈S∩Zd m m1 md with the notation x = (x1, . . . , xd), m = (m1, . . . , md) and x = x1 ····· xd .

When the set S is bounded, it makes sense to evaluate at x = (1,..., 1), and this evaluation gives us how many integer points lie in our set S.

When the set S is actually a simplicial cone, we have the following lemma:

Lemma 1.2.2. Let K be a simplicial cone with apex v and integer generators w1,..., wd, then

σΠ(x) σK (x) = , (1.2.1) (1 − xw1 ) ··· (1 − xwd )

where Π is the half-open fundamental parallelepiped, deﬁned as

Π = {v + λ1w1 + ··· + λdwd : 0 ≤ λ1, . . . , λd < 1}.

The geometric idea behind this proof is that we can tile our original cone K with copies of its fundamental parallelepiped Π in directions deﬁned by the generators wi’s. The proof, however, is purely algebraic.

We are interested in this decomposition because Π is bounded.

PROOF. Expanding the denominators in the RHS of Eq. (1.2.1) we get       X p X k w X k w  x   x 1 1  ···  x d d  . p∈Π∩Zd k1≥0 kN ≥0

p+k1w1+···+k w Therefore we have terms of the form x d d for some p ∈ Π and k1, . . . , kd nonnegative integers. On the other hand, in the LHS we have the terms of the form xm, where m ∈ K. By

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deﬁnition, points in K can be written as m = v + λ1w1 + . . . λdwd, with nonnegative real λi’s. Since the cone is simplicial, this decomposition is unique. Now we break down each λi in its ﬂoor integer part bλic and its fractional part {λi}. Therefore each point m ∈ K is uniquely written as

m = p + bλ1cw1 + ··· + bλ1cwd, where p = v + {λ1}w1 + ··· + {λd}wd ∈ Π.

Comparing with the previous expression and by the uniqueness of the coeﬃcients λi’s, we only need d to check that if a choice of λi’s produces a point m ∈ K ∩ Z , the corresponding p lies not only in Π d but in Π ∩ Z . This is indeed true because the vectors bλiciwi are integer vectors.

In the integer-point enumerator language, Remark 6 translates to

X t σcone(P)(x1, . . . , xd+1) = σtP (x1, . . . , xd)xd+1. t≥0

P d t If we evaluate this equation into x1 = ··· = xd = 1, we get t≥0 tP ∩ Z xd+1, a series whose coeﬃcients count the integer points in dilations of P. This is called in the literature the Ehrhart series of P.

Lemma 1.2.3. Let f : Z≥0 → R and g : R≥0 → R be functions satisfying g(x) X f(t)xt = . (1 − x)d+1 t≥0

Then f is a polynomial of degree d if g is a polynomial of degree at most d and g(1) 6= 0.

PROOF. d+1 1 Let g(x) = b0 + ··· + bd+1x , satisfying g(1) = b0 + ··· + bd+1 6= 0. Using the expansion (1−x)d+1 = P t+d t t≥0 d x , we can write g(x) = (1 − x)d+1 ! t + d X b + ··· + b xd+1 xt d 0 d+1 t≥0 ! d+1 t + d X X b xt+k. d k k=0 t≥0

Now we have a sum of d + 1 series, in each of them we perform a change of indexes τk = t + k. This way we get d+1 ! d+1 ! X X t + d X X τk − k + d b xt+k = b b xτk . d k k d k k=0 t≥0 k=0 τk≥k n Pd+1 P τ−k+d τ Finally, using k = 0 for n < k, we can write the previous sum as k=0 bk τ≥0 d x . When we regard this as a series in x, the coeﬃcient of degree d is ! g(x) d+1 τ − k + d [xτ ] = X b , (1 − x)d+1 k d k=0

25 of 62 1.3. ALGEBRA OF POLYHEDRA CHAPTER 1. POLYTOPES, POLYHEDRA, LATTICES which is clearly a polynomial in τ of degree at most d. It is indeed of degree d because the leading 1 term is d! (b0 + ··· + bd+1) 6= 0 by hypothesis.

Now we have all the ingredients to prove Ehrhart’s theorem for simplices.

PROOF. (Of Theorem 1.2.1 for simplices.) d Let ∆ be a d-dimensional simplex in R with vertices v1, . . . , vd+1. By Lemma 1.2.3 it suﬃces to prove that there exists some polynomial g of degree at most d and g(1) 6= 0 satisfying g(x) σ (1,..., 1, x) = . cone(∆) (1 − x)d+1

The cone over a simplex is simplicial with apex in the origin and generators of the form wi = (vi, 1), so we can apply Lemma 1.2.2. When we evaluate the ﬁrst d variables in order to count integer points, we σΠ(x) d+1 note that the denominator in w is indeed (1 − x ) , so we are only interested in (1−xw1 )···(1−x d+1 ) d+1 the numerator as a polynomial in the variable xd+1. In other words, we propose g(x) = σΠ(1,..., 1, x).

First we note that it is certainly a polynomial: since Π is bounded, it has ﬁnitely many integer points, thus its integer-point enumerator has a ﬁnite number of terms. Each term is of the form λ1w1+···+λ w x d+1 d+1 for some 0 ≤ λ1, . . . , λd+1 < 1. For each of them, we want to bound the exponent of xd+1. Since the (d+1)-th coordinate of all vectors wi’s is 1, the exponent of xd+1 is λ1 +···+λd+1 < d + 1. Since we are counting integer points, the exponent needs to be an integer, therefore it is at most d.

When we evaluate g(1) we are counting integer points in Π. This is non-zero, because the origin is an integer point and always belongs to Π. This concludes the proof.

Now we need to extend the result to general polytopes. To do this, we will use the fact that we can always triangulate any polytope into simplices (recall Theorem 1.1.18).

This is the tool we need to ﬁnish our ﬁrst proof of Ehrhart’s theorem.

PROOF. (Of Theorem 1.2.1) Let P be a d-dimensional polytope, and T a triangulation of P that adds no new vertices. Then by the inclusion-exclusion principle,

d X d X d |T |+1 d tP ∩ Z = t∆ ∩ Z − t(∆1 ∩ ∆2) ∩ Z + ··· + (−1) t(∆1 ∩ · · · ∩ ∆|T |) ∩ Z . ∆∈T (∆1,∆2∈T ) All the terms in the previous sum are polynomials of degree at most d by the simplex case, and only the ﬁrst one is of degree exactly d, so there can be no cancellations and thus the result is a polynomial of degree exactly d.

1.3 Algebra of polyhedra

We have already seen that in order to study integer points in polytopes it can be useful to introduce more general structures and then ﬁnd a way to specialize the results for the case we in which are interested.

26 of 62 1.3. ALGEBRA OF POLYHEDRA CHAPTER 1. POLYTOPES, POLYHEDRA, LATTICES

In this section we introduce the algebra of polyhedra, which is useful to think as a structure we give to polyhedra to deﬁne basic algebraic operations with them.

N Deﬁnition 1.3.1 (Indicator function). Let A ⊆ R be any set. Its indicator function is the N function [A]: R → R deﬁned by ( 1 if x ∈ A [A](x) = 0 if x ∈/ A

One interesting property of identifying sets with their indicator functions is that some algebraic operations with indicator functions do have geometric meaning. For multiplication we have [A]·[B] = [A∩B], and for addition we have [A] + [B] = [A ∪ B] + [A ∩ B]. In fact, for addition we actually have an inclusion-exclusion formula that will be useful later, namely

" #   [ X |J|+1 \ Ai = (−1)  Aj (1.3.1) i∈I J⊆I j∈J

With this concept we can deﬁne the linear spaces we will be working on for a major part of the thesis, as the span of indicator functions of certain families of sets.

N Deﬁnition 1.3.2 (Algebra of polyhedra). Given an ambient space R , the algebra of polyhedra is the linear span of indicator functions of polyhedra, we denote it by

N n N o P(R ) = span [A]: A ⊆ R polyhedron .

Similarly, the algebra of polytopes is the linear span of polytopes, we denote it with a subscript b, accounting for bounded:

N n N o Pb(R ) = span [A]: A ⊆ R polytope .

Note that the indicator functions of polytopes or polyhedra are far from being a basis of their corresponding algebras, since the property we have just described for addition of indicator functions is in itself a linear relation. Remark 7. Since all points in a polytope lie in the relative interior of a unique face, we have

[P ] = X [rel int F ] F : face of P

Also note that the algebra of polyhedra contains the indicator functions of open polyhedra as well. In particular, for a polyhedron P of dimension d we have

[rel int P ] = X (−1)d−dim F [F ]. (1.3.2) F : face of P Eq. (1.3.2) is a consequence of the inclusion-exclusion principle applied to faces of P .

27 of 62 1.3. ALGEBRA OF POLYHEDRA CHAPTER 1. POLYTOPES, POLYHEDRA, LATTICES

= − +

Figure 1.6: An open square as sum of polytopes.

N Remark 8. The algebra of polyhedra P(R ) is spanned by the indicator functions of polyhedra without lines.

This is easy to prove by induction. For N = 0 the result is clear. For N > 0, consider P a polyhedron, if it has a line, cut the polyhedron in two halves by a hyperplane. Repeating this process a ﬁnite amount of times we get some polyhedra without lines and in the intersection some polyhedra of smaller dimension, to which we apply the induction hypothesis.

1.3.1 Valuations

Now that we have given polyhedra the structure of a vector space, we may be able to ﬁnd useful results studying linear maps.

Deﬁnition 1.3.3 (Valuation). A valuation is a linear map T : P, Pb → V , where V is a vector space.

We typically think of valuations as a way of assigning value to a polyhedron. In that case, V = R. But it will be useful for more convoluted constructions to consider valuations that map to some functions spaces, so we keep a general vector space V in our deﬁnition.

N Theorem 1.3.1 (Euler characteristic). There exists a unique valuation χ : P(R ) → R such that χ([A]) = 1, for all polyhedron A. This is called the Euler valuation or Euler characteristic.

PROOF. First observe that we are giving the value of χ on a set of generators, so if it exists is necessarily unique.

We will prove by induction on N. If N = 0, clearly χ(f) = f(0) satisﬁes the theorem. Consider N > 0.

N Let a : R → R be a linear function, a 6= 0, and consider for all τ ∈ R the hyperplanes

N Hτ = {x ∈ R : a(x) = τ}.

By the induction hypothesis, there exists a valuation χτ : P(Hτ ) → R such that χτ (A) = 1 for all polyhedron A ∈ Hτ . We will ﬁrst deﬁne χ in the algebra of bounded polyhedra, and the we will extend

28 of 62 1.3. ALGEBRA OF POLYHEDRA CHAPTER 1. POLYTOPES, POLYHEDRA, LATTICES

N the deﬁnition to unbounded polyhedra too. Let us choose an element f ∈ Pb(R ). By construction, it can be written as X f = αi[Ai], where Ai are polytopes. i∈I P Consider also the restrictions fτ = f ∈ Pb(Hτ ). We can write them as fτ = αi[Ai ∩ Hτ ], so Hτ i∈I X χτ (fτ ) = αi, where Iτ = {i : Ai ∩ Hτ 6= ∅}. i∈Iτ

This value will be constant in τ except when we ﬁnd boundaries of sets Ai. In particular, observe that X lim χτ−ε (fτ−ε) = αi, where Jτ = i ∈ Iτ : min a(x) 6= τ . (1.3.3) ε→0+ x∈Ai i∈Jτ

This way, we can deﬁne χ : P ( N ) → as b R R τ4 τ4 − ε X χ(f) = χτ (fτ ) − lim χτ−ε (fτ−ε) . ε→0+ τ∈R τ3 (1.3.4) τ3 − ε

By Eq. (1.3.3) only finitely many terms of the τ2 sum are non-zero, so it is well defined. Linearity τ2 − ε is preserved because all the operations we have τ1 used to define χ (sums, limits, restriction to hy- τ1 − ε perplane) preserve linearity. Finally, for a poly- τ0 tope, only one term is non-zero, namely the one τ0 − ε corresponding to the only proper face of the form Figure 1.7: Sketch of the limit in Eq. (1.3.3) {a(x) = τ} (see Fig. 1.7). for polytopes.

N N N To extend χ to P(R ), consider for t ≥ 0, the dilated (±1)-cube t . Since 0 ∈ rel int , we have N S N that R = t≥0 t . We define h i χ(f) = lim χ f · t N . t→∞ N N N This is well defined because for any polyhedron [A], we have [A] · [t ] = [A ∩ (t )] ∈ Pb(R ), and N for sufficiently large t we have A ∩ (t ) 6= ∅, thus the limit is well defined.

Observe that the proof gives us a formula for computing χ in terms of hyperplanes Hτ . This will be useful later.

We know that the algebra of polyhedra contains also open polyhedra, see Fig. 1.6. We can recover the Euler-Poincar´eformula by studying χ on open polytopes. Theorem 1.3.2. Let P be a d-dimensional polytope. Then

χ([rel int P ]) = (−1)d.

PROOF. N Let a : R → R be a linear function, a 6= 0. Consider for all τ ∈ R the hyperplanes N Hτ = {x ∈ R : a(x) = τ}

29 of 62 1.3. ALGEBRA OF POLYHEDRA CHAPTER 1. POLYTOPES, POLYHEDRA, LATTICES

and the functions fτ = [rel int P ] · [Hτ ]. By Eq. (1.3.4) we know that X χ([rel int P ]) = χτ (fτ ) − lim χτ−ε (fτ−ε) , ε→0+ τ∈R and as in Fig. 1.7, only one term of the sum is non-zero, corresponding to τ such that minx∈Ai a(x) 6= τ. In this case, P ∩ Hτ = ∅ and P ∩ Hτ−ε is a (d − 1)-dimensional polytope, so we can ﬁnish the proof using the induction hypothesis: χ([rel int P ]) = 0 − (−1)d−1 = (−1)d.

Corollary 1.3.3 (Euler-Poincar´e formula). Let P be a d-dimensional polytope and let fk be the number of k-dimensional faces of P , for k = 0, . . . , d. Then

d X k (−1) fk = 1. k=0

PROOF. Recall Remark 7 and apply the previous theorem and deﬁnition of χ for polytopes:

d X X dim F X 1 = χ([P ]) = [rel int F ] = (−1) = fk. F : face of P F : face of P k=0

1.3.2 Basic properties

In this section we will basically study what can we say about the results we have seen through Chapter 1 when we regard not only polyhedra (resp. polytopes) but their whole algebra.

N M Proposition 1.3.4. Let T : R → R be a linear transformation. Then there exists a unique N M valuation T : P(R ) → P(R ) such that T ([P ]) = [T (P )] for every polyhedron P.

PROOF. As in Theorem 1.3.1, if such a valuation exists, it must be unique.

M −1 N N Given x ∈ R , its preimage T (x) ⊆ R is an aﬃne subspace, so for every polyhedron P ⊆ R , −1 −1 N N the set P ∩ T (x) is also a polyhedron. Therefore f · [T (x)] ∈ P(R ) for every f ∈ P(R ). With this observation, we will use the Euler valuation to construct T :

N M T : P R −→ P R M f 7−→ T (f): R −→ R x 7−→ h(x) = χ f · T −1(x) . Now note that if f = P , we have P ∩ T −1(x) 6= ∅ if and only if x ∈ P . Since it is always a polyhedron, this means that ( h i 1 if x ∈ P, χ P ∩ T −1(x) = 0 otherwise.

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N N N Proposition 1.3.5. There exists a unique bilinear operation ∗ : P(R ) × P(R ) → P(R ) such that [P1] ∗ [P2] = [P1 + P2] for every pair of polyhedra P1,P2.

PROOF. (Sketch only) As in Theorem 1.3.1 and previous proposition, if such a valuation exists, it must be unique. For N N N f, g ∈ P(R ), define the function f × g : R ⊕ R → R as (f × g)(x, y) = f(x)g(y). Note that N N N N N f × g ∈ P(R ⊕ R ). Let T : P(R ⊕ R ) → P(R ) be the linear transform of Proposition 1.3.4. Then define f ∗ g = T (f × g) and check that the properties are satisfied.

1.3.3 Decomposition modulo polyhedra with lines

One of the most powerful properties of working with the algebra of polyhedra is that we will be able to describe a polyhedron as a certain combinations of its cones from its vertices, modulo some equivalence relation.

N Deﬁnition 1.3.4 (Equivalence modulo polyhedra with lines). Let f, g ∈ P(R ), we say that f ≡ g modulo polyhedra with lines if f − g can be expressed as a linear combination of polyhedra that contain some line.

To understand why this deﬁnition is useful, we need to introduce tangent cones.

Deﬁnition 1.3.5 (Tangent cone, feasible cone). Let P be a polyhedron and v ∈ P a point. Then the tangent cone of P at v is

tcone(P, v) = {v + x : v + εx ∈ P, for some ε > 0}.

Analogously, the feasible cone or cone of feasible directions is

fcone(P, v) = {x : v + εx ∈ P, for some ε > 0}.

It is clear from the deﬁnition that tcone(P, v) = v+fcone(P, v). It is useful to characterize the tangent and feasible cones with respect to the deﬁning inequalities of a polyhedron.

Given a polyhedron P deﬁned by inequalities {ai(x) ≤ αi}i∈I and a point v ∈ P , we say that a particular inequality i0 ∈ I is active on v if ai0 (v) = αi0 . We denote the set of indices of active equations as at a given point as Iv. Then our cones have the following description:

tcone(P, v) = {x : ai(x) ≤ αi, for i ∈ Iv},

fcone(P, v) = {x : ai(x) ≤ 0, for i ∈ Iv}.

These cones are well behaved with respect to linear transforms and intersections:

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N M N Lemma 1.3.6. Let T : R → R be a linear transformation, P ⊆ R and polyhedron and v ∈ P . Then tcone T (P ),T (v) = T tcone(P, v).

PROOF. We know that the linear image of a polyhedron is also a polyhedron by Proposition 1.3.4. Let Q = T (P ) and w = T (v). (⊆) Let w + y ∈ tcone(Q, v). Then there exists ε > 0 such that w + εy ∈ Q = T (P ), so there exists v + x ∈ P such that T (v + x) = w + εy. But then T (εy) = x, so we have that v + ε−1x ∈ tcone(P, v) and T (v + ε−1x) = w + y. (⊇) Let x ∈ tcone(P, v). Then v + δx ∈ P for some δ, so w + δT (x) ∈ Q, and thus T (v + x) ∈ tcone(Q, w).

T Lemma 1.3.7. Given a ﬁnite collection of polyhedra {Pi}i∈I , and a point v ∈ i∈I Pi, the tangent cones satisfy ! \ \ tcone Pi, v = tcone(Pi, v). i∈I i∈I

PROOF. For each i ∈ I, let Ji denote the indices of the active inequalities of Pi on v. The tangent cone tcone(Pi, v) is defined precisely by the inequalities indexed by Ji, and the intersection of tangent cones in therefore defined by the inequalities indexed in ∪iJi. On the other hand, the inequalities defining ∩iPi are the reunion of the defining inequalities of all Pi’s, and an inequality is active on a point independently of the ambient polytope, so the set of defining inequalities of tcone[∩iPi, v] is precisely ∪iJi.

The following is the main result that motivates the introduction of this equivalence relation. It was essentially ﬁrst proved by Brion in [4] albeit in a diﬀerent setting, similar to Eq. (2.1.11).

Theorem 1.3.8 (Brion). Let P be a polyhedron and vert(P ) its set of vertices. Then

[P ] ≡ X [tcone(P, v)] modulo polyhedra with lines. v∈vert(P )

PROOF. We will prove the theorem first for simplices, then for polytopes and finally for a general polyhedron. For simplices, by Lemma 1.3.6 we can restrict ourselves to the standard simplex ∆d, defined by

d d ∆ = {x : x1 + ··· + xd = 1, x1, . . . , xd ≥ 0}, and equivalently as ∆ = conv(e1,..., ed),

d where e1,..., ed are the vectors of the canonical basis in R . Let H be the hyperplane deﬁned by the equation x1 + ··· + xd = 1, and let for i ∈ {1, . . . , d} let Hi ⊆ H be the half-hyperplane deﬁned by Sd the inequality {xi ≥ 0}. We have that i=1 Hi = H, so we can apply Eq. (1.3.1) to get   X |J|+1 \ [H] = (−1)  Hj . J⊆{1,...,d} j∈J

32 of 62 1.3. ALGEBRA OF POLYHEDRA CHAPTER 1. POLYTOPES, POLYHEDRA, LATTICES

Now we look at this equation with the modulo-polyhedra-with-lines glasses. The LHS vanishes, because T d a hyperplane clearly contains lines. In the RHS we have for J = {1, . . . , d} that j∈J Hj = ∆ . Also, by the characterization of tangent cones with active inequalities, for each i ∈ {1, . . . , d}, the term corresponding with J = {1, . . . , d}\{i} satisﬁes T H = tcone(∆d, e ). And then the rest i j∈Ji j i of the terms in the RHS vanish, because in those there are at least two indices, i1, i0 out of the intersection, and thus the aforementioned intersection contains the line that goes through ei1 and ei2 . This concludes the proof for simplices.

For a general polytope P , by Theorem 1.1.18 we can ﬁnd a triangulation T that adds no new vertices. Assume T = {∆ ,..., ∆ }. For each ∆ we know that [∆ ] ≡ P [tcone(∆ , v )]. Adding 1 n k k vk∈vert(∆k) k k all together we get n X X X [∆k] = [tcone(∆kv , v)], (1.3.5) k=1 v∈vert(P ) kv∈Kv where Kv is the set that indexes the simplices that have v as a vertex. Now we can expand the LHS of Eq. (1.3.5) using inclusion-exclusion:

n " n #   X [ X |J| \ [∆k] = ∆k + (−1)  ∆j . k=1 k=1 J⊆1,...,n,|J|>1 j∈J

The ﬁrst term is [P ], and we can expand the second term using the theorem for simplices:      X |J| \ X |J| X \ (−1)  ∆j = (−1) tcone  ∆j, vJ  . (1.3.6) J⊆1,...,n,|J|>1 j∈J J⊆1,...,n,|J|>1 vJ ∈vert(∩J ∆j ) j∈J

Now we apply the inclusion-exclusion principle to the RHS of Eq. (1.3.5). For each vertex v ∈ vert(P ):     X [ X |Jv| \ [tcone(∆kv , v)] =  tcone(∆kv , v) + (−1)  tcone(∆j, v) . kv∈Kv kv∈Kv Jv⊆Kv,|Jv|>1 j∈Jv

The ﬁrst term is just [tcone(P, v)], and adding for all v we get the desired result, so hopefully the second summation in the previous equation will vanish with Eq. (1.3.6). Indeed, if we compare them, the sums run over the same elements, they have the same sign with the (−1)|J|, and the indicator functions are the same thanks to Lemma 1.3.7. This concludes the proof for polytopes.

For an arbitrary polyhedron, it it contains a line, then it contains no vertices so the result is clear. If it contains a line, by Proposition 1.1.14 it can be decomposed as a Minkowski sum P = Q + K, where Q is the convex hull of the vertices of P and K is the recession cone of P . We already know P that [Q] ≡ v∈vert(Q)[tcone(Q, v)], and by Proposition 1.3.5 we know that Minkowski sum preserves linear relations, so we have

[Q + K] ≡ X [tcone(Q, v) + K] modulo polyhedra with lines. v∈vert(P )

Now we just need to note that for the recession cone tcone(Q, v) + K = tcone(Q + K, v) and we are done.

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Corollary 1.3.9. Let {Pi}i∈I be a family of polytopes such that 0 ∈ Pi for all i ∈ I and {αi}i∈I be real numbers. Then, modulo polyhedra with lines,

X X ◦ αi[Pi] ≡ 0 ⇐⇒ αi[Pi ] ≡ 0. i∈I i∈I

v3 v3 v3 v3 + + ≡ 1 1 tcone(P, v3) 3 tcone(P, v1) tcone(P, v2) v1 v2 v1 v2 v1 v2 v1 v2 1

−1

v3 v3 v3 −2 v3 + + ≡ −2 −1 −3 v v v v v v 1 2 1 2 1 2 v v2 1 −2 −1

Figure 1.8: Illustration of Brion’s theorem for a triangle. In the ﬁrst line we add the tangent cones. In the second line, we half-planes, that are polyhedra with lines. Adding both results together plus the indicator function of the whole plane (which is also a polyhedron with lines) yields the indicator of the triangle.

We have a similar result for the cones of feasible directions. Corollary 1.3.10 (Brion’s theorem with feasible cones). Let P be a polyhedron, vert(P ) its set of vertices, and KP its recession cone. Then X [KP ] ≡ [fcone(P, v)] modulo polyhedra with lines. v∈vert(P )

We also have an analogous result for open polyhedra, but we ﬁrst need to prove an auxiliary equation. Lemma 1.3.11. Let P be a d-dimensional cone. Then

[P ] ≡ (−1)d[− rel int K] modulo polyhedra with lines.

Corollary 1.3.12 (Brion’s theorem for open polytopes). Let P be a polytope and vert(P ) its set of vertices. Then

[rel int P ] ≡ X [rel int tcone(P, v)] modulo polyhedra with lines. v∈vert(P )

We will not prove these last results. A proof can be found in [1, Ch. 7].

34 of 62 Chapter 2

Local formula for Ehrhart polynomials

2.1 The exponential valuation

Recall that, informally, a valuation is a rule to assign a value to objects in our vector spaces. The exponential valuation we will deﬁne in this section, following [1, Ch. 8] is motivated by the idea of extending the concept of volume of a polytope to any element in the algebra of polyhedra, as deﬁned in Section 1.3.

The natural way to deﬁne the volume of a polytope is through its integral. This is useful for polytopes, and even for the algebra of polytopes, but we immediately reach a problem when the domains of the functions we want to measure are not bounded.

It turns out that the right extension is to assign a function instead of a number, that when evaluated at the origin, if the region is bounded, will let us recover the volume. Let’s get our hands dirty.

N ◦ Lemma 2.1.1. Let K ⊆ R be a pointed cone, then for every c ∈ rel int K , the integral Z ehc,xidx K converges absolutely and uniformly on compact subsets of rel int K◦ to a rational function of the form −1 φ(K, c) = X α Y , I hc, u i I i∈I i N where u1,..., un ∈ R are non-zero vectors, αI ∈ R+ and I ⊆ {1, . . . , n}.

PROOF. Without loss of generality, we can assume that dim K = N, otherwise the measure of K is null.

We will ﬁrst prove it for simplicial cones. If K is simplicial it can be written as K = co(u1,..., uN ) N for some u1,..., uN that for a basis of R . Let U be the corresponding basis transformation. In that case we can compute the integral via a change of variables. Let α = vol(u1,..., uN ), then

35 2.1. THE EXPONENTIAL VALUATION CHAPTER 2. LOCAL FORMULA

Z Z m Z ∞ m 1 hc,xi hc,Uyi Y hc,uiiyi Y e dx = α e dy = α e dyi = α . K N 0 −hc, u i R≥0 i=1 i=0 i In the last step we are using that the integrals converge precisely because c ∈ rel int K◦.

By Proposition 1.1.12 we can represent K as the cone over a (N − 1)-dimensional polytope. Let u1,..., um be the vertices of such a polytope, then K = co(u1,..., um). By Corollary 1.1.19 we can triangulate K using no new generators. Let I be the set indexing the vertices of such a triangulation. For every I ∈ I, the corresponding cone KI = co{ui : i ∈ I} is simplicial. We can perform the integral using the result for simplicial cones. Let αI = vol(ui : i ∈ I), then Z Z 1 ehc,xidx = X ehc,xidx = X α Y . I −hc, u i K I∈I KI I∈I i∈I i

Remark 9. If we want to allow this same result for a shifted cone K + v, with a change of variables we get Z Z ehc,xidx = ehc,vi ehc,xidx. K+v K

This observation, along with the previous lemma, justify the choice of the following space of functions: ( ) N −1 M( N ) = span f(c) = ehc,vi Y : u ,..., u form a basis and v ∈ N . R hc, u i 1 N R i=1 i

N N Theorem 2.1.2 (Exponential valuation). There exists a valuation Φ: P(R ) → M(R ) such that

(1) For any polyhedron P without lines, and for all c ∈ int K◦, where K is its recession cone, R hc,xi ◦ the integral P e dx converges absolutely and uniformly on compact subsets of int K to Φ([P ]).

(2) For any polyhedron with lines, Φ([P ]) = 0.

PROOF. We will prove the result by induction on N. If N = 0, the result is clear. Assume N > 0.

N (1) Since polyhedra without lines span P(R ), we just need to prove two diﬀerent things:

R hc,xi N (a) The integral P e dx converges to a function in M(R ). (b) The assignation deﬁned in (a) is linear.

(a) As we did before, we can assume dim P = N, otherwise the integral is null. By Proposition 1.1.14, we can write P = K + Q, where K is the recession cone of P and Q is the convex hull of the vertices

36 of 62 2.1. THE EXPONENTIAL VALUATION CHAPTER 2. LOCAL FORMULA of P . Let u ∈ int K◦ such that kuk = 1 and u is not orthogonal to any proper face of P . We partition N R into hyperplanes N Hτ = {x ∈ R : hx, ui = τ}, for τ ∈ R, and consider Pτ = P ∩ Hτ . The induction hypothesis holds for Pτ in Hτ .

To use it, ﬁrst note that there exists τ1 such that Pτ = ∅ for τ > τ1. This is true because given x ∈ P , we have x = y + v, where y ∈ Q and v ∈ K. Then

hx, ui = hy, ui + hv, ui ≤ hy, ui, where hv, ui ≤ 0 holds because u ∈ K◦ and Remark 4. Therefore

maxhx, ui = maxhx, ui. x∈P x∈Q

Since Q is compact, this max does exist. This max value is τ1 and thus we can write

Z Z τ1 Z ehc,xidx = ehc,yidy dτ, (2.1.1) P −∞ Pτ where dy is the standard Lebesgue measure in Hτ .

Since faces are deﬁned by valid inequalities, the faces of Pτ correspond to faces of P of greater or equal dimension. In particular, the vertices of Pτ correspond to 4 either vertices or edges of P .

Let v1,..., vm be the vertices of P , ordered in such a way that

τm ≥ · · · ≥ τ1, where τi = hvi, ui.

For notation purposes, we may consider τm+1 = −∞, and thus we can rewrite Eq. (2.1.1) as m Z Z τi Z ehc,xidx = X ehc,yidy dτ. (2.1.2) P i=1 τi+1 Pτ

The induction hypothesis holds in for P in H . Since K ⊆ K , we have K◦ ⊇ K◦ and thus by τ τ Pτ P Pτ P item (1) of the statement Z hc,yi Φ([Pτ ]) = e dy. (2.1.3) Pτ Item (2) of the statement is also part of our induction hypothesis, and it gives us permission to use Brion’s decomposition of Pτ into its tangent cones (see Theorem 1.3.8), so X Φ([Pτ ]) = Φ [tcone(Pτ , v)] . v∈vertPτ

We know that tcone(Pτ , v) = v + fcone(Pτ , v), and Lemma 2.1.1 applies to the feasible cone, so there c v exist functions fτ ∈ M(Hτ ) such that Φ([fcone(Pτ , v)]) = fτ (c), and by Remark 9 hc,vi v Φ([tcone(Pτ , v)]) = e fτ (c). This last equation jointly with Brion’s decomposition and Eq. (2.1.3) gives us for all τ: Z hc,yi X hc,vi v c e dy = e fτ (c), where fτ ∈ M(Hτ ). (2.1.4) Pτ v∈vertPτ

37 of 62 2.1. THE EXPONENTIAL VALUATION CHAPTER 2. LOCAL FORMULA

v The next step we need is to show that in the open intervals (τk+1, τk) the functions fτ remain constant with respect to τ. First observe that for τ ∈ (τk+1, τk), the vertices of Pτ come all from edges of P . This way, if a vertex v of some Pτ comes from the edge [vi, vj] of P , then its variation with respect to τ is τ − τj τi − τ vi,j(τ) = vi + vj. τi − τj τi − τj

For τ ∈ (τk+1, τk), the vertices vi,j(τ) of Pτ come from speciﬁc edges [vi, vj] of P .

The deﬁning inequalities of Pτ are the same of P plus hx, ui = τ. Therefore, the active inequalities of Pτ on the vertex vi,j(τ) are the union of the active inequalities of vi and vj. Hence, by the description of the feasible cone with active inequalities (see Deﬁnition 1.3.5) we conclude that indeed the functions v fτ remain constant with respect to τ in the open intervals (τk+1, τk). We will modify our notation vi,j (τ) i,j accordingly: we will denote fτ as fk when τ ∈ (τk+1, τk). As a convention, if the edge [vi, vj] i,j does not correspond to any vertex in τ or if said edge simply does not exist, we will say fk = 0. With this last observation and new notation, we can compute for the integrals in Eq. (2.1.2) as

m Z τk Z Z τk hc,yi X i,j hc,vi,j (τ)i e dy dτ = fk e dτ, (2.1.5) τk+1 Pτ i,j=1, i

D −τ v +τ v E hv −v ,ci D −τ v +τ v E hv −v ,ci hv −v ,ci ! Z τk j i i j Z τk i j j i i j i j i j ,c τ ,c τi − τj τk τk+1 ehc,vi,j (τ)idτ = e τi−τj e τi−τj dτ = e τi−τj e τi−τj − e τi−τj . τk+1 τk+1 hvi − vj, ci i,j Since fk are rational functions of degree −(d − 1) and the integrals in Eq. (2.1.5) add only one degree R hc,xi N in the denominator, we conclude that the integral P e dx converges to a function in M(R ). We will denote this function by φ(P, c). Note that φ(P, c) is deﬁned in all except for a ﬁnite number of R hc,xi ◦ values of c, while P e dx may diverge for values of c ∈/ int K . (b) To prove linearity of Φ, we only need to prove it with indicators of polyhedra without lines, since they span the whole algebra of polyhedra (see Remark 8). So we want to prove that given a linear relation X αi[Pi] = 0, (2.1.6) i∈I where αi ∈ R and Pi are polyhedra without lines, we have X αiΦ([Pi]) = 0. (2.1.7) i∈I Consider a decomposition h N i X R = βj [Qj] , j∈J where βj ∈ R and Qj are polyhedra without lines (such a decomposition exists on behalf of Remark 8). Then for every polyhedron Pi in Eq. (2.1.6) we have

h N i X [Pi] = [Pi] · R = βj[Pi ∩ Qj]. (2.1.8) j∈J

Now we can apply part (a) to each polyhedron Pi ∩ Qj because they do not contain lines, so for all c ∈ int K we have φ(P ∩ Q , c) = R ehc,xidx, and φ(P ∩ Q , c) ∈ M( N ). Since the Pi∩Qj i j Pi∩Qj i j R

38 of 62 2.1. THE EXPONENTIAL VALUATION CHAPTER 2. LOCAL FORMULA

recession cones satisfy KPi∩Qj ⊆ KPi and Z Z hc,xi X hc,xi e dx = βj e dx, Pi j∈J Pi∩Qj we have for all c ∈ int KP i X φ(Pi, c) = βjφ(Pi ∩ Qj, c). (2.1.9) j∈J Equation (2.1.9) is an equation between two meromorphic functions that holds in a set with non- empty interior, so it holds on all their domain. By an analogous argument, if instead of Eq. (2.1.8) we consider for a given j ∈ J the identity X 0 = αi[Pi ∩ Qj], i∈I we get X αiφ(Pi ∩ Qj, c) = 0 (2.1.10) i∈I for all c ∈ int K◦ , and thus for all c ∈ N . Qj R

Now combining Eqs. (2.1.9) and (2.1.10) we get what we wanted to prove (Eq. (2.1.7)):

  ! X X X X X αiΦ([Pi]) = αi  βjΦ([Pi ∩ Qj]) = βj αiΦ([Pi ∩ Qj]) = 0. i∈I i∈I j∈J j∈J i∈I

(2) Given a polyhedron without lines P , and a translation of it, P + v, observe that the recession cones are the same, so by Remark 9 we have for all c ∈ KP Z Z ehc,xidx = ehc,vi ehc,xidx. P +v P

Since we have constructed Φ as extending these integrals, we have Φ([P + v]) = ehc,viΦ([P ]) for all polyhedron P without lines and all vector v. Moreover, the identity holds also for polyhedra with lines because they can be expressed as linear combinations of polyhedra with lines (see Remark 8) and we know that part (1) holds.

If P contains a line in the direction of v 6= 0, then P = P + v, but then for the valuations we have

Φ([P ]) = Φ([P + v]) = ehc,viΦ([P ]).

We conclude the proof by noting that ehc,vi 6= 1 because v 6= 0.

Note that by the second statement of the theorem and Theorem 1.3.8, we can write

Φ([P ]) = X Φ([tcone(P, v)]). (2.1.11) v∈vert(P )

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2.2 The counting valuation

In the spirit of Sections 1.2 and 2.1, we want to extend the concept of counting integer points in a polytope to the algebra of rational polyhedra. In particular, we want to extend the integer-point enumerator σS(x) (Definition 1.2.1) to a valuation in the algebra of rational polytopes. Also, we want our valuation to assign values not in the integer or real numbers, but in a certain function space that we will define, similar to how we defined the exponential valuation, because our ultimate goal is to mix both valuations. We will follow mainly [1, Ch. 13] for this section.

N Recall that we deﬁned the integer-point enumerator of a set S ⊆ R as X m σS(x) = x . m∈S∩ZN

N Lemma 2.2.1. Let K ⊆ R be a pointed rational cone. Then K = co(w1,..., wn) where N N w w1,..., wn ∈ Z . Deﬁne the set WK = {x ∈ C : kx i k < 1 for i = 1, . . . , n}. Then the series X xm m∈K∩ZN

converges for every x ∈ WK and uniformly on compact subsets of WK to a function of the type v X x i f(K, x) = εi , (1 − xui,1 ) ····· (1 − xui,N ) i∈I

N where εi = ±1 and vi, ui,j ∈ Z for all i and j.

PROOF. (Sketch only) N First we prove the case K = R≥0. In this case using the geometric sum formula, when x ∈ WK we have X m X m1 mN 1 1 x = x1 ··· xN = ··· . 1 − x1 1 − xN m∈K∩Zd m1,...,mN ≥0 d PN Then, if K is a simplicial cone K = co(w1,..., wN ), for a given m ∈ Z , we have m = i=1 βiwi for some βi ∈ R≥0. We can decompose it into m = n + l by N N X X l = bβicwi, and l = {βi}wi. i=1 i=1 Thus getting

      N X n X m X l w +···+l w X n Y 1 x =  x   x 1 1 N N  =  x  . (2.2.1) 1 − xwi m∈K∩Zd n∈Π∩Zd l1,...,lN ≥0 n∈Π∩Zd i=1 Now, for a general (not necessarily simplicial) cone, we argue as in Lemma 2.1.1: triangulate K, apply the simplicial case to each element and add all together.

In order to use the same notation as in Section 2.1, we will consider the change of variables x = ec. ◦ With this notation, the condition x ∈ WK of Lemma 2.2.1 translates to c ∈ rel int K .

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Remark 10. In the spirit of Remark 9, if we want to allow this same result for a shifted cone K + v, we get X ehc,ui = X ehc,m+vi = ehc,vi X ehc,mi. u∈(K+v)∩ZN m∈K∩ZN m∈K∩ZN

This observation, along with the previous lemma, justify the choice of the following space of functions, N for a ﬁxed lattice Λ in R : ( N ) N hc,vi Y 1 N ( ) = span f(c) = e : v, u1,..., uN ∈ Λ . R hc,uii i=1 1 − e The main result of this section is the following:

N N Theorem 2.2.2 (Counting valuation). There exists a valuation F : P(Q ) → N (R ) such that

1. For any polyhedron P without lines, and for all c ∈ rel int K◦, where K is its recession cone, P hc,ui ◦ the series u∈P ∩Λ e converges absolutely and uniformly on compact subsets of int K to F([P ]).

2. For a polyhedron with lines, F([P ]) = 0.

PROOF. (Sketch only) The proof is, as usual, by induction on N. For N = 0 the result is clear. Assume N > 0.

First we have to prove that given a polyhedron P without lines and, K its recession cone and c ∈ int K◦, the series X ehc,ui u∈P ∩Λ N ◦ converges to a function in N (R ). Since P contains no lines, K is pointed and thus int K is a non-empty open set. Consider the dual lattice to Λ, namely

∗ Λ = {x : hx, yi ∈ Z for all y ∈ Λ} .

We can choose w ∈ − int K◦∩Λ∗. We choose it primitive, that is, such that w is a basis of Λ∗∩span(w). ∗ We can extend w to a basis of Λ . Let {w1 = w, w2,..., wN } be such basis. Then consider the basis {u1,..., uN } deﬁned by ( 1 if i + j = N + 1, hu , w i = i j 0 otherwise. Now we can deﬁne the hyperplanes

N Hk = {x ∈ R : hx, wi = k} for k ∈ Z, and a lattice Λk in each hyperplane

( N−1 ) X Λk = Λ ∩ Hk = kud + miui : mi ∈ Z . i=1

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As we argued in the proof of Theorem 2.1.2, the restricted polyhedra Pk = P ∩ Hk are bounded and there exists k0 such that Pk = ∅ for k < k0. Thus we can write

∞   X hc,ui X X hc,ui e =  e  (2.2.2) u∈P ∩Λ k=k0 u∈Pk∩Λ If P is bounded the result is clear, so we may assume P is unbounded. As in the proof of Theorem 2.1.2, in the previous sum, for sufficiently large k, all the vertices of Pk correspond to intersections with unbounded edges of P . This is the part that may diverge. To prove that it indeed converges, first we write the vertices of Pk for sufficiently large k as

vi(k) = ai + kbi, for some ai, bi Λ-rational vectors.

Being Λ-rational implies that there exists M ∈ Z such that Mbi ∈ Λ for all i, making vi(k + M) a lattice translation of vi(k). And thus, for suﬃciently large k, we can write the sum in Eq. (2.2.2) as

∞   X X hc,ui  e  . (2.2.3) j=0 u∈Pk+jM ∩Λ

Now applying the induction hypothesis in each polytope Pk+jM and Brion’s theorem we have

X hc,ui X e = F([tcone(Pk+jM , vi(k + jM))]). u∈Pk+jM ∩Λ i∈I By the translation property (see Remark 10) we have

hc,jzii F([tcone(Pk+jM , vi(k+jM))]) = e Fi(c), for some zi ∈ Λ, where Fi = F([tcone(Pk, vi(k))]), so ﬁnally we can write Eq. (2.2.3) as !   ∞ ∞ F (c) X X hc,jzii X X hc,jzii X i e Fi(c) =  Fi e  = . hc,zii j=0 i∈I i∈I j=0 i∈I 1 − e As in the proof of Theorem 2.1.2, we now need to prove that the correspondence that we have deﬁned respects linear relations. The proof is analogous as the one in Theorem 2.1.2.

Finally, to prove part (2) of the statement, we look again at the proof of Theorem 2.1.2, and argue analogously to get that if P contains a line, then for some vector u ∈ Λ \{0} we have F([P ]) = ehc,uiF([P ]), and thus F([P ]) = 0.

N Deﬁnition 2.2.1 (Unimodular cone). A cone K ⊆ R is unimodular with respect to a lattice Λ if it can be written as K = co(u1,..., uN ) with {u1,..., uN } being a basis of Λ.

An equivalent deﬁnition of unimodular cone is that the fundamental parallelepiped of K associated to {u1,..., uN }, that is ( N ) X Π = αiui : 0 ≤ αi < 1 , i=1 contains only one point of Λ.

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Remark 11. We want to point out two important properties of this cones.

1. If K is unimodular, then F(K) has a simple expression as in Eq. (2.2.1), because the only point in Π ∩ Λ is the origin with n = 0.

2. We can always decompose a lattice cone into unimodular cones. This is a non-trivial result that we will need in Section 2.4. Chapter 16 of [1] is dedicated to it.

2.3 Generalizing Ehrhart’s theorem

We do a little stop in our way on the land of valuations to present our second proof of Ehrhart’s theorem. In fact, we will present a more general result, that specializes to Ehrhart’s in a concrete case.

When we make dilations of a polytope, the vertices move, and so do the tangent cones, but the cones of feasible directions stay the same.

So now we ask ourselves what happens to the number of integer points in a polytope when we move the vertices in a way that the feasible cones are not modiﬁed. It turns out that the behavior is still polynomial, but as we can expect, a more complex kind of polynomial. We will follow mainly [1, Ch. 18].

We will also prove the result for open polytopes, using the results deduced in Section 1.3.

Theorem 2.3.1. Let {Pα : α ∈ Z} be a family of integer N-dimensional polytopes with n N vertices v1(α),..., vn(α) ∈ R , such that, for each family of vertices {vi(α)}α, the cone of feasible nN directions fcone(Pα, vi(α)) does not depend on α. Then there exists a polynomial p : R → R such that

N 1. Pα ∩ Z = p v1(α),..., vn(α) .

N N 2. int Pα ∩ Z = (−1) p − v1(α),..., −vn(α) . 3. p(0,..., 0) = 1.

PROOF.

Pn 1. Recall that by Theorem 1.3.8, we have that, modulo polyhedra with lines, [Pα] = i=1[tcone(Pα, vi(α))], Pn so by the deﬁnition of the counting valuation (see Theorem 2.2.2), F([Pα]) = i=1 F([tcone(Pα, vi(α))]). We know that the cone of feasible directions is not dependent on α, so we have the following decomposition tcone(Pα, vi(α)) = vi(α) + Ki, (2.3.1)

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where Ki is the cone of feasible directions. This translates into the counting valuation as n X hc,vi(α)i F([Pα]) = e F (Ki, c). (2.3.2) i=1 To actually count with the counting valuation, we have to evaluate at c = 0, so to do this, let N us ﬁx a generic c ∈ R and let us deﬁne for every vertex, the function fi : R → R as:

fi(τ) = F (Ki, τc), τ ∈ R.

By the space to which the counting valuation maps (see Theorem 2.2.2), each fi admits a Laurent expansion in a neighborhood of τ = 0 of the form ∞ X j fi(τ) = ai,jτ , j=−N

for some coeﬃcients ai,j ∈ R. We also have an expansion for the exponential around τ = 0: ∞ j X hc, vi(α)i ehτc,vi(α)i = τ j. j! j=0 With all this setup, number of integer points we want to count is the constant term in τ of     n ∞ j ∞ X X hc, vi(α)i j X j  τ   ai,jτ  , j! i=1 j=0 j=−N

which turns out to be a polynomial on vi(α)’s:

n N hc, v (α)ij P ∩ N = X X i a . (2.3.3) α Z j! i,−j i=1 j=1

2. By Corollary 1.3.12 we know that, modulo polyhedra with lines, n X [rel int Pα] ≡ [rel int tcone(Pα, vi(α))]. i=1 Using Eq. (2.3.1) we get an analogous result as Eq. (2.3.2). Now we apply that [rel int K] ≡ (−1)d[−K] modulo polyhedra with lines (see Lemma 1.3.11) and get that

n N X hc,vi(α) F (rel int Pα, c) = (−1) e F (Ki, −c). i=1 Using an analogous argument as in (1) we get the desired result. 3. Recall Brion’s theorem for the cones of feasible directions: n X [Ki] ≡ [0] modulo polyhedra with lines. i=1 Pn This together with Eqs. (1.1.8) and (2.3.1) implies that i=1 F (Ki, c) = 1, therefore for the coeﬃcients ai,j for all j = 0 we have a1,0 + ··· + an,0 = 1. Looking at Eq. (2.3.3), we check that the constant term corresponds precisely to a1,0 + ··· + an,0.

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The situation in which we study dilations of a polytope fits clearly in the hypothesis of the previous theorem. Take an original polytope P = conv(v1, . . . , vn), and move the vertices as vi(α) = α, vi, for α ∈ N. In this case, Ehrhart polynomial is, as in Eq. (2.3.3) n N hc, v ij p(α) = X X i a αj. j! i,−j i=1 j=1 With this proof, however, it seems much more difficult to prove that this polynomial has indeed degree N, without having to compute the coefficients ai,−N .

2.4 Berline-Vergne valuation

Our goal in this section is to use the exponential (Φ) and counting (F) valuations that we have previously deﬁned to construct a valuation that counts the number of integer points in a polyhedron, somehow decomposing the contribution of each face.

This result was ﬁrst proved by N. Berline and M. Vergne in [3]. We will follow the exposition on [1, Ch. 19]. N N Notation. Given a linear subspace L ⊆ R , we will denote R /L to the orthogonal complement of L N N N in R . In general, given a set S ⊆ R and a subspace L ⊆ R , we will denote S/L to the orthogonal N projection of S into R /L.

Note that, in particular, if Λ is a lattice and L a subspace, Λ0 = Λ/L is also a lattice, typically ﬁner N than Λ ∩ (L/R ). See Fig. 2.1.

RN /L

Figure 2.1: A lattice subspace L and the corresponding lattice Λ/L compared to the coarser lattice N Λ ∩ (L/R ).

We want to deﬁne ψ such that the following identity holds for a given rational polytope P : X F([P ]) = Ψ([P/L])ΦL([P ∩ L]), L⊆RN : lattice subspace

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where F is the counting valuation and ΦL is the exponential valuation with a measure dxL such that det(L ∩ Λ) = 1. This way we may be able to decompose the sum into subspaces L of each dimension. Since F encodes the number of integer points and Φ encodes the volume, that behaves as tdim L when dilating the polytope P , this seems a reasonable approach.

We will be able to define Ψ recursively from a formula similar to the one we have just written, because there is only one zero-dimensional subspace, and we will define that Ψ([A]) = 1 when dim A = 0. This way, we have it for dimension zero, and we will be able to define for dimension n using only the definition of smaller dimensions. It will become more clear in the proof of Theorem 2.4.2.

We will see that we cannot construct such valuation Ψ, but we will be able to construct a valuation in shifted cones around each vertex of P , and the use Brion’s theorem to get a formula for the whole polytope.

N Deﬁnition 2.4.1 (Algebra of v-shifted rational cones). Given a rational point v ∈ Q , we deﬁne the algebra of v-shifted rational cones as

Av = span {[K + v]: K rational cone} .

We will prove that there exists a valuation in Av for any v. First, we will prove that only the most essential identities need to be checked.

N N Lemma 2.4.1. Let v ∈ Q , W be a vector space. Let K+v = {K +v ⊆ R : K rational cone} N and θ : K → W . Given H ⊆ R be a hyperplane containing v, H+, H− the corresponding closed half-spaces and deﬁne A+ = A ∩ H+,A− = A ∩ H−,A0 = A ∩ H. If the identity

θ(A) = θ(A+) + θ(A−) − θ(A0) (2.4.1)

holds for every hyperplane H, then there exists a valuation Θ: Av → W that extends θ, that is, Θ([K + v]) = θ(K + v) for all K + v ∈ K + v.

PROOF. (Sketch only) Without loss of generality, we may assume that v = 0. We have to prove that given a collection of P cones Ki ∈ K and numbers αi ∈ R satisfying i∈I αi[Ki] = 0, we have that X αiθ(Ki) = 0. (2.4.2) i∈I

For every cone Ki, consider the supporting hyperplanes of its facets. Let Ji be a set of indices that indexes the facets of Ki, and Hi,j be the supporting hyperplane of the corresponding facet of Ki. Let Hi be the set containing such hyperplanes, so [ Hi = {Hij : j ∈ Ji}, and H = Hi. i∈I

N Considering all hyperplanes in H, they decompose R into ﬁnitely many polyhedral cones. Let C = {Cj : j ∈ J} be the set of all such cones and their non-empty faces.

First we observe that arguing about the dimensions of the diﬀerent cones, we have that the indicators {[Cj]: j ∈ J} are linearly independent.

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Next, observe that we can write, for some coeﬃcients βi,j: X X [Ki] = βi,j[Cj], and thus θ(Ki) = βi,jθ(Cj) (2.4.3) j∈J j∈J by iterating the identities

[K] = [K ∩ H+] + [K ∩ H−] − [K ∩ H], and θ(K) = θ(K ∩ H+) + θ(K ∩ H−) − θ(K ∩ H) for hyperplanes H ∈ H and cones K = Ki. P Using the fat that Cj’s are independent and Eq. (2.4.3) we get that for all j ∈ J, the sum i∈I αiβi,j = 0, and we apply it to prove Eq. (2.4.2).

We will deﬁne Ψ locally at every point v. Since we will be using the exponential and counting N N valuation, we need to pick a vector space that contains the functions in N (R ) and M(R ) in Sections 2.1 and 2.2. The easy choice is the space of meromorphic functions, that we will denote N Q(R ).

N N Theorem 2.4.2 (Berline-Vergne, 2007). Let v ∈ Q , Λ ∈ R a lattice there exists a valuation

N Ψ: Av → Q(R )

that satisﬁes, for every shifted rational cone A = K + v, the following properties:

1. The formula X F([A]) = Ψ([A/L])ΦL([A ∩ (L + v)]) (2.4.4) L N is satisﬁed, where the sum is taken over all subspaces L ⊆ R parallel to faces of A. 2. If A contains a line, then Ψ([A]) = 0.

N 3. The function Ψ([A]) ∈ Q(R ) is analytic at c = 0.

PROOF. (Sketch only) The proof is by induction on N. For N = 0 the result is clear, so we may assume N > 0. Since there is only one subspace of dimension zero, we can isolate Ψ([A]) in Eq. (2.4.4):

−hc,vi −hc,vi X Ψ([A]) = e F([A]) − e Ψ([A/L])ΦL([A ∩ (L + v)]). (2.4.5) L : dim L>0 We observe that if K contains a line, Ψ([A]) = 0, because both the counting and the exponential valuation vanish when K contains a line. This proves (2).

Next, we observe that we can allow the sum in Eq. (2.4.4) to range over all lattice subspaces of L of positive dimension. This is because

• If L lies in a face F of K with dim F > L, then A/L is a translation of a lower-dimensional cone K/L containing a line, thus Ψ([A/L]) = 0.

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• If dim(L ∩ K) < dim L, then by dimensions ΦL([A ∩ (L + v)]) = 0.

In conclusion, the only contributions correspond to spaces parallel to faces of P . Also, note that in such case, if L is parallel to a face F , we have A/L = tcone(A, F )/L.

Next we want to prove that Ψ as deﬁned by Eq. (2.4.4) satisﬁes the hypothesis of Lemma 2.4.1. Let H be a hyperplane through v and H+, H−, A+, A− and A0 as in Lemma 2.4.1. We want to prove that Ψ([A]) = Ψ([A+]) + Ψ(A−) − Ψ([A0]) (2.4.6) We already know that F([A]) = F([A+]) + F(A−) − F([A0]) and

ΦL([A ∩ (L + v)]) = ΦL([A+ ∩ (L + v)]) + ΦL(A− ∩ (L + v)) − ΦL([A0 ∩ (L + v)]) because F and ΦL are valuations. Also, since projections preserve linear relations among indicators of polyhedra, we have [A/L] = [A+/L] + [A−/L] − [A0/L]. We can thus apply the induction hypothesis to get

Ψ([A/L]) = Ψ([A+/L]) + Ψ(A−/L) − Ψ([A0/L]). (2.4.7) By Eq. (2.4.4), we need to check that, for every subspace L, the term

Ξ(A, L) = Ψ([A/L])ΦL([A ∩ (L + v)]) satisﬁes the relation Ξ(A, L) = Ξ(A+,L) + Ξ(A−,L) − Ξ(A0,L). (2.4.8) This is veriﬁed on a case-by-case basis. For a non-empty face F of A, either

(i) F lies in H, or

(ii) F lies in one of H+ or H− but not in both, or (iii) H passes through an interior point of F and is transversal to the span of F .

We note that every face A+ or A− is either the intersection of a face of A with H+ (or H−), or the intersecton of a face of A with H. Also, every face of A0 is the intersection of a face of A with H. Thus, to deﬁne Ψ([A]), Ψ([A+]), Ψ([A−]) and Ψ([A0]) we can reduce it to the following ﬁve cases:

(a) Let L be a subspace parallel to F such that F ⊂ H. Then

A ∩ (L + v) = A+ ∩ (L + v) = A− ∩ (L + v) = A0 ∩ (L + v), and by Eq. (2.4.7), Eq. (2.4.6) holds.

(b) Let L be a subspace parallel to F such that F lies in H+ but not in H−. Then A/L = A+/L and dim(A− ∩ (L + v)), dim(A0 ∩ (L + v)) < dim L, thus we have

Ξ(A, L) = Ξ(A+,L), and Ξ(A−,L) = Ξ(A0,L) = 0, so clearly Eq. (2.4.8) holds.

48 of 62 2.4. BERLINE-VERGNE VALUATION CHAPTER 2. LOCAL FORMULA

(c) Let L be a subspace parallel to F such that F lies in H− but not in H+. Then by an analogous argument as in (b) Eq. (2.4.8) holds. (d) Let L be a subspace parallel to F such that H passes through an interior point of F and intersects the span of F transversally. Then A/L = A+/L = A−/L and dim(A0 ∩ (L + v)) < dim L. Therefore Ψ([A/L]) = Ψ([A+/L]) = Ψ([A−/L]), and ΦL([A0 ∩ (L + v)]) = 0, so Eq. (2.4.8) holds.

(e) Let L be a subspace parallel to a face G = F ∩ H of A0 where F is a face of A, H passes through an interior point of F and intersects the aﬃne span of F transversally. Then

A ∩ (L + v) = A+ ∩ (L + v) = A− ∩ (L + v) = A0 ∩ (L + v), and therefore

ΦL([A ∩ (L + v)]) = ΦL([A+ ∩ (L + v)]) = ΦL([A− ∩ (L + v)]) = ΦL([A0 ∩ (L + v)]), concluding that Eq. (2.4.8) holds.

This concludes part (1) of the proof.

To prove part (3), we do it again by induction on N. By part (2) of Remark 11, it suﬃces to prove that ψ(A, c) = Ψ([A]) is analytic at c = 0 for A a full dimensional unimodular cone. Hence, we may suppose that A = K + v, with K = co(u1, . . . , uN ) and {u1,..., uN } constitute a basis of Λ. Since the cone is unimodular (recall part (1) of Remark 11) we can write

N 1 F([A]) = ehc,wi Y , hc,uii i=1 1 − e where w ∈ Λ is deﬁned as N N X X v = αiui, then w = dαieui. i=1 i=1 Also, for L parallel to faces of K such that

L = LI = span(ui : i ∈ I) where I ⊂ {1,...,N}, by Lemma 2.1.1 we have −1 Φ ([A ∩ (L + v)]) = ehc,vi Y . L hc, u i i∈I i By the induction hypothesis, all functions Ψ([A/L]) are analytic at c = 0. Therefore, the only poles that Ψ([A]) may have are located in the hyperplanes hc, uii = 0. So we have just to prove that for all i, the function hc, uiiΨ([A]) is identically zero on the hyperplane hc, uii = 0.

Note that by Taylor expansion, we have

hc, uii = −1, on the hyperplane hc, uii = 0. (2.4.9) 1 − ehc,uii

Therefore, on the hyperplane hc, uN i we have

N−1 hc,w−v Y 1 X Y −1 hc, uN iΨ([A]) = −e + Ψ([A/LI ]) . (2.4.10) 1 − ehc,uii hc, u i i=1 I⊂[N],I3N i∈I\{N} i

49 of 62 2.5. EHRHART’S THEOREM: THIRD PROOF CHAPTER 2. LOCAL FORMULA

0 0 Now we want to apply the induction hypothesis. Therefore we consider A = A/ span(uN ), that is A = 0 0 0 0 0 0 0 0 v +co(u1,..., uN−1), where v = v/ span(uN ) and ui = ui/ span(uN ). Let K = co(u1,..., uN−1). It 0 0 0 is a (d−1)-dimensional cone with respect to the lattice Λ = Λ/ span(uN ), generated by {u1,..., uN−1}. 0 0 We have hc, uii = hc, uii, so with w = w/ span(uN ) we can rewrite Eq. (2.4.10) as

N−1 0 hc,w0−v0 Y 1 X 0 Y −1 Ψ([A ]) − e 0 + Ψ([A /LI ]) 0 , hc,uii hc, u i i=1 1 − e I⊂[N−1],I6=∅ i∈I i which in terms of the original valuations is identically zero, since

0 hc,v0i 0 −hc,v0i X 0 0 Ψ([A ]) − e F([A ]) + e Ψ([A /L])ΦL([A ∩ (L + v)]) = 0. L : dim L>0

2.5 Ehrhart’s theorem: third proof

In this section we present our third and last proof of Ehrhart’s theorem (Theorem 1.2.1), using the Berline-Vergne valuation Ψ. In fact, we have already done all the work, we just need to interpret Theorem 2.4.2 in the correct way.

Since the valuation Ψ is analytic at c = 0, we can evaluate at c = 0. Note that for a given face δ F in L = span F , of dimension δ we have φL(tF, c = 0) = nvol(tF ) = t nvol(F ), and if we deﬁne α(P,F ) = ψ(tcone(P,F )/L, c = 0) we have

N X dim F tP ∩ Z = α(P,F ) nvol(F )t . (2.5.1) F : face of P

50 of 62 Chapter 3

Ehrhart positivity

3.1 Known results

Understanding the coeﬃcients of the Ehrhart polynomials is quite an open problem. We have proved in Section 2.3 that the constant coeﬃcient is 1.

d The leading term can also be computed easily. If we consider a d-dimensional polytope P ∈ R , its R volume is the integral vol P = P dx. We can approximate this integral with Riemannian sums. In 1 d particular, considering the lattices Λt = ( t Z ) for t ∈ N. The corresponding Riemannian sum in each is |Λt ∩ P |, and their ﬁneness increases arbitrarily with t, so in the limit we have 1 d 1 d vol P = lim Z ∩ P = lim Z ∩ P . t→∞ t t→∞ td And therefore, the leading term is vol P .

The second leading coeﬃcient can also be interpreted as counting something. In this case, the coeﬃ- cient equals one half of the sum of the normalized volumes of the facets of P . To prove it, consider Eq. (2.5.1) for faces of dimension dim P − 1. Then the second leading term of the Ehrhart polynomial is X α(P,F ) nvol(F ). F facet In this case, tcone(P,F )/L is 1-dimensional, so α(P,F ) = 1/2 by Eq. (3.2.4).

A more elementary proof, that is, without the need of constructing Ψ, can be found in [2, Sec. 5.3].

One natural question to ask is whether the rest of the coeﬃcients count something, and the ﬁrst step to an answer is to ask if they are at least positive.

d Deﬁnition 3.1.1 (Ehrhart positive). Let P be an integral polytope and p(t) = adt + ··· + a1t + 1 its Ehrhart polynomial. We say that P is Ehrhart positive if ai > 0 for all i.

51 3.2. COMPUTING Ψ CHAPTER 3. EHRHART POSITIVITY

The answer in general is no. Since we know the constant and the two leading coeﬃcients of the polynomial are positive, we need at least a polynomial with four coeﬃcients, which means a 3-dimensional polytope.

The ﬁrst known example was discovered by John Reeve in [10]. 3 Example 4 (Reeve’s tethrahedra). For m ∈ N, we deﬁne the m-th Reeve tetrahedron in R as

Tm = conv(0, e1, e2, e1 + e2 + me3).

3 2 Let pm(t) = a3t + a2t + a1t + 1 be its Ehrhart polynomial. We can check that the only integer points in Tm are its vertices, so we have for all m that pm(1) = 4. The volume is one sixth of the m corresponding cube, so a3 = 6 . Each facet has normalized volume 1/2, thus a2 = 1. Now using m 12−m pm(1) = 4, we get 6 + 1 + a1 + 1 = 4, and we can isolate a1 = 6 . Putting all the coeﬃcients together we have m 12 − m p (t) = t3 + t2 + t + 1. m 6 6

Clearly, a1 < 0 for m > 12, and in fact it can be arbitrarily small for t arbitrarily large.

m = 2 m = 3 m = 4 m = 5

Figure 3.1: Reeve’s tetrahedra for diﬀerent values of m.

There are many families of polyhedra that are known to be Ehrhart positive, and others that are known to have some negative coeﬃcients. We refer to [8] for a survey on what is currently known about Ehrhart positivity.

3.2 Computing the Berline-Vergne valuation

In this section we will compute the valuation Ψ for unimodular low-dimensional cones. In particular, we will compute up to 3-dimensional cones.

These computations serve two purposes. First, they will give an idea on how Ψ looks like and how diﬃcult it is to compute. Secondly, they motivate the results presented in Section 3.3, where we expose the work of Castillo and Liu in [5], conjecturing that generalized permutohedra are Ehrhart positive.

52 of 62 3.2. COMPUTING Ψ CHAPTER 3. EHRHART POSITIVITY

To make the notation more understandable, we will use a subscript to denote the dimension of the cone in which we are computing ψ. So, for example, if K is a two dimensional cone, we will denote ψ(K) as ψ2(K). Also, in this section we will make an exception to our general rule of writing vectors in boldface, because there appear a lot of vectors, and we ﬁnd more useful to use boldface characters to point out the most essential parts of our formulas.

3.2.1 General formulation

The formula that deﬁnes ψ in Theorem 2.4.2 for a unimodular cone K = co(u1, . . . , un) reads

n k 1 X X (−1) ψn(K, c) = n − ψn−k(K/uI ) (3.2.1) Q (1 − ehc,uii) Q hc, u i i=1 k=1 [n] i∈I i I∈( k )

⊥ where uI = {ui}i∈I , and K/uI is the orthogonal projection of K onto span(uI ) .

In general, this projection will also be a unimodular cone. The projected lattice will be generated by the orthogonal projection of the rest of vectors ui that are not in uI . Lemma 3.2.1. This projection is a linear mapping, with matrix

Id −A(AT A)−1AT (3.2.2)

where A is the matrix that contains vectors in uI as columns.

PROOF. Given a linear subspace F , any vector can be uniquely decomposed as x = xF + xF ⊥ , where xF ∈ F ⊥ T −1 T and xF ⊥ ∈ F . Here, F is the colspan of A and the matrix Π = A(A A) A is the orthogonal T −1 T projection of x into F , that is, xF = Πx. Then xF ⊥ = x − xF = x − A(A A) A x.

3.2.1.1 On the limit c → 0

We know that ψ is analytic at c = 0, and in fact, we are interested in the limit at said point.

To compute ψn recursively, we can see in Eq. (3.2.1) that we need the coeﬃcient of degree k of ψn−k. So, it may be interesting to compute ψn(K, c) as a power series and try to deduce some behavior from it.

3.2.2 Dimension 1 and 2

For dimension 1, Eq. (3.2.1) reads 1 1 ψ (K) = + . 1 hc,u i 1 − e 1 hc, u1i

53 of 62 3.2. COMPUTING Ψ CHAPTER 3. EHRHART POSITIVITY

1 1 We can see this as a function of τ = hc, u1i, ψ(τ) = 1−eτ + τ , and is very similar to the generating function of Bernoulli numbers: 1 τ ∞ B τ m = X m eτ − 1 m! m=0 so we know that 1 1 ∞ B τ m ψ (τ) = − X m . (3.2.3) 1 τ τ m! m=0

The ﬁrst Bernoulli numbers are B0 = 1 and B1 = −1/2, B2 = 1/6, B3 = 0, B4 = −1/30 and B5 = 0, so the ﬁrst terms of ψ1 are 1 1 1 ψ (τ) = − τ + τ 3 + o(τ 4). (3.2.4) 1 2 12 30

In particular, we know that ψ1(τ = 0) = 1/2.

For dimension 2, Eq. (3.2.1) reads for K = co(u1, u2) 1 1 1 1 ψ (K) = + ψ (co(v )) + ψ (co(v )) − , (3.2.5) 2 hc,u i hc,u i 1 1 1 2 (1 − e 1 )(1 − e 2 ) hc, u1i hc, u1i hc, u1ihc, u2i where v is the orthogonal projection of v onto hv i⊥, explicitly v = u − hu1,u2i u and analogous for 1 2 1 1 2 hu1,u1i 1 v2.

As before, we want to see this as a function of τ1 = hc, u1i and τ2 = hc, u2i. But ﬁrst we need to deal with ψ1, because we have them as a function of hc, vii, and not hc, uii.

3.2.2.1 Computation of ψ1

We deﬁne an auxiliary variable t = hc, v1i. We know how to express ψ1(co(v1)) in therms of t as in Eq. (3.2.3). Since we know that ψ is analytical at c = 0, when we express it as a power series, we know that coeﬃcients of negative degree will vanish, so we will not care of them. In some of the equations, negative degree terms will be ignored, we will use the notation ψ =0 φ, whenever the nonnegative degree terms of ψ and φ are equal.

For ψ1(t) we have: 1 ∞ B tm ∞ B tm ψ (t) =0 − X m =0 − X m+1 . 1 τ m! (m + 1)! m=0 m=0 Now using t = τ − hu1,u2i τ we get 2 hu1,u1i 1 ! m hu , u ik m tm = X(−1)k 1 2 τ kτ m−k. hu , u i k 1 2 k=0 1 1

In Eq. (3.2.5) we need 1 ψ (t). Mixing the two previous equations yields: τ1 1

1 For the Bernoulli numbers, we will be using the convention that B1 = −1/2.

54 of 62 3.2. COMPUTING Ψ CHAPTER 3. EHRHART POSITIVITY

! 1 B m hu , u ik+1 m + 1 ψ (co(v )) =0 X m+2 X(−1)k 1 2 τ kτ m−k. (3.2.6) τ 1 1 (m + 2)! hu , u i k + 1 1 2 1 m≥0 k=0 1 1

And analogously ! 1 B m hu , u im+1−k m + 1 ψ (co(v )) =0 X m+2 X(−1)m−k 1 2 τ kτ m−k. (3.2.7) τ 1 2 (m + 2)! hu , u i k 1 2 2 m≥0 k=0 2 2

With this notation, we can easily compute the sums in Eq. (3.2.5):

1 1 0 ψ1(co(v1)) + ψ1(co(v2)) = τ1 τ2 " ! !# B m hu , u ik+1 m + 1 hu , u im+1−k m + 1 X m+2 X (−1)k 1 2 + (−1)m−k 1 2 τ kτ m−k (m + 2)! hu , u i k + 1 hu , u i k 1 2 m≥0 k=0 1 1 2 2

Since all terms are multiplied by a Bernoulli number Bm+2, and for m odd they are zero, we may assume m even, and therefore (−1)k = (−1)m−k. With this and playing with the binomial coeﬃcients we get

1 1 0 ψ1(co(v1)) + ψ1(co(v2)) = τ1 τ2 !" # B m (−1)k m + 1 hu , u ik+1 hu , u im+1−k X m+2 X (m + 1 − k) 1 2 + (k + 1) 1 2 τ kτ m−k. (m + 2)! k + 1 k hu , u i hu , u i 1 2 m≥0 k=0 1 1 2 2

Finally, Eq. (3.2.5) reads:

ψ2(co(u1, u2) = ! 1 m m + 2 X X B B τ kτ m−k + (m + 2)! k + 1 k+1 m+1−k 1 2 m≥0 k=0 !" # B m (−1)k m + 1 hu , u ik+1 hu , u im+1−k X m+2 X (m + 1 − k) 1 2 + (k + 1) 1 2 τ kτ m−k. (m + 2)! k + 1 k hu , u i hu , u i 1 2 m≥0 k=0 1 1 2 2

We can check this formula for ﬁrst degrees:

! 1 2 B2 hu1, u2i hu1, u2i 1 1 1 1 [deg0]ψ2 = B1B1 + ·1· + = + hu1, u2i + . (3.2.8) 2! 1 2! hu1, u1i hu2, u2i 4 12 hu1, u1i hu1, u2i

55 of 62 3.2. COMPUTING Ψ CHAPTER 3. EHRHART POSITIVITY

! ! ! 1 3 3 −1 [deg1]ψ = B B τ + B B τ = (τ + τ ). (3.2.9) 2 3! 1 1 2 2 2 2 1 1 24 1 2

[deg2]ψ2 = ! ! ! ! ( !" 3# 1 4 2 4 4 2 B4 1 3 hu1, u2i hu1, u2i 2 B1B3τ2 + B2B2τ1τ2 + B3B1τ1 + 3 + τ2 + 4! 1 2 3 4! 1 0 hu1, u1i hu2, u2i !" 3 # !" 2 2# ) 1 3 hu1, u2i hu1, u2i 2 1 3 hu1, u2i hu1, u2i + + 3 τ1 − 2 + 2 τ1τ2 3 2 hu1, u1i hu2, u2i 2 1 hu1, u1i hu2, u2i (" 3 # τ1τ2 1 hu1, u2i hu1, u2i 2 = − + 3 τ1 + 144 720 hu1, u1i hu2, u2i " 3# " 2 2# ) hu1, u2i hu1, u2i 2 hu1, u2i hu1, u2i 3 + τ2 − 3 + τ1τ2 . hu1, u1i hu2, u2i hu1, u1i hu2, u2i

Observe that for the first degree term, only the first term in the sum is computed, because for m odd, the second term yields directly zero. The first term will also vanish for m odd, except for m = 1, because Bernoulli numbers Bm for m odd vanish all except for B1.

3.2.3 Dimension 3

First of all, we need to compute the orthogonal projections as in Lemma 3.2.1.

⊥ ⊥ For the projection onto huii the following vectors vij are the projection of uj onto huii :

hu1, u2i hu2, u3i hu3, u1i v12 = u2 − u1, v23 = u3 − u2, v31 = u1 − u3, hu1, u1i hu2, u2i hu3, u3i

hu1, u3i hu2, u1i hu3, u2i v13 = u3 − u1, v21 = u1 − u2, v32 = u2 − u3. hu1, u1i hu2, u2i hu3, u3i ⊥ ⊥ For the projections onto huj, uki , the following vectors wi are the projection of ui onto huj, uki (see Section 3.2.3.1 to check calculations):

1 w = u − (hu , u ihu , u i − hu , u ihu , u i)·u +(hu , u ihu , u i − hu , u ihu , u i)·u , 1 1 T T T 2 3 3 2 1 2 3 3 1 2 2 2 3 2 2 3 2 1 3 u2 u2u3 u3 − (u2 u3) 1 w = u − (hu , u ihu , u i − hu , u ihu , u i)·u +(hu , u ihu , u i − hu , u ihu , u i)·u , 2 2 T T T 2 1 1 3 2 3 1 1 2 3 3 3 1 2 3 1 3 2 1 u3 u3u1 u1 − (u3 u1) 1 w = u − (hu , u ihu , u i − hu , u ihu , u i)·u +(hu , u ihu , u i − hu , u ihu , u i)·u . 3 3 T T T 2 2 2 1 3 1 2 2 3 1 1 1 2 3 1 2 1 3 2 u1 u1u2 u2 − (u1 u2)

With this notation, ψ3(co(u1, u2, u3)) can be easily computed as

56 of 62 3.2. COMPUTING Ψ CHAPTER 3. EHRHART POSITIVITY

ψ3(co(u1, u2, u3)) = (3.2.10) 1 (3.2.11) (1 − ehc,u1i)(1 − ehc,u2i)(1 − ehc,u3i) 1 1 1 + ψ2(co(v31, v32)) + ψ2(co(v12, v13)) ψ2(co(v23, v21)) (3.2.12) hc, u3i hc, u1i hc, u2i 1 1 1 − ψ1(co(w1)) − ψ1(co(w2)) − ψ1(co(w3)) (3.2.13) hc, u2ihc, u3i hc, u3ihc, u1i hc, u1ihc, u2i 1 + . (3.2.14) hc, u1ihc, u2ihc, u3i

As before, we want to consider ψ3 = ψ3(τ1, τ2, τ3), where τi = hc, uii.

1 0 P Bm+1 m For Eq. (3.2.11) we use the expansion 1−et = − m≥0 (m+1)! t , that in our 3-dimensional case becomes ! 1 0 X 1 X m + 3 m1 m2 m3 = − Bm1+1Bm2+1Bm3+1τ1 τ2 τ3 . (1 − eτ1 )(1 − eτ2 )(1 − eτ3 ) (m + 3)! m + 1, m + 1, m + 1 m≥0 m : |m|=m 1 2 3 (3.2.15)

Before going to the full expansion of ψ3, observe that we can directly compute the constant term, which is the one we are most interested, and the one we can check with [5].

! 1 3 13 1 [deg0](Eq. (3.2.11)) = − − = . (3.2.16) 3! 1, 1, 1 2 8

Since ψ1 has vanishing degree 2 term,

[deg2](Eq. (3.2.13)) = 0.

So we are only left with [deg1](Eq. (3.2.12)), for which we use our computations in Eq. (3.2.9). Let us deﬁne the auxiliary variables σij := hc, viji. From deﬁnition of vij we can see that for all i, j holds hui,uj i σij = τj − τi, so we have: hui,uii 1 σ + σ σ + σ σ + σ [deg1](Eq. (3.2.12)) = − 12 13 + 21 23 + 31 32 (3.2.17) 24 τ1 τ2 τ3   τ − hu1,u2i τ + τ − hu1,u3i τ τ − hu2,u1i τ + τ − hu2,u3i τ τ − hu3,u1i τ + τ − hu3,u2i τ 1 2 hu1,u1i 1 3 hu1,u1i 1 1 hu2,u2i 2 3 hu2,u2i 2 1 hu3,u3i 3 2 hu3,u3i 3 = −  + +  24 τ1 τ2 τ3 (3.2.18) 1 hu , u i hu , u i hu , u i hu , u i hu , u i hu , u i τ + τ τ + τ τ + τ = 1 2 + 1 3 + 2 1 + 2 3 + 3 1 + 3 2 − 2 3 + 1 3 + 1 2 24 hu1, u1i hu1, u1i hu2, u2i hu2, u2i hu3, u3i hu3, u3i τ1 τ2 τ3 (3.2.19) 1 hu , u i hu , u i hu , u i hu , u i hu , u i hu , u i =0 1 2 + 1 3 + 2 1 + 2 3 + 3 1 + 3 2 . (3.2.20) 24 hu1, u1i hu1, u1i hu2, u2i hu2, u2i hu3, u3i hu3, u3i

57 of 62 3.3. GENERALIZED PERMUTOHEDRON CHAPTER 3. EHRHART POSITIVITY

3.2.3.1 Computation of wi’s

We will compute w3, and the rest are analogous. We will denote column vectors as ui, and row T T T u1 vectors as ui . With this notation, A = [u1 u2] and A = T . The scalar product is written u2 hu, vi = uT v = vT u. So: " # " # " # uT uT u uT u 1 uT u −uT u AT A = 1 [u u ] = 1 1 1 2 =⇒ (AT A)−1 = 2 2 1 2 . T 1 2 T T T T T 2 T T u2 u2 u1 u2 u2 u1 u1u2 u2 − (u1 u2) −u2 u1 u1 u1

To write less, we will denote the denominator det(A) as ζ.

" T T #" T # " T T # T −1 T 1 u2 u2 −u1 u2 u1 1 hu2, u2iu1 − hu1, u2iu2 (A A) A = T T T = T T . ζ −u2 u1 u1 u1 u2 ζ hu1, u1iu2 − hu1, u2iu1

" T T # T −1 T 1 hu2, u2iu1 − hu1, u2iu2 A(A A) A = [u1 u2] T T ζ hu1, u1iu2 − hu1, u2iu1 1 = hu , u iu uT − hu , u iu uT + hu , u iu uT − hu , u iu uT . ζ 2 2 1 1 1 2 1 2 1 1 2 2 1 2 2 1

1 A(AT A)−1AT ·u = (hu , u ihu , u i − hu , u ihu , u i)·u +(hu , u ihu , u i − hu , u ihu , u i)·u 3 ζ 2 2 1 3 1 2 2 3 1 1 1 2 3 1 2 1 3 2

3.3 Generalized permutohedron

In this section, we study Ehrhart positivity of a speciﬁc family of polyhedra, namely generalized permutohedra. We will mainly follow [5].

Deﬁnition 3.3.1 (Permutohedron). Let v = (v1, . . . , vd+1) with all its coordinates diﬀerent, the regular permutohedron in v is the polytope

d n o Πv = conv vσ(1), . . . , vσ(d+1) : σ ∈ Sd+1 .

In the case v = (1, 2, . . . , d) we call it the usual permutohedron. A generalized permutohedron is any polytope that can be obtained from a regular permutohedron by moving the vertices while preserving edge directions. Some faces may collapse into lower dimensional faces.

58 of 62 3.3. GENERALIZED PERMUTOHEDRON CHAPTER 3. EHRHART POSITIVITY

(b) Generalized permutohedron. (c) A square pyramid as a gen- One face has collapsed into a ver- eralized permutohedron. A lot of (a) Regular permutohedron tex. faces have collapsed.

Figure 3.2: It is well known that the 3-dimensional permutohedron is a truncated octahedron. In these ﬁgures we show a regular 3-permutohedron and two generalized permutohedra obtained from it.

Since we have in Eq. (2.5.1) a decomposition of the Ehrhart polynomial of a polytope in terms of the values α(P,F ), we can deﬁne the concept of α-positivity, that reﬁnes Ehrhart positivity, as having α(P,F ) > 0 for all face F .

In [5, Th. 1.4] Castillo and Liu prove that regular permutohedra being α-positive implies that generalized permutohedra being α-positive, and thus Ehrhart positive and conjecture that indeed regular permutohedra are α-positive. They prove it for permutohedra up to dimension 6, using the formulas for ψ1, ψ2 and ψ3 derived in the previous section. The problem they encounter is that the computations leading to ψn for larger values of n are too complicated. In fact, they claim that ψ4 has way more than 1000 terms. An important part of this thesis has been understanding and reproducing these computations, and trying to simplify the case of ψ4, but unfortunately we were not able to obtain a manipulable formula.

In this last section, we sketch the main ideas on how the values for α(P,F ) can be computed for a regular permutohedron, following [5].

First of all, observe that the tangent cones of a regular permutohedron indeed unimodular, so we may use the formulas from the previous section. The result is independent of which vector v we use to N N deﬁne the permutohedron Π = Πv , as long as all its coordinates are diﬀerent.

In [5, Lem. 4.9] prove that the problem is reduced to computing, for faces of dimension N − l, for every subset S ⊆ [N], |S| = N − l, with [N] \ S = {i1 < ··· < il}, the value of ψ(co(Ri1 ,..., Ril )), where      1 1 −1 −1  Ri = 0,..., 0, ,..., , ,..., , 0,..., 0  . j  i − i i − i i − i i − i  | {z } j j−1 j j−1 j+1 j j+1 j | {z }  ij−1 | {z } | {z } N+1−ij+1 ij −ij−1 ij+1−ij

Since we know that Ehrhart positivity is always satisﬁed for l = 0, l = 1 and l = N, we can check for l = 2, 3, using the formulas for ψ2 and ψ3 of the previous section.

59 of 62 3.3. GENERALIZED PERMUTOHEDRON CHAPTER 3. EHRHART POSITIVITY

1. For [N] \ S = {i, j} with i < j, we have

1 1 1 1 1 1 −1 −1 R = ,..., , ,..., , 0,..., 0 , R = 0,..., 0, ,..., , ,..., . i i i i − j i − j j j − i j − i N + 1 − j N + 1 − j Thus 1 1 1 1 1 hR , R i = + , hR , R i = + , hR , R i = − , i i i j − i j j j − i N + 1 − j i j j − i

applying Eq. (3.2.8) we get

1 1 1 j − i (N + 1 − j)(j − i) 1 1 1 N + 1 − j − + = − + > 0. 4 12 j − i i N + 1 − i 4 12 i N + 1 − i

2. For [N] \ S = {i, j, k} with i < j < k, we have

1 1 1 1 R = ,..., , ,..., , 0,..., 0 , i i i i − j i − j 1 1 −1 −1 R = 0,..., 0, ,..., , ,..., , 0,..., 0 , j j − i j − i k − j k − j 1 1 −1 −1 R = 0,..., 0, ,..., , ,..., . k k − j k − j N + 1 − k N + 1 − k

Computing the scalar products hRα, Rβi as in the previous case and applying this time Eqs. (3.2.16) and (3.2.20) we get 1 1 i N + 1 − k − + 1 + > 0. 8 24 j N + 1 − j

60 of 62 Bibliography

[1] Barvinok, A. Integer Points in Polyhedra. European Mathematical Society, 2008.

[2] Beck, M., and Robins, S. Computing the Continuous Discretely. Undergraduate Texts in Mathematics. Springer New York, New York, NY, 2015.

[3] Berline, N., and Vergne, M. Local euler–maclaurin formula for polytopes. Moscow Mathe- matical Journal 7, 3 (2007), 355–386.

[4] Brion, M. Points entiers dans les polyedres convexes. In Annales scientiﬁques de l’Ecole Normale Superieure (1988), vol. 21, pp. 653–663.

[5] Castillo, F., and Liu, F. Berline–Vergne Valuation and Generalized Permutohedra. Discrete & Computational Geometry 60, 4 (Dec. 2018), 885–908.

[6] Ehrhart, E. Sur les polyèdresrationnels homothétiquesà n dimensions. C. R. Acad. Sci. Paris 254 (1962), 616–618.

[7] Grunbaum,¨ B., Klee, V., Perles, M. A., and Shephard, G. C. Convex polytopes.

[8] Liu, F. On Positivity of Ehrhart Polynomials. In Recent Trends in Algebraic Combinatorics, H. Barcelo, G. Karaali, and R. Orellana, Eds., Association for Women in Mathematics Series. Springer International Publishing, Cham, 2019, pp. 189–237.

[9] Pick, G. Geometrisches zur zahlenlehre. Sitzenber. Lotos (Prague) 19 (1899), 311–319.

[10] Reeve, J. E. On the volume of lattice polyhedra. Proceedings of the London Mathematical Society 3, 1 (1957), 378–395.

[11] Ziegler, G. M. Lectures on Polytopes. Springer Science & Business Media, May 2012.

61 Index

cone Berline-Vergne valuation, 47 feasible cone, 31, 34 counting valuation, 41 over a polytope, 9, 15 Euler characteristic, 28, 29 pointed cone, 9, 14 exponential valuation, 36 rational cone, 40 recession cone, 9, 15 shifted cone, 9 tangent cone, 31 fundamental parallelepiped, 22 lattice, 20 basis, 21 determinant of a lattice, 22 fundamental parallelepiped, 22

Minkowski sum, 10, 15, 31 permutohedron, 58 generalized, 58 regular, 58 usual, 58 polarity, 10, 17 polyhedron, 6 face of a polyhedron, 7 proper face, 8 relative interior, 8 open polyhedron, 34 rational polyhedron, 6 polytope, 5 simplex, 18 standard simplex, 6 theorem of Berline-Vergne, 47 Brion, 32 Ehrhart, 23, 43 Euler-Poincar´e,30 Weyl-Minkowski, 13 valuation, 28