Integer Points in Polyhedra

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Author: Alexander Barvinok

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Zurich Lectures in Advanced Mathematics Edited by Erwin Bolthausen (Managing Editor), Freddy Delbaen, Thomas Kappeler (Managing Editor), Christoph Schwab, Michael Struwe, Gisbert Wüstholz Mathematics in Zurich has a long and distinguished tradition, in which the writing of lecture notes volumes and research monographs play a prominent part. The Zurich Lectures in Advanced Mathematics series aims to make some of these publications better known to a wider audience. The series has three main constituents: lecture notes on advanced topics given by internationally renowned experts, graduate text books designed for the joint graduate program in Mathematics of the ETH and the University of Zurich, as well as contributions from researchers in residence at the mathematics research institute, FIM-ETH. Moderately priced, concise and lively in style, the volumes of this series will appeal to researchers and students alike, who seek an informed introduction to important areas of current research. Previously published in this series: Yakov B. Pesin, Lectures on partial hyperbolicity and stable ergodicity Sun-Yung Alice Chang, Non-linear Elliptic Equations in Conformal Geometry Sergei B. Kuksin, Randomly forced nonlinear PDEs and statistical hydrodynamics in 2 space dimensions Pavel Etingof, Calogero-Moser systems and representation theory Guus Balkema and Paul Embrechts, High Risk Scenarios and Extremes – A geometric approach Demetrios Christodoulou, Mathematical Problems of General Relativity I Camillo De Lellis, Rectifiable Sets, Densities and Tangent Measures Michael Farber, Invitation to Topological Robotics Published with the support of the Huber-Kudlich-Stiftung, Zürich

Alexander Barvinok

Integer Points in Polyhedra

Author: Prof. Alexander Barvinok Department of Mathematics, East Hall 530 Church Street University of Michigan Ann Arbor, MI 48109-1043 USA E-mail: [email protected]

2000 Mathematics Subject Classification (primary; secondary): 52C07; 52B20, 05A15, 52B45, 52B55, 52C45, 11H06 Key words: Algebra of polyhedra, Berline–Vergne local formula, continued fractions, Ehrhart polynomial, Euler characteristic, generating functions, integer points, Lenstra–Lenstra–Lovász lattice reduction algorithm, Minkowski Convex Body Theorem

ISBN 978-3-03719-052-4 The Swiss National Library lists this publication in The Swiss Book, the Swiss national bibliography, and the detailed bibliographic data are available on the Internet at http://www.helveticat.ch. This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained.

© 2008 European Mathematical Society Contact address: European Mathematical Society Publishing House Seminar for Applied Mathematics ETH-Zentrum FLI C4 CH-8092 Zürich Switzerland Phone: +41 (0)44 632 34 36 Email: [email protected] Homepage: www.ems-ph.org Printed on acid-free paper produced from chlorine-free pulp. TCF °° Printed in Germany 987654321

Preface These are the lecture notes for a graduate course which the author taught twice, ﬁrst at the ETH, Z¨ urich in Spring - Summer 2007 as a part of the “Nachdiplom Lectures” series and then at the University of Michigan, Ann Arbor, in the Fall of 2007 as a regular graduate course with exercises and grades. One can view these lectures as a substantially extended version of [Ba07] (the course of [Ba07] consisted of ﬁve lectures only). The goal of the course is by starting from basic linear algebra, to walk the reader towards most recent advances in enumeration of integer points in polyhedra. As in [Ba07] and before that in [BP99], the structural theory is developed with an eye on algorithmic applications. Compared to [BP99] the present text is much more suitable for studying, as the author hopes. It contains ﬁgures, exercises, and the background material on lattices and polyhedra with more or less complete proofs. It also contains some new results, most notably recent remarkable “local” formulas by N. Berline and M. Vergne [BV07], and a new look at some old results, such as the role played by the classical construction of continued fractions or the “continuity” property of polynomials enumerating integer points in parametric polytopes as viewed through the prism of identities in the algebra of polyhedra. On the other hand, unlike in [BP99], there is no discussion in these notes of the relevant algebraic geometry of toric varieties. Exercises constitute an essential part of these notes. Problems marked with ◦ are relatively straightforward and solving them is important for understanding the material. Problems marked with ∗ are considered to be of the research level, they often come with a reference to the literature. The author acknowledges the National Science Foundation support via grant DMS 0400617. The author is grateful to the ETH, Z¨ urich, for hospitality.

Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

v

Chapter 1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . .

1

Chapter 2. The algebra of polyhedra . . . . . . . . . . . . . . . . . .

9

Chapter 3. Linear transformations and polyhedra . . . . . . . . . 19 Chapter 4. The structure of polyhedra . . . . . . . . . . . . . . . . . 27 Chapter 5. Polarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Chapter 6. Tangent cones. Decompositions modulo polyhedra with lines . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Chapter 7. Open polyhedra . . . . . . . . . . . . . . . . . . . . . . . . 57 Chapter 8. The exponential valuation . . . . . . . . . . . . . . . . . 63 Chapter 9. Computing volumes . . . . . . . . . . . . . . . . . . . . . 77 Chapter 10. Lattices, bases, and parallelepipeds . . . . . . . . . . . 81 Chapter 11. The Minkowski Convex Body Theorem . . . . . . . . 95 Chapter 12. Reduced basis . . . . . . . . . . . . . . . . . . . . . . . . 99 Chapter 13. Exponential sums and generating functions . . . . . 107 Chapter 14. Totally unimodular polytopes . . . . . . . . . . . . . . 121 Chapter 15. Decomposing a 2-dimensional cone into unimodular cones via continued fractions . . . . . . . . . . . . . . . 129 Chapter 16. Decomposing a rational cone of an arbitrary dimension into unimodular cones . . . . . . . . . . . . 137

viii

Chapter 17. Eﬃcient counting of integer points in rational polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 Chapter 18. The polynomial behavior of the number of integer points in polytopes . . . . . . . . . . . . . . . . . . . . . 155 Chapter 19. A valuation on rational cones . . . . . . . . . . . . . . 167 Chapter 20. A “local” formula for the number of integer points in a polytope . . . . . . . . . . . . . . . . . . . . . . . . . 183 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

CHAPTER 1

Introduction The two main objects of these notes are the standard integer lattice Zd ⊂ Rd consisting of points with integer coordinates and a polyhedron P ⊂ Rd consisting of points satisfying a ﬁnite set of linear inequalities. The unifying topic is how to count integer points in a polytope (bounded polyhedron). For example, we conclude by inspection that the polygon P in Figure 1 contains seven integer points, or, in other words, that |P ∩ Z2 | = 7.

m

P

Figure 1. The integer lattice Z2 ⊂ R2 , a polygon P ⊂ R2 , and an integer point m ∈ P . As the dimensions of our polytopes grow and their analytic descriptions become more complicated, “by inspection” no longer works and we need a theory. The ﬁrst step towards such a theory is to realize that the number of |P ∩ Zd | of integer points in a d-dimensional polytope P ⊂ Rd is a valuation, that is |P ∩ Zd | = |P1 ∩ Zd | + |P2 ∩ Zd | − |Q ∩ Zd | provided P = P1 ∪ P2 and Q = P1 ∩ P2 . This observation allows us to cut a given polytope into simpler pieces, enumerate integer points in those pieces and then obtain the total number of points by carefully accounting for various overlapping parts. This is indeed very useful, but not good enough: it turns out that for many polytopes there is no way to dissect them into a reasonably few simple pieces. We need more freedom in “cutting and pasting” of polyhedra.

2

INTEGER POINTS IN POLYHEDRA

What we need, is to be able to extend the valuation property further to unbounded polyhedra, since it turns out that only unbounded polyhedra (namely cones) are simple enough to deal with, as far as the integer point enumeration is concerned. This requires us to somehow make sense of the number of integer points in an unbounded polyhedron. Fortunately, a way of counting for inﬁnite sets has long been known under the name of generating functions. With an integer point m = (µ1 , . . . , µd ) in Rd we associate a monomial xm = xµ1 1 · · · xµd d in d variables x1 , . . . , xd . We consider the sum G (1.1) xm , m∈P ∩Zd

where P is a polyhedron and Zd ⊂ Rd is the standard integer lattice. We will show that for rational polyhedra P (that is, polyhedra deﬁned by linear inequalities with integer coeﬃcients) if the series (1.1) converges for some x, it converges to a rational function f (P, x). Moreover, we will be able to deﬁne a rational function f (P, x) even if the series (1.1) does not converge for any x. If P is bounded, we obtain the number |P ∩ Zd | of integer points in P by computing the value of f (P, x) at x1 = · · · = xd = 1. Let us see how this theory plays out in the familiar though admittedly not very exciting case of d = 1. Suppose that P+ = [0, +∞) is the positive ray. With every nonnegative integer m we associate a monomial xm and consider the sum over non-negative integer m, see Figure 2. x0 x1 x2 x3 x4 0

2

1

3

4

P+

Figure 2. Points on the positive ray P+ . The corresponding generating function is given by the formula for the inﬁnite geometric series: +∞ G m=0

xm =

1 1−x

provided |x| < 1,

so we say that f (P+ , x) =

1 . 1−x

1. INTRODUCTION

3

Similarly, for the negative ray P− = (−∞, 0], we get 0 G

xm =

m=−∞

1 1 − x−1

provided |x| > 1,

see Figure 3, so we say that f (P− , x) =

1 . 1 − x−1

x−4 x−3 x−2 x−1 x0 −4 −3 −2 −1 0 P−

Figure 3. Points on the negative ray P− . Let us consider integer points on the line R1 , see Figure 4. Although the series +∞ G xm m=−∞

does not converge for any x, we are able to ﬁnd the rational function f (R1 , x) using the valuation property. Indeed, we have R1 = P+ ∪ P−

and P+ ∩ P− = {0}.

x−3 x−2 x−1 x0 x1 x 2 x 3 −3 −2 −1 0 1 2 3 P−

P+

Figure 4. Points on the line R1 . Hence we must have 1 1 + −1 1 − x 1 − x−1 x 1 − − 1 = 0. = 1−x 1−x

f (R1 , x) = f (P+ , x) + f (P− , x) − x0 =

Now, let us do some counting. For some integers k < n let us consider the interval P = [k, n], P ⊂ R1 , which is a bona ﬁde one-

4

INTEGER POINTS IN POLYHEDRA

dimensional polytope. We do some cutting and pasting to represent the interval as a combination of unbounded polyhedra, see Figure 5. In terms of formal power series we have n G m=k

xm =

+∞ G

n G

xk +

m=k

k k

xm −

m=−∞

= + −

+∞ G

xm .

m=−∞

n

n

Figure 5. Representing the interval as a sum of two rays minus a line. The valuation property tells us that we must have f (P, x) = f (k + P+ , x) + f (n + P− , x) − f (R1 , x), where k + P+ and n + P− are the respective translations of the positive ray P+ by k and of the negative ray P− by n. It is not hard to convince ourselves that we must have f (k + P+ , x) = xk f (P+ , x) =

xk 1−x

and f (n + P− , x) = xn f (P− , x) =

xn . 1 − x−1

Together with the identity f (R1 , x) = 0 this gives us the familiar formula for the sum of a ﬁnite geometric series: (1.2)

n G m=k

xm = f (P, x) =

xn xk − xn+1 xk + − 0 = . 1 − x 1 − x−1 1−x

Finally, to compute the number |P ∩Z| of integer points on the interval, we substitute x = 1 in the expression for f (P, x). After a moment of hesitation, we apply l’Hospital’s rule and conclude that |P ∩ Z| = n − k + 1, which we knew all along.

1. INTRODUCTION

5

This one-dimensional exercise shows in a nutshell some important features of the general theory. First, higher-dimensional analogues of the strange identity 1

(1.3)

f (R , x) =

+∞ G

xm = 0

m=−∞

turn out to be indispensable in dealing with unbounded polyhedra in Rd . Second, the representation of Figure 5 turns out to be one degenerate case of a rich family of identities for higher-dimensional polyhedra. Finally, substituting x = (1, . . . , 1) in f (P, x) in higher dimensions also requires an appropriate version of l’Hospital’s rule, though the answer, in general, is far from obvious. There are, of course, important phenomena in higher dimensions that cannot be observed in dimension 1. Formula (1.2) writes a potentially long (for n − k large) polynomial in x as a short rational function in x. We will see that for a general polyhedron P the sum (1.1) can be written as a short rational function, with the term “short” deﬁned appropriately. Let us consider a 2-dimensional example: a triangle Δ in the plane with the vertices at A = (0, 0), B = (0, 100) and C = (100, 0). One can show that G

xm =

m∈Δ∩Z2

x100 1 2 + (1 − x1 )(1 − x2 ) (1 − x−1 )(1 − x1 x−1 2 2 ) +

x100 1 )(1 − x−1 (1 − x−1 1 x2 ) 1

and this formula will be seen as belonging to the same family of formulas as formula (1.2). Incidentally, the three terms of the right-hand side are the sums over the angles of the triangle, see Figure 6. B A

C B

A

C

Figure 6. A triangle and its angles.

6

INTEGER POINTS IN POLYHEDRA

Exercises below describe some interesting examples of short rational functions encoding long series. Problems 1. Let a and b be coprime positive integers and let > D S = m1 a + m2 b : m1 , m2 ∈ Z + be the set (semigroup) of all linear combinations of a and b with nonnegative integer coeﬃcients m1 and m2 . Prove that G

xm =

m∈S

1 − xab (1 − xa ) (1 − xb )

for |x| < 1.

2∗ . Let a, b, and c be coprime positive integers and let > D S = m1 a + m2 b + m3 c : m1 , m2 , m3 ∈ Z + be the set of all linear combinations of a, b, and c with non-negative integer coeﬃcients m1 , m2 , and m3 . Prove that there exist integers p1 , p2 , p3 , p4 and p5 , not necessarily distinct, such that G 1 − xp 1 − xp 2 − xp 3 + xp 4 + xp 5 for |x| < 1. xm = a ) (1 − xb ) (1 − xc ) (1 − x m∈S Hint: See [De03]. 3. Let a, b, c, and d be coprime positive integers and let > D S = m1 a + m2 b + m3 c + m4 d : m1 , m2 , m3 , m4 ∈ Z + be the set of all linear combinations of a, b, c, and d with non-negative integer coeﬃcients m1 , m2 , m3 , and m4 . a) Prove that G m∈S

xm =

(1 −

xa ) (1

p(x) − xb ) (1 − xc ) (1 − xd )

for some polynomial p and all |x| < 1. b∗ ) Prove that the number of monomials in the polynomial p above can be arbitrarily large, depending on a, b, c, and d. Hint: See [SW86].

1. INTRODUCTION

7

Problem 3 shows that the pattern of Problems 1 and 2 breaks down for semigroups with four or more generators. Nevertheless, the more general phenomenon that the series has a “short rational generating function representation” still holds, see [BW03].

CHAPTER 2

The algebra of polyhedra We introduce one of the main objects of the course. Let V be a ﬁnite-dimensional real vector space. Definition 2.1. A polyhedron P ⊂ V is a set deﬁned by ﬁnitely many linear inequalities: > P = x∈V :

Zi (x) ≤ αi ,

i∈I

D

for some ﬁnite (possibly empty) set I, where Zi : V −→ R are linear functions and αi ∈ R are real numbers. In particular, V itself is a polyhedron. A set A ⊂ V is called convex if for every two points x, y ∈ A, one has [x, y] ⊂ A, where >

[x, y] = αx + (1 − α)y :

D

0≤α≤1 ,

see Figure 7.

a)

b)

Figure 7. a) polyhedra (bounded and unbounded) and b) convex sets (bounded and unbounded).

10

INTEGER POINTS IN POLYHEDRA

Often, we identify V = Rd by choosing a basis. In this case, we may deﬁne a polyhedron P by 1 P =

x:

d G

, aij ξj ≤ bi

for x = (ξ1 , . . . , ξd )

and i ∈ I

,

j=1

where aij and bi are real numbers. We will be interested in cutting and pasting of polyhedra. This calls for the following deﬁnition. Definition 2.2. Let A ⊂ V be a set. The indicator function, or just indicator [A], of A is the function [A] : V −→ R deﬁned by 1 [A](x) =

1 if x ∈ A, 0 if x ∈ / A.

The algebra of polyhedra P(V ) is the vector space deﬁned as the span of the indicator functions of all polyhedra P ⊂ V . The algebra of closed convex sets C(V ) is the vector space deﬁned as the span of the indicator functions of all closed convex sets A ⊂ V . Similarly, the algebra of compact convex sets Cb (V ) is the vector space deﬁned as the span of the indicator functions of all compact convex sets A ⊂ V . Clearly, P(V ) ⊂ C(V ) and Cb (V ) ⊂ C(V ). We are mostly interested in P(V ) although some results we state in the greater generality of C(V ). Sometimes we consider the algebra Pb (V ) = P(V ) ∩ Cb (V ) of bounded polyhedra. The ground ﬁeld is not particularly important for us. It can be Q, R, or C. For deﬁniteness, let us choose R to be the ground ﬁeld. So far, we are interested in P(V ), C(V ), and Cb (V ) as vector spaces. They possess a natural algebra structure via the pointwise multiplication of functions: f = gh if f (x) = g(x)h(x) for x ∈ V . Since [A][B] = [A ∩ B] and the intersection of polyhedra is a polyhedron while the intersection of (bounded) closed convex sets is a (bounded) closed convex set, the spaces P(V ), C(V ), and Cb (V ) are closed under multiplication.

2. THE ALGEBRA OF POLYHEDRA

11

There is a more interesting algebra structure on P(V ) and Cb (V ), which will be described later. We note that as long as dim V > 0, the indicators [P ] of polyhedra do NOT form a basis of P(V ) as there are plenty of linear relations among them, see Figure 8.

=

−

+

Figure 8. Linear relations among indicators of polyhedra. We also note that although P(V ) is spanned by the indicators of closed polyhedra, it contains indicators of open polyhedra as well, as Figure 9 shows.

= −

+

Figure 9. The indicator of an open square lies in the algebra of polyhedra. Definition 2.3. A linear transformation T : P(V ), C(V ), Cb (V ) −→ W , where W is a vector space, is called a valuation. The main result of this chapter is the following theorem. Theorem 2.4. There exists a unique valuation χ : C(V ) −→ R, called the Euler characteristic, such that χ([A]) = 1 for every non-empty closed convex set A ⊂ V .

12

INTEGER POINTS IN POLYHEDRA

Proof. Since the indicators of non-empty closed convex sets span C(V ), the valuation χ, if it exists, is necessarily unique. Indeed, for f ∈ C(V ) we must have G G f= αi [Ai ] implies χ(f ) = αi , i

i:Ai 0=∅

where Ai ⊂ V are closed convex sets and αi are numbers. To establish the existence of χ, we ﬁrst establish the existence of χ : Cb (V ) −→ R with the required properties. We proceed by induction on d = dim V . If d = 0, we deﬁne χ(f ) = f (0) for all f ∈ P(V ). Suppose that d ≥ 1. Let us choose a non-zero linear functional Z : V −→ R and let us slice V into the aﬃne hyperplanes > D Hτ = x ∈ V : Z(x) = τ . We can view each Hτ as a (d − 1)-dimensional vector space (we choose the origin in Hτ arbitrarily). Hence there exists a valuation χτ : Cb (Hτ ) −→ R such that

χτ ([A]) = 1 for every non-empty convex compact A ⊂ Hτ . Let us choose a function f ∈ Cb (V ), so G (2.4.1) f= αi [Ai ] i∈I

for some ﬁnite set of non-empty compact convex sets Ai ⊂ V and some numbers αi . We note that the restriction fτ of f onto Hτ can be written as G fτ = αi [Ai ∩ Hτ ] . i∈I

Hence fτ ∈ Cb (Hτ ) and (2.4.2)

χτ (fτ ) =

G

αi .

i∈I: Ai ∩Hτ 0=∅

We claim that for all τ the one-sided limit lim χτ −O (fτ −O )

O−→0+

exists and that with a possible exception of ﬁnitely many τ ’s (the exceptional set depends on f ) the limit is equal to χτ (fτ ). Indeed, let us

2. THE ALGEBRA OF POLYHEDRA

13

deﬁne the set J ⊂ I of indices i in the decomposition (2.4.1) such that Ai ∩ Hτ >= ∅ and τ is not the minimum value of Z on Ai . It follows by (2.4.2) that G lim χτ −O (fτ −O ) = αi , O−→+0

i∈J

see Figure 10. τ τ−ε τ τ−ε A

τ τ−ε τ τ−ε τ τ−ε

Figure 10. A convex compact set and possible locations of the hyperplanes Hτ and Hτ −O . Now we are ready to deﬁne χ : Cb (V ) −→ R. Namely, we let J G% (2.4.3) χ(f ) = χτ (fτ ) − lim χτ −O (fτ −O ) . τ ∈R

O−→0+

Since for each f ∈ Cb (V ) all but ﬁnitely many summands are equal to 0, the sum is well-deﬁned. Furthermore, since all the operations used (sum, diﬀerence, one-sided limit, and the restriction of a function onto an aﬃne hyperplane) preserve linearity, χ : Cb (V ) −→ R is linear. It remains to check that χ([A]) = 1 if A ⊂ V is a non-empty compact convex set. This is indeed the case since the only non-zero term in (2.4.3) is 1 and corresponds to τ that is the minimum value of Z on A, see Figure 10. It remains to extend χ to C(V ). Let us choose a particular convex compact set B containing the origin in its interior. For a t > 1 let tB = {tx : x ∈ B} be the dilation of B. Clearly, for every f ∈ C(V ) we have f [tB] ∈ Cb (V ). We deﬁne (2.4.4)

χ(f ) = lim χ (f [tB]) . t−→+∞

14

INTEGER POINTS IN POLYHEDRA

Assuming that f=

G

αi [Ai ] ,

i

where Ai ⊂ V are (possibly unbounded) closed convex sets, we conclude that Ai >= ∅ implies Ai ∩ tB >= ∅ for all suﬃciently large t (see Figure 11), so the limit in (2.4.4) is well-deﬁned and equal to G αi , i: Ai 0=∅

as required.

A3

A2

B A1

tB

Figure 11. As t grows, the set tB eventually intersects every non-empty set Ai . It is immediate that (2.4.4) deﬁnes a linear functional χ.

!

Theorem 2.4 and its proof is due to Hadwiger, see [Ha55]. We note that while the Euler characteristic of (the indicator of) a line is 1 (since a line is a polyhedron), the Euler characteristic of (the indicator of) an open interval is −1, see Figure 12. Hence the Euler characteristic from Theorem 2.4 is not invariant under homeomorphisms. Similarly, one can observe that the Euler characteristic of (the indicator of) the open square is 1, cf. Figure 9. One can think of the Euler characteristic as of the combinatorialization of the Lebesgue integral. While the Lebesgue integral tells us how much of the set is there, the Euler characteristic just tells us whether the closed convex set is there at all. In the course of the proof of Theorem 2.4, we obtained a useful formula, namely, formula (2.4.3), which we state as a separate corollary.

2. THE ALGEBRA OF POLYHEDRA

χ

= =

−1

15

χ χ − −

χ

1 − 1 −1

Figure 12. The Euler characteristic of the indicator of an open interval is −1. Corollary 2.5. Let Z : V −→ R be a non-zero linear functional and let > D Hτ = x ∈ V : Z(x) = τ for τ ∈ R be aﬃne hyperplanes. Let f ∈ Cb (V ). Let fτ = f ·[Hτ ] be the restriction of f onto Hτ . Then J G% χ (fτ ) − lim χ (fτ −O ) . χ(f ) = O−→0+

τ ∈R

Problems 1◦ . Check that a polyhedron is a closed convex set. 2. Prove the inclusion-exclusion formula $ @ $ @n F G ' |I|−1 Xi (−1) Xi = i=1

I ⊂ {1, . . . , n} I '= ∅

i∈I

for sets Xi ⊂ V , i = 1, . . . , n. 3. Let

> X = (ξ1 , . . . , ξd ) :

0 < ξi < 1 for i = 1, . . . , d

D

be the open d-dimensional cube considered as a subset of Rd . Prove 2 dA that [X] ⊂ P R and show that χ([X]) = (−1)d . 4. Let us ﬁx a linear functional Z : V −→ R. Prove that there exists a valuation L : Cb (V ) −→ R such that L([A]) = max Z(x) x∈A

for every non-empty convex compact set A ⊂ V . 5m 5. Let Ai , i = 1, . . . , m, be closed convex sets such that the set i=1 Ai is convex. Suppose further that for some 1 < k < m the

16

INTEGER POINTS IN POLYHEDRA

intersection of every k of the sets Ai is non-empty. Prove that the intersection of some k + 1 sets of the sets Ai is non-empty. Hint: See [Kl63]. 6∗ . Let V be a ﬁnite-dimensional real vector space, let W be some other vector space and let φ be a map, which, with every compact convex set C ⊂ V associates an element φ(C) ∈ W . Suppose that C I Im C F G ' Ai , (−1)|I|−1 φ Ai = φ i=1

I ⊂ {1, . . . , m} I '= ∅

i∈I

5 whenever A1 , . . . , Am and m i=1 Ai are compact convex sets and that φ(∅) = 0. Prove that φ extends to a valuation Φ : Cb (V ) −→ W such that Φ([A]) = φ(A) for every compact convex set A ⊂ V . Hint: See [Gr78] and [PS70]. 7∗ . Let Z : V −→ R be a non-zero linear functional and let α ∈ R be a number. The set > D H = x ∈ V : Z(x) = α is called an aﬃne hyperplane while the sets > D > H + = x ∈ V : Z(x) ≥ α and H − = x ∈ V :

Z(x) ≤ α

D

are called the (closed) halfspaces bounded by H. Let W be a vector space, and let φ be a map, which, with every polyhedron P ⊂ V associates an element φ(P ) ∈ W and such that 2 A 2 A φ(P ) = φ P ∩ H + + φ P ∩ H − − φ(P ∩ H) for every polyhedron P and every aﬃne hyperplane H ⊂ V and φ(∅) = 0. Prove that φ extends to a valuation Φ : P(V ) −→ W such that Φ([P ]) = φ(P ) for every polyhedron P ⊂ V . Hint: See [Gr78] and [PS70]. 8. A function φ : R −→ R is called additive if φ(x + y) = φ(x) + φ(y) for all x, y ∈ R. Let φ : R −→ R be an additive function such that φ(π) = 0. a) Prove that if φ is continuous then φ(x) = 0 for all x. b∗ ) Show that the axiom of choice implies that there are (necessarily discontinuous) additive φ which satisfy φ(π) = 0 and which are not identically 0.

2. THE ALGEBRA OF POLYHEDRA

17

c∗ ) For a bounded polyhedron (polytope) P ⊂ R3 let us deﬁne the number dφ (P ) ∈ R, G 2 A 2 A length of l φ the dihedral angle at l . dφ (P ) = l is an edge of P

Show that there is a valuation Dφ : Pb (R3 ) −→ R, called the Dehn invariant such that Dφ ([P ]) = dφ (P ) 3 for every polytope P ⊂ R . Hint: See [Sa79].

CHAPTER 3

Linear transformations and polyhedra The main result of this chapter is the following theorem. Theorem 3.1. Let V and W be ﬁnite-dimensional real vector spaces and let T : V −→ W be a linear transformation. Then (1) For every polyhedron P ⊂ V the image T (P ) ⊂ W is a polyhedron; (2) There is a unique linear transformation (valuation) T : P(V ) −→ P(W ) such that T ([P ]) = [T (P )]

for every polyhedron P ⊂ V.

In other words, Theorem 3.1 asserts that linear transformations preserve linear relations among indicator functions of polyhedra, see Figure 13. P1

P2

= T ( P1 )

=

P

+ T (P2 )

P4

3

T (P3 )

+

− T(P4 )

−

Figure 13. The relation [P1 ] = [P2 ]+[P3 ]−[P4 ] projects onto the relation [T (P1 )] = [T (P2 )] + [T (P3 )] − [T (P4 )]. The following lemma, known as the Fourier – Motzkin elimination procedure is the crucial step for proving Part 1 of Theorem 3.1. Lemma 3.2. Let T : Rd −→ Rd−1 be the projection (ξ1 , . . . , ξd ) 9−→ (ξ1 , . . . , ξd−1 ) and let P ⊂ Rd be a polyhedron. Then T (P ) ⊂ Rd−1 is a polyhedron.

20

INTEGER POINTS IN POLYHEDRA

Proof. Let us suppose that P is deﬁned by the system of linear inequalities 1 P =

d G

(ξ1 , . . . , ξd ) :

, aij ξj ≤ bi

for i ∈ I

.

j=1

Let I0 = {i ∈ I : I− = {i ∈ I :

aid = 0} , I+ = {i ∈ I : aid < 0} .

aid > 0} ,

and

Writing x = (y, ξd ) for y = (ξ1 , . . . , ξd−1 ) we conclude that y ∈ T (P ) if and only if d−1 G

(3.2.1)

aij ξj ≤ bi

if i ∈ I0

j=1

and there is a ξd ∈ R satisfying the inequalities G aij bi − ξj aid j=1 aid

for all i ∈ I+

G aij bi ξd ≥ − ξj aid j=1 aid

for all i ∈ I− .

d−1

ξd ≤ and

d−1

For such a ξd to exist, it is necessary and suﬃcient that every lower bound for ξd does not exceed every upper bound for ξd , that is, G ai j G ai j bi b i1 1 2 − ξj ≥ 2 − ξj ai1 d j=1 ai1 d ai2 d j=1 ai2 d d−1

(3.2.2)

d−1

for every pair i1 ∈ I+ ,

i2 ∈ I − .

Thus the image T (P ) is deﬁned by the system of linear inequalities (3.2.1)– (3.2.2) and hence is a polyhedron. If the set I0 is empty, there are no inequalities (3.2.1) and if either of the sets I+ and I− is empty, there are no inequalities (3.2.2). !

3. LINEAR TRANSFORMATIONS AND POLYHEDRA

21

Now we are ready to prove Theorem 3.1. Proof of Theorem 3.1. To prove Part (1), we note that the result is obvious if T : V −→ W is an isomorphism. Similarly, if T : V −→ W is an injection, the result follows since the restriction T : V −→ image T is an isomorphism. For an arbitrary T , let us consider the composition 2 A V −→ V ⊕ W −→ W, x 9−→ x, T x 9−→ T x, where the ﬁrst map is an injection and the second is a projection. If P ⊂ V is a polyhedron, then its image Q in V ⊕W is also a polyhedron. Finally, the image T (P ) of Q under the projection V ⊕ W −→ W is a polyhedron by repeated application of Lemma 3.2. This concludes the proof of Part (1). To prove Part (2), we note ﬁrst that as in the proof of Theorem 2.4, it is clear that the map T is unique, once it exists. Let us show that T exists. For every x ∈ W , the set T −1 (x) ⊂ V is an aﬃne subspace. For a polyhedron P ⊂ V the product [P ][T −1 (x)] = [P ∩ T −1 (x)] is the indicator function of a polyhedron. Therefore, for every f ∈ P(V ), the product g = f [T −1 (x)] also lies in P(V ) and we can compute the Euler characteristic of g. Thus we deﬁne h = T (f ), h : W −→ R, by 2 A h(x) = χ f [T −1 (x)] for all x ∈ W. If f = [P ] then h(x) = χ([P ∩ T −1 (x)]) We note that the polyhedron P ∩ T −1 (x) is non-empty precisely when x ∈ T (P ). Thus T ([P ]) = [T (P )]. It is clear that the map T is linear and, in particular, T maps P(V ) into P(W ). ! Remark 3.3. In analysis, it is common to construct integral operators between function spaces of the type ; f 9−→ h, h(y) = G(x, y)f (x) dµ(x), X

where f : X −→ R is a function on a space X with a measure µ and h : Y −→ R is a function on a space Y . Here G : X × Y −→ R is often referred as the “Green function” or “kernel”. The syntax of the construction of the map T in Theorem 3.1 is quite similar. We deﬁne the “Green function” G : V × W −→ R by 1 1 if T (x) = y, G(x, y) = 0 if T (x) >= y and replace the integral by the Euler characteristic χ.

22

INTEGER POINTS IN POLYHEDRA

Now we can deﬁne a certain non-obvious algebra structure on P(V ). First, a deﬁnition. Definition 3.4. Let A, B ⊂ V be non-empty sets. The Minkowski sum A + B ⊂ V is deﬁned as > D A + B = a + b : where a ∈ A and b ∈ B , see Figure 14. A 0

0 B

A +B

Figure 14. Sets A and B and their Minkowski sum A + B. Theorem 3.5. Let V be a ﬁnite-dimensional real vector space. (1) If P1 , P2 ⊂ V are non-empty polyhedra then their Minkowski sum P1 + P2 ⊂ V is a polyhedron. (2) There exists a unique bilinear operation, ∗:

P(V ) × P(V ) −→ P(V ),

called convolution, such that [P1 ] ∗ [P2 ] = [P1 + P2 ] , if P1 , P2 ⊂ V are non-empty polyhedra. Proof. To prove Part (1), let us consider the set D > P = P1 × P2 = (x, y) : x ∈ P1 , y ∈ P2 . We note that P ⊂ V ⊕ V is a polyhedron, since to deﬁne P by inequalities, one can write separate inequalities for x and y coordinates. Let (3.5.1)

T : V ⊕ V −→ V,

(x, y) − 9 →x+y

be a linear transformation. Then P1 + P2 = T (P ) and hence P1 + P2 is a polyhedron by Part (1) of Theorem 3.1.

3. LINEAR TRANSFORMATIONS AND POLYHEDRA

23

To prove Part (2), we notice again that the operation ∗ with the required properties is clearly unique, if it exists. To establish existence, for f, g ∈ P(V ), let us deﬁne a function f × g : V ⊕ V −→ R by (f × g)(x, y) = f (x)g(y). One can observe that f × g ∈ P(V ⊕ V ) since for G G f= αi [Pi ] and g = βj [Qj ] , i

j

where αi , βj are numbers and Pi , Qj ⊂ V are polyhedra, we have G f ×g = αi βj [Pi × Qj ] . i,j

Let T : P(V ⊕ V ) −→ P(V ) be the linear transformation, cf. Part (2) of Theorem 3.1, associated with the projection (3.5.1). We deﬁne f ∗ g = T (f × g). It is immediate that so deﬁned ∗ satisﬁes the required properties.

!

Ordinarily, one calls the convolution of two functions f, g : V −→ R the function h deﬁned by ; h(x) = f (x − y)g(y) dµ(y), V

where µ is a measure on V . One can interpret the operation ∗ of Theorem 3.5 as the convolution of functions with respect to the Euler characteristic χ in lieu of the integral. The convolution ∗ has a number of curious properties. One can observe that the indicator [0] of the origin plays the role of the unit: f ∗ [0] = [0] ∗ f = f

for all f ∈ P(V ).

Furthermore, one can see that the indicator of the interval [0, 1] is invertible, the inverse being the negative indicator of the open interval (−1, 0), see Figure 15.

24

INTEGER POINTS IN POLYHEDRA

−1 −1

( +

0

−1

0

−1

0

* − +

=

1

0

−1

0

0

1

)* −

0

1

−1

= 1

=

0

Figure 15. The convolution of the negative indicator of the open interval (−1, 0) and the indicator of the closed interval [0, 1] is the indicator of the origin. More generally, one can show that the inverse of the indicator of a d-dimensional bounded polyhedron P is the indicator of the (relative) interior of −P multiplied by (−1)d . The indicator of a ray [0, +∞), on the other hand, is a zero-divisor. If we take its convolution with the indicator of a semi-open interval (0, 1], we get 0, see Figure 16. 0

(

0

1 1

* −

=

0 0

) * −

0

= 0

=

0 Figure 16. The convolution of the indicator of the ray [0, +∞) with the indicator of the semi-open interval (0, 1] is 0. Finally, we mention that the convolution ∗ commutes with the transformation T of Theorem 3.1, that is, T (f ∗ g) = T (f ) ∗ T (g). Problems 1. Let T : Rd −→ Rk be a linear transformation, let P ⊂ Rd be a polyhedron, and let Q = T (P ). Assuming that P is deﬁned by m linear

3. LINEAR TRANSFORMATIONS AND POLYHEDRA

25

inequalities, approximately how many linear inequalities are needed to deﬁne Q if one is to follow the construction of Theorem 3.1? 2◦ . Let T : V −→ W be a linear transformation. Prove that for every compact convex set A ⊂ V the set T (A) is compact and convex and that there exists a linear transformation T : Cb (V ) −→ Cb (W ) such that T ([A]) = [T (A)] for every compact convex set A. 3◦ . Construct an example of a closed convex set A ⊂ V and a linear transformation T : V −→ W such that T (A) is not closed. 4. Prove that the Minkowski sum A + B of compact convex sets A, B ⊂ V is compact and convex and that there exists a bilinear operation ∗ : Cb (V ) × Cb (V ) −→ Cb (V ) such that [A] ∗ [B] = [A + B] for any two compact convex sets A, B ⊂ V . 5◦ . Construct an example of closed convex sets A, B ⊂ V such that A + B ⊂ V is not closed. 6. Let T : P(V ) −→ P(W ) be the transformation of Theorem 3.1. Prove that T (f ∗ g) = T (f ) ∗ T (g). 7. Let P ⊂ V be an unbounded polyhedron. Prove that for some non-zero f ∈ P(V ) one has [P ] ∗ f = 0. 8. Let I = {(ξ1 , ξ2 ) : Check that

0 ≤ ξ1 , ξ2 ≤ 1} be the square in the plane. [I] ∗ [− int I] = [0],

where − int I = {(ξ1 , ξ2 ) :

−1 < ξ1 , ξ2 < 0}.

9∗ . Let P ⊂ Rd be a bounded polyhedron with the non-empty interior int P . Prove that [P ] ∗ [− int P ] = (−1)d [0], where −X = {−x : x ∈ X}. Hint: See [MS83] and [Mc93]. 10∗ . Prove that Pb (V ) = P(V ) ∩ Cb (V ) has no zero-divisors with respect to the convolution ∗. 2 A 11. Let us ﬁx two integers 0 < k < d. Let Gk Rd be the Grassmannian of all k-dimensional subspaces L ⊂ Rd , endowed with the Haar probability 2measure ν. For a compact convex set A ⊂ Rd and a A d subspace L ∈ Gk R let KL be the orthogonal projection of A onto L

26

INTEGER POINTS IN POLYHEDRA

and let volk be the k-dimensional volume in L. Let us deﬁne the k-th intrinsic volume wk (A) of A by ; wk (A) = volk (KL ) dν. Gk ( Rd ) In other words, wk (A) is the average k-dimensional volume of the orthogonal projection of K onto a k-dimensional subspace L ⊂ Rd . 2 A a) Prove that wk extends to a valuation on Cb Rd , that is, m G

αi wk (Ai ) = 0 whenever

i=1

m G

αi [Ai ] = 0

i=1

for convex compact sets Ai ⊂ Rd and that wk (T (A)) = wk (A) for any orthogonal transformation T ; b∗ ) Prove that wd−1 = cd vold−1 (∂K) for some constant cd , where vold−1 (∂K) is the surface area of K. Hint: For more information, see [KR97] and [Sc93]. 11∗ . Let Ki ⊂ V , i& = 1, . . . , m, be compact convex sets and let αi be numbers such that m i=1 αi [Ki ] = 0. Prove that G G α i Ki = (−αi )Ki , i:αi >0

i:αi <0

where the sums on both sides refer to the Minkowski sum of convex bodies. Hint: Use Problem 4 of Chapter 2 and the fact that max Z(x) = max Z(x) + max Z(x)

x∈A+B

x∈A

x∈B

for any sets A, B ⊂ V and any linear function Z : V −→ R.

CHAPTER 4

The structure of polyhedra The structure of polyhedra is determined by their relationship with three simple objects. Definition 4.1. Let V be a ﬁnite-dimensional real vector space and let v, u ∈ V be vectors such that u >= 0. The set N ( v + τu : τ ∈ R is called a line in V . The set

N

( v + τu : τ ≥ 0 is called a ray emanating from v in the direction of u. Finally, given two points v1 , v2 ∈ V , the set N ( τ v1 + (1 − τ )v2 : 0 ≤ τ ≤ 1

is called the interval with the endpoints v1 and v2 , see Figure 17.

Figure 17. A line, a ray, and an interval. Lemma 4.2. A polyhedron is bounded if and only if it does not contain any ray. Proof. Clearly, a polyhedron is unbounded if it contains a ray. Let > D P = x : Zi (x) ≤ αi for i ∈ I be a polyhedron. Let us introduce a Euclidean structure on the ambient space V . Suppose that P is unbounded, that is, there is a sequence

28

INTEGER POINTS IN POLYHEDRA

of points un ∈ P with Cun C −→ +∞. Let us consider the normalized sequence wn = un /Cun C. Then for a limit point w of wn we must have CwC = 1 and Zi (w) ≤ 0 for i ∈ I. Therefore, for every x ∈ P the ray emanating from x in the direction of w will lie in P . ! There is another important description of bounded polyhedra. Definition 4.3. A point v ∈ V is called a convex combination of points v1 , . . . , vm ∈ V if v=

m G

λi v i

where λi ≥ 0 for i = 1, . . . , m and

m G

λi = 1.

i=1

i=1

The set of all convex combinations of points from a given set A ⊂ V is called the convex hull of A and denoted by conv(A). A convex polytope is the convex hull of a ﬁnite set A ⊂ V of points, see Figure 18. G F

C E

A B D

Figure 18. A set of points and its convex hull. Next we prove the ﬁrst part of what is known as the Weyl – Minkowski Theorem. Theorem 4.4. Let P be a polytope. Then P is a bounded polyhedron. Proof. Let

/ M P = conv v1 , . . . , vm ,

where v1 , . . . , vm ∈ V are points. Let us consider the standard simplex Δ ⊂ Rm , 1 Δ=

(λ1 , . . . , λm ) :

λi ≥ 0 for i = 1, . . . , m and

m G i=1

, λi = 1 .

4. THE STRUCTURE OF POLYHEDRA

29

Clearly, Δ is a polyhedron. Let us consider a linear transformation m

T : R −→ V,

2

A

T λ1 , . . . , λm =

m G

λi v i .

i=1

Then P = T (Δ) and hence P is a polyhedron by Part (1) of Theorem 3.1. Since Δ is compact and T is continuous, P is bounded. ! The second part of the Weyl – Minkowski Theorem asserts, predictably, that if P is a bounded polyhedron then P is a polytope. We will prove it soon, but ﬁrst we need a deﬁnition. Definition 4.5. Let P ⊂ V be a polyhedron and let v ∈ P be a point. The point v is called a vertex of P if whenever v = (v1 + v2 )/2 with v1 , v2 ∈ P , one necessarily has v = v1 = v2 . In other words, v is a vertex of P if v is not the midpoint of an interval with distinct endpoints in P . For example, in Figure 18, the points A, D, E, F and G are vertices of the polygon while the points B and C are not. We deﬁne a face of F ⊂ P of a polyhedron P as follows. Let Z : V −→ R be a non-zero linear functional and let α ∈ R be a number such that Z(x) ≤ α for all x ∈ P . Then > D F = x ∈ P : Z(x) = α . A face F >= ∅, P is called a proper face of P . For example, the interval with the endpoints A and D is a proper face of the polygon in Figure 18. The following simple properties of a vertex turn out to be particularly useful. Lemma 4.6. Let P ⊂ V be a polyhedron, and let v ∈ P be a point. (1) Suppose that > D P = x ∈ V : Zi (x) ≤ αi for i ∈ I . Let

>

Iv = i ∈ I :

Zi (v) = αi

D

(the inequalities indexed by the elements of Iv are called active on v). Then v is a vertex of P if and only if M / span Zi : i ∈ Iv = V ∗ .

30

INTEGER POINTS IN POLYHEDRA

In particular, the set of vertices of a polyhedron is always ﬁnite, possibly empty; (2) Let F ⊂ P be a face of P and let v be a vertex of F . Then v is a vertex of P . Proof. To prove Part (1), suppose that the set Zi : i ∈ Iv spans the dual space V ∗ . Let us write v = (v1 + v2 )/2 for v1 , v2 ∈ P . Thus we must have Zi (v1 ), Zi (v2 ) ≤ αi for all i ∈ Iv . On the other hand, Zi (v1 ) + Zi (v2 ) = αi 2

for all i ∈ Iv .

Therefore,

Zi (v1 ) = Zi (v2 ) = αi for all i ∈ Iv and hence the only solution of this system of linear equations is v1 = v2 = v and so v is indeed a vertex. On the other hand, suppose that the set Zi : i ∈ Iv does not span V ∗ . Then there is a vector u ∈ V \ {0} such that Zi (u) = 0 for all i ∈ Iv . Then for

v1 = v + gu and v2 = v − gu, where g > 0 is suﬃciently small, we have v1 , v2 ∈ P and v = (v1 +v2 )/2, so v is not a vertex. We observe that for every vertex v of P the set of active inequalities contains at least dim V elements and uniquely determines the vertex. Therefore, if the polyhedron P is deﬁned by m linear inequalities 2in Aa d-dimensional space, the number of vertices of P does not exceed md . To prove Part (2), suppose Z(x) ≤ α for all x ∈ P and that > D F = x ∈ P : Z(x) = α . Suppose that v is a vertex of F . Let us write v = (v1 + v2 )/2 with v1 , v2 ∈ P . Then we must have Z(v1 ) + Z(v2 ) 2 and since Z(v1 ), Z(v2 ) ≤ α we must have α = Z(v) =

Z(v1 ) = Z(v2 ) = α so v1 , v2 ∈ F . But then we must have v1 = v2 = v.

!

4. THE STRUCTURE OF POLYHEDRA

31

Now we can complete the proof of the Weyl – Minkowski Theorem. Theorem 4.7. Let P be a bounded polyhedron. Then P is the convex hull of the set of its vertices. In particular, P is a polytope. Proof. Let P ⊂ V be a non-empty bounded polyhedron. We proceed by induction on dim V . Clearly, if dim V = 0, the result follows. Suppose that dim V > 0. Suppose further that P is deﬁned by a system of linear inequalities > D P = x ∈ V : Zi (x) ≤ αi , i ∈ I for some non-zero Zi . Clearly, the convex hull of the set of vertices of P lies in P . Let us choose an arbitrary y ∈ P . Suppose ﬁrst that there is a j ∈ I such that Zj (y) = αj . Let us consider a face > D F = x∈P :

Zj (x) = αj .

Then F lies in a proper aﬃne subspace of V , y ∈ F and by the induction hypothesis y must be a convex combination of vertices of F . By Part (2) of Lemma 4.6, every vertex of F is a vertex of P as well. Suppose now that Zi (y) < αi for all i ∈ I. Let us consider a line through y. Since P is bounded, the line has to intersect P along an interval with the endpoints z1 , z2 , where each zi lies on a proper face of P , cf. Figure 19. z1

P y

z2

Figure 19. Representing a point as a point of the interval with endpoints on the boundary. As above, we argue that by the induction hypothesis, both z1 and z2 are convex combinations of vertices of P . Hence y is also a convex combination of vertices of P . ! Here is another useful result. Theorem 4.8. A non-empty polyhedron contains a vertex if and only if it does not contain a line.

32

INTEGER POINTS IN POLYHEDRA

Proof. Suppose that P is given by a system of linear inequalities > D P = x ∈ V : Zi (x) ≤ αi for i ∈ I . Suppose that P contains a line {v + τ u : One can see that we must have

τ ∈ R} for some u >= 0.

Zi (u) = 0 for i ∈ I. Therefore, for every v ∈ P we can write v = (v1 + v2 )/2 where v1 = v + u ∈ P and v2 = v − u ∈ P , so there are no vertices in P . Suppose now that P does not contain a line. As in the proof of Theorem 4.7, we proceed by induction on the dimension V . If dim V = 0, the result is clear. As in the proof of Theorem 4.7, without loss of generality we may assume that there is a point x ∈ P such that Zi (x) < αi for all i ∈ I. Let us consider a line through x. Since P does not contain lines, the intersection of this line with P is either a ray emanating from a certain point z0 in a proper face F0 of P or an interval with the endpoints z1 ∈ F1 , z2 ∈ F2 , where F1 and F2 are proper faces of P . In either case, we apply the induction hypothesis to the appropriate face of P to claim that there is a vertex of F which must be a vertex of P , see Figure 20.

x

z F

P

Figure 20. A polytope P , a point x, a face F . ! Next, we turn to certain unbounded polyhedra. Definition 4.9. A polyhedron K ⊂ V is called a polyhedral cone or just a cone if 0 ∈ K and for every x ∈ K and every λ ≥ 0 we have λx ∈ K, cf. Figure 21. One can observe that a cone is deﬁned by a system of linear homogeneous inequalities: > D K = x ∈ V : Zi (x) ≤ 0 for i ∈ I .

4. THE STRUCTURE OF POLYHEDRA

B 0

33

C A

Figure 21. A typical cone in the plane and three points. Only the origin can possibly be a vertex of a cone. Theorem 4.8 implies that the origin is indeed the vertex if and only if the cone does not contain lines. Such a cone is called pointed. Similarly to the convex hull (Deﬁnition 4.3) we deﬁne the conic hull of points. Definition 4.10. A point v ∈ V is called a conic combination of points v1 , . . . , vm ∈ V if v=

m G

λi v i

where λi ≥ 0.

i=1

The set of all convex combinations of points from a given set A ⊂ V is called the conic hull of A and denoted by co(A). For example, the cone of Figure 21 is the conic combination of points A and B. The conic hull of a ﬁnite set of points is, necessarily, a (polyhedral) cone. The proof is completely analogous to that of Theorem 4.4 and therefore omitted. Cones without lines look like pyramids over polytopes, cf. Figure 22. 0

P K

Figure 22. A cone K without lines and a polytope P .

34

INTEGER POINTS IN POLYHEDRA

Theorem 4.11. Let K ⊂ V be a pointed cone, K = > {0}. Then there exists an aﬃne hyperplane H ⊂ V , 0 ∈ / H, and a polytope P ⊂ H such that K = co(P ). Proof. Suppose that > K= x∈V :

D Zi (x) ≤ 0 for i ∈ I .

Since 0 is a vertex of K, the functions Zi span the dual space V ∗ , cf. Part (1) of Lemma 4.6. Let Z=

G

Zi .

i∈I

We claim that Z >= 0. Indeed, if Z = 0 then for every x ∈ K we have Z(x) = 0 and hence Zi (x) = 0 for all i ∈ I, which is a contradiction. Let us deﬁne an aﬃne hyperplane >

H= x:

Z(x) = −1

D

and let P = H ∩ K. For every x ∈ K \ {0} we must have Z(x) < 0 and so λx ∈ H for some λ > 0. Hence K = co(P ), P ⊂ H is a polyhedron and it remains to show that P is bounded. Suppose that P is not bounded. Then it contains a ray in a direction u >= 0 parallel to the hyperplane H. Therefore, Z(u) = 0. On the other hand, since the ray is contained in K, we must have Zi (u) ≤ 0 for all i ∈ I. (Therefore, Zi (u) = 0 for all i ∈ I. But then the line N τ u : τ ∈ R is contained in K, which is a contradiction. ! Now we can describe polyhedra without lines as Minkowski sums of polytopes and pointed cones, see Figure 23.

= P

+ M

0 K

Figure 23. Representing a polyhedron P as a Minkowski sum of a polytope M and a pointed cone K.

4. THE STRUCTURE OF POLYHEDRA

35

Theorem 4.12. Let P ⊂ V be a non-empty polyhedron without lines. Let M be the convex hull of the set of vertices of P and let us deﬁne K = KP as follows: > D K = u ∈ V : x + τ u ∈ P for some x ∈ P and all τ ≥ 0 . Then K ⊂ P is a pointed cone, called the recession cone of P , and P = K + M. Proof. Suppose that > P = x∈V :

Zi (x) ≤ αi

D for i ∈ I .

If x + τ u ∈ P for all τ ≥ 0 we must have Zi (u) ≤ 0 for all i ∈ I. Hence > D K = x ∈ V : Zi (x) ≤ 0 for i ∈ I from which we deduce that K is a polyhedral cone, that x + K ⊂ P for all x ∈ P , and that K does not contain lines. Now it is clear that K + M ⊂ P , so it remains to show that P ⊂ K + M . To establish the latter inclusion, we proceed by induction on dim V . As usual, the case of dim V = 0 is clear. Let us pick a y ∈ P . Our goal is to write y = u + z, where u ∈ K and z ∈ M . If Zj (x) = αj for some j, we consider the face F , D > F = x ∈ P : Zj (x) = αi and argue by the induction hypothesis that y is represented as a sum of a convex combination of vertices of F , which are necessarily vertices of P , cf. Part (2) of Lemma 4.6, and a vector u ∈ KF ⊂ KP . Hence we may assume that Zi (y) < αi for all i. We note that if K = {0} then P contains no rays and by Lemma 4.2 is bounded, so the result follows by Theorem 4.7. Otherwise, let us choose u ∈ K \{0} and consider a line {y + τ u : τ ∈ R}. Since P contains no lines, the intersection of P with that line is a ray {z + τ u : τ ≥ 0} for some z on a proper face F of P . As above, by the induction hypothesis, z is a sum of a convex combination of vertices of F (which, necessarily, are vertices of P ) and u1 ∈ KF ⊂ K. Then y can be written as a sum y = z + u1 + λu2

for some λ2 ≥ 0.

Letting u = u1 + λu2 we complete the proof in this case, cf. also the proof of Theorem 4.8 and Figure 20. !

36

INTEGER POINTS IN POLYHEDRA

The next result of this chapter describes the structure of general polyhedra. Theorem 4.13. Let P ⊂ V be a non-empty polyhedron. Then P can be represented as the Minkowski sum P = M + K + L, where L ⊂ V is a subspace, K ⊂ V is a pointed cone without lines and M is a polytope. Proof. Suppose that P is deﬁned by a system of linear inequalities > D P = x ∈ V : Zi (x) ≤ αi for i ∈ I . We note that if P contains a line v + τ u, τ ∈ R, we necessarily have Zi (u) = 0 for all i ∈ I. Let us deﬁne > D L = u ∈ V : Zi (u) = 0 for i ∈ I . Clearly, L ⊂ V is a subspace. Let us decompose V =L⊕W and let us consider the projection T : V −→ W with the kernel L. By Part (1) of Theorem 3.1, the image Q = T (P ) ⊂ W is a polyhedron. Furthermore, Q does not contain lines. Hence, by Theorem 4.12, we can write Q = M + K, where M ⊂ W is a polytope and K ⊂ W is a cone without lines. It is seen now that P = L + K + M. ! Finally, we describe the structure of the faces of polyhedra. We need one more deﬁnition. Definition 4.14. Let P ⊂ V be a polyhedron. By the relative interior int P of P we mean its interior with respect to the smallest aﬃne subspace containing P . For example, on Figure 20, point x lies in the relative interior of P while point z lies in the relative interior of face F but not in the relative interior of P .

4. THE STRUCTURE OF POLYHEDRA

37

Theorem 4.15. Let P ⊂ V be a polyhedron, > D P = x ∈ V : Zi (x) ≤ αi , i ∈ I for a ﬁnite set of non-zero linear functions Zi : V −→ R and numbers αi . (1) Suppose that P is non-empty. Then either P contains an interior point or lies in a proper aﬃne subspace of V ; (2) Suppose that P has a non-empty interior. For a point v not in the interior of P , let > D Iv = i ∈ I : Zi (v) = αi be the set of inequalities active on v and let G G αi . Zi and αv = Zv = i∈Iv

i∈Iv

Then the set

> Fv = x ∈ P :

Zv (x) = αv

D

is a face of P containing v in its relative interior; (3) Every non-empty face F of P can be written as F = Fv for any v in the relative interior of F . Proof. To prove Part (1), suppose that for every i ∈ I there is a point xi ∈ P such that Zi (xi ) < αi . Then for the average 1 G x= xi |I| i∈I we have

Zi (x) < αi for all i ∈ I and x lies in the interior of P . If for some i ∈ I we have Zi (x) = αi for all i ∈ I, then P lies in the aﬃne hyperplane D > x : Zi (x) = αi and the proof of Part (1) follows. To prove Part (2), we note that Fv is obviously a face of P containing v. We claim that the smallest aﬃne subspace containing Fv is given by > D Av = x ∈ V : Zi (x) = αi for all i ∈ Iv .

38

INTEGER POINTS IN POLYHEDRA

Indeed, v ∈ Av and every point in a small neighborhood of v in Av lies in P and hence in Fv . To prove Part (3), let F ⊂ P be a non-empty face of P . Hence there is a non-zero linear function Z : V −→ R and a number α such that Z(x) ≤ α for all x ∈ P and > D F = x ∈ P : Z(x) = α . We observe that F is a non-empty polyhedron, so by Part (1) the relative interior of F is not empty. Suppose that v lies in the relative interior of F . We claim that F = Fv . To this end, we ﬁrst establish that Z ∈ Kv , where Kv = co (Zi : i ∈ Iv ). Indeed, since Kv is a polyhedral cone, if Z ∈ / Kv , one of the homogeneous linear inequalities deﬁning Kv is violated on Z and so we obtain a point u ∈ (V ∗ )∗ = V such that Zi (u) ≤ 0 for all i ∈ Iv but Z(u) > 0. But then for a suﬃciently small g > 0 the point v + gu ∈ P and Z(v + gu) > α, which is a contradiction. Hence G Z= λi Zi for some λi ≥ 0. i∈Iv

Substituting v we conclude that α=

G

λi α i

i∈Iv

and hence Fv ⊂ F . Suppose now that there is a point x ∈ F \ Fv . Then for some i ∈ Iv we have Zi (x) < αi . This shows that the set D > G = x ∈ F : Zi (x) = αi is a proper face of F and since v ∈ G we obtain a contradiction with the fact that v lies in the relative interior of F . ! For brevity, we often say “interior” instead of “relative interior”. Problems 1◦ . Prove that a polyhedron has ﬁnitely many faces. 2. Let A ⊂ V be a closed convex set. Let Z : V −→ R be a non-zero functional and > let α be a numberD such that Z(x) ≤ α for all x ∈ A. The set F = x ∈ A : Z(x) = α is called a face of A. Prove that if A has ﬁnitely many faces, then A is a polyhedron. 3◦ . Prove that a vertex of a polyhedron is a face of it.

4. THE STRUCTURE OF POLYHEDRA

39

4. Let P1 , P2 ⊂ V be non-empty polyhedra and let Q = P1 + P2 be their Minkowski sum. Prove that every face F of Q can be written in the form F = G + H, where G is a face of P1 and H is a face of P2 . 5. Let P1 , P2 ⊂ V be non-empty polyhedra and let Q = P1 ∩ P2 . Prove that every vertex v of Q can be written as v = F1 ∩ F2 , where F1 is a face of P1 , F2 is a face of P2 and dim F1 + dim F2 ≤ dim V . 6. How many faces does the d-dimensional cube > D Id = (ξ1 , . . . , ξd ) : 0 ≤ ξi ≤ 1 for i = 1, . . . , d have? 7. (The Birkhoﬀ – von Neumann Theorem). In the space of n × n matrices X = (xij ), let us consider the set P of matrices satisfying the equations n G

xij = 1 for i = 1, . . . , n,

j=1

n G

xij = 1 for j = 1, . . . , n

i=1

and inequalities

xij ≥ 0 for all i, j. Prove that P is a polytope of dimension (n − 1)2 and that its vertices are exactly the permutation matrices X having exactly one “1” entry and (n − 1) “0” entries in every row and column. Hint: Prove that among the entries of every vertex of P there have to be at least (n − 1)2 zeros. 8. Let us choose positive integers r1 , . . . , rm and c1 , . . . , cn such that m G

ri =

i=1

n G

cj .

j=1

In the space of m × n matrices X = (xij ), let us consider the transportation polytope P consisting of the matrices X = (xij ) satisfying the equations n G

xij = ri

j=1

and inequalities

for i = 1, . . . , m,

n G

xij = cj

for j = 1, . . . , n

i=1

xij ≥ 0 for all i, j. Prove that if X is a vertex of P , then the entries of X are integer.

CHAPTER 5

Polarity Let us ﬁx a scalar product 0·, ·8 in V , just making V Euclidean space. Definition 5.1. Let A ⊂ V be a non-empty set. The polar A◦ of A is deﬁned by > D A◦ = x ∈ V : 0x, y8 ≤ 1 for all y ∈ A . Instead of introducing the scalar product in V , we could have deﬁned A◦ as a subset of the dual space V ∗ , but we don’t gain anything working in this generality. Let us consider some examples. If L ⊂ V is a subspace then L◦ = L⊥ is the orthogonal complement to L. Indeed, if 0x, y8 = > 0 for some x and some y ∈ L then by taking a suitable multiple λy of y, we get 0x, λy8 > 1 and so x ∈ / L◦ . Similarly, if K ⊂ V is a cone, we must have > D K ◦ = x ∈ V : 0x, y8 ≤ 0 for all y ∈ K . Finally, if P = conv (v1 , . . . , vm ) is a polytope, then > D P ◦ = x ∈ V : 0x, vi 8 ≤ 1 for i = 1, . . . , m is a polyhedron, cf. Figure 24. 0

0

0

0

0

0

Figure 24. A line, an angle, a square and their polars.

42

INTEGER POINTS IN POLYHEDRA

Before we state the main result of this chapter, we need a relatively straightforward lemma, also known as one of the separation theorems. Lemma 5.2. Let A ⊂ V be a closed convex set and let a ∈ / A be a point. Then there exists a non-zero vector u ∈ V and a number α such that 0u, a8 > α

and

0u, x8 < α

for all x ∈ A.

Geometrically, a point that does not belong to a closed convex set can be separated from the set by an aﬃne hyperplane, see Figure 25. A

H

y a

B

Figure 25. A closed convex set A, a point a ∈ / A and a separating aﬃne hyperplane. Proof of Lemma 5.2. Without loss of generality, we assume that a = 0. Let C · = C be the norm associated with the Euclidean structure in V , so CxC = 0x, x8 for all x ∈ V . Let us establish ﬁrst that there is a point y ∈ A nearest to a. Indeed, choosing a ball centered at a of a suﬃciently large radius r, > B= x∈V :

D Cx − aC ≤ r ,

we observe that a point y ∈ A ∩ B closest to a (which necessarily exists because the intersection A ∩ B is compact) is also closest to a among all points from A, see Figure 25. Now we deﬁne u = y and pick any 0 < α < CyC2 . The halfspace > H− = x ∈ V :

0u, x8 ≤ α

D

5. POLARITY

43

cannot contain any point from A. Indeed, suppose that there is a point z ∈ A ∩ H − . Since A is convex, the point yO = (1 − g)y + gz belongs to A for any 0 ≤ g ≤ 1. On the other hand, we have CyO C2 = (1 − g)2 CyC2 + g2 CzC2 + 2(1 − g)g0y, z8. Since 0y, z8 < CyC2 , we conclude that for a suﬃciently small g > 0 the point yO will be closer to a than y, which is a contradiction. ! Here is the main result of this chapter. Theorem 5.3. Let us ﬁx Euclidean space V . (1) If A ⊂ V is a non-empty set, then A◦ ⊂ V is a closed convex set containing the origin; (2) Suppose that A ⊂ V is a closed convex set containing the origin. Then (A◦ )◦ = A; (3) There exists a unique linear transformation D : C(V ) −→ C(V ) such that D([A]) = [A◦ ] for all non-empty closed convex sets A; (4) If P ⊂ V is a polyhedron, then P ◦ ⊂ V is a polyhedron. Proof. Since A◦ is the intersection of closed halfspaces, it is closed and convex. Clearly, 0 ∈ A◦ , which proves Part (1). To prove Part (2) we note that the inclusion A ⊂ (A◦ )◦ follows directly from the deﬁnition (it does not require A to be closed or convex). Thus we have to prove that (A◦ )◦ ⊂ A. Suppose that there is a point a ∈ (A◦ )◦ such that a ∈ / A. Since A is closed and convex, by Lemma 5.2 there is a non-zero vector u and a number α such that 0u, x8 < α for all x ∈ A and 0u, a8 > α. Since 0 ∈ A, we conclude that α > 0, and, scaling u, if necessary, we can assume that α = 1. It follows then that u ∈ A◦ . Since a ∈ (A◦ )◦ , we must have 0u, a8 ≤ 1, which is a contradiction. This concludes the proof of Part (2). To establish Part (3), we obtain the operator D as a limit of some operators DO . For g > 0, let us deﬁne the “Green function” GO : V × V −→ R by 1 1 if 0x, y8 < 1 + g, GO (x, y) = 0 otherwise. For f ∈ C(V ), g > 0 and y ∈ V , let us deﬁne a function gO,y : V −→ R by gO,y (x) = f (x)GO (x, y). We claim that gO,y ∈ C(V ). Indeed, let us

44

INTEGER POINTS IN POLYHEDRA

suppose that f = [A], where A ⊂ V is a non-empty closed convex set. Let us deﬁne the halfspace > D HO,y = x : 0x, y8 ≥ 1 + g . Then

gy,O = [A] − [A ∩ HO,y ] ⊂ C(V ),

cf. Figure 26.

A

− =

A

Hε , y

Figure 26. A set, its intersection with a halfspace, and the diﬀerence. By linearity, gy,O ∈ C(V ) for any f ∈ C(V ). Thus we can deﬁne the Euler characteristic χ (gy,O ). Now we let hO = DO (f ) by hO (y) = χ (gO,y ) . Clearly, the transformation DO is a well-deﬁned linear transformation DO : C(V ) −→ image of DO . Furthermore, we observe that for f = [A], where A ⊂ V is a non-empty closed convex set, the function hO = DO (f ) is deﬁned by 1 1 if 0x, y8 < 1 + g for all x ∈ A, hO (y) = 0 if 0x, y8 ≥ 1 + g for some x ∈ A. Next, we observe that in the case of f = [A] the pointwise (over y ∈ V ) limit lim hO (y) O−→0+

exists and is equal to [A◦ ]. By linearity, this limit exists for any f ∈ C(V ) and allows us to deﬁne D by letting D(f ) = h, where h(y) = lim DO (f )(y). O−→0+

This completes the proof of Part (3).

5. POLARITY

45

To prove Part (4), we note that by Theorem 4.13, a non-empty polyhedron P can be represented as the Minkowski sum P = M + K + L, where M ⊂ V is a polytope, K ⊂ V is a polyhedral cone without lines and L ⊂ V is a subspace. Since M = conv (v1 , . . . , vm ), we have > D M ◦ = x : 0x, vi 8 ≤ 1 for i = 1, . . . , m , so M ◦ is a polyhedron. Furthermore, by Theorem 4.11 we may write K = co (u1 , . . . , un ) and hence > D ◦ K = x : 0x, ui 8 ≤ 0 for i = 1, . . . , n . In particular, K ◦ is a polyhedral cone. Finally, L◦ = L⊥ . Now we claim that (5.3.1)

P ◦ = M ◦ ∩ K ◦ ∩ L◦

and P ◦ is a polyhedron. Let us prove ﬁrst that P ◦ ⊂ M ◦ , K ◦ , L◦ . Let us choose any x ∈ P ◦ . Then 0x, y8 ≤ 1 for all y ∈ P and since M ⊂ P we conclude that P ◦ ⊂ M ◦ . Let us choose any u ∈ K and any y ∈ P . Then y + λu ∈ P for all λ ≥ 0 and hence for any x ∈ P ◦ we must have 0u, x8 ≤ 0 and so x ∈ K ◦ . This proves that P ◦ ⊂ K ◦ . Finally, let us choose any u ∈ L and any y ∈ P . Then y + λu ⊂ P for all λ ∈ R and hence for any x ∈ P ◦ we must have 0u, x8 = 0. This proves that P ◦ ⊂ L◦ . Thus we conclude that P ◦ ⊂ M ◦ ∩ K ◦ ∩ L◦ . On the other hand, let us pick any x ∈ M ◦ ∩ K ◦ ∩ L◦ . Then 0x, y8 ≤ 1 for all x ∈ M , 0x, u8 ≤ 0 for all u ∈ K and 0x, z8 = 0 for all u ∈ L. Since every point of P can be written as y + u + z for some y ∈ M , u ∈ K and z ∈ L, it follows that x ∈ P ◦ . By (5.3.1), P ◦ is the intersection of three polyhedra and hence is a polyhedron. ! Part (3) of Theorem 5.3 asserts that linear relations among indicators of polyhedra are preserved under polarity, see Figure 27 for a simple example.

46

INTEGER POINTS IN POLYHEDRA

0

0

= =

+

0

−

0

0

0

+

0

−

0

Figure 27. Relations among indicators of polyhedra and their polars. In some way, one can think of polarity as of the Fourier transform with respect to the Euler characteristic instead of the Lebesgue integral. Suppose that K1 , K2 ⊂ V are polyhedral cones. Then [K1 ] · [K2 ] = [K1 ∩ K2 ] and [K1 ] ∗ [K2 ] = [K1 + K2 ]. On the other hand, one can show that (K1 ∩ K2 )◦ = K1◦ + K2◦ and (K1 + K2 )◦ = K1◦ ∩ K2◦ . Hence for functions f and g which are linear combinations of indicators of polyhedral cones we have D(f g) = D(f ) ∗ D(g). In other words, the polarity map D of Theorem 5.3 transforms the product into the convolution, just what the Fourier transform is supposed to do. Remark 5.4. One important property of the polarity correspondence which we will use repeatedly is that if a set A contains a line (in the direction of u) then A◦ lies in a hyperplane (orthogonal to u) and vice versa, if A lies in a hyperplane (orthogonal to a non-zero vector u) then A◦ contains a line in the direction of u. Problems ◦ 1. Let A ⊂ V be a> non-empty set such D that A = A. Prove that A is the unit ball, A = x ∈ V : CxC ≤ 1 .

2. Let Id ⊂ Rd , > Id = (ξ1 , . . . , ξd ) :

−1 ≤ ξi ≤ 1 for i = 1, . . . , d

D

5. POLARITY

47

be a cube and let > Od = (ξ1 , . . . , ξd ) :

d G

D |ξi | ≤ 1

i=1

be a cross-polytope (octahedron). Prove that Id◦ = Od and that Od◦ = Id . 3. Find the eigenvalues of the operator D from Theorem 5.3. 4. Modify the function GO from the proof of Theorem 5.3 to its “simpler” version: 1 ˜ y) = 1 if 0x, y8 ≤ 1, G(x, 0 otherwise ˜ be the transformation deﬁned by and let D

A 2 ˜ y)f (x) . ˜ ) = h provided h(y) = χ G(x, D(f

˜ : C(V ) −→ image D ˜ is a linear transformation and comProve that D ˜ pute D([B]), where B ⊂ V is a ball of radius 1 centered at the origin. 5. Prove that (K1 ∩ K2 )◦ = K1◦ + K2◦ if K1 , K2 ⊂ V are polyhedral cones.

CHAPTER 6

Tangent cones. Decompositions modulo polyhedra with lines Let us look at polyhedra more closely. In particular, we will be interested in what a polyhedron looks like in a neighborhood of a point. Definition 6.1. Let P ⊂ V be a non-empty polyhedron and let v ∈ P be a point. We deﬁne the tangent cone of P at v by > D tcone(P, v) = v + y : v + gy ∈ P for some g > 0 , see Figure 28. We deﬁne the cone of feasible directions at v by > D fcone(P, v) = y ∈ V : v + gy ∈ P for some g > 0 . Thus tcone(P, v) = fcone(P, v) + v. A C

B

A B C

Figure 28. A polytope and its tangent cones. Note that the tangent cone is not a cone as deﬁned by Deﬁnition 4.9 but rather a translation of such a cone (namely, the cone of feasible directions), since the vertex of the tangent cone is not necessarily at the origin. Alternatively, one can deﬁne the tangent cone and the cone of feasible directions as follows.

50

INTEGER POINTS IN POLYHEDRA

Suppose that P ⊂ V is deﬁned by ﬁnitely many linear inequalities > D P = x ∈ V : Zi (x) ≤ αi , i ∈ I , where Zi : V −→ R are linear functions. For a point v ∈ P , let > D Iv = i ∈ I : Zi (v) = αi be the set of the inequalities active on v, cf. Lemma 4.6. Then > D tcone(P, v) = x ∈ V : Zi (x) ≤ αi for i ∈ Iv and > D fcone(P, v) = y ∈ V : Zi (y) ≤ 0 for i ∈ Iv . It is seen now that the tangent cone and the cone of feasible directions are polyhedra. Here is a simple but useful observation. Lemma 6.2. Let V and W be ﬁnite-dimensional real vector spaces, let T : V −→ W be a linear transformation, let P ⊂ V be a polyhedron and let v ∈ P be a point. Furthermore, let Q = T (P ), Q ⊂ W , be the image of P and let w = T (v), w ∈ Q, be the image of v. Then M / tcone(Q, w) = T tcone(P, v) . Proof. Without loss of generality, we assume that v = 0 and w = 0. If x ∈ tcone(Q, w) then gx ∈ Q for some g > 0. Hence there is y ∈ P such that T (y) = gx. Then g−1 y ∈ tcone(P, 0) and x = T (g−1 y). On the other hand, if y ∈ tcone(P, 0) then gy ∈ P for some g > 0. Therefore T (gy) ∈ Q and hence T (y) ∈ tcone(Q, 0) as desired. ! In what follows, polyhedra containing lines will be largely irrelevant. Hence we introduce the following deﬁnition. Definition 6.3. For f, g ∈ P(V ) we say that f ≡g

modulo polyhedra with lines

provided the diﬀerence f − g is a linear combination of indicators of polyhedra with lines. Also, we say f ≡g

modulo polyhedra in proper subspaces

provided the diﬀerence f − g is a linear combination of indicators of polyhedra lying in proper subspaces.

6. TANGENT CONES

51

Note that the functions f ≡ 0 modulo polyhedra with lines form an ideal in P(V ) with respect to the convolution ∗. The main result of this section is that modulo polyhedra with lines, every polyhedron is just the sum of its tangent cones, see Figure 29. A D

A B

+

C

+

B

+

C

D

Figure 29. A polygon is the sum of its tangent cones modulo polygons with lines. Theorem 6.4. Let P ⊂ V be a polyhedron and let Vert(P ) be the set of vertices of P . Then G [P ] ≡ [tcone(P, v)] modulo polyhedra with lines. v∈Vert(P )

Proof. Without loss of generality, we may assume that P is not empty. First, we prove the result for the standard simplex, cf. proof of Theorem 4.4. Namely, let Δ ⊂ Rm be the polyhedron , 1 m G λi = 1 and λi ≥ 0 for i = 1, . . . , m . Δ = (λ1 , . . . , λm ) : i=1

We consider Δ as a polyhedron in the aﬃne hyperplane A deﬁned by the equation λ1 + · · · + λm = 1. Let Hi ⊂ A be the halfspace deﬁned by the inequality λi ≥ 0. Let ei ⊂ A be the point such that λi = 1 and all other coordinates are equal to 0. We note that e1 , . . . , em are the vertices of Δ. We have m m F ' Δ= Hi and Hi = A. i=1

i=1

Using the inclusion-exclusion formula, we write G F (6.4.1) [A] = (−1)|I|−1 [HI ] , where HI = Hi I⊂{1,...,m}

i∈I

and the sum is taken over all non-empty subsets I ⊂ {1, . . . , m}.

52

INTEGER POINTS IN POLYHEDRA

We note that if I = {1, . . . , m} then HI = Δ and if I = {1, . . . , m}\ {i} then HI = tcone(Δ, ei ). Finally, if i, j ∈ / I for a pair of distinct i and j, then HI contains the line through the points ei and ej , cf. Figure 30.

= C

A

−

A

+

A

B

+ CB

− + C

B

− C

A B

Figure 30. The plane is the sum of the three halfplanes bounded by the sides of a triangle minus the angles of the triangle plus the triangle itself. Next, we prove Theorem 6.4 for polytopes. Let P = conv (v1 , . . . , vm ), P ⊂ V be the convex hull of m distinct points v1 , . . . , vm that are the vertices of P . Let us consider the standard simplex Δ ⊂ Rm and the linear transformation m

T : R −→ V,

T (λ1 , . . . , λm ) =

m G

λi v i .

i=1

Since linear transformations preserve linear dependencies among the indicator functions of polyhedra, (see Theorem 3.1), applying T to (6.4.1), we get G [T (Δ)] = (−1)|I|−1 [T (HI )] . I⊂{1,...,m}

Now, T (Δ) = P and for I = {1, . . . , m} \ {i} we have by Lemma 6.2: T (HI ) = T (tcone(Δ, ei )) = tcone(P, vi ). Furthermore, if I does not contain distinct i and j then T (HI ) contains the line through the points vi = T (ei ) and vj = T (ej ).

6. TANGENT CONES

53

Now, we prove the result for an arbitrary non-empty polyhedron P . If P contains no vertices, then P contains a line and the result is trivial. If P contains no lines, then by Theorem 4.12 we can write P = Q + K, where Q = conv (Vert(P )) and K is the recession cone of P . As we have already proved, G [Q] ≡ [tcone(Q, v)] modulo polyhedra with lines. v∈Vert(P )

Since Minkowski addition preserves linear relations among indicator functions of polyhedra, see Theorem 3.5, we have G [Q + K] ≡ [tcone(Q, v) + K] modulo polyhedra with lines. v∈Vert(P )

It remains to notice that tcone(Q, v) + K = tcone(P, v).

!

Recall (Remark 5.4) that if P contains a line, then P ◦ lies in a proper subspace (orthogonal to that line) and if P lies in a proper subspace, then P ◦ contains a line (orthogonal to that subspace). Since (P ◦ )◦ = P if P contains the origin, we come to the following useful corollary. Corollary 6.5. Let Pi ⊂ V , i ∈ I, be a ﬁnite family of polyhedra such that 0 ∈ Pi for all i ∈ I and let αi , i ∈ I, be numbers. Then G αi [Pi ] ≡ 0 modulo polyhedra with lines, i∈I

respectively modulo polyhedra in proper subspaces if and only if G αi [Pi◦ ] ≡ 0

modulo polyhedra in proper subspaces,

i∈I

respectively modulo polyhedra with lines. One can ask what is the analogue of Theorem 6.4 for decompositions modulo polyhedra in proper subspaces. It is a familiar decomposition of polyhedra into pyramids, see Figure 31.

54

INTEGER POINTS IN POLYHEDRA

0

+

0

+

0

+ 0

0

+

+

+

Figure 31. Above: representing the square as the sum of four triangles modulo polyhedra in proper subspaces. Below: representing the polar as the sum of four angles modulo polyhedra with lines. What happens if instead of the tangent cones we add the cones of feasible directions? Here is the answer in the case of a bounded polyhedra (for general polyhedra the question is stated as an exercise at the end of the section). Theorem 6.6. Let P ⊂ V be a non-empty polytope and let Vert(P ) be the set of its vertices. Then G [fcone(P, v)] ≡ [0] modulo polyhedra with lines. v∈Vert(P )

Proof. Let us consider the polar > D fcone(P, v)◦ = c ∈ V : 0c, x8 ≤ 0 for all x ∈ fcone(P, v) . We observe that

> fcone(P, v)◦ = c ∈ V :

D max0c, x8 = 0c, v8 , x∈P

see Figure 32. In words: the polar of the cone fcone(P, v) of feasible directions consists of all vectors c ∈ V for which the maximum value of the linear function Z(x) = 0c, x8 on P is attained at v. Since for every c ∈ V the maximum of the linear function 0c, x8 for x ∈ P is attained at some vertex v of P , we have ' fcone(P, v)◦ = V. v∈Vert(P )

6. TANGENT CONES

55

Moreover, for diﬀerent vertices v, u ∈ Vert(P ) the intersection fcone(P, v)◦ ∩ fcone(P, u)◦ lies in the hyperplane 0c, u − v8 = 0. Therefore, G [fcone(P, v)]◦ ≡ [V ] modulo polyhedra in proper subspaces. v∈Vert(P )

Applying Corollary 6.5, we obtain the desired result. u

fcone ( P, w )

!

fcone (P, v)

P v fcone (P, u )

w

0

o

fcone(P, u)

0 fcone(P, w)

o

fcone (P,v)o

Figure 32. A triangle P with the vertices v, u, and w, the cones of feasible directions at the vertices and their polars. Problems 1◦ . Let P be a non-empty polyhedron without lines, let Q be the convex hull of the set of vertices of P and let KP be the recession cone of P . Prove that for every vertex v of P we have tcone(P, v) = tcone(Q, v) + KP . 2. Let P be a non-empty polyhedron without lines and let KP be its recession cone. Prove that G [fcone(P, v)] ≡ [KP ] modulo polyhedra with lines. v∈Vert(P )

3◦ . Let P be a polyhedron, let F ⊂ P be a face of P and let v ∈ F be a point in the interior of F . Prove that the cones tcone(P, v) and fcone(P, v) do not depend on the choice of v in the interior of P . Hence

56

INTEGER POINTS IN POLYHEDRA

we can talk of the tangent cone tcone(P, F ) at a face and the cone of feasible directions fcone(P, F ) at a face. 4∗ . Let P be a polytope. Prove the following Gram – Brianchon Theorem: G [P ] = (−1)dim F [tcone(P, F )], F

where the sum is taken over all non-empty faces F of P including P itself. Hint: What is the dual identity?

CHAPTER 7

Open polyhedra We start with a useful formula. Theorem 7.1. Let P ⊂ V be a d-dimensional polytope. Then, for the (relative) interior of P , we have χ[int P ] = (−1)d . Proof. The proof is based on Corollary 2.5. We proceed by induction on the dimension d. The result holds for d = 1 as Figure 12 demonstrates. Suppose that d > 1. Let us consider the family of parallel hyperplanes Hτ consisting of the points with the last coordinate equal to τ . By Corollary 2.5, J G% χ([int P ∩ Hτ ]) − lim χ([int P ∩ Hτ −O ]) . χ([int P ]) = O−→0+

τ ∈R

Now we observe that for each τ the intersection Hτ ∩ int P is either the interior of a (d − 1)-dimensional polytope in Hτ or empty. The only non-zero contribution to the sum is the term corresponding to the largest value of the last coordinate on P and, by the induction hypothesis, it is equal to 0 − (−1)d−1 = (−1)d , as desired, see Figure 33. ! Hτ

0 −0 = 0

Hτ

0 −( −1)

Hτ

(−1) − ( −1 ) = 0

Hτ Hτ

d

d −1

d −1

P

= ( −1 ) d −1

0 −0 = 0 0 −0 = 0

Figure 33. The interior of a polytope P , the hyperplanes Hτ and their contributions to χ([int P ]). Here is a famous corollary, known as the Euler – Poincar´e formula.

58

INTEGER POINTS IN POLYHEDRA

Corollary 7.2. Let P ⊂ Rd be a d-dimensional polytope. For k = 0, . . . , d − 1 let fk be the number of k-dimensional faces of P and let fd = 1 (thus we agree that P is a face of itself ). Then d G

(−1)k fk = 1.

k=0

Proof. As follows by Theorem 4.15, every point x ∈ P lies in the relative interior of some face F of P , including, possibly, P itself. Hence we can write G [P ] = [int F ], F

where the sum is taken over all faces F of P including P , see Figure 34. Applying the Euler characteristic and Theorem 7.1, we conclude G G 1 = χ([P ]) = χ([int F ]) = (−1)dim F , F

F

and the proof follows.

!

=

+

+

Figure 34. Writing a square as the sum of the interiors of faces. Next, we turn to yet another identity in the algebra of polyhedra, which relates the indicator of a cone and the indicator of the interior of the opposite cone, see Figure 35. We call it the reciprocity relation. Theorem 7.3. Let K ⊂ V be a d-dimensional polyhedral cone. Then [K] ≡ (−1)d [− int K]

modulo polyhedra with lines.

Proof. First, we prove the identity in a very special situation of the cone K = Rn+ in Rn . For i = 1, . . . , n, let Hi+ be the closed halfspace deﬁned by the inequality xi ≥ 0, and let Hi− be the complementary open halfspace deﬁned by the inequality xi < 0. Then [Rn+ ]

=

n i=1

[Hi+ ]

n M / G [Rn ] − [Hi− ] = (−1)|I| [Hi− ], = i=1

I

i∈I

7. OPEN POLYHEDRA

59

where the sum is taken over all subsets I ⊂ {1, . . . , n}. It is not hard to see that if for some index k we have k ∈ / I then the corresponding term is a linear combination of indicators of polyhedra, each of which contains lines in the direction of the kth standard basis vector ek . Since n -

[Hi− ] = − int Rn+ ,

i=1

we get (7.3.1)

[Rn+ ] = (−1)n [− int Rn+ ] +

G

gI PI ,

I

where gI ∈ {−1, 1} and PI are polyhedra each of which contains a line in the direction of some basis vector ek of Rn . Next, suppose that K ⊂ V is a d-dimensional cone. Without loss of generality we assume that K is a pointed cone, so K = co (u1 , . . . , un ) , where uk ∈ V are the vertices of a polytope P = K ∩ H for some aﬃne hyperplane H, see Theorem 4.11. It is not hard to see that 1 n , G int K = λi ui : λi > 0 for i = 1, . . . , n . i=1

Let us consider the linear transformation (projection) T : Rn −→ V deﬁned by T (ek ) = uk for k = 1, . . . , n. Hence T (Rn+ ) = K. By Theorem 3.1, there is a linear transformation T : P (Rn ) −→ P (V ) such that T ([P ]) = [T (P )] for any non-empty polyhedron P ⊂ Rn . Moreover, as is described in the proof of Theorem 3.1, the transformation T is deﬁned as follows: we have 2 A T (f ) = h where h(y) = χ [T −1 (y)]f for all y ∈ V. In words: to compute the value of h = T (f ) at a given point y ∈ V , we consider the restriction of f onto the inverse image T −1 (y) and compute its Euler characteristic. Now let us choose f = [− int Rn+ ]. 4 B Then, the restriction of f onto T −1 (y) is T −1 (y) ∩ (−Rn+ ) , which is the 2indicator Aof the interior of an (n − d)-dimensional polytope if y ∈ T − int Rn+ and 0 otherwise. By Theorem 7.1, for h = T (f ) we have h = (−1)n−d [T (− int Rn+ )] = (−1)n−d [− int K].

60

INTEGER POINTS IN POLYHEDRA

Applying T to both sides of (7.3.1), we conclude G [K] = (−1)d [− int K] + gI [T (PI )], I

where each polyhedron T (PI ) contains a line in the direction of T (ek ) = uk for some basis vector ek of Rn , which completes the proof. !

−int K 0 K

Figure 35. Cones K and − int K. Let us look at the interior of a polytope more closely. As the indicator of a polytope is the sum of the indicators of its tangent cones modulo polyhedra with lines, the indicator of the interior of the polytope is the sum of the indicators of interiors of its tangent cones, see Figure 36. B

B C

+

A D

C

A D

Figure 36. Representing the interior of a polytope as the sum of the interiors of its tangent cones at the vertices modulo polyhedra with lines. Theorem 7.4. Let P ⊂ V be a polytope. Then G [int P ] ≡ [int tcone(P, v)] modulo polyhedra with lines. v∈Vert(P )

7. OPEN POLYHEDRA

61

Proof. The proof is a combination of the proof of Theorem 6.4 and Theorem 7.3. First, we establish the result for the simplex and then apply a suitable projection. ! Problems 1. Prove Theorem 7.4. 2. Let P ⊂ V be an unbounded d-dimensional polyhedron without lines. Let fi0 (P ) be the number of bounded i-dimensional faces of P , let fi∞ be the number of unbounded i-dimensional faces of P and let fi (P ) = fi0 (P ) + fi∞ (P ) be the total number of i-dimensional faces of P (we consider P as a face of itself). Prove that d d d−1 G G G i+1 ∞ i 0 (−1) fi (P ) = 1, and (−1)i fi (P ) = 0. (−1) fi (P ) = 1, i=0

i=0

i=1

3. Let P ⊂ V be a d-dimensional polytope. Prove that G (−1)dim F [F ], (−1)d [int P ] = F

where the sum is taken over all faces F of P , including P . 4. Let P be an unbounded polyhedron without lines. Prove that χ([int P ]) = 0.

CHAPTER 8

The exponential valuation Let V be Euclidean space with the scalar product 0·, ·8 and the Lebesgue measure dx. The volume of a polytope clearly extends to a valuation on the algebra Pb (V ) of bounded polyhedra in V , namely, ; f 9−→ f dx. V

Can this valuation be extended to the algebra P(V ) of all polyhedra, or, in other words, is there a way to deﬁne the “volume” of an unbounded polyhedron? In this section we provide a surprising answer to this question. It turns out that the correct extension of the volume to unbounded polyhedra is not a number but a function φ : V −→ R such that φ(0) = vol P if P is bounded. If P is unbounded, 0 may become a singular point of φ. Summarizing, we construct a valuation P(V ) −→

meromorphic functions on V

such that the restriction of this valuation to Pb (V ) specializes to volume at 0. We consider the integral ; (8.1) e%c,x5 dx, P

where P ⊂ V is a polyhedron and c ∈ V is a vector. We look at integral (8.1) as a function of c. Sometimes it is convenient to consider the integral for complex c as well, c = a + ib, where a, b ∈ V and 0c, x8 = 0a, x8 + i0b, x8. If P is bounded, the integral (8.1) converges for every c and for c = 0 its value is equal to vol P , the volume of P . If P is not bounded, the integral may or may not converge. Here is our basic example. Example 8.2. Let us identify V = Rd and let P = Rd+ be the non-negative orthant, consisting of the points with non-negative

64

INTEGER POINTS IN POLYHEDRA

coordinates. Then, for c = (c1 , . . . , cd ) we have ; (8.2.1)

Rd+

e

%c,x5

dx =

d ; -

+∞ 0

i=1

e

c i xi

d 1 , dxi = −ci i=1

provided c1 , . . . , cd < 0. Denoting by int Rd+ the interior of the non-negative orthant, we observe that the integral converges absolutely for all c ∈ − int Rd+ and uniformly on compact subsets of int −Rd+ . More generally, assuming d = dim V , let u1 , . . . , ud be a basis of V and let K = co (u1 , . . . , ud ). Let |u1 ∧ · · · ∧ ud | be the volume of the parallelepiped spanned by u1 , . . . , ud , that is, let 1 d , G |u1 ∧ · · · ∧ ud | = vol αi ui : 0 ≤ αi ≤ 1 for i = 1, . . . , d . i=1

Then, for all c ∈ int K ◦ we have ; (8.2.2)

K

e%c,x5 dx = |u1 ∧ · · · ∧ ud |

d i=1

1 , 0−c, ui 8

where the integral converges absolutely and uniformly on compact subsets of the interior of K ◦ . Integral (8.2.2) reduces to (8.2.1) by a change of variables. The calculation of Example 8.2 can be extended to an arbitrary pointed cone K. We remark that from the proof of Theorem 4.11, the interior of the cone K ◦ is non-empty. Lemma 8.3. Let K ⊂ V be a pointed cone. Then, for every c ∈ int K ◦ the integral ; e%c,x5 dx K

converges absolutely and uniformly on compact subsets of int K ◦ to a rational function φ(K, c) of the type φ(K, c) =

G I

αI

i∈I

1 , 0−c, ui 8

where u1 , . . . , um ∈ V are non-zero vectors, the sum is taken over some d-subsets I of the set {1, . . . , m} and αI > 0 are numbers.

8. THE EXPONENTIAL VALUATION

65

Proof. By Theorem 4.11 we can represent the cone K as K = co (u1 , . . . , um ) , where u1 , . . . , um are the vertices of the polytope P = K ∩ H for an aﬃne hyperplane H such that 0 ∈ / H. Without loss of generality we assume that span (u1 , . . . , um ) = V , since otherwise K is not a fulldimensional cone and the integral is identically 0. Now we triangulate P , that is, represent it as a union of (d − 1)dimensional simplices such that every two simplices intersect by their common face, cf. Figure 37. In our case, ΔI = conv (ui : i ∈ I) is a (d − 1)-dimensional simplex if |I| = d and the vectors {ui : i ∈ I} form a basis of V . One can establish the existence of such a triangulation by induction on the number m of vertices of P or by “dualizing” Theorem 6.4, see also Corollary 6.5 and Figure 37. 0

u4

u1

P u2

u2

u1 u3

u3

P

u

4

Figure 37. A pointed cone with a base P and a triangulation of P . Thus the cone K is represented as a union of the cones KI = co (ui : i ∈ I) , where for each I ⊂ {1, . . . , m} the vectors ui : i ∈ I form a basis of V . Furthermore, the dimension of the intersection of every two distinct cones KI is at most d − 1. This shows that ; G; %c,x5 e%c,x5 dx e = K

I

KI

and the proof follows by formula (8.2.2).

!

We observe that if we shift the cone K − 9 → K + v by a vector, the integral gets multiplied: ; ; %c,x5 %c,v5 e dx = e e%c,x5 dx. K+v

K

66

INTEGER POINTS IN POLYHEDRA

The main result of this section is the following theorem, proved by A. Khovanskii and A. Pukhlikov [KP92] and, independently, by J. Lawrence [L91a]. Theorem 8.4. Let V be Euclidean space, dim V = d, and let M(V ) be the space of functions on V spanned by functions of the type f (c) = e%c,v5

d i=1

1 0−c, ui 8

for

c ∈ V,

where v ∈ V and u1 , . . . , ud is a basis of V . There exists a linear transformation (valuation) Φ : P(V ) −→ M(V ) such that the following holds: (1) Let P ⊂ V be a non-empty polyhedron without lines and let K ⊂ V be its recession cone. Then for all c ∈ int K ◦ the integral ; e%c,x5 dx P

converges absolutely and uniformly on compact subsets of int K ◦ to the function φ(P, c) = Φ([P ]) ∈ M(V ). (2) If P contains a line, then Φ([P ]) = 0. Before we prove Theorem 8.4, let us consider some of its corollaries. Let P ⊂ V be a non-empty polyhedron without lines. Since by Theorem 6.4 we have G [P ] ≡ [tcone(P, v)] modulo polyhedra with lines, v∈Vert(P )

by Theorem 8.4 we should have G Φ[P ] =

Φ[tcone(P, v)].

v∈Vert(P )

This identity is known as Brion’s Theorem. It was ﬁrst proved for rational polytopes (to be discussed later) in [Br88], for general polytopes in [Ba91], and, independently, for general polyhedra in [KP92] and [L91a].

8. THE EXPONENTIAL VALUATION

67

Moreover, since the tangent cone tcone(P, v) is the translation v + fcone(P, v) of a pointed polyhedral cone of feasible directions fcone(P, v), we can use Lemma 8.3 to conclude that Φ[tcone(P, v)] = e%c,v5 fv (c), where fv (c) is a homogeneous rational function of c of degree −d. Moreover, fv (C) depends only on the cone of feasible directions at v. For example, in Figure 38, we must have ; e c1 e c2 1 + + for c = (c1 , c2 ) e%c,x5 dx = (−c1 )(−c2 ) c1 (c1 − c2 ) c2 (c2 − c1 ) Δ and ; 2Δ

e%c,x5 dx =

1 e2c1 e2c2 + + (−c1 )(−c2 ) c1 (c1 − c2 ) c2 (c2 − c1 )

for c = (c1 , c2 ).

2 1 0

2Δ

Δ 1

2

0

Figure 38. Triangle Δ with the vertices (0, 0), (1, 0), and (0, 1) and triangle 2Δ with the vertices (0, 0), (2, 0), and (0, 2). This calls for the following deﬁnition. Definition 8.5. Let P ⊂ V be a non-empty polyhedron and let Z : V −→ R be a linear functional such that max Z(x) < +∞. x∈P

Let

>

FP (P ) = x ∈ P :

Z(x) = mP

D

denote the corresponding face of P . We say that two polyhedra P1 , P2 ⊂ V are strongly combinatorially isomorphic if for any non-zero linear functional Z : V −→ R we have max Z(x) < +∞ if and only if x∈P1

max Z(x) < +∞ x∈P2

68

INTEGER POINTS IN POLYHEDRA

and dim FP (P1 ) = dim FP (P2 ) for all such Z. Figure 39 shows an example of two strongly combinatorially isomorphic polygons.

P2

P

1

Figure 39. Strongly combinatorially isomorphic polygons P1 and P2 . 8.6 Useful observations. If P1 and P2 are strongly combinatorially isomorphic polyhedra, then the cones of feasible directions fcone(P, v1 ) and fcone(P, v2 ) at their respective vertices coincide while the tangent cones tcone(P1 , v1 ) and tcone(P2 , v2 ) diﬀer only by a translation. Therefore, for any family Pτ of strongly combinatorially isomorphic polyhedra with the vertices v1 (τ ), . . . , vm (τ ) we must have (8.6.1)

Φ ([Pτ ]) =

m G

e%c,vi (τ )5 fi (c),

i=1

where fi (c) are rational functions which are independent of a particular polyhedron Pτ from the family. Here is another useful observation. Let us ﬁx a vector c ∈ V and suppose that a point v ∈ V moves with a constant speed v(τ ) = v0 (1 − τ ) + v1 τ

for 0 ≤ τ ≤ 1

from position v0 at τ = 0 to position v1 at τ = 1. Then ; 1 A 2 %c,v0 5 1 e − e%c,v1 5 e%c,v(τ )5 dτ = 0c, v0 − v1 8 0 if 0c, v0 − v1 8 = > 0 (8.6.2) and ; 1 e%c,v(τ )5 dτ = e%c,v0 5 = e%c,v1 5 if 0c, v0 − v1 8 = 0. 0

8. THE EXPONENTIAL VALUATION

69

Now we are ready to prove Theorem 8.4. Proof of Theorem 8.4. We proceed by induction on the dimension d = dim V . If d = 0, the result is clear. Suppose that d > 0. Let P ⊂ V be a non-empty polyhedron without lines. By Theorem 4.12 we have P = Q + K, where Q is the convex hull of the vertices of P and K = KP is the recession cone of P . First, we prove that for every c ∈ int K ◦ the integral ; e%c,x5 dx P

converges absolutely and uniformly on compact subsets of K ◦ to a function from M(V ). We note that every face of P is the Minkowski sum of a face of Q and a face of K. Let us choose a unit vector u ∈ − int K ◦ suﬃciently generic, so that u is not orthogonal to any face of P of positive dimension and let us slice V into aﬃne hyperplanes > D Hτ = x ∈ V : 0u, x8 = τ . Thus every hyperplane Hτ can be identiﬁed with a (d − 1)-dimensional space and hence Theorem 8.4 holds in every Hτ . Let Pτ = P ∩ Hτ . Because u ∈ − int K ◦ , every Pτ is bounded and Pτ = ∅ for all τ < τ0 , where τ0 = min0u, x8 = min0u, x8, x∈P

x∈Q

cf. Theorem 4.12. Thus we may write ; P

e

%c,x5

; dx =

+∞ τ0

%;

J Pτ

e

%c,y5

dy

dτ,

where dy is the Lebesgue measure in Hτ . The faces of every polytope Pτ are obtained by intersecting faces of P with Hτ , cf. Theorem 4.15. Next, we observe that there are ﬁnitely many points τ0 , τ1 , . . . , τn such that on every open interval (τk , τk+1 ) and on the ray (τn , +∞) the polytopes Pτ remain strongly combinatorially isomorphic, cf. Figure 40. Indeed, the points τk are deﬁned by the condition that Hτ passes through a vertex of P . To make notation uniform, let us agree that τn+1 = +∞.

70

INTEGER POINTS IN POLYHEDRA

P Hτ Hτ

0

Figure 40. A polyhedron P and hyperplanes Hτ . Let us choose a particular interval (τk , τk+1 ). For τ ∈ (τk , τk+1 ), let vi (τ ), i = 1, . . . , m, be the vertices of Pτ . As we remarked, the cone fcone (Pτ , vi (τ )) of feasible directions at vi (τ ) does not depend on τ and hence by the induction hypothesis, see also (8.6.1), there are rational functions fi (c) of degree −d + 1 such that for all τ ∈ (τk , τk+1 ), we have ; m G e%c,y5 dy = e%c,vi (τ )5 fi (c). Pτ

i=1

Moreover, since the vertices vi (τ ) of Pτ are intersections of edges of P and Hτ , each vertex vi (τ ) changes linearly with τ . Therefore, by (8.6.2), the integral J ; τk+1 ; τk+1 %; m G %c,y5 e%c,vi (τ )5 dτ fi (c) e dy dτ = τk

Pτ

τk

i=1

as a function of c belongs to the space M(V ). It remains to consider the integral J ; ; +∞ %; m G %c,y5 e dy dτ = fi (c) τn

Pτ

i=1

+∞

τn

e%c,vi (τ )5 dτ.

If P is not bounded, then for τ > τn , the vertices vi (τ ) of Pτ are the intersection of unbounded edges of P and Hτ . Hence for each such a vertex, we have vi (τ ) = vi (0) + (τ − τn )w

for all τ ≥ τn

and some w ∈ K. Therefore, ; +∞ e%c,vi (0)5 e%c,vi (τ )5 dτ = , 0−c, w8 τn where 0c, w8 < 0 for all c ∈ int K ◦ , which shows that ; e%c,x5 dx P

8. THE EXPONENTIAL VALUATION

71

indeed converges absolutely for all c ∈ int K ◦ and uniformly on compact subsets of K ◦ . Next, we construct the valuation Φ. So far, for every polyhedron P ⊂ V without lines, we deﬁned a function φ(P, c) ∈ M(V ) such that ; φ(P, c) = e%c,x5 dx P

int KP◦ ,

where KP is the recession cone of P and the intefor every c ∈ gral converges absolutely and uniformly on compact subsets of int KP◦ . Let us show that this deﬁnition respects linear relations among indicator functions of polyhedra without lines. In other words, suppose that G αi [Pi ] = 0, (8.7) i∈I

where Pi ⊂ V are polyhedra without lines and αi are numbers. Our goal is to prove that G αi φ(Pi , c) = 0. (8.8) i∈I

For that, let us choose a decomposition G βj [Qj ], [V ] = j∈J

where Qj are polyhedra without lines and βj are numbers (for example, we can cut V into orthants by the coordinate hyperplanes). Then, for every polyhedron Pi we have G βj [Pi ∩ Qj ]. (8.9) [Pi ] = [Pi ][V ] = j∈J

Now, every polyhedron [Pi ∩Qj ] is a polyhedron without lines (possibly empty) and hence the function φ(Pi ∩ Qj , c) is deﬁned. Moreover, we have the obvious inclusion KPi ∩Qj ⊂ KPi for the recession cones, and hence ; G ; %c,x5 e dx = βj e%c,x5 dx Pi

j∈J

Pi ∩Qj

for all c ∈ int KP◦i , where all the involved integrals converge absolutely and uniformly on compact subsets of int KP◦i . Therefore G (8.10) φ(Pi , c) = βj φ(Pi ∩ Qj , c) j∈J

72

INTEGER POINTS IN POLYHEDRA

for all c ∈ int KP◦i . Since (8.10) is an identity among meromorphic functions which holds on a non-empty open subset of V , the same identity must hold for all c ∈ V . Let us pick a j ∈ J and let us multiply (8.7) by [Qj ]. Thus we get G

αi [Pi ∩ Qj ] = 0.

i∈I

Since we have an obvious inclusion KPi ∩Qj ⊂ KQj for the recession cones, we have G ; e%c,x5 dx = 0 αi Pi ∩Qj

i∈I

for all c ∈ int KQ◦ j . Therefore, G

(8.11)

αi φ(Pi ∩ Qj , c) = 0

i∈I

for all c ∈ int KQ◦ j . Since (8.11) is an identity among meromorphic functions which holds on a non-empty open subset of V , the same identity must hold for all c ∈ V . Now, from (8.10) and (8.11) we conclude G

αi φ(Pi , c) =

i∈I

G

I αi

i∈I

=

G j∈J

G

I βj

C βj φ(Pi ∩ Qj , c)

j∈J

G

C αi φ(Pi ∩ Qj , c)

= 0,

i∈I

which proves (8.8). Thus we conclude that the correspondence [P ] 9−→ φ(P, c) on polyhedra without lines gives rise to a linear transformation of the span of the indicators of polyhedra without lines into the space M(V ) of functions. On the other hand, by (8.9) the indicators of polyhedra without lines span the whole space P(V ), so we get a valuation Φ : P(V ) −→ M(V ). It remains to show that we must have Φ([P ]) = 0 provided P contains a line.

8. THE EXPONENTIAL VALUATION

73

First, we observe that if P is a polyhedron without lines and P + u is a translation of P , then the recession cones of P and P + u coincide and so we have ; ; %c,x5 %c,u5 e%c,x5 dx e dx = e P

P +u

for all c ∈ int KP◦ . Therefore, we have φ(P +u, c) = e%c,u5 φ(P, c) and hence Φ([P +u]) = %c,u5 e Φ([P ]) for all polyhedra without lines. Since the indicators of polyhedra without lines span P(V ), we must have Φ([P + u]) = e%c,u5 Φ([P ]) for all polyhedra P and all u ∈ V . Now, if P is a polyhedron with a line, then there is a vector u ∈ V \ {0} such that P + u = P , cf. Theorem 4.13. Thus we must have Φ([P ]) = Φ([P + u]) = e%c,u5 Φ([P ]), which implies that Φ([P ]) = 0 provided P contains a line. This completes the proof of Theorem 8.4.

!

One illustration of Theorem 8.4 is the continuous version of the phenomenon described by (1.3). Thus we have ; +∞ 1 ecx dx = provided c < 0 and −c 0 ; 1 ec provided c > 0. ecx dx = c −∞ Moreover,

;

1 0

ecx dx =

which agrees with the formula ; ; 1 ; 1 cx cx e dx + e dx = 0

−∞

ec 1 − , c c

+∞ 0

e

cx

provided we assume that ;

+∞

−∞

ecx dx = 0.

; dx −

+∞ −∞

ecx dx,

74

INTEGER POINTS IN POLYHEDRA

Problems 1◦ . Let P ⊂ V be a polytope. Prove that G Φ([fcone(P, v)]) = 0. v∈Vert(P )

2. Let P ⊂ V be a polytope with non-empty interior. With every facet (a face of codimension 1) F of P , let us associate the unit outer normal vector nF . Let us choose a vector u = > 0 and let dxF be the Lebesgue measure on the facet F . Prove that for all c such that 0c, u8 >= 0 we have ; ; 1 G %c,x5 e dx = 0u, nF 8 e%c,x5 dxF , 0c, u8 P F F where the sum is taken over all facets F of P . Hint: This is, essentially, the Stokes formula applied to P and the diﬀerential form ω(x) = e%c,x5 u1 ∧ · · · ∧ ud−1 , where vectors u1 , . . . , ud−1 are chosen in such a way that 0u, ui 8 = 0 for i = 1, . . . , d − 1 and u ∧ u1 ∧ · · · ∧ ud is the standard volume form on V , see [Ba91] for detail. 3. Let ρ : V −→ R be a polynomial. Prove the analogue of Theorem 8.4 for valuation Φρ which is deﬁned on polyhedra without lines by ; Φρ ([P ]) = φρ (P, c) where φρ (P, c) = e%c,x5 ρ(x) dx P

and the integral converges for all c ∈ int KP◦ , where KP is the recession cone of P and on polyhedra with lines by Φρ ([P ]) = 0. Hint: Apply the diﬀerential operator J % ∂ ∂ ,..., ρ ∂c1 ∂cd to valuation Φ, see [KP92] for detail. 4. Let Av (V ) be the span of the indicator functions of translations A = K + v of polyhedral cones K ⊂ V . Suppose that there is a valuation Ψv : Av (V ) −→ W such that Ψv ([K + v]) = 0 if K + v is a

8. THE EXPONENTIAL VALUATION

75

translation of a cone with a line. Prove that one can deﬁne valuation Ψ : P(V ) −→ W by G Ψ([P ]) = Ψv ([tcone(P, v)]) v∈Vert(P )

if P contains no lines and by Ψ([P ]) = 0 if P contains a line. Hint: See [L91a].

CHAPTER 9

Computing volumes Let dim V = d and let P ⊂ V be a polytope. One way to compute the volume of P is to let c = 0 in the exponential integral (8.1). Suppose, however, we are to use the formula ; G e%c,x5 dx = e%c,v5 fv (c), P

v∈Vert(P )

where fv (c) are the rational functions associated with the cones fcone(P, v) of feasible directions at v, ; 2 A e%c,x5 dx. fv (c) = φ fcone(P, v), c = fcone(P,v)

Then c = 0 is the pole of all fv (c), the simplest example being the formula ; 1 ec 1 − . ecx = c c 0 There is a way around that diﬃculty. Let us choose a suﬃciently generic c ∈ V which is not a pole of any fv (c). Let t be a real parameter and let us consider the integral ; e%tc,x5 dx P

as a function of t. Clearly, the value of the integral is an analytic function of t and we are interested in its value at t = 0. On the other hand, by Lemma 8.3, each of the functions fv (c) is a rational function of degree −d and hence e%tc,v5 fv (tc) is a meromorphic function of t which has the Laurent expansion around t = 0: e%tc,v5 fv (tc) =

+∞ G n=0

tn−d

0c, v8n fv (c). n!

We are interested in the constant term of this expansion, which is equal to 0c, v8d fv (c). d!

78

INTEGER POINTS IN POLYHEDRA

Thus we obtain the formula (9.1)

vol P =

G v∈Vert(P )

0c, v8d fv (c). d!

The interesting feature of this formula is that the left-hand side is a constant, independent of c, while the right-hand side is a sum of rational functions of c. Applied to the interval [0, 1], formula (9.1) results in 1 1=0+c· , c while applied to the triangle Δ on Figure 38, formula (9.1) results in c21 c22 1 =0+ + . 2 2c1 (c1 − c2 ) 2c2 (c2 − c1 ) Suppose that {Pτ } is a family of strongly combinatorially isomorphic polytopes with the vertices v1 (τ ), . . . , vn (τ ). Then (9.1) implies that vol Pτ is a polynomial of degree d in v1 (τ ), . . . , vn (τ ). If for every vertex v of P the cone fcone(P, v) of feasible directions is simple, that is, of the form co (u1 (v), . . . , ud (v)), where ui (v) form a basis of V , then (9.1) transforms into (9.2)

vol P =

G v∈Vert(P )

d !0c, v8d !! 1 , u1 (v) ∧ · · · ∧ ud (v)! d! 0−c, u (v)8 i i=1

see Figure 41. u (v) 3

v

u1(v )

v

P u (v) 2

Figure 41. A polytope P , a vertex v and a simple cone at v. Such polytopes are called simple and formula (9.2) is due to J. Lawrence [L91b] who proved it by a diﬀerent method.

9. COMPUTING VOLUMES

79

Summarizing, the exponential valuation allows one to use the whole power of the algebra of polyhedra, including identities involving unbounded polyhedra and polyhedra with lines, to compute volumes. Problems 1◦ . Write formula (9.2) if P is the unit cube in Rd consisting of the points x = (x1 , . . . , xd ) with 0 ≤ xi ≤ 1 for i = 1, . . . , d. 2. Let a ∈ Rd be a vector with distinct coordinates and let P be the convex hull of all the d! points obtained from a by permutations of the coordinates. Prove that P is a simple (d − 1)-dimensional polytope which lies in the aﬃne hyperplane A deﬁned by the equation x1 + · · · + xd = α1 + · · · + αd , where a = (α1 , . . . , αd ), and write formula (9.2) for P as a polytope in the (d − 1)-dimensional space A. 3. Let P ⊂ V be a d-dimensional polytope. Prove that G 0c, v8k fv (c) = 0 for k = 0, . . . , d − 1, v∈Vert(P )

where fv (c) = Φ([fcone(P, v)]).

CHAPTER 10

Lattices, bases, and parallelepipeds Let us introduce yet another structure. Definition 10.1. Let V be Euclidean space. A subset Λ ⊂ V is called a lattice, if Λ is an additive subgroup of V : for all x, y ∈ Λ we have x ± y ∈ Λ; Λ is discrete, that is, for every bounded set B ⊂ V , the intersection B ∩ Λ is ﬁnite; Λ spans V : span(Λ) = V . An equivalent form of the second condition (that Λ is discrete) is that there is a neighborhood of the origin that does not contain any lattice point other than the origin. All lattices look the same if we are not interested in Euclidean structure. Once a Euclidean structure is introduced, lattices are starting to look diﬀerent, see Figure 42.

Figure 42. Two lattices in the plane. Our main example of a lattice is the lattice Zd ⊂ Rd consisting of the points with integer coordinates. Other important examples of lattices can be obtained as follows: let L ⊂ Rd be a subspace spanned by points from Zd . Then Λ = L ∩ Rd is a lattice in L. Let us deﬁne the norm and the distance in V in the usual way: = CxC = 0x, x8 and dist(x, y) = Cx − yC.

82

INTEGER POINTS IN POLYHEDRA

For a subset L ⊂ V , let dist(x, L) = inf dist(x, y). y∈L

We introduce a notation which we are going to use frequently. Let α ∈ R be a number. We deﬁne 7α6 to be the largest integer not exceeding α and {α} = α − 7α6 to be the fractional part of α, so that 0 ≤ {α} < 1. Thus we have α = 7α6 + {α},

where 7α6 ∈ Z and 0 ≤ {α} < 1.

Lemma 10.2. Let Λ ⊂ V be a lattice and let L ⊂ V be a subspace spanned by some points from Λ. Then, among all the lattice points that are not in L there exists a point v closest to L. That is, there exists a point v such that v ∈Λ\L

and

dist(v, L) ≤ dist(w, L)

for all w ∈ Λ \ L.

Proof. Let k = dim L and let u1 , . . . , uk be a basis of L consisting of lattice points, so ui ∈ Λ for i = 1, . . . , k. Let Π be the parallelepiped spanned by u1 , . . . , uk : 1 Π=

k G

, λ i ui :

0 ≤ λi ≤ 1 for i = 1, . . . , k

.

i=1

We claim that among the lattice points that are not in L there is a point closest to Π. Indeed, let us choose a suﬃciently large ρ so that the ρ-neighborhood Πρ of Π, > Πρ = x ∈ V :

dist(x, Π) ≤ ρ

D

contains a lattice point which is not in L. Since Λ is discrete, Πρ ∩ Λ is ﬁnite. Among the points of Πρ ∩ (Λ \ L), let us choose a point v that is closest to Π. Clearly, (10.2.1)

dist(v, Π) ≤ dist(a, Π) for all a ∈ Λ \ L.

We claim now that dist(v, L) ≤ dist(w, L) for all w ∈ Λ \ L.

10. LATTICES, BASES, AND PARALLELEPIPEDS

83

Indeed, suppose that for some w ∈ Λ\L we have dist(w, L) < dist(v, L). Let x ∈ L be the point closest to w, so that dist(w, x) = dist(w, L). Since x ∈ L, it can be written as a linear combination x=

k G

α i ui = u + y

where

i=1

u=

k G

7αi 6ui

k G and y = {αi }ui .

i=1

i=1

Now, u ∈ Λ ∩ L and y ∈ Π. Moreover, w − u ∈ Λ \ L and dist(w−u, Π) ≤ dist(w−u, x−u) = dist(w, x) < dist(v, L) ≤ dist(v, Π), which contradicts (10.2.1) with a = w − u.

!

Note that if L is not spanned by lattice points, there can be lattice points not in L but arbitrarily close to L, cf. Figure 43.

Figure 43. For a line with a rational slope there is the smallest distance from a grid point not on the line to the line. For a line with an irrational slope, grid points not on the line may get arbitrarily close to the line. Here is an important corollary. Corollary 10.3. Let Λ ⊂ V be a lattice and let L ⊂ V be a subspace spanned by lattice points. Let us consider a decomposition V = L ⊕ W and let pr : V −→ W be the projection with the kernel L. Then pr(Λ) is a lattice in W . Proof. Clearly, Λ1 = pr(Λ) is an additive subgroup in W which spans W as a vector space. By Lemma 10.2, there exists the minimum positive distance from a point in Λ\L to L. Hence there exists the minimum

84

INTEGER POINTS IN POLYHEDRA

positive distance from a point in Λ1 \{0} to the origin in W . This proves that Λ1 is discrete. ! Again, if L is not spanned by lattice points, then Λ1 = pr(Λ) may get everywhere dense in W . Theorem 10.4. Suppose that dim V = d > 0. (1) Let Λ ⊂ V be a lattice. Then there exist vectors u1 , . . . , ud in Λ such that every point u ∈ Λ admits a unique decomposition u=

d G

m i ui

where mi ∈ Z

for

i = 1, . . . , d.

i=1

The set {u1 , . . . , ud } is called a basis of Λ; (2) Let u1 , . . . , ud be a basis of V and let 1 d , G Λ= mi ui where mi ∈ Z . i=1

Then Λ ⊂ V is a lattice. Proof. To prove Part (1), we proceed by induction on d. Suppose that d = 1, so we identify V = R. Since Λ is discrete, there is the smallest positive number a ∈ Λ. We claim now that every point x ∈ Λ can be written as x = ma for some m ∈ Z. Suppose that x > 0. Then we can write x = µa for some α > 0. Then x = µa = 7µ6a + {µ}a. Since x ∈ Λ and 7µ6a ∈ Λ, we have {µ}a ∈ Λ. Since {µ} < 1 we must have {µ} = 0, because otherwise we would have obtained a positive point in Λ which is smaller than a. This proves that {µ} = 0, so µ, is, in fact, an integer. Similarly, if x < 0 then −x ∈ Λ, so as we proved −x = ma for some positive integer m. Suppose now that d > 1. Let us choose some d−1 linearly independent points and let L be a subspace spanned by those points. Hence dim L = d − 1 and Λ1 = Λ ∩ L is a lattice in L. By the induction hypothesis, there is a basis u1 , . . . , ud−1 of Λ1 . By Lemma 10.2, there is a vector ud ∈ Λ \ L such that dist(ud , L) ≤ dist(u, L) for all u ∈ Λ \ L.

10. LATTICES, BASES, AND PARALLELEPIPEDS

85

We claim that u1 , . . . , ud−1 , ud is a basis of Λ. To check that, let us pick any u ∈ Λ. Then there exists a unique decomposition (10.4.1)

u=

d G

α i ui

for some αi ∈ R.

i=1

We claim that αd ∈ Z. Indeed, suppose that αd ∈ / Z, or, equivalently, that {αd } > 0. Let us consider the point v = u − 7αd 6ud = {αd }ud +

d−1 G

α i u1 .

i=1

Then v ∈ Λ \ L and dist(v, L) = dist ({αd }ud , L) = {αd } dist(ud , L) < dist(ud , L), which is a contradiction. This proves that αd ∈ Z in (10.4.1). Then for w = u − α d ud =

d−1 G

α i ui

i=1

we have w ∈ Λ and w ∈ L, so w ∈ Λ1 . Then α1 , . . . , αd−1 have to be integer as well, which completes the proof of Part (1). To prove Part (2), let us consider the map T : Rd −→ V,

T (α1 , . . . , αd ) =

d G

α i ui .

i=1

2 A Hence Λ = T Zd . Since T is an invertible linear transformation, Λ is a discrete additive subgroup of V which spans V . ! Remark 10.5. In the course of the proof of Theorem 10.4, we proved the following useful statement. Let Λ ⊂ V be a lattice and let L ⊂ V be a subspace spanned by lattice points, so that Λ1 = Λ ∩ L is a lattice in Λ1 . Then any basis u1 , . . . , uk of Λ1 can be appended to a basis u1 , . . . , uk , uk+1 , . . . , ud of Λ. Definition 10.6. Let Λ ⊂ V be a lattice and let u1 , . . . , ud be a basis of Λ. The semi-open parallelepiped , 1 d G αi ui : 0 ≤ αi < 1 for i = 1, . . . , d Π = Π(u1 , . . . , ud ) = i=1

is called a fundamental parallelepiped of Λ, see Figure 44.

86

INTEGER POINTS IN POLYHEDRA

0

Figure 44. The lattice Z2 and two of its fundamental parallelepipeds. A fundamental parallelepiped is a fundamental domain of the action of Λ on V , which is emphasized in the following result. Lemma 10.7. Let Λ ⊂ V be a lattice and let Π be a fundamental parallelepiped of Λ. Then every point x ∈ V can be uniquely written as x = y + u, where y ∈ Π and u ∈ Λ. Proof. Let u1 , . . . , ud be a basis of Λ that spans the parallelepiped Π. Let x ∈ V be a point. Thus we may write x=

d G

α i ui

for some αi ∈ R.

i=1

We let y=

d G

{αi }ui

i=1

and u =

d G

7αi 6ui ,

i=1

which proves that the decomposition x = y + u exists. Suppose that x = y 1 + u1 = y 2 + u2

where y1 , y2 ∈ Π and u1 , u2 ∈ Λ.

Then y1 − y2 = u2 − u1 , so y1 − y2 ∈ Λ. On the other hand, writing y1 =

d G i=1

α i ui ,

y2 =

d G

β i ui

i=1

and y1 − y2 =

d G

(αi − βi ) ui ,

i=1

we observe that we must have 0 ≤ αi , βi < 1 and αi − βi ∈ Z for i = 1, . . . , d, which implies that αi = βi for all i and that y1 = y2 and u1 = u2 , which concludes the proof of uniqueness. !

10. LATTICES, BASES, AND PARALLELEPIPEDS

87

Now we are ready to introduce perhaps the most important numerical invariant of a lattice. Theorem 10.8. Let Λ ⊂ V be a lattice. Then all the fundamental parallelepipeds Π of Λ have the same volume, called the determinant of Λ and denoted det Λ. Furthermore, det Λ can be obtained as follows. Let Bρ be the ball of radius ρ centered at the origin. Then lim

ρ−→+∞

|Bρ ∩ Λ| 1 = . vol Bρ det Λ

More generally, if a ∈ V is an arbitrary point, |Bρ ∩ (a + Λ) | 1 = . ρ−→+∞ vol Bρ det Λ lim

Proof. Let us choose a particular fundamental parallelepiped Π of Λ. Lemma 10.7 implies that the translates Π + u : u ∈ Λ cover the space V without overlapping. Suppose that Π ⊂ Bα for some α > 0. Let us choose a ρ > α and consider the union Xρ of the shifts Π + u, where u ∈ Bρ : / 'M Π+u . Xρ = u∈Bρ

Then

Bρ−α ⊂ Xρ ⊂ Bρ+α . While the second inclusion is obvious, the ﬁrst inclusion follows since every point x ∈ Bρ−α has to be covered by some translation Π + u for u ∈ Λ, from which it follows that we must have u ∈ Bρ . Thus vol Bρ−α ≤ vol Xρ = |Bρ ∩ Λ| vol Π ≤ vol Bρ+α . Since

vol Bρ±α = 1, ρ−→+∞ vol Bρ lim

we get

1 |Bρ ∩ Λ| = . vol Bρ vol Π More generally, for any a ∈ V and α = CaC, we have lim

ρ−→+∞

a + (Bρ−α ∩ Λ)

⊂

Bρ ∩ (a + Λ)

⊂

a + (Bρ+α ∩ Λ) ,

from which it follows that |Bρ−α ∩ Λ| ≤ |Bρ ∩ (a + Λ)| ≤ |Bρ+α ∩ Λ|

88

INTEGER POINTS IN POLYHEDRA

and

|Bρ ∩ (a + Λ)| 1 = . ρ−→+∞ vol Bρ det Λ lim

45.

!

In other words, det Λ is “the volume per lattice point”, see Figure

Figure 45. The number of lattice points in a ball of a large radius is approximately the ratio of the volume of the ball to the volume of the fundamental parallelepiped. Let Λ0 , Λ ⊂ V be lattices such that Λ0 ⊂ Λ. Then Λ0 is a subgroup of Λ and we can consider the quotient Λ/Λ0 and cosets of Λ modulo Λ0 . The following useful result characterizes the index of Λ0 in Λ. Theorem 10.9. Let Λ0 ⊂ Λ be lattices and let Π be a fundamental parallelepiped of Λ0 . Then the set Π ∩ Λ contains each coset Λ/Λ0 representative exactly once. Furthermore, |Π ∩ Λ| = |Λ/Λ0 | =

det Λ0 . det Λ

Proof. By Lemma 10.7, every x ∈ Λ has a unique representation x = y + u where u ∈ Λ0 and y ∈ Π. Then y ∈ Λ and y ≡ x mod Λ0 , so y is the unique coset representative of x in Π. This proves that |Π ∩ Λ| = |Λ/Λ0 |. It remains to show that |Λ/Λ0 | = (det Λ0 )/(det Λ). To this end, let us choose a set S of the coset Λ/Λ0 , so |S| = |Λ/Λ0 |. Since / 'M Λ= a + Λ0 , a∈S

10. LATTICES, BASES, AND PARALLELEPIPEDS

89

we have (10.9.1)

|Bρ ∩ Λ| =

/ 'M Bρ ∩ (a + Λ0 ) . a∈S

By Theorem 10.8, |Bρ ∩ Λ| 1 = ρ−→+∞ vol Bρ det Λ lim

and 1 |Bρ ∩ (a + Λ0 )| = , ρ−→+∞ vol Bρ det Λ0 lim

and hence the proof follows by (10.9.1).

!

We will often use Theorem 10.9 in the situation of V = Rd , Λ = Z and Λ0 being the lattice generated by some linearly independent integer vectors u1 , . . . , ud . Then, the number of integer points in the parallelepiped d

1 Π=

d G

, α i ui :

0 ≤ αi < 1 for i = 1, . . . , d

i=1

is equal to the volume of Π, see Figure 46.

Figure 46. The number of integer points in the semiopen parallelogram spanned by integer points is equal to the area of the parallelogram. In dimension 2, this fact leads to a nice formula for the number of integer points in a convex polygon with integer vertices.

90

INTEGER POINTS IN POLYHEDRA

Pick’s formula 10.10. Let P ⊂ R2 be a convex polygon with integer vertices and non-empty interior. ! 1! |P ∩ Z2 | = vol P + !∂P ∩ Z2 ! + 1, 2 see Figure 47.

Figure 47. Pick’s formula: the number of integer points in the polygon with integer vertices and nonempty interior is equal to the area of the polygon plus one half of the points on the boundary plus one. First, we prove the formula for a triangle. Let us consider a triangle Δ with integer vertices at 0, a, b as a part of the closed parallelogram Π with vertices at 0, a, b, and a + b, see Figure 48. a+b

a Δ 0

Δ1

_ Π

b

Figure 48. A triangle with integer vertices and a parallelogram. We have |Δ ∩ Z2 | + |Δ1 ∩ Z2 | − |[a, b] ∩ Z2 | = |Π ∩ Z2 |. The crucial observation is that there is a bijection between the integer points in Δ and the integer points in the (closed) triangle Δ1 with the vertices (a + b), a, and b via the map u 9−→ (a + b) − u. Hence

1 1 |Δ ∩ Z2 | = |Π ∩ Z2 | + |[a, b] ∩ Z2 |. 2 2

10. LATTICES, BASES, AND PARALLELEPIPEDS

91

Now, |Π ∩ Z2 | = |Π ∩ Z2 | + |[a, a + b] ∩ Z2 | + |[b, a + b] ∩ Z2 | − 1, where Π is the semi-open parallelogram spanned by a and b. By Theorem 10.9, |Π ∩ Z2 | = vol Π. Using that vol Δ = 21 vol Π and that the integer points on the interval [a, a + b] are in a bijection with the integer points on the interval [0, b] while the integer points on the interval [a + b, b] are in a bijection with the integer points on the interval [0, a], we get 1 1 1 1 |Δ ∩ Z2 | = vol Δ + |[a, b] ∩ Z2 | + |[0, a] ∩ Z2 | + |[0, b] ∩ Z2 | − 2 2 2 2 1 = vol Δ + |∂Δ ∩ Z2 | + 1, 2 as desired. Now we obtain the formula for an arbitrary convex polygon P with integer vertices by induction on the number of vertices. We represent a polygon P with n vertices as a union of a polygon Q with (n − 1) vertices and a triangle Δ, intersecting on an interval [a, b] see Figure 49. b Δ Q

a P

Figure 49. Representing a polygon with n vertices as a union of a polygon with (n − 1) vertices and a triangle. Thus we have |P ∩ Z2 | = |Q ∩ Z2 | + |Δ ∩ Z2 | − |[a, b] ∩ Z2 | 1 1 = vol Q + vol Δ + |∂Q ∩ Z2 | + |∂Δ ∩ Z2 | 2 2 2 + 2 − |[a, b] ∩ Z | 1 = vol P + |∂P ∩ Z2 | + 1, 2 which completes the proof for a general polygon.

92

INTEGER POINTS IN POLYHEDRA

Problems √ 1. Let us consider the line y = 2x in the plane R2 . Prove that a point a ∈ Z2 may lie arbitrarily close to the line but not on the line. 2. A set S ⊂ V is called a semigroup provided x + y ∈ S for any two x, y ∈ S. A set X ⊂ S is called a set of generators of S if every s ∈ S can be written as a (ﬁnite) linear combination of elements of X with non-negative integer coeﬃcients. Prove that every semigroup S ⊂ Z possesses a ﬁnite set of generators and construct an example of a semigroup S ⊂ Z2 which does not have a ﬁnite set of generators. 3. Let Λ ⊂ R2 be a lattice. Prove that there exists a basis u1 , u2 of Λ such that the angle between u1 and u2 is between 60◦ and 90◦ . 4. Let a1 , . . . , ad be coprime integers and let n be a positive integer. Let us consider a set Λ ⊂ Zd of points m = (m1 , . . . , md ) deﬁned by the equation d G ai m i ≡ 0 mod n. i=1 d

Prove that Λ ⊂ R is a lattice and that det Λ = n. 5. Let a1 , . . . , ad be coprime integers. Let us identify V with the hyperplane H ⊂ Rd deﬁned by the equation a1 x1 + · · · + ad xd = 0 and let Λ = V ∩ Zd . Prove that Λ ⊂ V is a lattice and that det Λ = = a21 + · · · + a2d . 6. Let P ⊂ R2 be a convex polygon with integer vertices and nonempty interior. For a point a ∈ P let us deﬁne the angle α(a, P ) by vol (P ∩ (a + BO )) , O−→0+ vol BO

α(a, P ) = lim

where BO is the ball of radius g. Deduce from Pick’s formula that G α(a, P ) = vol P. a∈P ∩Z2

7. Lattices Λ1 , Λ2 ⊂ V are called similar (denoted Λ1 ∼ Λ2 ) if Λ1 = T (λ2 ), where T : V −→ V is a composition of an orthogonal transformation and dilation x 9−→ αx for some α >= 0. Here are some special lattices. Lattice Dn ⊂ Zn consists of the integer vectors (m1 , . . . , md ) such that m1 + · · · + md ≡ 0 mod 2.

10. LATTICES, BASES, AND PARALLELEPIPEDS

For n even, let Dn+ ⊂ Rn be the set Dn+

= Dn ∪ (Dn + w)

%

where w =

1 1 ,..., 2 2

93

J .

Prove that Dn+ is a lattice (lattice D8+ is called E8 ). Let V ⊂ Rn+1 be the hyperplane deﬁned by the equation x1 + · · · + xn+1 = 0. Lattice An ⊂ V is deﬁned as An = V ∩ Zn+1 . Prove that D2 ∼ Z2 , D3 ∼ A3 , and D4+ ∼ Z4 . 8. Let Λ ⊂ V be a lattice. Let us deﬁne the Voronoi region by > D P = x ∈ V : dist(x, 0) ≤ dist(x, u) for all u ∈ Λ . Prove that P is a polytope and that vol P = det Λ. 9. Let Λ0 ⊂ Λ be lattices. Prove that there exists a basis u1 , . . . , ud of Λ and a basis v1 , . . . , vd of Λ0 such that vi = mi ui for some positive integers mi and i = 1, . . . , d. 10◦ . A set of vectors u1 , . . . , uk ∈ Λ is called primitive if u1 , . . . , uk is a basis of the lattice Λ ∩ L where L = span (u1 , . . . , uk ). Prove that a primitive set can be appended to a basis of Λ. 11. Prove that a set of integer vectors u1 = (m11 , . . . , m1d ) , . . ., uk = (mk1 , . . . , mkd ) is primitive in Zd if and only if the greatest common divisor of all k × k minors of the matrix (mij ) is 1. 12. Show that there are integer tetrahedra (convex hulls of four aﬃnely independent integer points) in R3 of an arbitrarily large volume that do not contain any integer point other than their vertices. 13. Let P ⊂ Rd be a polytope with integer vertices and containing no integer points other than its vertices. Prove that the number of vertices of P does not exceed 2d .

CHAPTER 11

The Minkowski Convex Body Theorem The following classical result is known as the Minkowski (First) Convex Body Theorem. Theorem 11.1. Let V be a d-dimensional Euclidean space and let Λ ⊂ V be a lattice. Let B ⊂ V be a Lebesgue measurable set such that the following holds: for every x, y ∈ B we have (x + y)/2 ∈ B; for every x ∈ B we have −x ∈ B; vol B > 2d det Λ. Then B contains a non-zero point from Λ. Moreover, if B is compact, then the last condition can be replaced by vol B ≥ 2d det Λ. Minkowski’s Theorem is deduced from the following Blichfeldt’s Theorem. Theorem 11.2. Let V be a d-dimensional Euclidean space and let Λ ⊂ V be a lattice. Let X ⊂ V be a Lebesgue measurable set such that vol X > det Λ. Then there exist two distinct points x, y ∈ X such that x − y ∈ Λ. Proof. Let us choose a fundamental parallelepiped Π of Λ, so vol Π = det Λ. For every point u ∈ Λ let us deﬁne > D Xu = x ∈ Π : x + u ∈ X . In words: we consider the translates Π + u : u ∈ Λ, which cover V without overlapping according to Lemma 10.7. For every translate, we consider the intersection (Π + u) ∩ X and obtain Xu by translating

96

INTEGER POINTS IN POLYHEDRA

it back into Π by −u. Since the sets (Π + u) ∩ X cover X without overlapping, we must have / G G M vol Xu = vol (Π + u) ∩ X = vol X. u∈Λ

u∈Λ

Now we claim that some sets Xu and Xv must intersect for u = > v. Indeed, if Xu and Xv are disjoint for all u >= v, then C I G ' vol Xu = vol X > vol Π, vol Xu = u∈Λ

u∈Λ

which is a contradiction since '

Xu ⊂ Π.

u∈Λ

This shows that there are two lattice points u >= v and a point a such that a ∈ Xu and a ∈ Xv . In other words, a + u ∈ X and a + v ∈ X. Denoting x = a + u and y = a + v we obtain two distinct points from X such that x − y = u − v is a non-zero lattice point. ! Proof of Theorem 11.1. Let 9 1 1 X= B= x: 2 2

6 x∈B .

Then

1 vol B > 1. 2d Therefore, by Theorem 11.2, there are two points x, y ∈ B such that vol X =

1 1 u= x− y 2 2 is a non-zero lattice point. On the other hand, 1 1 u = x + (−y). 2 2 We note that −y ∈ B and hence u ∈ B as well. Suppose now that B is compact and vol B = 2d det Λ. Then, for every g > 0 the set > D BO = (1 + g)x : x ∈ B satisﬁes the conditions of the theorem, and, additionally, vol BO = (1 + g)d vol B > 2d det Λ. Hence BO contains a non-zero lattice point uO . A limit point u of {uO } is a non-zero lattice point in B. !

11. THE MINKOWSKI CONVEX BODY THEOREM

97

As an application we sketch a proof, due to H. Davenport, of the famous Lagrange Theorem. Theorem 11.3. Let n be a positive integer. Then n can be represented as a sum of four squares of integers: n = m21 + m22 + m23 + m24 for some integer m1 , m2 , m3 , m4 . Sketch of proof. One can check that the product of two sums of four squares of integers is a sum of four squares of integers (multiplication of quaternions provide a hint here). Hence it suﬃces to prove the result when n is a prime. If n is a prime, one can show that the congruence a2 + b 2 + 1 ≡ 0

mod n

has a solution. Let us consider the lattice Λ ⊂ Z4 consisting of the integer vectors (m1 , m2 , m3 , m4 ) such that m1 ≡ am3 + bm4

mod n and m2 ≡ bm3 − am4

mod n.

The number of cosets Z4 /Λ does not exceed n2 and hence (det Λ ≤ n2 . N Let B ⊂ R4 be the open ball, B = x : CxC2 < 2n . One can check that vol B = 2n2 π 2 > 16n2 ≥ 24 det Λ and hence the ball contains a non-zero point (m1 , m2 , m3 , m4 ). We must have m21 + m22 + m23 + m24 ≡ 0 from which

mod n and m21 + m22 + m23 + m24 < 2n,

m21 + m22 + m23 + m24 = n.

!

(11.4) The dual lattice. Let Λ ⊂ V be a lattice and let > D Λ∗ = x ∈ V : 0x, y8 ∈ Z for all y ∈ Λ . It is not hard to see that Λ∗ is, in fact, a lattice. Indeed, if u1 , . . . , ud is a basis of Λ, then the vectors v1 , . . . , vd deﬁned by 1 1 if i = j, 0ui , vj 8 = 0 if i >= j constitute a basis of Λ∗ . Furthermore, we have (det Λ)(det Λ∗ ) = 1. The lattice Λ∗ is called dual or reciprocal to Λ.

98

INTEGER POINTS IN POLYHEDRA

Problems 1. Fill in the gaps in the proof of Theorem 11.3. 2. Let n be a positive integer. Prove that if there is a solution to the congruence x2 + 1 ≡ 0 mod n, then n is the sum of two squares of integers. Deduce that every prime number p ≡ 1 mod 4 is the sum of two squares of integers. 3◦ . Let Λ ⊂ Rd be a lattice and let N ( CxC∞ = max |xi | : i = 1, . . . , d . Prove that there exists a vector u ∈ Λ\{0} such that CuC∞ ≤ (det Λ)1/d . 4. Let V be Euclidean space with norm C · C and let Λ ⊂ V be a lattice. Prove that there exists a vector u ∈ Λ \ {0} such that CuC ≤ √ d (det Λ)1/d . 5. Prove that E8∗ = E8 , cf. Problem 7 of Chapter 10. 6. Let Λ ⊂ V be a lattice, dim V = d. Prove that there exist vectors u ∈ Λ \ {0} and v ∈ Λ∗ \ {0} such that CuC · CvC ≤ d.

CHAPTER 12

Reduced basis In this section, we describe a computationally eﬃcient procedure due to A.K. Lenstra, H.W. Lenstra, Jr., and L. Lov´asz [L+82] which, given an arbitrary basis of a lattice Λ ⊂ V , produces a particular basis u1 , . . . , ud of the same lattice that has many convenient properties. For example, vectors u1 , . . . , ud are almost orthogonal, that is, d -

Cui C ≤ γd det Λ,

i=1

where γd is a constant depending on the dimension d of the ambient space V alone. The basis is called Lenstra–Lenstra–Lov´ asz reduced or LLL-reduced. The algorithm starts with an arbitrary basis u1 , . . . , ud of Λ and then gradually improves it. It is important that any given moment the vectors of the basis are ordered. 12.1 Subspaces and projections. Given a basis u1 , . . . , ud , let us deﬁne subspaces L0 , L1 , . . . , Ld as follows: L0 = {0},

Lk = span (u1 , . . . , uk )

for k = 1, . . . , d.

Moreover, let us deﬁne vectors wk for k = 1, . . . , d as the orthogonal complement of the projection of uk onto Lk−1 , see Figure 50. uk wk

0

L k −1

Figure 50. A basis vector uk , the space Lk−1 , and the vector wk . Often, the vectors w1 , . . . , wd are referred to as the Gram – Schmidt orthogonalization of u1 , . . . , ud (without normalization). We note that dist(uk , Lk−1 ) = Cwk C for k = 1, . . . , d

100

INTEGER POINTS IN POLYHEDRA

and that det Λ =

d -

Cwk C.

k=1

We deﬁne Λ k = Λ ∩ Lk

for k = 1, . . . , d.

Hence Lk is a lattice in Lk and det Λk =

k -

Cwi C.

i=1

12.2 Weakly reduced basis. We can write uk = w k +

k−1 G

αki wi

for k = 1, . . . , d

i=1

and some real numbers αki We say that the basis u1 , . . . , ud is weakly reduced if (12.2.1)

|αki | ≤

1 2

for all i, k.

Given a basis u1 , . . . , ud of Λ we can modify it to enforce (12.2.1) as follows: assuming that |αki | > 1/2 for some k and i, let us replace uk by u(k by letting (12.2.2)

u(k = uk − mki ui ,

where mki is the integer such that |αki − mki | ≤ 1/2. It is easy to see that u1 , . . . , uk−1 , u(k , uk+1 , . . . , ud is still a basis of Λ. By replacing uk with u(k we don’t change the subspaces L0 , . . . , Ld and we don’t change the vectors w1 , . . . , wd . Therefore, the coeﬃcients αji do not change unless j = k and the only coeﬃcients αkj that do change ( αkj 9−→ αkj ( | = |αki − mki | ≤ 1/2. are those with j ≤ i. Moreover, we have |αki Hence to enforce (12.2.1) for a particular k, we apply transformation (12.2.2) at most k times, each time for the largest i for which (12.2.1) is violated. All in all, to make a basis weakly reduced, we have to make not more than d(d − 1)/2 transformations (12.2.2).

12. REDUCED BASIS

101

12.3 Reduced basis. We call a basis u1 , . . . , ud reduced if it is weakly reduced and if 4 dist2 (uk , Lk−1 ) ≤ dist2 (uk+1 , Lk−1 ) 3 (12.3.1) for k = 1, . . . , d − 1. In fact, the constant 4/3 can be replaced by any number 1 < τ < 4 and 4/3 is just the most convenient choice. Geometrically, condition (12.3.1) means that uk+1 is not much closer to Lk−1 than uk , see Figure 51. u k +1 uk 0

L k −1

Figure 51. Vector uk+1 should not be much closer to Lk−1 than vector uk . 12.4 The algorithm. Now we can describe the algorithm to produce the reduced basis. We start with an arbitrary basis u1 , . . . , ud of Λ. We check whether the basis is weakly reduced and modify it as in Section 12.2 to make it weakly reduced, if it was not. Then we check if the resulting basis is reduced. If it is reduced, we stop and output the basis. If (12.3.1) is violated for some k, we change the order of the basis vectors by switching uk and uk+1 This may produce a basis which is no longer weakly reduced so we repeat the procedure again. If the algorithm ever stops, it clearly outputs a reduced basis. To see that the algorithm indeed stops, let us associate with a basis u1 , . . . , ud of Λ the quantity D (u1 , . . . , ud ) =

d -

det Λk

k=1

and let us see how D(u1 , . . . , ud ) changes when we apply the algorithm. Replacing uk by u(k as in Section 12.2 does not change the subspaces Li and lattices Λi , so it does not change the value of D(u1 , . . . , ud ). Switching uk and uk+1 does not change subspaces L0 , . . . , Lk−1 and Lk+1 , . . . , Ld . The subspace Lk and the lattice Λk does change though: we replace Lk = span (u1 , . . . , uk−1 , uk ) by L(k = span (u1 , . . . , uk−1 , uk+1 )

102

INTEGER POINTS IN POLYHEDRA

and consequently, we replace vector wk by wk( that is the orthogonal complement to the projection of uk+1 onto Lk−1 . Since the switch occurs if dist2 (uk+1 , Lk−1 ) < we have

3 dist2 (uk , Lk−1 ) , 4

√ Cwk( C

<

3 Cwk C 2

and hence lattice Λk is replaced by a lattice Λ(k with √ det Λ(k

<

3 det Λk . 2

Therefore, the value of D(u1 , . . . , ud ) decreases by at least a factor of √ 3/2. Hence to show that the algorithm stops it suﬃces to show that the value of D (u1 , . . . , ud ) remains bounded from below by a positive number which depends on the lattice Λ alone. Let λ = min CuC u∈Λ\{0}

be the minimum length of a non-zero vector in Λ. Clearly, λ≥

min CuC for k = 1, . . . , d.

u∈Λk \{0}

Applying, for example, the result of Problem 4, Chapter 11, we conclude that % Jk λ det Λk ≥ √ for k = 1, . . . , d. k Therefore, D (u1 , . . . , ud ) ≥ λ

d(d+1) 2

d -

k −k/2

k=1

for any basis u1 , . . . , ud of Λ. The above analysis shows that the algorithm runs in polynomial time. It is very eﬃcient in practice as well.

12. REDUCED BASIS

103

Here is a useful property of the reduced basis. Lemma 12.5. Let u1 , . . . , ud be a reduced basis of Λ and let wk , k = 1, . . . , d, be the orthogonal complement to the projection of uk onto Lk−1 = span (u1 , . . . , uk−1 ). Then (1) 1G Cwk C ≤ Cuk C ≤ Cwk C + Cwi C2 4 i=1 2

(2)

2

k−1

2

1 Cwk+1 C2 ≥ Cwk C2 2

for

for

k = 1, . . . , d;

k = 1, . . . , d − 1.

Proof. We have uk = w k +

k−1 G

αki wi

where |αki | ≤

i=1

Since 2

2

Cuk C = Cwk C +

for i = 1, . . . , k − 1.

2 αki Cwi C2 .

i=1

Part (1) follows. To prove Part (2), we note dist2 (uk , Lk−1 ) = Cwk C2

k−1 G

1 2

and

1 2 Cwk C2 ≤ Cwk+1 C2 + Cwk C2 . dist2 (uk+1 , Lk−1 ) = Cwk+1 C2 + αk+1,k 4 Since 3 dist2 (uk+1 , Lk−1 ) ≥ dist2 (uk , Lk−1 ) 4 we conclude that 1 Cwk+1 C2 ≥ Cwk C2 2 as desired. ! An immediate corollary is that the reduced basis is nearly orthogonal. Corollary 12.6. Let u1 , . . . , ud be a reduced basis of Λ. Then d i=1

Cui C ≤ 2

d(d−1) 4

det Λ.

104

INTEGER POINTS IN POLYHEDRA

Proof. Applying Lemma 12.5, we obtain M &k−1 &k−1 k−i / Cuk C2 ≤ Cwk C2 + 41 i=1 Cwi C2 ≤ Cwk C2 1 + 41 i=1 2 2k−1 Cwk C2 .

≤ Since

det Λ =

d -

Cwk C,

k=1

the proof follows.

!

We will need a technical result which shows that short vectors in Λ must be integer combinations of the reduced basis vectors with small coeﬃcients. Lemma 12.7. Let Λ be a lattice and let u1 , . . . , ud be a reduced basis of Λ. Let λ = min CuC u∈Λ\{0}

be the minimum length of a non-zero vector from Λ. Suppose that CuC ≤ βλ for some u ∈ Λ and some β ≥ 1. Then, in the representation u=

d G

m i ui

i=1

one must have |mk | ≤ 2

d−1 2

% Jd−k 3 β ≤ 3d β 2

Proof. Let us write u=

d G

for k = 1, . . . , d.

µi w i .

i=1

Since Cw1 C = Cu1 C ≥ λ and by Part (2) of Lemma 12.5 we have Cwk C ≥ 2 we must have

|µk | ≤ 2

k−1 2

1−k 2

β

Cw1 C ≥ 2

1−k 2

λ,

for k = 1, . . . , d.

We have µd = m d

and G µk = m k + αik mi i>k

where |αik | ≤

1 2

for k < d.

12. REDUCED BASIS

105

Therefore, |md | ≤ 2

d−1 2

β

and k−1 1G |mk | ≤ 2 2 β + |mi | 2 i>k d−1 1G < 2 2 β+ |mi | for k < d. 2 i>k The desired inequality follows by induction.

!

The immediate corollary of Lemma 12.7 is that if we want to ﬁnd the shortest non-zero vector in a d-dimensional lattice, we have to inspect only const(d) integer linear combinations of the reduced basis vectors. Problems 1. Let u1 , . . . , ud be a basis of lattice Λ and let w1 , . . . , wd be vectors constructed in Section 12.1. Show that min Cwk C ≤ min CuC.

k=1,...,d

u∈Λ\{0}

2. Let u1 , . . . , ud be a reduced basis of Λ. Prove that Cu1 C ≤ 2

d−1 2

min CuC.

u∈Λ\{0}

3. For real α let 0 ≤ {{α}} ≤ 1/2 be the distance to the nearest integer. Let Λ ∈ V be a lattice and let x ∈ V be a point. Prove that dist(x, Λ) ≥

{{0w, v8}} CvC

for all v ∈ Λ∗ \ {0}.

4∗ . Let Λ ⊂ V be a lattice, let u1 , . . . , ud be a reduced basis of Λ, let v1 , . . . , vd be the basis of Λ∗ such that 1 1 if i = j, 0ui , vj 8 = 0 if i >= j, and let x ∈ V be a point. Let us write x=

d G i=1

α i ui

106

INTEGER POINTS IN POLYHEDRA

and let u ∈ Λ be the point u=

d G

m i ui

where mi ∈ Z and |αi −mi | ≤

i=1

Prove that

%

1 2

for i = 1, . . . , d.

Jd 3 {{0vi , x8}} . Cx − uC ≤ √ max i=1,...,d Cvi C 2 Hint: See [Ba86] and [Lo86].

CHAPTER 13

Exponential sums and generating functions Let Λ ⊂ V be a lattice and let P ⊂ V be a polyhedron. Similarly to exponential integrals of Chapter 8, we consider exponential sums G (13.1) e%c,m5 for c ∈ V. m∈P ∩Λ

Suppose that V = Rd and Λ = Zd is the standard integer lattice. Let c = (c1 , . . . , cd ). Then, for m = (µ1 , . . . , µd ) we can write e%c,m5 = xµ1 1 · · · xµd d

where xi = eci

for i = 1, . . . , d.

Hence in this case we can write (13.1) as the generating function G xm where xm = xµ1 1 · · · xµd d (13.2) m∈P ∩Zd

for m = (µ1 , . . . , µd ). We will switch freely between (13.1) and (13.2). If xi = eci for i = 1, . . . , d, we write simply x = ec

where x = (x1 , . . . , xd )

and c = (c1 , . . . , cd ) .

The diﬀerence between the generating function (13.2) and the exponential sum (13.1) is that the former may be well-deﬁned if some xi = 0, which corresponds to the limit case of ci = −∞ in the latter. 13.3 Rational polyhedra. Let us ﬁx Euclidean space Rd and the standard integer lattice Zd ⊂ Rd . A polyhedron P ⊂ Rd is called rational if it can be deﬁned by a system of ﬁnitely many linear inequalities d G

αij ξj ≤ βi ,

for i ∈ I

j=1

with integer coeﬃcients αij , βi ∈ Z for all i, j.

108

INTEGER POINTS IN POLYHEDRA

A linear transformation T : Rn −→ Rm is called rational if the m matrix of T (in the standard bases of R2n and A R ) is rational. d The algebra of rational polyhedra P Q is deﬁned as the span of the indicators [P ] of rational polyhedra P ⊂ Rd . One can easily check that the results of Chapters 2 – 7 are adjusted in a straightforward way for rational polyhedra. Namely, if T : Rn −→ Rm is a rational linear transformation and P ⊂ Rn is a rational polyhedron, then T (P ) ⊂ Rm is a rational polyhedron, so a rational linear transformation T : Rn −→ Rm gives rise to a valuation T : P (Qn ) −→ P (Qm ). Furthermore, faces of a rational polyhedron are rational polyhedra, the recession cone of a rational polyhedron is a rational polyhedral cone, and if a rational polyhedron contains a line, it contains a rational line (a line which is a rational polyhedron). If P ⊂ Rd is a rational polyhedron then G [tcone(P, v)] modulo rational polyhedra with lines. [P ] ≡ v∈Vert(P )

Finally, if P ⊂ Rd is a rational polyhedron, then P ◦ (where the polar is computed under the standard scalar product in Rd ) is a rational polyhedron as well. More generally, let us ﬁx Euclidean space V and a lattice Λ ⊂ V . A polyhedron P ⊂ V is called Λ-rational or just rational if it can be deﬁned by a system of ﬁnitely many linear inequalities 0ui , x8 ≤ αi

for i ∈ I,

where ui ∈ Λ∗ and αi ∈ Z for all i ∈ I. A point v ∈ V is Λ-rational or just rational if mv ∈ Λ for some positive integer m. The algebra of Λ-rational polyhedra or just the algebra of rational polyhedra P (QV ) is deﬁned as the span of the indicators [P ] of rational polyhedra P ⊂ V . Our main motivating example of series (13.1)–(13.2) is the multiple geometric series. 13.4 The non-negative orthant. Let Rd+ be the non-negative orthant, that is, the set of all points in Rd with non-negative coordinates. Then G m∈Rd+ ∩Zd

m

x =

d i=1

1 1 − xi

for x = (x1 , . . . , xd ) ,

13. EXPONENTIAL SUMS AND GENERATING FUNCTIONS

109

where the series converges absolutely in the region |xi | < 1 for i = 1, . . . , d and uniformly on compact subsets of the region. Similarly, G

e

%c,m5

=

d i=1

m∈Rd+ ∩Zd

1 1 − e ci

for c = (c1 , . . . , cd ) ,

where the series converges absolutely in the region ci < 0 (or, more generally, Re ci < 0 for i = 1, . . . , d) and uniformly on compact subsets of the region. This is, of course, a straightforward extension of the formula for the inﬁnite geometric series +∞ G

xm =

m=0

1 1−x

where |x| < 1.

Next, we consider simple, not necessarily full-dimensional, rational cones. Lemma 13.5. Let u1 , . . . , uk ∈ Zd be linearly independent vectors and let K = co (u1 , . . . , uk ) be the cone spanned by u1 , . . . , uk . Let 1 k G Π= α i ui : 0 ≤ α i < 1

, for i = 1, . . . , k

i=1

be the semi-open parallelepiped spanned by u1 , . . . , uk . Let us deﬁne a Laurent polynomial p(x) by G p(x) = xn . n∈Π∩Zd

Then

G m∈K∩Zd

xm = p(x)

k i=1

1 , 1 − x ui

where the series converges absolutely for all x in the non-empty open set ( N WK = x ∈ Cd : |xui | < 1 for i = 1, . . . , k and uniformly on compact subsets of WK .

110

INTEGER POINTS IN POLYHEDRA

Proof. Let us choose any m ∈ K ∩ Zd , so m=

k G

β i ui

for some βi ≥ 0 for i = 1, . . . , k.

i=1

Letting l=

k G

7βi 6ui

and n =

i=1

k G

{βi }ui ,

i=1

we observe that l is a non-negative integer combination of u1 , . . . , uk , that m = l + n and that n ∈ Π ∩ Zd . By Lemma 10.7, for any point m ∈ K, the representation m = n + l, where n ∈ Π and l is an integer linear combination of u1 , . . . , uk , is unique, so in terms of formal power series we may write:  I C G G xm = xn  xµ1 u1 +...+µk uk  . m∈K∩Zd

n∈Π∩Zd

µ1 ,...,µk ∈Z+

While the ﬁrst factor is the Laurent polynomial p(x), the second factor is a multiple geometric series, see Example 13.4. We note that WK is obviously open and it is non-empty since it contains vectors x = ec for c ∈ int K ◦ . ! The corresponding result for exponential sums states that if u1 , . . . , uk ∈ Λ are linearly independent vectors, K = co (u1 , . . . , uk ), and Π is the semi-open parallelepiped spanned by u1 , . . . , uk , then I C k G G 1 e%c,u5 = e%c,u5 1 − e%c,ui 5 i=1 u∈K∩Λ u∈Π∩Λ for all c ∈ int K ◦ . Figure 52 shows a cone K in the plane spanned by vectors u1 = (1, 1) and u2 = (2, −1). We have G m∈K∩Z2

xm =

1 + x1 + x21 2 A, (1 − x1 x2 ) 1 − x21 x−1 2

where x = (x1 , x2 ) .

Not all simple cones look the same when it comes to their interaction with the lattice. Some cones are simpler than others, see Figure 53.

13. EXPONENTIAL SUMS AND GENERATING FUNCTIONS

u1

111

K Π

0 u

2

Figure 52. A cone and a parallelepiped. Definition 13.6. Let u1 , . . . , uk ∈ Zd be linearly independent vectors and let K = (u1 , . . . , uk ) be the cone they span. The cone K is called unimodular if u1 , . . . , uk is a basis of the lattice Λ = Zd ∩ span (u1 , . . . , uk ) , or, equivalently, if the parallelepiped 1 k , G Π= αi ui : 0 ≤ αi < 1 for i = 1, . . . , k i=1

contains no integer points other than the origin, cf. Theorem 10.9.

0

0

Figure 53. A unimodular (left) and a non-unimodular (right) cone. If K ⊂ Rd is a unimodular cone, the generating function looks especially simple, k G 1 m x = , 1 − x ui d i=1 m∈K∩Z

provided K = co (u1 , . . . , uk ), where {u1 , . . . , uk } is a basis of the lattice Zd ∩ span(K).

112

INTEGER POINTS IN POLYHEDRA

More generally, we say that K ⊂ V is unimodular if K = co (u1 , . . . , uk ) where u1 , . . . , uk is a basis of the lattice Λ0 = Λ ∩ span (u1 , . . . , uk ) in span (u1 , . . . , uk ). If K is unimodular, then G u∈K∩Λ

e%c,u5 =

k i=1

1 1 − e%c,ui 5

◦

for all c ∈ int K . Our next step is to extend Lemma 13.5 to an arbitrary pointed rational cone. Lemma 13.7. Let K ⊂ Rd be a pointed rational cone. Then K = co (w1 , . . . , wn ) for some w1 , . . . , wn ∈ Zd \ {0}. Let us deﬁne N ( WK = x ∈ Cd : |xwi | < 1 for i = 1, . . . , n . Then WK is a non-empty open set and for every x ∈ WK , the series G xm m∈K∩Zd

converges to a rational function f (K, x) of the type G x vi , gi f (K, x) = (1 − xui1 ) · · · (1 − xuid ) i∈I where gi ∈ {1, −1}, vi ∈ Zd and uij ∈ Zd \ {0} for all i and j. Furthermore, the convergence is uniform on compact subsets of WK . Proof. We assume that K >= {0}. We need a rational version of Theorem 4.11. Namely, there is a rational aﬃne hyperplane H such that 0 ∈ / H and a rational polytope P ⊂ H such that K = co(P ). Scaling, if necessary, we assume that P has integer vertices w1 , . . . , wn . Now we proceed as in the proof of Lemma 8.3. We triangulate P and using the inclusion-exclusion formula, write G gj [Kj ], [K] = j∈J

where Kj are simple cones spaned by subsets of linearly independent vectors of {w1 , . . . , wn } and gj = ±1. Next, we use Lemma 13.5 for each cone Kj (note that by multiplying the numerators and denominators of fractions by additional binomials (1 − xu ) we can always assume that each denominator is a product of exactly d binomials).

13. EXPONENTIAL SUMS AND GENERATING FUNCTIONS

113

The set WK is open and non-empty, since it contains x = ec for c ∈ int K ◦ . ! Similarly, if K ⊂ V is a pointed Λ-rational cone, then K = co (w1 , . . . , wn ) for some wi ∈ Λ \ {0} for i = 1, . . . , n and G e%c,u5 u∈K∩Λ ◦

converges for all c ∈ int K to a meromorphic function F (K, c) of the type G e%c,vi 5 F (K, c) = , gi (1 − e%c,ui1 5 ) · · · (1 − e%c,uid 5 ) i∈I where gi ∈ {1, −1}, vi ∈ Λ and uij ∈ Λ \ {0} for all i and j. The main result of this section is the following theorem, proved by A. Khovanski and A. Pukhlikov [KP92], and, independently, by J. Lawrence [L91a], which extends Lemma 13.7 to arbitrary rational polyhedra. 2 A Theorem 13.8a. Let R Cd be the space of rational functions on Cd spanned by the functions of the type xv , (1 − xu1 ) · · · (1 − xud ) where v ∈ Zd and ui ∈ Zd \ {0} for i = 1, . . . , d. Then there exists a linear transformation (valuation) 2 A 2 A F : P Qd −→ R Cd such that the following holds: (1) Let P ⊂ Rd be a non-empty rational polyhedron without lines and let K = KP ⊂ Rd be its recession cone. Let us write K = co (w1 , . . . , wn ) for some wi ∈ Zd \ {0} and let us deﬁne N ( WK = x ∈ Cd : |xwi | < 1 for i = 1, . . . , n . Then WK is a a non-empty open subset and for all x ∈ WK the series G xm m∈P ∩Zd

converges absolutely and uniformly on subsets of WK A 2 compact to a function f (P, x) = F ([P ]) ∈ R Cd . (2) If P contains a line, then F ([P ]) = 0.

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INTEGER POINTS IN POLYHEDRA

Before we prove Theorem 13.8a, we would like to extract some useful corollaries. Let P ⊂ Rd be a non-empty rational polyhedron without lines. Since by the rational version of Theorem 6.4, we have [P ] ≡

G

[tcone(P, v)] modulo rational polyhedra with lines,

v∈Vert(P )

by Theorem 13.8a, we must have G

F ([P ]) =

F ([tcone(P, v)]).

v∈Vert(P )

This identity is known as Brion’s Theorem [Br88]. Another useful observation: if P is a rational polyhedron without lines and a ∈ Zd is an integer vector, we have F ([P + a]) = xa F ([P ]). Indeed, for a formal power series we have: G

xm =

m∈(P +a)∩Zd

G

G

xm+a = xa

m∈P ∩Zd

xm .

m∈Zd

Since there is a non-empty open set of x ∈ Cd for which the series converges absolutely, we have the desired identity. It turns out to be more convenient to prove Theorem 13.8a in terms of exponential sums rather than in terms of generating functions. Theorem 13.8b. Let us ﬁx Euclidean space V , dim V = d, and a lattice Λ ⊂ V . Let M(V ) be the space of functions on V spanned by functions of the type e%c,v5 , (1 − e%c,u1 5 ) · · · (1 − e%c,ud 5 ) where v ∈ Λ and ui ∈ Λ \ {0} for i = 1, . . . , d. Then there exists a linear transformation (valuation) F : P (QV ) −→ M (V )

13. EXPONENTIAL SUMS AND GENERATING FUNCTIONS

115

such that the following holds: (1) Let P ⊂ V be a non-empty rational polyhedron without lines and let K = KP ⊂ V be its recession cone. Then for all c ∈ int K ◦ the series G e%c,u5 u∈P ∩Λ

converges absolutely and uniformly on compact subsets of int K ◦ to a function F (P, c) = F([P ]) ∈ M(V ). (2) If P contains a line, then F ([P ]) = 0. Proof of Theorem 13.8. We prove version 13.8b of the Theorem. We proceed by induction on the dimension d. For d = 0 the result is clear. Let us assume that d ≥ 1. First, we prove that if P ⊂ V is a non-empty rational polyhedron without lines and K = KP is its recession cone, then the series G e%c,u5 u∈P ∩Λ

converges absolutely, for all c ∈ int K ◦ and uniformly on compact subsets of int K ◦ , to a meromorphic function F (P, c) ∈ M(V ). Since P does not contain lines the cone KP is pointed and hence KP◦ has a non-empty interior. Let us choose a vector w ∈ − int KP◦ ∩ Λ∗ . Furthermore, we choose w to be primitive, that is, to form a basis of the 1-dimensional lattice Λ∗ ∩ span(w). We can append w to a basis w1 = w, w2 , . . . , wd of Λ∗ , see Problem 10 of Chapter 10. Let u1 , . . . , ud be the basis of Λ deﬁned by 1 0ui , wj 8 =

1 if i + j = d + 1, 0 otherwise.

Let us consider a family Hk of aﬃne hyperplanes Hk = {x ∈ V :

0x, w8 = k}

for k ∈ Z.

We view Hk as a (d − 1)-dimensional Euclidean space with the origin at kud . Then 1 , d−1 G Λk = Λ ∩ Hk = kud + mi ui : mi ∈ Z for i = 1, . . . , d − 1 i=1

is a lattice in Hk .

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INTEGER POINTS IN POLYHEDRA

Since w ∈ − int K ◦ , Pk = P ∩ Hk is a rational bounded polyhedron (rational polytope), and Pk = ∅ for k < k0 for some k0 , cf. Figure 54.

Hk

P

Figure 54. A polyhedron P and its sections by lattice hyperplanes Hk . We can write (13.8.1)

G

e%c,u5 =

u∈P ∩Λ

+∞ G k=k0

I

G

C e%c,u5 .

u∈Pk ∩Λ

Now, for any ﬁnite set of indices k, the contribution to the sum (13.8.1) is just a ﬁnite sum of terms of the type e%c,u5 , so we are interested in the sum for k ≥ k1 where k1 is suﬃciently large. Without loss of generality, P is unbounded, so for any k ≥ k0 the section Pk is a non-empty Λ-rational polytope. For all suﬃciently large k ≥ k1 , the vertices of Pk are intersections of Hk with inﬁnite edges of P . Let vi (k) for i ∈ I be the vertices of Pk , so vi (k) change linearly with k: vi (k) = ai + kbi

for all k ≥ k1

and all i ∈ I.

Here ai are Λ-rational points and bi are Λ-rational vectors which span edges of the recession cone KP of P . The crucial observation is that since bi are Λ-rational vectors, there is a positive integer M so that M bi ∈ Λ for all i and hence vi (k + M ) is a lattice translation of vi (k) for all k ≥ k1 . Thus vi (k+M ) = vi (k)+zi for some zi ∈ Λ and all k ≥ k1 and i ∈ I. Let us ﬁx some particular k ≥ k1 and consider the subsum in (13.8.1) over the polytopes Pk , Pk+M , Pk+2M , . . .:   +∞ G G  (13.8.2) e%c,u5  . j=0

u∈Pk+jM ∩Λ

13. EXPONENTIAL SUMS AND GENERATING FUNCTIONS

117

We apply the induction hypothesis for each polytope Pk+jM . Namely, G G B 4 e%c,u5 = F tcone(Pk+jM , vi (k + jM )) . u∈Pk+jM ∩Λ

i∈I

Let

Fi (c) = F ([tcone(Pk , vi (k))]), so by the induction hypothesis Fi is a meromorphic function in c. It remains to notice that the tangent cone of Pk+jM at vi (k + jM ) is a translation of the tangent cone at vi (k) by a vector jzi ∈ Λ, so 2 A F [tcone(Pk+jM , vi (k + jM ))] = e%c,jzi 5 Fi (c) and (13.8.2) can be written as +∞ G G j=0 i∈I

e%c,jzi 5 Fi (c) =

G

Fi (c)

I +∞ G

C e%c,jzi 5

j=0

i∈I

=

G i∈I

Fi (c) . 1 − e%c,zi 5

Since zi span edges of the recession cone KP of P , we have 0c, zi 8 < 0 and all the series converge absolutely for c ∈ int K ◦ and uniformly on compact subsets of int K ◦ . Since the series (13.8.1) splits into M disjoint subseries of the type k, k + M, k + 2M, . . . , we proved that for every c ∈ int K ◦ the series G e%c,u5 u∈P ∩Λ

converges absolutely for c ∈ int K ◦ and uniformly on compact subsets of int K ◦ to a meromorphic function F (P, c) ∈ M (V ). The proof is completed as in Theorem 8.4. For every rational polyhedron P ⊂ Λ we deﬁned a function F (P, c) ∈ M(V ) and our immediate goal is to show that the correspondence P −→ F (P, c) respects linear identities among indicators of rational polyhedra P . In other words, if G (13.8.3) αi [Pi ] = 0 i∈I

then (13.8.4)

G

αi F (Pi , c) = 0.

i∈I

Let us choose a decomposition [V ] =

G j∈J

βj [Qj ],

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INTEGER POINTS IN POLYHEDRA

where Qj ⊂ V are rational polyhedra without lines. Then G [Pi ] = [Pi ][V ] = βj [Pi ∩ Qj ]. j∈J

Since the recession cone KPi ∩Qj of Pi ∩ Qj lies in the recession cone KPi of Pi , we have   G G G e%c,u5 = βj  e%c,u5  u∈Pi ∩Λ

j∈J

u∈(Pi ∩Qj )∩Λ

for all c ∈ int K ◦ and all the series converge absolutely for all such c. Therefore, G βj F (Pi ∩ Qj , c) F (Pi , c) = j∈J ◦

on a non-empty open set of K . Since the functions F (Pi , c) and F (Pi ∩ Qj , c) are meromorphic, the identity holds for all c ∈ V . Similarly, multiplying (13.8.3) by [Qj ] we get G αi [Pi ∩ Qj ] = 0. i∈I

Since the recession cone KPi ∩Qj of Pi ∩ Qj lies in the recession cone KQj of Qj , we have   G G αi  e%c,u5  = 0 i∈I

u∈(Pi ∩Qj )∩Λ

for all c ∈ int K ◦ , where all the series converge absolutely for all such c. Therefore, G αi F (Pi ∩ Qj , c) = 0 i∈I ◦

for all c ∈ int K . Since F (Pi ∩ Qj , c) are meromorphic functions, the identity holds for all c. Summarizing, I C G G G βj F (Pi ∩ Qj , c) αi F (Pi , c) = αi i∈I

i∈I

=

G j∈J

and (13.8.4) is established.

I βj

j∈J

G i∈I

C αi F (Pi ∩ Qj , c)

=0

13. EXPONENTIAL SUMS AND GENERATING FUNCTIONS

119

Since the algebra P (QV ) of rational polyhedra is spanned by indicators [Pi ] of rational polyhedra without lines, we get a valuation F : P (QV ) −→ M (V ). It remains to show that F ([P ]) = 0 provided P contains a line. First, we notice that for u ∈ Λ we must have F ([P + u]) = e%c,u5 F ([P ]). This identity holds for rational polyhedra without lines and by linearity for all rational polyhedra. Next, we observe that if P is a rational polyhedron containing a line, then there is a vector u ∈ Λ \ {0} such that P = P + u. Then we must have F ([P ]) = F ([P + u]) = e%c,u5 F ([P ]), from which F ([P ]) = 0.

! Problems

1◦ . Let V be Euclidean space and let Λ ⊂ V be a lattice. Let K ⊂ V be a unimodular cone (with respect to Λ) with non-empty interior. Prove that K ◦ ⊂ V is a unimodular cone with respect to Λ∗ . 2. Let u1 , . . . , ud ∈ Zd be linearly independent vectors. Let K = co (u1 , . . . , ud ). Let 1 d , G ˜ = Π αi ui : 0 < αi ≤ 1 for i = 1, . . . , d . i=1

Prove that

 f (int K, x) = 

G

 x

u

˜ u∈Π∩Λ

d i=1

1 1 − x ui

and deduce that f (int K, x−1 ) = (−1)d f (K, x). 3. Let ρ : V −→ R be a polynomial. Prove the analogue of Theorem 13.8b for valuation Fρ which is deﬁned on rational polyhedra without lines by G Fρ ([P ]) = Fρ (P, c) where Fρ (P, c) = e%c,u5 ρ(u) u∈P ∩Λ

and the sum converges for all c ∈ int KP◦ , where KP is the recession cone of P , and on rational polyhedra with lines by Fρ ([P ]) = 0.

120

INTEGER POINTS IN POLYHEDRA

Hint: Cf. Problem 3 of Chapter 8. 4. Write a proof of Theorem 13.8b along the following lines: consider the embedding Rd ⊂ Rd+1 as the aﬃne hyperplane ξd+1 = 1. For a rational polyhedron P ⊂ Rd , let K = co(P ) be a cone, K ⊂ Rd+1 . Then by Lemma 13.7 the series G xm m∈K∩Zd+1

deﬁnes a rational function f (K, x). Show that we must have ! ∂ f (P, x) = f (K, x)!x =0 . d+1 ∂xd+1

CHAPTER 14

Totally unimodular polytopes We start with a deﬁnition. Definition 14.1. Let us consider the function F (τ ; ξ1 , . . . , ξd ) =

d i=1

τ ξi . 1 − e−τ ξi

Then F is analytic in the neighborhood of τ = ξ1 = · · · = ξd = 0 and hence admits an expansion F (τ, ξ1 , . . . , ξd ) =

+∞ G

τ k tdk (ξ1 , . . . , ξd ) ,

k=0

where tdk is a homogeneous polynomial of degree k, called the Todd polynomial. Thus tdk (ξ1 , . . . , ξd ) is a symmetric polynomial with rational coeﬃcients. In particular, 1G ξi , td1 (ξ1 , . . . , ξd ) = 2 i=1 d

td0 (ξ1 , . . . , ξd ) = 1,

and

1 G 2 1G td2 (ξ1 , . . . , ξd ) = ξi ξj . ξ + 12 i=1 i 4 i<j d

Todd polynomials turn out to be relevant for integer point counting. 14.2 Counting in integral totally unimodular polytopes. Let us consider a polytope P ⊂ Rd of a particularly nice structure. Suppose that the polytope has integer vertices v1 , . . . , vn and that the cone of feasible directions at every vertex vi is a unimodular cone spanned by integer vectors ui1 , . . . , uid , so tcone(P, vi ) = vi + co (ui1 , . . . , uid ) and for every i the vectors ui1 , . . . , uid constitute a basis of Zd .

122

INTEGER POINTS IN POLYHEDRA

Such polytopes are called (integral) totally unimodular polytopes. More generally, given a lattice Λ ⊂ V , we call a polytope P ⊂ V (lattice) totally unimodular or Λ-totally unimodular if the vertices of P belong to Λ and if the cone of feasible directions at every vertex of P is unimodular. Using Theorem 13.8 and the identity [P ] ≡

n G

[tcone(P, vi )] modulo rational polyhedra with lines,

i=1

we write

G

xm =

n G i=1

m∈P ∩Zd

x vi . (1 − xui1 ) · · · (1 − xuid )

Equivalently, G

(14.2.1)

e

%c,m5

n G

=

i=1

m∈P ∩Zd

e%c,vi 5 . (1 − e%c,ui1 5 ) · · · (1 − e%c,uid 5 )

Let us choose a particular c ∈ Rd such that 0c, uij 8 = > 0 for all i, j and let us consider what happens when we scale c 9−→ τ c for some real or complex τ . Expanding eτ %c,m5 =

+∞ G

τk

k=0

we obtain G

eτ %c,m5 =

+∞ G

I

G

τk

k=0

m∈P ∩Zd

0c, m8k , k!

m∈P ∩Zd

0c, m8k k!

C .

In particular, the 0th term of the series is the number |P ∩Zd | of integer points in P . Let us denote ηi = 0c, vi 8 and ξij = −0c, uij 8 for all i, j. Expanding e

τ ηi

=

+∞ G k=0

τk

ηik , k!

14. TOTALLY UNIMODULAR POLYTOPES

123

and 1 (1 −

e−τ ξi1 ) · · · (1

−

e−τ ξid )

(τ ξi1 ) · · · (τ ξid ) 1 −τ ξi1 ) · · · (1 − e−τ ξid ) i1 . . . ξid (1 − e +∞ G 1 τ k tdk (ξi1 , . . . , ξid ) , = d τ ξi1 · · · ξid k=0

=

τ dξ

we conclude that the constant term of the right-hand of (14.2.1) is given by the formula I d C n G G ηi k 1 tdd−k (ξi1 · · · ξid ) . ξ · · · ξid k=0 k! i=1 i1 Hence we obtain the formula (14.2.2)

|P ∩ Zd | =

n G i=1

1 ξi1 · · · ξid

I

d G ηk i

k=0

k!

C tdd−k (ξi1 , . . . , ξid ) ,

where ηi = 0c, vi 8 and ξij = −0c, uij 8 for all i, j, where c is an arbitrary, suﬃciently generic vector (so that ξij >= 0 for all i, j). Let us consider, for example, a triangle P in the plane with the vertices at (0, 0), (n, 0), and (0, n), where n is a positive integer, see Figure 55. (0,n )

0

P

(n ,0 )

Figure 55. A totally unimodular polygon. Let us pick a vector c = (c1 , c2 ), where c1 , c2 >= 0 and c1 >= c2 . We observe that the tangent cone at (0, 0) is unimodular since it is spanned by a basis e1 = (1, 0) and e2 = (0, 1) of Z2 . We compute η1 = 0,

ξ11 = −c1 ,

and ξ12 = −c2 .

The tangent cone at (n, 0) is unimodular since it is spanned by a basis −e1 = (−1, 0) and −e1 + e2 = (−1, 1) of Z2 . We compute η2 = nc1 ,

ξ21 = c1 ,

and ξ22 = c1 − c2 .

124

INTEGER POINTS IN POLYHEDRA

The tangent cone at (0, n) is unimodular since it is spanned by a basis −e2 = (0, −1) and e1 − e2 = (1, −1) of Z2 . We compute η3 = nc2 ,

ξ31 = c2 ,

and ξ32 = c2 − c1 .

Hence formula (14.2.2) says in this case that J % 2 c1 1 c22 c1 c2 2 |P ∩ Z | = + + c1 c2 12 12 4 % 2 c1 (c1 − c2 )2 c1 (c1 − c2 ) 1 + + + c1 (c1 − c2 ) 12 12 4 % J J 2 2 c1 c1 − c2 n c1 + +(nc1 ) + 2 2 2 % 2 (c2 − c1 )2 c2 (c2 − c1 ) 1 c2 + + + c2 (c2 − c1 ) 12 12 4 J J % 2 2 n c2 c2 c2 − c1 + . + +(nc2 ) 2 2 2 Surprising as it may seem, the last expression simpliﬁes to just 1+

3n n2 + , 2 2

which is indeed the number of integer points in the triangle with vertices (0, 0), (n, 0), and (0, n). 14.3 Counting in rational totally unimodular polytopes. Let us consider what happens if the polytope P of Section 14.2 has rational vertices v1 , . . . , vn . We still assume that tcone(P, vi ) = vi + co (ui1 , . . . , uid ) , where ui1 , . . . , uid form a basis of Zd for all i, but we no longer assume that vi are integer points. Such polytopes are called (rational) totally unimodular polytopes. More generally, given a lattice Λ ⊂ V , a polytope P ⊂ V is called (rational) totally unimodular or Λ-rational totally unimodular if for every vertex v of P for some positive integer m = m(v) we have mv ∈ Λ and the cone of feasible directions of at v is unimodular. It turns out that the situation changes very little. Namely, one can easily ﬁnd an integer point wi such that the sets of integer points in wi + co (ui1 , . . . , uid )

and vi + co (ui1 , . . . , uid )

14. TOTALLY UNIMODULAR POLYTOPES

125

coincide. In particular, for generating functions we have f (tcone(P, vi ), x) = xwi f (Ki , x),

for Ki = co (ui1 , . . . , uid ) .

Indeed, let us consider the model case of uij = ej , the standard basis vectors. Thus Ki = K = Rd+ and K + v is a rational translation of the non-negative orthant. Let w be the integer point obtained by rounding the coordinates of v up to the nearest integer (we write w = ;v=). Then the set of integer points in K + w and K + v coincide so we have f (K + v, x) = x

4v.

f (K, x) = x

4v.

d i=1

1 , 1 − xi

see Figure 56.

v

w

Figure 56. The set of integer points in the non-negative orthant shifted by a rational vector v coincides with the set of integer points in the non-negative orthant shifted by an integer vector w. If K = co (u1 , . . . , ud ) is a general full-dimensional unimodular cone, the case reduces to that of Rd+ by a change of the coordinates. Given a point v, we compute its rounding w = ;v=K as follows. We write v=

d G

α i ui

for some αi ∈ R for i = 1, . . . , d

i=1

and let w=

d G ;αi =ui . i=1

Then f (K + v, x) = xw f (K, x) =

xw . (1 − xu1 ) · · · (1 − xud )

126

INTEGER POINTS IN POLYHEDRA

Summarizing, if P is a rational totally unimodular polytope, formula (14.2.2) still holds with the correction that ηi = 0c, wi 8 where wi = ;vi =Ki and Ki + vi is the tangent cone of P at vi . We note that rounding cannot be deﬁned with respect to a nonunimodular cone, see Figure 57.

Figure 57. For a rational translation of a nonunimodular cone, there may be no integer translation containing the same set of integer points. Since unimodular cones turn out to be quite handy, we consider the general problem of decomposing a given rational cone into a combination of unimodular cones. The following simple example shows that there may be several diﬀerent ways to do it. Example 14.4: decomposing a cone into unimodular cones. Let us consider the cone K in the plane spanned by u1 = (1, 0) and u2 = (1, n). On one hand we can dissect K into n unimodular cones by drawing a ray through (1, k) for all integers 1 < k < n. Indeed, it easy to see that the cone spanned by (1, k) and (1, k + 1) is unimodular since the area of the fundamental parallelepiped is ! % J! ! ! k ! = 1. !det 1 ! 1 k+1 ! On the other hand, we can write [K] = [K1 ] − [K2 ] + [K3 ], where K1 is the cone spanned by (0, 1) and (1, 0), K2 is the cone spanned by (0, 1) and (1, n) and K3 is the cone spanned by (1, n), see Figure 58.

14. TOTALLY UNIMODULAR POLYTOPES

(1 , n )

(1 , n )

K 0

127

1

1

K

0

1

Figure 58. A cone K spanned by (1, 0) and (1, n) can be dissected into n unimodular cones or represented as a diﬀerence of just two unimodular cones (not counting the boundary). All three cones are unimodular, so we can write f (K, x) =

1 1 1 − + . (1 − x1 )(1 − x2 ) (1 − x1 xn2 )(1 − x2 ) (1 − x1 xn2 )

Our goal is to decompose rational cones into unimodular cones eﬃciently. Problems 1◦ . Compute td3 (ξ1 , . . . , ξd ). 2. Let P ⊂ Rd be an integral totally unimodular polytope with the vertices v1 , . . . , vn so that tcone(P, vi ) = vi + co (ui1 , . . . , uid )

for i = 1, . . . , n.

Let c ∈ Rd be a vector and let ηi = 0c, vi 8 for i = 1, . . . , n and

ξij = −0c, uij 8 for all i, j. Suppose that ξij >= 0 for all i, j. Prove that for all 0 ≤ s < d we have I s C n G ηk G 1 i tds−k (ξi1 , . . . , ξid ) = 0. ξ · · · ξ k! i1 id i=1 k=0

3. Let H ⊂ Rd+1 be the aﬃne hyperplane deﬁned by the equation ξ1 + · · · + ξd+1 = 1 and let Λ = Zd+1 ∩ H. Let us consider H as a (d − 1)-dimensional Euclidean space V and Λ ⊂ V as a lattice by choosing the origin at any point u ∈ Λ. Let Δ = conv (e1 , . . . , ed+1 ),

128

INTEGER POINTS IN POLYHEDRA

where e1 , . . . , ed+1 is the standard basis of Rd+1 . Prove that Δ is a Λ-totally unimodular polytope. 4. Let a = (a1 , . . . , ad ), a ∈ Rd , be a point with distinct integer coordinates a1 , . . . , ad and let P be the convex hull of all d! points obtained from a by permutations of the coordinates. Let us consider the aﬃne hyperplane H ⊂ Rd deﬁned by the equation ξ 1 + · · · + ξ d = a1 + · · · + ad as a (d − 1)-dimensional Euclidean space V and Λ = Zd ∩ H as a lattice Λ ⊂ V by choosing the origin at a point u ∈ Λ. Prove that P ⊂ V is Λ-totally unimodular. Hint: See, for example, Section VI.2 of [Ba02] for the facial description of P . 5. Let us choose a positive integer m-vector R = (r1 , . . . , rm ) and a positive integer n-vector C = (c1 , . . . , cn ) such that m G i=1

ri =

n G

cj = N.

j=1

mn

In the space R of m × n matrices X = (xij ), let us consider a polyhedron P deﬁned by the equations n G

xij = ri

j=1

and inequalities

for i = 1, . . . , m,

m G

xij = cj

for j = 1, . . . , n

i=1

xij ≥ 0 for all i, j. Let us consider the aﬃne span A of P as (m − 1)(n − 1)-dimensional Euclidean space V and Λ = Zmn ∩ A as a lattice in V by choosing a point u ∈ Λ as the origin. Suppose that P is simple, that is, the cone of feasible directions at every vertex of P is spanned by exactly (m − 1)(n − 1) vectors. Prove that P is Λ-totally unimodular. Hint: See Chapter 19 of [Sc86].

CHAPTER 15

Decomposing a 2-dimensional cone into unimodular cones via continued fractions Let us choose a number a ∈ R. To produce the continued fraction expansion [a0 ; a1 , . . . , an , . . .] of a, we iterate the following procedure. We write a = 7a6 + {a} and let a0 = 7a6. If {a} = 0 we stop. Otherwise, we let b = 1/{a}, so b > 1. We write b = 7b6 + {b} and let a1 = 7b6. If {b} = 0, we stop. Otherwise, we update b by b := 1/{b} and continue as above. In the end, we get a potentially inﬁnite expansion 1

a = a0 +

1

a1 +

a2 +

1 ...

For example, if a = 164/31, we write 9 164 =5+ =5+ 31 31

Hence we write

1 3+

4 9

1

=5+ 3+

1 2+

1 4.

164 = [5; 3, 2, 4]. 31 It is not hard to see that the continued fraction expansion of a is ﬁnite if and only if a is rational.

130

INTEGER POINTS IN POLYHEDRA

Next, we deﬁne the k-th convergent of a as the fraction [a0 ; a1 , . . . , ak ], so 1

a = a0 + a1 +

1 .. . ak−1 +

1 ak .

In our example: 1

[5; 3, 2] = 5 +

3+

1 2

=

37 , 7

[5; 3] = 5 +

1 16 = , 3 3

5 and [5] = . 1

Continued fractions are relevant for unimodular cone decompositions. Theorem 15.1. Let K ⊂ R2 be a cone spanned by (1, 0) and (q, p) where q and p are coprime positive integers. Let p = [a0 ; a1 , . . . , an ] q be the continued fraction expansion of p/q. For i = −1, . . . , n let us deﬁne the cone Ki as follows: we consider the i-th convergent pi = [a0 ; a1 , . . . , ai ] qi

for i = 0, . . . , n.

We deﬁne K−1 as the cone spanned by (1, 0) and (0, 1), K0 as the cone spanned by (0, 1) and (1, p0 ), and Ki as the cone spanned by (qi−1 , pi−1 ) and (qi , pi ) for i = 1, . . . , n. Then Ki are unimodular cones for i = −1, 0, . . . , n. If n is odd, then [K] =

n G

(−1)i+1 [Ki ].

i=−1

If n is even, then [K] = [R] +

n G

(−1)i+1 [Ki ],

i=−1

where R is the ray emanating from the origin in the direction of (qn , pn ).

15. DECOMPOSING A 2-DIMENSIONAL CONE

131

Theorem 15.1, follows from the following technical lemma. Lemma 15.2. Let a = [a0 ; a1 , . . . , an ] be a continued fraction expansion of a rational number a for n ≥ 2. Then, for 0 ≤ i ≤ n one can ﬁnd integers pi and qi such that pi /qi = [a0 ; a1 , . . . , ai ]

for 0 ≤ i ≤ n

and the following identities hold: (1) pi = ai pi−1 + pi−2

and qi = ai qi−1 + qi−2

for

2 ≤ i ≤ n,

(2) pi−1 qi − pi qi−1 = (−1)i

for

1 ≤ i ≤ n.

Proof. We proceed by induction on the length n of the continued fraction expansion of a. For n = 2 we have p2 = a0 a1 a2 + a0 + a2 , q2 = a1 a2 + 1, p1 = a1 a0 + 1, q1 = a1 , p0 = a0 , q0 = 1, and the identities (1) and (2) check. Suppose that n ≥ 2. Let b = [a1 ; a2 , . . . , an ]. Since the continued fraction expansion for b is shorter by 1, we apply the induction hypothesis to claim that for some integers si and ti we have si = [a1 ; a2 , . . . ai ] ti and si = ai si−1 + si−2 and ti = ai ti−1 + ti−2 . We note that a = a0 +

1 b

and, similarly, [a0 ; a1 , . . . , ai ] = a0 + [a1 ; a2 , . . . , ai ]−1 ,

so we deﬁne pi and qi by p i = a0 s i + t i

and qi = ti .

132

INTEGER POINTS IN POLYHEDRA

Therefore, ai pi−1 + pi−2 = ai a0 si−1 + ai ti−1 + a0 si−2 + ti−2 = a0 si + ti = pi , ai qi−1 + qi−2 = ai ti−1 + ti−2 = ti = qi . This proves Part (1). We have already checked Part (2) for i = 1. For i > 1, pi−1 qi − pi qi−1 = pi−1 (ai qi−1 + qi−2 ) − (ai pi−1 + pi−2 ) qi−1 = pi−1 qi−2 − pi−2 qi−1 = −(−1)i−1 = (−1)i . ! It follows from Part (2) of Lemma 15.2 that pi and qi are coprime for all i. Proof of Theorem 15.1. The cone K−1 is obviously unimodular. The cone K0 is unimodular since the area of the fundamental parallelepiped is ! % J! ! ! !det 0 1 ! = 1. ! 1 a0 ! The cone Ki for i = 1, . . . , n is unimodular because the volume of the fundamental parallelogram is ! % J! ! ! !det qi−1 pi−1 ! = |qi−1 pi − pi−1 qi | = 1 ! qi pi ! by Part (2) of Lemma 15.2. Moreover, by Lemma 15.2, the basis consisting of the vectors ui−1 = (qi−1 , pi−1 ) and ui = (qi , pi ) is co-oriented with the basis (0, 1) and (1, 0) for i odd and counter-oriented for i even. Therefore, for any odd k the sum k G

(−1)i−1 [Ki ]

i=−1

is the indicator of the cone spanned by (1, 0) and (qk , pk ), while for k even the sum k G (−1)i−1 [Ki ] i=−1

is the indicator of the cone spanned by (1, 0) and (qk , pk ) minus the indicator of a ray emanating through the origin in the direction of (qk , pk ), see Figure 59. !

15. DECOMPOSING A 2-DIMENSIONAL CONE

133

Hence we compute n G

f (K, x) =

(−1)i+1 f (Ki , x) if n is odd and

i=−1 n G

f (K, x) = f (R, x) +

(−1)i+1 f (Ki , x) if n is even.

i=−1

In particular, for the cone spanned by (1, 0) and (31, 164), we get f (K, x) =

1 1 − (1 − x1 )(1 − x2 ) (1 − x2 )(1 − x1 x52 ) 1 1 + − 5 3 16 3 16 (1 − x1 x2 )(1 − x1 x2 ) (1 − x1 x2 )(1 − x71 x37 2 ) 1 + , 31 164 (1 − x71 x37 2 )(1 − x1 x2 )

cf. Figure 59. 1

−

0

1

1

(1, 5)

+

0

(31, 164)

(3, 16)

−

(7, 37) 0

(3, 16)

+ 0

0

(1, 5)

(31, 164)

=

(7, 37)

0

1

Figure 59. Cutting and pasting unimodular cones to get from the cone spanned by (1, 0) and (0, 1) to the cone spanned by (1, 0) and (31, 164). Some remarks are in order. Remark 15.3. First, we remark that Theorem 15.1 provides an eﬃcient procedure for computing the generating function f (K, x) for an arbitrary 2-dimensional rational cone K. Indeed, we can represent an arbitrary 2-dimensional rational cone K as a combination of two 2dimensional cones, each of which is spanned by (1, 0) and some other integer vector, and one 1-dimensional cone, see Figure 60. Hence it suﬃces to consider the case of a cone K spanned by (1, 0) and some other integer vector (q, p) for coprime q and p.

134

INTEGER POINTS IN POLYHEDRA

= 0

0

= 0

0

−

+ 0

0

−

+ 0

0

Figure 60. Representing a 2-dimensional cone as a combination of two 2-dimensional cones spanned by (1, 0) and some other integer vector and a 1-dimensional cone. If q < 0, we let K1 be the cone spanned by (1, 0) and (−q, −p). We notice that the indicators [K] and [K1 ] diﬀer by a halfplane and some boundary rays. Hence computing f (K, x) reduces to computing f (K1 , x) in a straightforward way. Thus we may assume that K is spanned by (1, 0) and (q, p) with q ≥ 0. If p < the cone K1Aspanned by (1, 0) and (q, −p). 2 0, let us consider A Then f K1 , (x1 , x2 ) = f (K2 , (x1 , x−1 2 ) . Thus we may assume that p, q ≥ 0 in the deﬁnition of K and this case is covered by Theorem 15.1. Second, we remark that the procedure of Theorem 15.1 is polynomial time. Namely, the number of operations performed by the algorithm of the theorem is bounded by a polynomial in the bit size of p and q, which is about log p + log q, see Chapter 2 and Section 6.1 of [Sc86]. A classical reference for the theory of continued fractions is [Kh97]. Problems 1◦ . Compute the generating function f (K, x) for the cone K ⊂ R2 spanned by vectors (1, 0) and (337, 1001). In Problems 2–4 below, let a and pi , qi be numbers as in Lemma 15.2. 2◦ . Prove that qi pi−2 − pi qi−2 = (−1)i−1 ai for i ≥ 2. 3. Prove that

! ! ! ! !a − pi ! ≤ 1 ! qi ! qi qi+1

for i ≥ 0.

15. DECOMPOSING A 2-DIMENSIONAL CONE

135

4∗ . Prove that for i ≥ 2 at least one of the three inequalities ! ! ! ! ! ! ! ! ! ! ! pi−2 !! 1 1 !a − pi ! ≤ 1√ , !a − pi−1 ! ≤ ! √ , or !a − ≤ 2 √ ! ! ! ! ! 2 2 qi qi−1 qi−2 qi 5 qi−1 5 qi−2 5 must hold. Hint: See [Kh97].

√ A 2 5. Compute the continued fractions expansion of a = 1 + 5 /2, interpret convergents in terms of Fibonacci numbers and show that one √ cannot replace 5 by a bigger constant in Problem 4 above. 6. For coprime positive integers p and q, let A ⊂ R2 be the cone spanned by (1, 0) and (q, p) and let B ⊂ R2 be the cone spanned by (1, 0) and (p, q). Prove that 2 A 2 A 1 1 − f A, (x2 , x1 ) + . f B, (x1 , x2 ) = (1 − x1 )(1 − x2 ) 1 − xq1 xp2

CHAPTER 16

Decomposing a rational cone of an arbitrary dimension into unimodular cones In Chapter 15, we presented an eﬃcient algorithm to write a 2dimensional rational cone as a combination of unimodular cones. In this section, we treat cones of arbitrary dimensions d ≥ 2. For the moment, it is convenient to work with an arbitrary lattice, not necessarily Zd . Thus we ﬁx a d-dimensional Euclidean space V and a lattice Λ ⊂ V there. Since every cone can be dissected into simple cones, we assume that our cone K ⊂ V is a simple rational cone, that is spanned by linearly independent vectors u1 , . . . , ud ∈ Λ. Our immediate goal is a computationally eﬃcient procedure to express [K] as a combination [K] =

G

gi [Ki ],

i∈I

where gi = ±1 and Ki ⊂ V are unimodular cones. Thus Ki = co (ui1 , . . . , uiki ) for some ki ≤ d, where ui1 , . . . , uiki is a basis of the lattice Λ ∩ span (ui1 , . . . , uiki ). Our ultimate goal is computing the generating function f (K, x). We start with a deﬁnition. Definition 16.1. Let K = co (u1 , . . . , ud ) be the cone spanned by primitive linearly independent vectors u1 , . . . , ud ∈ Λ and let 1 Π=

d G

, α i ui :

0 ≤ αi < 1 for i = 1, . . . , d

i=1

be the corresponding fundamental parallelepiped of the cone. The index of K (denoted ind K) is the number of lattice points in Π, ind K = |Π ∩ Λ|.

138

INTEGER POINTS IN POLYHEDRA

Thus ind K = 1 if and only if u1 , . . . , ud is a basis of Λ, in which case K is a unimodular cone. By Theorem 10.9, vol Π . det Λ For a subset I ⊂ {1, . . . , k} let us deﬁne ind K =

KI = co (ui :

i ∈ I) .

Thus KI is a face of K. We note that ind KI ≤ ind K, since the fundamental parallelepiped of KI is a subset of the fundamental parallelepiped of K. The main ingredient of our decomposition procedure is the following step of reducing ind K. 16.2 Decomposing a given cone into cones of smaller indices. Given a cone K = co (u1 , . . . , ud ), where ui ∈ Λ are linearly independent primitive vectors, let us consider the parallelepiped Π0 deﬁned by 1 d , G Π0 = αi ui : |αi | ≤ (ind K)−1/d for i = 1, . . . , d . i=1

Obviously, Π0 is centrally symmetric, convex and compact, see Figure 61. Moreover, vol Π = 2d det Λ. vol Π0 = 2d ind K Hence by Theorem 11.1, there is a non-zero vector w ∈ Π0 ∩ Λ. Replacing w 9−→ −w, if necessary, we assume that w and u1 , . . . , ud lie in the same open halfspace, cf. Figure 61. Furthermore, we assume that w is primitive. Π

u1

0

0 w

u

2

Figure 61. The parallelepiped Π0 and the vector w.

16. DECOMPOSING A RATIONAL CONE

139

Let us introduce d new cones Ki = co (u1 , . . . , ui−1 , w, ui+1 , . . . , ud ) (we replaced ui in the deﬁnition of Ki by w). Writing w=

d G

α i ui

i=1

and denoting, as usual, the volume of the parallelepiped spanned by a1 , . . . , ad by |a1 ∧ · · · ∧ ad |, we conclude that ind Ki = (det Λ)−1 |u1 ∧ · · · ∧ ui−1 ∧ w ∧ ui+1 ∧ · · · ∧ ud | = (det Λ)−1 |αi ||u1 ∧ · · · ∧ ui−1 ∧ ui ∧ ui+1 ∧ · · · ∧ ud | = |αi | (ind K) ≤ (ind K)(d−1)/d . We note that the index of every face F of Ki does not exceed (ind K)(d−1)/d as well. Next, we observe that [K] can be written as a linear combination of Ki and their faces F : d G gi [Ki ] + a combination of proper faces of Ki , (16.2.1) [K] = i=1

where  1 if replacing ui by w in u1 , . . . , ud does not change the     orientation, gi =  −1 if replacing ui by w in u1 , . . . , ud changes the orientation,    0 if u1 , . . . , ui−1 , w, ui+1 , . . . , ud are linearly dependent. Indeed, let H ⊂ V be an aﬃne hyperplane such that the rays emanating from 0 in the direction of u1 , . . . , ud and w intersect H at the respective points u(1 , . . . , u(d and w( (the existence of such an H follows from the fact that u(i and w( lie in the same open halfspace). Writing 2 A P = conv (u(1 , . . . , u(d ) and Pi = conv u(1 , . . . , u(i−1 , w( , u(i+1 , . . . , u(d , we observe that d G gi [Pi ] [P ] =

+

a combination of lower-dimensional faces of Pi

i=1

which gives rise to (16.2.1), see Figure 62. One can construct vector w eﬃciently, that is, in polynomial time when the dimension d is ﬁxed in advance. The construction is based on the Lenstra–Lenstra–Lov´asz basis reduction algorithm of Chapter 12.

140

INTEGER POINTS IN POLYHEDRA

0

0

=

0

+

−

u2

u1

u’ u’

1

2

=

w’

w

u1’

u2 w

u’

u’

u ’3

2

u

3

u2

+

w’

2

w’

u1

0

− +

u

u3 u1

0

0

w

u’

3

− u’ w’ 1

u

3

u

2 w

u’3 u’

−

2

+

u

w’ u 1’

3

1

u’

3

Figure 62. A decomposition of a 3-dimensional cone. Let us identify V = Rd and let us construct a linear transformation T : V −→ V such that T (ui ) = ei for i = 1, . . . , d, where e1 , . . . , ed is the standard basis of Zd . Let Λ0 = T (Λ) and let us construct an LLL reduced basis of Λ0 . Let CxC∞ = max {|ξi |,

for i = 1, . . . , d}

where x = (ξ1 , . . . , ξd ) .

Our goal is to construct a vector w0 ∈ Λ0 \ {0} with the smallest value of Cw0 C∞ , cf. also Problem 3 of Chapter 11. We can then let w = T −1 (w0 ). Since the length of the shortest√ non-zero vector from Λ0 cannot be smaller than Cw0 C∞ and Cw0 C ≤ dCw0 C∞ , to construct w0 , by Lemma 12.7 we have to inspect not more that 2O(d) integer combinations of the vectors of the LLL reduced basis of Λ0 . 16.3 Iterating the decomposition. Let us see what happens when we iterate the decomposition of Section 16.2. We assume that the cone K we start with is not unimodular, so ind K ≥ 2. The number of terms in the right-hand side of (16.2.1) is bounded by d2d and the indices of the cones in the right-hand side do not exceed (ind K)(d−1)/d . Iterating the construction n times, we obtain a decomposition of K with at most d−1 n ( ) d n (d2 ) cones with indices not exceeding (ind K) d . Let us choose 3 7 ln ln ind K − ln ln 2 2 d A +1 n = ln d−1 M / = O d ln ln ind K .

16. DECOMPOSING A RATIONAL CONE

141

Then the indices of all the cones in the n-th iteration of the decomposition are strictly less than 2, so we have obtained a unimodular decomposition of K, G [K] = gi [Ki ] for gi = ±1. i∈I

The number |I| of cones in the decomposition is about (ln ind K)O(d

2

ln d)

.

Hence in any particular dimension d the number of cones in a unimodular decomposition of K is bounded by a polynomial in the logarithm of ind K. Consequently, for the generating function f (K, x) we obtain a representation G 1 gi (16.3.1) f (K, x) = u i1 (1 − x ) · · · (1 − xuiki ) i∈I with the number |I| of terms bounded by a polynomial in ln ind K and ki ≤ d for all i ∈ I. Moreover, for all i ∈ I, vectors ui1 , . . . , uiki form a basis of the lattice Λ ∩ span (ui1 , . . . , uiki ). This should be contrasted with Lemma 13.5, which represents f (K, x) as a sum G x vj (16.3.2) f (K, x) = (1 − xu1 ) · · · (1 − xud ) j∈J of |J| = ind K terms. Thus in any ﬁxed dimension d, formula (16.3.2) is exponentially longer than (16.3.1). The procedure of Sections 16.2–16.3 was ﬁrst described in [Ba94]. Paper [DK97] contained some improvements, in particular, the idea to use the Lenstra–Lenstra–Lov´asz algorithm directly to construct vector w (in [Ba94] a more complicated algorithm of integer programming, see [Sc86],was used). 16.4 Getting rid of lower-dimensional terms via dualization. The presence of lower-dimensional cones in the decomposition (16.2.1) is a nuisance, especially since these cones tend to accumulate. In this section we describe a simple way to ignore lower-dimensional cones. Suppose that V = Rd , Λ = Zd . Let K ⊂ Rd be a d-dimensional simple rational cone. Let us consider its polar K ◦ and let us apply the construction of Sections 16.2–16.3 to K ◦ ignoring all lower-dimensional cones.

142

INTEGER POINTS IN POLYHEDRA

Hence after

n = O (d ln ln ind K ◦ ) iterations we obtain a decomposition G [K ◦ ] ≡ gi [Ki ] modulo lower-dimensional rational cones, i∈I

where Ki are unimodular cones and the number |I| of unimodular cones is about (ln ind K ◦ )O(d ln d) . Corollary 6.5 allows us to write G [K] ≡ gi [Ki◦ ] modulo rational cones with lines i∈I

and hence by Theorem 13.8a, we have G gi f (Ki◦ , x). f (K, x) = i∈I

It remains to notice that if Ki ⊂ Rd is a d-dimensional unimodular cone, then Ki◦ ⊂ Rd is also a d-dimensional unimodular cone, cf. Problem 1 of Chapter 13. Summarizing, we obtain a representation G 1 gi , (16.4.1) f (K, x) = u (1 − x i1 ) · · · (1 − xuid ) i∈I where for all i we have gi ∈ {−1, 1}, vectors ui1 , . . . , uid form a basis of Zd and the number |I| of terms is of the order of |I| = (ln ind K ◦ )O(d ln d) . Formula (16.4.1) is preferable to formula (16.3.1) when the cone K is given by a set of homogeneous linear inequalities. Indeed, if > D K = x ∈ Rd : 0x, ui 8 ≤ 0 for i = 1, . . . , d for some linearly independent integer vectors u1 , . . . , ud , then K ◦ = co (u1 , . . . , ud ) and the algorithm is especially straightforward. If a d-dimensional rational cone K ⊂ Rd is not simple, we ﬁrst dissect K ◦ into d-dimensional simple cones Kj : G [K ◦ ] ≡ [Kj ] modulo lower-dimensional rational cones, j∈J

16. DECOMPOSING A RATIONAL CONE

143

write [Kj ] as a combination of d-dimensional unimodular cones ignoring lower-dimensional cones G gi [Kj,i ] modulo lower-dimensional rational cones [Kj ] ≡ i∈Ij

and dualize again GG ◦ [K] ≡ gi [Kj,i ] modulo rational cones with lines. j∈J i∈Ij

Finally, we get f (K, x) =

GG

◦ , x). gi f (Kj,i

j∈J i∈Ij

This version of the algorithm was ﬁrst described in [BP99]. It was based on an observation from [Br88]. 16.5 Getting rid of lower-dimensional terms via semi-open decompositions. Here is another way that allows us to discard lowerdimensional terms in (16.2.1). We start with a useful result of [KV08], which allows us to replace an “approximate” identity modulo lower-dimensional polyhedra by an “exact” identity for “semi-open” polyhedra, see Figure 63.

0

0

=

=

−

+

0

0

+ 0

0

Figure 63. Replacing identities modulo lowerdimensional polyhedra by identities with “semi-open” polyhedra. Theorem 16.5.1. Let Pi , i ∈ I, be a ﬁnite family of d-dimensional polyhedra in a d-dimensional vector space V , and let gi , i ∈ I, be numbers such that G gi [Pi ] ≡ 0 modulo lower-dimensional polyhedra. i∈I

144

INTEGER POINTS IN POLYHEDRA

Suppose that Pi = {x ∈ V :

Zj (x) ≤ βj

for

j ∈ Ji } ,

where Zj : V −→ R are non-zero linear functions and βj are numbers. Let u ∈ V be a vector such that Zj (u) >= 0 for all j. Let us deﬁne 9 6 Zj (x) ≤ βj if j ∈ Ji and Zj (u) < 0 # Pi = x ∈ V : . Zj (x) < βj if j ∈ Ji and Zj (u) > 0 Then

G

0 O gi P#i = 0.

i∈I

Proof. For some polyhedra Pi , i ∈ I0 , and numbers gi , i ∈ I0 , we can write G G gi [Pi ] + gi [Pi ] = 0 where dim Pi < d provided i ∈ I0 . i∈I

i∈I0

Without loss of generality we may assume that u is not parallel to the aﬃne hulls of polyhedra Pi with i ∈ I0 . Let us pick an arbitrary x ∈ V . Then, for all suﬃciently small τ > 0 the point xτ = x + τ u does not lie in the aﬃne hull of Pi for i ∈ I0 and does not satisfy any of the equations Zj (xτ ) = βj . Therefore, for all suﬃciently small τ > 0 we have G G G 0 O gi [Pi ] (xτ ) + gi [Pi ] (xτ ) = 0. gi P#i (xτ ) = i∈I

i∈I

i∈I0

Next, we are going to prove that 0 O 0 O P#i (xτ ) = P#i (x) for all i ∈ I and all suﬃciently small τ > 0, which will complete the proof. Indeed, suppose that Zj (x) ≤ βj and Zj (u) < 0. Then, for any τ > 0 we have Zj (xτ ) = Zj (x) + τ Zj (u) ≤ βj . Similarly, if Zj (x) > βj and Zj (u) < 0 then for all suﬃciently small τ > 0 we still have Zj (xτ ) > βj .

16. DECOMPOSING A RATIONAL CONE

145

Suppose that Zj (x) < βj and Zj (u) > 0. Then, for all suﬃciently small τ > 0 we still have Zj (x) (xτ ) < βj . Similarly, if Zj (x) ≥ βj and Zj (u) > 0, then for any τ > 0 we have Zj (xτ ) = Zj (x) + τ Zj (u) ≥ βj , see Figure 64.

u

u

Figure 64. Keeping a closed halfspace closed or replacing it by an open halfspace depending on the direction of vector u. Therefore, for all suﬃciently small τ > 0 we have G 0 O G 0 O # gi P#i (x) = 0, gi Pi (xτ ) = i∈I

i∈I

which completes the proof.

!

Theorem 16.5.1 allows us to iterate the decomposition of Sections 16.2–16.3 disregarding lower-dimensional cones at every step. Suppose again that V = Rd and Λ = Zd . Let us choose a suﬃciently generic u ∈ int K. Then, by Theorem 16.5.1, we obtain a decomposition G 0 O Ei , gi K (16.5.2) [K] = i∈I

Ei are “semi-open” unimodular cones. A typical where gi = ±1 and K Ei is of the form K , 1 d G αj > 0 for j ∈ Ji and Ei = , αi uij , where K αj ≥ 0 for j ∈ / Ji j=1

146

INTEGER POINTS IN POLYHEDRA

for some set Ji ⊂ {1, . . . , d}, where ui1 , . . . , uid is a basis of Zd for all i. One can show that M / G x vi Ei , x = where v = uij , f K i (1 − xui1 ) · · · (1 − xuid ) j∈J i

see Figure 65.

0

K

# Figure 65. Integer points in a semi-open cone K. Hence, from (16.5.2) we obtain (16.5.3)

f (K, x) =

G

gi

i∈I

x vi , (1 − xui1 ) · · · (1 − xuid )

where g ∈ {−1, 1}, vi ∈ Zd , and ui1 , . . . , uid is a basis of Zd for all i. The number |I| of terms in (16.5.3) is about |I| = (ln ind K)O(d ln d) , which is essentially smaller than in (16.3.1). This version of the algorithm was suggested in [KV08]. Summarizing, we obtain the following result. Theorem 16.6. Let us ﬁx the dimension d and let K ⊂ Rd be the cone spanned by d linearly independent integer vectors. Let G f (K, x) = xm m∈K∩Zd

be the generating function of K. Then f (K, x) admits a representation (16.6.1)

f (K, x) =

G i∈I

gi

1 (1 −

xui1 ) · · · (1

− xuid )

,

16. DECOMPOSING A RATIONAL CONE

147

where gi = ±1, ui1 , . . . , uid is a basis of Zd for all i ∈ I, and the number |I| of terms satisﬁes |I| = (ln ind K ◦ )O(d ln d) , and a representation (16.6.2)

f (K, x) =

G i∈I

gi

x vi , (1 − xui1 ) · · · (1 − xuid )

where gi = ±1, ui1 , . . . , uid is a basis of Zd for all i ∈ I, vi ∈ Zd for all i and the number |I| of terms satisﬁes |I| = (ln ind K)O(d ln d) . Moreover, both representations (16.6.1) and (16.6.2) can be computed in polynomial time. 16.7 Finding an “approximate” decomposition. In Section 16.2, it may be computationally more eﬃcient not to look for the “perfect” vector w but for a “reasonably good” vector w˜ instead. As in Section 16.2, let us identify V = Rd and let us construct a linear transformation T : V −→ V such that T (ui ) = ei for i = 1, . . . , d. Let Λ0 = T (Λ). Let us construct an LLL reduced basis v1 , . . . , vd of Λ0 and let w˜ = T −1 (v1 ). Thus w˜ ∈ Λ \ {0} and writing w˜ =

d G

β i ui ,

i=1

one can show that |βi | ≤ 2

d−1 2

√

d (ind K)−1/d

for i = 1, . . . , d.

If we iterate the decomposition as in Section 16.3, using each time an approximate vector w, ˜ we obtain a decomposition [K] =

G

gi [Ki ] where

ind Ki ≤ 2

d(d−1) 2

dd/2 .

i∈I

Consequently, f (K, x) =

G

gi f (Ki , x),

i∈I

where every generating function f (Ki , x) is computed as in Lemma 13.5.

148

INTEGER POINTS IN POLYHEDRA

Problems 1◦ . Let u1 , . . . , ud be a basis of Zd , let I ⊂ {1, . . . , d} and let 1 d , G αi > 0 for i ∈ I #= . K αi ui , where α / I i ≥ 0 for i ∈ i=1 Prove that / M # x = f K,

xv (1 − xu1 ) · · · (1 − xud )

where v =

G

ui .

i∈I

2◦ . Let u1 , . . . , ud be a basis of Zd , let K = co (u1 , . . . , ud ), let # be deﬁned as in Problem 1 above. Let a ∈ Rd I ⊂ {1, . . . , d} and let K be a rational point. Prove that M / G xb+v # ui where v = f a + K, x = (1 − xu1 ) · · · (1 − xud ) i∈I and b = ;a=K is rounding of a with respect to K, see Section 14.3. 3◦ . Let K = co(u1 , . . . , ud ) for some linearly independent vectors u1 , . . . , ud ∈ Zd , let T : Rd −→ Rd be a linear transformation such that T (ui ) = ei for i = 1, . . . , d, where e1 , . . . , ed is the standard basis of Zd . Let Λ = T (Zd ) and let v1 , . . . , vd be a LLL reduced basis of Λ. Let w = T −1 (v1 ) and let us write w=

d G

α i ui .

i=1

Prove that

√ |αi | ≤ 2d−1 d (ind K)−1/d

Hint: See Problem 2 of Chapter 12.

for i = 1, . . . , d.

CHAPTER 17

Eﬃcient counting of integer points in rational polytopes Let P ⊂ Rd be a rational polytope. The polytope may be given by the list of its vertices or by a list of deﬁning inequalities and our goal is to compute the number |P ∩ Zd | of integer points in P . We describe some variations of the general approach to this problem. Without loss of generality, we assume that dim P = d. All versions of the algorithm compute ﬁrst the generating function G xm . f (P, x) = m∈P ∩Zd

There are several ways to do that. Each has its own advantages depending on circumstances. 17.1 Using the tangent cone decomposition (dual). We compute the generating function f (P, x) as a rational function of the form (17.1.1)

f (P, x) =

G i∈I

gi

x vi , (1 − xui1 ) · · · (1 − xuid )

where gi ∈ {−1, 1}, vi ∈ Zd and ui1 , . . . , uid is a basis of Zd for all i ∈ I. To obtain (17.1.1), we compute the set Vert(P ) of vertices of P , for every vertex v ∈ Vert(P ), we compute the cone Kv = fcone(P, v) of feasible directions at v, and then decompose every cone Kv into a linear combination of unimodular cones Ki modulo rational cones with lines, as in Section 16.4: G [Kv ] ≡ gi [Ki ] modulo rational cones with lines. i∈Iv

This gives us the decomposition of the tangent cones tcone(P, v) = v + Kv , G tcone(P, v) = gi [Ki + v] modulo rational cones with lines. i∈Iv

150

INTEGER POINTS IN POLYHEDRA

Since [P ] ≡

G

tcone(P, v) modulo rational cones with lines,

v∈Vert(P )

by Theorem 13.8b we obtain f (P, x) =

G

G

gi f (Ki + v, x).

v∈Vert(P ) i∈Iv

Now, let vi be the rounding of v with respect to the cone Ki , vi = ;v=Ki , see Section 14.3. Assuming that Ki = co(ui1 , . . . , uid ), we obtain f (Ki + v, x) =

x vi (1 − xui1 ) · · · (1 − xuid )

and hence (17.1.1). Note that if the vertices v of P are integral, each vi in (17.1.1) is a vertex of P . 17.2 Using the tangent cone decomposition (primal). We compute the generating function f (P, x) as a rational function of the form G xvi +wi , gi (17.2.1) f (P, x) = (1 − xui1 ) · · · (1 − xuid ) i∈I where gi ∈ {−1, 1}, wi ∈ Zd and ui1 , . . . , uid is a basis of Zd for all i ∈ I. Expression (17.2.1) looks identical to (17.1.1), but the meaning of the exponents vi + wi is diﬀerent. As in Section 17.1, we compute the set Vert(P ) of vertices of P and for every vertex v ∈ Vert(P ) we compute the cone Kv = fcone(P, v) of feasible directions at v. Now, as in Section 16.5 we compute the “semi-open” unimodular decomposition G Ei ]. gi [ K [Kv ] = i∈Iv

Here for each i ∈ I there is a basis ui1 , . . . , uid of Zd and a subset Ji ⊂ {1, . . . , d} such that 1 d , G αj > 0 for j ∈ Ji Ei = K αj uij : . αj ≥ 0 for j ∈ / Ji j=1

17. EFFICIENT COUNTING OF INTEGER POINTS

151

Letting wi =

G

uij

and vi = ;v=K

for K = co (ui1 , . . . , uid )

j∈Ji

we obtain (17.2.1), cf. Problem 2 of Chapter 16. 17.3 Using homogenization. Let us consider Rd embedded in Rd+1 as the aﬃne hyperplane ξd+1 = 1 and let K = co(P ) be the cone based on P , see Figure 66.

P K 0

Figure 66. A polytope P ⊂ Rd and a cone K ⊂ Rd+1 based on P . Thus K is a rational cone and we compute the generating function f (K, x). We note that f (P, x) =

! ∂ ! f (K, x)! . ∂xd+1 xd+1 =0

If P is deﬁned by its vertices, it is convenient to use the “semi-open” decomposition of Section 16.5 to compute f (K, x). If v1 , . . . , vn ∈ Rd are the vertices of P , then K is spanned by vectors (vi , 1) for i = 1, . . . , n. If P is deﬁned by a system of linear inequalities, it is convenient to use the dual decomposition of Section 16.4. If P is deﬁned by the system of linear inequalities d G j=1

aij ξj ≤ bi

for i = 1, . . . , m,

152

INTEGER POINTS IN POLYHEDRA

then K is deﬁned by the inequalities −bi ξd+1 +

d G

aij ξj ≤ 0 for i = 1, . . . , m and

j=1

ξd+1 ≥ 0, and K ◦ is spanned by vectors (0, . . . , 0, −1) and (ai1 , . . . , aid , −bi ) for i = 1, . . . , m. The algorithm along the lines of Section 17.1 was suggested in [BP99], after a more complicated, “non-implementable” version was suggested in [Ba94]. The algorithm along the lines of Section 17.2 was suggested in [KV08], whereas the algorithm along the lines of Section 17.3 was suggested in [D+04] with some of the ideas going back to [DK97]. The advantage of the approach of Section 17.3 is that it does not require us to compute the vertices of P and the tangent cones at those vertices. The advantage of the approaches of Sections 17.1 and 17.2 is that they can simultaneously treat the whole family of polytopes with the given set of cones of feasible directions at the vertices. After the generating function f (P, x) is computed, we substitute x = (1, . . . , 1) as in Chapter 14. Suppose, for example, we compute f (P, x) as in Section 17.1, f (P, x) =

G i∈I

gi

x vi . (1 − xui1 ) · · · (1 − xuid )

We ﬁnd a vector c ∈ Rd such that ξij = −0c, uij 8 = > 0 for all i and j. Letting ηi = 0c, vi 8 we compute the number |P ∩ Zd | of integer points in P as (17.4)

|P ∩ Zd | =

G i∈I

1 gi ξi1 · · · ξid

I

d G ηk i

k=0

k!

C tdd−k (ξi1 , . . . , ξid ) .

If the dimension d is ﬁxed in advance, we obtain a polynomial time algorithm to count integer points in a rational polytope P . By now, several computer packages are publicly available to count integer points. These are LattE [D+04], LattE macchiato [Ko07], and barvinok [V+07].

17. EFFICIENT COUNTING OF INTEGER POINTS

153

Problem 1. Let us ﬁx dimension d. Construct a polynomial time algorithm, which, given a rational polytope P ⊂ Rd and a polynomial ρ : Rd −→ R computes G ρ(m). m∈P ∩Zd

Hint: Use that J % ! G ∂ ∂ ! f (P, x)! , , . . . , xd ρ(m) = ρ x1 ∂x ∂x x=1 1 d d m∈P ∩Z

see [Ba06] for detail.

CHAPTER 18

The polynomial behavior of the number of integer points in polytopes Let us consider how the number of integer points in a polytope changes if the polytope evolves in such a way that the cones of feasible directions at the vertices remain intact, see Deﬁnition 8.5. Our main result is as follows. Theorem 18.1. Let {Pα : α ∈ A} be a family of d-dimensional polytopes Pα = conv (v1 (α), . . . , vn (α)) , d where vi (α) ∈ Z and the cones of feasible directions at vi (α) do not depend on α: fcone (Pα , vi (α)) = Ki

for

i = 1, . . . , n.

Then there exists a polynomial d p:R · · × RKd −→ R < × ·H: n times

with the following properties: (1) For all α ∈ A we have |Pα ∩ Zd | = p (v1 (α), . . . , vn (α)) , that is, the number of integer points in Pα is a polynomial in the coordinates of the vertices of Pα ; (2) For all α ∈ A we have | int Pα ∩ Zd | = (−1)d p (−v1 (α), . . . , −vn (α)) , that is, the number of integer points in the interior of P is equal, up to a sign, to the value of the polynomial at the points −v1 (α), . . . , −vn (α) opposite to the vertices of Pα ; (3) We have p(0, . . . , 0) = 1, that is, the constant term of p is 1.

156

INTEGER POINTS IN POLYHEDRA

Proof. We have [Pα ] ≡

n G

[tcone (Pα , vi (α))]

modulo rational polyhedra with lines,

i=1

cf. Theorem 6.4 and Section 13.3. Therefore, by Theorem 13.8b, for the exponential sum G F (Pα , c) = e%c,m5 m∈Pα ∩Zd

we have the identity F (Pα , c) =

n G

F (tcone (Pα , vi (α)) , c) .

i=1

Since tcone (Pα , vi (α)) = vi (α) + Ki and vi (α) ∈ Zd , we obtain F (tcone (Pα , vi ) , c) = e%c,vi (α)5 F (Ki , c). Thus we have F (Pα , c) =

n G

e%c,vi (α)5 F (Ki , c).

i=1

Let us ﬁx a suﬃciently generic c ∈ Rd and let us consider functions fi : R −→ R deﬁned by fi (τ ) = F (Ki , τ c). As in Chapters 14 and 17, fi (τ ) admits a Laurent expansion in a neighborhood of τ = 0, (18.1.1)

fi (τ ) = F (Ki , τ c) =

+∞ G

aj (Ki , c)τ j .

j=−d

Combining (18.1.1) with the Taylor expansion e

%τ c,vi (α)5

=

+∞ G 0c, vi (α)8j j=0

j!

τ j,

18. THE POLYNOMIAL BEHAVIOR

2

157

we obtain that the constant term of the Laurent expansion of F Pα , τ c in a neighborhood of τ = 0 is I d C n G 0c, vi (α)8j G d (18.1.2) |Pα ∩ Z | = a−j (Ki , c) . j! j=0 i=1

A

Hence we deﬁne

I d C n G 0c, vi 8j G p (v1 , . . . , vn ) = a−j (Ki , c) , j! j=0 i=1

which proves Part (1). To prove Part (2) we note that by Theorem 7.4, G [int Pα ] ≡ [int tcone (Pα , vi (α))] modulo rational polyhedra i=1n with lines. As above

int tcone (Pα , vi (α)) = vi (α) + int Ki

and hence F (int Pα , c) =

n G

e%c,vi (α)5 F (int Ki , c) .

i=1

By Theorem 7.3, [int Ki ] ≡ (−1)d [−Ki ]

modulo polyhedra with lines

and hence F (int Ki , c) = (−1)d F (−Ki , c) = (−1)d F (Ki , −c) . Summarizing, F (int Pα , c) = (−1)

d

n G

e%c,vi (α)5 F (Ki , −c) .

i=1

From (18.1.1) we get the following Laurent expansion of F (Ki , −τ c) in a neighborhood of τ = 0: F (Ki , −τ c) =

+∞ G

aj (Ki , c)(−τ )j .

j=−d

Comparing the constant terms of Laurent expansions, we obtain C I d n j G G d d j 0c, vi (α)8 a− j(Ki , c) . | int Pα ∩ Z | = (−1) (−1) j! i=1 j=0

158

INTEGER POINTS IN POLYHEDRA

In other words, | int Pα ∩ Zd | = (−1)d p (−v1 (α), . . . , −vn (α)) and Part (2) follows. To prove Part (3), we remark that by Theorem 6.6, n G

[Ki ] ≡ [0] modulo rational polyhedra with lines,

i=1

which implies

n G

F (Ki , c) = 1,

i=1

and, consequently,

n G

a0 (Ki , c) = 1

i=1

in (18.1.1). On the other hand, p(0, . . . , 0) =

n G

a0 (Ki , c),

i=1

which completes the proof of Part (3).

!

The identity of Part (2) of Theorem 18.1 is often called the reciprocity relation. Example 18.2. Let Pm,n ⊂ R1 be the interval with integer endpoints m < n. Then |Pm,n ∩ Z| = n − m + 1, so we deﬁne p(m, n) = n − m + 1. We observe that p(−m, −n) = −n + m + 1 = −(n − m − 1) = −| int Pm,n ∩ Z|. Moreover, p(0, 0) = 1, just as predicted by Theorem 18.1. One can observe that Part (3) essentially says that the polynomial formula for the number of integer points in Pα still holds if the polytope contracts to the origin in Rd . One can ask if the polynomial formula still holds under milder degenerations, such as the one in Figure 67. For instance, if in Example 18.2 the interval [m, n] contracts to the point m = n, then the formula p(m, n) = n − m + 1 still predicts the correct number of integer points in the degenerate polyhedron.

18. THE POLYNOMIAL BEHAVIOR

159

v =v 3 4 v3 v

1

v4 v

2

v

1

v

2

Figure 67. A degeneration of a lattice polygon. Theorem 18.3. Under conditions of Theorem 18.1, let P = (v1 , . . . , vn ) be a polytope, where vi ∈ Zd for i = 1, . . . , n are not necessarily distinct points and such that in an arbitrarily small neighborhood of vi there is a point vi( such that, for the polytope P ( = conv (v1( , . . . , vn( ) , one has fcone (P ( , vi( ) = Ki

for i = 1, . . . , n.

Then (1) We have |P ∩ Zd | = p (v1 , . . . , vn ) . In other words, the polynomial expression for the number of integer points in Pα still holds when Pα degenerates into P ; (2) Let us assume that dim P = k. Then | int P ∩ Zd | = (−1)k p (−v1 , . . . , −vn ) , where the interior is taken with respect to the aﬃne hull of P . In other words, the reciprocity relation with the modiﬁed sign still holds when Pα degenerates into P . Proof. Let w1 , . . . , wm be the distinct vertices of P . Then the set {1, . . . , n} can be represented as a disjoint union of the subsets Ij for j = 1, . . . , m such that in arbitrarily small neighborhoods of wj one can choose points vi( , i ∈ Ij so that conv (v1( , . . . , vn( ) is strongly combinatorially isomorphic to Pα , see Deﬁnition 8.5.

160

INTEGER POINTS IN POLYHEDRA

Now we are going to use a trick similar to that of the proof of Theorem 6.6. Let us consider the cone fcone(P, wj ) of feasible directions of P at wj . We have 9 6 ◦ d fcone(P, wj ) = c ∈ R : max0c, x8 = 0c, wj 8 , x∈P

that is, the polar of fcone(P, wj ) consists of the linear functions whose maximum on P is attained at wj . Similarly, the polar Ki◦ of the cone of feasible directions of P ( at vi( consists of the linear function whose maximum on P ( is attained at vi( . It is then not hard to see that G [fcone (P, wj )◦ ] ≡ [Ki◦ ] modulo rational polyhedra in proper i∈Ij subspaces, see Figure 68.

K

o

o

K4

3

v

w3

v4

3

v

v

1

v

1

2

v

2

Figure 68. As vertices v3 and v4 merge, the polars of the cones of feasible directions at v3 and v4 also merge. By Corollary 6.5, (18.3.1)

[fcone(P, wj )] ≡

G

[Ki ]

i∈Ij

modulo rational polyhedra with lines. By Theorem 7.3, [int fcone(P, wj )] ≡ (−1)k [− fcone(P, wj )]

modulo rational polyhedra with lines

and [int Ki ] ≡ (−1)d [−Ki ]

modulo rational polyhedra with lines.

18. THE POLYNOMIAL BEHAVIOR

161

Therefore, by (18.3.1), (18.3.2)

[int fcone(P, wj )] ≡(−1)d−k

n G

[int Ki ]

i=1

modulo rational polyhedra with lines. Therefore, by Theorem 13.8b we deduce from (18.3.1) and (18.3.2) G F (fcone(P, wj ), c) = F (Ki , c) and i∈Ij

F (int fcone(P, wj ), c) = (−1)

d−k

n G

F (int Ki , c) .

i=1

Now, F (P, c) =

m G

e%c,wj 5 F (fcone(P, wj ), c) =

j=1

m G

  G e%c,wj 5  F (Ki , c)

j=1

i∈Ij

and F (int P, c) =

m G

e%c,wj 5 F (fcone(int P, wj ), c)

j=1

= (−1)d−k

m G

  G e%c,wj 5  F (int Ki , c)

j=1

i∈Ij

and the proof proceeds as in Theorem 18.1.

!

Let P ⊂ Rd be a polytope with integer vertices. Then the family {Pn = nP } of polytopes for positive integer n satisﬁes the conditions of Theorem 18.1. The corresponding polynomial p(n) = |nP ∩ Zd | is called the Ehrhart polynomial of P , after E. Ehrhart who discovered the existence of such polynomials [Eh67]. For example, if Δ ⊂ R2 is the triangle with vertices at (0, 0), (1, 0), and (0, 1), then p(n) = 1 + as follows from Section 14.2.

3n n2 + , 2 2

162

INTEGER POINTS IN POLYHEDRA

Theorem 18.1 in a more general situation was proved by P. McMullen, see [Mc77] and especially [Mc78]. Theorem 18.3 can also be deduced from [Mc78]. One can view Theorem 18.3 as a certain “continuity property” of polynomials of Theorem 18.1. Often, it is convenient to restate Theorems 18.1 and 18.3 in terms of normal fans of polytope. Let V be Euclidean space, let P ⊂ V be a polytope with the vertices v1 , . . . , vn , and let Ki be the cone of feasible directions of P at vi . The polar Ki◦ ⊂ Rd consists of all vectors c ∈ Rd such that the maximum of the linear function 0c, x8 on P is attained at vi . The cones Ki◦ form a dissection of V , which is called the normal fan of P (see Theorem 6.6). Part (1) of Theorem 18.1 states that as long as the normal fan is ﬁxed, the number of integer points in an integer polytope is a polynomial in the vertices of the polytope. Suppose that the normal fan of a polytope Q ⊂ V is a coarsening of the normal fan of P , that is, every cone in the normal fan of Q is a union of some cones of the normal fan of P . Part (1) of Theorem 18.3 states that the polynomial formula for the number of integer points in P still holds for such an integer polytope Q. Finally, we discuss the case of rational polytopes. Theorem 18.4. Let {Pα : topes

α ∈ A} be a family of d-dimensional poly-

Pα = conv (v1 (α), . . . , vn (α)) , where vi (α) ∈ Qd and the cones of feasible directions at vi (α) do not depend on α: fcone (Pα , vi (α)) = Ki

for

i = 1, . . . , n.

Suppose further, that there are vectors u1 , . . . , un ∈ Qd such that vi (α) − ui ∈ Zd

for all α ∈ A

and

i = 1, . . . , n.

Then there exists a polynomial d p:R · · × RKd −→ R < × ·H: n times

such that for all α ∈ A.

! ! !Pα ∩ Zd ! = p (v1 (α), . . . , vn (α))

18. THE POLYNOMIAL BEHAVIOR

163

Proof. As in the proof of Theorem 18.1, we obtain F (P, c) = =

n G i=1 n G

F (Ki + vi (α), c) =

n G

e%c,

vi (α)−ui 5

F (Ki + ui , c)

i=1

e%c,

vi (α)−ui 5

F (Ki( , c) ,

i=1

where Ki( = Ki + ui are rational shifts of the cones Ki . As before, for a generic c and τ in a neighborhood of the origin, we have the Laurent expansion F

(Ki( , τ c)

=

+∞ G

aj (Ki( , c)τ j .

j=−d

As in the proof of Theorem 18.1, we let I d C n G 0c, vi − ui 8j G ( p (v1 , . . . , vn ) = a−j (Ki , c) . j! j=0 i=1 ! Let P ⊂ Rd be a rational polytope. Then, for some positive integer k the dilated polytope kP has integer vertices. Let us pick an integer 0 ≤ j < k. The family of polytopes jP, (j + k)P, (j + 2k)P, . . . satisﬁes the conditions of Theorem 18.4. Thus for any ﬁxed j there is a polynomial pj such that |(j + kmP ) ∩ Zd | = pj (m).

(18.5) In other words, |P ∩ mP | =

d G

bj (m)mj

for all positive integers m,

j=0

where bj (m) are periodic: bj (m + k) = bj (m) for all positive integers m. Expression (18.5) is called the Ehrhart quasi-polynomial after E. Ehrhart who discovered the phenomenon [Eh67]. Theorem 18.4 can be easily deduced from results of [Mc78].

164

INTEGER POINTS IN POLYHEDRA

Problems 1◦ . Let P ⊂ Rd be a d-dimensional polytope with integer vertices and let p be its Ehrhart polynomial. Prove that p is a polynomial of degree d with the constant term 1, the highest coeﬃcient equal to the volume of P and such that p(−n) = |n int P ∩ Zd | for a positive integer n. 2. Let P ⊂ R3 be the tetrahedron with the vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, a), where a > 0 is an integer parameter. Prove that the Ehrhart polynomial p of P satisﬁes a 12 − a n + 1. p(n) = n3 + n2 + 6 6 3. Let P1 , . . . , Pk ⊂ Rd be polytopes with integer vertices. Prove that there exists a k-variate polynomial p such that ! ! p (m1 , . . . , mk ) = !(m1 P1 + · · · + mk Pk ) ∩ Zd ! for non-negative integers m1 , . . . , mk . Hint: Notice that for typical m1 , . . . , mk the normal fan of m1 P1 + · · · + mk Pk is a common reﬁnement of the normal fans of P1 , . . . , Pk . 4. Let us ﬁx a positive integer k. Prove that there exists a univariate polynomial p of degree (k − 1)2 such that: (a) for a positive integer m the value of p(m) is equal to the number of k × k non-negative integer matrices with the row and column sums equal to m; (b) p(0) = 1, p(−1) = · · · = p(−k + 1) = 0, and p(−m) = (−1)k−1 p(m − k) for integer m ≥ k. Hint: Cf. Problem 7 of Chapter 4. This is a conjecture of H. Anand, V.C. Dumir, and H. Gupta proved by R.P. Stanley, see [St73]. 5. Let us ﬁx a polynomial ρ : Rd −→ R and let {Pα } be a family of polytopes as in Theorem 18.1. Prove that there exists a polynomial p : Rd × · · · × Rd −→ R such that G ρ(m) = p (v1 (α), . . . , vn (α)) m∈Pα ∩Zd

for all α. Hint: See Problem 3 of Chapter 13.

18. THE POLYNOMIAL BEHAVIOR

165

6◦ . Let P ⊂ Rd be a d-dimensional rational polytope and let p(m) =

d G

bj (m)mj

j=0

be its Ehrhart quasi-polynomial. Prove that bd (m) = vol P . 7. Construct an example of a rational but not integer polytope P ⊂ Rd such that its Ehrhart quasi-polynomial is actually a genuine polynomial. Hint: For ways to construct such polytopes, examples, and conjectures of what such polytopes may look like, see [HM08].

CHAPTER 19

A valuation on rational cones Our next goal is to connect the exponential valuation Φ (Chapter 8) and the counting valuation F (Chapter 13) with the ultimate goal to express the number |P ∩ Zd | of integer points in a rational polytope P in terms of the volumes of faces of P and a certain valuation on the tangent cones at those faces. These results were quite recently obtained by N. Berline and M. Vergne [BV07]. We follow their exposition. We have to deal with an arbitrary lattice, not necessarily Zd . 19.1 Valuation ψ. Let V be Euclidean space with a lattice Λ ⊂ V , let K ⊂ V be a rational polyhedral cone and let A = K +v be a rational translation of K. We deﬁne a meromorphic function ψ(A, V, Λ; c) where c ∈ V or, more generally, c ∈ CV = V ⊕ iV . We often suppress the dependence on V and Λ and just write ψ(A, c). We deﬁne ψ recursively on d = dim V . If d = 0 we let ψ(A, c) = 1. Suppose that d > 0. Let L ⊂ V be a lattice subspace, that is, a subspace spanned by points from Λ, see Chapter 10. Assuming that dim L < d, let dxL be the Lebesgue measure on L normalized by the condition that det(Λ ∩ L) = 1. Let ΦL be the exponential valuation on L (see Chapter 8) deﬁned as the extension of the integral ; φL (P, c) = e%c,x5 dxL , P

where P ⊂ L is a rational (with respect to the lattice Λ∩L )polyhedron and c ∈ V . Let us ﬁx a rational point v ∈ V and let us consider a rational translation L + v of a lattice subspace L ⊂ V . We denote the exponential valuation on L + v also by ΦL . This should not lead to confusion since the choice of v will always be clear.

168

INTEGER POINTS IN POLYHEDRA

Let L⊥ ⊂ V be the orthogonal complement to L which we also denote V /L and let Λ/L be the orthogonal projection of Λ onto V /L. We note that by Corollary 10.3, the set Λ/L is a lattice in V /L and that Λ/L is generally ﬁner than the lattice L⊥ ∩ Λ, see Figure 69.

L 0

V/L

Figure 69. A lattice subspace L, the quotient V /L, a lattice Λ (black dots) and the lattice Λ/L (black and white dots on V /L). If A ⊂ V is a rational shift of a rational cone, then its orthogonal projection A/L onto L⊥ is a rational shift of a rational cone in V /L and hence ψ(A/L, V /L, Λ/L; c) is deﬁned for c ∈ V /L. We extend this function to c ∈ V by letting ψ(A/L, V /L, Λ/L; c) = ψ(A/L, V /L, Λ/L; c/L), where c/L is the orthogonal projection of c onto L⊥ , see Figure 70. V/L c/L

c

L

0 A A/L

Figure 70. A shifted cone A, its projection A/L, a vector c and its projection c/L. Finally, let F be the valuation on V deﬁned as the extension of exponential sums G F (A, c) = e%c,m5 , see Theorem 13.8b.

m∈A∩Λ

19. A VALUATION ON RATIONAL CONES

169

Assuming that A = K + v, we deﬁne ψ(A, c) so that the following identity holds, (19.1.1)

G

F (A, c) =

ψ(A/L, c)φL (A ∩ (L + v), c),

L

where the sum ranges over all lattice subspaces of V . In particular, if K is a pointed cone, the identity (19.1.1) reads G

e

%c,m5

m∈A∩Λ

=

G

; ψ(A/L, c)

L

(L+v)∩A

e%c,x5 dxL .

The sum over all lattice subspaces L is inﬁnite, but we will see soon that only ﬁnitely many terms in the sum are not zero. They correspond to the subspaces L parallel to the faces of K of the same dimension, see Figure 71. We say that a subspace L ⊂ V is parallel to a face F if the aﬃne hull of F is a translation of L.

v

0 L2

A

L1

Figure 71. A shifted cone A and the two 1-dimensional subspaces L1 and L2 that count. Since there is only one zero-dimensional subspace, we can rewrite (19.1.1) as ψ(A, c)e%c,v5 = F (A, c) −

G

ψ(A/L, c)φL (A ∩ (L + v), c),

L: dim L>0

or, equivalently, (19.1.2)

ψ(A, c) = e−%c,v5 F (A, c) G −e−%c,v5

ψ(A/L, c)φL (A ∩ (L + v), c),

L: dim L>0

which serves as the deﬁnition of ψ(A, c).

170

INTEGER POINTS IN POLYHEDRA

Again, if K is a pointed cone, (19.1.2) reads as G ψ(A, c) = e−%c,v5 e%c,m5 m∈A∩Λ

;

G

−e−%c,v5

ψ(A/L, c)

L: dim L>0

A∩(L+v)

e%c,x5 dxL .

Before we prove anything about function ψ (before we even prove that it is well-deﬁned), let us compute some examples. Example 19.2: one-dimensional cones. Suppose that V = R1 with the standard scalar product and Λ = Z1 . Let K = [0, +∞) and let v = α ∈ Q, so A = K + v = [α, +∞). In the sum (19.1.2) there is only one subspace of a positive dimension, that is, L = V . We have F (A, c) =

G

e

cm

=

m∈A∩Z

+∞ G

ecm =

m=4α.

ec4α. , 1 − ec

where ;α= is the smallest integer greater than or equal to α, and ; +∞ ecα ecx dx = − . φL (A, c) = c α Therefore,

% ψ(A, c) = e

−cα

ecα ec4α. + 1 − ec c

J =

1 ecβ + , c 1−e c

where β = ;α= − α. An important feature of ψ(A, c) is that it is analytic at c = 0. Indeed, ecβ 1 ecβ c + 1 − ec + = . 1 − ec c c(1 − ec ) Expanding in the neighborhood of c = 0, we get cecβ = c + c2 β + · · · , 1 − ec = −c − c2 /2 + · · · , and c(1 − ec ) = −c2 + · · · , where dots stand for higher-order terms, we conclude that ψ(A, 0) =

1 − β. 2

19. A VALUATION ON RATIONAL CONES

171

Example 19.3: a two-dimensional unimodular cone. Let V = R2 with the standard scalar product, let Λ = Z2 , and let K = co(v1 , v2 ) where v1 and v2 is a basis of Z2 . Hence K is a unimodular cone. Our goal is to compute ψ(K, c), and, in particular, ψ(K, 0). We have 1

F (K, c) =

(1 −

e%c,v1 5 ) (1

− e%c,v2 5 )

.

In the formula (19.1.2) we should choose 1-dimensional subspaces L = span(v1 ), L = span(v2 ) and the 2-dimensional subspace L = V . Let L = span(v1 ). Then V /L = L⊥ is the line perpendicular to v1 and a basis of the lattice Z2 /L consists of the vector w 1 = v2 −

0v1 , v2 8 v1 , 0v1 , v1 8

see Figure 72. L

v

1

V/L

v

2

0 w

1

Figure 72. Vectors v1 , v2 , and w1 . From (slightly modiﬁed) Example 19.2, we have ψ(K/L, c) =

1 1 + α 1−e α

for α = 0c, w1 8 = 0c, v2 8 −

0v1 , v2 8 0c, v1 8. 0v1 , v1 8

Moreover,

1 . 0c, v1 8 Similarly, for L = span(v2 ), the space V /L = L⊥ is the line perpendicular to v2 and a basis of the lattice Z2 /L consists of the vector φL (K ∩ L, c) = −

w 2 = v1 −

0v1 , v2 8 v2 . 0v2 , v2 8

We have ψ(K/L, c) =

1 1 + β 1−e β

for β = 0c, w2 8 = 0c, v1 8 −

0v1 , v2 8 0c, v2 8 0v2 , v2 8

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INTEGER POINTS IN POLYHEDRA

and φL (K ∩ L, c) = −

1 . 0c, v2 8

Finally, for L = V we have ψ(K/V, c) = 1 and φL (K, c) =

1 . 0c, v1 80c, v2 8

Then by formula (19.1.2), we have ψ(K, c) =

1 (1 − e%c,v1 5 )(1 − e%c,v2 5 ) % J J % 1 1 1 1 1 1 + + + + 1 − eα α 0c, v1 8 1 − eβ β 0c, v2 8 1 − . 0c, v1 80c, v2 8

Using expansions 1 1 τ 1 + = − + ··· τ 1−e τ 2 2 and 1 1 = τ 1−e τ

%

τ τ2 1− + + ··· 2 12

J

about τ = 0, we conclude that 1 0v1 , v2 8 ψ(K, 0) = + 4 12

%

1 1 + 0v1 , v1 8 0v2 , v2 8

J .

Definition 19.4. Let V be a ﬁnite-dimensional vector space, let Λ ⊂ V be a lattice and let v ∈ V be a rational point. We deﬁne the vector space Av (V ) (over Q, R or C) to be the span of the indicator functions [A] = [K + v], where K ⊂ V is a rational polyhedral cone. Our goal is to prove that the correspondence ψ : A 9−→ ψ(A, c) extends to a valuation on Av (V ). As the ﬁrst step in this direction we prove a useful lemma which asserts that only the simplest identities need to be checked, namely that ψ remains additive when A is subdivided into two cones by an aﬃne rational hyperplane H through v, cf. Figure 73 and Problem 7 of Chapter 2.

19. A VALUATION ON RATIONAL CONES

A− A v

173

0

A+ A

Figure 73. To prove that ψ extends to a valuation it suﬃces to prove that ψ(A) = ψ(A+ ) + ψ(A− ) − ψ(A0 ). The proof below follows [Br95], cf. also [Gr78]. Lemma 19.5. Let us ﬁx rational point v ∈ V , let W be a vector space and let θ be a map which associates an element θ(A) ⊂ W with every translation A = K + v of a rational cone K ⊂ V . Let H ⊂ V be a rational aﬃne hyperplane such that v ∈ H and let H+ and H− be corresponding closed halfspaces. Let A+ = A ∩ H+ ,

A− = A ∩ H−

and

A0 = A ∩ H.

Suppose that the identity θ(A) = θ (A+ ) + θ (A− ) − θ (A0 ) holds for all translations A = K + v of rational cones K ⊂ V and all rational aﬃne hyperplanes H containing v. Then the map θ extends to a linear transformation (valuation) Θ : Av −→ W such that Θ([A]) = θ(A) for all A = K + v, where K ⊂ V is a rational cone. Proof. Without loss of generality, we assume that v = 0. Let Ki ⊂ V , i ∈ I, be a ﬁnite set of rational cones and let αi , i ∈ I, be a set of numbers such that G (19.5.1) αi [Ki ] = 0. i∈I

We have to show that (19.5.2)

G

αi θ (Ki ) = 0.

i∈I

This would imply that the map θ extends to the valuation Θ.

174

INTEGER POINTS IN POLYHEDRA

For every cone Ki let us list rational hyperplanes Hij : j ∈ Ji which bound halfspaces whose intersection is Ki . Let H = {Hij :

j ∈ Ji }

be the set of all hyperplanes Hij . The hyperplanes from H cut the space V into ﬁnitely many closed full-dimensional polyhedral regions, each of which is a rational cone. Let {Cj : j ∈ J} be the set of all such cones and their non-empty faces, see Figure 74. Alternatively, the cones Cj can be described as the intersections of halfspaces bounded by hyperplanes from H such that the intersection of Cj with any other halfspace bounded by a hyperplane from H has dimension strictly smaller than that of Cj . First, we observe that the indicators [Cj ], j ∈ J, are linearly

Hi C3

j

C1 C2

Figure 74. In R3 , some of the bases of the cones Cj (with the origin outside of the plane). independent. Indeed, suppose that G

γj [Cj ] = 0

j∈J

for some coeﬃcients γj . Since the intersection of every two cones Cj1 and Cj2 is a face of both, the coeﬃcients γj of the cones Cj of the maximum dimension dim V must be 0. Discarding those cones from the linear combination and repeating the argument, we conclude that the coeﬃcients γj of the cones Cj with dim Cj = dim V − 1 must be 0 and so forth, rendering all the coeﬃcients γj to be 0. Next, we observe that by iterating the identities [K] = [K ∩ H + ] + [K ∩ H − ] − [K ∩ H] and θ(K) = θ(K ∩ H + ) + θ(K ∩ H − ) − θ(K ∩ H),

19. A VALUATION ON RATIONAL CONES

175

where H are hyperplanes from the set H, we can write G G [Ki ] = βij [Cj ] and θ(Ki ) = βij θ(Cj ) j∈J

j∈J

for all i ∈ I and some βij . From (19.5.1), we have GG

αi βij [Cj ] = 0

i∈I j∈J

and hence

G

αi βij = 0 for all j ∈ J.

i∈I

Therefore,

G

αi θ(Ki ) =

GG

αi βij θ(Ki ) = 0

i∈I j∈J

i∈I

and (19.5.2) follows.

!

We recall that M(V ) is the space of meromorphic functions on V , that F : P(QV ) −→ M(V ) is the valuation that extends exponential sums G e%c,m5 m∈P ∩Λ

for rational polyhedra P ⊂ V without lines and that for a lattice subspace L ⊂ V the valuation ΦL :

P(L + v) −→ M(V )

extends the exponential integral ; e%c,x5 dxL P

for rational polyhedra P ⊂ L + v without lines. Now we are ready to prove the main result of this chapter, the Berline – Vergne Theorem. Theorem 19.6. With every triple consisting of a Euclidean space V , a lattice Λ ⊂ V, and a rational point v ∈ V we can associate a linear transformation (valuation) Ψ : Av (V ) −→ M(V ),

176

INTEGER POINTS IN POLYHEDRA

such that the following properties are satisﬁed: (1) Let K ⊂ V be a rational cone, let A = K + v be its rational translation, and let L ⊂ V be a lattice subspace. Then the projection A/L ⊂ V /L is a rational (with respect to the lattice Λ/L) translation of a rational cone K/L ⊂ V /L, so that the functions Ψ([A]), Ψ([A/L]) ∈ M(V ) are well-deﬁned and the following identity holds: G F ([A]) = Ψ([A/L])ΦL ([A ∩ (L + v)]), L

where the sum is taken over all subspaces L ⊂ V parallel to the faces of A. (2) If A = K +v is a rational translation of a rational cone K ⊂ V which contains a line, then Ψ([A]) = 0. (3) If A = K+v is a rational translation of a rational cone K ⊂ V , then the function Ψ([A]) ∈ M(V ) is analytic at c = 0. Proof. We proceed by induction on d. For d = 0 we have Ψ(f ) = f (0) for all f ∈ Av (V ), so the result follows. Suppose that d ≥ 1. As in Section 19.1, the identity of Part (1) allows us to deﬁne recursively (19.6.1)

Ψ([A]) = e−%c,v5 F ([A]) G −e−%c,v5

Ψ([A/L])ΦL ([A ∩ (L + v)]),

L: dim L>0

where A = K + v is a rational translation of a rational cone and the sum is taken over the lattice subspaces L parallel to the faces of A. Next, we observe that if K contains a line, then (19.6.1) implies that Ψ([A]) = 0. Indeed, in that case A contains a line, so F ([A]) = 0. Furthermore, every face A∩(L+v) of A of a positive dimension contains a line, so ΦL ([A∩(L+v)]) = 0, and so indeed we must have Ψ([A]) = 0. Our next observation is that in (19.6.1) we can formally allow the sum to be taken over all lattice subspaces L ⊂ V of a positive dimension. Indeed, 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, so Ψ([A/L]) = 0 by the induction hypothesis. If dim(L ∩ K) < dim L then ΦL ([A ∩ (L + v)]) = 0, cf. Figure 75.

19. A VALUATION ON RATIONAL CONES

V/L

v K

K a)

L +v

L 0

L 0

177

A

b)

Figure 75. If L lies “inside” K (a) then the projection K/L contains a line and if L lies “outside” of K (b) then the integral over L + v is 0. Thus a rational subspace L ⊂ V of a positive dimension does not contribute to the sum (19.6.1) unless dim(K ∩ L) = dim L and L does not lie in a face F of K with dim F > dim L. In other words, the contribution of L is non-zero if and only if L is the span of a face of K. Another useful observation is that if L is parallel to a face F of A, then A/L = tcone(A, F )/L, where tcone(A, F ) is the tangent cone of A at an interior point of F , see Problem 3 of Chapter 6. Now we prove that (19.6.1) extends to a valuation. By Lemma 19.5 it suﬃces to check that Ψ([A]) = Ψ([A+ ]) + Ψ([A− ]) − Ψ([A0 ]), where A = K + v is the translation of a rational cone K ⊂ V , H ⊂ V is an aﬃne rational hyperplane through v and A0 = A∩H, A+ = A∩H+ , A− = A ∩ H− where H+ and H− are closed halfspaces bounded by H, cf. Figure 76. We use (19.6.1) to deﬁne Ψ([A]), Ψ([A+ ]), Ψ([A− ]), and Ψ(A0 ). First, we note 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)], since F and Φ are valuations. Moreover, [A/L] = [A+ /L] + [A− /L] − [A0 /L]

178

INTEGER POINTS IN POLYHEDRA

since projections preserve linear relations among indicators of polyhedra, see Theorem 3.1. Hence by the induction hypothesis, (19.6.2)

Ψ([A/L]) = Ψ([A+ /L]) + Ψ([A− /L]) − Ψ([A0 /L]).

For a rational subspace L ⊂ V , let Ξ(A, L) = Ψ([A/L])ΦL ([A ∩ (L + v)]) denote the corresponding term of (19.6.1). Clearly, it is enough to check that (19.6.3)

Ξ(A, L) = Ξ(A+ , L) + Ξ(A− , L) − Ξ(A0 , L)

for all subspaces L. We verify this on a case-by-case basis. Let F be a non-empty face of A. Then either F lies in H or F lies in one of the closed halfspaces H+ and H− but not in both, or H passes through an interior point of F and is transversal to the aﬃne span of F. (a) Let L be a subspace parallel to a face F of A such that F ⊂ H. Then A ∩ (L + v) = A+ ∩ (L + v) = A− ∩ (L + v) = A0 ∩ (L + v). Hence by (19.6.2) we conclude that (19.6.3) holds. (b) Let L be a subspace parallel to a face F of A such that F lies in the closed halfspace H+ but not in H− . Then A/L = A+ /L and dim (A− ∩ (L + v)) , dim (A0 ∩ (L + v)) < dim L. Therefore, Ξ(A, L) = Ξ(A+ , L) and Ξ(A− , L) = Ξ(A0 , L) = 0, so (19.6.3) holds. (c) Let L be a subspace parallel to a face F of A such that F lies in the closed halfspace H− but not in H+ . Then A/L = A− /L and dim (A+ ∩ (L + v)) , dim (A0 ∩ (L + v)) < dim L. Therefore, Ξ(A, L) = Ξ(A− , L) and Ξ(A+ , L) = Ξ(A0 , L) = 0, so (19.6.3) holds.

19. A VALUATION ON RATIONAL CONES

179

(d) Let L be a subspace parallel to a face F of A such that H passes through an interior point of F and intersects the aﬃne span of F transversally. Then A/L = A+ /L = A− /L and

dim (A0 ∩ (L + v)) < dim L.

Hence Ψ([A/L]) = Ψ([A+ /L]) = Ψ([A− /L]),

ΦL ([A0 ∩ (L + v)]) = 0

and (19.6.3) holds by the valuation property of Φ. (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). Then ΦL ([A ∩ (L + v)]) = ΦL ([A+ ∩ (L + v)]) = ΦL ([A0 ∩ (L + v)]) and (19.6.3) follows from (19.6.2). We note that every face of A+ (respectively A− ) is either the intersection of a face of A with H+ (respectively H− ) or the intersection of a face of A with H and that every face of A0 is the intersection of a face of A with H. Hence we conclude that to deﬁne Ψ([A]), Ψ([A+ ]), Ψ([A− ]), and Ψ([A0 ]) by (19.6.1), it suﬃces to consider the sum over subspaces L of types (a)–(e) only, cf. also Figure 76. Hence we conclude that Ψ([A]) = Ψ([A+ ]) + Ψ([A− ]) − Ψ([A0 ]), so Ψ is indeed a valuation. F

F F

H

H

(a)

(b)−(c)

H

G

(d)−(e)

v A

Figure 76. A 3-dimensional cone A (below) with a triangular base (above) and possible positions of a hyperplane H, of a face F of A, and of a face G of A ∩ H.

180

INTEGER POINTS IN POLYHEDRA

It remains to show that Ψ([A]) is analytic at c = 0. Again, we proceed by induction on dim V . Without loss of generality, we assume that A = K + v is a rational translation of a full-dimensional rational pointed cone K ⊂ V , since if K is not full-dimensional, we can consider K as a cone in span(K). Since every rational full-dimensional pointed cone can be represented as a linear combination of full-ﬁmensional unimodular cones modulo rational cones with lines (see Section 16.4) and since Ψ is a valuation, it suﬃces to prove that Ψ(A) = ψ(K + v, c) is analytic at c = 0, where K = co (u1 , . . . , ud ) and u1 , . . . , ud is a basis of Λ. We use formula (19.6.1). We have F ([A]) = e%c,w5

d i=1

1 , 1 − e%c,ud 5

where w ∈ Λ is the rounding w = ;v=K of v, see Section 14.3. That is, if v=

d G

α i ui

then w =

d G

;αi =ui .

i=1

i=1

Furthermore, the subspaces L parallel to the faces of K are of the type L = LI = span (ui : i ∈ I)

where I ⊂ {1, . . . , d}

is a non-empty subset. We have ΦL ([A ∩ (L + v)]) = e%c,v5

i∈I

1 , 0−c, ui 8

see Example 8.2. By the induction hypothesis, functions Ψ([A/L]) are analytic at c = 0. Thus Ψ([A]) is a meromorphic function with possible poles in a neighborhood of c = 0 located on the hyperplanes 0c, ui 8 = 0 for i = 1, . . . , d. Thus to prove that Ψ([A]) is, in fact, regular at c = 0 it suﬃces to prove that for every i = 1, . . . , d the meromorphic function 0c, ui 8Ψ([A]) is identically 0 on the hyperplane 0c, ui 8 = 0. Without loss of generality we assume that i = d. We note that 0c, ud 8 = −1 on the hyperplane 0c, ud 8 = 0, 1 − e%c,ud 5 cf. Deﬁnition 14.1.

19. A VALUATION ON RATIONAL CONES

181

Therefore, the value of 0c, ud 8Ψ([A]) on the hyperplane 0c, ud 8 = 0 is equal to −e%c,w−v5 (19.6.2)

d−1 i=1 G

+

1 1 − e%c,ui 5 Ψ([A/LI ])

i∈I\{d}

I ⊂ {1, . . . , d} d∈I

1 . 0−c, ui 8

Let A( = A/ span(ud ), so

A 2 A( = v ( + co u(1 , . . . , u(d−1 ,

where v ( = v/ span(ud ) and u(i = ui / span(ud ) for i = 1, . . . , d − 1. Let K ( = co (u(1 , . . . , u(d ). Thus K ( is a (d − 1)-dimensional unimodular cone with respect to the lattice Λ( generated by u(1 , . . . , u(d−1 . Since c is orthogonal to ud , we have 0c, ui 8 = 0c, u(i 8 for i = 1, . . . , d − 1 and hence we can rewrite (19.6.2) as (

Ψ(A )−e

%c,w! −v ! 5

d−1 i=1

1 ! + 1 − e%c,ui 5

G I ⊂ {1, . . . , d − 1} I '= ∅

Ψ([A( /LI ])

i∈I

1 , 0−c, u(i 8

where w( = w/ span(ud ) is the rounding w( = ;v ( =K ! . The last expression in turn can be rewritten as G ! ! Ψ([A( ]) − e−%c,v 5 F ([A( ]) + e−%c,v 5 Ψ([A( /L])ΦL ([A( ∩ (L + v)]), L: dim L>0

which is identically 0 by (19.6.1). This concludes the proof.

!

Remark 19.7. It follows from the proof that as long as dim V is ﬁxed in advance, given a rational translation K = A + v of a rational cone K ⊂ V , the function ψ(A, c) = Ψ([A]) can be computed in polynomial time as a linear combination of functions of the type - 1 1 , e%c,v5 1 − e%c,ui 5 j∈J 0c, uj 8 i∈I where v, ui , uj are rational vectors, ui , uj >= 0, and |I| + |J| ≤ d. Indeed, since valuations F and Φ are computable in polynomial time (cf. Chapter 17), the recurrence relation (19.6.1) results in an algorithm of a polynomial time complexity. Another, faster, way is to write K as a combination of unimodular cones (cf. Chapter 16), use the valuation property of Ψ and apply the recurrence (19.6.1) for the unimodular

182

INTEGER POINTS IN POLYHEDRA

cones, where it becomes particularly straightforward, since only unimodular cones are produced in the recurrence (see Example 19.3). Problems 1◦ . Let A ⊂ V be a rational translation of a rational cone and let u ∈ Λ be a lattice vector. Prove that ψ(A, c) = ψ(A + u, c). 2. Let u1 , u2 , u3 ∈ Z3 be a basis of Z3 and let K = co (u1 , u2 , u3 ). Compute ψ(K, c) and ψ(K, 0).

CHAPTER 20

A “local” formula for the number of integer points in a polytope As N. Berline and M. Vergne observed [BV07], the fact that Ψ([A]) is regular at c = 0 allows one to obtain an expression for the number of lattice points in a rational polytope in terms of the volumes of faces of the polytope and certain explicit valuations on the tangent cones at the faces. Theorem 20.1. Let us ﬁx Euclidean space V and a lattice Λ ⊂ V and let P ⊂ V be a rational polytope. For a face F of P let us deﬁne the volume vol(F ) and the value α(P, F ) as follows. Let L = LF ⊂ V be the subspace parallel to F . We normalize the volume form in L and hence in the aﬃne span of F in such a way that det(Λ ∩ L) = 1 and let vol F be the the volume of F with respect to that form. Let tcone(P, F ) be the tangent cone of P at an interior point of F and let A = AF = tcone(P, F )/L be the orthogonal projection of the tangent cone onto L⊥ . We deﬁne α(P, F ) = ψ(A, 0). Then G vol(F )α(P, F ), |P ∩ Λ| = F

where the sum is taken over all non-empty faces F of P including P itself. Note that if dim F = k, then vol(F ) is homogeneous of degree k under dilations F 9−→ tF for t > 0. The terms in the formula of Theorem 20.1 are “local”, meaning that α(P, F ) depends only on the tangent cone tcone(P, F ). The tangent cone tcone(P, F ) has the k-dimensional aﬃne subspace that is the aﬃne span of F as its apex, that is the nonempty face of minimum dimension. Since Ψ is a valuation, α(P, F ) is also a valuation on rational cones whose apex is the aﬃne span of F . Thus, in some sense, α(P, F ) measures “angles” (of course, values of α(P, F ) can be negative). We also note that α(P, F ) = α(Q, G) if tcone(Q, G) is a lattice translation of tcone(P, F ). The formula of Theorem 20.1 easily implies the existence of the Ehrhart (quasi-) polynomial, see Chapter 18. In fact, because of the local character of the

184

INTEGER POINTS IN POLYHEDRA

formula, we can deduce more. Namely, let P ⊂ V be a rational polytope. Then |nP ∩ Λ| =

d G

ak (n)nk

for all positive integers n,

k=0

where ak (n) are periodic functions in n. In particular, ak (n+t) = ak (n) for a positive integer t such that tP is a lattice polytope. From Theorem 20.1 one can easily deduce that ak (n) = ak (n + t) for a positive integer t such that the aﬃne span of every k-dimensional face of tP contains a lattice point. Finally, we remark that values α(P, F ) are computable in polynomial time as long as the codimension k = dim P − dim F is ﬁxed in advance (P and F can be deﬁned by the lists of their vertices or by systems of linear inequalities). Proof of Theorem 20.1. For a vertex v of P , let Av = tcone(P, v) be the tangent cone of P at v. Hence Av is a rational translation of a rational cone. Moreover, G G e%c,m5 . F ([Av ]) = F (P ) = (20.1.1) m∈P ∩Λ

v∈Vert(P )

Let us substitute F ([Av ]) =

G

Ψ([Av /L])Φ[Av ∩ (L + v)],

L

where the sum is taken over all subspaces parallel to the faces of Av . Rearranging the terms in (20.1.1), we obtain G G G e%c,m5 = Ψ([tcone(P, F )/LF ]) ΦLF ([tcone(F, v)]). m∈P ∩Λ

F

v∈Vert(F )

On the other hand (see Chapter 8), ; G ΦLF ([tcone(F, v)]) = e%c,x5 dxL , F

v∈Vert(F )

so we get G m∈P ∩Λ

e

%c,m5

=

G F

; Ψ([tcone(P, F )/LF ])

F

e%c,x5 dxL ,

where the sum is taken over all non-empty faces F of P , including P . Substituting c = 0 we complete the proof. !

20. A “LOCAL” FORMULA

185

Example 20.2. Let P ⊂ R2 be a convex polygon with a non-empty interior and integer vertices. Theorem 20.1 provides a reﬁnement of Pick’s formula (see Section 10.10). Indeed, Pick’s formula states 1 |P ∩ Z2 | = vol P + |∂P ∩ Z2 | + 1. 2 The ﬁrst term vol P , the area of P corresponds to the face F = P in Theorem 20.1. We note that α(P, P ) = 1. The second term, one half of the number of integer points on the boundary, corresponds to the sum over the edges F of P in Theorem 20.1. It is not hard to see that the sum of properly computed lengths of the edges of P (measured with respect to the lattice) is exactly the number of integer points on the boundary of P , while α(P, F ) = 1/2 for any edge F of P , cf. Example 19.2. Finally, for every vertex v of P we have vol v = 1 and the sum of α(P, v) over all vertices of v is equal to 1, since G G α(P, v) = ψ(tcone(P, v), 0) v∈Vert(P )

v∈Vert(P )

G

=

ψ(fcone(P, v), 0) = 1.

v∈Vert(P )

Indeed, ψ is a valuation and G [fcone(P, v)] ≡ [0] modulo rational polyhedra with lines, v∈Vert(P )

cf. Theorem 6.6. The way “1” is apportioned among the vertices of P depends on the cones of feasible directions of the vertices, see Figure 77 and Example 19.3. 3/8 3/10

1/4

3/8

1/4

9/20

Figure 77. Values of α(P, v) at the vertices. Problem 1. Check the values at the vertices on Figure 77.

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Index algebra of bounded polyhedra, 10 of closed convex sets, 10 of compact convex sets, 10 of polyhedra, 10 of rational polyhedra, 108 cone, 32 of feasible directions, 49 pointed, 33 recession, 35 tangent, 49 unimodular, 111 conic combination, 33 hull, 33 convex set, 9 combination, 28 hull, 28 convergent, 130 convolution, 22 Euler characteristic, 11 halfspace, 16 hyperplane, 16 indicator function, 10 interior, 36 interval, 27 lattice, 81 basis of, 84 determinant of, 87 dual, 97 fundamental parallelepiped of, 85 reduced basis of, 101

line, 27 Minkowski sum, 22 polar, 41 polyhedron, 9 face of, 29 rational, 107 strongly combinatorially isomorphic, 67 vertex of, 29 polynomial Ehrhart, 161 Todd, 121 polytope, 28 normal fan of, 162 simple, 78 totally unimodular, 121, 124 transportation, 39 ray, 27 simplex standard, 28 Theorem of Berline – Vergne, 175, 183 of Birkhoﬀ – von Neumann, 39 of Blichfeldt, 95 of Brion, 66, 114 of Gram – Brianchon, 56 of Khovanskii – Pukhlikov and Lawrence, 66, 113, 114 of Lagrange, 97 of Minkowski, Convex Body, 95 of Weyl – Minkowski, 28, 31 valuation, 11 exponential, 63