Lecture Notes on Geometry of Numbers

Martin Henk

May 12, 2015 Contents

Once upon a time... iii 1 Basic (and some more) facts from Convexity 1 Exercises 13 Notes 15 2 Basic facts about Lattices 17 3 Minkowski’s successive minima 31 References 41

i ii Once upon a time...

One of the main research directions of number theory around the middle of the 19th century was devoted to the study of general quadratic forms q(x) n in n variables x = (x1, . . . , xn)| and its values on integral points z ∈ Z . In particular, for the class of positive deﬁnite quadratic forms, i.e., we can write n×n q(x) = x|Qx with a positive deﬁnite matrix Q ∈ R , one was interested n in upper bounds on its minimum on Z \{0}, i.e., minz∈Zn\{0} q(z). Of course, such a bound has to depend on Q and since

Acknowledgement. For helpful remarks, corrections and discussions I would like to thank Andr´eApitzsch, Eva Linke, Mar´ıade los Angeles Hern´andez Cifre, Eva Linke,....

iii

Basic (and some more) facts from Convexity

In this chapter we will brieﬂy collect some basic results and concepts concerning convex bodies, which may be considered as the main ”continuous” component/ingredient of the Geometry of Numbers. For proofs of the results we will refer to the literature, mainly to the books [2, 5, 6, 7, 9, 11, 13, 14], which are also excellent sources for further information on the state of the art concerning convex bodies, polytopes and their applications. As usual we will start with setting up some basic notation; more speciﬁc ones will be introduced when needed.

n Let R = {x = (x1, . . . , xn)| : xi ∈ R} be the n-dimensional Euclidean Pn space, equipped with the standard inner product hx, yi = i=1 xi yi, x, y ∈ n R . n For a subset X ⊆ R , let lin X and aff X be its linear and affine hull, respectively, i.e., with respect to inclusions, the smallest linear or affine subspace containing X. If X = ∅ we set lin X = {0}, and aff X = ∅. The dimension of a set X is the dimension of its affine hull and will be denoted n by dim X. By X +Y we mean the vectorwise addition of subsets X,Y ⊆ R , i.e., X + Y = {x + y : x ∈ X, y ∈ Y }.

n The multiplication λ X of a set X ⊆ R with λ ∈ R is also declared vectorwise, i.e., λ X = {λ x : x ∈ X}. We write −X = (−1) X, X − Y = X + (−1) Y and x + Y instead of {x} + Y . n A subset C ⊆ R is called convex, iﬀ for all c1, c2 ∈ C and for all λ ∈ [0, 1] hold λ c1 + (1 − λ) c2 ∈ C. A convex compact (i.e., bounded and n closed) subset K ⊂ R is called a convex body. The set of all convex bodies n n n in R is denoted by K . K ∈ K is called o-symmetric, i.e., symmetric with respect to the origin, if K = −K. The set of all o-symmetric convex bodies n n in R is denoted by Ko . There is a simple one-to-one correspondence between n-dimensional o- n n symmetric convex bodies and norms on R : given a norm | · | : R → R≥0

1 2 1. Basic (and some more) facts from Convexity

n we consider its unit ball {x ∈ R : |x| ≤ 1}, which is a o-symmetric convex body by the deﬁnition of a norm. For the reverse correspondence we need n the notion of a distance function. For K ∈ Ko , dim K = n, the function n |·|K : R → R≥0 given by |x|K := min{ρ ∈ R≥0 : x ∈ ρ K} is called the distance function of K.

By the convexity and symmetry of K it is easily veriﬁed that |·|K is indeed a norm. A well-studied family of o-symmetric convex bodies are the p n unit balls Bn := {x ∈ R : |x|p ≤ 1} associated to the p-norms

n !1/p X p |x| := |xi| , p ∈ [1, ∞), and |x| := max {|xi|}. p ∞ 1≤i≤n i=1 In the Euclidean case, i.e., p = 2, we denote the Euclidean norm just by |·| and its unit ball by Bn.

0 −1 1 1 1 1

−1 1 0 0 0 0 0 0

0 −1 1 −1 −1 −1 p = 1 p = 2 p = 4 p = ∞

Figure 1.1: Unit balls of p-norms

∞ n Bn = {x ∈ R : −1 ≤ xi ≤ 1} is the o-symmetric (regular) cube with 1 n Pn edge length 2, and Bn = {x ∈ R : i=1 |xi| ≤ 1} is called the (regular) crosspolytope. One big advantage of convex sets K is their property that points not belonging to K can easily be separated from K. In order to formulate this n precisely and a bit more generally, we denote for a ∈ R \{0}, b ∈ R n by H(a, b) = {x ∈ R : ha, xi = b} the hyperplane with normal vector a and right hand side b. The associated two closed halfspaces are denoted by n H≤(a, b) and H≥(a, b), respectively, i.e., H≤(a, b) = {x ∈ R : ha, xi ≤ b} and H≥(a, b) = H≤(−a, −b).

n 1.1 Theorem [Separation Theorem]. Let C1,C2 ⊂ R be convex with C1 ∩ C2 = ∅.

i) Then there exists a separating hyperplane H(a, b), i.e., C1 ⊆ H≤(a, b) and C2 ⊆ H≥(a, b). n ii) If C1 is compact, i.e., C1 ∈ K and C2 is closed then there exists a strictly separating hyperplane H(a, b), i.e., C1 ⊂ H≤(a, b) and C2 ⊆ H≥(a, b) and C1 ∩ H(a, b) = ∅ = C2 ∩ H(a, b). 1. Basic (and some more) facts from Convexity 3

H≤(a, b) H≥(a, b)

H(a, b)

Figure 1.2: Strictly separating hyperplane of two convex bodies

A hyperplane H(a, b) passing through the boundary of a closed convex set C and containing C entirely in one of the two halfspaces, i.e., C ∩ H(a, b) 6= ∅ and C ⊆ H≤(a, b), say, is called a supporting hyperplane of C. The intersection of F = C ∩ H(a, b) is called a face of C, or more precisely, a k-face if dim F = k. If k = 0 then F is called an extreme point of C. Faces themselves are closed convex sets, and the interesection of faces of C is again a face of C, where we regard the empty set as the (−1)-dimensional face of C. The boundary of C, denoted by bd C, is the union of its faces. For a convex body K ∈ Kn and a given normal vector (direction) a ∈ n R \{0} the supporting hyperplane in direction a is given by H(a, hK (a)) where n hK : R → R with hK (a) := max{ha, xi : x ∈ K} is called the support function of K. The support function is positive homogeneous, i.e., hK (λ a) = λ hK (a), λ ∈ R≥0, and sub-additive, i.e., hK (x + y) ≤ n hK (x) + hK (y) for all x, y ∈ R . Indeed, these two properties characterize n support functions of convex bodies among the real-valued functions on R . A quite general concept in diﬀerent branches of mathematics is that of n duality and/or polarity. In our setting it is deﬁned as follows: for X ⊆ R , the set ? n X := y ∈ R : hx, yi ≤ 1 for all x ∈ X is called the polar set of X. As the intersection of convex closed sets containing the origin, the polar set is always a convex closed set containing 0. Its actual shape, however, depends on its position relative to the origin. For instance, X? is bounded if and only if 0 is an interior point of X. On the other hand, as stated in the next proposition, its shape be- haves properly with respect to linear transformations and a polar set of a o-symmetric convex body is again an o-symmetric convex body.

1.2 Proposition.

n ? − ? i) Let X ⊆ R , and let M ∈ GL(n, R). Then (MX) = M |X . Here GL(n, R) denotes the general linear group of all invertible real n × n 4 1. Basic (and some more) facts from Convexity

1/2 ? ? 1/2

0 0 0 0 1/2 −1/2

1 1 1 Figure 1.3: Polar bodies of B2 and 0 + B2

matrices.

n ? ? ii) Let K ⊆ R be convex and closed with 0 ∈ K. Then (K ) = K. n ? n iii) Let K ∈ Ko . Then K ∈ Ko and |x|K? = maxy∈K hx, yi = hK (x).

By Proposition 1.2 iii) and H¨older’sinequality for sums,

n X n |xk yk| ≤ |x|p |y|p/(p−1) , x, y ∈ R , p ∈ (1, ∞), i=1 p−1 with equality if and only if |xi| = c |yi| for a constant c, the polar bodies of the p-norms are given by

p p ? p−1 1 ? ∞ ∞ ? 1 (Bn) = Bn , 1 < p < ∞, and (Bn) = Bn and thus (Bn ) = Bn. Since we are mainly dealing with simple structured bounded convex sets, we deﬁne the volume of sets via the Jordan-Peano measure: A subset D ⊂ n n R is called (Jordan-Peano) measurable if its indicator function χD : R → {0, 1} given by ( 1, x ∈ D, χD(x) := 0, x ∈/ D, is Riemann integrable, i.e., the Riemann integral Z vol(D) := χD(x) dx Rn exists. vol(D) is called the volume of D. So the volume of a set D is based on the principle of exhausting or approximating D by ”small” n-dimensional n boxes {x ∈ R : αi ≤ xi ≤ βi, 1 ≤ i ≤ n}, αi < βi ∈ R, for which the Qn volume is deﬁned as i=1(βi − αi). In particular, we have V(X) = 0 for a measureable set of dim X < n, and

n 1.3 Proposition. Let X ⊂ R be a measurable set. Then 1 n #(X ∩ Z ) vol(X) = lim m , m→∞ mn 1. Basic (and some more) facts from Convexity 5

n n where Z = {z ∈ R : zi ∈ Z} denotes the set of all points with integral coordinates.

1 n Figure 1.4: Approximation by ”small” boxes centered at points of m Z

∞ The unit ball Bn , i.e., the cube (box) of edge length 2 centered at the ∞ n p origin has vol(Bn ) = 2 . In general, the volume of the unit balls Bn can be R ∞ x−1 −t expressed via the Γ-function Γ(x) := 0 t e dt as (cf. Exercise 1.3) Γ(1 + 1/p)n vol(Bp) = 2n , (1.1) n Γ(1 + n/p) where the p = ∞ case is covered by the limit of the formula above. n As a function vol : K → R≥0 on the set of convex bodies, the volume shares the following properties: first of all it is valuation, i.e., it is an additive function on the space of all convex bodies, which means that for K,L ∈ Kn with K∪L ∈ Kn it is vol(K∪L) = vol(K)+vol(L)−vol(K∩L). The volume is translation invariant, i.e., vol(t+X) = vol(X) for any measurable set X and n t ∈ R , and since vol(AX) = | det A|vol(X) for A ∈ GL(n, R) the volume is also invariant with respect to rotations, i.e., vol(AX) = vol(X) for any orthogonal matrix A of determinant 1. Moreover, the volume is homogeneous of degree n, i.e., vol(λ K) = |λ|n vol(K) and (strictly) monotone in the sense that for K ⊆ L ∈ Kn we have vol(K) ≤ vol(L) with equality iff dim L < n or dim L = n and K = L. Finally, we mention that the volume is also a continuous function with respect to the Hausdorff metric induced by the Hausdorff distance which for K,L ∈ Kn is defined as

dH(K,L) := min{ρ ∈ R≥0 : K ⊆ L + ρ Bn and L ⊆ K + ρ Bn}. p For instance, for the convex bodies Bn we have (cf. Exercise 1.4)

p q √ −1/p −1/q dH(Bn,Bn) = n |n − n |. n n dH(·, ·) deﬁnes a metric on K which makes K a complete space. The next theorem, known as Blaschke’s selection theorem, is a very useful tool in guaranteeing the existence of solutions of extremum problems.

1.4 Theorem [Blaschke selection theorem]. Any bounded sequence of convex bodies in Kn contains a convergent subsequence. 6 1. Basic (and some more) facts from Convexity

1 B2 √ 1 2 B2 + ( 2/2) B2 ∞ B2

√ 1 ∞ Figure 1.5: dH(B2 ,B2 ) = 2/2

Another theorem with an enormous aftermath is about the so-called L¨owner- John ellipsoids. In its basic form it says

1.5 Theorem [L¨owner-Johnellipsoids]. For each convex body K ∈ Kn, dim K = n, there is an unique ellipsoid a + ABn, A ∈ GL(n, R) of maximal volume contained in K and it holds

a + ABn ⊆ K ⊂ a + n A Bn.

n √ If K ∈ Ko then we have a = 0 and the factor n can be replaced by n. √ The dilatation factors n or n are also the smallest possible. For instance, in ∞ the symmetric case the bound is attained by the cube Bn or the crosspoly- 1 tope Bn whose maximal volume ellipsoid is a ball. In the general setting a simplex (see Figure 1.8) is an extremal case. In some cases the L¨owner-Johnellipsoids provide an easy way to gener- alize inequalities valid for ellipsoids to arbitrary o-symmetric convex bodies. As an example we consider the so-called Mahler volume M(K) := ? n vol(K) vol(K ) for K ∈ Ko , dim K = n. In view of Proposition 1.2 i) we know that the Mahler volume is invariant with respect to linear transformations, i.e., M (AK) = M (K) for any A ∈ GL(n, R), and hence, in order to bound M (K) we may assume that B is the volume maximal n √ ellipsoid contained in K. By Theorem 1.5 we have Bn ⊆ K ⊂ nBn √ ? and thus (1/ n) Bn ⊂ K ⊆ Bn (cf. Exercise 1.1). So by the mono- ? √ tonicity of the volume we get vol(K) vol(K ) ≤ vol( n Bn)vol(Bn) and ? √ vol(K) vol(K ) ≥ vol(Bn) vol((1/ n) Bn), or

√ −n √ n n M(Bn) ≤ M(K) ≤ n M(Bn).

Of course, these bounds are not the best possible, but they show that M (K) is bounded. In order to improve these bounds, we firstly observe that convex bodies mini- and maximizing M (K) really exist. This, however, is a consequence of Theorem 1.4 and the fact that it suffices to consider convex n √ bodies K ∈ Ko fulfilling Bn ⊆ K ⊂ nBn. 1. Basic (and some more) facts from Convexity 7

It was shown by Blaschke and Santal´othat the maximum of M (K) is 2 only attained for ellipsoids, and so we have M (K) ≤ M(Bn) = vol(Bn) . The lower bound leads to the so called Mahler conjecture claiming that

4n 4 n M(K) ≥ M(B∞) = vol(B∞) vol(B1) = = + o(1) M(B ). n n n n! 2π n

This conjecture is open for n ≥ 3, and it is also known that there is a whole ∞ family of (aﬃnely inequivalent) convex bodies with M (K) = M (Bn ). The n best known lower bounds are of the type (constant ) · M(Bn), and the ﬁrst who established such a bound were Bourgain and Milman

1.6 Theorem [Bourgain-Milman]. There exists a positive absolute con- ? n 2 stant C such that vol(K) vol(K ) ≥ C vol(Bn) for any convex body K ∈ n Ko , dim K = n.

Most of the (classical) results in Geometry of Numbers are dealing with o-symmetric bodies because of their interpretation as unit balls of norms. On the other hand, most of the underlying geometric problems and structures can be formulated for arbitrary convex bodies and so there are many problems which have a symmetric and non-symmetric side. In particular, since there are already many results for o-symmetric bodies, one is interested in transformations making non-symmetric bodies symmetric such that the geometric sizes in question, e.g., volume, keeps controllable. A standard pro- cedure in this respect is building the central symmetral cs (K) of K ∈ Kn deﬁned by 1 cs : Kn → Kn with cs (K) := (K − K) . (1.2) o 2

2 2 3/2 0 3/2 0 1 1 3/2 1/2 2 cs 0 0 0 1 −1/2

−1 −1

Figure 1.6: Central symmetrization of a convex body

n Of course, K ∈ Ko remains invariant under this map and cs (t + K) = n cs (K) for t ∈ R . The following classical theorem by Rogers and Shephard completely describes the behavior of the volume with respect to this map. 8 1. Basic (and some more) facts from Convexity

1.7 Theorem [Rogers-Shephard]. Let K ∈ Kn. Then

1 2 n vol(K) ≤ vol(cs (K)) ≤ vol(K). 2n n

n In the lower bound equality is attained iﬀ K ∈ Ko and in the upper bound iﬀ K is a simplex. In fact, the lower bound is just an application of the famous Brunn-Minkowski theorem

1.8 Theorem [Brunn-Minkowski]. Let K,L ∈ Kn and λ ∈ [0, 1]. Then

vol(λ K + (1 − λ) L)1/n ≥ λvol(K)1/n + (1 − λ) vol(L)1/n (1.3) with equality iﬀ K and L are contained in parallel hyperplanes, or K and n L are homothetic, i.e., there exists a µ ∈ R≥0 and t ∈ R with L = t + µ K. In words, the n-th root of the volume is a concave function on Kn. Due to the (weighted) arithmetic-geometric mean inequality (1.3) also implies

vol(λK + (1 − λ)L) ≥ vol(K)λ vol(L)1−λ, (1.4) i.e., the volume is a log-concave function. In fact, if (1.4) holds for all K,L ∈ Kn then it also implies (1.3). Next we turn to some more discrete aspects of convex sets. Let x1,..., n Pm xm ∈ R and let λi ∈ R≥0, 1 ≤ i ≤ m, with i=1 λi = 1. Then x = Pm i=1 λi xi is called a convex combination of x1,..., xm. As linear or affine sets are closed with respect to taking finite linear or affine combinations, convex sets are closed with respect to taking finite convex combinations. n More precisely (cf. Exercise 1.5), K ⊆ R is convex if and only if

( m m ) X X K = λi xi : m ∈ N, xi ∈ K, λi ≥ 0, λi = 1 . i=1 i=1

n In analogy to linear and aﬃne hulls of a subset X ⊆ R we deﬁne the convex hull, conv X, of X as the smallest – with respect to inclusions – convex set containing X, i.e., \ conv X := C. n C⊆R ,C convex, X⊆C

As in the linear and affine setting, the convex hull of a set X is the set of all finite convex combinations which can be formed by elements of X. In fact, due to a classical result of Carathéodory it suffices to consider convex combinations consisting of at most dim X + 1 elements (cf. Exercise 1.6). 1. Basic (and some more) facts from Convexity 9

n 1.9 Theorem [Carath´eodory]. Let X ⊆ R . Then

(dim X+1 dim X+1 ) X X conv X = λi xi : xi ∈ X, λi ≥ 0, λi = 1 . i=1 i=1

By Caratheodory’s theorem one can easily argue that the convex hull of a compact set X is again compact, but observe, convex hulls of closed sets do n not necessarily need to be closed. We also note that for X,Y ⊆ R

conv X + conv Y = conv (X + Y ).

n If #X < ∞, i.e., X ⊂ R is ﬁnite, the convex body conv X is called a polytope, and we shall denote the set of all polytopes and all o-symmetric n n polytopes by P and Po , respectively. In order to emphasize the dimension of a polytope P ∈ Pn, P will be called an m-polytope if dim P = m. In the case of a polytope P ∈ Pn, the extreme points, i.e., the 0-dimensional faces are called vertices, the 1-dimensional faces are called edges and factes are (dim P − 1)-dimensional faces. Each polytope is the convex hul of its

∞ 1 Figure 1.7: Polytopes: The 3-cube B3 , the 3-crosspolytope B3 , and the ”soccer ball” (truncated icosahedron) vertices, in fact, each convex body is the convex hull of its extreme points ∞ (cf. Exercise 1.9). The vertices of the cube Bn , for instance, are given by the 2n vectors whose coordinate are plus or minus 1, and so we have

∞ | Bn = conv {(1, . . . , n) , i ∈ {−1, 1}}. (1.5)

n Given a polytope P = conv {x1,..., xm} ∈ P , the polar set of P is given ? n by the system of inequalities P = {y ∈ R : hxi, yi ≤ 1, 1 ≤ i ≤ m}. In general, such a set, i.e., a set given by ﬁnitely many linear (weak) inequalities is called a polyhedron. It was ﬁrstly shown by Minkowski and Weyl that bounded polyhedra are polytopes and vice versa.

n 1.10 Theorem [Minkowski-Weyl]. C ⊂ R is a convex polytope if and only if it is a bounded polyhedron. 10 1. Basic (and some more) facts from Convexity

So we always have two representations of polytopes, either as the convex hull of some points, e.g., the vertices or as the intersection of finitely many linear inequalities, e.g., the inequalities corresponding to facet defining supporting hyperplanes. For instance, the cube has also the presentation ∞ n Bn = {x ∈ R : h±ei, xi ≤ 1, 1 ≤ i ≤ n}, (1.6) where ei = (0,..., 0, 1, 0,..., 0)| with the 1 at the i-th position denotes the i-th unit vector. In comparison with the representation as the convex hull of 2n points we only need 2 n linear inequalities here for describing n the cube. The convex hull of n + 1 affinely independent points in R is called an n-simplex or just a simplex, e.g., Tn = conv {0, e1,..., en} is the so-called standard simplex. By definition, any other simplex is an affine

Figure 1.8: 2-simplex (triangle), 3-simplex (tetrahedron), and Schlegel- diagram of a 4-simplex transformation of Tn. If the distance between any two points of a simplex is equal, the simplex will be called regular. Denoting by fi(P ) the number of n n+1 i-faces of a polytope P ∈ P , we ﬁnd for an n-simplex that fi(Tn) = i+1 , ∞ 1 i = 0, . . . , n − 1, and for the cube Bn and the crosspolytope Bn we have n f (B1) = f (B∞) = 2n−i , i = 0, . . . , n − 1. n−1−i n i n i Here the ﬁrst equality follows from the general fact that the polarity map induces an inclusion reversing bijection between the i-faces of a polytope P and the (n − 1 − i)-faces of P ?, where we assume that 0 ∈ int P , i.e., the origin is an interior point of P .

∞ 1 Figure 1.9: ”Polarity” between the faces of B3 and B3

In particular, for those n-polytopes P we have: if P = conv {v1,..., vm} ? n with vertices vi, 1 ≤ i ≤ m, then P = {x ∈ R : hvi, xi ≤ 1, 1 ≤ i ≤ m} 1. Basic (and some more) facts from Convexity 11

? with facets P ∩ H(vi, 1), 1 ≤ i ≤ m, and vice versa. Thus, in view of (1.6) and (1.5) we get the representations for the crosspolytope 1 Bn = conv {±ei : 1 ≤ i ≤ n} n | = {x ∈ R : h(1, . . . , n) , xi ≤ 1, i ∈ {−1, 1}}. In principle, it is an easy task to calculate the volume of a polytope. To this end we note that the volume of an n-dimensional pyramid P = conv {Q, w} over an (n − 1)-dimensional polytope Q ⊂ H(a, b), say, and with apex w ∈/ H(a, b) is given by vol(P ) = (1/n) voln−1(Q) dist (Q, w). Here voln−1(·) denotes the (n − 1)-dimensional volume and dist (Q, w) = | ha, vi − b|/ |a| is the distance between w and aﬀ Q = H(a, b).

H(a, b)

Figure 1.10: Pyramid over Q with apex w

Hence, since the volume is additive we can dissect (without gaps) a polytope P into pyramids over its facets F1,...,Fm, and a common apex Pm w ∈ P , say, and get vol(P ) = (1/n) i=1 voln−1(Fi) dist (Fi, w). Since the volume is translation invariant the apex w can also lie outside of P , we just have to take the signed distances. For instance, let aﬀ Fi = H(ai, bi) with |ai| = 1 and P ⊂ H≤(ai, bi), then with respect to the point w = 0 we get m m 1 X 1 X vol(P ) = vol (F ) b = vol (F ) h (a ). (1.7) n n−1 i i n n−1 i P i i=1 i=1 This recursive formula, however, is apparently inapplicable for practical computations. In fact, it is known that computing the volume of a polytope is a #P- problem. Another ad hoc approach for computing the volume of a polytope is to dissect the polytope into simplices, since the volume of an n-simplex S = conv {v0,..., vn} can easily be calculated as 1 1 v0 vn+1 vol(S) = | det(v1 − v0,..., vn − v0)| = det ,..., . n! n! 1 1 12 1. Basic (and some more) facts from Convexity

F3(+)

F4(+)

F2(+)

F5(−)

F1(−)

a4 a3

a2 0

a5 a1

Figure 1.11: ”Pyramid” formula for computing volume of polytopes

Here, however, the question is how to dissect P in as few as possible simplices, which is not as easy as it might look. For instance, ”even” for the ∞ cube Bn the minimal number of simplices in a dissection is only known in small dimensions and the best bounds diﬀer by an exponential factor. n A dissection of a polytope P ∈ P , dim P = n, into n-simplices Si, m 1 ≤ i ≤ m, i.e., P = ∪i=1Si and int (Si) ∩ int (Sj) = ∅, i 6= j, is called a triangulation if Si ∩ Sj is a face of both, Si and Sj. The theorem of Carath´eodory (Theorem 1.9) implies that a polytope can be covered by the simplices generated by its vertices, the next theorem strengthens this statement in the sense that we can even trinagulate it with its vertices.

1.11 Theorem. P ∈ Pn, dim P = n, can be triangulated into simplices without introducing new vertices, i.e., the vertices of each simplex of the triangulation are a subset of the vertices of P .

Figure 1.12: A triangulation of the 3-cube into 5 simplices 1. Basic (and some more) facts from Convexity 13

Exercises

1.1 Show that

n ? ? i) If X1 ⊆ X2 ⊆ R then X2 ⊆ X1 . n S ? T ? ii) If Xi ⊆ R , i ∈ I, then i∈I Xi = i∈I Xi . n ? ? iii) If X ⊆ R then X ⊆ (X ) . n ? iv) If X ⊆ R then X = X if and only if X = Bn.

n n 1.2 Let X ⊆ R and t ∈ R such that 0 ∈ int X ∩ int (t + X). Show that 1 (t + X)? = y : y ∈ X? . 1 + ht, yi

n 1.3 i) Let K ∈ Ko , dim K = n, and let p ∈ (1, ∞). Show that Z −|x|p e K dx = Γ(n/p + 1) vol(K). Rn ii) Show for 1 ≤ p < ∞

Γ(1 + 1/p)n vol(Bp) = 2n , n Γ(1 + n/p)

∞ n and vol(Bn ) = 2 .

1.4 Show that for 1 ≤ p, q ≤ ∞

p q √ −1/p −1/q dH(Bn,Bn) = n |n − n |.

n 1.5 Show that K ⊆ R is convex if and only if

( m m ) X X K = λi xi : m ∈ N, xi ∈ K, λi ≥ 0, λi = 1 . i=1 i=1

n 1.6 Let X ⊆ R . Show that i)

( m m ) X X conv X = λi xi : m ∈ N, xi ∈ X, λi ≥ 0, λi = 1 . i=1 i=1

ii) Show that the integer m in the identity above can be replaced by dim X + 1. 14 1. Basic (and some more) facts from Convexity

n 1.7 Let P ∈ P be an n-dimensional polytope with facets F1,...,Fm, n and let ai ∈ R , |ai| = 1, bi ∈ R, such that aﬀ Fi = H(ai, bi) and P ⊂ H≤(ai, bi), 1 ≤ i ≤ m. Show that

m X voln−1(Fi) ai = 0. i=1 (Hint: Formula (1.7) might help.)

1.8 Let K,L ∈ Kn, dim K = dim L = n. Show that int K + int L = int (K + L).

1.9 Show that each convex body is the convex hull of its extreme points.

∞ 1.10 Show that the number of simplices in any dissection of the cube Bn into simplices can not be smaller than 2n n!/(n + 1)(n+1)/2. 1. Basic (and some more) facts from Convexity 15

Rough Notes

n+1 1) The number of i-faces of an n-simplex, i+1 , is also the minimum number of i-faces which an n-polytope can have. In particular we have Pn−1 n i=0 fi(P ) ≥ 2 − 2 with equality if and only if P is an n-simplex. There is a very appealing conjecture by Gil Kalai claiming that for o-symmetric n-polytopes the lower bound 2n − 2 can be replaced by 3n − 1 as it is attained by the cube. The cube, however, is as in the case of the Mahler conjecture not the only potential minimizer. In fact, the known bodies for which the Mahler volume is minimal are (up to combinatorial equivalence) precisely those polytopes for which the sum of the i-faces seems to be minimal (see Kalai’s paper [8]). The conjecture of Kalai has been conﬁrmed in dimensions ≤ 4 [12], and the only result in the spirit of the conjecture for general dimensions is due to Figiel, Lindenstrauss and Milman [3]

ln f0(P ) ln fn−1(P ) ≥ c n,

where c is an absolute constant. The interesting proof utilizes the L¨owner-Johnellipsoid of o-symmetric convex bodies.

2) The Mahler conjecture has also a “non-symmetric side”. Here we consider for K ∈ Kn, dim K = n, the product

M(K) = min vol(K) vol((K − z)?). z∈int K It is known that, as in the o-symmetric case, it is maximized only for ellipsoids, but the lower bound is open for n ≥ 3. Simplices are conjectured to be the only minimizer, i.e.,

(n + 1)n+1 M(K) ≥ M(T ) = , n (n!)2

with equality if and only if K is an n-simplex. literature

3) The survey of Gardner [4] gives excellent access to the huge amount of classical and modern aspects of the Brunn-Minkowski inequality. A highly recommended article on modern aspects of convex geometry (in general) is (still) [1] by Ball.

4) For more information on the surprising and marvellous world of triangulations&friends we refer to the recent book [10] of De Loera, Rambau and Santos.

Basic facts about Lattices

In this chapter we will start to introduce the second and discrete ingredient of the Geometry of Numbers: lattices. Although (or should one say because) trivial to deﬁne, they have a surprisingly rich geometric structure and even basic theoretical as well as elementary algorithmic questions are still unsolved.

n 2.1 Definition [Lattice]. Let b1,..., bn ∈ R be linearly independent. The set Λ = {z1b1 + z2b2 + ··· + zkbk : zi ∈ Z, 1 ≤ i ≤ n} is called an (n-dimensional) lattice. The set of generating vectors {b1,..., bn} or the matrix B = (b1,..., bn) with columns bi is called the basis of Λ. An element b ∈ Λ is called the lattice point of Λ. The set of all n-dimensional n n lattices in R is denoted by L . n In other words, a lattice is of the form Λ = B Z , where B is a regular n × n-matrix, i.e., B ∈ GL(n, R). In the case that B is the identity matrix n we obtain the lattice Z itself, which is also called integral lattice or (standard n lattice). The unit vectors e1,..., en are a (canonical) basis of Z , but, of n course, there are many (infinitely many) more bases of Z . For instance, the vectors b1 = (1, 1)| and b2 = (−1, 0)| or b1 = (25, 16)| and b2 = (64, 41)| 2 are also bases of Z (since e1 and e2 are integral linear combinations of n these vectors). In order to characterize all bases of Z we need the following definition.

n×n 2.2 Deﬁnition [Unimodular matrix]. An integral matrix U ∈ Z is called unimodular iﬀ | det U| = 1. The group of all unimodular matrices is denoted by GL(n, Z). Oberserve that a matrix is unimodular if and only if the matrix and its inverse are integral matrices, which implies that GL(n, Z) is precisely the n set of all bases of Z as pointed out in the next proposition.

17 18 2. Basic facts about Lattices

n×n n n 2.3 Proposition. GL(n, Z) = {U ∈ R : U Z = Z }. −1 n×n Proof. U ∈ GL(n, Z) if and only if U, U ∈ Z , which is equivalent n n −1 n n to U Z ⊆ Z and U Z ⊆ Z . Since the last inclusion is equivalent to n n n n Z ⊆ U Z we are done. Observe that Z ⊆ U Z implies that U is a regular matrix. By this proposition we can easily describe all bases of a given lattice.

n n 2.4 Lemma. Let Λ = B Z ∈ L . A = (a1,..., an) is a basis of Λ if and only if there exists a U ∈ GL(n, Z) such that A = BU. n n Proof. A is a basis of Λ if and only if A Z = Λ = B Z . Since B is regular −1 n n this identity is equivalent to B AZ = Z which by Proposition 2.3 is −1 equivalent to B A ∈ GL(n, Z). So two diﬀerent bases of a lattice Λ are related by an unimodular transformation, and thus | det B| is independent of the chosen basis B of Λ. Hence it makes sense to associate this size to the lattice.

2.5 Deﬁnition [Determinant, fundamental cell]. Let Λ ∈ Ln with basis B = (b1,..., bn). i) det Λ = | det B| is called determinant of Λ.

n ii) PB = {ρ1b1 + ··· + ρnbn : 0 ≤ ρi < 1, 1 ≤ i ≤ n} = B [0, 1) is called fundamental cell or fundamental parallelepiped of Λ (with respect to the basis B).

n n n Since vol(B [0, 1) ) = | det B|vol([0, 1) ) = | det B| we have for Λ = B Z ∈ Ln that vol(PB) = det Λ, (2.5.1) i.e., the determinant is the volume of a fundamental cell. Another apparent, but useful property of a fundamental cell is expressed via the identity

(PB − PB) ∩ Λ = {0}. (2.5.2)

n Observe that PB − PB = B (−1, 1) and thus, the only vector in PB − PB which has a representation as an integral combination of the columns of B is 0. Before we state the main property of such a fundamental cell, we introduce a notation which can be interpreted as a kind of rounding of a vector n x ∈ R with respect to a lattice, or more precisely with respect to a basis Pn n B of a lattice Λ. To this end let B ∈ GL(n, R) and for x = i=1 ρibi ∈ R , ρi ∈ R, we write n X bxcB = bρicbi, i=1 2. Basic facts about Lattices 19

n where bρc is the largest integer not bigger than ρ. In particular, bxcB ∈ BZ and x − bxcB ∈ PB.

n n 2.6 Proposition. Let Λ = BZ ∈ L . Then n S· R = b∈Λ (b + PB) ,

n i.e., R is the pairwise disjoint union of the lattice translates b + PB. n Proof. Each x ∈ R can be decomposed as x = (x − bxcB) + bxcB, where the ﬁrst summand is in PB and the latter is a lattice point of Λ. To show that the union is pairwise disjoint, we note that the intersection of two lattice translates b + PB, b + PB, b, b ∈ Λ, is non-empty, if and only if b − b ∈ (PB − PB) ∩ Λ. By observation (2.5.2) this is equivalent to b = b. The next theorem gives a characterization of lattices as the discrete n n subgroups S of R which spans R as a vectorspace. Here ”discrete” just means that there exists an > 0 such that |s1 − s2| ≥ for any s1, s2 ∈ S. The main part of the proof of this theorem is covered by the next lemma which will also be used later on.

n 2.7 Lemma. Let S ⊂ R be a discrete subgroup containing n linearly independent elements s1,..., sn ∈ S. There exist b1,..., bn ∈ S such that for 1 ≤ k ≤ n

lin {s1,..., sk} ∩ S = {z1b1 + ··· + zkbk : zi ∈ Z} . (2.7.1)

Proof. We will construct the vectors b1,..., bn inductively. For k = 1 let b1 6= 0 be the (Euclidean) shortest vector in conv {0, s1} ∩ S. Since S is discrete such a choice is possible, and since S is a subgroup we certainly have {z1b1 : z1 ∈ Z} ⊆ lin {s1} ∩ S. For the reverse inclusion, let s ∈ lin {s1} ∩ S and let λ ∈ R such that s = λ b1. Then s−bλc b1 = (λ−bλc)b1, and by the minimality of b1 we conclude λ = bλc ∈ Z. Hence (2.7.1) holds for k = 1. Let us assume that we have already found b1,..., bm, m ≥ 1, satisfying (2.7.1) for 1 ≤ k ≤ m < n. Next we consider the (m + 1)-dimensional parallelepiped

( m ) X Pm+1 = αi bi + αm+1sm+1 : 0 ≤ αi ≤ 1 . i=1

Let bm+1 ∈ Pm+1 ∩ S having minimum positive distance to lin {b1,..., bk}, i.e., m X bm+1 = αi bi + αm+1sm+1, (2.7.2) i=1 20 2. Basic facts about Lattices and αm+1 > 0 is minimal among all points in Pm+1 ∩ S. Obviously, by our inductive construction we have

lin {b1,..., bm+1} = lin {s1,..., sm+1}, (2.7.3) and together with the discrete subgroup property of S we get {z1b1 + ··· + zm+1bm+1 : zi ∈ Z} ⊆ lin {s1,..., sm+1} ∩ S. Again, in order to establish the reverse inclusion, let s ∈ lin {s1,..., sm+1} ∩ S which, by (2.7.3), can be Pm+1 written as s = i=1 βi bi for some scalars βi ∈ R, and we have to show that βi ∈ Z. On account of (2.7.2), we can write

m+1 m+1 X X s − bβic bi = (βi − bβic) bi i=1 i=1 m X = (βi − bβic) + αi(βm+1 − bβm+1c) bi i=1

+ αm+1(βm+1 − bβm+1c) bm+1.

For abbreviation we write µi for all of these scalars in front of the vectors bi in the last sum. Then 0 ≤ µm+1 < αm+1 ≤ 1, and by the discrete subgroup property of S and the deﬁnition of Pm+1 we also know

m+1 m m X X X s − bβic bi − bµic bi = (µi − bµic) bi + µm+1 bm+1 ∈ S ∩ Pm+1. i=1 i=1 i=1

By the minimality of αm+1 we must have µm+1 = 0 and so βm+1 = bβm+1c ∈ . Thus Z m X s − βm+1 bm+1 = βi bi ∈ lin {s1,..., sm} ∩ S, i=1 and by our inductive hypothesis, (2.7.1) implies the integrality of the other scalars βi, 1 ≤ i ≤ m.

n 2.8 Theorem. S ⊂ R is a lattice if and only if S a discrete subgroup of n R and it contains n linearly independent points. n n Proof. Obviously, a lattice Λ = B Z is a subgroup of R containing n linearly independent points. Since 0 is an interior point of the convex set n PB −PB = B(−1, 1) , there exists an > 0 such that Bn ⊂ PB −PB. Since (PB − PB) ∩ Λ = {0} (cf. (2.5.2)) and since Λ is a subgroup we conclude |b1 − b2| ≥ for all b1 6= b2 ∈ Λ, i.e., Λ is discrete. For the reverse implication, we observe that Lemma 2.7, more precisely equation (2.7.1) for k = n, implies that there exists n linearly independent b1,..., bn ∈ S such that

S = {z1b1 + ··· + znbn : zi ∈ Z}, 2. Basic facts about Lattices 21 i.e., S is a lattice. Vectors b1,..., bk of a lattice Λ fulﬁlling 2.7.1, i.e., they generate all lattice point in a certain plane (here lin {s1,..., sk}) are of particular interest and we deﬁne

2.9 Deﬁnition [Primitive set of lattice vectors]. Let b1,..., bk ∈ Λ ∈ n L be linearly independent. The set {b1,..., bk} is called a primitive set of lattice vectors if

k lin {b1,..., bk} ∩ Λ = (b1,... bk)Z .

So lattice vectors are primitive if each lattice point in their linear hull can be represented as an integral combination of these vectors.

2.10 Lemma. Let Λ ∈ Ln.

i) Each primitive set of vectors can be supplemented to a basis of the lattice, and each subset of a basis is a primitive set. Pn ii) Let b1,..., bn be a basis of Λ, let c = i=1 zibi ∈ Λ, zi ∈ Z, and let j ∈ {1, . . . , n}. Then it holds:

{b1,..., bj−1, c} is primitive ⇔ gcd(zj, . . . , zn) = 1.

Proof. In order to show that a primitive set of vectors {b1,..., bk} can be supplemented to a basis we first supplement it to a set {b1,..., bk, sk+1,..., sn} of linearly independent lattice vectors. Since the first k vectors of this set satisfy (2.7.1) the inductive proof of of Lemma 2.7 finds basis containing {b1,..., bk}. If b1, . . . , bn is a basis of Λ, then any b ∈ Λ has an unique and integral representation as linear combination of {b1, . . . , bk} ∩ Λ. Hence any subset of a basis is primitive. For ii) assume first that gcd(zj, . . . , zn) 6= 1. If the gcd is ∞, i.e., all zi = 0, i = j . . . , n, then {b1,..., bj−1, c} are linearly dependent. So let Pn Pj−1 gcd(zj, . . . , zn) = m ≥ 2 and let a = i=j zi bi = c − i=1 zi bi. Then

n j−1 1 X zi 1 X zi a = b = c − b . m m i m m i i=j i=1

1 1 The ﬁrst identity shows that m a ∈ Λ and the second says that m a does not have an integral representation by the the linearly independent vectors {b1,..., bj−1, c}. Hence these vectors are not primitive. For the reverse Pj−1 direction let a ∈ lin {b1,..., bj−1, c} ∩ Λ, i.e., a = i=1 µi bi + µj c for 22 2. Basic facts about Lattices

Pn some µi ∈ R and a = i=1 zi bi with zi ∈ Z. We have to show µi ∈ Z, Pn 1 ≤ i ≤ j. Substituting c = i=1 zibi yields

j−1 n n X X X (µi + µj zi)bi + µj zi bi = zi bi. (2.10.1) i=1 i=j i=1

Hence µj zi = zi, i = j, . . . , n, and since gcd(zj, . . . , zn) = 1 we conclude µj ∈ Z. Comparing the frist (j − 1) scalars in (2.10.1) then also shows µi ∈ Z, 1 ≤ i ≤ j − 1. Lemma 2.7 applied to a lattice Λ, in particular, shows that for any set of n linearly independent vectors a1,..., an ∈ Λ we can ﬁnd a basis b1,..., bn Pk such that ak = i=1 zi,k bi for some integral scalars zi,k and 1 ≤ k ≤ n, n×n or in matrix notation A = BZ, where Z ∈ Z is an upper triangular matrix. Via some ”integral Yoga” on the basis vectors bi one can achieve the following unique representation of this type which is the theory of matrices and is also known as the Hermite Normalform.

2.11 Theorem [Hermite Normalform]. Let A = (a1,..., an) be linearly independent lattice vectors of a lattice Λ ∈ Ln. Then there exists a basis n×n B = (b1,..., bn) of Λ and an upper triangular matrix Z ∈ Z such that A = BZ and for 1 ≤ k ≤ n

0 ≤ zi,k < zk,k, 1 ≤ i ≤ k − 1. (2.11.1)

In words, Z is a non-negative upper triangular integral matrix and for each column of Z the unique maximal element is the diagonal element. Moreover, for given A such a basis B is uniquely determined.

Proof. First we show the existence: As mentioned above, on account of Lemma 2.7 there exists a basis b1,..., bn of Λ such that

k X ak = zi,k bi, zi,k ∈ Z, 1 ≤ k ≤ n. i=1

In the following we modify successively this basis in order to ﬁnd a basis satisfying (2.11.1). If z1,1 < 0, we replace b1 by −b1 and (2.11.1) is fulﬁlled for k = 1. Let us assume that (2.11.1) holds for 1 ≤ i ≤ l < n, say. Obviously, for arbitrary scalars zel+1,l+1 ∈ {−1, 1}, zei,l+1 ∈ Z, 1 ≤ i ≤ l, the vectors l+1 ! X b1,..., bl, zei,l+1 bi , bl+2,..., bn (2.11.2) i=1 2. Basic facts about Lattices 23

Pl+1 form a basis of Λ. Substituting bl+1 by i=1 zei,l+1 bi yields for al+1

l+1 X al+1 = zi,l+1 bi i=1 l l+1 l ! X X X = zi,l+1 bi + zl+1,l+1zel+1,l+1 zei,l+1bi − zei,l+1bi i=1 i=1 i=1 l X = (zi,l+1 − zl+1,l+1zel+1,l+1zei,l+1) bi i=1 l+1 ! X + zl+1,l+1zel+1,l+1 zei,l+1bi . i=1

Now we choose the sign zel+1,l+1 such that zl+1,l+1zel+1,l+1 > 0 and then we choose zei,l+1 such that

zi,l+1 − zl+1,l+1zel+1,l+1zei,l+1 < zl+1,l+1zel+1,l+1, 1 ≤ i ≤ l. The new basis (2.11.2) satisfies now (2.11.1) for 1 ≤ k ≤ l + 1, and, of course, in this iterative way we obtain the desired basis. It remains to show the uniqueness of such a basis. To this end let B0 be another basis of Λ and Z0 be another upper triangular integral matrix satisfying (2.11) and such that A = B0Z0. Then 0 0 B = B U for some U ∈ GL(n, Z) and thus Z = UZ . Hence, the rows of Z and Z0 generate the same lattice Λ¯ ∈ Ln, say. Suppose there Let k be a row index such that the kth row of Z and Z0 differ and let l be the 0 minimal column index such that zk,l 6= zk,l, l ∈ {k, . . . , n}. We may assume 0 zk,l > zk,l. Therefore the difference vector h of the kth row of Z and the kth 0 row of Z has its frist nonzero coordinate in hl > 0. Since it also belongs to the lattice Λ¯ we find that

0 zk,j − zk,l = hl = m zl,l for a non-negative integer m. In view of (2.11) we conclude m = 0 which 0 shows B = B . n n×n With respect to the lattice Z and an integral matrix A ∈ Z the theorem implies that there exists an unimodular matrix U ∈ GL(n, Z) and n×n a uniquely determined upper triangular matrix Z ∈ Z with entries zi,k satisfying (2.11.1) such that A = UZ. Hence, if we start with A| then n×n we ﬁnd that for any integral matrix A ∈ Z there exists a unique lower triangular matrix Z such that the entries of Z| satisﬁes (2.11.1) and

A = Z U. 24 2. Basic facts about Lattices

Z is usually called the Hermite Normalform of A. In particular, we see that the columns of Z and A generate the same lattice (cf. Lemma (2.4)), and sometimes, it is preferable to work with a basis in triangular form.

2.12 Deﬁnition [Sublattice, Index of a sublattice]. Let Λ ∈ Ln and let a1,..., an ∈ Λ be linearly independent. The lattice

Λ0 = {z1a1 + ··· + znan : zi ∈ Z} is called a sublattice with basis A = (a1,..., an). The number of cosets of the subgroup Λ0 in Λ is called the index of Λ0 in Λ, and it is denoted by |Λ:Λ0|. Of course, the index is a positive integer and the next lemma states that it can easily be determined, either by counting lattice points or by the ratio of the determinants of the involved lattices.

n 2.13 Lemma. Let Λ0 ⊆ Λ ∈ L be a sublattice of Λ. Then

i) |Λ:Λ0| = #(PA ∩ Λ) for any basis A of Λ0.

ii) |Λ:Λ0| = det Λ0/ det Λ.

Proof. By the deﬁnition of cosets, for i) we have to show that Λ is the disjoint union of the cosets c + Λ0, c ∈ PA ∩ Λ, i.e., Λ = S· (c + Λ ) . c∈PA∩Λ 0

Let b ∈ Λ, then bbcA ∈ Λ0 ⊆ Λ and so (b − bbcA) ∈ PA ∩ Λ. Thus

b = (b − bbcA) + bbcA ∈ (PA ∩ Λ) + Λ0, and it remains to show that the cosets are diﬀerent. Let b1, b2 ∈ PA ∩ Λ such that (b1 + Λ0) ∩ (b2 + Λ0) 6= ∅. Then b1 − b2 ∈ (PA − PA) ∩ Λ0 = {0} (cf. (2.5.2)) and i) is shown. For ii) we ﬁrstly observe that

m P = S· (m a + ··· + m a + P ) , A 0≤mi

Finally, since PA is measureable we may write (cf. Proposition 1.3) 1 det Λ #(mPA ∩ Λ) det Λ0 = vol(PA) = lim # PA ∩ Λ = det Λ lim m→∞ m mn m→∞ mn

= det Λ |Λ:Λ0|. 2. Basic facts about Lattices 25

A simple and useful application of the last lemma is the fact that the determinant of an integral matrix is equal to the number of lattice points contained in the associated halfopen parallelepiped and, by deﬁniton, it is also equal to the volume of this parallelepiped.

n 2.14 Corollary. Let A = (a1,..., an) ∈ GL(n, Z), and let PA = A[0, 1) be the half open parallelepiped generated by the columns of A. Then | det A| = n #(PA ∩ Z ) = vol(PA).

n Proof. We just apply Lemma 2.13 i), ii) with Λ = Z and Λ0 being the lattice generated by A. n In some cases we are interested in subsets Z fullﬁling certain (homoge- n neous) linear congruences, for instance, {z ∈ Z : z1 + ··· + zn ≡ 0 mod 2}. The next application of Theorem 2.8 shows that those sets are lattices.

n 2.15 Corollary. Let u1,..., um ∈ Z and let ki ∈ N≥1, 1 ≤ i ≤ m. The set n Λ = {z ∈ Z : hui, zi ≡ 0 mod ki, 1 ≤ i ≤ m} n is a sublattice of Z with det Λ ≤ k1k2 · ... · km.

n n Proof. By definition Λ is a discrete subgroup of R , actually of Z . Since the n linearly independent vectors (k1 ··· km) ei, 1 ≤ i ≤ n, belong to Λ, n Theorem 2.8 shows that Λ is a lattice. Since Λ is a sublattice of Z we may n n consider the different cosets of Λ w.r.t. Z . Two points z1, z2 ∈ Z belong to different cosets if and only if (z1 − z2) ∈/ Λ, i.e., there exists an i such that hui, z1 − z2i 6≡ 0 mod ki. In other words, for some ui the numbers hui, z1i and hui, z2i belong to different residue classes mod ki. For each ui there are at most ki different residue classes and thus the number of different cosets is bounded by the product k1 · ... · km. Hence we get by Lemma (2.13)

n n det Λ = |Z :Λ| det Z ≤ k1k2 · ... · km.

n Given linearly independent vectors A = (a1,..., an) of a lattice Λ ∈ L we can easily decide when these vectors form a basis of Λ. Namely by Lemma 2.13 we know that

n A is basis of Λ ⇔ |Λ: AZ | = 1 ⇔ Λ ∩ PA = {0}. (2.15.1)

So we just have to check if the parallelepiped PA contains a non-trivial lattice point of Λ. In the planar case we even have 26 2. Basic facts about Lattices

2 2.16 Proposition. Let Λ ∈ L and let a1, a2 ∈ Λ be linearly independent. Then

a1, a2 is basis of Λ ⇔ conv {0, a1, a2} ∩ Λ = {0, a1, a2}.

Proof. The main and only observation is that in the plane each parallelogram P can be dissected into two triangles which are reﬂections of each other with respect to the midpoint of the main diagonal. With A = (a1, a2) this becomes for the halfopen parallelogram PA

PA = conv {0, a1, a2}\{a!, a2} ∪ ((a1 + a2) − int conv {0, a1, a2}) .

Hence PA ∩Λ = {0} if and only if conv {0, a1, a2}∩Λ = {0, a1, a2} and with (2.15.1) we are done. As one can guess from the arguments in the proof of Proposition 2.16 an analogous statement does not exist in dimension ≥ 3. For instance, for n n ≥ 3 and m ∈ N let b(m) = (1,..., 1, m)| ∈ R and let

Tn(m) = conv {0, e1,..., en−1, b(m)}.

Then the only integral points of the simplices Tn(m) are the vertices, i.e., n Tn(m) ∩ Z = {0, e1,..., en−1, b(m)}, but e1,..., en−1, b(m) do not build n a basis of Z for m ≥ 2.These simplices are called Reeve simplices and will also play a role later on. Next we want to extend the deﬁnition of lattices to k-dimensional structures, more precisely, to k-dimensional linear subspaces.

2.17 Deﬁnition [Lattice planes, k-dimensional (sub)lattices]. Let Λ ∈ Ln.

n i) A linear subspace L ⊆ R is called a lattice plane of Λ if dim(L ∩ Λ) = dim L. The set of all k-dimensional lattice planes L of Λ, i.e., dim L = k, is denoted by L(k, Λ).

ii) For L ∈ L(k, Λ), the set Λk = L ∩ Λ is called a (k-dimensional) (sub)lattice of Λ.

We note that Theorem 2.8, applied to the Euclidean space Lk ∈ L(k, Λ), shows that Λk is indeed a (k-dimensional) lattice. Thus there exists a basis b1,..., bk of Λk, which can also be seen directly from (2.7.1). As in the full dimensional case, the determinant of such a k-dimensional lattice Λk, denoted by det Λk, is the k-dimensional volume of a fundamental cell {ρ1 b1+ ··· + ρk bk : ρi ∈ [0, 1)} which can be calculated as p det Λk = det ((b1,..., bk)|(b1,..., bk)). (2.17.1) 2. Basic facts about Lattices 27

n 2.18 Theorem [Smith Normalform]. Let Λ0 ⊆ Λ ∈ L be a sublattice of Λ. Then there exists a Basis A = (a1,..., an) of Λ0 and a basis B = (b1,..., bn) of Λ such that ai = zi bi with zi ∈ Z, 1 ≤ i ≤ n, i.e., A = BZ and Z is a diagonal matrix.

Proof. We use induction on the dimension n, and the case n = 1 is obvious. So let n > 1. First we show the assertion under the assumption that

there exists a b1 ∈ Λ0 which is primitive w.r.t. Λ. (2.18.1)

Then we can ﬁnd a basis b1,..., bn of Λ (cf. Lemma 2.10 i)) and let L = lin {b2,..., bn}. Then Λ = Z b1 + (L ∩ Λ) and since Λ0 ⊆ Λ and b1 ∈ Λ0 we also have Λ0 = Z b1 + (L ∩ Λ0). Applying our inductive argument to the (n − 1)-dimensional sublattice L ∩ Λ0 of L ∩ Λ gives the assertion. Next we will scale our lattice Λ0 such that (2.18.1) holds. To this end we consider for any a ∈ Λ the unique α ∈ such that 1 a ∈ Λ is primitive. 0 a N αa Let a ∈ Λ0 having minimal factor α = αa, say, and let a = α b, with b ∈ Λ primitive. In the following we will show

α|αa for every a ∈ Λ0. (2.18.2)

Let a ∈ Λ0 and we may assume that a is linearly independent of a. Hence 0 2 0 2 Λ0 = (a, a)Z is a 2-dimensional sublattice of Λ = lin {a, a} ∩ Λ = (b, b)Z for a suitable b ∈ Λ (cf. Lemma 2.10 i)). Hence we may write

a = αa (z1 b1 + z2 b) with gcd(z1, z2) = 1, and it is enough to show that α|(αa z1) and α|(αa z2). To this end let γ ∈ Z such that 0 ≤ (αa z1 − α γ) < α. Then

c = a − γ a = (αa z1 − α γ) b + αa z2 b = αc (s1 b + s2 b) for suitable relatively prime integers s1, s2. By the choice of γ and α we have 0 ≤ αc s1 < α ≤ αc, and thus s1 = 0. Hence αa z1 − α γ = 0 and so

α|(αa z1). (2.18.3)

Moreover we hve c = αa z2 b and so a + c = α b + αa z2 b ∈ Λ0 \{0}. Hence αa+c|α and by the minimality of α we get αa+c = α and thus

α|(αa z2). (2.18.4)

1 (2.18.3) and (2.18.4) imply (2.18.2), which shows that Λ0 = α Λ0 is a sublattice of Λ. Since b ∈ Λ0 is primitive w.r.t. Λ, (2.18.1) is satisﬁed. Hence Λ0, Λ, and thus Λ0, Λ, have bases with the desired property. 28 2. Basic facts about Lattices

n 2.19 Deﬁnition [Polar lattice]. Let Λ ∈ L with basis b1,..., bn. Let ? n bi ∈ R , 1 ≤ i ≤ n be given by ( ? 0, i 6= j hbj, b i = δi j = , 1 ≤ j ≤ n. i 1, i = j,

? ? ? The lattice Λ with basis b1,..., bn is called the polar lattice of Λ.

2.20 Remark.

? ? −| i) If B = (b1,..., bn) is a basis of Λ then (b1,..., bn) = B . In particular, the deﬁnition of the polar lattice is independent of the basis B of Λ.

ii) det Λ? = det(Λ)−1.

n 2.21 Proposition. Let Λ ⊂ R be a lattice. Then ? n Λ = {y ∈ R : hb, yi ∈ Z for all b ∈ Λ} . − n n Proof. Let B be a basis of Λ and let y = B |x ∈ R with x ∈ R . Then ? n n y ∈ Λ ⇔ x ∈ Z ⇔ x|z ∈ Z for all z ∈ Z − n n ⇔ (B |x)|Bz ∈ Z for all z ∈ Z ⇔ (y)|Bz ∈ Z for all z ∈ Z .

2.22 Lemma. Let L ∈ L(k, Λ) and let L⊥ be the (n − k)-dimensional orthogonal complement of L. Then i) the orthogonal projection of Λ onto L⊥, denoted by Λ|L⊥, is an (n−k)- dimensional lattice with

det(Λ|L⊥) · det(L ∩ Λ) = det Λ. (2.22.1)

ii) L⊥ ∈ L(n − k, Λ?) and it holds (L⊥ ∩ Λ?) = (Λ|L⊥)? (w.r.t. the space L⊥). In particular we have

det(L ∩ Λ) = det Λ · det(L⊥ ∩ Λ?). (2.22.2)

Proof. Let b1,..., bn be a basis of Λ such that b1,..., bk is a basis of L∩Λ (cf. Lemma (2.10) i)). Furthermore let bk+1,..., bn be the orthogonal ⊥ ⊥ projection of bk+1,..., bn onto L , i.e., bi = bi|L , i = k + 1, . . . , n. Then Pn for every b = i=1 zibi ∈ Λ we have n n ⊥ X ⊥ X b|L = zi(bi|L ) = zibi. i=1 i=k+1 2. Basic facts about Lattices 29

⊥ This shows that Λ|L is an (n−k)-dimensional lattice with basis bk+1,..., bn. Now let Bk = (b1,..., bk) and Bn−k = (bk+1,..., bn). Then we can write

? ? ? ? Next let b1,..., bn be the associated basis of Λ , i.e., hbj, bi i = δi j. This ? ? ⊥ ? immediately implies that bk+1,..., bn ∈ L ∩ Λ and since

? ? bk+i, bk+j = bk+i, bk+j = δi j we see that L⊥ ∩ Λ? is an (n − k)-dimensional lattice, which is polar to Λ|L⊥. Thus we have det(L⊥ ∩ Λ?) = det(Λ|L⊥)−1 and from (2.22.1) we get (2.22.2).

Minkowski’s successive minima

In this chapter we start to study the interaction of convex bodies with lattices, and the beginning we address the problem to bound the volume of a 0-symmetric convex body K in terms of functionals related to the lattices points of K. The ﬁrst lemma, although it is more or less just an observation on so volume of so-called packing and covering sets, is of particular interest.

n n 3.1 Lemma. Let X ⊂ R be a bounded Jordan-measurable set and Λ ∈ L .

i) If (b1 + X) ∩ (b2 + X) = ∅, for all b1, b2 ∈ Λ, b1 6= b2, then vol(X) ≤ det Λ.

n ii) If Λ + X = R then vol(Λ) ≥ det Λ.

n Proof. Let P be the fundamental cell of Λ. Since Λ + P is a tiling of R (cf. Proposition 2.6) and the volume is a translation invariant functional we may write X vol(X) = vol ((Λ + P ) ∩ X) = vol ((b + P ) ∩ X) b∈Λ X = vol (P ∩ (X − b)) . b∈Λ

In the first case i) we have that two different translates P ∩ (X − b1), P ∩ (X − b2), b1 6= b2, do not overlap and thus X vol (P ∩ (X − b)) = vol (∪b∈Λ(P ∩ (X − b))) ≤ vol(P ) = det Λ. b∈Λ n In the second case ii), where Λ + X = R we find X vol (P ∩ (X − b)) ≥ vol (∪b∈Λ(P ∩ (X − b))) = vol(P ) = det Λ. b∈Λ

31 32 3. Minkowski’s successive minima

n n 3.2 Corollary. Let X ⊂ R with vol(X) > det Λ and let Λ ∈ L . Then (X − X) ∩ Λ \{0}= 6 ∅.

n Proof. By Lemma 3.1 i) there exist b1, b2 ∈ Λ, b1 6= b2, and a x ∈ R such that x ∈ (b1 + X) ∩ (b2 + X). Thus x − b1, x − b2 ∈ X and so b2 − b1 = (x − b1) − (x − b2) ∈ (X − X) ∩ Λ \{0}. Another way to state the conclusion of the corollary above is to say that n there exists a t ∈ R such that #((t + X) ∩ Λ) ≥ 2.

n n 3.3 Theorem [Minkowski]. Let K ∈ Ko , Λ ∈ L , and let vol(K) ≥ 2n det Λ. Then K ∩ Λ \{0}= 6 ∅, i.e., a 0-symmetric convex body of volume at least 2n det Λ contains a non- trivial lattice point.

n 1 Proof. First we assume vol(K) > 2 det Λ. Then vol( 2 K) > det Λ and by 1 1 Corollary 3.2 there exists b ∈ Λ \{0} with b ∈ 2 K − 2 K = K. Now let vol(K) = 2n det Λ. Since K compact and Λ discrete there exists a λ > 1 with λ K ∩ Λ = K ∩ Λ. However vol(λK) > 2n and thus the conclusion follows from the ﬁrst case. The cube [−1, 1]n shows that the statement in Minkowski’s theorem is best possible.

3.4 Corollary [Theorem on linear forms]. Let l1(x), . . . , ln(x) be n homogeneous linear forms given by

li(x) = ai 1x1 + ··· + ai nxn, ai j ∈ R, 1 ≤ i, j ≤ n.

n×n Let A ∈ R be the matrix with entries ai j and let det A 6= 0. For any choice of positive numbers τi, 1 ≤ i ≤ n, satisfying τ1τ2 · ... · τn ≥ | det A| n there exists a z ∈ Z \{0} with

|li(z)| ≤ τi, 1 ≤ i ≤ n.

n Proof. Let P = {x ∈ R : |li(x)| ≤ τi, 1 ≤ i ≤ n}. In order to calculate the volume of that 0-symmetric parallelepiped we observe that −1 n P = A {x ∈ R : |xi| ≤ τi, 1 ≤ i ≤ n}. Thus τ · ... · τ vol(P ) = 2n 1 n ≥ 2n. | det A|

The assertion follows from Theorem 3.3. 3. Minkowski’s successive minima 33

3.5 Corollary [Dirichlet, 1842]. Let α1, . . . , αn ∈ R and let 0 < < 1. −n Then there exist p1, . . . , pn, q ∈ Z with 1 ≤ q ≤ such that

pi αi − < , 1 ≤ i ≤ n. q q

Proof. Let li(x) = αixn+1 − xi, 1 ≤ i ≤ n, and ln+1(x) = xn+1. By Corol- lary 3.4 there exists for every τ > 0 an integral point z = (p1, . . . , pn, q)| ∈ n+1 Z \{0} (depending on τ) satisfying the system of linear forms

−1/n |li(z)| = |αi q − pi| ≤ τ , 1 ≤ i ≤ n, and |ln+1(z)| = |q| ≤ τ.

Note that the product of the right hand sides as well as the absolute value of the determinant of the system is 1. Now we choose a τ > −n such that bτc ≤ −n and we get

−n |li(z)| = |αi q − pi| < , 1 ≤ i ≤ n, and |ln+1(z)| = |q| ≤ bτc ≤ .

Finally, we observe that q 6= 0, because otherwise pi = 0 for i = 1, . . . , n since < 1 and we would get the contradiction z = 0. Thus we may assume q ≥ 1.

3.6 Proposition. Let p be prime. Then there exist a, b ∈ N with

a2 + b2 + 1 ≡ 0 mod p.

Proof. For p = 2 the statement is certainly true. So let p be odd. For 1 2 0 ≤ a ≤ 2 (p − 1) the numbers a belong to pairwise distinct residue classes mod p, because

a2 ≡ a¯2 mod p ⇔ (a − a¯)(a +a ¯) ≡ 0 mod p.

The same is true if we look at the residue classes of −b2 − 1 for 0 ≤ b ≤ 1 2 (p − 1). Since there are precisely p diﬀerent residue classes mod p we can 1 2 2 ﬁnd integers 0 ≤ a, b ≤ 2 (p − 1) such that a and −(b + 1) belong to the same residue class modp which proves the proposition.

3.7 Corollary [Fermat, Lagrange]. Every positive number m ∈ N can be written as the sum of four integer squares, i.e., there exist m1, m2, m3, m4 ∈ N such that 2 2 2 2 m = (m1) + (m2) + (m3) + (m4) . 34 3. Minkowski’s successive minima

Proof. First we observe that it suﬃces to prove the theorem for integers m which are not divisible by a square other than 1. Hence let m = p1 · ... · pk with distinct primes pi, 1 ≤ i ≤ k. For each pi we choose ai, bi, 1 ≤ i ≤ k, according to Proposition 3.6 such that

2 2 (ai) + (bi) + 1 ≡ 0 mod pi, 1 ≤ i ≤ k. (3.7.1)

With this notation we set

4 Λ = z ∈ Z : z1 ≡ (aiz3 + biz4) mod pi, z2 ≡ (biz3 − aiz4) mod pi, 1 ≤ i ≤ k .

4 2 By Corollary 2.15 we know that Λ ⊂ R is a lattice with det Λ ≤ (p1) · ... · 2 2 (pk) = m . Thus we get √ π2 vol( 2m B ) = (2m)2κ = (2m)2 > 24m2 ≥ 24 det Λ. 4 4 2 Hence by Theorem 3.3 there exists a z ∈ Λ with

2 2 2 2 0 < (z1) + (z2) + (z3) + (z4) < 2m.

In order to prove the theorem it suﬃces to show that m is a divisor of the 2 2 2 2 sum (z1) + (z2) + (z3) + (z4) . By the choice of Λ we get for 1 ≤ i ≤ k

2 2 2 2 (z1) + (z2) + (z3) + (z4) 2 2 2 2 ≡ (aiz3 + biz4) + (biz3 − aiz4) + (z3) + (z4) mod pi 2 2 2 2 2 2 ≡ (z3) ((ai) + (bi) + 1) + (z4) ((ai) + (bi) + 1) mod pi

≡ 0 mod pi, where the last relation is a consequence of (3.7.1). Thus all the distinct pi 2 2 2 2 are divisors of (z1) + (z2) + (z3) + (z4) and so is m.

n 3.8 Theorem. Let k ∈ N and let X ⊂ R be a Jordan measurable set with vol(X) > k. Then there exist x1,..., xk+1 ∈ X, xi 6= xj, 1 ≤ i 6= j ≤ k + 1, n n such that xi − xj ∈ Z . (In other words: there exists a t ∈ R such that n t + X contains at least k + 1 lattice points of Z ). Proof. Since we have a Jordan-measurable set we know 1 n 1 k < vol(X) = lim # X ∩ Z . m→∞ m mn

1 n n Thus there exists an m ∈ N such that # X ∩ m Z > k m . Since there n n 1 n are m cosets of the sublattice Z with respect to m Z there exist at least 1 n (k + 1) diﬀerent b1,..., bk+1 ∈ X ∩ m Z belonging to the same coset and n thus bi − bj ∈ Z . 3. Minkowski’s successive minima 35

n n n 3.9 Corollary. Let K ∈ Ko , Λ ∈ L and let vol(K) ≥ k 2 det Λ. Then #(K ∩ Λ) ≥ 2k + 1.

n n Proof. Without loss of generality let Λ = Z and vol(K) > k 2 (cf. the arguments in Corollary ?? and Theorem 3.3). By Theorem 3.8 there exist 1 n (k + 1) diﬀerent points x1,..., xk+1 ∈ 2 K with xi − xj ∈ Z . Let us assume that x1 is one with maximal Euclidean length among these (k + 1) points and let zi = xi+1 − x1, 1 ≤ i ≤ k. Then we have zi 6= zj, i 6= j, and n zi ∈ K ∩ Z \{0}. Furthermore, by the choice of x1 all these points satisfy hx1, zii < 0 which implies that the 2k points ±zi, 1 ≤ i ≤ k, are pairwise distinct. Together with the point 0 this gives the desired lower bound.

n n 3.10 Deﬁnition [Successive minima]. Let K ∈ Ko , Λ ∈ L . For 1 ≤ i ≤ n, λi(K, Λ) = min {λ > 0 : dim (λK ∩ Λ) ≥ i} is called the i-th successive minimum (of K w.r.t. Λ).

3.11 Remark.

i) λi(K, Λ) ≥ λi−1(K, Λ), 2 ≤ i ≤ n.

n×n ii) λi(K, Λ) = λi(AK, AΛ), A ∈ GL(n, R), i.e., A ∈ R with det A 6= 0. 1 1 iii) λi(µ K, Λ) = µ λi(K, Λ) = λi(K, µ Λ), µ ∈ R>0.

iv) int K ∩ Λ \{0} = ∅ ⇔ λ1(K, Λ) ≥ 1. v) λ1(K, Λ) = min |b|K : b ∈ Λ \{0} .

n n 3.12 Proposition. Let K ∈ Ko , Λ ∈ L , and let a1,..., an ∈ Λ be linearly independent such that ai ∈ λi(K, Λ) K, 1 ≤ i ≤ n. Then

int (λi(K, Λ) K) ∩ Λ ⊂ lin {a1,..., ai−1}, 1 ≤ i ≤ n, where we set lin ∅ = {0}.

Proof. For short we write λi = λi(K, Λ), and let k ≤ i be the minimal index with λk = λi. Due to the deﬁnition of the successive minima, int (λi K) = int (λk K) contains at most k −1 linearly independent lattice points and due to teh choice of k, it contains the lattice points ai, 1 ≤ i ≤ k.

3.13 Theorem [Minkowski’s ﬁrst theorem on successive minima]. Let K ∈ n n Ko and Λ ∈ L . Then n n λ1(K, Λ) vol(K) ≤ 2 det Λ. 36 3. Minkowski’s successive minima

Proof. By the deﬁnition of λ1(K, Λ) it is int (λ1(K, Λ) K) ∩ Λ \{0} = ∅ n and so by Theorem 3.3 we get vol (λ1(K, Λ) K) ≤ 2 det Λ. Actually, the last statement is equivalent to Theorem 3.3.

3.14 Theorem [Minkowski’s second theorem on successive minima].Let n n K ∈ Ko and Λ ∈ L . Then 2n det Λ ≤ λ (K, Λ) λ (K, Λ) · ... · λ (K, Λ) vol(K) ≤ 2n det Λ. n! 1 2 n

n Proof. Without loss of generality let Λ = Z . For convenience we write n n λi = λi(K, Z ), and let z1,..., zn ∈ Z be n linearly independent lattice points with n zi ∈ λi K ∩ Z , 1 ≤ i ≤ n. (3.14.1)

λi We start with the proof of the upper bound. Let Ki = 2 K. After a suitable unimodular transformation we may assume (cf. e.g., Theorem 2.11) that zi ∈ Li, where Li denotes the i-dimensional coordinate plane Li = {x ∈ n n n R : xi+1 = ··· = xn = 0}. For an integer q ∈ N let Mq = {z ∈ Z : |zi| ≤ n j n q} = q Cn ∩ Z and for 1 ≤ j ≤ n − 1 let Mq = Mq ∩ Lj. Since K is a bounded set there exists a positive constant γ, only depending on K, such that n n vol(Mq + Kn) ≤ (2q + 2 γ) . (3.14.2)

By the deﬁnition of λ1 we have (z + int (K1)) ∩ (z + int (K1)) = ∅ for two diﬀerent lattice point z, z, because otherwise we would get the contradiction z − z ∈ int (K1) − int (K1) = int (K1 − K1) = int (λ1 K). Thus we have

λ n vol(M n + K ) = (2q + 1)nvol(K ) = (2q + 1)n 1 vol(K). (3.14.3) q 1 1 2

In the following we will show that for 1 ≤ i ≤ n − 1

n−i n λi+1 n vol(Mq + Ki+1) ≥ vol(Mq + Ki). (3.14.4) λi

n To this end we may assume λi+1 > λi and z, z ∈ Z with (zi+1,..., zn) 6= (zi+1,..., zn). Then

z + int (Ki+1) ∩ z + int (Ki+1) = ∅.

Otherwise, in view of Proposition 3.12 we get the contradiction

z − z ∈ int λi+1 K ⊂ lin {z1,..., zi} ⊂ Li. 3. Minkowski’s successive minima 37

n Hence, for z = (0,..., 0, zi+1, . . . , zn)| 6= z = (0,..., 0, zi+1,..., zn)| ∈ Z i i we have (z + Mq + int (Ki+1)) ∩ (z + Mq + int (Ki+1)) = ∅. Thus n n−i i vol Mq + Ki+1 = (2q + 1) vol Mq + Ki+1 , n n−i i vol Mq + Ki = (2q + 1) vol Mq + Ki , and in order to verify (3.14.4) it suﬃces to show n−i i λi+1 i vol Mq + Ki+1 ≥ vol(Mq + Ki). (3.14.5) λi n n Now let f1, f2 : R → R be the linear maps given by | λi+1 λi+1 f1(x) = x1,..., xi, xi+1,..., xn , λi λi | λi+1 λi+1 f2(x) = x1,..., xi, xi+1,..., xn . λi λi i i Since Mq + Ki+1 = f2(Mq + f1(Ki)) we get n−i i λi+1 i vol Mq + Ki+1 = vol(Mq + f1(Ki)) λi and for the proof of (3.14.5) we have to show i i vol Mq + f1(Ki) ≥ vol Mq + Ki . (3.14.6) ⊥ To this end let Li be the (n − i)-dimensional orthogonal complement of Li. ⊥ Then it is easy to see that for every x ∈ Li there exists a t(x) ∈ Li with Ki ∩ (x + Li) ⊂ (f1(Ki) ∩ (x + Li)) + t(x) and so i i Mq + Ki ∩ (x + Li) ⊂ Mq + f1(Ki) ∩ (x + Li) + t(x). Thus we get Z i i vol(Mq + Ki) = voli (Mq + Ki) ∩ (x + Li) d x ⊥ x∈Li Z i ≤ voli (Mq + f1(Ki)) ∩ (x + Li) d x ⊥ x∈Li i = vol(Mq + f1(Ki)), where voli(·) denotes the i-dimensional volume. This shows (3.14.6) and so we have veriﬁed (3.14.4). Finally, it follows from (3.14.2), (3.14.3) and (3.14.4) n n n λn n (2q + 2 γ) ≥ vol Mq + K ≥ vol Mq + Kn−1 ≥ ... λn−1 2 n−1 λn λn−1 λ2 n ≥ · ... · vol Mq + K1 λn−1 λn−2 λ1 vol(K) = λ · ... · λ (2q + 1)n n 1 2n 38 3. Minkowski’s successive minima and so 2q + 2 γ n λ · ... · λ vol(K) ≤ 2n . n 1 2q + 1 Since this holds for all q ∈ N the upper bound is proven. The lower bound follows just by inclusion. To this end we observe that (3.14.1) implies that the cross-polytope conv {± 1 z : 1 ≤ i ≤ n} ⊆ K. λi i Thus n 1 2 1 1 vol(K) ≥ volconv ± zi : 1 ≤ i ≤ n = det z1,..., zn λi n! λ1 λn 2n 1 ≥ . n! λ1 · ... · λn

Obviously, both bounds of Minkowski’s second theorem are best possible. ? For instance, the lower bound is attained by the regular cross-polytope Cn and the upper bound by any o-symmetric box with edges parallel to the coordinate edges.

n n 3.15 Proposition. Let K ∈ Ko and Λ ∈ L . Then 2 n #(K ∩ Λ) ≤ + 1 . λ1(K, Λ)

n n In particular: Let K ∈ Ko with int K ∩ Λ = {0}. Then #(K ∩ Λ) ≤ 3 . n Proof. Without loss of generality let Λ = Z and we write λ1 instead of n n λ1(K, Z ). Let k = b2/λ1 + 1c. Suppose there exist z, z¯ ∈ K ∩ Z such that z ≡ z¯ mod k. Then 1 2 1 1 2 n 3 (z − z¯) = z − z¯ ∈ K ⊂ int (λ K) , Z k k 2 2 k 1 since 2/k < λ1. By definition of λ1 we conclude z = z¯. This shows that for 1 n two different lattice points z, z¯ ∈ K belongs to different cosets mod k Z . n Thus the number of lattice points in K can not exceed k .

n n 3.16 Conjecture. Let K ∈ Ko and Λ ∈ L . Then

n Y 2 #(K ∩ Λ) ≤ + 1 . λ (K, Λ) i=1 i

3.17 Remark. The upper bounds on the lattice points in Proposition 3.15 and in Conjecture 3.16 imply the corresponding bounds on the volume as 3. Minkowski’s successive minima 39

n stated in the Theorems 3.13, 3.14. For instance, for Λ = Z and with n λi(K) = λi(K, Z ) we (would) get by Conjecture 3.16 n 1 1 n vol(K) = lim # K ∩ Z m→∞ m m n n 1 Y 2 ≤ lim + 1 m→∞ m λ (m K) i=1 i n Y 2 1 = lim + m→∞ λ (K) m i=1 i n Y 1 = . λ (K) i=1 i

The second theorem of Minkowski can easily be extended to arbitrary (not necessarily o-symmetric) convex body, if the successive minima are mea- 1 sured with respect to central symmetral cs (K) = 2 (K − K) of K (cf. (1.2)).

3.18 Theorem. Let K ∈ Kn, Λ ∈ Ln. Then

2n det Λ ≤ λ (cs (K), Λ) λ (cs (K), Λ) · ... · λ (cs (K), Λ) vol(K) ≤ 2n det Λ. n! 1 2 n

n Proof. Without loss of generality Λ = Z and for short, we write λi = n λi(cs (K), Z ), 1 ≤ i ≤ n. We begin with the upper bound. Due to Theorem 3.14 we have n λ1 · ... · λnvol(cs (K)) ≤ 2 . By the Brunn-Minkowski inequality we also know (cf. e.g. (1.3))

vol(cs (K)) ≥ vol(K), which shows the upper bound. For the lower bound let zi ∈ λics (K), 1 ≤ i ≤ n, be linearly independent and let u , w ∈ 1 K such hat v = 1 z = u − w , 1 ≤ i ≤ n. Let i i 2 i λi i i i 1 C = conv {ui, wi : 1 ≤ i ≤ n} ⊆ 2 K. In the following we show 1 vol(C) ≥ | det(v ,..., v )| n! 1 n which implies the assertion. For the proof we use induction on n. For n = 1 n nothing is to show and so let n > 1. Let L = {x ∈ R : hv1, xi = 0} and for n x ∈ R we denote by x¯ its orthogonal projection onto L. Application of the Steiner symmetrization to C with respect to the hyperplane H gives a convex body CS with vol(CS) = vol(C), and by deﬁnition of this symmetrization 40 3. Minkowski’s successive minima

1 1 CS contains the points u¯1 + 2 v1, u¯1 − 2 v1, u¯2, w¯ 2,..., u¯n, w¯ n. Hence we have |v | vol(C) = vol(C ) ≥ 1 · vol (conv {u¯ , w¯ ,..., u¯ , w¯ }). S n n−1 2 2 n n Hence, via our inductive approach we get

|v1| 1 vol(C) = vol(CS) ≥ · detf |(u¯2 − w¯ 2, ··· , u¯n − w¯ n)| n (n − 1)!

|v1| 1 = detf |(v¯2, ··· , v¯n)| = | det(v1,..., vn)|. n! n!

Here det()f denotes the determinant in the (n − 1)-dimensional setting, i.e., the (n − 1)-dimensional volume of the parallelepiped generated by the n − 1 vectors. Bibliography

[1] Keith Ball. An elementary introduction to modern convex geometry. Cambridge University Press. Math. Sci. Res. Inst. Publ., 31:1 – 58, 1997.

[2] Alexander Barvinok. A course in convexity. Graduate Studies in Math- ematics, 54, 2002.

[3] T Figiel, J Lindenstrauss, and V.D Milman. The dimension of almost spherical sections of convex bodies. Acta Math, 139:53–94, 1977.

[4] R. J Gardner. The brunn-minkowski inequality. Bull. Amer. Math. Soc. (N.S.), 39(3):355–405 (electronic), 2002.

[5] Richard J Gardner. Geometric Tomography. 1995.

[6] Peter Manfred Gruber. Convex and Discrete Geometry, volume 336. 2007.

[7] Branko Gr¨unbaum. Convex polytopes. 2003.

[8] G Kalai. The number of faces of centrally-symmetric polytopes. Graphs and Combinatorics, pages 389–391, Sep 1989.

[9] Alexander Koldobsky. Fourier Analysis in Convex Geometry, volume 116. 2005.

[10] Jesus De Loera, J¨orgRambau, and Francisco Santos. Triangulations: Structures for algorithms and applications. Book, pages 1–545, Apr 2010.

[11] Jiri Matousek. Lectures on Discrete Geometry, volume 212.

[12] R Sanyal, A Werner, and G.M Ziegler. On kalai’s conjecture concerning centrally symmetric polytopes. Discrete & Computational Geometry, 41:183—198, Sep 2009. doi: 10.1007/s0054-008-9104-8.

[13] R Schneider. Convex bodies: The Brunn-Minkowski theory, volume 44. 1993.

41 42 3. Bibliography

[14] G M Ziegler. Lectures on polytopes, volume 152. 1995. Revised sixth printing 2006.