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Chapter 2 Geometry of Numbers Chapter 2 Geometry of numbers Literature: W.M. Schmidt, Diophantine approximation, Lecture Notes in Mathematics 785, Springer Verlag 1980, Chap.II, xx1,2, Chap. IV, x1 J.W.S. Cassels, An Introduction to the Geometry of Numbers, Springer Verlag 1997, Classics in Mathematics series, reprint of the 1971 edition C.L. Siegel, Lectures on the Geometry of Numbers, Springer Verlag 1989 2.1 Introduction Geometry of numbers is concerned with the study of lattice points in certain bodies n in R , where n > 2. We discuss Minkowski's theorems on lattice points in central symmetric convex bodies. In this introduction we give the necessary definitions. Lattices. A (full) lattice in Rn is an additive group L = fz1v1 + ··· + znvn : z1; : : : ; zn 2 Zg n where fv1;:::; vng is a basis of R , i.e., fv1;:::; vng is linearly independent. We call fv1;:::; vng a basis of L. The determinant of L is defined by d(L) := j det(v1;:::; vn)j; that is, the absolute value of the determinant of the matrix with columns v1;:::; vn. 7 We show that the determinant of a lattice does not depend on the choice of the basis. Recall that GL(n; Z) is the multiplicative group of n×n-matrices with entries in Z and determinant ±1. Lemma 2.1. Let L be a lattice, and fv1;:::; vng, fw1;:::; wng two bases of L. Then there is a matrix U = (uij) 2 GL(n; Z) such that n X (2.1) wi = uijvj for i = 1; : : : ; n: j=1 Consequently, j det(v1;:::; vn)j = j det(w1;:::; wn)j. Proof. Let U be the matrix relating v1;:::; vn to w1;:::; wn, that is, the matrix given by (2.1). A priori, U is just a non-singular matrix, but since w1;:::; wn lie in the lattice generated by v1;:::; vn, it must have its entries in Z. −1 ij Pn ij Let U = (u ). Then from linear algebra we know that vi = j=1 u wj for −1 i = 1; : : : ; n. Now U has its entries in Z since v1;:::; vn lie in the lattice generated −1 by w1;:::; wn. Since both det U and det U are integers and must be multiplicative inverses of one another, we have det U = ±1, i.e., U 2 GL(n; Z). Finally, we observe that j det(w1;:::; wn)j = j det Uj · j det(v1;:::; vn)j = j det(v1;:::; vn)j: n Let L; M be two lattices in R with M ⊆ L. Choose bases fv1;:::; vng of L, fw1;:::; wng of M. Let U = (uij) be the matrix relating v1;:::; vn to w1;:::; wn. Then U has its entries in Z. We define the index of M in L by [L : M] := j det Uj: The relation det(w1;:::; wn) = det U · det(v1;:::; vn) easily translates into d(M) = [L : M] · d(L) and this shows that [L : M] does not depend on the choices of the bases of L and M. Convex bodies. Recall that a subset C of Rn is convex if for any two points 8 x; y 2 C, also the line segment connecting them, i.e., fλx + (1 − λ)y : 0 6 λ 6 1g, is contained in C.A central symmetric convex body in Rn is a closed, bounded, convex subset C of Rn having 0 as an interior point, and which is symmetric about 0, i.e. if x 2 C then also −x 2 C. Examples. (i). If C is a central symmetric convex body, then so is its dilation with factor λ > 0, λC := fλx : x 2 Cg: (ii). If C is a central symmetric convex body and φ a linear transformation of Rn, then φ(C) is also a central symmetric convex body. (iii). Parallelepipeds and ellipsoids: Let n n 2 2 Kn = fx 2 R : max jxij 6 1g;Bn = fx 2 R : x1 + ··· + xn 6 1g 16i6n be the n-dimensional unit cube, and the n-dimensional Euclidean unit ball, respec- n n tively, where x = (x1; : : : ; xn) 2 R . A (symmetric) parallelepiped in R is the n n image of Kn under a linear transformation of R , and a (symmetric) ellipsoid in R n the image of Bn under a linear transformation of R . Both are central symmetric convex bodies. n n Exercise 2.1. Recall that a norm on R is a function k · k : R ! R>0 such that • kλxk = jλj · kxk for all x 2 Rn; λ 2 R; n • kx + yk 6 kxk + kyk for all x; y 2 R ; • kxk = 0 () x = 0. n Prove that the unit ball Bk·k = fx 2 R : kxk 6 1g is a central symmetric convex body. Hint. You may use that all norms on Rn induce the usual topology, that is, a subset n U of R is open if and only for every x0 2 U there is δ > 0 such that n fx 2 R : kx − x0k < δg ⊆ U. In fact, every central symmetric convex body arises from a norm. On Rn we define the function kxkC := minfλ 2 R>0 : x 2 λCg: 9 Lemma 2.2. (i) k · kC is well defined. n (ii) k · kC defines a norm on R . n (iii) λC = fx 2 R : kxkC 6 λg for λ > 0. Proof. (i). Clearly, k0kC = 0. Suppose x 6= 0. Consider the set S := fξx : ξ 2 R>0; ξx 2 Cg: Then S 6= f0g. For since 0 is an interior point of C there is δ > 0 with δx 2 C. Further, since C is compact and S is a closed subset of C, the set S is compact. S is homeomorphic to I = fξ 2 R>0 : ξx 2 Cg. Hence I is compact. So I has a maximum, call it µ0. Then µ0 > 0 since S 6= f0g. −1 −1 We claim that kxkC = µ0 . First µ0x 2 C, hence x 2 µ0 C. Further, if x 2 λC, −1 −1 −1 then λ x 2 C hence λ 6 µ0. So λ > µ0 . This proves our claim, hence (i). −1 (ii). We have shown above that kxkC = µ0 > 0 if x 6= 0. The proofs of the other two norm properties are left to the reader. (iii). Left to the reader. Exercise 2.2. Prove (ii) and (iii). 2.2 Minkowski's first convex body theorem Using Lebesgue theory, one can define an n-dimensional volume vol(S) 2 R>0 [f1g (the so-called Lebesgue measure) for subsets S of Rn from a large class, the so-called measurable subsets of Rn. We do not need the precise definition of Lebesgue mea- sure or measurable set. What is important to us is that all open sets and all closed sets are measurable, bounded measurable sets have finite volume, and the empty R set has volume 0. The volume of S is equal to the Riemann integral S dx1 ··· dxn for every set S for which this integral is defined. However, there are measurable sets S for which the Riemann integral is not defined. We mention some important properties of the volume: 1. Let S be a measurable subset of Rn. Then every translate a+S := fa+x : x 2 Sg is also measurable and vol(a + S) = vol(S). Further, if φ is a linear transformation of Rn, then φ(S) is measurable and vol(φ(S)) = j det φj · vol(S). 10 2. Let S ⊂ Rn be measurable. Then Sc = Rn n S is measurable. n 3. Let Sn (n = 1; 2; 3;:::) be a countable collection of measurable subsets of R . S1 Then S = n=1 Sn is measurable. Moreover, if the sets Sn are pairwise disjoint, P1 then vol(S) = n=1 vol(Sn). Theorem 2.3. (Minkowski's first convex body theorem, 1896). Let C be a central symmetric convex body in Rn and L a lattice in Rn of rank n. Suppose that n vol(C) > 2 d(L). Then C contains a point from L n f0g. Choose a basis fv1;:::; vng of L. We call F := fx1v1 + ··· + xnvn : xi 2 R; 0 6 xi < 1 for i = 1; : : : ; ng a fundamental parallelepiped for L. Notice that F has volume d(L), the translates u + F (u 2 L) are pairwise disjoint, and n [ R = (u + F ): u2L n 1 1 Proof. We first assume that vol(C) > 2 d(L). Then the set 2 C = f 2 x : x 2 Cg has 1 volume > d(L). For u 2 L, define Su := 2 C \ (u + F ). Then the sets Su (u 2 L) 1 are pairwise disjoint and their union is precisely 2 C. Hence X 1 vol(Su) = vol( 2 C) > d(L): u2L We shift the sets Su into F , that is, we define ∗ 1 Su := −u + Su = (−u + 2 C) \ F for u 2 L: ∗ Since Su has the same volume as Su, we have X ∗ X vol(Su) = vol(Su) > d(L) = vol(F ): u2L u2L ∗ That is, we have a collection of subsets Su (u 2 L) of F , the sum of whose volumes is ∗ ∗ larger than the volume of F . So there are two distinct u; v 2 L such that Su\Sv 6= ;. ∗ ∗ 1 Pick a point a 2 Su \Sv. Then for certain x; y 2 2 C we have x−u = y −v = a. Hence x − y = u − v 2 L n f0g. 11 Now 2x; 2y 2 C, by the symmetry of C we have −2y 2 C, and by the convexity 1 of C we have 2 (2x − 2y) = x − y 2 C.
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