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1 Basic Notions and Tools from Lattice Theory RFG on Applications of Convex Geometry Lecturer: Amitabh Basu Lectures on Modern Approaches to Cutting Planes Date: Fall 2010 1 Basic notions and tools from lattice theory This lecture surveys the basic definitions and important tools from lattice theory that we will need in this course. Throughout this class, we will be concerned with the n-dimensional Euclidean space Rn. For any real number r 2 R, we denote the greatest integer smaller than or equal to r by brc, the smallest integer greater than or equal to r by dre and frg will be used to denote r − brc. Definition 1 Given a set S ⊆ Rn, span(S) is the linear space spanned by the vectors in S. span(S) is equivalent to the set of all finite linear combinations of vectors from S. Definition 2 A subset Λ ⊆ Rn is a subgroup under addition if the following three properties hold : i 0 2 Λ. ii For any x; y 2 Λ, x + y 2 Λ. iii For any x 2 Λ then −x 2 Λ. A subgroup Λ is called a discrete subgroup if there exists ϵ > 0, such that B(0; ϵ) \ Λ = f0g. Definition 3 Given a linear subspace L of Rn, a discrete subgroup Λ is called a lattice of L, if span(Λ) = L. A basis for a lattice Λ in L is a set of linearly independent vectors a1; : : : ; ak 2 L such that Λ = fµ1a1 + ::: + µkak j µ1; : : : µk 2 Zg. Λ is then said to be generated by a1; : : : ; ak. Note that it is not a priori obvious that every lattice Λ of a linear subspace L will have a basis. However, note that if Λ does have a basis A = fa1; : : : ; akg then span(A) = L. We give some examples of lattices. Example 4 1. The set Zn, i.e. all points in Rn with integral coordinates, is a lattice of Rn. n n 2. Consider the hyperplane H = f(x1; : : : ; xn) j x1 + ::: + xn = 0g in R . Then H \ Z is a lattice of H. Try to draw this lattice for n = 3. n 3. Consider any set of linearly independent vectors fb1; : : : ; bkg in R . Let L = span(b1; : : : ; bk) be the linear subspace spanned by these vectors. Then the set Λ = fµ1b1 + ::: + µkbk j µ1; : : : µk 2 Zg is a lattice of L. Prove this ! The following fact will be useful, we leave the proof for the reader. Proposition 5 Let Λ be a discrete subgroup. Consider any bounded set A. Then jA \ Λj is finite, i.e. there are finitely many points from a discrete subgroup in any bounded set. n n n We define dist(x; y) = kx − yk2 for any two points x; y 2 R . Given a set A ⊆ R and a point x 2 R define dist(x; A) = inffdist(x; y) j y 2 Ag. The following result from finite dimensional analysis will be used (it follows from Weierstrauss’ theorem on minimizing continuous functions over compact sets). 1 Proposition 6 If A ⊆ Rn is a compact set (which is equivalent to saying it is closed and bounded), and x 62 A is a point in Rn, dist(x; A) > 0 and there exists a y 2 A such that dist(x; A) = dist(x; y). The following lemma is an important property of lattices. It says that if we consider a lattice Λ of a subspace W and a subspace L ( W , such that L is spanned by lattice vectors from Λ, then there is a non-zero distance ϵ > 0 between L and any lattice point y 2 Λ n L. Lemma 7 Let Λ be a lattice of a linear subspace W . Let b1; : : : ; bk 2 Λ, k < dim(W ) be linearly independent vectors in the lattice Λ and let L = span(b1; : : : ; bk). Then there exists v 2 Λ n L such that dist(v; L) ≤ dist(w; L) for all w 2 Λ n L. Note that the crucial property of the subspace is that it is spanned by vectorsp from the lattice. For example, consider the lattice Z2 of R2 and the subspace L spanned by the ray (1; 2). The only lattice point in L is f0g. Then for any v 2 Z2 n f0g, there exists a w 2 Z2 n f0g such that dist(w; L) < dist(v; L). (Check this !) Proof:[Proof of Lemma 7] P f k j ≤ ≤ g 2 n Consider the set Π = i=1 µibi 0 µi 1 for all i . Consider any ρ Λ Π (why does such a ρ exist ?) and by Proposition 6, d = dist(ρ, Π) > 0 since Π is closed and bounded. n Consider the bounded set Πd = fp 2 R j dist(p; Π) ≤ dg. Note that Π ⊆ Πd and ρ 2 Πd. Moreover, Fact 5 shows that there are finitely many points from Λ in Πd. In particular, there are finitely many points in Λ n L in Πd (ρ is one such point). Therefore, there exists v 2 Λ n L such that dist(v; Π) ≤ dist(w; Π) for all w 2 Λ n L (why ?): (1) We now claim that this implies dist(v; L) ≤ dist(w; L) for all w 2 Λ n L Consider any w P2 ΛnL and considerP any y 2 L. Since y 2 L, one can express y as a linear combination k k b c 2 of b1; : : : ; bk : y = i=1 λibi. Let z = i=1 λi bi. Notice that since bi Λ, an integral combination of the bi’s are also in Λ by the subgroup property (ii) and hence z 2 Λ. Then, y − z 2 Π and w − z 2 Λ n L (why ?). We then have the following relations, dist(w; y) = dist(w − z; y − z) ≥ dist(w − z; Π) ≥ dist(v; Π) ≥ dist(v; L) The first equality is simply the invariance of distance under translation. The second inequality follows because y − z 2 Π. The third inequality follows from (1). The fourth inequality follows simply because Π ⊆ L. This shows that dist(w; y) ≥ dist(v; L) for any w 2 Λ n L and any y 2 L. Taking an infimum over all y 2 L, we get that dist(w; L) ≥ dist(v; L) for any w 2 Λ n L. 2 We now prove the basis theorem for lattices. Theorem 8 Every lattice Λ of a linear subspace L has a basis. Proof: We will iteratively build up a basis for Λ. At iteration k ≥ 0, we will have a set Bk = fb1; : : : bkg of linearly independent vectors in L and we will work with the subspace Lk ⊆ L spanned by Bk. We set B0 = fg and so L0 = f0g. The property that we will satisfy at every iteration step k ≥ 0 is that Lk \ Λ = fµ1b1 + ::: + µkbk j µ1; : : : µk 2 Zg: (2) 2 Hence, if we can show that we reach iteration k = dim(L) then we will be done. Suppose we are at iteration k < dim(L). So we have a set Bk = fb1; : : : bkg with the above property (2) satisfied. We now show how to find bk+1. Consider w¯ 2 Λ n Lk (why does such a w¯ exist ?). Let 0 0 0 0 0 L = span(b1; : : : ; bk; w¯) and consider the lattice Λ = L \ Λ of L (why is this a lattice of L ?). By 0 0 Lemma 7, there exists v 2 Λ n Lk such that dist(v; Lk) ≤ dist(w; Lk) for all w 2 Λ n Lk. Note that 0 0 since v 2 Λ n Lk, span(b1; : : : ; bk; w¯) = span(b1; : : : ; bk; v). So L = span(b1; : : : ; bk; v) and we will set 0 0 0 bk+1 := v and Lk+1 = L . Consequently, we have that Λ = L \ Λ = Lk+1 \ Λ. We then need to show the following claim. 0 Claim 1 Lk+1 \ Λ = Λ = fµ1b1 + ::: + µkbk + µk+1v j µ1; : : : µk+1 2 Zg: 2 0 \ f g P Consider any p Λ = Lk+1 Λ. Since Lk+1 is the span of b1; : : : ; bk; v , we can write p = k − b c 0 0 i=1 λibi + λv. Then p¯ = p λ v belongs to Λ (sinceP p and v are both inPΛ , the subgroup property (ii) − b c 0 k f g k implies that p λ v is also in Λ ). Note that p¯ = i=1 λibi + λ v and i=1 λibi is in Lk. Therefore, dist(¯p; Lk) = dist(fλgv; Lk) = fλgdist(v; Lk) < dist(v; Lk), since fλg < 1. This implies that p¯ is in ≤ 2 0 n f g Lk, because dist(v; Lk)P dist(w; Lk) for all w Λ Lk and therefore λ = 0. Therefore, λ is an k 2 0 ⊆ 2 \ integer. Moreover, p¯ = i=1 λibi and p¯ Λ Λ. So in fact, p¯ Λ Lk. But then property (2) holds for iteration k and therefore λi are all integer (note that this uses the standard fact from linear algebra that there is a unique representation of any vector using linearly independent vectors). We have thus shown that p can be expressed as an integral combination of fb1; : : : ; bk; vg and the claim is proved (actually this proves only one inclusion, the other inclusion is trivial to prove - check this !). 2 Combined with Example 3 above, this theorem says that the concept of a lattice and the concept of taking integral combinations of linearly independent vectors are equivalent.
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