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Dual Lattices Scribe: Gillat Kol

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Dual Lattices Scribe: Gillat Kol Tel Aviv University, Fall 2004 Lecture 8 Lecturer: Oded Regev Lattices in Computer Science Dual Lattices Scribe: Gillat Kol In this lecture we define the notion of the dual of a lattice and see some if its applications. DEFINITION 1 For a full-rank lattice ¤ we define its dual lattice (sometimes known as the reciprocal lattice) ¤¤ = fy 2 Rn j 8x 2 ¤; hx; yi 2 Zg: In general, we define ¤¤ = fy 2 span(¤) j 8x 2 ¤; hx; yi 2 Zg: In words, the dual of ¤ is the set of all points (in the span of ¤) whose inner product with any of the points in ¤ is integer. As we will show later, ¤¤ is indeed a lattice, as the name suggests. EXAMPLE 1 The lattice of integer points satisfies (Zn)¤ = Zn (and hence can be called self-dual). Simi- n ¤ 1 n larly, (2Z ) = 2 Z , and this gives some justification to the name reciprocal lattice. From the above definition, we have the following geometrical interpretation of the dual lattice. For any vector x, the set of all points whose inner product with x is integer forms a set of hyperplanes perpendicular to x and separated by distance 1=kxk. Hence, any vector x in a lattice ¤ imposes the constraint that all points in ¤¤ lie in one of the hyperplanes defined by x. See the next figure for an illustration. 1 kxk 1 kyk y x Λ Λ∗ Figure 1: A lattice and its dual m£n m£n DEFINITION 2 For a basis B = (b1; : : : ; bn) 2 R , define the dual basis D = (d1; : : : ; dn) 2 R as the unique basis that satisfies ² span(D) = span(B) ² BT D = I The second condition can be interpreted as saying that hbi; dji = ±ij where ±ij = 1 if i = j and 0 otherwise. It is not hard to check that D is indeed unique. In fact, for the case of a full-rank lattice, D is given by (BT )¡1; in general, we get D = B(BT B)¡1 (and we could use this as our definition of a dual basis). In the next claim, we show that if B is a basis of a lattice ¤, then the dual basis of B is a basis of ¤¤. In particular, this shows that ¤¤ is indeed a lattice. CLAIM 1 If D is the dual basis of B then (L(B))¤ = L(D). 1 ¤ P PROOF: We first show that L(D) is contained in (L(B)) . Any x 2 L(B) can be written as aibi for some ai 2 Z. Therefore, for any j we have X hx; dji = aihbi; dji = ai 2 Z i and we get D ⊆ (L(B))¤. It is easy to check that (L(B))¤ is closed under addition, hence, L(D) ⊆ ¤ ¤ ¤ (L(B)) . To complete the proof, we show that (L(PB)) is contained in L(D). Take any y 2 (L(B)) . PSince y 2 span(B) = span(D), we can write y = aidi for some ai 2 R. Now for all j, Z 3 hy; bji = aihdi; bji = aj. Hence, y 2 L(D) and the proof is complete. 2 CLAIM 2 For any lattice ¤, (¤¤)¤ = ¤. PROOF: Let B be a basis of ¤. Then B(BT B)¡1 is a basis of ¤¤ and ¡ ¢ ¡ ¢ B(BT B)¡1 ¢ (B(BT B)¡1)T ¢ B(BT B)¡1 ¡1 = B is a basis of (¤¤)¤. 2 The next claim says that the volume of the basic parallelepiped of ¤¤ is the reciprocal of that of ¤. For example, it implies that the volume of the basic parallelepiped of a self-dual lattice must be 1 (as is the case with Zn). CLAIM 3 For any lattice ¤, det(¤¤) = 1= det(¤). PROOF: For full-rank lattices, ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 ¯ ¯ 1 ¯ 1 det(¤¤) = ¯det((BT )¡1)¯ = ¯ ¯ = ¯ ¯ = : ¯det(BT )¯ ¯det(B)¯ det(¤) In general, q det(¤¤) = det(DT D) q = det(((BT B)¡1)T BT ¢ B(BT B)¡1) q = (det(BT B)¡1) 1 1 = p = : det(BT B) det(¤) 2 The following two claims give some relations between properties of a lattice and that of its dual. Such properties are known as transference theorem. In a few lectures, we will see a considertable strengthening ¤ of Claim 4 by Banaszczyk (showing that ¸1(¤) ¢ ¸n(¤ ) · n). ¤ CLAIM 4 For any rank n lattice ¤, ¸1(¤) ¢ ¸1(¤ ) · n. PROOF: By Minkowski’s bound, p 1 ¸1(¤) · n ¢ (det(¤)) n and p p 1 n ¤ ¤ n ¸1(¤ ) · n ¢ (det(¤ )) = 1 : (det(¤)) n 2 2 ¤ CLAIM 5 For any rank n lattice ¤, ¸1(¤) ¢ ¸n(¤ ) ¸ 1. PROOF: Let v 2 ¤ be such that kvk = ¸1(¤). Take any set x1; : : : ; xn of n linearly independent vectors in ¤ ¤ . Not all of them are orthogonal to v. Hence, there exists an i such that hxi; vi 6= 0. By the definition of 1 the dual lattice, we have hxi; vi 2 Z and hence kxik ¸ kvk . 2 ? For a basis b1; : : : ; bn, let ¼i denote the projection on the space span(b1; : : : ; bi¡1) . In particular, ¼1(b1); : : : ; ¼n(bn) is the Gram-Schmidt orthogonalization of b1; : : : ; bn. 0 0 CLAIM 6 Let B; D be dual bases. Then, for all i, B = (¼i(bi); : : : ; ¼i(bn)) and D = (di; : : : ; dn) are also dual bases. 0 ? PROOF: First, notice that span(B ) = span(b1; : : : ; bi¡1) . Moreover, since di; : : : ; dn are orthogonal to 0 ? 0 0 b1; : : : ; bi¡1 and linearly independent, span(D ) = span(b1; : : : ; bi¡1) . Hence, span(B ) = span(D ). Finally, we have that for any j; k ¸ i, hdj; ¼i(bk)i = hdj; bki = ±jk ? where the first equality holds since dj 2 span(b1; : : : ; bi¡1) . 2 CLAIM 7 Let b1; : : : ; bn be some basis and let b~1;:::; b~n be its Gram-Schmidt orthogonalization. Let dn; : : : ; d1 be the dual basis of b1; : : : ; bn in reverse order and let d~n;:::; d~1 be its Gram-Schmidt orthogo- nalization (using this order). Then for all i, b~i d~i = : 2 kb~ik PROOF: The proof is by induction on n. Assume the claim holds for lattices of rank n ¡ 1 and let us prove ? it for lattices of rank n. First, notice that b~1 = b1 and d~1 is the projection of d1 on span(d2; : : : ; dn) = span(b1). Hence, d~1 2 span(b1) and hd~1; b1i = hd1; b1i = 1. This implies that b1 b~1 d~1 = = : 2 2 kb1k kb~1k We can now complete the proof by applying the inductive hypothesis to the bases (¼2(b2); : : : ; ¼2(bn)) and d2; : : : ; dn. Indeed, Claim 6 says that these are dual bases, and moreover, the Gram-Schmidt orthogonaliza- tion of the former is b~2;:::; b~n. 2 1 Korkine-Zolotarev bases In this section we define the notion of a Korkine-Zolotarev (KZ) basis. This gives one way to formalize the idea of a ‘shortest possible’ basis. DEFINITION 3 For a rank n lattice ¤, we define its Korkine-Zolotarev (KZ) basis b1; : : : ; bn recursively as 0 follows. We let b1 be the shortest vector in ¤. We then let ¤ be the lattice given by the projection of ¤ on 0 the subspace of span(¤) orthogonal to b1. Let c2; : : : ; cn be the KZ basis of ¤ . Define bi = ci + ®ib1 where 1 1 ®i 2 (¡ 2 ; 2 ] is the unique number such that bi 2 ¤. 3 b2 c2 b1 b1 Λ Λ′ KZ basis Figure 2: A lattice and its KZ basis 0 1 It is not too difficult to verify that ¤ is indeed a lattice. Moreover, b1 is a primitive vector in ¤ (since it is a shortest vector) and hence the vectors b1; : : : ; bn defined above indeed form a basis of ¤. The definition is illustrated in Figure 2. As a first application of Korkine-Zolotarev bases, we prove the following lemma by Lagarias, Lenstra, and Schnorr. Recall that for any basis b1; : : : ; bn, we have that min(kb~1k;:::; kb~nk) · ¸1(¤). The lemma says that any lattice has a basis where this lower bound is not far from being tight. LEMMA 8 ([2]) For any lattice ¤, there exists a basis b1; : : : ; bn such that 1 min(kb~ k;:::; kb~ k) ¸ ¢ ¸ (¤) 1 n n 1 ¤ PROOF: Let d1; : : : ; dn be a KZ basis of ¤ and let bn; : : : ; b1 be its dual basis in reverse order. We claim ~ ~ di that bn; : : : ; b1 satisfies the lemma. By Claim 7, we know that bi = 2 . Hence, its enough to show that kd~ik max(kd~ k;:::; kd~ k) · n . First, d~ is the shortest vector in ¤¤. By Claim 4, we have kd~ k · n . 1 n ¸1(¤) 1 1 ¸1(¤) Next, d~2 = ¼2(d2) is the shortest vector in L(¼2(d2); : : : ; ¼2(dn)). By Claim 6, the dual of this lattice is L(b2; : : : ; bn). But ¸1(L(b2; : : : ; bn)) ¸ ¸1(¤) (since L(b2; : : : ; bn) is a sublattice of ¤) and hence n ¡ 1 n kd~2k · · : ¸1(L(b1; : : : ; bn)) ¸1(¤) We continue similarly for all i. 2 COROLLARY 9 GapSVPn 2 coNP PROOF: Recall that an instance of GapSVPn is given by (B; d); it is a YES instance if ¸1(L(B)) · d and a NO instance if ¸1(L(B)) > n¢d. The verifier expects a witness of the form v1; : : : ; vn. It accepts if and only if v1; : : : ; vn form a basis of L(B) (recall that this can be verified efficiently) and min(kv~1k;:::; kv~nk) > d. If ¸1(L(B)) > n ¢ d then such a basis exists by Lemma 8. If ¸1(L(B)) · d then no such basis exists since min(kv~1k;:::; kv~nk) · ¸1(L(B)) · d for any basis.
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