The Optimal Isodual Lattice Quantizer in Three Dimensions J. H. Conway Mathematics Department Princeton University Princeton, NJ 08544 7 N. J. A. Sloane 0 0 AT&T Shannon Labs 2 180 Park Avenue n Florham Park, NJ 07932-0971 a J Email: [email protected], [email protected] 3 ] Jan 02 2006. T N Abstract . h t a The mean-centered cuboidal (or m.c.c.) lattice is known to be the optimal packing and m covering among all isodual three-dimensional lattices. In this note we show that it is also the best [ quantizer. It thus joins the isodual lattices Z, A and (presumably) D , E and the Leech lattice 2 4 8 1 in being simultaneously optimal with respect to all three criteria. v 0 8 Keywords: quantizing, self-dual lattice, isodual lattice, f.c.c. lattice, b.c.c. lattice, m.c.c. 0 lattice 1 0 7 AMS 2000 Classification: 52C07 (11H55, 94A29) 0 / h 1. Introduction t a m : An isodual lattice [10] is one that is geometrically similar to its dual. Let Λ be an n-dimensional v i lattice andlet V denotethe Voronoi cell containing the origin. We may assumeΛ is scaled sothat X r V has unit volume. Three important parameters of Λ are the packing radius (the in-radius of V), a the covering radius (the circum-radius of V) and its quantization error, which is the normalized second moment 1 G := x x dx (1) n Z · V (see [8], also [11], [12], [13]). It is known that the face-centered cubic (or f.c.c.) lattice A has the largest packing radius of 3 any three-dimensional lattice (Gauss), while its dual, the body-centered cubic (or b.c.c.) lattice A has both the smallest covering radius [1] and the smallest quantization error [2]. In [10] it was ∗3 shown that among all isodual three-dimensional lattices, the mean-centered cuboidal (or m.c.c.) lattice has the largest packing radius and the smallest covering radius. The m.c.c. lattice, denoted here by M , has Gram matrix 3 1+√2 1 1 − − 1 1 1+√2 1 √2 , 2 − − 1 1 √2 1+√2 − − determinant 1, packing radius 1 1 + 1 , center density δ = 0.1657... (which is between the 2q2 √2 values for the f.c.c. and b.c.c. lattices), covering radius 30.52 1.25 (again between the values for − the f.c.c. and b.c.c. lattices), kissing number 8 and automorphism group of order 16. The m.c.c. lattice received a brief mention in [3] and was studied in more detail in [10]. It also arises from the period matrix of the hyperelliptic Riemann surface w2 = z8 1 [4]. − The purpose of this note is to prove: Theorem 1. The m.c.c. lattice M is the optimal quantizer among all isodual three-dimensional 3 lattices. In higher dimensions, less is known. The lattice D is optimal among all four-dimensional lat- 4 tice packings, andhas alower quantization errorthanany other four-dimensionallattice presently known (see [8] for references). It is not the best four-dimensional lattice covering (A is better), ∗4 but it is isodual and may be optimal among isodual lattices with respect to all three criteria. Similar remarks apply to the isodual eight-dimensional lattice E (again A is a better covering 8 ∗8 but is not isodual). In 24 dimensions the isodual Leech lattice is known to be the best lattice packing [6], and may well also be the optimal covering and quantizer. 2. Proof of Theorem 1 We will specify three-dimensional lattices by giving Gram matrices and conorms (or Selling pa- rameters) — cf. [7], [9], [10]. Itwas shown in [10] that, upto equivalence, indecomposable isodual three-dimensional lattices of determinant 1 have Gram matrices of the form 2α αβ α(2 β) β − − − 1 αβ 2β 2β(1−α) , (2) 2 αβ − α − α − −α(2−β) −2β(α1−α) α2β+2α+α2β−4αβ 2 where α, β are any real numbers satisfying 0 < α < 1, 0 < β < 1; and decomposable lattices have Gram matrices 1 0 0 0 α h , (3) − 0 h β − where α, β, h are any real numbers satisfying 0 2h α β, αβ h2 = 1. ≤ ≤ ≤ − In the indecomposable case the nonzero conorms are: α(2 β) αβ 2α(1 β) p = − , p = , p = − , 01 02 03 γ γ βγ 2β(1 α) 2(1 α)(1 β) β(2 α) p = − , p = − − , p = − , (4) 12 13 23 αγ γ γ where γ = 2 αβ; in the decomposable case they are: − p = 1, p = α h, p = β h, p = 0, p = 0, p = h. (5) 01 02 03 12 13 23 − − The m.c.c. lattice corresponds to the case α= β = 2 √2 of Eqs. (2) and (4). − From [2] we know that the normalized second moment (1) for a decomposable or indecompos- able three-dimensional lattice Λ is given by DS +2S +K 1 2 G = , (6) 36D4/3 where D = det Λ, S = p +p +p +p +p +p , 1 01 02 03 12 13 23 S = p p p p +p p p p +p p p p , 2 01 02 13 23 01 03 12 23 02 03 12 13 K = p p p (p +p +p )+p p p (p +p +p ) 01 02 03 12 13 23 01 12 13 02 03 23 +p p p (p +p +p )+p p p (p +p +p ). 02 12 23 01 03 13 03 13 23 01 02 12 For our lattices, D = 1. We first consider the indecomposable case. From (4), (6) we find that f(α,β) G = (7) 36αβ(2 αβ)4 − where f(α,β) = 3α5β5 8α4β4(α+β)+4α3β3(α2+10αβ +β2) − 48α3β3(α+β)+8α2β2(3α2 +4αβ +3β2)+32α2β2(α+β) − 8αβ(5α2 +6αβ +5β2)+16(α2 +αβ +β2). − 3 It turns out that there is exactly one choice for (α,β) in the range 0 < α < 1, 0 < β < 1 for which both partial derivatives ∂G and ∂G vanish, namely α = β = 2 √2, corresponding to the ∂α ∂β − m.c.c. lattice. At all other points in the interior of this region, one of the two partial derivatives does not vanish and so the point cannot be a local minimum. To show this we used the computer algebra system Maple [5] to compute the numerators of ∂G and ∂G, giving a pair of simultaneous ∂α ∂β equations in α and β, symmetrical in α and β. By eliminating β, we obtain a single equation for α: α(α 1)(α 2)(α2 +2α 2)(α2 4α+2)(α4 4α3+6α2 4) − − − − − − × (57α6 220α5 102α4 +1448α3 1860α2 +832α 152) = 0. × − − − − Theonlyrootsintherange0< α < 1are2 √2,√3 1andtheroot0.9894... ofthesixth-degree − − factor. By symmetry, β must also take one of these three values. At only one of these nine points do both partial derivatives vanish, namely α = β = 2 √2. At that point the matrix of second − partial derivatives is positive definite, showing that m.c.c. is a local minimum. The resulting value of G is 17+4√2 = 0.0786696... , 288 between the values for the b.c.c. and f.c.c. lattices, which are respectively 19 21/3 22/3 = 0.0785432... and = 0.0787450... . 384 16 Ontheboundaryoftheregionthelatticesareeitherdegenerate(ifαorβ is0)ordecomposable (if α or β is 1). In the latter case we assume α β and find that there is a unique point where ≤ ∂G and ∂G vanish, when α = √3 1, β = 1. This is the lattice Z 3 1/4A , for which ∂α ∂β − ⊕ − 2 G = 5√3 + 1 = 0.0812361.... It is not a local minimum. 162 36 It remains to consider the decomposable case. From (5), (6), we find that 1 G = αβ(α+β)+2(αβ 1)3/2 +2α+2β +1 . (8) 36 n − o Again we solve ∂G = ∂G = 0, and find that the only possibilities in the range 0 α β are ∂α ∂β ≤ ≤ α = β = 1, G = 1 ; α = 1, β = 2, G = 1 ; and α = β = √3 1 (the decomposable lattice 12 12 − mentioned above). None of these are local minima. This completes the proof. 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