Sphere Packings Ji Hoon Chun Thursday, July 5, 03 Abstract The density of a (point) lattice sphere packing in n dimensions is the volume of a sphere in R n divided by the volume of a fundamental region of the (point) lattice We will give examples of packings where the centers of the spheres are points on the Z n, A n, and D n lattices, calculate their densities, center densities, and covering radii, and state the densest known lattice packings in through dimensions We will then explain how the Z and D 8 packings have enough room for a second copy of the respective packing to be placed next to the first one without any sphere intersections, resulting in lattice packings with twice the densities of the originals In addition, we will mention Rogers s upper bound and Minkowski s lower bound regarding sphere packing densities, and also prove Mordell s inequality Sphere packings Note Unless otherwise noted, information is from [Conway and Sloane] All decimal numbers will be rounded to decimal places Definitions and notes The volume of a unit n-dimensional sphere will be denoted by V n Definition A (point) lattice in R n is a discrete subgroup of R n under addition containing the origin [MathWorld, Point lattice ] A fundamental region of L is a subset E of R n that satisfies E + L = R n and (E + l ) (E + l ) = for any l, l L, l l v = (v,, v m ) Example For a lattice L in R n, there exist n vectors in R m, where m n, such v n = (v n,, v nm ) that for any x L, x = n i= c iv i where c i Z The region {a v + + a n v n : 0 a i < } In this talk the only fundamental regions that will be discussed are of this type M = v v n is called the generator matrix for L The Gram matrix A is defined as A = MM T The determinant of a lattice L is det L = det A If M is square, det L = (det M) det L can be considered to be square of the volume of the fundamental region, regardless of the particular region used
The density of a lattice sphere packing is = where r is the radius of each sphere In other words, = = V n r n Volume of the fundamental region, Vnrn det L For a lattice with Gram matrix A, the definition can be rewritten as Vnrn det A We also define n to be the maximal density of any lattice sphere packing in R n The center density δ of a sphere packing is δ = V n, In other words, δ = rn det A for lattices We also define δ n to be the maximal center density of any lattice sphere packing in R n A deep hole of a lattice L is a point x R n such that dist (x, L) = max y R n {y, L} The covering radius R of a lattice L is half the distance between a point of L and its nearest deep hole Various lattices = denotes the equivalence of two lattices Two lattices are equivalent if one can be transformed into the other using scaling, rotations, and/or reflections More precisely, L and L are two equivalent lattices if and only if their generator matrices M and M satisfy M = cumb, where c is a nonzero constant, U has integer entries and det U {, }, and B is an orthogonal matrix with real entries
The cubic lattices Z n Lattice Definition {(x,, x n ) Z n }, n v = (, 0, 0,, 0, 0) v = (0,, 0,, 0, 0) Basis vectors v n = (0, 0, 0,, 0, ) Generator M Z n = matrix n n Gram matrix A Z n = Z n Volume of fundamental region det AZ n = n n Radius of sphere r = Center density δ = ( ) n Covering radius R = n Dimension 3 5 6 7 8 Density 0785 0536 0308 065 00807 00369 0059 50 0 0 Densities calculated from [MathWorld, Hypersphere ] Sphere packings from the Z and Z 3 lattices can be seen in [Figure ] and [Figure 3] In n dimensions, consider a cube of side length surrounding a cube of spheres (each of radius ) such that the cube just touches all the spheres Draw a sphere centered at the center of the cube with the largest possible ( radius such that it does not intersect the other spheres The radius of this center sphere is r n,center = ( n )) n + = n (The covering radius is simply this value plus the radius of a lattice sphere) For n = and n = 3, one gets r,center 007 and r 3,center 03660, both less than 05 However, r,center = 05 and r 9,center =, which means that in 9 dimensions, the center sphere will touch the sides of the surrounding cube! Since {r n,center } n N is monotone increasing and unbounded as n, r n,center > in dimensions 0 and above, which results in parts of the sphere protruding from the surrounding cube The specific case n = will be discussed later 3
The lattices A n Lattice A n { Definition (x0, x,, x n ) Z n+ : x 0 + x + + x n = 0 }, n v = (,, 0, 0,, 0, 0, 0) v = (0,,, 0,, 0, 0, 0) Basis vectors v n = (0, 0, 0, 0,, 0,, ) Generator M An = matrix n (n+) Gram matrix A An = n n Volume of det AAn = n + fundamental region Radius of sphere r = ( ) n Center density δ = n+ Covering radius R = n+ (n+ n+ ) n+ Dimension 3 5 6 7 8 Density 09069 0705 0557 03799 0 076 0086 97 0 8 A is the hexagonal lattice which gives the densest possible circle packing [Figure ] and [Figure ]
3 The checkerboard lattices D n Lattice Definition {(x,, x n ) Z n : x + + x n is even}, n 3 v = (,, 0, 0,, 0, 0, 0) v = (,, 0, 0,, 0, 0, 0) Basis vectors v 3 = (0,,, 0,, 0, 0, 0) v n = (0, 0, 0, 0,, 0,, ) Generator M Dn = matrix n n 0 0 Gram matrix A Dn = Volume of det ADn = fundamental region Radius of sphere Center density δ D n ( ) n { n = 3 Covering radius R = n n Dimension 3 5 6 7 8 Density 0705 0669 0653 0330 0088 068 355 0 7 The D 3 lattice is known as the face-centered cubic lattice It can be seen in [Figure 5] As discussed earlier, the covering radius of Z is, which allows for another copy of Z to fit in the deep holes in the lattice without modifying the existing lattice or the sphere radius In precise terms, this configuration is the set ( Z ) + := Z ( Z + (,,, ( )) ) [Weaire and Aste] Since Z + consists of two ( copies of Z in the same space, it has double the density of Z, in other words 0669 ) Z + is in fact equivalent to D It is also true that D + := D ( D + (,,, )) = Z, although this time the sphere radius changes In addition, D n + := D n ( D n + (,, )) is a lattice iff n is even Note that when n = 8, R =, so one can fit in another copy of D 8 into the deep holes of the existing copy of D 8 without modifying it This lattice D 8 + is known as the E 8 lattice Theorem (Gauss) D 3 is the densest lattice packing in 3 dimensions This lattice is unique up to reflections and rotations Proof See [Zong] n n 5
3 Some theorems 3 Sphere packing density bounds Theorem (Rogers s upper bound) Consider a regular n-dimensional simplex of side length in R n Draw n + n-dimensional spheres of unit radius, each centered at a vertex of the simplex Let σ n be the proportion of the simplex that the interiors of the spheres fill Then the density of any sphere packing in R n satisfies σ n This bound coincides with the hexagonal packing in n = Theorem (Minkowski) In R n, lattices of density ζ (n) n exist, where ζ (n) is the Riemann zeta function, ζ (n) = k= k n The general form of this theorem, valid for centrally symmetric convex bodies, is called the Minkowski- Hlawka Theorem [Zong] The best known packing densities in some dimensions are shown in the following table Dimension 3 5 6 7 8 Upper bound (Rogers) 09069 07796 0678 0557 09 0398 0568 0005 Best known A D 3 = A3 D Λ 5 = D5 Leech lattice packing 09069 0705 0669 0653 03730 0953 0537 0009 Lower bound 080 03005 0353 0068 0038 0058 00078 9 0 7 (Minkowski) Z n 0785 0553 0308 065 00807 00369 0059 50 0 0 The densest known lattice packing in dimensions is called the Leech lattice It is 0 million times as dense as Z! Lastly, we have an inequality that places an upper bound on the density of a lattice in R n given the highest possible density of any lattice in R n Definition For a lattice L, the minimal norm µ of L is µ = min {x x} x L,x 0 Claim For a lattice L, the following equation holds: δ = ( ) n µ det L Proof Since the radius of the sphere is equal to half the distance between the origin and the closest nonzero ) point in L, δ = rn n det L = det L ( µ ) n det L = ( µ Definition For a lattice L, the dual lattice L is defined as L = {x R n : x y Z for all y Λ} Since the Gram matrix of L is the inverse of the Gram matrix of L, det L = (det L) Theorem Let L n be a lattice in R n (not necessarily integral), and let S be a k-dimensional subspace of R n Let E k = L n S and F n k = L n S Then det F = det E det L n 6
Proof Let {e,, e n } be an orthonormal basis for R n, where {e,, e k } is a basis for S and {e k+,, e n } is a basis for S There exists an integral basis {v,, v n } for L such that {v,, v k } is an integral basis for E k Let {w,, w n } be the dual basis, where v i w j = δ ij Then {w,, w n } is ( an integral ) A 0 basis for L n and {w k+,, w n } is an integral basis for F n k Then there s a matrix M = so B C v = M v n e e n det F = (det C) and w w n = ( M ) T e e n Now det L n = (det A) (det C), det E = (det A), and Corollary (Mordell s inequality) δ n δ n n n Proof Let L n be an optimal lattice in n dimensions, so δ (L n ) = δ n, and choose the scaling such that det L n = Then det L n =, which implies that δ (L n) δ (L n ) and µ (L n) µ (L n ) Then by the previous theorem, µ (L n) = the determinant of the densest -dimensional section of L n = the determinant of the ( ) n densest (n )-dimensional section of L n Now using the above claim, δ n µ(ln ) and det Ln δ n = ( µ(ln) ) n, so δ n ( µ (Ln ) ( µ (Ln ) = ( µ (Ln ) = ( µ (Ln ) ( (µ (Ln ) = ) n det Ln ) n µ (Ln ) ) n ) n ( ) µ (L n ) ) n ) n n = δ n n n Example Recall that the center density of D 3 is δ (D 3 ) = Since D 3 is the optimal packing in R 3, we have δ 3 = Then by Mordell s inequality, δ (δ 3 ) = 8 Note that δ (D ) = 8, so D is optimal in R 7
Figures All of these figures were created by the author of these notes [Figure ] [Figure ] [Figure 3] [Figure ] 3 [Figure 5] References Conway, J H and Sloane, N J Sphere Packings, Lattices and Groups 3rd ed Springer-Verlag Insall, Matt; Rowland, Todd; and Weisstein, Eric W Point Lattice From MathWorld A Wolfram Web Resource http://mathworldwolframcom/pointlatticehtml Weaire, D and Aste, T The Pursuit of Perfect Packing nd ed CRC Press Weisstein, Eric W Hypersphere From MathWorld A Wolfram Web Resource http://mathworldwolframcom/hypers Zong, C Sphere Packings Springer-Verlag 8