Sphere Packings, Coverings and Lattices

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1 Sphere Packings, Coverings and Lattices Anja Stein Supervised by: Prof Christopher Smyth September, 06 Abstract This article is the result of six weeks of research for a Summer Project undertaken at the University of Edinburgh School of Maths The article discusses the problems of packing spaces in Euclidean space: how densely can we pack an infinite number of spheres in space? The article then talks about the covering problems and kissing problems The Sphere Packing Problem Consider a large empty region or container What is the greatest number of identical objects that can be packed into this region? This class of optimization problem is easy to solve for identical cubes However when considering the general problem, which asks for the densest packing of spheres in n-dimensional space, it is not so straight-forward In fact, this problem dated back to the late 590s and took over 400 years to solve Since we extend the problem to an infinitely large space, calculating the area is not possible Instead we look at the proportion of space occupied by the spheres This quantity is the density, which we use to determine the efficiency of the packing The goal of sphere packing is to find a certain packing in a given number dimensions, which achieves the highest possible density So how do we pack spheres in a infinitely large box? Well when we look at each n-dimensional space in turn, we first restrict our attention to spheres whose centres form a lattice, which we will now define: Definition (Lattice) A lattice Λ is a discrete additive subgroup of R n ie it is a subset Λ R n satisfying: Λ is closed under addition and subtraction ϵ > 0 st two distinct lattice points x y Λ are at a distance at least x y ϵ Definition (Basis of a Lattice) Let B = [b,, b k ] R n k be linearly independent vectors in R n The lattice generated by B is L (B) = {Bx : x Z k } = { k x i b i : x i Z} i=

2 So B is a basis for the lattice L (B), which generates the space using integer linear combinations We can write a basis for a lattice in matrix form by taking each basis vector as a row of a matrix M of size n and let M be the generator matrix So all the row vectors in the matrix generate all the points in a given lattice Definition 3 (Minimum Distance) For any lattice Λ = L (B) the minimum distance of Λ is the smallest distance between any two lattice points denoted as λ Formally: λ(λ) = inf{ v : v Λ \ {0}} The packing radius ρ > 0 is exactly half of the value of the minimum distance The packing radius is the radius of the sphere which we will try and pack into the given space Knowing the packing radius of a lattice Λ, we are then able to define the packing density of sphere packing Definition 4 (Packing Density and Fundamental Region) Given a arrangement of points P in R n with packing radius ρ, the packing density of P is: = volume of occupied space in the fundamental region volume of the fundamental region The fundamental region of Λ is a subset E of R n that satisfies E + Λ = R n and (E + l ) (E + l ) = for any l, l Λ, l l Each packing problem has a dual covering problem, which asks how many of the same objects are required to completely cover every region of the chosen space, where objects are allowed to overlap Definition 5 (Covering Radius) The covering radius of a lattice λ in Euclidean space, denoted by ρ(λ), is defined as the smallest radius ρ such that the closed spheres of radius ρ centered at all lattice points cover the entire space In other words, the covering radius R of the sphere is the smallest number r such that spheres of radius r centered around all of the arrangement of points cover the entire space Hence, overlap is allowed The Kissing Problem In Cambridge in 694, two famous scientists Isaac Newton and David Gregory argued about the number of spheres of the same radius that could be brought into contact with a central ball in space This was known as the Kissing Problem as the verb kiss is used in billiards to means two balls that touch each other According to Newton, the maximum Kissing Number in three dimensions was, whereas Gregory thought the maximum Kissing Number was 3 Leech proved that the solution in three dimensions was more than 50 years later We define the kissing number as follows:

3 Definition 6 (Kissing Number): Given an arrangement of points P in R n with packing radius ρ, the kissing number of P denoted as τ is the highest number of equal nonoverlapping spheres in R n that can touch another sphere of the same size Finding the maximum τ for a given n is quite difficult There are only five known values of n for which τ is known Investigating Sphere Packings Best sphere packing in R When we talk about n-dimensional space we mainly refer to points in space R n as a string of n real numbers: x = (x, x,, x n ) For n = this is trivial for a one-dimensional sphere is a line segment within the Euclidean space consisting of a single line If we give our spheres a radius of one unit and place the center of the spheres on the even integers, we see that the density of the packing is Figure : Sphere packing in one-dimension where the dots represent the centres of units spheres Best sphere packing in R In two-dimensional case we are effectively packing circles into a plane and there will always be unused space For example consider the square packing Figure : A square packing Assume that the radius of each circle is r Then observing the covering in each fundamental region: We see that in each square of side r there are four quarter circles of radius r So the packing density is: (r) = πr (r) = π 4 =

4 Figure 3: The square is the fundamental region of the square packing The dot is the furthest point from centres of each sphere Now in order to cover all of the fundamental region we need the radii of the circles in the relevant region to cover the point in the region furthest away for each center of each circle The furthest point is the dot in the figure and so the covering radius R = r Hence the covering density is (R) = π( r) (r) = π = 57 The square packing provides a decent density but the best way to pack spheres is using a hexagonal packing, with circles centered on the points of the root lattice A Figure 4: A hexagonal packing Using the same assumptions, consider each hexagonal fundamental region: The point in the figure is one of the points furthest away from the centre of circles surrounding the point The distance from the centre of the circle to the point is the covering radius R By using simple geometrical arguments, for a circle of radius ρ the covering radius R = ρ 3 So the packing density is therefore = π = In A the hexagonal lattice generated by the basis of vectors {(, 0),(, 3)} Best Sphere Packing in R 3 A couple of ways one would think to pack in R 3 are as follows: The face centered cubic lattice (fcc) The integer vertex cubic lattice 4

5 Figure 5: The fundamental region of the hexagonal packing To get an idea of what the packing and covering densities are in each packing, we can take a look and see what is happening in each unit cube of length Figure 6: a D image of the fundamental region of the integer vertex packing Looking at the integer vertex packing we see that ρ = In each unit cube there are eight eighths of a sphere since they are eight vertices in a cube So the volume of one sphere fills the fundamental region So the density is as follows: (ρ) = 4 volume of occupied space volume of the fundamental region = 3 π( )3 3 = π 6 With regards to the covering radius, the point inside the unit cube, which is furthest away from any of the vertices is at the center of the cube So the covering radius is R = 3 the covering density is 4 3 (R) = π( 3 )3 3 = π 3 Considering only lattice packings, the densest packing is that of a face centered cubic lattice (fcc), which is a simple cubic lattice with the addition of a 5

6 lattice point in the center of each of the faces of each cube Figure 7: A D image of the fundamental region of the face centred cubic lattice Again looking at the slice of the cubic formation we see that ρ = Six half spheres of radius ρ per unit cube so in total three spheres of radius ρ occupy the unit cube So the density is therefore: (ρ) = 4 3 π( )3 3 3 = π 4 The covering radius is R = so the covering density is therefore (R) = 4 3 π( )3 3 3 = π Best Sphere Packings in Higher Dimensions When talking about dimensions greater the 3, these are very difficult to visualise Recall that when we talk about n-dimensional space we mainly refer to points in space R n as a string of n real numbers: x = (x, x,, x n ) Also, a lattice is set of basis vectors which generate the space using integer linear combinations We can take each basis vector as a row of a matrix M of size n and called it the generator matrix So all the row vectors in the matrix generate all the points in a given lattice Using generator matrices, we can obtain information needed to calculate the densities The most efficient packing in R 4 occurs in D 4 which has generator matrix: This matrix tells us that the packing radius is because the shortest distance between the origin and a base vector point is ( ) = and the radius is half of the distance 6

7 In dimensions eight and 4, there exist symmetric sphere packings called E 8 and the Leech lattice, respectively In the even coordinate system E 8 consists of the points: {(x,, x 8 ) : x i Z + x i Z x i = (mod )} In other words, the points in R 8 are such that either all coordinates are integers or all coordinates are half-integers, and the sum of all coordinates is even The Minimum Covering Radius Problem The minimum covering radius is also a large topic with useful applications in error correcting coding The problem is what is the minimum distance between any point in an n-dimensional space and its corresponding nearest lattice point such that every point in space is covered by spheres of the resulting radius? Again, there are many answers to this problem depending on what lattice we consider The n-dimensional cubic lattice Z n The set of integers is denoted by Z and Z n = {(x,, x n ) : x i Z} Each generator matrix for the Z n is the identity matrix of size n The maximum kissing number τ = n for each n since the following minimum vectors (ie the points in the lattice closest to the origin) are (0,, ±,, 0) where we can one entry of the co-ordinates to be either or - Since each lattice point is distance one away from their closest neighbour, in Z n the ρ = Upon inspection of Z n, we conclude that the minimal covering radius R for when n is n The n-dimensional lattice A n For n A n = {(x 0, x,, x n ) Z n+ : x x n = 0} So n + coordinates are used to define an n-dimensional lattice The standard generator matrix is as follows: M = We see that the kissing number τ = n(n+) as the minimal vectors from the origin are all permutations of (,-,0,,0) The packing radius ρ = and the covering radius is R = ρ{ a(n+ a) n+ } where a is the integer part of (n + )/ 7

8 The n-dimensional lattice D n For n 3 D n = {(x 0, x,, x n ) Z n : x x n even}, D n is sometimes referred to as the checkerboard lattice The standard generator matrix is as follows: M = We see that the kissing number τ = n(n ) as the minimal vectors from the origin are all permutations of (±,±,0,,0) The packing radius ρ = Corollary The minimum covering radius for D 4 is The following text below is an indication of understanding the corollary: Let x = (a, b, c, d) be a point in R n and let D 4 = L = {(x 0, x,, x n ) Z n : x x n even For any lattice point l in L, L ± l = L We need to show that we can add and subtract lattice points to find that x l i Since we can use the basis vectors of D 4 to add and subtract integer multiples of the basis vectors, we can reduce the chosen point in space down to the decimal points of its co-ordinates If the decimals of the co-ordinates of the point lie within [-05, 05] then we know that the minimum distance from the point to the nearest lattice is at most as ( ) + ( ) + ( ) + ( ) = Now we consider the other cases, which are the following permutations of the coordinates with the number of decimals in the range [-05, 05] and the remaining number of decimals in the range (-,05] and [05,) No matter what the sign (±) used in each coordinate is, we can still add and subtract integer combinations of the lattice points in D 4 such that (a, b, c, d ) where (a, b, c, d ) = x l i For example: Let x = (05, 099, 05, 099) be the decimals of the coordinate of a point in R 4 (05, 099, 05, 099) (, 0, 0, ) = ( 05, 099, 05, 00) ( 05, 099, 05, 00) + (,, 0, 0) = (05, 00, 05, 00) (05, 00, 05, 00) We can use the same arguments to show that for n 4, ρ = and the minimum covering radius R = ρ n 8

9 The n-dimensional lattice E n In the even coordinate system E 8 consists of the points: {(x,, x 8 ) : x i Z + or all x i Z, x i = 0(mod )} The odd coordinate system is obtained by changing the sign of any coordinate: the points are {(x,, x 8 ) : x i Z + or all x i Z, x i = x 8 (mod )} With E 8 the generator matrix in the even coordinate system is M = The kissing number τ = 40 as the minimal vectors are permutations of (±, 0 6 ), together with the vectors ( 8 ) that have an even number of minus signs in the even coordinate system, or an odd number of minus signs in the odd coordinate system The packing radius is ρ = and the covering radius is R = ρ = Conclusion In this report we have investigated the best sphere packings in up to eight dimensions, defined various lattices and attempted to prove why the minimum covering radius for D 4 is There is a lot more to be investigated including looking at the 4-dimensional Leech Lattice which has applications in errorcorrecting coding 9

10 References [] Leech, J (964) Some Sphere Packings in Higher Space Canadian Journal of Mathematics 9 [] Conway, J H, Sloane, N J A (988) Sphere Packings, Lattices and Groups New York: Springer-Verlag [3] Zong, C (999) Sphere Packings New York: Springer-Verlag [4] Aste, T, Weaire, D (000) The Pursuit of Perfect Packing IoP Publishing Ltd [5] Roberts, A (006) Available at: Current/Journal_Spring_006/ARoberts_LeechLatticepdf (Accessed: 3 June 06) [6] (Accessed June 06) [7] [8] 0

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