Sphere Packings
Chuanming Zong
Springer
To Peter M. Gruber and David G. Larman
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Sphere Packings
Chuanming Zong
Springer
To Peter M. Gruber and David G. Larman
This page intentionally left blank
Preface
Sphere packings is one of the most fascinating and challenging subjects in mathematics. Almost four centuries ago, Kepler studied the densities of sphere packings and made his famous conjecture. Several decades later, Gregory and Newton discussed the kissing numbers of spheres and proposed the GregoryNewton problem. Since then, these problems and related ones have attracted the attention of many prominent mathematicians such as Blichfeldt, Dirichlet, Gauss, Hermite, Korkin, Lagrange, Minkowski, Thue, Voronoi, Watson, and Zolotarev, as well as many active today. As work on the classical sphere packing problems has progressed many exciting results have been obtained, ingenious methods have been created, related challenging problems have been proposed and investigated, and surprising connections with other subjects have been found. Thus, though some of its original problems are still open, sphere packings has developed into an important discipline. There are several books dealing with various aspects of sphere packings. For example, Conway and Sloane [1], Erd¨ os, Gruber, and Hammer [1], L. Fejes T´ oth [9], Gruber and Lekkerkerker [1], Leppmeier [1], Martinet [1], Pach and Agarwal [1], Rogers [14], Siegel [2], and Zong [4]. However, none of them gives a full account of this fascinating subject, especially its local aspects, discrete aspects, and proof methods. The purpose of this book is try to do this job. It deals not only with the classical sphere packing problems, but also the contemporary ones such as blocking light rays, the holes in sphere packings, and ﬁnite sphere packings. Not only are the main results of the subject presented, but also, its creative methods from areas such as geometry, number theory, and linear programming are described.
viii
Preface
In addition, the book contains short biographies of several masters of this discipline and also many open problems. I am very much indebted to Professors P.M. Gruber, T.C. Hales, M. Henk, E. Hlawka, D.G. Larman, C.A. Rogers, J.M. Wills, G.M. Ziegler, and the reviewers for their helpful information, suggestions, and comments to the manuscript, and to J. Talbot for his wonderful editing work. Nevertheless, I assume sole responsibility for any remaining mistakes. The staﬀ at SpringerVerlag in New York have been courteous, competent, and helpful, especially M. Cottone, K. Fletcher, D. Kramer, and Dr. I. Lindemann. Also, I am very grateful to my wife, Qiaoming, for her understanding and support. This work is supported by The Royal Society, the Alexander von Humboldt Foundation, and the National Scientiﬁc Foundation of China.
Berlin, 1999
C. Zong
Basic Notation
En x o x, y x, y x d(X) conv{X} K int(K) rint(K) bd(K) v(K) s(K) C x, yC Sn ωn x, yg In Z
Euclidean ndimensional space. A point (or a vector) of E n with coordinates (x1 , x2 , . . . , xn ). The origin of E n . The inner product of two vectors x and y. The Euclidean distance between two points x and y. The Euclidean norm of x. The diameter of a set X. The convex hull of X. An ndimensional convex body. The interior of K. The relative interior of K in the considered space. The boundary of K. The volume of K. The surface area of K. An ndimensional centrally symmetric convex body centered at o. The Minkowski distance between x and y with respect to C. The ndimensional unit sphere centered at o. The volume of the ndimensional unit sphere. The geodesic distance between two points x and y in bd(Sn ). The ndimensional unit cube {x : xi  ≤ 12 }. The set of all integers.
x Zn Λ det(Λ) Q(x) γn ℵ x, yH D(x) k(Sn ) k ∗ (Sn ) b(Sn ) δ(Sn ) δ ∗ (Sn ) θ(Sn ) θ∗ (Sn ) r(Sn ) r∗ (Sn ) ρ∗ (Sn ) h(Sn ) h∗ (Sn ) (Sn )
Basic Notation The ndimensional integer lattice {z : zi ∈ Z}. An ndimensional lattice. The determinant of Λ. A positive deﬁnite quadratic form. The Hermite constant. A code. The Hamming distance between x and y. The DirichletVoronoi cell at x. The kissing number of Sn . The lattice kissing number of Sn . The blocking number of Sn . The maximum packing density of Sn . The maximum lattice packing density of Sn . The minimum covering density of Sn . The minimum lattice covering density of Sn . The maximum radius of a sphere that can be embedded into every packing of Sn . The maximum radius of a sphere that can be embedded into every lattice packing of Sn . The maximum radius of the spherical base of an inﬁnite cylinder that can be embedded into every lattice packing of Sn . The Hornich number of Sn . The lattice Hornich number of Sn . The L. Fejes T´ oth number of Sn .
Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Basic Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.
2.
vii ix
The Gregory–Newton Problem and Kepler’s Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.1. 1.2. 1.3. 1.4. 1.5.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Packings of Circular Disks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The GregoryNewton Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . Kepler’s Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Some General Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 7 10 13 18
Positive Deﬁnite Quadratic Forms and Lattice Sphere Packings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
2.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2. The LagrangeSeeberMinkowski Reduction and a Theorem of Gauss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3. Mordell’s Inequality on Hermite’s Constants and a Theorem of Korkin and Zolotarev . . . . . . . . . . . . . . . . . . 2.4. Perfect Forms, Voronoi’s Method, and a Theorem of Korkin and Zolotarev . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5. The KorkinZolotarev Reduction and Theorems of Blichfeldt, Barnes, and Vet˘cinkin . . . . . . . . . . . . . . . . . . . . . . 2.6. Perfect Forms, the Lattice Kissing Numbers of Spheres, and Watson’s Theorem . . . . . . . . . . . . . . . . . . . . . . .
23 25 31 33 36 41
xii
Contents 2.7. Three Mathematical Geniuses: Zolotarev, Minkowski, and Voronoi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.
4.
Lower Bounds for the Packing Densities of Spheres . . . . 3.1. The MinkowskiHlawka Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 3.2. Siegel’s Mean Value Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3. Sphere Coverings and the CoxeterFewRogers Lower Bound for δ(Sn ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4. Edmund Hlawka . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Lower Bounds for the Blocking Numbers and the Kissing Numbers of Spheres . . . . . . . . . . . . . . . . . . . . . 4.1. The Blocking Numbers of S3 and S4 . . . . . . . . . . . . . . . . . . . . . 4.2. The ShannonWyner Lower Bound for Both b(Sn ) and k(Sn ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3. A Theorem of SwinnertonDyer . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4. A Lower Bound for the Translative Kissing Numbers of Superspheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42 47 47 51 55 62
65 65 71 72 74
5.
Sphere Packings Constructed from Codes . . . . . . . . . . . . . . . 5.1. Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2. Construction A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3. Construction B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4. Construction C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5. Some General Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
79 79 82 84 85 89
6.
Upper Bounds for the Packing Densities and the Kissing Numbers of Spheres I . . . . . . . . . . . . . . . . . . . 91 6.1. Blichfeldt’s Upper Bound for the Packing Densities of Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 6.2. Rankin’s Upper Bound for the Kissing Numbers of Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 6.3. An Upper Bound for the Packing Densities of Superspheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 6.4. Hans Frederik Blichfeldt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
7.
Upper Bounds for the Packing Densities and the Kissing Numbers of Spheres II . . . . . . . . . . . . . . . . . . 7.1. Rogers’ Upper Bound for the Packing Densities of Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2. Schl¨ aﬂi’s Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3. The CoxeterB¨ or¨ oczky Upper Bound for the Kissing Numbers of Spheres . . . . . . . . . . . . . . . . . . . . . . 7.4. Claude Ambrose Rogers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
103 103 107 111 122
Contents 8.
9.
Upper Bounds for the Packing Densities and the Kissing Numbers of Spheres III . . . . . . . . . . . . . . . . . 8.1. Jacobi Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2. Delsarte’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3. The KabatjanskiLeven˘stein Upper Bounds for the Packing Densities and the Kissing Numbers of Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
xiii
125 125 127
132
The Kissing Numbers of Spheres in Eight and Twenty–Four Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1. Some Special Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2. Two Theorems of Leven˘stein, Odlyzko, and Sloane . . . . . . . 9.3. Two Principles of Linear Programming . . . . . . . . . . . . . . . . . . . 9.4. Two Theorems of Bannai and Sloane . . . . . . . . . . . . . . . . . . . . .
139 139 141 142 143
10. Multiple Sphere Packings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2. A Basic Theorem of Asymptotic Type . . . . . . . . . . . . . . . . . . . 10.3. A Theorem of Few and Kanagasahapathy . . . . . . . . . . . . . . . 10.4. Remarks on Multiple Circle Packings . . . . . . . . . . . . . . . . . . . .
153 153 154 157 162
11. Holes in Sphere Packings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1. Spherical Holes in Sphere Packings . . . . . . . . . . . . . . . . . . . . . . 11.2. Spherical Holes in Lattice Sphere Packings . . . . . . . . . . . . . . 11.3. Cylindrical Holes in Lattice Sphere Packings . . . . . . . . . . . .
165 165 176 178
12. Problems of Blocking Light Rays . . . . . . . . . . . . . . . . . . . . . . . . . 12.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2. Hornich’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3. L. Fejes T´ oth’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4. L´ aszl´ o Fejes T´ oth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
183 183 185 189 198
13. Finite Sphere Packings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2. The Spherical Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3. The Sausage Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4. The Sausage Catastrophe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
199 199 200 204 214
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
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1. The Gregory–Newton Problem and Kepler’s Conjecture
1.1. Introduction Let K denote a convex body in ndimensional Euclidean space En . In other words, K is a compact subset of E n with nonempty interior such that λx + (1 − λ)y ∈ K, whenever both x and y belong to K and 0 < λ < 1. We call K a centrally symmetric convex body if there is a point p such that x ∈ K if and only if 2p − x ∈ K. For example, both the ndimensional unit sphere Sn =
n x = (x1 , x2 , . . . , xn ) : (xi )2 ≤ 1 i=1
and the ndimensional unit cube 1 In = x = (x1 , x2 , . . . , xn ) : max xi  ≤ 1≤i≤n 2 are centrally symmetric convex bodies. As usual, the interior, boundary, volume, surface area, and diameter of K are denoted by int(K), bd(K), v(K), s(K), and d(K), respectively. Denoting the volume of Sn by ωn , it is wellknown that π n/2 ωn = , Γ(n/2 + 1) and therefore s(Sn ) = nωn =
nπ n/2 , Γ(n/2 + 1)
2
1. The GregoryNewton Problem, Kepler’s Conjecture
where Γ(x) is the gamma function. If ai = (ai1 , ai2 , . . . , ain ), i = 1, 2, . . . , n, are n linearly independent vectors in E n , then the set Λ=
n
zi ai : zi ∈ Z ,
i=1
where Z is the set of all integers, is called a lattice, and we call {a1 , a2 , . . . , an } a basis for Λ. As usual, the absolute value of the determinant aij  is called the determinant of the lattice and is denoted by det(Λ). Geometrically speaking, det(Λ) is the volume of the fundamental parallelepiped P =
n
λi ai : 0 ≤ λi ≤ 1
i=1
of Λ. Let X be a set of discrete points in En . We shall call K +X a translative packing of K if (int(K) + x1 ) ∩ (int(K) + x2 ) = ∅ whenever x1 and x2 are distinct points of X. In particular, we shall call it a lattice packing of K when X is a lattice. Let k(K) and k∗ (K) denote the translative kissing number and the lattice kissing number of K, respectively. In other words, k(K) is the maximum number of nonoverlapping translates K + x that can touch K at its boundary, and k∗ (K) is deﬁned similarly, with the restriction that the translates are members of a lattice packing of K. For every convex body K, it is easy to see that k ∗ (K) ≤ k(K). (1.1) The following result provides a general upper bound for k∗ (K) and k(K). Theorem 1.1 (Minkowski [7] and Hadwiger [1]). Let K be an ndimensional convex body. Then k ∗ (K) ≤ k(K) ≤ 3n − 1, where equality holds for parallelepipeds. Proof. The parallelepiped case is easy to verify. We omit the veriﬁcation here. Let D(A) denote the diﬀerence set of a convex set A. In other words, D(A) = {x − y : x, y ∈ A}.
1.1. Introduction
3
By convexity, it is easy to see that (A + x1 ) ∩ (A + x2 ) = ∅ if and only if
1
2 D(A)
+ x1 ∩ 12 D(A) + x2 = ∅.
Therefore, by considering the two cases A = int(K) and A = K it can be easily deduced that k(K) = k(D(K)). Since D(K) is centrally symmetric, in order to get an upper bound for k(K), we assume that K itself is centrally symmetric and centered at o. If a translate K + x touches K at y ∈ bd(K), then for any point z ∈ K + x, we have 3y ∈ 3K and 3 2 (z
− y) ∈ 3K.
Hence we have z=
1 3
× 3y +
2 3
× 32 (z − y) ∈ 3K,
and therefore (see Figure 1.1) K + x ⊆ 3K.
3K K
K +x
Figure 1.1 Let K + xi , i = 1, 2, . . . , k(K), be nonoverlapping translates that touch K at its boundary. Then
k(K)
i=0
(K + xi ) ⊆ 3K,
4
1. The GregoryNewton Problem, Kepler’s Conjecture
where x0 = o, and therefore (k(K) + 1)v(K) ≤ v(3K) = 3n v(K), which implies that k(K) ≤ 3n − 1, 2
and hence using (1.1), Theorem 1.1 is proved.
Remark 1.1. In fact, ndimensional parallelepipeds are the only convex bodies in En such that k ∗ (K) = k(K) = 3n − 1. For a proof see Hadwiger [1] and Groemer [2]. Let l be a positive number and let m(K, l) be the maximum number of translates K + x that can be packed into the cube lIn . We deﬁne δ(K) = lim sup l→∞
m(K, l)v(K) , v(lIn )
the density of the densest translative packings of K in En . Similarly, the density δ∗ (K) of the densest lattice packings of K is deﬁned by restricting the translative vectors to those in a lattice. In this case, it can be deduced that v(K) δ ∗ (K) = sup , Λ det(Λ) where the supremum is over all lattices Λ such that K + Λ is a packing. For every convex body K, it is easy to see that δ ∗ (K) ≤ δ(K) ≤ 1.
(1.2)
The density of a given translative packing K + X is deﬁned by δ(K, X) = lim sup l→∞
card{X ∩ lIn }v(K) . v(lIn )
Then simple analysis yields that δ(K) = sup δ(K, X), X
where the supremum is over all sets X such that K + X is a packing. In fact, as the following theorem shows, the density δ(K) can be deﬁned using any convex body containing the origin, not just the cube. Theorem 1.2 (Hlawka [4]). Let A be an arbitrary convex body that contains the origin o as an interior point, and let m(K, lA) be the maximum number of translates K + x that can be packed into lA. Then, lim
l→∞
m(K, lA)v(K) = δ(K). v(lA)
1.1. Introduction
5
Proof. Let m∗ (K, l) be the maximum number of nonoverlapping translates K + x that intersect lIn and let d be the diameter of K. Since v((l + 2d)In ) − v((l − 2d)In ) = 0, l→∞ v(lIn ) lim
(1.3)
from the deﬁnition of δ(K) it follows that for each > 0 there is a t such that m(K, t )v(K) > δ(K) − v(t In ) and
m∗ (K, t )v(K) < δ(K) + . v(t In )
On the other hand, for any convex body A, there exists an l > 0 such that for each l > l there are two families of nonoverlapping cubes, t l In + xi : i = 1, 2, . . . , g and
t l
such that
In + yi : i = 1, 2, . . . , h , g t
In + xi ⊆ A, l v ∪gi=1 tl In + xi v(A) ≤ , 1−
h t In + yi , A⊆ l i=1 i=1
v ∪hi=1 tl In + yi v(A) ≥ . 1+ Hence, when l > l , one has
and
(1 − )(δ(K) − )
0, there exist a packing K + X and a positive number c such that d(D(xi )) < c for every point xi ∈ X, and card{X ∩ lIn }v(K) ≥ δ(K) − . l→∞ v(lIn ) lim
Then, by (1.3), one has δ(K) ≤ for suﬃciently large l, where λ(X, l) =
v(K) + 2 λ(X, l)
(1.4)
v(D(xi )) . card{X ∩ lIn }
xi ∈X∩lIn
In particular, since all the DirichletVoronoi cells are congruent if X is a lattice, one has v(K) δ ∗ (K) = sup , (1.5) Λ v(D(o)) where the supremum is over all lattices Λ such that K + Λ is a packing. Clearly, (1.4) and (1.5) provide reasonable ways to approximate δ(K) and δ ∗ (K) using local computation.
1.2. Packings of Circular Disks
7
1.2. Packings of Circular Disks Lemma 1.1. Let P be a polygon with p edges containing the unit disk S2 . Then, π v(P ) ≥ p tan , p where equality holds if and only if P is regular and its edges are tangent to bd(S2 ). Proof. Without loss of generality, we may assume that the p edges L1 , L2 , . . . , Lp of P are tangent to bd(S2 ) at t1 , t2 , . . . , tp in a circular order (see Figure 1.2). Letting θi be the angle between ti and ti+1 , where tp+1 = t1 , it follows that p θi = 2π i=1
and v(P ) =
p
tan
i=1
θi . 2
t3
S2
θ2 θ1
t2
t1 Figure 1.2 Since the function f (x) = tan x is strictly concave on [0, π/2), by Jensen’s inequality it follows that
p θi π v(P ) ≥ p tan i=1 = p tan , 2p p where equality holds if and only if P is a regular polygon. Lemma 1.1 is proved. 2 √ Let Λ2 be the twodimensional lattice with basis {(2, 0), (1, 3)}. Then S2 + Λ2 is a lattice packing of S2 . From this example it follows that k ∗ (S2 ) ≥ 6
(1.6)
8
1. The GregoryNewton Problem, Kepler’s Conjecture
and
v(S2 ) π =√ . (1.7) det(Λ2 ) 12 On the other hand, if S2 +x1 and S2 +x2 are two nonoverlapping circular disks that touch S2 at its boundary, then the angle between x1 and x2 is at least π/3. Therefore, 2π k(S2 ) ≤ = 6. (1.8) π/3 δ ∗ (S2 ) ≥
From (1.1), (1.6), and (1.8) it follows that k ∗ (S2 ) = k(S2 ) = 6. The following result determines the exact values of δ∗ (S2 ) and δ(S2 ). Theorem 1.3 (Lagrange [1] and Thue [2]). π δ ∗ (S2 ) = δ(S2 ) = √ . 12 Proof. Let X be a set such that S2 + X is a packing and d(D(x)) ≤ c for every point x ∈ X, where c is a suitable positive number. By (1.4) v(S2 ) + 2 λ(X, l)
δ(S2 ) ≤
(1.9)
for suﬃciently large l, where
λ(X, l) =
v(D(x)) . card{X ∩ lI2 } x∈X∩lI2
(1.10)
For convenience, we write X ∗ = X ∩ (l + 2c)I2 and deﬁne D∗ (x) = {y : y, x ≤ y, w for all w ∈ X ∗ } for each x ∈ X ∗ . It is easy to see that D∗ (x) = D(x),
x ∈ X ∩ lI2 ,
(1.11)
and the D∗ (x) form a tiling of E2 . Let e, f , and v be the number of edges, polygons, and vertices of this tiling. By Euler’s formula, f + v = e + 2.
(1.12)
1.2. Packings of Circular Disks
9
Let V be the set of vertices of this tiling, and let p(x) and q(v) denote the number of edges of D∗ (x) and the number of edges into v ∈ V , respectively. Since every vertex belongs to at least three edges and every edge joins two vertices, we have 3v ≤ q(v) = 2e. (1.13) v∈V
Then from (1.12) and (1.13) it follows that e + 6 ≤ 3f. Writing
(1.14)
p(X, l) =
x∈X∩lI2 p(x) , card{X ∩ lI2 }
(1.3), (1.11), and (1.14) imply that p(X, l) ≤
2e 12 + ≤6− + f f
for suﬃciently large l, and hence lim sup p(X, l) ≤ 6.
(1.15)
l→∞
By (1.10) and Lemma 1.1 it follows that λ(X, l) ≥
π
x∈X∩lI2
p(x) π
π tan p(x)
card{X ∩ lI2 }
.
Since the function
x π tan π x is concave when x ≥ 3, by Jensen’s inequality and (1.15) it follows that f (x) =
λ(X, l) ≥ π Hence by (1.9) one has
√ p(X, l) π tan ≥ 12. π p(X, l) π δ(S2 ) ≤ √ . 12
Then Theorem 1.3 follows from (1.2), (1.7), and (1.16).
(1.16) 2
Remark 1.3. From the proof of Theorem 1.3 it follows that the lattice with which π δ ∗ (S2 ) = √ 12 can be realized is unique with respect to rotation and reﬂection.
10
1. The GregoryNewton Problem, Kepler’s Conjecture
1.3. The GregoryNewton Problem √ √ √ Write a1 = (2, 0, 0), a2 = (1, 3, 0), and a3 = (1, 3/3, 2 6/3), and let Λ3 be the lattice generated by them. Usually, Λ3 is known as the facecentered cubic lattice. It is easy to see that S3 + Λ3 is a packing of S3 , in which every sphere touches 12 others. This observation implies k ∗ (S3 ) ≥ 12.
(1.17)
In 1694, during a famous conversation, D. Gregory and I. Newton discussed the following problem. The GregoryNewton Problem. Can a sphere touch 13 spheres of the same size? Newton thought “no, the maximum number is 12,” while Gregory believed the answer to be “yes.” In the literature this problem is sometimes referred to as the thirteen spheres problem. In S3 + Λ3 , locally speaking, the kissing conﬁguration is stable. In other words, none of the twelve spheres that touch S3 can be moved around (see Figure 1.3). However, if twelve unit spheres are placed at positions corresponding to the vertices of a regular icosahedron concentric with the central one, the twelve outer spheres do not touch each other and may all be moved around freely (see Figure 1.4).
Figure 1.3
Figure 1.4
Let S3 + xi , i = 1, 2, . . . , k(S3 ), be nonoverlapping spheres that touch S3 at its boundary, and deﬁne √ Ωi = x ∈ bd(S3 ) : x, xi ≥ 3 .
1.3. The GregoryNewton Problem
11
The k(S3 ) congruent caps Ωi form a cap packing on the surface of S3 , and therefore, comparing the area of bd(S3 ) with the area of Ωi , it follows that k(S3 ) ≤
s(S3 ) = 14.9282 . . . , s(Ω1 )
which together with (1.17) implies that 12 ≤ k(S3 ) ≤ 14. However, this is not enough to solve the GregoryNewton problem. These facts show the inherent diﬃculty of this fascinating problem. Theorem 1.4 (Hoppe [1], Sch¨ utte and van der Waerden [1], and Leech [1]). k ∗ (S3 ) = k(S3 ) = 12. Proof. Let X = {x1 , x2 , . . . , xv } be a subset of bd(S3 ) such that S3 + {o, 2x1 , 2x2 , . . . , 2xv } is a kissing conﬁguration, and let xi , xj g be the geodesic distance between xi and xj . Clearly, we have xi , xj g ≥ π/3 whenever xi and xj are distinct points of X. Joining xi and xj by a geodesic arc if and only if xi , xj g < arccos 17 , we obtain a network on the sphere. No two arcs of this network cross, since any four points of the network form a spherical quadrilateral with side length at least π/3 whose diagonals cannot both be less than π/2. Therefore, bd(S3 ) is divided into polygons whose vertices are points of X. Clearly, we may assume that every point of X is connected to at least two others by arcs. An easy computation yields that every angle of these polygons is larger than π/3. Thus at most ﬁve arcs of the network meet in any point of X. Let Pn be a polygon of this network with n sides. By routine computation one can verify the following statements: s(P3 ) ≥ 0.5512 . . . ,
(1.18)
where equality holds if P3 is an equilateral triangle with side length π/3. s(P4 ) ≥ 1.3338 . . . ,
(1.19)
where equality holds if P4 is an equilateral quadrilateral with side length π/3, and one of its diagonals is of length arccos 17 . s(P5 ) ≥ 2.2261 . . . ,
(1.20)
12
1. The GregoryNewton Problem, Kepler’s Conjecture
where equality holds if P5 is an equilateral pentagon with side length π/3 having two coterminous diagonals of length arccos 17 . Also, when n ≥ 4, it is clear that the excess area of an ngon over the minimum area of n − 2 triangles increases with n, since if on a side of any ngon a triangle of minimum area is abutted, the resulting (n + 1)gon has to be deformed so that the original common side becomes a diagonal of length at least arccos 17 ; consequently, the area is increased. By (1.12), Euler’s formula, one has 2v − 4 = 2e − 2f = 3f3 + 4f4 + 5f5 + · · · − 2(f3 + f4 + f5 + · · ·) = f3 + 2f4 + 3f5 + · · · , where e is the number of the arcs, f is the number of the polygons, and fi is the number of igons in the network. Comparing the surface area of S3 and the area of the network, by (1.18), (1.19), and (1.20) it follows that 4π ≥ 0.5512f3 + 1.3338f4 + 2.2261f5 + · · · = 0.5512(f3 + 2f4 + 3f5 + · · ·) + 0.2314f4 + 0.5725(f5 + · · ·) = 0.5512(2v − 4) + 0.2314f 4 + 0.5725(f5 + · · ·). Thus we have 2v − 4 ≤
4π = 22.79, 0.5512
and therefore v ≤ 13. Suppose now that v = 13. Then 2v − 4 = 22, and we have 4π ≥ 0.5512 × 22 + 0.2314f 4 + 0.5725(f5 + · · ·), 0.44 ≥ 0.2314f 4 + 0.5725(f5 + · · ·), and therefore f4 = 0 or 1 and f5 = f6 = · · · = 0. In other words, the network has to divide bd(S3 ) into triangles except possibly for one quadrilateral. Now, we deal with two cases. Case 1. f4 = 0. Then we have f3 = 2e/3, 13 + 2e/3 = e + 2, and therefore e = 33. Consequently, the average number of arcs meeting at each point xi ∈ X is 66/13 > 5, which contradicts the fact that at most ﬁve arcs meet at any of the v points. Case 2. f4 = 1. Then it follows from Euler’s formula that f3 = 20, e = 32, and 4 arcs meet at one point and 5 at every other. In fact, it is impossible to construct such a network. This can be veriﬁed by starting from the quadrilateral and adding triangles one by one. Hence we have proved that k(S3 ) ≤ 12.
(1.21)
1.4. Kepler’s Conjecture
13
Then, by (1.1), (1.17), and (1.21) Theorem 1.4 is proved.
2
1.4. Kepler’s Conjecture √ Routine computation yields that the packing density of S3 + Λ3 is π/ 18, which implies that π (1.22) δ ∗ (S3 ) ≥ √ . 18 Based on this fact and (1.2), J. Kepler in 1611 made the following conjecture. Kepler’s Conjecture.
π δ(S3 ) = √ . 18
By (1.4), if the volume of every DirichletVoronoi cell of any packing √ S3 + X were greater than or equal to 4 2, one would be able to deduce π δ(S3 ) ≤ √ 18 and hence, by (1.22), π δ ∗ (S3 ) = δ(S3 ) = √ . 18 Unfortunately, the √ volumes of some DirichletVoronoi cells of some packings are less than 4 2. For example, let D be one of the smallest regular dodecahedra circumscribed to S3 . Take X = {o, 2x1 , 2x2 , . . . , 2x12 } , where xi ∈ bd(S3 ) ∩ bd(D). Then, S3 + X is a packing (see Figure 1.4), and the DirichletVoronoi cell D(o), deﬁned with respect to X, is the dodecahedron D itself. By routine computation, one obtains
√ 2π π v(D) = 20 1 − 2 cos tan < 4 2. 5 5 This example shows the fundamental diﬃculty of Kepler’s conjecture. Clearly, the density of any sphere packing may be improved by adding spheres as long as there is suﬃcient room to do so. When there is no longer room to add additional spheres we say that the sphere packing is
14
1. The GregoryNewton Problem, Kepler’s Conjecture
saturated. Without loss of generality, we assume that the sphere packings considered in the following subsections are saturated. We now introduce two approaches to Kepler’s conjecture.
1.4.1. L. Fejes T´ oth’s Program and Hsiang’s Approach Consider the system of DirichletVoronoi cells associated with √ a sphere packing. If the volume of a DirichletVoronoi cell is less than 4 2, it seems √ likely that the volume of some of its neighbors will be larger than 4 2. Therefore, it is reasonable to consider a locally averaged density, which is what L. Fejes T´ oth [9] proposed in 1953 to attack Kepler’s conjecture. However, he was unable to realize this program. In 1993, Hsiang [2] announced a proof of Kepler’s conjecture. Unfortunately this has been found to contain errors and has not been accepted by experts (see Hales [3]). In fact, the fundamental ideas of both L. Fejes T´ oth’s program and Hsiang’s approach are the same. We only attempt to sketch their key ideas here. Let h be a ﬁxed number. Two spheres S3 + xi and S3 + xj in a given packing S3 + X are called hneighbors of each other if xi , xj ≤ h. Take Xi = {x ∈ X : x, xi ≤ h} and, for convenience, enumerate it using double indices, namely setting Xi = {xij : j = 1, 2, . . . , mi } , where mi = card{Xi}. Then the hlocally averaged density of S3 +X around S3 + xi is deﬁned by
mi j=1 µij σ(xij , X)
mi σ(xi , X) = , j=1 µij where µij =
v(D(xij )) mi
and σ(xij , X) =
v(S3 ) . v(D(xij ))
Lemma 1.2 (Hsiang [2]). Let S3 + X be a saturated packing, and set µi =
mi j=1
µij
1.4. Kepler’s Conjecture
15
to be the weight assigned to S3 + xi . Then,
∈lI3 µi σ(xi , X) lim sup xi = δ(S3 , X). l→∞ xi ∈lI3 µi Proof. Since S3 + X is saturated, the diameter of any DirichletVoronoi cell is bounded by 4 from above. Thus, by (1.4) and the deﬁnitions of µi and σ(xi , X) it follows that
µi σ(xi , X) =
xi ∈lI3
=
mi xi ∈lI3 j=1 mi xi ∈lI3 j=1
µij σ(xij , X) v(S3 + xij ) mi
= card{X ∩ lI3 }v(S3 ) + O l2 and xi ∈lI3
Hence,
lim sup
µi =
mi v(D(xij ))
mi 2 = v(lI3 ) + O l . xi ∈lI3 j=1
µi σ(xi , X) card{X ∩ lI3 }v(S3 ) = lim sup µ v(lI3 ) l→∞ xi ∈lI3 i
xi ∈lI3
l→∞
= δ(S3 , X). 2
Lemma 1.2 is proved. From Lemma 1.2 it follows that for any ﬁxed number h, δ(S3 ) ≤ sup sup σ(xi , X), X xi ∈X
where the outer supremum is over all sets X such that S3 + X is a packing. Thus, if it can be proved that for a suitable number h, π σ(xi , X) ≤ √ 18
(1.23)
holds for every point xi of any set X such that S3 +X is a packing, Kepler’s conjecture will follow. In this way, the original global problem has been converted into a local one. Since S3 + X is saturated, the number of combinatorial types of the DirichletVoronoi cells is ﬁnite. Thus one can try to verify (1.23) by checking a ﬁnite number of cases. Clearly, the number of cases depends on the
16
1. The GregoryNewton Problem, Kepler’s Conjecture
choice of h. In dealing with the individual cases linear programming plays a very important role. L. Fejes T´ oth [9] suggested h = 2.0523 . . . in his program. Hsiang [2] took h = 2.18 in his approach.
1.4.2. Delone Stars and Hales’ Approach Let X be a set of points such that S3 + X is a saturated packing. Then E 3 can be decomposed into simplices such that their vertices belong to X and the interior of any sphere circumscribing such a simplex contains no point of X. This decomposition is called the Delone triangulation, the simplices are called Delone simplices, and the union of the simplices with a common vertex x is called a Delone star (it is denoted by D (x), or simply D ). To attack Kepler’s conjecture by studying Delone stars, Hales [4] took the following approach. For convenience, we write
1 11π − 12 arccos √ , β= 3 3 √ −3π + 12 arccos(1/ 3) √ , γ= 8 and F (D ) = −v(D )γ + v (D ∩ (S3 + x)) . x∈X
Let D (x) be a Delone star with vertices x1 , x2 , . . ., xl , and denote the intersection of xxi with bd(S3 ) + x by pi . Joining pi and pj by a geodesic arc if and only if max {x, xi , x, xj , xi , xj } ≤ 2.51, we get a network on bd(S3 )+x that divides the surface into a set of regions, say A1 , A2 , . . ., Am . Let Vi be the cone with vertex x over Ai , and write Ci = Vi ∩ D (x). The Delone star is divided into m clusters C1 , C2 , . . ., Cm . Assign a score σ(Ci ) to every cluster, and deﬁne
σ(D (x)) =
m
σ(Ci )
i=1
(the exact score for an individual cluster is based on its structure and is very complicated). σ(D (x)) is called the score of D (x). It has the following properties.
1.4. Kepler’s Conjecture
17
1. The score of a cluster depends only on the cluster, and not on the way it sits in a Delone star or in the Delone triangulation of the space. 2. If D is a Delone star of the facecentered cubic packing or the hexagonal close packing of S3 , then σ(D ) = 8β. 3. For any saturated packing S3 + X, we have σ(D (x)) = F (D (x)) + O r2 . x∈X∩rS3
(1.24)
x∈X∩rS3
Lemma 1.3 (Hales [4]). Let S3 + X be a saturated packing. If max {σ(D (x))} = η ≤ x∈X
then δ(S3 , X) ≤
16π , 3
16πγ . 16π − 3η
In particular, if η = 8β, then π δ(S3 , X) ≤ √ . 18 Proof. Since S3 + X is a saturated packing, d(D (x)) ≤ 4 holds for every point x ∈ X. Thus, the number of points x ∈ X such that D (x) meets the boundary of rS3 has order O(r2 ). For convenience, we write m(r) = card {X ∩ rS3 } . Since the Delone stars give a fourfold cover of E 3 , by (1.24) we have
4π 4 −v(rS3 )γ + m(r) = − v(D (x))γ 3 x∈X∩rS3
+ v(D (x) ∩ (S3 + y)) + O r2 y∈X
=
x∈X∩rS3
=
F (D (x)) + O r2 σ(D (x)) + O r2
x∈X∩rS3
≤ m(r)η + O r2 . In other words,
m(r)
4π η − 3 4
≤ v(rS3 )γ + O r2 .
18 Thus,
1. The GregoryNewton Problem, Kepler’s Conjecture
m(r)v(S3 ) 16πγ 1 ≤ +O . v(rS3 ) 16π − 3η r
The ﬁrst part of the lemma follows. Clearly, the second part is the special case η = 8β. Lemma 1.3 is proved. 2 In order to use this lemma to prove Kepler’s conjecture, Hales [4] proposed a program to verify the following assertions. Assertion 1.1. If all the regions are triangles, then σ(D ) ≤ 8β. Assertion 1.2. If the corresponding region of a cluster C has more than three sides, then σ(C ) ≤ 0. Assertion 1.3. If all the regions are triangles and quadrilaterals (excluding the case of pentagonal prisms), then σ(D ) ≤ 8β. Assertion 1.4. If one region has more than four sides, then σ(D ) ≤ 8β. Assertion 1.5. If D is a pentagonal prism, then σ(D ) ≤ 8β. Remark 1.4. A Delone star is called a pentagonal prism if its regions consist of ten triangles and ﬁve quadrilaterals, with the ﬁve quadrilaterals in a band around the equator, capped on both ends by ﬁve triangles. Proofs of these assertions were announced in Hales [4–7] and Ferguson [1], respectively. In fact, Hales [6, 7] and Ferguson [1] employed a diﬀerent decomposition and a diﬀerent score system formulated in Ferguson and Hales [1]. However, the basic idea is the same as Hales [4]. Their proofs are extraordinarily complicated. First a computer is used to list all the possibilities for the regions. Then optimization is used to deal with each case. In the course of this proof the computer plays a very important role.
1.5. Some General Remarks Kepler’s conjecture was made in 1611 based on the observation of S3 + Λ3 , while the GregoryNewton problem was proposed in 1694 during a famous discussion. Over the course of the centuries, these problems and their generalizations have attracted the attention of many mathematicians. This section reviews related results that will not be discussed in other chapters. As with many other problems in geometry, the ﬁrst signiﬁcant progress concerning packing densities was made in E 2 . In 1773, by studying the
1.5. Some General Remarks
19
minimum of positive binary quadratic forms (this method will be discussed in Chapter 2), Lagrange [1] deduced π δ ∗ (S2 ) = √ . 12
(1.25)
In 1831, Gauss [1] used similar methods to prove a conjecture of Seeber [1] that implies π δ ∗ (S3 ) = √ . (1.26) 18 Besides the facecentered cubic lattice packing, for which this density is attained, Barlow [1] found inﬁnitely many nonlattice packings of spheres with the same density. They are the laminations of hexagonal layers of spheres. Removing the lattice restriction, the ﬁrst proof of π δ(S2 ) = √ 12
(1.27)
was achieved by Thue [1] and [2]. Roughly speaking, Thue’s idea was to compute the area left uncovered by the circular disks in certain ﬁnite packings. His method was developed further by Groemer [1] in 1960. Later, diﬀerent proofs of (1.27) were discovered by L. Fejes T´ oth [1], Segre and Mahler [1], Davenport [3], and Hsiang [1]. The proof of Theorem 1.3 is essentially that of L. Fejes T´ oth. Clearly (1.25), (1.26), and (1.27) all support Kepler’s conjecture. In 1900, at the International Congress of Mathematicians in Paris, D. Hilbert (see [1]) listed this conjecture as the third part of his 18th problem. Thus it became one of the most popular problems among mathematicians. C.A. Rogers once commented, “Many mathematicians believe, and all physicists √ know” that the density of the densest sphere packings in E3 is π/ 18. In 1976, in a review of Hilbert’s 18th problem, Milnor [1] wrote on this conjecture, “This is a scandalous situation since the (presumably) correct answer has been known since the time of Gauss (compare Hilbert and CohnVossen). All that is missing is a proof.” In 1929, Blichfeldt [1] obtained the ﬁrst upper bound 0.835 for δ(S3 ). This has been successively improved by Rankin [1], Rogers [10], Lindsey [1], and Muder [1] and [2] to 0.773055. In 1953, in his wellknown book, L. Fejes T´ oth suggested a program to attack Kepler’s conjecture by checking a ﬁnite number of cases. In 1987, Dauenhauer and Zassenhaus [1] made a local approach and suggested the possibility of verifying this conjecture with the help of a computer. In 1993, Hsiang [2] announced a proof using a computer. Unfortunately, it has been found to contain errors (see Hales [3], Hsiang [3], and the preface of Conway and Sloane [1]). Recently, Hales [4–7] and Ferguson [1] announced another proof. To deal with the thousands of cases considered in their proof, even an eﬃcient computer has to run for years.
20
1. The GregoryNewton Problem, Kepler’s Conjecture
H. Minkowski was the ﬁrst mathematician to study the packings of general convex bodies, followed by E. Hlawka, C.A. Rogers, L. Fejes T´ oth, and many others. In 1904, Minkowski [5] developed a method by which one can determine the value of δ∗ (K) for any threedimensional convex body K. As examples he considered the cases of the octahedron and the tetrahedron. Unfortunately, he made a mistake in the tetrahedral case (see Groemer [3]). Later, this mistake was corrected by Hoylman [1]. Although Minkowski’s method is important in theory, it was considered not to be practical. However, recently, Betke and Henk [2] used Minkowski’s work as a starting point for an algorithm by which one can determine the value of δ∗ (K) for any threedimensional polytope. As an application, they calculated the value of δ ∗ (K) for all regular and Archimedean polytopes. Around 1950, L. Fejes T´ oth [5] and Rogers [4] proved δ ∗ (K) = δ(K) oth for an arbitrary convex domain in E 2 . Thirty years later, L. Fejes T´ [12] presented an elegant new proof of this result. In high dimensions, it is commonly believed (see Rogers [14], Gruber and Lekkerkerker [1]) that there exist convex bodies K such that δ ∗ (K) < δ(K). Perhaps some highdimensional spheres have this property. However, no example has been conﬁrmed as yet. As early as 1896, Minkowski [7] studied lattice packings of a general ndimensional convex body K, and proved that k ∗ (K) ≤ 3n − 1, where equality holds if K is a parallelepiped. In 1957 Hadwiger [1] gave an elegant proof (see the proof of Theorem 1.1) of k(K) ≤ 3n − 1, improving the above result. A few years later, Groemer [2] produced a more detailed proof. On the other hand, it was conjectured by Gr¨ unbaum [1] that the best lower bound for k(K) is n(n + 1), which can be attained by ndimensional simplices. In 1996, Zong [3] disproved the second part of this conjecture. Very recently, exponential lower bounds for k(K) were discovered by Talata [1] and Larman and Zong [1]. In the plane case, conﬁrming a conjecture of Hadwiger [1], Gr¨ unbaum [1] proved that 8 if K is a parallelogram, k(K) = k∗ (K) = (1.28) 6 otherwise. For a survey on the kissing numbers of general convex bodies we refer to Zong [7].
1.5. Some General Remarks
21
In the literature R. Hoppe has often been cited as the ﬁrst mathematician to solve the GregoryNewton problem. In fact, his proof is not complete. In 1953, using arguments from graph theory, Sch¨ utte and van der Waerden [1] completely solved this problem, in favor of Newton. In 1956, Leech [1] gave a new proof. Very recently, Leech’s proof was further improved by Aigner and Ziegler [1].
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2. Positive Deﬁnite Quadratic Forms and Lattice Sphere Packings
2.1. Introduction There is a remarkable relationship between lattice sphere packings and positive deﬁnite quadratic forms. This relationship plays an important role in determining the values of δ∗ (Sn ) and k∗ (Sn ) for small n. Let Λ be a lattice with a basis {a1 , a2 , . . . , an }, where ai = (ai1 , ai2 , . . . , ain ), and write a11 a12 · · · a1n a21 a22 · · · a2n A= . .. .. . .. .. . . . an1 an2 · · · ann For convenience, we denote the ndimensional integer lattice by Zn . There is a positive deﬁnite quadratic form related to the lattice Λ, Q(x) = xA, xA = xAA x , where A and x indicate the transposes of A and x, respectively. Let r(Λ) be the largest number such that r(Λ)Sn + Λ is a packing, let dis(Q) be the discriminant of the quadratic form Q(x), and write m(Q) =
min
z∈Zn \{o}
It is easy to see that r(Λ) =
1 2
Q(z).
m(Q)
24
2. Quadratic Forms, Lattice Sphere Packings
and dis(Q) = det(Λ)2 . Hence, δ(r(Λ)Sn , Λ) =
ωn r(Λ)n ωn m(Q)n/2 = . det(Λ) 2n dis(Q)
On the other hand, for each positive deﬁnite quadratic form Q(x) = xSx , where S is a positive deﬁnite symmetric matrix with entries sij , there is a corresponding lattice, Λ = {zA : z ∈ Zn }, such that S = AA . Therefore, there is also a corresponding lattice sphere packing r(Λ)Sn + Λ. Hence, letting F be the family of all positive deﬁnite quadratic forms in n variables, one has ωn m(Q)n/2 δ ∗ (Sn ) = sup . (2.1) Q(x)∈F 2n dis(Q) Similarly, letting M (Q) be the set of points z ∈ Zn \ {o} at which Q(z) attains its minimum, we have k ∗ (Sn ) = max {card{M (Q)}} . Q(x)∈F
(2.2)
Thus, the geometric problems of determining the values of δ∗ (Sn ) and k ∗ (Sn ) are reformulated as arithmetic ones. Write
m(Q) f (Q) = , n dis(Q)
and regard it as a function of the coeﬃcients of Q(x). Then Q(x) is called respectively a stable form or an absolutely stable form if f (Q) attains a local maximum or an absolute maximum at Q(x). If, in addition, m(Q) = 1, then Q(x) is called an extreme form or a critical form, respectively. Usually, the number m(Q) (2.3) γn = sup f (Q) = sup n dis(Q) Q(x)∈F Q(x)∈F is called Hermite’s constant. From (2.1) and (2.3) it follows that δ ∗ (Sn ) = ωn
γ n/2 n
4
.
(2.4)
Moreover, it is easy to see that the density of r(Λ)Sn + Λ is respectively a local maximum or an absolute maximum if and only if the corresponding form Q(x) is stable or absolutely stable. Therefore, the two problems, of determining the densest lattice sphere packings and determining the critical positive deﬁnite quadratic forms, are equivalent. Let U be a unimodular matrix and write Q(x) = xU SU x .
2.2. The LagrangeSeeberMinkowski Reduction
25
We say that Q(x) is equivalent to Q(x). Since the map z → zU is an automorphism in Zn , one has = m(Q), m(Q) = card{M (Q)}, (2.5) card{M (Q)} dis(Q) = det(U SU ) = dis(Q). Let Q be the subfamily of positive deﬁnite quadratic forms that are equivalent to Q(x). In other words, Q = {xU SU x : U is a unimodular matrix} . Then, the family F can be represented as a union of diﬀerent subfamilies Q. So, by (2.5), if in each subfamily Q a particular form can be chosen, the problem of determining the values of δ∗ (Sn ), k∗ (Sn ), and γn using (2.1), (2.2), and (2.3) can be simpliﬁed. This is the basic idea of reduction theory. In this chapter the values of δ∗ (Sn ), k∗ (Sn ), and γn for small n will be determined using (2.1), (2.2), and (2.3) together with the assistance of diﬀerent reductions.
2.2. The LagrangeSeeberMinkowski Reduction and a Theorem of Gauss Deﬁnition 2.1. As usual, we denote the greatest common divisor of k integers z1 , z2 , . . . , zk by (z1 , z2 , . . . , zk ) (here we use bold parentheses). A positive deﬁnite quadratic form Q(x) = xSx is said to be LSM reduced if s1j ≥ 0,
j = 2, 3, . . . , n,
and Q(z) ≥ sii
(2.6)
for all integer vectors z = (z1 , z2 , . . . , zn ) such that (zi , zi+1 , . . . , zn ) = 1,
i = 1, 2, . . . , n.
Lemma 2.1 (Minkowski [6]). Every positive deﬁnite quadratic form is equivalent to an LSM reduced one. Proof. Let Q(x) = xSx be a positive deﬁnite quadratic form. Using basic algebra we know that there exists a nonsingular matrix A such that S = AA . Denote the lattice {zA : z ∈ Zn } by Λ. Then Q(z)1/2 is the Euclidean norm of the vector zA in Λ. Hence, Lemma 2.1 can be proved by showing that the following assertion holds.
26
2. Quadratic Forms, Lattice Sphere Packings
Assertion 2.1. Every lattice Λ has a basis {a1 , a2 , . . . , an } such that n j=1
zj aj ,
n
≥ ai , ai
zj aj
(2.7)
j=1
for all integer vectors z = (z1 , z2 , . . . , zn ) with (zi , zi+1 , . . . , zn ) = 1,
i = 1, 2, . . . , n,
and a1 , aj ≥ 0,
j = 2, 3, . . . , n.
(2.8)
We now prove this assertion. First, we take a1 to be a point of Λ satisfying a1 , a1 = min a, a. a∈Λ\{o}
Let B2 be the set of points a ∈ Λ such that {a1 , a} can be extended to a basis for Λ. From linear algebra we know that the set B2 is nonempty. Then, take a2 to be a point of B2 satisfying a1 , a2 ≥ 0 and a2 , a2 = min a, a. a∈B2
Inductively, assume that i < n and a1 , a2 , . . . , ai have been chosen. Let Bi+1 be the set of points a ∈ Λ such that {a1 , a2 , . . . , ai , a} can be completed to a basis for Λ. Note that again we know that Bi+1 is nonempty. Take ai+1 to be a point of Bi+1 satisfying a1 , ai+1 ≥ 0 and ai+1 , ai+1 = min a, a. a∈Bi+1
(2.9)
It is easy
n to see that {a1 , a2 , . . . , an } is a basis of Λ that satisﬁes (2.8). Let a = j=1 zj aj be a point of Λ such that (zi , zi+1 , . . . , zn ) = 1. From linear algebra we know that {a1 , a2 , . . . , ai , a} can be extended to a basis for Λ. In other words, a ∈ Bi+1 . Then, (2.7) follows from (2.9). Therefore, Assertion 2.1 and Lemma 2.1 are proved. 2 In fact, by a theorem of Minkowski [6], in order to verify that a positive deﬁnite quadratic form is LSM reduced one only need check that (2.6) holds for a ﬁnite number of integer vectors. Here we introduce two practical criteria for the LSM reduced binary forms and the LSM reduced ternary forms. Lemma 2.2 (Lagrange [1]). A positive deﬁnite binary quadratic form Q(x) = xSx is LSM reduced if and only if s11 ≤ s22 , (2.10) 0 ≤ 2s12 ≤ s11 .
2.2. The LagrangeSeeberMinkowski Reduction
27
Proof. Let Q(x) be a form satisfying (2.10). Then, for z = (z1 , z2 ) ∈ Z2 , Q(z) = s11 (z1 )2 + 2s12 z1 z2 + s22 (z2 )2 ≥ s11 (z1 )2 − s11 z1 z2  + s11 (z2 )2 + (s22 − s11 )(z2 )2 ! = s11 (z1 )2 − z1 z2  + (z2 )2 + (s22 − s11 )(z2 )2 . Clearly, for z = o,
(z1 )2 − z1 z2  + (z2 )2 ≥ 1.
Therefore, since s22 ≥ s11 , we have s11 , if z = o, Q(z) ≥ s22 , if z2 = 0. Then, it follows from Deﬁnition 2.1 that Q(x) is LSM reduced. On the other hand, if Q(x) is LSM reduced, the inequalities of (2.10) follow from Q(1, 0) ≤ Q(0, 1) and Q(0, 1) ≤ Q(1, −1). Hence, Lemma 2.2 is proved. 2 Remark 2.1. Let Q(x) be an LSM reduced positive deﬁnite binary quadratic form. By (2.10), we have dis(Q) s11 s22 − (s12 )2 s11 s22 − (s11 )2 /4 = ≥ s11 s22 s11 s22 s11 s22 s11 3 ≥ 1− ≥ . 4s22 4 Hence, m(Q) = s11 ≤
√
s11 s22 ≤
"
4 3 dis(Q).
Then, by (2.1) and certain example it follows that π δ ∗ (S2 ) = √ . 12 Thus we obtain an arithmetic proof for the lattice case of Theorem 1.3. Examining the above argument it is easy to see that every critical positive deﬁnite binary quadratic form is equivalent to Q(x) = (x1 )2 + x1 x2 + (x2 )2 , which implies the uniqueness of the densest lattice circle packing. This idea goes back to J.L. Lagrange. Lemma 2.3 (Seeber [1]). A positive deﬁnite ternary quadratic form Q(x) = xSx is LSM reduced if and only if s11 ≤ s22 ≤ s33 , 0 ≤ 2s12 ≤ s11 , 0 ≤ 2s13 ≤ s11 , (2.11) 0 ≤ 2s  ≤ s , 23 22 −2s23 ≤ s11 + s22 − 2(s12 + s13 ).
28
2. Quadratic Forms, Lattice Sphere Packings
Proof. For convenience, let (2.11.i) indicate the ith inequality of (2.11). Assume that Q(x) satisﬁes (2.11), we proceed to show that it is LSM reduced. In other words, s11 , if (z1 , z2 , z3 )= 1, s22 , if (z2 , z3 )= 1, (2.12) Q(z) ≥ s33 , if z3 = 0. If z3 = 0, then by (2.11.1) and (2.11.2), Q(z1 , z2 , 0) is LSM reduced. Then, the ﬁrst two inequalities of (2.12) follow from Lemma 2.2. Now we proceed to show the last inequality of (2.12) by dealing with two cases. Case 1. z3 = 0 and z1 = 0. Deﬁne θ = 1 if s23 ≥ 0, and θ = −1 if s23 < 0. From (2.11.1) and (2.11.4) it follows that Q(0, z2 , θz3 ) is LSM reduced. Thus, by Lemma 2.2, Q(0, z2 , z3 ) ≥ s33 . Similarly, it can be proved that Q(z1 , 0, z3 ) ≥ s33 . Case 2. z1 z2 z3 = 0. Routine computation yields that Q(z) = (s11 − s12 − s13 )(z1 )2 + (s22 − s12 − s23 )(z2 )2 + (s33 − s13 − s23 )(z3 )2 + Σ,
(2.13)
where Σ = s12 (z1 + z2 )2 + s13 (z1 + z3 )2 + s23 (z2 + θz3 )2 . From the ﬁrst four inequalities of (2.11) it follows that all the coeﬃcients of (2.13) and Σ are nonnegative. Assume that s12 ≤ s13 ≤ s23 . We consider three subcases. i. At least two of the three terms z1 + z2 , z1 + z3 , and z2 + θz3 are nonzero. Then Σ ≥ 2s12 , and therefore, by (2.13), (2.11.3), and (2.11.4), Q(z) ≥ s11 + s22 + s33 − 2(s12 + s13 + s23 ) + 2s12 = (s11 − 2s13 ) + (s22 − 2s23 ) + s33 ≥ s33 . ii. Exactly two of the three terms z1 + z2 , z1 + z3 , and z2 + θz3 are zero. Then, the third term must be a multiple of two, and therefore Σ ≥ 2s12 . Similarly to the previous subcase we have Q(z) ≥ s33 . iii. z1 + z2 = z1 + z3 = z2 + θz3 = 0. Then θ = −1 and s23 < 0. Thus, by (2.13) and (2.11.5), Q(z) = s11 + s22 + s33 − 2(s12 + s13 − s23 ) = s33 + 2s23 + [s11 + s22 − 2(s12 + s13 )] ≥ s33 . On the other hand, if Q(x) is an LSM reduced ternary form, then (2.11) follows from the inequalities Q(1, 0, 0) ≤ Q(0, 1, 0) ≤ Q(0, 0, 1),
2.2. The LagrangeSeeberMinkowski Reduction
29
Q(0, 1, 0) ≤ Q(1, −1, 0), Q(0, 0, 1) ≤ Q(1, 0, −1), Q(0, 0, 1) ≤ Q(0, 1, ±1), and Q(0, 0, 1) ≤ Q(−1, 1, 1). Hence, Lemma 2.3 is proved. 2 Theorem 2.1 (Gauss [1]). π δ∗ (S3 ) = √ . 18
(2.14)
Furthermore, the densest lattice packing of S3 is unique up to rotation and reﬂection. Proof. By (2.1), (2.5), Lemma 2.1, and Lemma 2.3, in order to prove (2.14) it is suﬃcient to show that dis(Q) ≥ 12 s11 s22 s33
(2.15)
for every positive deﬁnite ternary quadratic form Q(x) = xSx that satisﬁes (2.11). It is wellknown that dis(Q) = s11 s22 s33 + 2s12 s13 s23 − s11 (s23 )2 − s22 (s13 )2 − s33 (s12 )2 . (2.16) Fixing s11 , s22 , and s33 , and regarding dis(Q) = f (s12 , s13 , s23 ) as a function of s12 , s13 , and s23 , we proceed to prove (2.15) by dealing with two cases. Case 1. s23 < 0. Since dis(Q) is a convex function of s23 , its minimum, under the constraints of (2.11), is attained at s23 = min s12 + s13 − 12 (s11 + s22 ), − 12 s22 . Thus, this case can be dealt with in two subcases. i. s23 = s12 + s13 − (s11 + s22 )/2. Then, it follows from (2.16) that dis(Q) = a1 (s13 )2 + b1 s13 + c1 ,
(2.17)
where a1 , b1 , and c1 are independent of s13 , and a1 = −[(s11 − 2s12 ) + s22 ] < 0. In addition, by (2.11.4) and the assumption of this subcase, s11 − 2s12 ≤ 2s13 .
(2.18)
By (2.11.3) the minimal value of (2.17) is attained either at s13 = 0 or at s13 = s11 /2. When s13 = 0, it follows from (2.11.2), (2.18), and the
30
2. Quadratic Forms, Lattice Sphere Packings
assumption of this subcase that s12 = s11 /2 and s23 = −s22 /2, which will be dealt with in the following subcase. When s13 = s11 /2 we have dis(Q) = a2 (s12 )2 + b2 s12 + c2 ,
(2.19)
where a2 , b2 , and c2 are independent of s12 , and a2 = −s33 < 0. By (2.11.2) the minimal value of (2.19) is attained either at s12 = 0 or at s12 = s11 /2. Then, since s11 ≤ s22 ≤ s33 , routine computation yields s11 s22 s33 1 − s11 − s22 ≥ 1 s11 s22 s33 , if s12 = 0, 4s33 4s33 2 dis(Q) = s11 s22 s33 1 − s11 − s22 ≥ 1 s11 s22 s33 , if s12 = 1 s11 . 4s22 4s33 2 2 ii. s23 = −s22 /2. In this case, it follows from (2.16) that dis(Q) = s11 s22 s33 − s22 s12 s13 − 14 s11 (s22 )2 − s22 (s13 )2 − s33 (s12 )2 . (2.20) Also, by (2.11.5), 2s13 ≤ s11 − 2s12 .
(2.21)
We calculate the derivative ∂dis(Q) = −s22 (s12 + 2s13 ) ≤ 0. ∂s13 Hence, in order to minimize the value of dis(Q) we shall choose the maximum possible value for s13 under the conditions 2s13 ≤ s11 and (2.21), namely s13 = s11 /2 − s12 . Then, it follows from (2.20) that dis(Q) = a3 (s12 )2 + b3 s12 + c3 , where a3 , b3 , and c3 are independent of s12 , and a3 = −s33 < 0. Thus by a similar argument to the previous subcase we obtain dis(Q) ≥ 12 s11 s22 s33 . Case 2. s23 ≥ 0. Then, the last condition in (2.11) follows from the others. Arguing as before, it follows that the minimal value of dis(Q) will be attained either at s23 = 0 or at s23 = s22 /2. Hence, by (2.16), dis(Q) ≥ min{f (s12 , s13 , 0), f (s12 , s13 , s22 /2)} s s 11 22 = s11 s22 s33 − s22 − s12 s13 4 − s22 (s13 )2 − s33 (s12 )2 . Considering the righthand side of this inequality as a function of s12 and s13 , and denoting it by g(s12 , s13 ), simple analysis yields dis(Q) ≥
min
0≤s12 ≤s11 /2, 0≤s13 ≤s22 /2
g(s12 , s13 ) = g(s11 /2, s22 /2)
s11 s22 = s11 s22 s33 1 − − 4s22 4s33 1 ≥ 2 s11 s22 s33 .
2.3. Mordell’s Inequality on Hermite’s Constants
31
Hence (2.15) follows from Case 1 and Case 2. Also, tracing back through the above argument, it can be veriﬁed that every critical positive deﬁnite ternary quadratic form is equivalent to Q(x) = (x1 )2 + (x2 )2 + (x3 )2 + x1 x2 + x1 x3 + x2 x3 , which implies the second part of our theorem. Theorem 2.1 is proved.
2
2.3. Mordell’s Inequality on Hermite’s Constants and a Theorem of Korkin and Zolotarev Lemma 2.4 (Mordell [1]). n−1
γn ≤ (γn−1 ) n−2 . Proof. Let Xi =
n
aij xj ,
i = 1, 2, . . . , n,
(2.22)
j=1
be n linear forms with real coeﬃcients and determinant 1, and let bij be the cofactor of aij of A, the n × n matrix with entries aij . Deﬁne n linear forms Y1 , Y2 , . . . , Yn in variables y1 , y2 , . . . , yn by means of the identity n
Xi Yi =
i=1
For example, Yi =
n
n
xi yi .
(2.23)
i=1
bij yj ,
i = 1, 2, . . . , n.
j=1
Obviously, the n linear forms Yi also have determinant 1. Letting Q(x) = xSx = xAA x be a positive deﬁnite quadratic form with dis(Q) = 1, we can write 2
2
Q(x) = (X1 ) + (X2 ) + · · · + (Xn )
2
with appropriate linear forms Xi satisfying (2.22). Then let Q∗ (y) = (Y1 ) + (Y2 ) + · · · + (Yn ) 2
2
2
(2.24)
32
2. Quadratic Forms, Lattice Sphere Packings
be the form associated with Q(x) using (2.23). By applying a unimodular substitution in (2.24) and changing Q∗ (y) into an equivalent form, we may assume that the minimum m(Q∗ ) is given by (y1 , y2 , . . . , yn ) = (0, 0, . . . , 1). In other words, m(Q∗ ) =
n
2
(bin ) .
(2.25)
i=1
In this case we must then also
n apply a corresponding unimodular substitution to the xi s such that i=1 xi yi remains unaltered, and then Q(x) is replaced by an equivalent form. Now we estimate the values of Q(z) for zn = 0. In this case, Q(z) becomes a quadratic form in n − 1 variables, namely
2 n n−1 Q (z) = aij zj . i=1
j=1
For convenience, we denote by S the (n − 1) × (n − 1) matrix with entries sij =
n
aki akj
k=1
for 1 ≤ i ≤ n − 1 and 1 ≤ j ≤ n − 1, and denote by M the n × (n − 1) matrix with entries aij for 1 ≤ i ≤ n and 1 ≤ j ≤ n − 1. Then, by (2.25), we have dis(Q ) = det(S ) = det(M M ) =
n
(bin )2
i=1
= m(Q∗ ) ≤ γn , and therefore 1
1
m(Q) ≤ m(Q ) ≤ γn−1 dis(Q ) n−1 ≤ γn−1 (γn ) n−1 . Since the righthand side is independent of the form Q(x), it follows from (2.3) that 1 γn ≤ γn−1 (γn ) n−1 and consequently n−1
γn ≤ (γn−1 ) n−2 . Lemma 2.4 is proved.
2
From (2.4), Lemma 2.4, Theorem 2.1, and their proofs it is easy to deduce the density of the densest lattice sphere packings in E4 , which was determined by Korkin and Zolotarev by a much more complicated method.
2.4. Perfect Forms, Voronoi’s Method
33
Theorem 2.2 (Korkin and Zolotarev [1]). δ ∗ (S4 ) =
π2 . 16
Also, the densest lattice packing of S4 is unique up to rotation and reﬂection.
2.4. Perfect Forms, Voronoi’s Method, and a Theorem of Korkin and Zolotarev To determine the value of δ∗ (Sn ) by studying extreme forms, we introduce the concept of a perfect form. A positive deﬁnite quadratic form Q(x) is called perfect if it is determined uniquely by the equations Q(zi ) = m(Q). Extreme forms and perfect forms have the following important relationship. Lemma 2.5 (Korkin and Zolotarev [2]). Every extreme positive deﬁnite quadratic form is perfect. Proof. Let Q(x) = xSx be an extreme positive deﬁnite quadratic form in n variables, and assume that M (Q) = {±z1 , ±z2 , . . . , ±zk } . If, on the contrary, Q(x) is not perfect, there is a nonzero symmetric matrix S ∗ such that zi S ∗ zi = 0, i = 1, 2, . . . , k. Deﬁne
Qλ (x) = x(S + λS∗ )x .
It is easy to see that for a suitable positive number α, Qλ (x) is positive deﬁnite and m(Qλ ) = m(Q) = 1 (2.26) whenever λ ≤ α. It is wellknown from linear algebra that there is an orthogonal matrix U such that both U SU and U S ∗ U are diagonal, with diagonal entries si and s∗i , respectively. Hence, dis(Qλ ) =
n # (si + λs∗i ). i=1
Writing f (λ) = log dis(Qλ ),
34
2. Quadratic Forms, Lattice Sphere Packings
simple analysis yields that f (λ) is a strictly concave function of λ. Therefore, one has (2.27) dis(Qλ ) < dis(Q) for λ > 0, or for λ < 0. Then, (2.26) and (2.27) together yield, m(Q) m(Qλ ) > , n n dis(Qλ ) dis(Q) which contradicts the assumption that Q(x) is extreme. Lemma 2.5 is proved. 2 Using Lemma 2.5, we can determine all the extreme forms if we can determine all the perfect forms. For this reason, since a perfect form can be determined by its minimum integer points, Voronoi [1] developed the following method. The method is very elegant. However, its proof is very complicated. So, here we only introduce the method itself. Voronoi’s Method. For convenience, we write d=
n(n + 1) . 2
Let Q1 (x) be a perfect form in n variables with m(Q1 ) = 1 and assume that M (Q1 ) = {±z1 , ±z2 , . . . , ±zk } , where zi = (zi1 , zi2 , . . . , zin ). Then, corresponding to zi and Q1 (x), respectively, in E d we introduce a point z∗i = (zi1 zi1 , zi1 zi2 , . . . , zi1 zin , zi2 zi2 , zi2 zi3 , . . . , zi2 zin , . . . , zin zin ) and a polyhedral convex cone V1 =
k
αi z∗i : αi ≥ 0 .
i=1
Assume that V1 has l1 − 1 facets Fi , i = 2, 3, . . . , l1 , which are determined by o∗ , z∗i1 , z∗i2 , . . . , z∗if (i) . Let Q∗i (x) be nonzero quadratic forms in n variables such that Q∗i (zij ) = 0, Deﬁning λi =
j = 1, 2, . . . , f (i).
Q1 (z) − 1 z∈Zn , Qi (z) 0. Then, averaging the inner sum on the righthand side with respect to (y1 , y2 , . . . , yn−1 ) and substituting kyi = xi , we have %
%
1
1
··· 0
0
z∈Λ∗ % k
1
=
0
%
%
1
··· 0
%
%
···
= −∞
z∈Λ∗
%
1
0
0 ∞
f (z + x) dx1 · · · dxn−1
···
z ∈Λ∗ 0≤z ≤k−1 i
1
%
k n−1
k
0
1
=
%
···
k n−1
=
f (z + ky) dy1 · · · dyn−1
0
1
f (z + z + x) dx1 · · · dxn−1
z∈Λ∗
f (z + x) dx1 · · · dxn−1
z∈Λ∗ ∞
−∞
f (x)dx1 · · · dxn−1 = I(kα).
Therefore, %
%
1
··· 0
0
1
f (u) dy1 · · · dyn−1 =
u∈Λ, un =0
I(kα).
k∈Z\{0}
Then there is a point y such that (3.1) holds with det(Λ∗ ) = 1, which proves Lemma 3.1. 2 Lemma 3.2 (Hlawka [1]). Let f (x) be a bounded Riemann integrable function vanishing outside a bounded region, and let be a positive number. Then, there exists a lattice Λ with determinant 1 such that % f (u) < f (x)dx + . u∈Λ\{o}
En
3.1. The MinkowskiHlawka Theorem
49
Proof. Without loss of generality, we may assume that f (x) is continuous. Let α and β be positive numbers such that β = α1/(1−n) , and denote by Λ∗ the lattice βZn−1 , in the subspace xn = 0. Then det(Λ∗ ) = α−1 .
(3.2)
Since f (x) = 0, it follows that if x is outside a bounded region, f (u) = 0
(3.3)
holds for every point u ∈ Λ∗ \ {o} when α is suﬃciently small. Also, by the deﬁnition of I(λ) in the previous lemma it follows that % lim αI(kα) = f (x)dx. (3.4) α→0
En
k∈Z\{0}
Thus, by (3.2), (3.3), (3.4), and Lemma 3.1, for any > 0 there exist a suitable number α and a suitable point y = (y1 , . . . , yn−1 , α) such that the corresponding lattice, Λ = {zy + Λ∗ : z ∈ Z} , satisﬁes both det(Λ) = 1 and f (u) =
f (u) ≤
u∈Λ, un =0
u∈Λ\{o}
%
αI(kα)
k∈Z\{0}
f (x)dx + .
< En
2
Lemma 3.2 is proved. Proof of Theorem 3.1. It is suﬃcient to show that if v(C) < 2ζ(n),
(3.5)
then there is a lattice Λ with determinant 1 such that 12 C + Λ is a packing. In other words, C ∩ Λ = {o}. (3.6) Let χ(x) be the characteristic function of C and deﬁne f (x) =
∞ k=1
µ(k)χ(kx),
(3.7)
3. Lower Bounds for δ ∗ (Sn ) and δ(Sn )
50
where µ(k) is the M¨ obius function. It is well known that ∞ µ(k) k=1
and
kn
µ(k) =
km
1 ζ(n)
(3.8)
1 if m = 1, 0 otherwise.
(3.9)
=
It follows from (3.7), (3.8), and (3.5) that % f (x)dx = En
=
∞ k=1 ∞ k=1
=
% µ(k)
χ(kx)dx En
µ(k) v(C) kn
v(C) < 2. ζ(n)
(3.10)
Furthermore, denoting the set of primitive points of Λ by Λ , it follows from (3.7) and (3.9) that
f (u) =
∞
f (lu)
l=1 u∈Λ
u∈Λ\{o}
= = =
∞ ∞
µ(k)χ(klu)
u∈Λ l=1 k=1 ∞
u∈Λ m=1
km
χ(mu)
µ(k)
χ(u).
(3.11)
u∈Λ
Then, by Lemma 3.2, (3.10), and (3.11) there is a suitable lattice Λ with determinant 1 such that % χ(u) = f (u) < f (x)dx < 2. u∈Λ
u∈Λ\{o}
En
This implies (3.6), and hence Theorem 3.1 is proved.
2
Remark 3.1. The MinkowskiHlawka theorem is one of the most important results in the geometry of numbers. In the past ﬁve decades, diﬀerent proofs and improvements were achieved by Bateman [1], Cassels [1], Davenport and Rogers [1], Lekkerkerker [1], Macbeath and Rogers [1], [2], [3], Mahler
3.2. Siegel’s Mean Value Formula
51
[1], Maly˘ sev [1], Rogers [6], [7], [9], Sanov [1], Schmidt [1], [4], Schneider [1], Siegel [1], Weil [1], and others. However, none were able to improve the asymptotic order of the lower bound.
3.2. Siegel’s Mean Value Formula A positive deﬁnite quadratic form Q(x) = xSx is called properly reduced if sij ≥ 0 for all index pairs j = i + 1 and if Q(z) ≥ sii whenever (zi , zi+1 , . . . , zn ) = 1. As in the LSM reduced case, it can be proved that every positive deﬁnite quadratic form is equivalent to a properly reduced one. Let A be the family of n × n matrices A with determinant 1, and let A1 be the subfamily such that Q(x) = xAA x is properly reduced and the trace of A is nonnegative. Then, denote by P and P1 the sets of points (a11 , a12 , . . . , a1n , a21 , a22 , . . . , a2n , . . . , ann ) 2
in E n such that A ∈ A and A ∈ A1 , respectively. It follows from the deﬁnition of the properly reduced form that A ∈ rint(P1 ) if and only if the corresponding form Q(x) = xAA x = xSx satisﬁes both sij > 0 for j = i + 1 and Q(z) > sii for z ∈ Zn such that (zi , zi+1 , . . . , zn ) = 1 and (zi , zi+1 , . . . , zn ) = (±1, 0, . . . , 0). Thus, routine argument yields that P1 is a fundamental domain of P. In other words, let P1 (U ) be the subset of P corresponding to {AU : A ∈ A1 } and let U be the family of all n × n unimodular matrices U with determinant 1. Then P1 (U ) U ∈U
is a tiling in P. For any Jordan measurable subset J of P, let J be the 2 cone in E n with base J and vertex o. Then, in P we deﬁne a measure σ by σ(J ) = v(J ). Let f (x) be a bounded Riemann integrable function vanishing outside a bounded region and, for convenience, write % F = f (x)dx. En
3. Lower Bounds for δ ∗ (Sn ) and δ(Sn )
52 Also, for any matrix A we write
Φ(A) =
f (zA ).
z∈Zn \{o}
With these deﬁnitions and notations, in 1945 C.L. Siegel proved the following mean value formula. Lemma 3.3 (Siegel [1]). n 1 # ζ(k) σ(P1 ) = n k=2
%
and
P1
Φ(A)dσ = σ(P1 )F.
Proof. Let I be the (n − 1)dimensional cube {y = (y2 , y3 , . . . , yn ) : 0 ≤ yi < 1}, and let X be a bounded Lebesgue measurable set in En such that v(X) = F . For convenience, we denote the analogues of A, A1 , P, P1 , and σ for the (n − 1) × (n − 1) matrices by A∗ , A∗1 , P ∗ , P1∗ , and σ∗ , respectively. In addition, we assume that x1 = 0 for any point x ∈ X. Let N be the family of the matrices x1 0 · · · 0 x2 A= . .. D xn
1 y2 0 .. .
··· A∗
yn
,
0
where x ∈ X, y ∈ I, A∗ ∈ A∗1 , and D is a diagonal matrix with diagonal elements x1 −1/(n−1) , . . . , x1 −1/(n−1) , x1 −1/(n−1) sign x1 , and let G be 2 its corresponding set of points in E n . Clearly, G is a subset of P. Let χ(x) be the characteristic function of X and let X (A) be the characteristic function of G. Then, % σ(G) = X (A)dσ. P
By routine computation it follows that
% n−1 ∗ ∗ σ (P1 ) χ(x)dx n En n−1 ∗ ∗ = σ (P1 )F. (3.12) n Next, let z be a primitive point of Zn and let U(z) be the set of all unimodular matrices with determinant 1 having ﬁrst column z . For any matrix A∈ rint(P1 (U )), σ(G) =
U ∈U
3.2. Siegel’s Mean Value Formula
53
by the deﬁnitions of I and P1∗ and routine argument it follows that if zA ∈ X, then AU ∈ N for exactly one U ∈ U(z); if zA ∈ X, then AU ∈ N for any U ∈ U(z). Therefore, one has X (AU ) = χ(zA ), (3.13) z∈Z
U ∈U
where Z indicates the set of primitive points of Zn . Since P1 is a fundamental domain in P with respect to U and σ is invariant under the transformation A → AU for any ﬁxed U ∈ U, we have % % σ(G) = X (A)dσ = X (AU )dσ. P
P1 U ∈U
Then, subsitituting (3.12) and (3.13) in this formula, we obtain % n−1 ∗ ∗ χ(zA )dσ = σ (P1 )F. n P1 z∈Z
Since χ(kx) is the characteristic function of k1 X, we have %
χ(zA )dσ =
P1 z∈Z \{o} n
∞ % k=1
= and therefore % n P1
χ( zA )dσ =
z∈Zn \{o}
χ(kzA )dσ
P1 z∈Z
n−1 ζ(n)σ∗ (P1∗ )F n n−1 ζ(n)σ∗ (P1∗ )F n
(3.14)
(3.15)
for every > 0. For any lattice Λ with determinant 1 and any point x, card {Λ ∩ (In + x)} ≤ card {Λ ∩ 2In } . Hence, by approximation it follows that χ( zA ) ≤ c −n
z∈Zn \{o}
z∈Zn \{o}
χ (zA ),
where c is a constant depending only on X and n, and χ (x) is the characteristic function of the unit cube In . Clearly, lim n χ( zA ) = v(X) = F. →0
z∈Zn \{o}
3. Lower Bounds for δ ∗ (Sn ) and δ(Sn )
54
In addition, c χ (zA ) is integrable over P1 , on account of (3.14) with χ replaced by χ . Hence, letting → 0 in (3.15) and applying Lebesgue’s dominated convergence theorem, we have σ(P1 )F =
n−1 ζ(n)σ∗ (P1∗ )F n
and therefore σ(P1 ) =
(3.16)
n 1 # ζ(k), n k=2
which proves the ﬁrst assertion of our lemma. In addition, it follows from (3.14) and (3.16) that % χ(zA )dσ = σ(P1 )F. P1 z∈Z \{o} n
By a suitable approximation process, one ﬁnds that this formula remains true if χ(x) is replaced by an arbitrary integrable function f (x), which proves the second assertion of our lemma. Lemma 3.3 is proved. 2 Remark 3.2. Siegel’s mean value formula is an improvement of Lemma 3.2. So, one can deduce the MinkowskiHlawka theorem from it. Remark 3.3. Let f (x1 , x2 , . . . , xk ) be a nonnegative Borel measurable function deﬁned in the space Ekn that vanishes outside a bounded region, and for any n × n matrix A write f (z1 A, z2 A, . . . , zk A). Ψ(A) = zi ∈Zn
By studying the integral
% Ψ(A)dσ, P1
Rogers and Schmidt have successively improved the MinkowskiHlawka theorem to n log 2 + β , δ ∗ (C) ≥ 2n where β is a suitable constant. So far, this is the best known lower bound of this type. Remark 3.4. For the most interesting case C = Sn , the best known lower bound (n − 1)ζ(n) δ ∗ (Sn ) ≥ 2n−1 is due to Ball [2].
3.3. Sphere Coverings
55
3.3. Sphere Coverings and the CoxeterFewRogers Lower Bound for δ(Sn ) Let Y be a discrete set of points in En . We will call Sn + Y a sphere covering of En if En = (Sn + y). y∈Y
In addition, if Y is a lattice, it will be called a lattice sphere covering of E n . Let Y be the family of sets Y such that Sn + Y is a covering of E n , and let L be the family of lattices Λ such that Sn + Λ is a lattice covering of E n . Then we deﬁne θ(Sn ) = lim inf l→∞ Y ∈Y
and θ∗ (Sn ) = lim inf l→∞ Λ∈L
card{lIn ∩ Y }ωn ln card{lIn ∩ Λ}ωn . ln
Usually, θ(Sn ) and θ∗ (Sn ) are called the density of the thinnest sphere coverings of E n and the density of the thinnest lattice sphere coverings of E n , respectively. It is easy to see that as with the lattice packing density, θ∗ (Sn ) = lim inf Λ∈L
ωn . det(Λ)
Let Sn + X be a sphere packing in E n with the maximal density δ(Sn ). Without loss of generality, we assume that for any point y ∈ En , int(Sn + y) ∩ (Sn + X) = ∅. Then, 2Sn + X is a sphere covering in E n . By the deﬁnitions of δ(Sn ) and θ(Sn ), one has card{lIn ∩ X}v(2Sn ) ln l→∞ ≥ θ(2Sn ) = θ(Sn )
2n δ(Sn ) = lim sup
and therefore
θ(Sn ) . (3.17) 2n Besides the interest in θ(Sn ) itself, by (3.17) a lower bound for θ(Sn ) will imply a lower bound for δ(Sn ). So, in this section we introduce both an upper bound and a lower bound for θ(Sn ). δ(Sn ) ≥
Theorem 3.2 (Rogers [8]). When n ≥ 3, θ(Sn ) < n log n + n log log n + 5n.
3. Lower Bounds for δ ∗ (Sn ) and δ(Sn )
56
Proof. For convenience, in this proof let Sn be the ndimensional sphere of volume 1. Let l be a large positive number and let m be a large positive integer. Also, write I = {(x1 , x2 , . . . , xn ) : 0 ≤ xi < l} and Λ = lZn . It is easy to see that Sn + Λ is a packing. Let X = {x1 , x2 , . . . , xm } be a subset of I. We will study the system Sn + X + Λ.
(3.18)
Let χ(x) be the characteristic function of Sn . Then the characteristic function of the set E = E n \ (Sn + X + Λ) is
m # χ(x − xi − u) . χ (x) = 1− i=1
u∈Λ
Since this function is periodic in each coordinate with period l, E is a set with asymptotic density equal to v(I ∩ E )/v(I), where % v(I ∩ E ) = χ (x)dx. I
Let M be the mean value of v(I ∩ E ), taken over all distributions of x1 , x2 , . . . , xm throughout the cube I. Then
% % % % 1 M = mn ··· χ (x) dx dx1 dx2 · · · dxm l I I I I
' % &# m % 1 = mn 1− χ(x − xi − u) dxi dx l I i=1 I u∈Λ
' % &# m % 1 = mn ln − χ(−y)dy dx l I i=1 u∈Λ I−x+u
m % 1 = mn ln ln − χ(−y)dy l En
m 1 = ln 1 − n . l Hence, there is a set X such that the asymptotic density of the corresponding E does not exceed (1 − l−n )m . Let η be any number satisfying 0 < η ≤ 1, and let m be the maximal cardinality of the set Y = {y1 , y2 , . . . , ym } such that ηSn + Y + Λ is a packing in E . Then, one has
m 1 m η n ≤ ln 1 − n l
(3.19)
3.3. Sphere Coverings
57
and therefore m ≤
m ln 1 . 1 − ηn ln
(3.20)
Since m is maximal and 0 < η ≤ 1, it follows from the choice of the two systems (3.18) and (3.19) that (1 + η)Sn + X ∪ Y + Λ is a covering of E n . The density of this covering is (1 + η)n (m + m ) . ln Therefore, by (3.20) and the deﬁnition of θ(Sn ), &
m ' m 1 1 . θ(Sn ) ≤ min (1 + η)n n + n 1 − n l, m, η l η l Then, taking m 1 = n log(1/η) and η = n l n log n successively, one has (1 + η)n n log(1/η) + η −n e−n log(1/η) 0 k(K2 ).
4.3. A Theorem of SwinnertonDyer Despite the regularity of lattices, our knowledge of the asymptotic behavior of k ∗ (Sn ), when n is large, is very limited. In this section we discuss a simple result concerning k∗ (K) and the densest lattice packings of a general convex body K in E n . For this purpose we denote by k(K, Λ) the number of translates K + u, u ∈ Λ, that touch K at its boundary. Theorem 4.4 (SwinnertonDyer [1]). If K + Λ is one of the densest lattice packings of K, then k(K, Λ) ≥ n(n + 1). Proof. Clearly, K + Λ is a packing if and only if 12 D(K) + Λ is a packing, and k(K, Λ) = k( 12 D(K), Λ). Without loss of generality, we may assume that K is centrally symmetric and centered at o. Further we may assume that Λ = Zn . Let ±u1 , ±u2 , . . . , ±um be the points of Zn such that K + uk , k = 1, 2, . . . , m, touch K at its boundary. Then, uk = (u1k , u2k , . . . , unk ) ∈ bd(2K) for k = 1, 2, . . . , m. Let I be the n × n unit matrix. Suppose that m < n(n + 1)/2. We proceed to show that for suitable choices of an n × n matrix A and a small number η, Λ = (I + ηA)Zn is a packing lattice of K and det(Λ ) < 1 = det(Λ).
4.3. A Theorem of SwinnertonDyer
73
Let vk = (v1k , v2k , . . . , vnk ) be a unit exterior norm of 2K at uk . Since m + 12 n(n + 1) < n2 , there is a nonzero symmetric matrix A with elements aij such that uk A, vk = aij ujk vik = 0 i, j
for k = 1, 2, . . . , m. Writing wk = uk (I + ηA), it is easy to see that wk ∈ int(2K),
k = 1, 2, . . . , m.
Thus, K + Λ is a packing of K. To calculate det(Λ ), we have det(Λ ) = 1 + c1 η + c2 η 2 + · · · + cn η n , where c1 =
n
aii ,
c2 =
i=1
aii ajj − a2ij , i<j
and 2c2 − c21 = −2
i<j
a2ij −
n
a2ii < 0.
i=1
Thus, we can choose a small η with proper sign such that det(Λ ) < 1 = det(Λ). By this contradiction Theorem 4.4 is proved.
2
Remark 4.3. As a counterpart of SwinnertonDyer’s theorem, Gruber [1] proved that in the sense of Baire category most ndimensional convex bodies K satisfy k(K, Λ) ≤ 2n2 for all of their densest packing lattices. Clearly, by Theorem 2.6 and Remark 2.5 both S7 and S8 do not belong to this category. Remark 4.4. As a consequence of SwinnertonDyer’s theorem, we have k ∗ (Sn ) ≥ n(n + 1). This lower bound is too small, even when n = 4. Inductively, it can be deduced that k∗ (Sn ) ≥ k∗ (Sn−1 ) + 2n.
74
4. Lower Bounds for b(Sn ) and k(Sn )
For more lower bounds for k∗ (Sn ) and k(Sn ), especially the constructive ones, we refer to the next chapter.
4.4. A Lower Bound for the Translative Kissing Numbers of Superspheres Let α be a number satisfying α ≥ 1. The set
1/α n n α Sn (α) = x ∈ E : xi  ≤1 i=1
is called an ndimensional supersphere. By Minkowski’s inequality,
1/α
1/α
1/α n n n θxi + (1 − θ)yi α ≤θ xi α + (1 − θ) yi α i=1
i=1
i=1
whenever 0 ≤ θ ≤ 1. It follows that Sn (α) is a centrally symmetric convex body. Also, it can be veriﬁed that Sn (1) is a cross polytope, Sn (2) is a sphere, and Sn (∞) is a unit cube. In this section we prove the following result. Theorem 4.5 (Larman and Zong [1]). 30.1072...n(1+o(1)) ≤ k(Sn (α)) ≤ 3n . To prove this theorem, ﬁrst we introduce a basic concept and a simple lemma. Let C be an ndimensional centrally symmetric convex body centered at o. Then, the Minkowski metric given by C is deﬁned by ) 0, if x = y, x, yC =
x−y , otherwise,
p(x−y) where p(x) denotes the boundary point of C in the direction of x, and x denotes the Euclidean norm of x. Lemma 4.3 (Zong [5]). The translative kissing number k(C) of C is the maximal number of points xi ∈ bd(C) such that xi , xj C ≥ 1,
i = j.
Proof. Suppose X = {x1 , x2 , . . . , xk } is a subset of bd(C) such that xi , xj C ≥ 1
4.4. A Lower Bound for k(Sn (α))
75
for every pair of distinct points xi and xj . Then, by symmetry it follows that C touches each of the k translates C + 2xi , xi ∈ X, at its boundary, and (int(C) + 2xi ) ∩ (int(C) + 2xj ) = ∅ whenever xi = xj . Therefore, k(C) ≥ card{X}. On the other hand, if X is a set of points such that C + {o} ∪ {2X} is an optimal kissing conﬁguration of C, then X ⊂ bd(C), and x, yC ≥ 1 holds for every pair of distinct points x and y of X. Thus, Lemma 4.3 follows. 2 Proof of Theorem 4.5. It is easy to see that x, ySn (α) =
n
1/α xk − yk 
α
.
(4.13)
k=1
In order to prove Theorem 4.5, ﬁrst let us introduce a basic combinatorial result. Assertion 4.2. Let m be a positive integer with m ≤ n. Also let Z∗ be a n set of integer points zi = (zi1 , zi2 , . . . , zin ) with zik  ≤ 1, k=1 zik  = m, and n zik − zjk  ≥ m (4.14) k=1
whenever i = j. Then, k(Sn (α)) ≥ card{Z ∗ }. Taking
Y = yi = m−1/α zi : zi ∈ Z ∗ ,
we have n k=1
1/α yik 
α
1/α n 1 α = zik  m k=1
1/α n 1 = zik  = 1, m k=1
76
4. Lower Bounds for b(Sn ) and k(Sn )
which means that yi ∈ bd(Sn (α)). On the other hand, by (4.13) and (4.14) it follows that
1/α n 1 yi , yj Sn (α) = zik − zjk α m k=1
1/α n 1 ≥ zik − zjk  ≥1 m k=1
whenever i = j. Hence, by Lemma 4.3, k(Sn (α)) ≥ card{Y } = card{Z ∗ }, which proves Assertion 4.2. Writing
n Z = z ∈ Zn : zi  ≤ 1, zi  = m ,
i=1
a simple combinatorial argument yields that
n m card{Z } = 2 . m
(4.15)
Similarly, writing h(m) = m/2 + 1, for any point z ∈ Z there are at most
m n − h(m) m−h(m) g(n, m) = 2 (4.16) h(m) m − h(m) points w ∈ Z such that n
zi − wi  < m.
i=1
Deﬁning
f (n, m) = max {card{Z ∗ }} ,
by (4.15) and (4.16) it follows that
n m 2 − f (n, m)g(n, m) ≤ 0 m and therefore n
2m m g(n, m)
−1
−1 n m n − h(m) = 2h(m) . m h(m) m − h(m)
f (n, m) ≥
(4.17)
4.4. A Lower Bound for k(Sn (α))
77
Writing l = n/m, by Stirling’s formula and detailed computation we have n(1+o(1)) f (n, n/l) ≥ 21−3/2l (2l − 1)1/2l−1 l .
(4.18)
It can be shown by routine computation that the function f (x) = 21−3/2x (2x − 1)1/2x−1x attains its maximum (4.18) it follows that
9 8
at x = 92 . Therefore, by Assertion 4.2, (4.17), and
k(Sn (α)) ≥ f (n, 2n/9) ≥
n(1+o(1)) 9 8
= 30.1072...n(1+o(1)) . Theorem 4.5 follows from this together with Theorem 1.1.
2
Remark 4.5. In 1985 Milman [1] proved the following theorem: Let > 0 and let C be an ndimensional centrally symmetric convex body. Then there are subspaces El ⊂ E m of E n with l ≥ c( )n and an ellipsoid E in El such that E ⊆ p(C ∩ E m ) ⊆ (1 + )E, where p(x) indicates the orthogonal projection from En onto E l . As a corollary to this result, Talata [1] deduced that there exists a positive constant c such that k(K) ≥ 3cn holds for every ndimensional convex body K. In fact, Bourgain knew this result much earlier, but he did not publish it.
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5. Sphere Packings Constructed from Codes
5.1. Codes Let F be a ﬁnite ﬁeld. A code ℵ over F of length n is simply a subset of F n . A typical point, or codeword, of ℵ has the form u = (u1 , u2 , . . . , un ), where ui ∈ F . For convenience, we say that a codeword or point is of type [µk ν l  · · ·] if µ = ui  for k choices of ui , ν = ui  for l choices of ui , etc. The Hamming distance between two codewords u and v of ℵ is the number of coordinates at which they diﬀer, and is denoted by u, vH . The weight of a codeword u, w(u), is the number of its nonzero coordinates. Thus, u, vH = w(u − v). The signal transmission process can be described as encoder m
6
m is mapped into u
channel u

error v is added
decoder u +v
message
most likely v can be eliminated
m6 estimate
Thus, in this process the number of errors that can be corrected by ℵ is determined by its minimum Hamming distance d(ℵ) = min u, vH . u,v∈ℵ u=v
80
5. Sphere Packings Constructed from Codes
In coding theory, one of the main problems is to determine m(F, n, d), the maximum number of codewords in any code over F of length n and minimal Hamming distance d. When F = {0, 1}, determining m(F, n, d) is also an interesting geometric problem. In fact, in this case m(F, n, d) is the maximal number of vertices of the unit cube In with √ the property that any two of them are at Euclidean distance at least d apart. Usually, a code of length n containing m codewords and with minimal Hamming distance d is said to be an (n, m, d) code. For convenience, we assume that every code contains the zero codeword. A code ℵ over F is called linear if µu + νv ∈ ℵ holds whenever u, v ∈ ℵ and µ, ν ∈ F. In other words, a linear code is an F module. Then, it is wellknown from algebra that ℵ can be spanned by the rows of a generator matrix A = (IB), where I is the k × k unit matrix, and card{ℵ} = q k , where q = card{F }. For convenience, a kdimensional linear code of length n with minimal Hamming distance d is said to be an [n, k, d] code (or sometimes an [n, k] code). A code ℵ over F is called cyclic if whenever (u0 , u1 , . . . , un−1 ) is a codeword so is (un−1 , u0 , u1 , . . . , un−2 ). Unless otherwise stated a cyclic code is assumed to be linear. It is convenient to represent a codeword u = (u0 , u1 , . . . , un−1 ) by the polynomial f (x) = u0 + u1 x + · · · + un−1 xn−1 in the ring Fn [x] of polynomials modulo xn − 1 with coeﬃcients from F . Then xf (x) represents a cyclic shift of u, and so a linear cyclic code is represented by an ideal in Fn [x]. It is wellknown from algebra that this ideal can be generated by a single polynomial g(x), a generator polynomial for the code. It is easily seen that g(x) divides xn − 1 over F and that the dimension of ℵ is k = n − deg g(x). To end this section, we introduce several important codes as examples. First of all, we observe a general principle for cyclic codes. Let n be relatively prime to q, and let ξ be a primitive nth root of unity, so that xn − 1 =
n−1 # i=0
x − ξi .
5.1. Codes
81
If l is the multiplicative order of q modulo n, then ξ ∈ F where F is a suitable ﬁnite ﬁeld of ql elements. Also, # x − ξj g(x) = j∈J
for some set J ⊆ {0, 1, . . . , n − 1}. Example 5.1 (Binary Hamming Codes). Let n = 2l − 1, k = n − l, d = 3, and J = 1, 2, 22, . . . , 2l−1 . Then the corresponding g(x) generates a binary Hamming code Hn over F = {0, 1}. Based on this code, the corresponding extended binary Hamming code Hn+1 is deﬁned by Hn+1
n+1 = (u1 , u2 , . . . , un+1 ) : (u1 , u2 , . . . , un ) ∈ Hn , ui = 0 . i=1
Example 5.2 (Quadratic Residue Codes). Let p and n be primes such that p is a square modulo n. For p = 2, the most important case, this means that n is a prime of the form 8l ± 1. Let F be the ﬁeld {0, 1, . . . , p − 1} and let J = {j : j = 0 is a square modulo n} . Then the corresponding g(x) generates a quadratic residue code over F . In particular, the binary Golay code G23 is the quadratic residue code over F = {0, 1} of length 23. Based on this code, the extended binary Golay code G24 is deﬁned by G24 =
24 (u1 , u2 , . . . , u24 ) : (u1 , u2 , . . . , u23 ) ∈ G23 , ui = 0 . i=1
Example 5.3 (BoseChaudhuriHocquenghem Codes). The BCH code Bn,d of length n and distance d over F is the cyclic code whose generator polynomial g(x) has roots exactly ξ, ξ2 , . . . , ξ d−1 , and their conjugates. Then, routine computation yields that d(Bn,d ) ≥ d. The special BCH codes over F of length q − 1, where q = card{F }, are usually called ReedSolomon codes. In these cases, d(Bn,d ) = d.
82
5. Sphere Packings Constructed from Codes
Example 5.4 (Justesen Codes). Let R be a ReedSolomon code over F , card{F } = 2l , with length n = 2l − 1, dimension k, and minimal Hamming distance d(R) = n − k + 1. Let ν be a primitive element of F . If u = (u0 , u1 , . . . , un−1 ) is a codeword of R, we deﬁne u∗ to be the vector u∗ = u0 , u0 , u1 , νu1 , . . . , un−1 , ν n−1 un−1 of length 2n over F . Since F is a vector space of dimension l over {0, 1}, we may set up a onetoone correspondence between F and {0, 1}l . Let u be obtained from u∗ by replacing each component by the corresponding binary ltuple. Then u is a binary vector of length 2ln. The linear code J = {u : u ∈ R} is called a Justesen code. It can be easily shown that its binary dimension is kl. Example 5.5 (ReedMuller Codes). Let m(k) be the number of ones in the binary expansion of k. For 1 ≤ j ≤ l − 2, the jthorder binary punctured ReedMuller code Mn of length n = 2l − 1 is the cyclic code whose generator polynomial has as roots those ξ k such that 1 ≤ k ≤ 2l − 2 and 1 ≤ m(k) ≤ l − j − 1, where ξ is a primitive element of F . Then, we deﬁne Mn+1 =
(u1 , u2 , . . . , un+1 ) : (u1 , u2 , . . . , un ) ∈ Mn ,
n+1
ui = 0
i=1
and call it the jthorder ReedMuller code of length n + 1 = 2l .
5.2. Construction A Theorem 5.1 (Leech and Sloane [2]). Let ℵ be an (n, m, d) binary code, and let h(t) be the number of its codewords u such that u, oH = t. Deﬁne X = {x ∈ E n : x (mod 2) ∈ ℵ} ,
(5.1)
5.2. Construction A
83 √ r=
and
1
d/2 if d ≤ 4, if d ≥ 4,
d if d < 4, 2 h(d) τ= 2n + 24 h(4) if d = 4, 2n if d > 4.
Then rSn + X is a periodic packing with density δ(rSn , X) =
mωn rn , 2n
in which the sphere rSn touches τ others. In particular, X is a lattice if and only if ℵ is a linear code. Suppose ℵ is a linear code of dimension k. Then the corresponding lattice sphere packing has density ωn r n δ(rSn , X) = n−k . 2 Proof. By its deﬁnition, X is a periodic discrete set and can be represented as X = ℵ + 2Zn , (5.2) where the codewords are regarded as points in E n . If d < 4, the points closest to the origin o are the 2d h(d) points of type [1d 0n−d ] obtained √ from the codewords of weight d. Such points are at Euclidean distance d from the origin. If d > 4, the points closest to the origin are the 2n points of type [21 0n−1 ], at Euclidean distance 2. Finally, if d = 4, both sets of points are at the same distance 2 from the origin. Thus, by (5.2) and routine computation, rSn + X forms a periodic packing with density mωn rn δ(rSn , X) = , (5.3) 2n in which rSn touches τ others. Since ℵ is a code over F = {0, 1}, routine argument shows that X is a lattice if and only if ℵ is linear. When ℵ is a linear code of dimension k, m = 2k . Thus, the last assertion follows from (5.3). Theorem 5.1 is proved.
2
The construction deﬁned by (5.1) is usually called Construction A. Example 5.6. Let ℵ be a linear binary code with codewords u0 = (0, 0, 0), u1 = (0, 1, 1), u2 = (1, 0, 1), and u3 = (1, 1, 0). Then the lattice given by Construction A is the facecentered cubic lattice that attains both δ ∗ (S3 ) and k∗ (S3 ).
84
5. Sphere Packings Constructed from Codes
Example 5.7. Let ℵ be the extended Hamming code H8 deﬁned in Example 5.1. By Theorem 5.1 one can construct a lattice sphere packing S8 + Λ8 that attains both δ∗ (S8 ) and k∗ (S8 ). Example 5.8. In 1954, Golay [1] constructed a nonlinear (9, 20, 4) code. Applying Construction A to this code, Conway and Sloane [1] deduced that k(S9 ) ≥ 306. Thus, comparing with Theorem 2.6, we have k(S9 ) > k∗ (S9 ). Remark 5.1. With a (mod p) version of Construction A, Rush [1] was able to obtain δ ∗ (Sn ) ≥ 2−n(1+o(1)), which is asymptotically the MinkowskiHlawka lower bound. As shown in Chapter 3, there are several proofs for the MinkowskiHlawka theorem. However, none can be used to construct lattice sphere packings with such densities, since all the known proofs are based on mean value arguments. Rush’s method also shares this drawback, since the codes he has used cannot be clearly constructed.
5.3. Construction B Theorem 5.2 (Leech and Sloane [2]). Let ℵ be an (n, m, d) binary code with the property that the weight of each codeword is even, and let h(t) be the number of its codewords u such that u, oH = t. Deﬁne n n X = x ∈ E : x (mod 2) ∈ ℵ, xi ∈ 4Z , (5.4) i=1
√ √d/2 if d ≤ 8, r= 2 if d ≥ 8, and
d−1 if d < 8, 2 h(d) τ= 2n(n − 1) + 27 h(8) if d = 8, 2n(n − 1) if d > 8.
Then rSn + X is a periodic packing with density δ(rSn , X) =
mωn rn , 2n+1
5.4. Construction C
85
and the sphere rSn touches τ others. In particular, X is a lattice if and only if ℵ is linear. Suppose ℵ is a linear code of dimension k. Then the corresponding lattice sphere packing has density ωn r n δ(rSn , X) = n+1−k . 2 This theorem can be proved in a way similar to Theorem 5.1, so its proof is omitted here. Usually, (5.4) is called Construction B. With this method one can construct some sphere packings with high densities. Here we give two examples. Example 5.9. Let ℵ be the ﬁrstorder ReedMuller code M16 deﬁned in Example 5.5. By Theorem 5.2 one can construct a lattice Λ16 such that S16 + Λ16 is a packing with density ω16 δ(S16 , Λ16 ) = 4 2 in which each sphere touches 4320 others. This is the densest lattice sphere packing known in E 16 . Example 5.10. Let ℵ be the extended Golay code G24 deﬁned in Example 5.2 and let X be the set constructed by (5.4). Then the set Λ24 = √12 X u + √12 X , where u = √18 (1, 1, . . . , 1, −3), is a lattice known as the Leech lattice. This lattice has many important properties. For instance, up to isometry, S24 + Λ24 is the only packing in which every sphere touches k(S24 ) = k∗ (S24 ) = 196560 others (see Theorem 9.4). In addition, δ(S24 , Λ24 ) =
π 12 , 12!
which is the densest lattice sphere packing known in E24 .
5.4. Construction C Let x = (x1 , x2 , . . . , xn ) be a point in E n with integer coordinates. If the component xi of x can be expanded as a binary series xi =
∞ j=0
xji 2j ,
86
5. Sphere Packings Constructed from Codes
where complementary notation is used for negative integers, we call the matrix x01 x02 · · · x0n x11 x12 · · · x1n A(x) = x21 x22 · · · x2n .. .. . . .. . . . . the coordinate array of x. For example, the coordinate array of (−2, −1, 1, 2) in E 4 is 0 1 1 0 1 1 0 1 1 1 0 0 . .. .. .. .. . . . . For convenience, letting u and v be codewords of a binary code ℵ, we say that v is contained in u if ui = 1 whenever vi = 1. Theorem 5.3 (Leech and Sloane [2]). Suppose ℵ0 , ℵ1 , . . . , ℵk are binary codes of types (n, m0 , d0 ), (n, m1 , d1 ), . . . , (n, mk , dk ), respectively, where di = η4k−i and η = 1 or 2. Deﬁne X = {x ∈ Zn : (xj1 , xj2 , . . . , xjn ) ∈ ℵj , j = 0, 1, . . . , k} , r= and τ=
√
(5.5)
η 2k−1 ,
k
τj (u),
j=0 u∈ℵj
where τj (u) is the number of codewords in ℵj+1 ∩ ℵj+2 ∩ · · · ∩ ℵk that are contained in the codeword u of ℵj . Then rSn + X is a periodic packing with density k ωn η n/2 # δ(rSn , X) = mj 22n j=0 in which the sphere rSn touches τ others. Proof. Let x and y be distinct points in X, and assume that j is the ﬁrst index such that (xj1 , xj2 , . . . , xjn ) = (yj1 , yj2 , . . . , yjn ). We now consider two cases. Case 1. j > k. Then the Euclidean distance between x and y is at least 2k+1 .
5.4. Construction C
87
Case 2. 0 ≤ j ≤ k. Since ℵj is an (n, mj , dj ) code with dj = η4k−j , the two points x and y diﬀer in at least dj coordinates, and therefore the Euclidean distance between them is at least " √ dj 4j = η 2k . So in both cases we have min x, y ≥
x,y∈X x=y
√ k η2
and rSn + X is a periodic packing. As in the proof of Theorem 5.1, it is easy to see that there are m=
k #
mj
j=0
points of X in the cube I = x ∈ E n : 0 ≤ xi < 2k+1 . Thus, by the periodic property of X it can be deduced that the packing density of rSn + X is k mv(rSn ) ωn η n/2 # δ = (k+1)n = mj . 22n j=0 2
Let X be the set of points x ∈ X such that rSn + x touches rSn at its boundary, and let Xj be the set of points x ∈ X of [(2j )dj 0n−dj ] type. For a point x ∈ X, suppose that j is the ﬁrst index such that (xj1 , xj2 , . . . , xjn ) = (0, 0, . . . , 0). Then, by routine argument it follows that both (xj1 , xj2 , . . . , xjn ) and x are of [(2j )dj 0n−dj ] type if x ∈ X . Thus, we have card{X } =
k
card{Xj }.
j=0
By the deﬁnition of the coordinate array we have 0 if xj = 2j , xli = 1 if xj = −2j for all indices l ≥ j. Thus, if x ∈ X is a point such that (xj1 , xj2 , . . . , xjn ) ∈ ℵj ,
(5.6)
88
5. Sphere Packings Constructed from Codes
then the minus signs of its coordinates must be at the locations of the ones in some codeword v ∈ ℵj+1 ∩ ℵj+2 ∩ · · · ∩ ℵk . (5.7) On the other hand, for any codeword of (5.7) that is contained in a codeword u ∈ ℵj there is a point x ∈ X such that x, o = 2r. Thus, card{Xj } =
τj (u).
(5.8)
u∈ℵj
Then (5.6) and (5.8) together yield card{X } =
k
τj (u) = τ,
j=0 u∈ℵj
which implies the last assertion of Theorem 5.3.
2
Usually, (5.5) is called Construction C. Remark 5.2. For general codes, the set X constructed by (5.5) is not a lattice. However, if the codes ℵ0 , ℵ1 , . . . , ℵk are linear and nested, i.e., ℵj ⊆ ℵj+1 for all indices 0 ≤ j ≤ k − 1, then the set is a lattice. Example 5.11. Let ℵj be the 2jth ReedMuller code of length n = 2m . In addition, take η = 1 and k = m/2 if m is even; and take η = 2 and k = (m − 1)/2 if m is odd. Then, by applying Construction C, Leech [2] was able to construct a lattice packing Sn + Λn in which every sphere touches m # 2 + 2i ∼ 4.768 . . . n(m+1)/2 τ= i=1
others. So far this is the best known lower bound for k∗ (Sn ) when n = 2m is large. Remark 5.3. Construction C can be used to obtain dense sphere packings in high dimensions. For example, by applying this construction to certain Justesen codes, Sloane [1] has shown that δ(Sn ) ≥ 2−6n(1+o(1)) for integers n of k2k type. This result is weaker than the MinkowskiHlawka theorem. Since the Justesen codes used cannot be clearly constructed, Sloane’s lower bound is not eﬃciently constructed.
5.5. Some General Remarks
89
5.5. Some General Remarks Besides the previous constructions, there are many other ingenious methods for constructing sphere packings. For example, applying ideas from algebraic curves, Litsyn and Tsfasman [1], [2] obtained a nonlattice packing Sn + X with δ(Sn , X) ≥ 2−1.31n(1+o(1)) and a lattice packing Sn + Λ with δ(Sn , Λ) ≥ 2−2.30n(1+o(1)). For more about sphere packing constructions, we refer to Conway and Sloane [1]. To end this chapter, based upon Example 5.11 we propose the following problems. Problem 5.1. Does there exist a positive number c such that k ∗ (Sn ) ≥ 3cn holds for every suﬃciently high dimension? If so, what is the largest c? Very probably the answer to this problem is “no.” If so, the following modiﬁcation would be interesting. Problem 5.2. Does there exist a positive number c such that k ∗ (Sn ) ≥ 3cn holds for inﬁnitely many dimensions? If so, what is the largest c?
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6. Upper Bounds for the Packing Densities and the Kissing Numbers of Spheres I
6.1. Blichfeldt’s Upper Bound for the Packing Densities of Spheres In 1929 H.F. Blichfeldt discovered a remarkable method for determining an upper bound for δ(Sn ). Roughly speaking, his idea runs as follows. Let Sn + X be a packing and let r be a number such that r > 1. Then, replace the spheres Sn + x, where x ∈ X, by rSn + x and ﬁll each of these new spheres with a certain amount of mass, of variable density, such that the total mass at any point of E n does not exceed 1. Hence, the total mass of the spheres rSn + x, where x ∈ X ∩ (l − r)In , does not exceed the volume of the large cube lIn . In this way, Blichfeldt obtained the ﬁrst signiﬁcant upper bound for δ(Sn ). Theorem 6.1 (Blichfeldt [1]). δ(Sn ) ≤
n + 1 −0.5n 2 . 2
In order to prove this theorem, we introduce two fundamental lemmas. Lemma 6.1. Let m be a positive integer and let x, x1 , . . . , xm be arbitrary points in E n . Then, one has m j, k=1
xj , xk 2 ≤ 2m
m j=1
x, xj 2 .
92
6. Upper Bounds for δ(Sn ) and k(Sn ), I
Proof. Let a1 , a2 , . . . , am be m real numbers. Routine analysis yields m
m 2 aj + a2k − 2aj ak
2
(aj − ak ) =
j, k=1
= ≤
j, k=1 m j, k=1 m
2 m 2 2 aj + ak − 2 aj j=1
2 aj + a2k
j, k=1 m
= 2m
a2j .
j=1
Thus, writing yj = x − xj = (yj1 , yj2 , . . . , yjn ), we have m
m
xj , xk 2 =
j, k=1
=
yj , yk 2
j, k=1 m n
2
(yji − yki )
i=1 j, k=1 m n
≤ 2m
(yji )2
i=1 j=1
= 2m
= 2m
n m
2
(yji )
j=1 i=1 m
yj 2
j=1
= 2m
m
x, xj 2 .
j=1
2
Lemma 6.1 is proved.
Lemma 6.2 (Blichfeldt [1]). Let r be a positive number, and let σ(x) be a function that is nonnegative and continuous for 0 ≤ x ≤ r and vanishes for x > r. Suppose that for each packing Sn + X, σ(x, xi ) ≤ 1 (6.1) xi ∈X
holds for every point x ∈ En . Then δ(Sn ) ≤ J −1 ,
6.1. Blichfeldt’s Upper Bound for δ(Sn ) %
where
93
r
xn−1 σ(x)dx.
J =n 0
Proof. Let l be a large number and suppose that there are exactly m spheres Sn + x1 , Sn + x2 , . . . , Sn + xm of the packing Sn + X that are entirely contained in the large cube lIn . It is obvious that rSn + xi ⊂ (l + 2r)In , and therefore σ(x, xi ) = 0
(6.2)
holds for every index i = 1, 2, . . . , m if x ∈ (l + 2r)In . Then, by (6.1) and (6.2) it follows that %
m
v((l + 2r)In ) ≥ =
σ(x, xi )dx
(l+2r)In i=1 m %
σ(x, xi )dx
i=1
%
En
=m
σ(x)dx n %Er
σ(x)d(ωn xn ) % r = mnωn xn−1 σ(x)dx =m
0
0
= mωn J. Hence, by the deﬁnition of δ(Sn ), we have mωn r→∞ v(lIn ) X v((l + 2r)In ) ≤ sup lim sup Jv(lIn ) r→∞ X
n 2r = sup lim sup 1 + J −1 l r→∞ X
δ(Sn ) = sup lim sup
= J −1 . Lemma 6.2 is proved.
2
With these preparations, we proceed to prove Theorem 6.1. Proof of Theorem 6.1. Let Sn + X be a packing and, for convenience, enumerate the points of X by x1 , x2 , . . . . Since Sn + X is a packing, we have xj , xk ≥ 2
94
6. Upper Bounds for δ(Sn ) and k(Sn ), I
whenever j = k and therefore, for each ﬁnite set of distinct indices i1 , i2 , . . . , im , m xij , xik 2 ≥ 4m(m − 1). j, k=1
Thus, for each point x ∈ En , by Lemma 6.1 we have m
x, xij ≥ 2(m − 1).
(6.3)
j=1
Deﬁning
one has r =
√
σ(x) = max 0, 1 − 12 x2 , 2. In addition, for every point x ∈ En , it follows that ∞
σ(x, xi ) =
i=1
=
m
σ(x, xij )
j=1 m j=1
1 1 − x, xij 2 2 1 x, xij 2 , 2 j=1 m
= m−
where ij are the indices i such that x, xi < ∞
√
2. Hence, by (6.3),
σ(x, xi ) ≤ m − (m − 1) = 1.
i=1
This means that the function σ(x) satisﬁes the condition of Lemma 6.2. Then, routine computation yields that √ 2
% J =n 0
1 2 xn−1 1 − x2 dx = 20.5n . 2 n+2
Thus, it follows from Lemma 6.2 that δ(Sn ) ≤ J −1 = Theorem 6.1 is proved.
n + 2 −0.5n 2 . 2 2
Remark 6.1. Blichfeldt’s method was improved by Rankin [1], [6], Sidelnikov [1], [2], Leven˘ s tein [1], [2], and others. In this way, Theorem 6.1 was improved to δ(Sn ) ≤ 2−0.5237n(1+o(1)).
6.2. Rankin’s Upper Bound for k(Sn )
95
By studying the number of spheres that can be packed in an ndimensional torus, Yudin [1] gave a new proof of this result.
6.2. Rankin’s Upper Bound for the Kissing Numbers of Spheres Let m(n, α) be the maximal number of caps of geodesic radius α that can be packed in bd(Sn ). It is easy to see that k(Sn ) = m(n, π/6) and δ(Sn ) = lim
α→0
m(n + 1, α)s(α) , (n + 1)ωn+1
(6.4)
(6.5)
where s(α) indicates the area of a cap of geodesic radius α. Generalizing Blichfeldt’s method from En to bd(Sn ) and applying (6.4) and (6.5), R.A. Rankin obtained a new proof of Theorem 6.1. In addition, he obtained the following upper bound for k(Sn ). Theorem 6.2 (Rankin [6]). √ k(Sn )
πn3 0.5n 2 , 2
where f (x) g(x) means lim sup x→∞
f (x) ≤ 1. g(x)
First, let us introduce a set of fundamental lemmas. Lemma 6.3 (Rankin [6]). Let xj = (xj1 , xj2 , . . . , xjn ), j = 1, 2, . . . , m, be m distinct points in bd(Sn ), and deﬁne d=
min
1≤j v(C2 ) and s(C1 ∩ H) < s(C2 ∩ H) for every hyperplane H that contains o. The ﬁrst example completely solves a problem of W. Blaschke, and the second solves a problem of H. Busemann and C.M. Petty in the corresponding dimensions. He is also the author of the classic book Packing and Covering. Professor Rogers has been a fellow of the Royal Society since 1959, and he was the president of the London Mathematical Society for the term 1970– 1972. For his distinguished contributions, he was awarded a De Morgan medal by the London Mathematical Society in 1977.
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8. Upper Bounds for the Packing Densities and the Kissing Numbers of Spheres III
8.1. Jacobi Polynomials The Jacobi polynomials are a family of wellknown special functions. They are deﬁned, for k = 0, 1, 2, . . . , as
k 1 k+α k+β α,β (t + 1)i (t − 1)k−i , Pk (t) = k 2 i=0 i k−i where α > −1 and β > −1 are parameters. These polynomials play an important role in obtaining the KabatjanskiLeven˘stein upper bounds for δ(Sn ) and k(Sn ). In this section we introduce some of their fundamental properties. Since these properties are wellknown and easy to deduce (see Szeg¨ o [1]), we omit their proofs.
Assertion 8.1. Pkα,β (1) =
k+α . k
Assertion 8.2. % 1 Pkα,β (t)Plα,β (t)(1 − t)α (1 + t)β dt = δkl (k, α, β), −1
where δkl is the Kronecker symbol and (k, α, β) =
2α+β+1 Γ(k + α + 1)Γ(k + β + 1) . k!(2k + α + β + 1)Γ(k + α + β + 1)
126
8. Upper Bounds for δ(Sn ) and k(Sn ), III
Conversely, if P0 (t), P1 (t), P2 (t), . . . is a sequence of polynomials such that %
1
−1
Pk (t)Pl (t)(1 − t)α (1 + t)β dt = δkl (k, α, β),
where the degree of Pk (t) is k, then Pk (t) = ±Pkα,β (t),
k = 0, 1, 2, . . . .
Assertion 8.3. The Jacobi polynomials y = Pkα,β (t) satisfy the diﬀerential equation (1 − t2 )y + [β − α − (α + β + 2)t]y + k(k + α + β + 1)y = 0. Assertion 8.4 (The ChristoﬀelDarboux Formula). α,β α,β Pk+1 (t)Pkα,β (s) − Pkα,β (t)Pk+1 (s) t−s
(2k + α + β + 2)(2k + α + β + 1) (k, α, β) α,β Pi (t)Piα,β (s). 2(k + 1)(k + α + β + 1) (i, α, β) i=0 k
=
Clearly, by taking s → t one obtains the following result. Corollary 8.1.
α,β α,β (t) Pkα,β (t) − Pk+1 (t) Pkα,β (t) > 0. Pk+1
Assertion 8.5. Let t1 (k, α, β), t2 (k, α, β), . . . , tk (k, α, β) be the k zeros (in decreasing order) of the polynomial Pkα,β (t). Then, for k = 1, 2, . . . and i = 1, 2, . . . , k, ti+1 (k + 1, α, β) < ti (k, α, β) < ti (k + 1, α, β) < 1. Assertion 8.6. When α = β = (n − 3)/2, Pkα,β (t)Plα,β (t) =
k+l
ai (k, l)Piα,β (t)
i=0
holds for some suitable nonnegative coeﬃcients ai (k, l).
8.2. Delsarte’s Lemma
127
8.2. Delsarte’s Lemma A spherical code ℵ of dimension n, cardinality m, and minimum angle ϕ is a set of m points of bd(Sn ) such that x, y ≤ cos ϕ holds for any two distinct points x and y of ℵ. In other words, the m caps Ω(x, ϕ/2), x ∈ ℵ, form a packing in bd(Sn ). For convenience, we call ℵ an {n, m, ϕ} code. Then, we deﬁne m[n, ϕ] to be the maximum size, m, of any such spherical code. Clearly, m(n, ϕ/2) = m[n, ϕ].
(8.1)
As an upper bound for m[n, ϕ], we have the following remarkable result. Lemma 8.1 (Delsarte [1]). Write α = (n − 3)/2. Let f (t) =
k
fi Piα,α (t)
i=0
be a real polynomial such that f0 > 0, fi ≥ 0 for i = 1, 2, . . . , k, and f (t) ≤ 0 for −1 ≤ t ≤ cos ϕ. Then m[n, ϕ] ≤
f (1) . f0
Let L be the linear space of square integrable real functions over bd(Sn ). For two functions f1 (x) and f2 (x) of L we deﬁne the inner product % 1 f1 ◦ f2 = f1 (x)f2 (x)dω(x), nωn bd(Sn ) where ω(x) indicates the surface element of bd(Sn ) at x. Let Pk be the subspace of L consisting of spherical polynomials of degree at most k, and let Hk be the subspace of L consisting of homogeneous harmonic polynomials of degree k. Here, we say f (x) is a harmonic if ∆f =
∂2f ∂2f ∂2f + + · · · + = 0, ∂x21 ∂x22 ∂x2n
where ∆ is the Laplace operator. Assertion 8.7. Pk = Hk ⊕ Hk−1 ⊕ · · · ⊕ H0 . Proof. A polynomial pk (x) in Pk is a linear combination of monomials m(x) = xk11 xk22 · · · xknn
128
8. Upper Bounds for δ(Sn ) and k(Sn ), III
with k1 + k2 + · · · + kn = l ≤ k. Since x ∈ bd(Sn ), we have x, x = 1 and therefore m(x) = x, xi m(x). Take i = (k − l)/2. Then x, xi m(x) is a homogeneous polynomial, of degree k if k − l is even and of degree k − 1 if k − l is odd. Denote the subspace of L consisting of homogeneous polynomials of degree k by Hk∗ . Then pk (x) = h∗k (x) + h∗k−1 (x), ∗ where h∗k (x) ∈ Hk∗ and h∗k−1 (x) ∈ Hk−1 . Also, since
h∗k ◦ h∗k−1 = (−1)2k−1 h∗k ◦ h∗k−1 , we have
h∗k ◦ h∗k−1 = 0.
Thus,
∗ . Pk = Hk∗ ⊕ Hk−1
(8.2)
By routine argument it is easy to see that ∆ is a linear operator on Hk∗ ∗ with kernel Hk and image Hk−2 . On the other hand, it is wellknown from real analysis (see Groemer [5]) that % % f (x)∆g(x)dω(x) = g(x)∆f (x)dω(x) bd(Sn )
bd(Sn )
holds for every pair of twice continuously diﬀerentiable functions f (x) and ∗ , we g(x) on bd(Sn ). In particular, for hk (x) ∈ Hk and h∗k−2 (x) ∈ Hk−2 have % 1 hk ◦ h∗k−2 = g ∗ (x)∆hk (x)dω(x) = 0, nωn bd(Sn ) k where gk∗ (x) is a suitable function satisfying h∗k−2 (x) = ∆gk∗ (x). Thus, we have
∗ Hk∗ = Hk ⊕ Hk−2 .
(8.3)
Applying (8.2) and (8.3) inductively, it follows that Pk = Hk ⊕ Hk−1 ⊕ · · · ⊕ H0 . Assertion 8.7 is proved.
2
Next we introduce a special class of spherical harmonics, which play an important role in our considerations. Since Hk is a ﬁnitedimensional
8.2. Delsarte’s Lemma
129
linear space equipped with a nontrivial inner product, it follows from a fundamental result of functional analysis that for any realvalued linear functional F (hk ) deﬁned on Hk there is a unique g ∈ Hk such that F (hk ) = hk ◦ g. In particular, for each point x ∈ bd(Sn ), by taking F (hk ) = hk (x) it is easy to see that there is a unique function Qk (x, y) ∈ Hk such that % 1 hk (x) = hk (y)Qk (x, y)dω(y). (8.4) nωn bd(Sn ) Usually, Qk (x, y) is called a zonal spherical harmonic. Assertion 8.8. For every isometry σ of En that leaves the origin ﬁxed, Qk (σx, σy) = Qk (x, y). Proof. Assume that x = σy = yA, where A is an n × n matrix satisfying AA = I, and assume hk (x) ∈ Hk . Then, routine computation yields n ∂ 2 hk ∆hk (σy) = aik ajk ∂xi ∂xj i, j k=1
= ∆hk (x) = 0.
(8.5)
This means that hk (σx) ∈ Hk if hk (x) ∈ Hk . In particular, Qk (σx, σy), as a function of y, is a spherical harmonic. Write % 1 Jσ = hk (y)Qk (σx, σy)dω(y). nωn bd(Sn ) By substituting z = σy and using (8.4), we have % 1 Jσ = hk (σ−1 z)Qk (σx, z)dω(z) nωn bd(Sn ) = hk (σ−1 σx) = hk (x) % 1 = hk (y)Qk (x, y)dω(y). nωn bd(Sn ) Then, Assertion 8.8 follows from the uniqueness of Qk (x, y).
2
Assertion 8.9 (Addition Formula). Let hk,1 (x), hk,2 (x), . . . , hk,d(k) (x) be an orthonormal basis of Hk , where d(k) is the dimension of Hk . Then,
d(k)
Qk (x, y) =
i=1
hk,i (x)hk,i (y).
130
8. Upper Bounds for δ(Sn ) and k(Sn ), III
Proof. Any function hk (x) ∈ Hk can be expanded as
d(k)
hk (x) =
hk ◦ hk,i hk,i (x).
i=1
In particular, using (8.4),
d(k)
Qk (x, y) =
Qk (x, y) ◦ hk,i (y)hk,i (y)
i=1
d(k)
=
hk,i (x)hk,i (y).
i=1
2
Assertion 8.9 is proved.
By Assertion 8.8 it follows that the value of Qk (x, y) is determined by k and x, y. So, for convenience, we write Gk (x, y) = Qk (x, y). Assertion 8.10. Write α = (n − 3)/2. Then Gk (t) = ck Pkα,α (t) holds for some positive number ck . Proof. By Assertion 8.7, for k = l, we have % J= Qk (x, y)Ql (x, y)dω(y) = 0. bd(Sn )
Then, substituting t = x, y and dω(y) = (n − 1)ωn−1 (1 − t2 )(n−3)/2 dt, we obtain % J = (n − 1)ωn−1
1
−1
Gk (t)Gl (t)(1 − t2 )(n−3)/2 dt = 0.
Thus, by the second part of Assertion 8.2 we have Gk = ck Pkα,α (t),
k = 0, 1, . . . .
To determine the sign of ck , we observe that % 1 Gk (1) = Gk (1)dω(x) nωn bd(Sn )
(8.6)
8.2. Delsarte’s Lemma
131 1 = nωn
%
d(k)
h2k,i (x)dω(x)
bd(Sn ) i=1
1 % h2k,i (x)dω(x) nω n bd(S ) n i=1
d(k)
=
d(k)
=
1 = d(k).
(8.7)
i=1
Then, it follows from (8.6), (8.7), and Assertion 8.1 that ck > 0 holds for k = 0, 1, . . . . Assertion 8.10 is proved. 2 We can now proceed to prove Lemma 8.1. Proof of Lemma 8.1. Let X be a discrete subset of bd(Sn ) such that card{X} = m[n, ϕ], and Ω(x, ϕ/2), x ∈ X, form a cap packing in bd(Sn ). Then, x, y ≤ cos ϕ holds for any two distinct points x and y of X. Let k f (t) = fi Piα,α (t)
(8.8)
i=0
be a real polynomial such that f0 > 0, fi ≥ 0 (i = 1, 2, . . . , k), and f (t) ≤ 0 for −1 ≤ t ≤ cos ϕ. For convenience, we apply f (t) =
k
gi Gi (t)
i=0
rather than (8.8), where g0 = f0 , G0 (t) = 1, and gi = fi /ci ≥ 0 for i = 1, 2, . . . , k. Write J= f (x, y). x, y∈X
Let W be the set of numbers w such that w = x, y holds for some pair of distinct points x and y of X. For w ∈ W let p(w) be the number of ordered pairs (x, y), x, y ∈ X, such that x, y = w. Then, we have J = m[n, ϕ]f (1) +
w∈W
p(w)f (w).
(8.9)
132
8. Upper Bounds for δ(Sn ) and k(Sn ), III
On the other hand, since G0 (w) = Q0 (x, y) = 1, the addition formula implies that J =
k i=0
gi
k
Gi (x, y) =
i=0
x, y∈X
= g0 m[n, ϕ]2 +
2
= f0 m[n, ϕ] +
k
gi
x, y∈X
k
d(i)
gi
j=1
Qi (x, y)
x, y∈X
Qi (x, y)
i=1
i=1
gi
2 hi,j (x) .
(8.10)
x∈X
Comparing (8.9) and (8.10), since f (w) ≤ 0 for w ∈ W , we have f (1) m[n, ϕ] ≥ f0 m[n, ϕ]2 and therefore m[n, ϕ] ≤
f (1) . f0 2
This proves Lemma 8.1.
Remark 8.1. This fundamental lemma was ﬁrst proved by Delsarte in [1] and [2]. Alternative proofs can be found in Delsarte, Goethals, and Seidel [1], Kabatjanski and Leven˘ stein [1], Odlyzko and Sloane [1], Lloyd [1], and Seidel [1]. Our proof follows Seidel’s presentation.
8.3. The KabatjanskiLeven˘stein Upper Bounds for the Packing Densities and the Kissing Numbers of Spheres For convenience, we write α = (n − 3)/2 and denote by F(n, ϕ) the family of polynomials k f (t) = fi Piα,α (t), i=0
such that f0 > 0, fi ≥ 0 for i = 1, 2, . . . , k, and f (t) ≤ 0 for −1 ≤ t ≤ cos ϕ. Then, we deﬁne f (1) m∗ [n, ϕ] = inf . f ∈F (n,ϕ) f0 It is obvious that
m[n, ϕ] ≤ m∗ [n, ϕ].
(8.11)
8.3. The KabatjanskiLeven˘ stein Upper Bounds
133
As an upper bound for m[n, ϕ], we have the following fundamental result. Lemma 8.2 (Kabatjanski and Leven˘stein [1]). and n → ∞,
When 0 < ϕ < π/2
log m[n, ϕ] 1 + sin ϕ 1 + sin ϕ 1 − sin ϕ 1 − sin ϕ log − log . n 2 sin ϕ 2 sin ϕ 2 sin ϕ 2 sin ϕ The proof of this deep lemma requires some careful preparation. Assertion 8.11. Write s = cos ϕ and τ =−
α,α k + 1 Pk+1 (s) . k + α + 1 Pkα,α (s)
When t1 (k, α, α) < s < t1 (k + 1, α, α) (see Assertion 8.5), m∗ [n, ϕ] ≤
k+2α+1 (1 + τ )2 k . (1 − s)τ
Proof. Consider the polynomial (k + α + 1)(2k + 2α + 1) α,α α,α f (t) = Pk+1 (t)Pkα,α (s) − Pkα,α (t)Pk+1 (s) (k + 1)(k + 2α + 1) ×
k (k, α, α) i=0
(i, α, α)
Piα,α (s)Piα,α (t)
(8.12)
of degree 2k + 1. By Assertion 8.5 and our assumption it follows that α,α Pkα,α (s)Piα,α (s) > 0 and Pk+1 (s)Piα,α (s) < 0
for all i ≤ k. Hence, by Assertions 8.1 and 8.6 we have f0 > 0 and fi ≥ 0 for i = 1, 2, . . . , 2k + 1. Also, by Assertion 8.4, we have f (t) =
2 α,α α,α Pk+1 (t)Pkα,α (s) − Pkα,α (t)Pk+1 (s)
(8.13)
t−s
and therefore f (t) ≤ 0 when −1 ≤ t ≤ s. Thus, f (t) ∈ F(n, ϕ). By Assertion 8.1 and (8.13) it follows that 2
f (1) =
(Pkα,α (s)) 1−s
k+α k
2
α,α k + α + 1 Pk+1 (s) − α,α k+1 Pk (s)
2 .
134
8. Upper Bounds for δ(Sn ) and k(Sn ), III
On the other hand, (8.12) and the ﬁrst part of Assertion 8.2 together yield % f0 (0, α, α) =
1
−1
f (t)(1 − t2 )α dt
α,α = − Pk+1 (s)Pkα,α (s)
(k + α + 1)(2k + 2α + 1)(k, α, α) . (k + 1)(k + 2α + 1)
Hence α,α f0 = −Pk+1 (s)Pkα,α (s)
(k + α + 1)(2k + 2α + 1)(k, α, α) . (k + 1)(k + 2α + 1)(0, α, α)
Then, by Lemma 8.1, the deﬁnition of (i, α, α), and routine computation we obtain k+2α+1 (1 + τ )2 f (1) ∗ k m [n, ϕ] ≤ ≤ . f0 (1 − s)τ 2
Assertion 8.11 is proved. Assertion 8.12. When s = cos ϕ ≤ t1 (k, α, α), 4 k+2α+1 ∗ k m [n, ϕ] ≤ . 1 − t1 (k + 1, α, α)
Proof. Regard τ as a function of s. When s decreases from t1 (k + 1, α, α) to t1 (k, α, α), it follows from Corollary 8.1 that τ increases from 0 to ∞. So, τ = 1 holds at a certain s = cos ϕ with t1 (k, α, α) < s < t1 (k + 1, α, α). Since m∗ [n, ϕ] is an increasing function of s, it follows from Assertion 8.11 that 4 k+2α+1 ∗ ∗ k m [n, ϕ] ≤ m [n, ϕ ] ≤ 1 − s k+2α+1 4 k . ≤ 1 − t1 (k + 1, α, α) 2
Assertion 8.12 is proved. Assertion 8.13. If c is a positive number such that lim
k→∞
then
α n−3 1 = lim = , k→∞ k 2k 2c
c(1 + c) lim t1 (k, α, α) = . k→∞ 1 + 2c 2
(8.14)
8.3. The KabatjanskiLeven˘ stein Upper Bounds
135
Proof. Consider the function y = (1 − t2 )(α+1)/2 Pkα,α (t),
(8.15)
which has the same zeros in −1 < t < 1 as Pkα,α (t). From Assertion 8.3, it can be deduced that this function satisﬁes the diﬀerential equation y + g(t)y = 0, where g(t) =
(8.16)
η(γ 2 − t2 ) , (1 − t2 )2
η = (k + α)(k + α + 1), and γ=
[k(k + 2α + 1) + α + 1]/η.
Since γ < 1, g(t) is negative in γ < t < 1, and y → 0 when t → 1. Thus, by (8.16) it follows that y cannot vanish for γ ≤ t < 1. Consequently, t1 (k, α, α) < γ.
(8.17)
On the other hand, for any positive number b the equation y + b2 y = 0 has a solution y = sin(bt + d), where d is a constant, that has a zero between t1 = γ−2π/b and t2 = γ−π/b. If we choose b such that t1 ≥ −1 and g(t) ≥ b2 for t1 ≤ t ≤ t2 , the function (8.15) will have a zero in this interval, and hence t1 (k, α, α) ≥ γ − In fact, taking b=
2π . b
πγη 1/4
, 4 it can be veriﬁed that t1 ≥ −1, γ ≥ 2π/b, and g(t) ≥
(4πγb − 4π 2 )η b2 ≥ b 2 [b2 − (γb)2 + 2πγb − π 2 ]2
holds for t1 ≤ t ≤ t2 when k is suﬃciently large. Then, by (8.17), (8.18), and routine computation we obtain 2 c(1 + c) lim t1 (k, α, α) = γ = . k→∞ 1 + 2c
(8.18)
136
8. Upper Bounds for δ(Sn ) and k(Sn ), III 2
Assertion 8.13 is proved. Proof of Lemma 8.2. Take c=
1 − sin ϕ 2 sin ϕ
(8.19)
and apply Assertion 8.13. This gives lim t1 (k, α, α) = cos ϕ.
k→∞
Then, by (8.11) and Assertion 8.12 it follows that 4 k+2α+1 k m[n, ϕ] ≤ 1 − t1 (k + 1, α, α)
k + 2α + 1 4 . 1 − cos ϕ k
(8.20)
Applying (8.14), Stirling’s formula, and (8.19) to (8.20) we obtain log m[n, ϕ] (1 + c) log(1 + c) − c log c n 1 + sin ϕ 1 + sin ϕ 1 − sin ϕ 1 − sin ϕ = log − log . 2 sin ϕ 2 sin ϕ 2 sin ϕ 2 sin ϕ 2
Lemma 8.2 is proved.
The special case ϕ = π/3 of Lemma 8.2 yields the following upper bound for the kissing numbers of spheres. Theorem 8.1 (Kabatjanski and Leven˘stein [1]). k(Sn ) ≤ 20.401n(1+o(1)). To get a corresponding upper bound for the packing densities of spheres, we need another lemma. Lemma 8.3. δ(Sn ) ≤
sin
ϕ n m[n + 1, ϕ] 2
holds for a certain small ϕ. Proof. Let r be a large number and let H be a hyperplane through the center of rSn+1 . In H we construct an ndimensional packing of unit spheres with density δ(Sn ). By shifting the packing slightly we may assume that the portion of H inside rSn+1 contains at least rn δ(Sn ) centers. We project these centers, perpendicularly to H, “upwards” onto bd(rSn+1 ). The Euclidean distance between the new points is at least 2, and their geodesic distance is at least ϕ, where sin
ϕ 1 = . 2 r
8.3. The KabatjanskiLeven˘ stein Upper Bounds
137
Thus, we have rn δ(Sn ) ≤ m[n + 1, ϕ] and therefore δ(Sn ) ≤
sin
ϕ n m[n + 1, ϕ]. 2 2
Lemma 8.3 is proved.
Remark 8.2. This lemma ﬁrst appeared in the appendix of the Russian edition of L. Fejes T´ oth [9] (translated by I.M. Yaglom). Here, we follow the proof of Sloane [4]. Theorem 8.2 (Kabatjanski and Leven˘stein [1]). δ(Sn ) ≤ 2−0.599n(1+o(1)). Proof. Routine computation yields 1 + sin ϕ 1 + sin ϕ 1 − sin ϕ 1 − sin ϕ log − log 2 sin ϕ 2 sin ϕ 2 sin ϕ 2 sin ϕ 1 ≤ − log(1 − cos ϕ) − 0.099 2 when 0 < ϕ ≤ ϕ0 ≈ 63◦ . Then, it follows from Lemma 8.2 that m[n, ϕ] ≤ (1 − cos ϕ)−0.5n 2−0.099n ϕ −n −0.599n = sin 2 . 2 Consequently, by Lemma 8.3 we have ϕ n ϕ −(n+1) −0.599(n+1) 2 δ(Sn ) ≤ sin sin 2 2 ϕ −1 −0.599n(1+o(1)) = sin 2 2 for some suitable ϕ. Theorem 8.2 follows.
(8.21)
2
To end this chapter let us propose two open problems. Problem 8.1. What is the behavior of f (n) = k(Sn ) − k∗ (Sn )? Does f (n) = 0 (or = 0) hold for inﬁnitely many dimensions? What is the asymptotic order of lim f (n)? Problem 8.2. What is the behavior of g(n) =
δ(Sn ) ? δ ∗ (Sn )
Does g(n) = 1 (or = 1) hold for inﬁnitely many dimensions? What is the asymptotic order of lim g(n)?
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9. The Kissing Numbers of Spheres in Eight and Twenty–Four Dimensions
9.1. Some Special Lattices Let Λ be an ndimensional lattice. As usual, we call it an integral lattice if u, u ∈ Z,
u ∈ Λ.
In particular, we call it an even integral lattice if u, u ∈ 2Z,
u ∈ Λ,
and an odd integral lattice otherwise. There are some special integral lattices that play important roles in diﬀerent areas of mathematics. For example, Zn = z ∈ E n : z i ∈ Z , n+1 An = z ∈ Zn+1 : zi = 0 , i=1
n Dn = z ∈ Zn : zi ∈ 2Z , i=1
8 1 E8 = u ∈ Z8 : ui − uj ∈ Z, ui ∈ 2Z , 2 i=1 8 E7 = u ∈ E8 : ui = 0 , i=1
140
9. The Kissing Numbers of S8 and S24
E6 =
6 u ∈ E8 : ui = u 7 + u 8 = 0 , i=1
and the√wellknown √ Leech lattice Λ24 , which is determined √ by the basis √ a1 = 2√ 2e1 , a2 = 2(e1 √ + e2 ), a3 = 2(e1 + e ), a = 2(e1 + e4 ), 4 √ 3 + e ), a = 2(e + e ), a = 2(e + e ), a a5 = 2(e 5 √6 1 6 √ 7 1 7 √ 8 = (e1 + √1 · · · + e8 )/ 2, a9 = 2(e1 + e9 ), a10 = √ 2(e1 + e10 ),√a11 = 2(e1 + e11 ), a12 = (e1 + · · · + e4 + e9 + · · · + e12 )/√ 2, a13 = 2(e1 + e13 ), a14 = (e1 + e2 + e5 + e6 √ + e9 + e10 + e13 + e14 )/ 2, a15 = (e1 + e3 + e5 + e7 + e√ 9+ e11 + e√ + e )/ 2, a = (e + e + e + e + e + e + e + e )/ 13 15 16 1 4 5 8 9 12 13 16 √2, a17 = 2(e1 + e17 ), a18 = (e1 + e3 + e5 + e8√+ e9 + e10 + e17 + e18 )/ 2, a19 = (e1 +e4 +e5 +e 7+ √6 +e9 +e11 +e17 +e19 )/ 2, a20 = (e1 +e2 +e5 +e√ e9 + e12 + e17 + e20 )/ 2, a21 = (e2 + e3 + e4 + e5 + e9 √ + e13 + e17 + e21 )/ 2, a22 = (e9 + e10 + e13 + e14 + e17 +√ e18 + e21 + e22 )/ 2, a23 = (e9 + e11 √+ e13 + e15 + e17 + e19 + e21 + e23 )/ 2, a24 = (−3e1 + e2 + · · · + e24 )/ 8, where e1 , e2 , . . . , e24 is an orthonormal basis of E 24 . It can be veriﬁed that An (n ≥ 1), Dn (n ≥ 4), En (n = 6, 7, 8), and Λ24 are even integral lattices, and Zn (n ≥ 1) are odd integral lattices. Writing r = min
1
2 u
: u ∈ Λ \ {o}
and M (Λ) = {u ∈ Λ : u = 2r}, it is easy to see that rSn + Λ is a packing and rSn + u touches rSn at its boundary if u ∈ M (Λ). Thus, k∗ (Sn ) = max {card{M (Λ)}} ,
(9.1)
where the maximum is over all ndimensional lattices. In particular, routine computation yields card {M (E8 )} = 240 (9.2) and card {M (Λ24 )} = 196560.
(9.3)
To end this section we cite a wellknown result about the representations of integral lattices that will be useful in Section 4 of this chapter. We omit its extremely complicated proof. Lemma 9.1 √ (Kneser [1]). Every integral lattice generated by vectors of norm 1 and 2 can be written as a direct sum of the special lattices Zn (n ≥ 1), An (n ≥ 1), Dn (n ≥ 4), and En (n = 6, 7, 8).
9.2. Two Theorems of Leven˘stein, Odlyzko, and Sloane
141
9.2. Two Theorems of Leven˘stein, Odlyzko, and Sloane Theorem 9.1 (Leven˘stein [2], Odlyzko and Sloane [1]). k ∗ (S8 ) = k(S8 ) = 240. Proof. Abbreviate the Jacobi polynomial Pk2.5,2.5 (t) to Pk , and deﬁne 320 1 2 2 1 (t + 1) t + t t− 3 2 2 16 200 832 = P0 + P1 + P2 + P3 7 63 231 1216 5120 2560 + P4 + P5 + P6 . 429 3003 4641
f (t) =
Clearly, f (t) satisﬁes the conditions of Lemma 8.1 for ϕ = π/3. Thus, by (8.1) and Lemma 8.1 it follows that k(S8 ) = m(8, π/6) ≤
f (1) = 240. f0
(9.4)
Then, (9.1), (9.2), and (9.4) together yield k ∗ (S8 ) = k(S8 ) = 240. 2
Theorem 9.1 is proved. Theorem 9.2 (Leven˘stein [2], Odlyzko and Sloane [1]). k∗ (S24 ) = k(S24 ) = 196560. Proof. Abbreviate the Jacobi polynomial Pk10.5,10.5 (t) to Pk , and deﬁne 1490944 1 2 1 2 2 1 2 1 (t + 1) t + t+ t t− t− 15 2 4 4 2 48 1144 12992 73888 = P0 + P1 + P2 + P3 + P4 23 425 3825 22185 2169856 59062016 4472832 + P5 + P6 + P7 687735 25365285 2753575 23855104 7340032 7340032 + P8 + P9 + P10 . 28956015 20376455 80848515
f (t) =
Then, f (t) satisﬁes the conditions of Lemma 8.1 for ϕ = π/3. Thus, by (8.1) and Lemma 8.1 we have k(S24 ) = m(24, π/6) ≤
f (1) = 196560. f0
(9.5)
142
9. The Kissing Numbers of S8 and S24
Then, (9.1), (9.3), and (9.5) together yield k ∗ (S24 ) = k(S24 ) = 196560. 2
Theorem 9.2 is proved. Remark 9.1. By constructing a suitable polynomial one can prove that k(S4 ) ≤ 25. On the other hand, by Theorem 2.6, we know that k∗ (S4 ) = 24.
Thus, k(S4 ) is either 24 or 25. However, the correct answer is still unknown. Remark 9.2. By choosing suitable polynomials, numerical upper bounds for δ(Sn ) and k(Sn ) can be obtained in special dimensions. For results of this kind we refer to Conway and Sloane [1].
9.3. Two Principles of Linear Programming For convenience, for two points u and v of En we write u v if ui ≤ vi holds simultaneously for i = 1, 2, . . . , n. Usually, a linear programming problem is to determine p = max {a, x : xA b, o x} ,
(9.6)
where a ∈ E n , b ∈ E m , and A is a m × n matrix. Its dual is to determine q = min {b, y : a yA, o1 y} ,
(9.7)
where o1 is the origin of E m . Now we introduce two fundamental principles concerning these problems that will be useful in the next section. Since they are wellknown and can be found in any book on linear programming (see Gr¨ otschel, Lov´ asz, and Schrijver [1] or Nemhauser and Wolsey [1]), we omit their proofs here. Principle 9.1. If p or q is ﬁnite, then p = q. Principle 9.2. If x∗ is an optimal solution of (9.6) and y∗ is an optimal solution of (9.7), writing s∗ = b − x∗ A and t∗ = y∗ A − a, then x∗i t∗i = 0,
i = 1, 2, . . . , n,
9.4. Two Theorems of Bannai and Sloane and
yj∗ s∗j = 0,
143
j = 1, 2, . . . , m.
Remark 9.3. In fact, Principle 9.2 can be easily deduced from Principle 9.1.
9.4. Two Theorems of Bannai and Sloane Let ℵ be an {n, m, ϕ} code. For u ∈ ℵ the inner product distribution of ℵ with respect to u is the set of numbers {a(u, t) : −1 ≤ t ≤ 1}, where a(u, t) = card {v ∈ ℵ : u, v = t} , and the inner product distribution of ℵ is the set of numbers {a(t) : −1 ≤ t ≤ 1}, where 1 a(t) = a(u, t). m u∈ℵ
In particular, writing γ = cos ϕ, we have a(1) = 1, a(t) = 0 for γ < t < 1, and a(t). m−1= −1≤t≤γ
Also, by Assertions 8.10 and 8.9, when α = (n − 3)/2 we have
a(t)Pkα,α (t) =
−1≤t≤1
1 Qk (x, y) mck x, y∈ℵ
=
d(k) 1 hk,i (x)hk,i (y) mck i=1 x, y∈ℵ
d(k)
1 = mck i=1
2 hk,i (x) ≥ 0.
(9.8)
x∈ℵ
Thus, for ﬁxed n and ϕ an upper bound to m is given by the following linear programming problem: p(n, ϕ) = max 1 + a(t) : − a(t)Pkα,α (t) ≤ Pkα,α (1) −1≤t≤γ
−1≤t≤γ
for k ≥ 1, a(t) ≥ 0 .
(9.9)
144
9. The Kissing Numbers of S8 and S24
Its dual can be formulated as l f (1) q(n, ϕ) = min : f (t) = fk Pkα,α (t), f0 > 0, fk ≥ 0 for k ≥ 1; f0 k=0
f (t) ≤ 0 for − 1 ≤ t ≤ γ .
(9.10)
Remark 9.4. Formulae (9.9), (9.10), and Principle 9.1 together imply Lemma 8.1. Theorem 9.3 (Bannai and Sloane [1]). There is a unique way (up to isometry) of arranging 240 nonoverlapping unit spheres in E8 so that they all touch another unit sphere. Proof. Let ℵ∗ = {u∗1 , u∗2 , . . . , u∗240 } be an {8, 240, π/3} code, and let {a∗ (t) : −1 ≤ t ≤ 1} be its inner product distribution. Then {a∗ (t) : −1 ≤ t ≤ 1} is an optimal solution to the linear programming problem (9.9) for n = 8 and ϕ = π/3, and the polynomial f (t) deﬁned in the proof of Theorem 9.1 is an optimal solution to its dual problem (9.10). Since the dual variables f1 , f2 , . . . , f6 are nonzero, by Principle 9.2, (9.8) holds with equality for k = 1, 2, . . . , 6. On the other hand, since the optimal polynomial f (t) does not vanish, except for t = −1, − 12 , 0, and 1 ∗ 2 , the primal variables must vanish everywhere except perhaps for a (−1), 1 ∗ ∗ ∗ 1 a (− 2 ), a (0), and a ( 2 ). From (9.8), these numbers satisfy the equations a∗ (t)Pkα,α (t) = −Pkα,α (1) −1≤t . 8 2 Case 3. 3 ≤ p2 ≤ 4. By (10.15) and (10.14), √ √ 3 3 3 27 3 3 det(Λ) ≥ p ≥ > . 8 8 2 √ Case 4. 1.3 ≤ p ≤ 2.1 and r2 ≥ 3. Let w2 be the lattice point in the plane {x : x3 = c2 } that is nearest to o. Then, o, w2 2 ≤ c22 +
p2 p2 q2 r2 + 2 ≤ c22 + + . 4 4 4 4
Hence, based on the choice of q2 , c22 +
p2 r2 + ≥ r2 , 4 4
3r2 − p2 9 − p2 ≥ . 4 4 Therefore, by (10.12) and routine analysis we have c22 ≥
√ p 3 3 2 2 (4 − p )(9 − p ) > . det(Λ) = pq2 c2 ≥ 2 2 √ Case 5. 1.3 ≤ p ≤ 2.1 and r2 ≤ 3. The triangle oq2 u2 is acuteangled and oq2 is the shortest side. Hence, by elementary geometry and (10.11), o, u2 2 + u2 , q2 2 = 2c22 + o, u∗2 2 + q2 , u∗2 2 = 2c22 + 2u∗2 , 12 q2 2 + 12 r2 ≥ 6.
(10.16)
We now need to prove that c22 ≥
9 − p2 . 4
(10.17)
If u∗2 , 12 q2 ≤ p/2, then (10.17) follows from (10.16). If u∗2 , 12 q2 > p/2, it follows from Lemma 10.4 that u∗2 , 12 p ≤ 12 r.
(10.18)
162
10. Multiple Sphere Packings
If the angle u2 op is obtuse, then the angle sop, where s = u2 + p, is acute and s∗ , 12 p ≤ u∗2 , 12 p ≤ 12 r,
(10.19)
where s∗ = u∗2 + p. Hence applying (10.11) to the triangle opt, where t = u2 or s, o, t2 + t, p2 = 2c22 + o, t∗ ∗ + p, t∗ 2 = 2c22 + 2t∗ , 12 p2 + 12 p2 ≥ 6,
(10.20)
where t∗ = u∗2 or s∗ . In this case, (10.17) follows from (10.18), (10.19), and (10.20). Then, by (10.12) and (10.17), √ p 3 3 2 2 det(Λ) = pq2 c2 ≥ (4 − p )(9 − p ) > . 2 2 Hence, (10.10) is proved. Clearly, (10.9) and (10.10) together yield Theorem 10.2. 2 Remark 10.5. Theorem 10.2 is the only known result about the exact values of δk (Sn ) and δk∗ (Sn ) when n ≥ 3 and k ≥ 2.
10.4. Remarks on Multiple Circle Packings In E 2 , much more explicit results are available. Since their proofs are elementary and technical, we only quote the results. k
δk∗ (S2 )
Author
2
π √ 3 3π 2
Heppes [1]
2π √ 3 2π √ 7 35π √ 8 6 8π √ 15 3969π √ √ √ 4 (220−2 193)(449+32 193)
Heppes [1]
3 4 5 6 7 8
Heppes [1]
Blundon [2] Blundon [2] Bolle [2] Yakovlev [1]
10.4. Remarks on Multiple Circle Packings
163
Several Vienna dissertations dealt with problems of this kind. Unfortunately, it is not easy to obtain them. Also, Linhart [1] and Temesv´ ari, Horv´ ath, and Yakovlev [1] developed algorithms with which one can determine δk∗ (S2 ) for each k to any prescribed accuracy. Multiple packings of general 2dimensional convex domains have been studied by many authors. For results of this kind we refer to G. Fejes T´ oth and W. Kuperberg [1].
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11. Holes in Sphere Packings
11.1. Spherical Holes in Sphere Packings In order to study the eﬃciency of a sphere packing, it is both important and interesting to investigate its holes, especially spherical holes. Let Sn +X be a sphere packing in E n , and let r(Sn , X) be the supremum of the radii of all spheres disjoint from any sphere of the packing. The number 1/r(Sn , X) is called the closeness of the packing Sn + X. We deﬁne r(Sn ) = inf {r(Sn , X)} , X
where the inﬁmum is taken over all sets X such that Sn + X is a packing. By routine argument it follows that there exists a discrete set X such that r(Sn ) = r(Sn , X). The corresponding packing Sn + X is called the closest sphere packing. Similarly, one can deﬁne r∗ (Sn ) by assuming that X is a lattice. Clearly, we have r(Sn ) ≤ r∗ (Sn ). These concepts were ﬁrst introduced by L. Fejes T´oth [11], who suggested the problems of determining the number r(Sn ) and the corresponding closest sphere packings. These problems, like Kepler’s conjecture and the GregoryNewton problem, are both natural and important. However, our knowledge about them is comparatively limited. In this section we will prove the following main result. Theorem 11.1 (B¨ or¨ oczky [2]). r(S3 ) = r∗ (S3 ) =
5/3 − 1.
166
11. Holes in Sphere Packings
Also, up to isometry, r(S3 , X) = r(S3 ) = r∗ (S3 ) if and only if X is a suitable spacecentered cubic lattice. To prove this theorem, we need several technical lemmas. Lemma 11.1 (B¨ or¨ oczky [2]). Let T = v1 v2 v3 v4 be a tetrahedron with circumradius r such that vi , vj ≥ 2, i = j, and the dihedral angle at edge v1 v2 is at most π/3. Then r ≥ 5/3, where equality holds if and only if T is congruent to the tetrahedron T = v1 v2 v3 v4 for which √ v1 , v3 = v2 , v4 = 4/ 3 and v1 , v2 = v2 , v3 = v3 , v4 = v4 , v1 = 2. Proof. Let S be the circumsphere of T centered at o. Lemma 11.1 will be proved by showing that any tetrahedron T inscribed in S and satisfying the conditions of the lemma is congruent to T .
v1 w1
v4 w2
v3
w3
w4 v2 Figure 11.1
Assume that v1 v2 is parallel to v1 v2 , and the four points v1 , v1 , v2 , and v2 are contained in a half great circle of S . Denote by G the part of bd(S ) cut out by the dihedral angle of T at the edge v1 v2 , by H the spherical set {x ∈ bd(S ) : x, v1 ≥ 2, x, v2 ≥ 2}, and by Q the intersection of G and H. Clearly, we have {v3 , v4 } ⊂ Q,
(11.1)
11.1. Spherical Holes in Sphere Packings
167
and Q is a spherical domain with four vertices, say w1 , w2 , w3 , and w4 in a circular order. Also, d(Q) = w1 , w3 = w2 , w4 ,
(11.2)
where {w1 , w3 } and {w2 , w4 } are the only solutions (see Figure 11.1). Since (11.1) and (11.2) hold, we may modify the considered tetrahedron as follows. Step 1. Replace v1 and v2 by v1 and v2 , respectively. Then, replace v3 and v4 by the new points w1 and w3 , respectively. Step 2. Move v3 and v4 along the boundary of H until the dihedral angle of T at v1 v2 is π/3. Clearly, d(Q) strictly increases during this process. v1
v4
m
o4
o3 v3
v2 Figure 11.2 Let T be a modiﬁed tetrahedron, m the midpoint of v1 v2 , oi the center of the circumcircle of v1 v2 vi , and write x = m, o3 and y = m, o4. Then, elementary but complicated computation (see Figure 11.2) yields that √ 3+1 1 ≤ xy ≤ 6 2 and 96(1 + xy) v3 , v4 2 = 36 − 2 2 ≤ 4, 2x y + 2xy + 3 where equality holds if and only if T is congruent to T . Hence, Lemma 11.1 is proved. 2 √ Write b = 4/ 3 and let Λ be the spacecentered cubic lattice with basis u1 = (b, 0, 0), u2 = (0, b, 0), and u3 = (b/2, b/2, b/2). Then, T is congruent to the tetrahedron with vertices u1 , u1 + u2 , u3 , and u3 + u1 . Also, since the DirichletVoronoi cells of Λ are regular truncated octahedra, the solid
168
11. Holes in Sphere Packings
angle formed by the three planes perpendicular to the edges of T meeting at a vertex is congruent to the solid angle at a vertex of a regular truncated octahedron. Lemma 11.2 (B¨ or¨ oczky [2]). There is no convex polyhedron bounded by one quadrangle and ten pentagons such that in each vertex of the polyhedron exactly three facets meet. Proof. Suppose there is such a polyhedron, say P . We proceed to deduce a contradiction. Let v1 , v2 , v3 , and v4 be the vertices of the quadrangular facet of P , in a circular order (see Figure 11.3). Let wi be the endpoint of the third edge of P emanating from vi , i = 1, . . . , 4. The points wi are diﬀerent from one another, and diﬀerent from the points vi , i = 1, . . . , 4. For convenience, we write v5 = v1 and w5 = w1 . Then, for 1 ≤ i ≤ 4, the points wi , vi , vi+1 , and wi+1 are four consecutive vertices of a pentagonal facet of P . Let xi be the ﬁfth vertex of this facet. Again, the points vi , wi , xi , i = 1, . . . , 4, are all distinct. In the part of P considered so far the points vi and wi have valency 3, while xi has valency 2. Thus, there is a third vertex, say yi , that is connected by an edge to xi . Again for convenience, we write x5 = x1 and y5 = y1 . It is easy to see that yi = xi+1 and yi = yi+1 , since otherwise we would obtain a triangular or quadrangular facet. If we have yi = xi+2 for i = 1 or 2, then yi+1 has valency 1; if we have yi = yi+2 for i = 1 or 2, then yi = yi+1 = yi+2 . In the second case, xi wi+1 xi+1 yi and yi xi+1 wi+2 xi+2 are quadrangular facets of P . Hence, the points yi are diﬀerent from one another and from the points vi , wi , and xi , i = 1, . . . , 4. y1
y2 w2
x1
x2
v2 w1
v3
v1
w3
v4 x4
w4
x3 y3
y4 Figure 11.3
Now, yi , xi , wi+1 , xi+1 , and yi+1 are ﬁve consecutive vertices of a facet of P ; thus the points yi and yi+1 are connected by an edge. This means that all the vertices considered so far are trivalent, and our polyhedron has only ten facets, two quadrangles, and eight pentagons. By this contradiction Lemma 11.2 is proved. 2
11.1. Spherical Holes in Sphere Packings
169
It is wellknown (see Coxeter [5]) that there are eight diﬀerent types of regular polytopes in E 4 , one of which, say P , is bounded by 120 regular dodecahedra such that at each vertex four of them meet. A cell complex {F, ≺} is a family F of cells (polytopes and their faces) with an incidence relation ≺ (for example, the relation ⊂). Two cell complexes {F1 , ≺1 } and {F2 , ≺2 } are isomorphic if there is a onetoone mapping f : F1 → F2 such that f (X) ≺2 f (Y ) if and only if X ≺1 Y for X, Y ∈ F1 . Lemma 11.3 (B¨ or¨ oczky [2]). There is no tetravalent facettofacet tiling in E 3 such that each tile has exactly 12 pentagonal facets. Proof. Let ℘ be the family of all faces of P . Then, {℘, ⊂} is a cell complex. Without loss of generality, we label the facets of P as F1 , F2 , . . . , F120 such that for k = 1, 2, . . . , 119, the set Uk =
k
Fi
i=1
is homeomorphic to S3 and the set Uk = Fk ∩
k−1
Fi
i=1
is homeomorphic to S2 . For convenience, we write ℘k = {X ∈ ℘ : X ⊂ Uk } , ℘k = X ∈ ℘ : X ⊂ Uk+1 , and ℘120 = ℘. Suppose, on the contrary, there is a tetravalent facettofacet tiling in E3 such that each tile has exactly 12 pentagonal facets. Let P1 be an arbitrary tile of the tiling, let 1 be the family consisting of P1 and all its faces, and let ≺1 be the relation ⊂. Clearly, the two cell complexes {℘1 , ⊂} and {1 , ≺1 } are isomorphic. Let f1 be an isomorphism between them. Now, for k = 1, 2, . . . , 120, we proceed to deﬁne cell complexes {k , ≺k } and isomorphisms fk between {℘k , ⊂} and {k , ≺k } by induction. In general, the cell complex {k , ≺k } has the following properties: 1. k is a family consisting of some tiles of the tiling and all their faces. 2. If X ≺k Y , then X ⊂ Y . Suppose that we have a cell complex {k , ≺k } with the above properties and an isomorphism fk between {℘k , ⊂} and {k , ≺k }. Let {k , ≺k } be
170
11. Holes in Sphere Packings
the subcomplex of {k , ≺k } corresponding to {℘k , ⊂} at the isomorphism fk , let Qk ∈ k be a facet of some tile, say Pk ∈ k , and let Pk+1 be the tile such that Qk = Pk ∩ Pk+1 . Since Uk+1 is homeomorphic to S2 , all the facets in k are connected by edges of the tiles. On the other hand, since the tiling is tetravalent and is facettofacet, at every edge three tiles join properly together. Thus, Pk+1 is independent of the choice of Qk . Let k+1 be the family consisting of k , Pk+1 , and all the faces of Pk+1 , and deﬁne the relation ≺k+1 by X ≺k Y if X, Y ∈ k , X ≺k+1 Y = X ⊂Y if Y ∈ k+1 \ k . Then, it is easy to show that the cell complex {k+1 , ≺k+1 } satisﬁes the required properties and is isomorphic to {℘k+1 , ⊂}. As a result of this process, we obtain a cell complex {120 , ≺120 } that satisﬁes properties 1 and 2 and is isomorphic to {℘, ⊂}. Obviously, this contradicts the fact that P1 , P2 , . . . , P120 are tiles of the tiling. Thus, Lemma 11.3 is proved. 2 Now we introduce a fundamental result concerning cap packings and cap coverings in bd(S3 ) that will be useful in the proof of Theorem 11.1. Lemma 11.4 (L. Fejes T´ oth [9]). Let γ(m) be the maximum spherical radius such that m caps of such radius can be packed into bd(S3 ). Let γ (m) be the minimum spherical radius such that m caps of such radius can cover bd(S3 ), and write m θm = π. 6(m − 2) Then, γ(m) ≤ arccos and γ (m) ≥ arccos
1
2 cosec θm
"
1 3 cot θm
,
where equalities hold only if m = 3, 4, 6, or 12. Proof. Let Ω1 , Ω2 , . . . , Ωm be m caps, of spherical radius γ, centered at o1 , o2 , . . . , om , respectively, and let D1 , D2 , . . . , Dm be the corresponding spherical DirichletVoronoi cells. It is easy to see that Ωi ⊂ Di if the caps form a packing in bd(S3 ), Di ⊂ Ωi if they form a covering in bd(S3 ), and that the DirichletVoronoi cells form a tiling in bd(S3 ). For convenience, we assume that the Ωi are inscribed in the Di in the packing case and that the Di are inscribed in the Ωi in the covering case. Let k be the number of edges of the tiling. By Euler’s formula and the fact that at least three edges meet at every vertex, it can be deduced that k ≤ 3(m − 2).
(11.3)
11.1. Spherical Holes in Sphere Packings
171
Joining each oi to the vertices of Di and its perpendicular feet on the edges of Di by arcs, bd(S3 ) is divided into 4k spherical triangles. Consider one of them, say T , and assume that its spherical angle at the corresponding oi is α. It is wellknown that s(T ) = α − arcsin(cos γ · sin α)
(11.4)
in the packing case, and s(T ) = α − arctan(cos γ · tan α)
(11.5)
in the covering case. Since (11.4) is a convex function of α and (11.5) is a concave function of α when 0 < α < π/2, by Jensen’s inequalities we have mπ mπ 4π = T ≥ 4k − arcsin cos γ · sin 2k 2k and therefore cos γ ≥
sin (m−2)π 2k sin mπ 2k
in the packing case. Also, mπ mπ 4π = T ≤ 4k − arctan cos γ · tan , 2k 2k and hence cos γ ≤
tan (m−2)π 2k tan mπ 2k
in the covering case. Thus, by (11.3) we have cos γ(m) ≥
sin(π/6) 1 = cosec θm sin θm 2
(11.6)
tan (π/6) 1 = √ cot θm tan θm 3
(11.7)
in the packing case, and cos γ (m) ≤
in the covering case. Checking the cases of equality in every step, it is easy to see that equalities hold in (11.6) and (11.7) only if m = 3, 4, 6, or 12. Thus, Lemma 11.4 follows from (11.6) and (11.7). 2 With these preparations, Theorem 11.1 can be proved as follows. Proof of Theorem 11.1. For convenience, we write τ = 5/3. In E 3 , let X = {x1 , x2 , . . . } be a discrete set such that S3 + X is a packing and τ S3 + X is a covering, and let Di be the DirichletVoronoi cell associated with xi . Obviously, we have S3 + xi ⊂ Di ⊂ τ S3 + xi ,
i = 1, 2, . . . .
(11.8)
172
11. Holes in Sphere Packings
First, we proceed to show that the tiling formed by the DirichletVoronoi cells is tetravalent. In other words, at every vertex four DirichletVoronoi cells meet. Suppose, on the contrary, a vertex v belongs to ﬁve cells, say D1 , D2 , . . . , D5 . Then xi ∈ bd(dS3 ) + v,
i = 1, 2, . . . , 5,
where d is the distance between v and x1 . It follows from routine combinatorial argument that two of these ﬁve points, say x1 and x2 , satisfy
x1 vx2 ≤ π/2.
Thus, since x1 , x2 ≥ 2, we obtain d = v, x1 ≥ cosec (π/4) =
√
2>
5/3 = τ,
which contradicts (11.8). As a simple consequence, at every vertex of every DirichletVoronoi cell three edges meet. x2
v3 F v2 v1 x1 Figure 11.4 Second, we claim that none of the DirichletVoronoi cells has a facet with more than six vertices. Suppose, on the contrary, that one of the cells, say D1 , has a facet F with more than six vertices v1 , v2 , . . . , vl , and assume that v2 is also a vertex of D2 , D3 , and D4 ,
v1 v2 v3 > 2π/3,
and x1 x2 is perpendicular to F . Then the dihedral angle of the tetrahedron x1 x2 x3 x4 at the edge x1 x2 is (see Figure 11.4) π − v1 v2 v3 < π/3. Thus, Lemma 11.1 implies x1 , v2 > τ,
11.1. Spherical Holes in Sphere Packings
173
which contradicts (11.8). Next, we are going to show that one of the DirichletVoronoi cells, say D1 , has a hexagonal facet. Assume that D1 has k facets with e1 , e2 , . . . , ek edges, respectively. It follows from Euler’s formula and the fact that three edges meet at every vertex that k
ei = 6k − 12.
i=1
Thus, keeping the second claim in mind, the assertion can be easily shown if k > 12. We proceed to deduce the assertion by excluding the following cases. Case 1. k ≤ 10. It follows from Lemma 11.4 that there is a vertex v of D1 such that v, x1 ≥ sec γ (10) > τ, which contradicts (11.8). Case 2. k = 11 and ei ≤ 5 for i = 1, 2, . . . , 11. By Euler’s formula we have, without loss of generality, e1 = 4 and e2 = · · · = e11 = 5. Then, by Lemma 11.2 this is impossible. Case 3. Each of the DirichletVoronoi cells has exactly 12 facets, each of which has at most 5 edges. Then, Euler’s formula implies that every facet of every cell is a pentagon. This case is excluded by Lemma 11.3. Let F1 = v1 v2 . . . v6 be a hexagonal facet of D1 . Since no angle of F1 can be greater than 2π/3, every angle of F1 is 2π/3. Without loss of generality, we may assume that F1 = D1 ∩ D2 and let x2+i , i = 1, 2, . . . , 6, be the points such that vi vi+1 is the radical axis of x1 , x2 , and xi+2 , where v7 = v1 . Then, the six tetrahedra x1 x2 x3 x4 , x1 x2 x4 x5 , . . . , x1 x2 x8 x3 satisfy the condition of Lemma 11.1, and their circumradii are not greater than τ . Thus all of them are congruent to T and, in particular, x1 , x2 = 2 and √ x1 , x3 = x1 , x5 = x1 , x7 = 4/ 3. (11.9) For convenience, we denote the midpoint of x1 xi by mi. Then, by routine computation it follows that m2, vi = 2/3, i = 1, 2, . . . , 6, and therefore, since every angle of F1 is 2π/3, F1 is a regular hexagon. By the observation just above Lemma 11.2, we can draw a regular truncated octahedron G of side length 2/3 and centered at x1 , say x1 = o, such that F1 is a facet of G and the facets of G and D1 adjacent to F1 lie in the same planes. Without loss of generality, we may assume that the three facets of G along the edges v1 v2 , v3 v4 , and v5 v6 are squares.
174
11. Holes in Sphere Packings
We proceed to show that D1 = G. First, we prove that the three facets F2 , F4 , and F6 of D1 that join F1 at edges v1 v2 , v3 v4 , and v5 v6 , respectively, are squares (see Figure 11.5). By (11.9) and Lemma 11.1, every angle of F2 is less than 2π/3. This, together with the fact that the angles of F2 at v1 and v2 are π/2, implies that F2 is a quadrangle, say v1 v2 w2 w1 . Let x9 be the point such that the line w1 w2 is the radical axis of x1 , x1 x3 x9 . Since x1 , x9 ≥ 2, x3 , and x9 . Now, we consider the triangle √ x3 , x9 ≥ 2, and x1 , x3 = b = 4/ 3, we have x1 x9 x3 < π/2. Thus, routine analysis yields that the distance from the center of the circumcircle of x1 x3 x9 to the line x1 x3 attains its minimum 1/6 when x1 , x9 = x3 , x9 = 2. This means that the line w1 w2 does not intersect the open circle C = x ∈ H2 : x, m3 < 1/6 , where H2 indicates the plane determined by F2 and mi indicates the midpoint of x1 xi . On the other hand, the distance between x1 and any point x ∈ H2 that does not belong to the circle C = x ∈ H2 : x, m3 ≤ 1/3 is greater than τ . Thus, both w1 and w2 belong to C . We observe that C and C are, in fact, the incircle and circumcircle of the corresponding square facet F2 of G. Therefore, we have F2 = F2 . x2 x3 F1 v3
m2 v2
v1 m3
F2 x9
m9
x1 = o Figure 11.5
Let F3 be the facet of D1 joining F1 at v2 v3 , and let v2 , v3 , u1 , u2 , u3 , and w2 be its consecutive vertices. Then, v2 , w2 ∈ F2 and v3 , u1 ∈ F4 . By considering the angles at u1 and w2 , it follows that the edges u1 u2 and u3 w2 lie on the lines determined by the corresponding edges of G. If
11.1. Spherical Holes in Sphere Packings
175
u2 = u3 , routine computation yields that x1 , u2 > τ. Hence, F3 is a regular hexagon. Repeating this process, it follows that D1 = G. Then, by considering the neighbors of D1 , we obtain that Di = G + xi and therefore X is a spacecentered cubic lattice. Theorem 11.1 follows. 2 Clearly, we have r(Sn ) < 1 in every dimension. We also have the following lower bound for r(Sn ). Theorem 11.2 (Ry˘skov and Horv´ ath [1]). 8 r(Sn ) ≥
n
θ(Sn ) − 1 0.51466. δ(Sn )
Proof. Let Sn + X be a sphere packing in which the maximal radius of the spherical holes is r(Sn ). Then (1 + r(Sn ))Sn + X is a covering in E n . From the deﬁnitions of δ(Sn ) and θ(Sn ) it follows that card{X ∩ lIn }v((1 + r(Sn ))Sn ) v(lIn ) card{X ∩ lIn }v(Sn ) n = lim (1 + r(Sn )) l→∞ v(lIn ) n ≤ (1 + r(Sn )) δ(Sn ).
θ(Sn ) ≤ lim
l→∞
In other words,
8 r(Sn ) ≥
n
θ(Sn ) − 1. δ(Sn )
Also, by Theorem 3.4, Corollary 7.2, and Theorem 8.2 we have 8 θ(Sn ) r(Sn ) ≥ n − 1 0.51466. δ(Sn ) 2
Theorem 11.2 is proved. To end this section, let us list a couple of open problems. Problem 11.1. Is it true that for all n ≥ 2, r(Sn+1 ) ≥ r(Sn )? Problem 11.2. Determine the value of lim r(Sn ). n→∞
176
11. Holes in Sphere Packings
11.2. Spherical Holes in Lattice Sphere Packings Let Sn + Λ be a lattice sphere packing in E n , and write µ(Λ) = r(Sn , Λ) + 1.
(11.10)
Usually, µ(Λ) is called the depth of the deepest holes of Λ. Clearly, we have µ(Λ) = maxn min {x, z} = max {o, x} . x∈E
z∈Λ
x∈D(o)
(11.11)
In this section we will introduce several results about r∗ (Sn ) by studying µ(Λ). Theorem 11.3 (Rogers [3]). r∗ (Sn ) < 2. Proof. More generally, we prove the following constructive assertion. Assertion 11.1. If Sn + Λ is a lattice sphere packing with µ(Λ) ≥ 3, then there exists a new lattice sphere packing Sn + Λ such that Λ ⊂ Λ and det(Λ ) = 13 det(Λ). It follows from (11.11) that there is a point x ∈ En such that x, z ≥ µ(Λ),
z ∈ Λ.
(11.12)
Also, it is clear that µ( 13 Λ) = 13 µ(Λ), and therefore there exists a lattice point u ∈ Λ such that x, 13 u ≤ 13 µ(Λ).
(11.13)
Clearly, 13 u ∈ Λ and thus det(Λ ) = 13 det(Λ), where Λ = Λ
1
3u
+Λ
2
3u
+Λ .
Then it follows from (11.12), (11.13), and the assumption of Assertion 11.1 that 13 u, z ≥ x, z − x, 13 u ≥ 23 µ(Λ) ≥ 2 holds for every point z ∈ Λ, and therefore u , v ≥ 2
11.2. Spherical Holes in Lattice Sphere Packings
177
for any distinct points u and v of Λ . In other words, Sn + Λ is a packing. Clearly, Theorem 11.3 is a consequence of (11.10) and Assertion 11.1. 2 Using a very complicated existence method, based upon the work of C.A. Rogers and C.L. Siegel, G.J. Butler was able to improve the previous theorem as follows. Theorem 11.4 (Butler [1]). r∗ (Sn ) ≤ 1 + o(1). Remark 11.1. Both Rogers’ original result and Butler’s improvement dealt not only with spheres, but also with general centrally symmetric convex bodies. By modifying Rogers’ constructive method for spheres, Henk [1] obtained √ r∗ (Sn ) ≤ 21/2 − 1. Remark 11.2. Analogously to Theorem 11.2 we have the follwing lower bound for r∗ (Sn ): 8 θ∗ (Sn ) ∗ r (Sn ) ≥ n ∗ − 1 0.51466. δ (Sn ) By studying diﬀerent patterns of the DirichletVoronoi cells D(o) of ath proved the following the lattice sphere packings in E4 and E 5 , J. Horv´ result. Theorem 11.5 (Horv´ ath [2]). " √ √ 3−1 −1 r∗ (S4 ) = 2 3 8
and r∗ (S5 ) =
√ 3 13 + − 1. 2 6
The proofs of both Theorem 11.4 and Theorem 11.5 are very complicated. We omit them here. We list three more problems. Problem 11.3. Is it true that for all n ≥ 2, r∗ (Sn+1 ) ≥ r∗ (Sn )? Problem 11.4. Determine the value of lim r∗ (Sn ). n→∞
Problem 11.5. Is there a dimension n such that r∗ (Sn ) ≥ 1?
178
11. Holes in Sphere Packings
It is conjectured (see Conway and Sloane [1], Gruber and Lekkerkerker [1], and Rogers [14]) that the answer to Problem 11.5 is yes. If so, δ(Sn ) > δ ∗ (Sn ) holds in the corresponding dimension.
11.3. Cylindrical Holes in Lattice Sphere Packings In 1960, in a short note A. Heppes proved the following result: In every threedimensional lattice sphere packing there is a straight line of inﬁnite length that does not intersect any of the spheres. In fact, his proof implied the existence of cylindrical holes (of inﬁnite length) in every threedimensional lattice sphere packing. Let ρ∗ (Sn ) be the maximum number such that in every ndimensional lattice sphere packing Sn + Λ there is a cylindrical hole with an (n−1)dimensional spherical base of radius ρ∗ (Sn ). Improving Heppes’ result, I. Hortob´ agyi determined the value of ρ∗ (S3 ). Also, J. Horv´ ath and S.S. Ry˘skov obtained a lower bound for ρ∗ (Sn ) in high dimensions. In this section we will introduce the main results of this kind. For convenience, we denote by H(x1 , x2 , . . . , xm ) the hyperplane determined by x1 , x2 , . . . , xm . A Seeber set of an ndimensional lattice Λ is a set of n points {u1 , u2 , . . . , un } such that for i = 1, 2, . . . , n, ui is a point of Λ \ H(o, ui−1 , ui−2 , . . .) with minimal norm. We now introduce a basic lemma concerning the structure of a lattice. Lemma 11.5 (Horv´ ath [3]). Let Λ be a packing lattice of Sn and let {u1 , u2 , . . . , un } be a Seeber set of Λ. Denote by Hi the hyperplane {x : x, ui = 0} and by pi (x) the projection of x onto Hi . Then √ o, pi (u) ≥ 3 holds for every point u ∈ Λ \ H(o, ui ). Proof. First we proceed to show that for any point u ∈ Λ, o, ui ≤ max {o, u, u, ui} .
(11.14)
If u ∈ H(o, . . . , ui−1 ), then from the deﬁnition of a Seeber set it follows that o, ui ≤ o, u. If u ∈ H(o, . . . , ui−1 ), then u − ui ∈ H(o, . . . , ui−1 ) and hence o, ui ≤ u, ui .
11.3. Cylindrical Holes in Lattice Sphere Packings
179
These two cases imply (11.14).
ui u
o
pi (u)
Hi Figure 11.6 Now, assuming u ∈ Λ \ H(o, ui ), we consider the triangle ouui . It follows from (11.14) that oui is not the longest side of the triangle, and therefore (see Figure 11.6) max { oui u, uoui } ≥ π/3. On the other hand, since pi (u) = pi (u + kui ) for any integer k, we may assume that max { oui u, uoui } ≤ π/2. Thus, since o, u ≥ 2 and u, ui ≥ 2, o, pi(u) ≥ 2 sin (π/3) =
√ 3. 2
Lemma 11.5 is proved.
Although it is simple, this lemma plays an important role in the work of Heppes, Hortob´ agyi, Horv´ ath, and Ry˘skov on the cylindrical holes in lattice sphere packings. Theorem 11.6 (Heppes [2] and Hortob´ agyi [1]). ρ∗ (S3 ) =
√ 3 2 − 1. 4
In other words, every lattice sphere packing S3 + Λ has a cylindrical hole √ with a circular base of radius 3 2/4 − 1. Proof. Let Λ be a packing lattice of S3 , and let {u1 , u2 , u3 } be a Seeber set of √ Λ. We aim to show that if rS2 + p1 (Λ) is a covering in H1 , then r ≥ 3 2/4. In other words, letting T = v1 v2 v3 (where v1 = o) be a
180
11. Holes in Sphere Packings
triangle with vi ∈ p1 (Λ) and letting r(T ) be its circumradius, we proceed to show that √ 3 2 r(T ) ≥ . (11.15) 4 For convenience, we denote the circumcircle of T by S, and denote the cylinder with base S by Q.
u1 u2
u3 v2
o H1
v3 Figure 11.7
√ By Lemma 11.5, every side of T has√length not less than 3. Thus, simple computation yields that r(T ) ≥ 3 2/4 if one of its sides is greater than 2. So, without loss of generality, we may assume that o, u1 = 2, vi = p1 (ui ), i = 1, 2, 3, √ 3 ≤ vi , vj ≤ 2, i = j, and that the dihedral angle of ou1 u2 u3 at the edge ou3 is not greater than π/2. We deal with two cases (see Figure 11.7). Case 1. Both u2 and u3 lie on the same side of the plane H1 + 12 u1 . Without loss of generality, we may assume that o, u2 , and u3 lie on the same side of the plane H1 + 12 u1 and that u3 is the nearest to it. Then, we adjust the positions of u2 and u3 using the following process: Step 1. Replace u3 by its projection onto the plane H1 + 12 u1 . Step 2. Move u3 along the boundary of S + 12 u1 towards 12 u1 until u1 , u3 = o, u3 = 2. Step 3. Move u2 along the line parallel with ou1 until we have u2 , u3 = 2. Step 4. Move u2 along the curve bd(2S3 ) + u3 ∩ bd(Q) towards the
11.3. Cylindrical Holes in Lattice Sphere Packings
181
plane H1 + 12 u1 until we have o, u2 = u2 , u3 = 2.
√ Step 5. Move u2 along the circle {x : x − 12 u3 , u3 = 0, x, 12 u3 = 3} in the direction from u2 to u1 so that u1 , u2 = 2. Then we obtain a regular tetrahedron ou1 u2 u3 . Simple analysis shows that in this process the corresponding lattices are packing lattices of S3 and the corresponding circumradii r(T ) do not increase. Thus, (11.15) is true in this case. Case 2. The plane H1 + 12 u1 separates u2 and u3 . Without loss of generality, we may assume that o and u2 lie on the same side of H1 + 12 u1 . Then, we adjust the positions of u2 and u3 as follows: Step 1. Move u2 along the line parallel with ou1 towards H1 until o, u2 = 2. Then until u2 , u3 = 2. Step 2. Move u3 along the curve move u3 similarly bd(2S3 ) + u2 ∩ bd(Q) towards the plane H1 + 12 u1 so that u1 , u3 = 2. Then, we have o, u1 = o, u2 = u2 , u3 = u1 , u3 = 2, o, u3 > 2, and u1 , u3 > 2. Step 3. Move u2 along the circle {x : x− 12 u3 , u3 = 0, x, 12 u3 = c}, where c is a constant, in the direction from u2 to u1 so that u1 , u2 = 2. Then, adjust the position of u3 accordingly. At the end of this process we obtain a regular tetrahedron ou1 u2 u3 . Just as in Case 1, in this process the corresponding lattices are packing lattices of S3 and the corresponding circumradii r(T ) do not increase. Therefore, (11.15) is true in this case. Hence, Theorem 11.6 is proved.
2
Remark 11.3. By looking at the proof more closely, it seems that the extreme lattice sphere packing is unique up to isometry. Between ρ∗ (Sn ) and r∗ (Sn ) we have the following relationship, which provides a lower bound for ρ∗ (Sn ). Theorem 11.7 (Ry˘skov and Horv´ ath [1]). √ 3(1 + r∗ (Sn−1 )) ρ∗ (Sn ) ≥ − 1 0.3117. 2 Proof. Let Sn + Λ be a lattice sphere packing with the thinnest cylinder hole, and let {u1 , u2 , . . . , un } be a Seeber set of Λ. Then, it follows from Lemma 11.5 that √ ( 3/2)Sn−1 + pi (Λ) is a lattice sphere packing in Hi . From the deﬁnition of r∗ (Sn−1 ) we have √ 3(1 + r∗ (Sn−1 )) r≥ 2
182
11. Holes in Sphere Packings
if rSn−1 + pi (Λ) is a covering in Hi . On the other hand, we have pi (Sn + u) = Sn−1 + pi (u) for every point u ∈ Λ. Thus, using Remark 11.2, √ 3(1 + r∗ (Sn−1 )) ∗ ρ (Sn ) ≥ − 1 0.3117. 2 2
Theorem 11.7 is proved.
Remark 11.4. From the proof of Theorem 11.7 it follows that every lattice sphere√packing Sn + Λ has cylindrical holes, with spherical bases of radii at least 3(1 + r∗ (Sn−1 ))/2 − 1, in n independent directions. Analogous to Problems 11.2 and 11.4, we have the following problem concerning ρ∗ (Sn ). Problem 11.6. Determine the value of lim ρ∗ (Sn ). n→∞
12. Problems of Blocking Light Rays
12.1. Introduction We call a straight line with one end extending to inﬁnity a light ray. Let h(Sn ) be the smallest number such that there exists a ﬁnite sphere packing Sn + X, with X = {o, x1 , x2 , . . . , xh(Sn ) }, such that every light ray starting from o is blocked by one of the translates Sn + xi , xi ∈ X \ {o}. Similarly, when X is restricted to being a subset of a lattice, we denote the corresponding number by h∗ (Sn ). Hornich proposed the following problem (see L. Fejes T´oth [7]). Hornich’s Problem: Determine the numbers h(Sn ) and h∗ (Sn ). In 1959, L. Fejes T´ oth [7] obtained the ﬁrst result in this direction, proving h(S3 ) ≥ 19. He also proposed another related problem. Let (Sn ) be the smallest number such that there is a ﬁnite packing Sn + Y , where Y = {o, y1 , y2 , . . . , y (Sn ) }, and such that every light ray starting from any point of Sn is blocked by one of the translates Sn + yj , yj ∈ Y \ {o}. His problem can be formulated as follows. L. Fejes T´ oth’s Problem: Determine the number (Sn ). Remark 12.1. Let ∗ (Sn ) be the corresponding number in the lattice case. It follows from Theorem 11.6 and Theorem 11.7 that when n ≥ 3, ∗ (Sn ) = ∞.
184
12. Problems of Blocking Light Rays
It is obvious that
h(Sn ) ≤ h∗ (Sn ),
(12.1)
h(Sn ) ≤ (Sn ),
(12.2)
and, in particular, h(S2 ) = h∗ (S2 ) = (S2 ) = ∗ (S2 ) = 6. Also, relating k(Sn ) and h(Sn ) we have the following simple result. Theorem 12.1. If n ≥ 3, then k(Sn ) < h(Sn ). Proof. Let Sn , Sn +x1 , . . . , Sn +xk(Sn ) be an optimal kissing conﬁguration, and let Sn , Sn + y1 , . . . , Sn + yh(Sn ) be a ﬁnite packing such that every light ray starting from o is blocked by one of the translates Sn + yj , j = 1, 2, . . . , h(Sn ). For any point x, let Ω(x) be the intersection of bd(Sn ) and the smallest cone with vertex o, containing Sn +x. Clearly, Ω(x) is a cap on bd(Sn ). For convenience, we denote the surface area of a cap of geodesic radius π/6 by µn . From the deﬁnitions of k(Sn ) and h(Sn ) it follows that the caps Ω(xi ), i = 1, 2, . . . , k(Sn ), form a packing in bd(Sn ), the caps Ω(yj ), j = 1, 2, . . . , h(Sn ), form a covering in bd(Sn ), s(Ω(xi )) = µn , s(Ω(yj )) ≤ µn , and therefore
k(Sn )
k(Sn )µn =
s(Ω(xi )) < nωn
i=1
h(Sn )
δ(C, Λ) 2n = δ(C), =
which is impossible. Therefore, int 12 C + i1 v + u1 ∩ int 12 C + i2 v + u2 = ∅ holds for certain 0 ≤ i1 < i2 ≤ m, u1 ∈ Λ, and u2 ∈ Λ. In other words, (i2 − i1 )v ∈ int(C) + u1 − u2 . Thus, taking j = i2 − i1 and u = u1 − u2 , Lemma 12.1 is proved.
2
Theorem 12.2 (Zong [8]). 2
δ(C)n 2n (1+o(1)) h(C) ≤ h (C) ≤ , δ ∗ (C)n ∗
h(Sn ) ≤ h∗ (Sn ) ≤ 21.401n
2
(1+o(1))
.
Proof. It is wellknown from convex geometry (see John [1] or Gruber and Lekkerkerker [1]) that for any ndimensional centrally symmetric convex body C there is a linear transformation L such that √ Sn ⊆ L(C) ⊆ nSn . Thus, without loss of generality, we may assume that √ Sn ⊆ C ⊆ nSn .
(12.3)
Let C + Λ be a lattice packing with density δ∗ (C). By Lemma 12.1, for √ every vector v with v = n + 1 there is a positive integer j≤
2n δ(C) δ ∗ (C)
and a point u ∈ Λ \ {o} such that jv ∈ int(C) + u.
(12.4)
Let D be a DirichletVoronoi cell associated with Λ. By Assertion 11.1, Remark 11.1, and (12.3) we have √ d(D) ≤ 3( n + 1). (12.5)
12.2. Hornich’s Problem Thus, writing
187 6
7 2n δ(C) m= , δ ∗ (C)
by (12.3), (12.4), and (12.5) we have √ h∗ (C) ≤ card m( n + 1)Sn ∩ Λ √ n ωn (m( n + 1) + d(D)) δ ∗ (C) ≤ v(C) √ √ n ωn (m( n + 1) + 3( n + 1)) δ ∗ (C) ≤ ωn 2
δ(C)n 2n (1+o(1)) ≤ . δ ∗ (C)n This proves the ﬁrst part of Theorem 12.2. Then, taking C = Sn , the second part follows from Theorem 3.1 and Theorem 8.2. 2 Remark 12.4. The original result of Zong [8], which is based on the numbertheoretic version of Dirichlet’s approximation theorem, is weaker than Theorem 12.2. As a counterpart to Theorem 12.2 we have the following lower bound for h(Sn ). Theorem 12.3 (B´ ar´ any and Leader [1]). 2
h(Sn ) ≥ 20.275n
(1+o(1))
.
Proof. For convenience we write r = 20.275n , µ = 1/n, and ri = 2 + iµ. Let X be a subset of rSn such that Sn + X ∪ {o} is a ﬁnite packing, and write Xi = {x ∈ X : ri ≤ x < ri+1 } . Clearly we have Xi = ∅ if i > nr, and X=
nr
Xi .
(12.6)
i=0
Suppose that x ∈ Xi , and denote by ϕ/2 the spherical radius of the cap (Sn + x) ∩ bd(ri Sn ). Then (see Figure 12.1), for large n, 8 $ 2
2 2 ri + ri+1 −1 ϕ ϕ 2 sin = 1 − cos ≥ 1− 2 2 2ri ri+1 8
1 15 n − 4 ≥ . ri 16 n
188
12. Problems of Blocking Light Rays
Thus, by (8.1) and (8.21) we have card{Xi} ≤ m(n, ϕ/2) = m[n, ϕ] −n ≤ 20.599 sin(ϕ/2) ≤ rn 2−0.552n(1+o(1)).
(12.7)
Sn + x ri Sn
V (p)
Sn
Sn + p
Figure 12.1 Let p be an arbitrary point in E n . Let V (p) be the cone with vertex o over Sn + p, denote by Ω(p) the cap V (p) ∩ bd(Sn ), and let β be the spherical radius of Ω(p). Then, % s(Ω(p)) (n − 1)ωn−1 β ≤ (sin θ)n−2 dθ s(Sn ) nωn 0 ωn−1 (sin β)n−1 ≤ np1−n . ≤ nωn
(12.8)
By (12.6), (12.7), and (12.8) we have
nr s(Ω(x)) x∈X s(Ω(x)) = s(Sn ) nωn i=0 x∈Xi
≤ 2−0.552n(1+o(1))
nr
ri
i=0 2 −0.552n(1+o(1))
≤r 2 = 2−0.002n(1+o(1)).
This means that Sn + X can block at most half of the light rays starting from o when n is large. Therefore, considering the light rays not blocked by Sn + X and using (12.8) we have h(Sn ) ≥
2 rn−1 ≥ 20.275n (1+o(1)) . 2n
12.3. L. Fejes T´ oth’s Problem
189 2
Theorem 12.3 is proved. To end this section, we propose the following problem. Problem 12.1. Do there exist absolute constants c and c∗ such that 2
h(Sn ) = 2cn
(1+o(1))
or
h∗ (Sn ) = 2c
∗
n2 (1+o(1))
?
If so, determine them.
12.3. L. Fejes T´ oth’s Problem From the physical point of view, L. Fejes T´ oth’s problem is more natural, more fascinating, and more challenging than Hornich’s problem. In 1996, B¨ or¨ oczky and Soltan [1] proved that (Sn ) < ∞, which shows the fundamental diﬀerence between (Sn ) and ∗ (Sn ) (see Remark 12.1). Almost simultaneously, Zong [6] obtained the following upper bound for (Sn ) using a constructive method. Theorem 12.4 (Zong [6]). 2
(Sn ) ≤ (8e)n (n + 1)n−1 n(n
+n−2)/2
.
√ Proof. For convenience, we consider S = ( n/2)Sn instead of Sn and assume that n ≥ 3. Clearly, we have √ In ⊂ S ⊂ nIn . (12.9) Let α be a positive number strictly less than one, and take 9√ : k = αn + 1 .
(12.10)
Then, αIn ⊂ S ⊂ kαIn . Denote the intersections of αIn and kαIn with the hyperplane {x ∈ En : xi = 0} by J(i) and J (i), respectively. Let Xi = {x1 , x2 , . . . , xkn−1 } be a set of points such that J(i) + Xi is a tiling in J (i), and write √ Yi = xj + j nei : xj ∈ Xi , (12.11)
190
12. Problems of Blocking Light Rays
where ei indicates the ith unit basis vector. Then, it follows from (12.9) and (12.11) that S + Yi ∪ {o} is a ﬁnite packing. Let w be a point with wi > 0. We introduce another set of points √ wi + j n Zi (w) = (w + xj ) : xj ∈ Xi . (12.12) wi Let xj , yj , and zj be three geometrically corresponding points in Xi + w, Yi + w, and Zi (w), respectively. Then, zj is the intersection of the hyperplane {x ∈ E n : xi = yji } and the light ray of direction xj . By (12.9) and (12.12) it is clear that S + Zi (w) ∪ {o} is another ﬁnite packing. First, let us prove the following statement. Assertion 12.1. When w is a point with wi ≥ 3k n−1 n/2(1 − α), the spheres S+Zi(w) block all the light rays starting from S and passing through J (i) + w. Vj∗
αIn + w + xj
wj R
Vj
x
√
nIn Figure 12.2
Let w be a point with √ √ k n−1 n( n + α) wi ≥ , 1−α
(12.13)
and let xj be a point of Xi . Take √ wj =
α+
n √ (w + xj ) n
∗ and √ denote by Vj , Vj the two cones with the same vertex wj and over nIn , αIn + w + xj , respectively. Then (see Figure 12.2), we have
Vj∗ − wj = −Vj + wj
(12.14)
12.3. L. Fejes T´ oth’s Problem and Vj∗
191
β + √n √ (w + xj ) + βIn . = α+ n
(12.15)
β≥0
If R is a light ray starting from a point x ∈ Vj and passing through a point y ∈ Vj∗ , x = y, then the rest of the light ray is contained in Vj∗ . Otherwise, let H be the twodimensional plane containing both R and wj . By considering the intersection H ∩ (Vj ∪ Vj∗ ) and applying the symmetry indicated by (12.14) one can easily obtain a contradiction. On the other hand, using (12.13) and since j ≤ kn−1 , it is easy to get β ≤ 1 when √ √ wi + j n β+ n √ = . α+ n wi √ Since J(i) ⊂ αIn , by (12.9) and (12.14) the light rays starting from nIn and passing through J(i) + w + xj can be blocked by the translate S+
√ wi + j n (w + xj ). wi
Then, Assertion 12.1 follows from the fact that J(i) + Xi + w is a tiling in J (i) + w.
F3
S
F2 F1
Figure 12.3 Write
P (ξ1 , ξ2 , . . . , ξn ) = x ∈ E n : xi  ≤ 12 ξi , i = 1, 2, . . . , n
√ and denote its facet with xi = ξi /2 by Fi (ξ1 , ξ2 , . . . , ξn ). Take l = kα ≥ n and 6 n−1 7 3k n m=2 +1 . (12.16) 2(1 − α)
192
12. Problems of Blocking Light Rays
Abbreviate F1 (ml, ml, . . . , ml), F2 (2ml, (m + kn−1 )l, ml, . . . , ml), . . . , F (2ml, 2ml, . . . , 2ml, (m + kn−1)l) by F1 , F2 , . . . , Fn , respectively. Clearly, we have ml 3k n−1 n ≥ . (12.17) 2 2(1 − α) We proceed to prove the following assertion. Assertion 12.2. Every light ray starting from S intersects at least one of the 2n facets F1 , −F1 , F2 , −F2 , . . . , Fn , −Fn (see Figure 12.3). For convenience, we write Fi∗ = Fi (l, l, . . . , l). Then Fi blocks all the light rays starting from lIn and passing through a ﬁxed point p ∈ bd(mlIn ) if and only if Fi blocks all the light rays starting from Fi∗ and passing through p, in other words, as showed in Figure 12.4, if and only if
(m + kn−1 )l − 2pi l l ∗ Fi − ei + ei 2pi − l 2 2
(m + kn−1 − 1)l l + (12.18) p − ei ⊂ Fi . 2pi − l 2
mlIn
lIn p
Fi
Figure 12.4 Deﬁne qi reductively by
l m n−1 qn = m+k −1 +1 , 2 2m − 1 m + k n−1 − 1 l qi = qi+1 + m−1 2 for i = n − 1, n − 2, . . . , 2, and q1 = ml/2. Then, by (12.10), (12.16), and the assumption n ≥ 3, (m + kn−1 + 1)l ≤ qn ≤ qn−1 ≤ · · · ≤ q2 4
12.3. L. Fejes T´ oth’s Problem
193
n−2
m + k n−1 − 1 (n − 2)l qn + m−1 2
n−2
1 1 1 ml ≤ 1+ + 2n 2 3n 2
n 1 + 2/3n ml 1 1 ≤ 1+ 2 2n (1 + 1/2n)2 2 √ e ml ml ≤ < . 2 2 2
≤
(12.19)
From (12.18), (12.19), and routine computation based upon elementary geometry illustrated by Figure 12.4, it can be shown that every light ray starting from Fi∗ and passing through a point p = (p1 , p2 , . . . , pn ) ∈ bd(mlIn ) with l m + k n−1 − 1 pi ≥ max qj + = qi i+1≤j≤n m−1 2 and pj  ≤ qj ,
j = i + 1, . . . , n,
can be blocked by Fi . On the other hand, it follows from (12.19) that for each point p ∈ bd(mlIn ) there is an index i such that pi  ≥ qi
and pj  ≤ qj
for j = i + 1, . . . , n. Therefore, by (12.9) all the light rays starting from S can be blocked by the union of the 2n facets F1 , −F1 , F2 , −F2 , . . . , Fn , −Fn . Assertion 12.2 follows. Let Wi = {wi,1, wi,2 , . . . , wi,τi } be a set of points such that J (i) + Wi is a tiling in Fi , where τi = 2i−1 mn−1 . (12.20) We deﬁne i =
Zi (w)
w∈Wi
and =
n i (−i ) . i=1
Based on the periodic distribution of Yi + Wi and the relationship between Yi + Wi and i , it is easy to see that int(S + v1 ) ∩ int(S + v2 ) = ∅ whenever v1 and v2 are diﬀerent points in i . Also, from the deﬁnition of Fi and by looking at the (j + t)th coordinates of vj and vj+t , it can be veriﬁed that int(S + vj ) ∩ int(S + vj+t ) = ∅
194
12. Problems of Blocking Light Rays
whenever vj ∈ j , vj+t ∈ j+t , and t is a positive integer. Therefore, S + ∪ {o} is a ﬁnite packing. On the other hand, by Assertion 12.1, Assertion 12.2, and (12.17) the spheres S + block all the light rays starting from S. Hence, by (12.10), (12.16), and (12.20), (Sn ) = (S) ≤ card{} = 2
n
card{i }
i=1
= 2k n−1
n i=1
τi = 2 (2n − 1) (km)n−1
3k n n ≤ 2 (2 − 1) +k 2(1 − α) (n+2)/2 n−1 n ≤ 8n αn (1 − α) n
n−1
n
holds for every number α with 0 < α < 1. Since the function g(α) = αn (1 − α) n 1 n n attains its maximum n+1 at α = n+1 , we have n+1
n2 2 1 (Sn ) ≤ 8 1 + (n + 1)n−1 n(n +n−2)/2 n n
2
≤ (8e)n (n + 1)n−1 n(n
+n−2)/2
. 2
Theorem 12.4 is proved.
Remark 12.5. As a generalization of John’s result, Leichtweiß [1] proved that for every ndimensional convex body K there is a linear transformation L such that Sn ⊆ L(K) ⊆ nSn (for a proof of this result, we refer to Assertion 13.2). Applying Leichtweiß and John’s results, Zong [6] proved, in a similar manner to Theorem 12.4, that 2 (K) ≤ (8e)n (n + 1)n−1 n3(n −1)/2 for every ndimensional convex body K, and that 2
(C) ≤ (8e)n (n + 1)n−1 nn
−1
for every ndimensional centrally symmetric convex body C. Improving the upper bound for (C) with a nonconstructive method, we have the following theorem.
12.3. L. Fejes T´ oth’s Problem
195
Theorem 12.5. 2
(C) ≤ 48n
(1+o(1))
.
Remark 12.6. Using a modiﬁcation of Zong’s method, Talata [3] obtained a similar bound for general convex bodies. Meanwhile B´ ar´ any discovered a proof for the sphere case based on the ideas of Section 2. Our proof here is based on his method. In order to prove Theorem 12.5, as well as Henk’s lemma (Lemma 12.1), we require another basic result. Lemma 12.2 (Rogers and Zong [1]). Let K1 and K2 be two convex bodies in En . Then K2 can be covered by v(K1 − K2 )θ(K2 )/v(K1 ) translates of K1 . In particular, when K1 is centrally symmetric and K2 = γK1 , K2 can be covered by (1 + γ)n θ(K1 ) translates of K1 . Proof. For convenience, we assume that both K1 and K2 contain o. Let X be a set of points such that K1 + X is a covering in E n with covering density θ(K1 ), and let l be a large number. For any point y ∈ En we deﬁne f (y) = card {x ∈ X : (K2 + y) ∩ (K1 + x) = ∅} . It is easy to show that (K2 + y) ∩ (K1 + x) = ∅ if and only if y − x ∈ K1 − K2 . Thus, letting χ(y) be the characteristic function of K1 − K2 , we have % % f (y)dy = χ(y − x)dy lIn
x∈X
lIn
≤ v(K1 − K2 ) card {X ∩ (l + 2d(K1 )) In } v((l + 2d(K1 ))In )(θ(K1 ) + ) ≤ v(K1 − K2 ) . v(K1 ) Therefore,
6 f (y) ≤
7 v(K1 − K2 ) θ(K1 ) v(K1 )
holds for some point y ∈ En . The general case is proved. In the symmetric case, it is easy to see that K1 − K2 = (1 + γ)K1 . Thus, the assertion follows from the general case.
2
196
12. Problems of Blocking Light Rays
Proof of Theorem 12.5. Assume that C is centered at o, and let Λ be a lattice such that C + Λ is a packing with density δ∗ (C). For convenience, we write r = 14 , 6 7 δ(C)2n m= , rn δ ∗ (C) m = 3n θ(C) , ri = 14(m + i − 1)r, and L(i, j) = {u ∈ Λ : (2rC + u) ∩ bd(jri C) = ∅} . Clearly, we have 2rC + L(i, j) ⊂ (jri + 4r)C \ (jri − 4r)C.
(12.21)
By Lemma 12.1, for every point v ∈ bd(ri C) there is a j ≤ m and a lattice point u ∈ Λ such that jv ∈ rC + u and therefore (see Figure 12.5) rC + jv ⊂ 2rC + u. This means that all the light rays starting from rC can be blocked by m
(2rC + L(i, j)).
j=1
o
rC + u rC 2rC + u Figure 12.5
By Lemma 12.2, we assume that 2rC can be covered by m translates rC + x1 , rC + x2 , . . . , rC + xm . Then we have xi ∈ 3rC. Deﬁning L (i, j) =
jm + i − 1 u : u ∈ L(i, j) j(m + i − 1)
(12.22)
(12.23)
12.3. L. Fejes T´ oth’s Problem and
197
L (i, j) = L (i, j) + xi ,
(12.24)
it follows that all the light rays starting from rC + xi can be blocked by m
(2rC + L (i, j)).
j=1
Therefore, all the light rays starting from 2rC can be blocked by
m m
(2rC + L (i, j)).
(12.25)
i=1 j=1
Since C + Λ is a packing and 1 jm + i − 1 ≤ ≤ 1, 2 j(m + i − 1) 2rC + L (i, j) is a ﬁnite packing of 2rC. Also, by (12.21), (12.22), (12.23), and (12.24) it follows that 2rC + L (i, j) ⊂ (jr1 + 7(2i − 1)r)C \ (jr1 + 7(2i − 3)r)C. Denoting the set on the righthand side of the above formula by C(i, j), it can be veriﬁed that the int(C(i, j)) are pairwise disjoint. Thus, the system (12.25) is a ﬁnite packing of 2rC in (m + 1)r1 C. Hence, by Theorem 3.1 and Remark 3.5, (C) ≤
v((m + 1)r1 C) (m + 1)n r1n ≤ v(2rC) 2n r n 2
≤ 48n
(1+o(1))
. 2
Theorem 12.5 is proved.
Based on (12.2), Theorem 12.3, and Theorem 12.5 we propose the following problems to end this section. Problem 12.2. Does there exist an absolute constant c such that 2
(Sn ) = 2cn
(1+o(1))
?
If so, determine it. Problem 12.3. Let Sn +X be a packing in En , let ς(Sn , X) be the maximal length of the segments contained in En \ {int(Sn ) + X}, and let ς(Sn ) = min ς(Sn , X). X
198
12. Problems of Blocking Light Rays
Does there exist a constant c such that ς(Sn ) = 2cn(1+o(1)) ?
12.4. L´ aszl´ o Fejes T´ oth L´ aszl´ o Fejes T´ oth was born in 1915 in Szeged, Hungary. In 1933 he entered the University of Budapest to study mathematics and physics, where L. Fej´er was one of his teachers. In 1938 he obtained a doctorate in mathematics. Soon afterwards, L. Fejes T´ oth served in the Hungarian army for two years. In 1941 he left the army and became an assistant at the University of Kolozsv´ ar. Four years later, he moved to Budapest and worked as a middleschool teacher there. At the same time he lectured on mathematics at the University of Budapest. In 1949 he became a professor of mathematics at the Technical University of Veszpr´em. In 1965, Professor L. Fejes T´ oth moved to the Mathematical Institute of the Hungarian Academy of Sciences, where he served as the director from 1970 to 1985. He retired in 1985. Professor L. Fejes T´ oth was a guest of many foreign universities such as the University of Freiburg, University of Wisconsin, University of Washington, Ohio State University, and the University of Salzburg. Professor L. Fejes T´ oth mainly worked in the areas of convex and discrete geometry, in which he published about 180 papers and two books. He has never had a Ph.D student. However, his work and open problems have inﬂuenced almost every contemporary discrete geometer, both domestic and foreign. Professor L. Fejes T´ oth is a member of the Hungarian Academy of Sciences and a corresponding member of the S¨ achsische Akademie der Wissenschaften zu Leipzig. He has been awarded honorary doctorates by the University of Salzburg and Veszpr´em University. In addition, he was honored by several Hungarian national prizes and a Gauss medal of the Braunschweigische Wissenschaftliche Gesellschaft.
13. Finite Sphere Packings
13.1. Introduction How can one pack m unit spheres in En such that the volume or the surface area of their convex hull is minimal ? Let m be a positive integer, and let u be a unit vector in En . Then, we deﬁne Lm = {2iu : i = 0, 1, . . . , m − 1} and Sm,n = conv {Sn + Lm } . Clearly, Sn + Lm is a packing in Sm,n . Writing µm (Sn ) =
mv(Sn ) , v(Sm,n )
it can be regarded as a local density of Sn + Lm in Sm,n . By routine computation it follows that µm (Sn ) =
mωn δ(Sn ). ωn + 2(m − 1)ωn−1
Based on this observation, in 1975 L. Fejes T´ oth [10] made the following conjecture about the volume case of the above problem. Sausage Conjecture: In En , n ≥ 5, the volume of the convex hull of m nonoverlapping unit spheres is at least ωn + 2(m − 1)ωn−1 .
200
13. Finite Sphere Packings
Also, equality is attained only if their centers are equally spaced on a line, a distance 2 apart. On the other hand, for the surface area case, Croft, Falconer, and Guy [1] formulated the following counterpart. Spherical Conjecture: When packing m unit spheres into a convex set so that the surface area is minimal, it seems likely that the optimum shape is roughly spherical if m is large. From an intuitive point of view, it is hard to imagine that both conjectures are correct. Nevertheless, the sausage conjecture was proved by Betke, Henk, and Wills [1] for suﬃciently large n, while the spherical conjecture was conﬁrmed independently by B¨ or¨ oczky Jr. [2] and Zong [1]. In this chapter we will deal mainly with the work of U. Betke, K. B¨ or¨ oczky Jr., M. Henk, J.M. Wills, and C. Zong.
13.2. The Spherical Conjecture The following theorem provides a positive solution for the spherical conjecture. ◦ Theorem 13.1 (B¨ or¨ oczky Jr. [2] and Zong [1]). Let Km,n be the family ◦ of convex bodies Km,n that contain m nonoverlapping ndimensional unit spheres and at which the surface area attains its minimum. Write
ηm =
max
min
◦ ◦ rSn +x⊆Km,n ⊆r Sn +x Km,n ∈K◦ m,n x∈E n
r . r
Then lim ηm = 1.
m→∞
To prove this theorem, in addition to Theorem 1.2 and the isoperimetric inequality we need the following lemma. Lemma 13.1. For every ndimensional convex body K there is a rectangular parallelepiped P such that P ⊆ K ⊆ n3/2 P. Proof. First, let us quote two fundamental results from Leichtweiß [1]. These results are not only useful in the proof of this theorem, but are also interesting and important.
13.2. The Spherical Conjecture
201
Assertion 13.1. Let α and β be two ﬁxed numbers with −1 ≤ α ≤ 1 and −1 ≤ β ≤ 1. If among all the ellipsoids
n 2 x ∈ E : λ(x1 − α)(x1 + β) + (xi ) ≤ 1 , n
Eλ =
i=1
where λ ≥ 0, the unit sphere takes the minimum volume, then αβ ≥ 1/n. It follows from routine computation that v(Eλ ) =
(α + β)2 2 1 + (1 + αβ)λ + λ 4
and therefore
n/2
(1 + λ)−(n+1)/2 ωn
* ∂(v(Eλ )) ** nαβ − 1 = ωn . ∂λ *λ=0 2
Thus, the assumption v(Eλ ) ≥ ωn for all λ ≥ 0 implies * ∂(v(Eλ )) ** ≥ 0, ∂λ *λ=0 and consequently αβ ≥
1 n.
Assertion 13.1 follows. Assertion 13.2. For every ndimensional convex body K there exists an ellipsoid E such that 1 nE
⊆ K ⊆ E.
Assume that E is a minimal circumscribed ellipsoid of K; for convenience we take E to be centered at o. Let L be a linear transformation of E n such that L(E) = Sn . Then, Sn is a minimal circumscribed ellipsoid of L(K). If H1 = {x ∈ E n : x1 = α} and H2 = {x ∈ E n : x1 = −β} , where 0 ≤ α ≤ 1 and 0 ≤ β ≤ 1 are two supporting hyperplanes of L(K), then every point x ∈ L(K) satisﬁes both n (xi )2 ≤ 1 i=1
and (x1 − α)(x1 + β) ≤ 0.
202
13. Finite Sphere Packings
Thus, it follows that L(K) ⊆ Eλ for all λ ≥ 0 and therefore ωn is the minimal value of v(Eλ ). Then, by Assertion 13.1 we have 1/n ≤ α ≤ 1, 1 n Sn
⊆ L(K) ⊆ Sn ,
and ﬁnally 1 nE
⊆ K ⊆ E.
This proves Assertion 13.2. By Assertion 13.2, if a1 , a2 , . . . , an are the n axes of E with lengths 1 , 2 , . . . , n , respectively, and if P = x ∈ E n : x, ai ≤ n−3/2 i , i = 1, 2, . . . , n , then we have P ⊆ n1 E ⊆ K ⊆ E ⊆ n3/2 P. 2
Lemma 13.1 is proved.
◦ Proof of Theorem 13.1. By Lemma 13.1, for every convex body Km,n there is a rectangular parallelepiped P such that ◦ P ⊆ Km,n ⊆ n3/2 P.
(13.1)
Let l1 , l2 , . . . , ln be the lengths of the n edges of P . It follows from (13.1) that n # ◦ mωn ≤ v(Km,n ) ≤ v(n3/2 P ) = n3n/2 li (13.2) i=1
and ◦ s(Km,n ) ≥ s(P ) =
n n # 2 li . l i=1 j=1 j
(13.3)
On the other hand, by Theorem 1.2, m unit spheres can be packed into a large sphere of radius $ m r=2n δ(Sn ) when m is suﬃciently large. Then, by (13.2), (13.3), and the minimum ◦ assumption of s(Km,n ) we have ◦ s(P ) ≤ s(Km,n ) ≤ s(rSn ),
13.2. The Spherical Conjecture
203
$
n−1 n n # 2 m li ≤ nωn 2 n l δ(Sn ) i=1 j=1 j ; 8 0. Meanwhile, since the sum of the three angles of any triangle is π, it follows that φ < π/3 and u1 , u2 > 0 cannot hold at three points of Xm,n . This fact plays a very important role in the proof of Theorem 13.2. Deﬁnition 13.2. Write G = conv{o, 2u1 , 2u2 }∩Sn , and denote by f (G, x) the map given by f (G, x) = y ∈ G, where x, y = min {x, w} . w∈G
Then, we deﬁne D1 = {x ∈ D∗ : f (G, x) ∈ rint(G)} , D2 = {x ∈ D∗ : f (G, x) ∈ (o, u1 ) ∪ (o, u2 )} , D3 = {x ∈ D∗ : f (G, x) = o} , and
D4 = {x ∈ D∗ : f (G, x) ∈ [2u1 , 2u2 ]} .
206
13. Finite Sphere Packings
It is clear that v(D∗ ) ≥
4
v(Di ).
(13.11)
i=1
By (13.8) and (13.11), the proof of Theorem 13.2 depends on complicated estimates of the values of v(Di ). Lemma 13.2. Let u and v be two orthogonal unit vectors, and let α and be two positive numbers such that α(α + ) ≥ 1 and (α + )v ∈ D∗ . Then [o, h1 (α, )u] + αv ∈ D∗ , where h1 (α, ) = / (α + )2 − 1. Proof. It follows from the assumption and the convexity of D∗ that conv {Sn , (α + )v} ⊂ D∗ .
(13.12)
u h1 (α, )u + αv
Sn
o
v
αv
(α + )v
Figure 13.1 With the help of Figure 13.1, since α ≥ 1/(α + ), routine argument yields h1 (α, )u + αv ∈ conv {Sn , (α + )v} , which implies [o, h1 (α, )u] + αv ⊂ conv {Sn , (α + )v} . Thus, Lemma 13.2 follows from (13.12) and (13.13). Lemma 13.3. v(G) ≥ φ/2. Proof. Write γ = u1 , u2 , θ = arccos(γ) and V = {αu1 + βu2 : α ≥ 0, β ≥ 0} .
(13.13) 2
13.3. The Sausage Conjecture
207
2u2 M u2
G
o
u1
2u1
Figure 13.2 Now, as illustrated in Figure 13.2, we deal with two cases. Case 1. γ ≥ − 12 . Then, we have θ = φ, V ∩ Sn ⊂ G, and therefore v(G) ≥ φ/2.
(13.14)
Case 2. γ < − 12 . Writing M = V ∩ Sn \ G, routine computation yields that v(G) = v(V ∩ Sn ) − v(M ) π−θ = − arccos x + x 1 − x2 , 2 where x = 2 sin(θ/2). Thus, π 3θ − arccos x − + x 1 − x2 2 2 x = arcsin x − 3 arcsin + x 1 − x2 2 ≥ 0.
v(G) − θ =
Then, by (13.10) it follows that v(G) ≥ θ ≥ φ/2. From (13.14) and (13.15) Lemma 13.3 is proved.
(13.15) 2
Before estimating the values of v(Di ) we introduce some new notation. H = {x = αu1 + βu2 : −∞ ≤ α, β ≤ +∞} , H ∗ = {x ∈ E n : u1 , x = u2 , x = 0} , Hi = {x ∈ E n : ui , x = 0} . We now proceed to estimate the values of v(Di ).
208
13. Finite Sphere Packings
Lemma 13.4. v(D1 ) ≥
h1 (1, cosec φ − 1)2 φ ωn−2 . 2
Proof. It follows from the deﬁnition of φ that ui , uj  ≥ cos φ for i = 1, 2 and j = 1, 2, . . . , m − 1, which implies uj , v ≤ sin φ whenever v ∈ H ∗ ∩ bd(Sn ). Hence, from the deﬁnition of D∗ we obtain cosec φ (Hi ∩ Sn ) ⊂ D∗
(13.16)
for i = 1, 2, and cosec φ (H ∗ ∩ Sn ) ⊂ D∗ . Then, applying Lemma 13.2 with u ∈ H ∩ bd(Sn ),
v ∈ H ∗ ∩ bd(Sn ),
α = 1, and = cosec φ − 1, we have h1 (1, cosec φ − 1)G + H ∗ ∩ Sn ⊂ D1 and therefore, applying Lemma 13.3, v(D1 ) ≥
h1 (1, cosec φ − 1)2 φ ωn−2 . 2
Lemma 13.4 is proved.
2
Lemma 13.5. v(D2 ) ≥ h1 (1, cosec φ − 1)ωn−1 . Proof. In a similar way to the proof of Lemma 13.4, it follows from (13.16) and Lemma 13.2 that [o, h1 (1, cosec φ − 1)ui ] + Hi ∩ Sn ⊂ D∗
(13.17)
for i = 1, 2. For convenience, we denote the convex body on the lefthand side of (13.17) by Ji . Let v1 and v2 be unit vectors of H determined by ui , vi = 0 and ui , vj < 0 (see Figure 13.3), where {i, j} = {1, 2}. Then {x ∈ E n : vi , x ≥ 0, x ∈ int(Ji )} ⊆ D2 .
(13.18)
13.3. The Sausage Conjecture
209
u2 o u1
v2 v1 Figure 13.3 Hence, it follows from (13.18) and the deﬁnitions of Ji and D2 that v(D2 ) ≥ h1 (1, cosec φ − 1)ωn−1 . 2
Lemma 13.5 is proved. Lemma 13.6. If φ < π/3 and u1 , u2 > 0, then v(D3 ) ≥
π−φ ωn . 2π
Proof. Writing V ∗ = {x = αu1 + βu2 : α ≤ 0, β ≤ 0, x ∈ Sn } , routine argument shows that v(V ∗ ) = and
π−φ 2
(V ∗ + H ∗ ∩ Sn ) ∩ Sn ⊂ D3 .
Thus, we have v(D3 ) ≥
v(V ∗ ) π−φ v(Sn ) = ωn , v(S2 ) 2π 2
which proves our lemma. Lemma 13.7. If φ ≤ π/4 and u1 , u2 < 0, then v(D4 ) ≥
cos φ − sin φ ωn−1 . cos(φ/2)
Proof. Write u = (u1 − u2 )/u1 , u2 and H ∗∗ = {x ∈ E n : u, x = 0} .
210
13. Finite Sphere Packings
u1 o u2 uj
Figure 13.4 Since φ ≤ π/4 < π/3 it is easy to see that for 1 ≤ j ≤ m−1, u1 , uj ≥ cos φ if and only if u2 , uj ≤ − cos φ (see Figure 13.4). It follows that u, uj  ≥
u1 , uj − u2 , uj  ≥ cos φ, u1 , u2
which implies uj , v ≤ sin φ ∗∗
for any unit vector v ∈ H . Thus, for any λ ∈ [0, 1] we obtain 2λu1 + 2(1 − λ)u2 + v, uj 2λ(1 + cos φ) + 2 cos φ + sin φ, ≤ −2λ(1 + cos φ) + 2 + sin φ,
if u1 , uj ≥ cos φ, if u1 , uj ≤ − cos φ.
So, taking h2 (φ) =
1 + sin φ 2(1 + cos φ)
and keeping our assumption in mind, for λ ∈ [h2 (φ), 1 − h2 (φ)] we have 2λu1 + 2(1 − λ)u2 ∈ Sn , 2λu1 + 2(1 − λ)u2 + H ∗∗ ∩ Sn ⊂ D, and therefore
2λu1 + 2(1 − λ)u2 + H ∗∗ ∩ Sn ⊂ D∗ .
(13.19)
Let w be a unit vector determined by w = αu1 + βu2 for nonnegative α and β and satisfying u, w = 0. Also, let N = {x ∈ H ∗∗ ∩ Sn : w, x ≥ 0} . From (13.19) and the deﬁnition of D4 we have [2u1 , 2u2 ] + N ∩ D∗ ⊂ D4
13.3. The Sausage Conjecture
211
and therefore (1 − 2h2 (φ))2u1 , 2u2 ωn−1 2 cos φ − sin φ ≥ ωn−1 . cos(φ/2)
v(D4 ) ≥
2 √ Lemma 13.8. If φ > 0, then for every with 0 < < 2/ 3 − 1 we have Lemma 13.7 is proved.
v(D1 ) ≥
h1 (1, )2 φ ωn−2 , 2(1 + h3 (1 + , n))
√ where, for 1 ≤ β < 2/ 3, %
=%
1
(1 − x )
2 (n−5)/2
h3 (β, n) =
dx
1/β
(1 √ 3/2
1/β
− x2 )(n−5)/2 dx.
Proof. Using polar coordinates we have % % 1 v(D1 ) = dx dy. n − 2 G x∈H ∗ ∩bd(Sn ), y+x∈D∗
(13.20)
We now proceed to show that for a certain set G∗ ⊂ G with v(G∗ ) > 0, the above integral is of order ωn−2 . For this purpose we write Mρ = {x ∈ H ∗ ∩ bd(Sn ) : ρx ∈ D∗ } and
Mρ∗ = {x ∈ H ∗ ∩ bd(Sn ) : ρx ∈ D∗ } ,
and consider the inner integral at y = o. If Mρ = ∅, then H ∗ ∩ bd(ρSn ) intersects the aﬃne hull of certain facets Fj , say j = 1, 2, . . . , k, of the DirichletVoronoi cell D. Let vj ∈ H ∗ be the outer unit normal of aﬀ{Fj }∩ H ∗ . It follows from a special case of Lemma 7.1 that aﬀ{Fj } ∩ (H ∗ ∩ bd(ρSn )) ⊂ int(Fj ∩ H ∗ )
√ for 1 ≤ ρ < 2/ 3 and there exists an αj ∈ [1, ρ] such that αj vj ∈ int(Fj ∩ H ∗ ). Writing
Mj = {x ∈ H ∗ ∩ bd(Sn ) : vj , x > αj /ρ} ,
we have Mρ =
k j=1
Mj .
212
13. Finite Sphere Packings
Deﬁning
√ Mj∗ = x ∈ H ∗ ∩ bd(Sn ) : 3αj /2 ≤ vj , x ≤ αj /ρ
and, for x ∈ Mj∗ ,
αj ≥ ρ, vj , x
γ(x) = we have
γ(x)x ∈ aﬀ{Fj } ∩ H ∗
and γ(x)x, αj vj 2 ≤
4 3
− α2j .
This implies γ(x)x ∈ Fj ∩ H ∗ , int(Mi∗ ) ∩ int(Mj∗ ) = ∅, and
k
i = j,
Mj∗ ⊂ Mρ∗ .
j=1
Hence, we have % dx = Mρ∗
+ Mρ∗
1+
+
dx +
Mρ
+
dx/
Mρ
+
dx
Mρ∗
dx
≥
(n − 2)ωn−2 + + 1 + Mj dx/ M ∗ dx
(13.21)
j
for a suitable index j ∈ {1, 2, . . . , k}. Let g(x) = (1 − x2 )(n−5)/2 , and for ξ ∈ [1, ρ] deﬁne %
1
f1 (ξ) =
g(x)dx ξ/ρ
and
% f2 (ξ) =
ξ/ρ
g(x)dx. √ 3ξ/2
Using polar coordinates and the monotonic property of f1 (ξ)/f2 (ξ) we have + dx f1 (αj ) f1 (1) + Mj = ≤ . f (α ) f2 (1) dx ∗ 2 j M j
On the other hand, it follows from (13.21) that % (n − 2)ωn−2 dx ≥ . 1 + h3 (ρ, n) ∗ Mρ
(13.22)
13.3. The Sausage Conjecture
213
Then, applying Lemma 13.2 with α = 1 and = ρ − 1 we obtain h1 (1, )G + Mρ∗ ⊆ D∗ .
(13.23)
√ Hence, for 0 < < 2/ 3 − 1, it follows from (13.20), (13.22), and (13.23) that h1 (1, )2 φ v(D1 ) ≥ ωn−2 . 2(1 + h3 (1 + , n)) 2
This proves Lemma 13.8.
Proof of Theorem 13.2. Using Lemmas 13.4–13.8, the proof is a consequence of the fact that ωn−1 lim = ∞. n→∞ ωn First we consider three cases depending on the value of φ and the sign of u1 , u2 . Case 1. 0 < φ < π/4 and u1 , u2 > 0. By Lemmas 13.4, 13.5, and 13.6 we have v(D∗ ) ≥ v(D1 ) + v(D2 ) + v(D3 )
ωn ωn (1 − sin(π/4))2 ≥ ωn−1 + ωn−2 − ωn−1 − +φ 2 2π 2(1 − sin2 (π/4)) ωn > ωn−1 + 2 for suﬃciently large n. Case 2. 0 < φ < π/4 and u1 , u2 < 0. By Lemmas 13.4, 13.5, 13.6, and the relation cos φ ≥ 1 − φ2 /2 we have v(D∗ ) ≥ v(D1 ) + v(D2 ) + v(D4 )
(1 − sin(π/4))2 πωn−1 ≥ 2ωn−1 + φ ω − 2ω − n−2 n−1 8 2(1 − sin2 (π/4)) > 2ωn−1 for suﬃciently large n. Case 3. φ ≥ π/4. Choose a suitable such that the assumption of Lemma 13.8 holds. Then, by Lemma 13.8 we have v(D∗ ) ≥ v(D1 ) ≥ > 2ωn−1 for suﬃciently large n.
h1 (1, )2 π ωn−2 8(1 + h3 (1 + , n))
214
13. Finite Sphere Packings
By Remark 13.2, Case 1 holds for at most two DirichletVoronoi cells. Thus, when n is suﬃciently large and Km,n is not congruent to Sm,n , we have v(Km,n ) > ωn + 2(m − 1)ωn−1 . 2
Theorem 13.2 is proved. With explicit computation, Theorem 13.2 can be restated as follows.
Theorem 13.2∗ (Betke, Henk, and Wills [1]). When n ≥ 13387, L. Fejes T´ oth’s sausage conjecture is true. Remark 13.4. Using a more complicated method, considering a threedimensional section instead of a twodimensional one, Theorem 13.2∗ was improved by Henk [1] and by Betke and Henk [1] to n ≥ 45 and n ≥ 42, respectively. Remark 13.5. Besides Betke, Henk, and Wills, many authors such as G. Fejes T´ oth, Gritzmann, Kleinschmidt, and Pachner have made contributions towards the sausage conjecture. For example, it was proved by Gritzmann [2] that v(Km,n ) >
ωn + 2(m − 1)ωn−1 √ 2+ 2
for every ndimensional convex body Km,n that contains m nonoverlapping unit spheres.
13.4. The Sausage Catastrophe In lower dimensions, E 3 and E 4 , it seems that the ﬁnite sphere packing problem is much more complicated than its highdimensional counterpart. In 1983, Wills [1] proposed the following phenomenon. The Sausage Catastrophe. In En , n = 3 or 4, the sausage arrangements are optimal when the number of spheres is small; then, for a certain number kn of spheres the extremal conﬁgurations become fulldimensional without going through intermediatedimensional arrangements ﬁrst. In E 3 and E 4 it was proved by Betke, Gritzmann, and Wills [1] that intermediatedimensional arrangements are not extremal. As for the values of k3 and k4 , it was conjectured by Wills [1] that k3 ≈ 56 and k4 ≈ 75000.
13.4. The Sausage Catastrophe
215
In this section we demostrate several examples to support this phenomenon. Example 13.1. Let K3,n be an ndimensional convex body that contains three nonoverlapping unit spheres. Then v(K3,n ) ≥ ωn + 4ωn−1 , where equality holds if and only if K3,n is congruent to S3,n . Veriﬁcation. Suppose K3,n contains three nonoverlapping unit spheres Sn + v1 , Sn + v2 , and Sn + v3 , and v(K3,n ) is minimal. Without loss of generality, we may assume that v2 , v3 ≥ v1 , v2 = v1 , v3 = 2 and K3,n = conv {Sn + v1 , Sn + v2 , Sn + v3 } = Sn + T, where T is the triangle with vertices v1 , v2 , and v3 . Clearly, since Si is the largest section of Si+1 , we have ωi+1 < 2ωi . Then, writing θ = min { v1 v2 v3 , v1 v3 v2 } , we have 0 ≤ θ ≤ π/3 and v(K3,n ) = s(T )ωn−2 + 12 (4 + v2 , v3 )ωn−1 + ωn = 4 sin θ cos θ ωn−2 + 2(1 + cos θ)ωn−1 + ωn > 2(1 + cos θ + sin θ cos θ)ωn−1 + ωn . It is easy to see that f (x) = 1 + cos x + sin x cos x attains its minimum, 2, only at x = 0, for 0 ≤ x ≤ π/3. Thus, we have v(K3,n ) ≥ ωn + 4ωn−1 , and equality can be attained if and only if K3,n is congruent to S3,n . The example is veriﬁed. 2 Example 13.1 shows that the sausage arrangements are optimal for three spheres in E n , n ≥ 3. In E 3 , this result was extended to four spheres by B¨ or¨ oczky Jr. [1]. On the other hand, Gandini and Wills [1] showed that in E 3 for m = 56, 59–62, and all m ≥ 65 some fulldimensional packings of spheres are best possible and conjectured that in all remaining cases
216
13. Finite Sphere Packings
the sausage packing is optimal. The following example considers the case m = 56. Example 13.2 (Gandini and Wills [1]). There is a threedimensional convex body K56,3 such that v(K56,3 ) < v(S56,3 ) = ω3 + 110ω2 . √ √ √ √ √ √ Veriﬁcation. Write u1 = ( 2, 2, 0), u2 = ( 2, 0, 2), u3 = (0, 2, 2), and deﬁne 3 Λ= zi ui : zi ∈ Z i=1
and Tk = conv {o, ku1 , ku2 , ku3 } . Then, S3 + Λ is a lattice packing and card {Tk ∩ Λ} =
k+3 . 3
T3
(13.24)
T6
T3
T2
T2
Figure 13.5 Let K be the convex polytope obtained by cutting oﬀ two translates of T2 and two translates of T3 from T6 (see Figure 13.5). Then, by (13.24) and routine computation we obtain card{K ∩ Λ} = 56. On the other hand, by routine computation we have v(S3 + K) < ω3 + 110ω2 = v(S56,3 ).
13.4. The Sausage Catastrophe
217
Thus, taking K56,3 = S3 + K, 2
Example 13.2 follows.
In E 4 , applying a similar method to a 24cell with side length 34, we obtain the following result. Example 13.3 (Gandini and Zucco [1]). There is a fourdimensional convex body K375769,4 such that v(K375769,4 ) < v(S375769,4 ) = ω4 + 2 × 375768ω 3. Based on this result, Gandini and Zucco [1] gave the following modiﬁed version of Wills’ conjecture: k4 ≈ 375769.
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Index
Aﬄerbach, L. 40 Aigner, M. 21 automorphism 25 Babai, L. 40 Baire category 73 Ball, K. 54, 61 Bambah, R.P. 61 Bannai, E. 144, 146, 148 Baranovskii, E.P. 40, 45, 61 B´ ar´ any, I. 187, 195 Barlow, W. 19 Barnes, E.S. 40, 42, 45, 61 Bateman, P.T. 50 Bertrand, J.L.F. 43 Betke, U. 20, 200, 204, 214 Blaschke, W. 123 Blaschke’s selection theorem 203 Blichfeldt, H.F. 19, 40, 43, 91, 92, 94, 101, 103, 107, 156 blocking number 65, 71 Blundon, W.J. 162 Bolle, U. 156 Boltjanski, V.I. 184 Bonnet, P.O. 43
B¨ or¨ oczky, K. 111, 117, 165, 169, 189 B¨ or¨ oczky Jr, K. 110, 200, 204, 215 Bosanquet, L.S. 122 BoseChaudhuriHocquenghem code 81 Bouquet, J.C. 43 Bourgain, J. 77 BrunnMinkowski inequality 44 Busemann, H. 123 Butler, G.J. 177 cap packing 11, 131 Cassels, J.W.S. 50, 62 cell complex 169 Chebyshev, P.L. 43 ChristoﬀelDarboux formula 126 closeness 165 closest sphere packing 165 cluster 16, 18 code 79, 83, 88, 143 codeword 79, 150 convex body 1, 65, 73 convex hull 199 Conway, J.H. 19, 84, 89, 142, 178 Cooke, R.G. 122 coordinate array 86
238 coorthosimplex 115, 117 Corput, J.G. van der 99 covering density 55, 57, 195 Coxeter, H.S.M. 58, 61, 111, 169 critical form 24, 27 Croft, H.T. 200 cross polytope 69, 74, 112 Cs´ oka, G. 185 cyclic code 80, 82 cylindrical hole 178, 181 Dalla, L. 66 Daniel’s asymptotic formula 107, 110 Danzer, L. 185 Dauenhauer, M.H. 19 Davenport, H. 19, 47, 50, 62, 99, 122 deepest hole 176 DehnSommerville equation 70 Delone, B.N. 40, 43, 45, 61 Delone simplex 16, 61 Delone star 16, 18 Delone triangulation 16, 61 Delsarte, P. 127, 132, 145 densest lattice packing 4, 24, 29, 32, 35, 40, 72 densest translative packing 4 density 4, 13, 24, 32, 35, 40, 153, 157, 186 determinant 2 diﬀerence body 65 diﬀerence set 2 dihedral angle 107, 110, 121, 166, 172 Dirichlet, G.L. 43 Dirichlet’s approximation theorem 185 DirichletVoronoi cell 5, 13, 15, 60, 103, 112, 119, 167, 170, 173, 186, 204, 211 discriminant 23
Index dodecahedron 13 Euler’s formula 8, 12, 173 extreme form 24, 33 facecentered cubic lattice 10, 17, 19 facet 34, 67, 69, 104, 168, 191, 211 Falconer, H.J. 200 Fej´er, L. 198 Fejes T´ oth, G. 163, 214 Fejes T´ oth, L. 14, 16, 19, 137, 165, 170, 183, 198, 199, 214 L. Fejes T´ oth number 185 L. Fejes T´ oth’s problem 183, 189 Ferguson, S.P. 18, 19 Few, L. 58, 61, 157 ﬁnite ﬁeld 79 F module 80 fundamental domain 51, 53 fundamental parallelepiped 2 gamma function 2 Gandini, P.M. 215 Gauss, C.F. 19, 29, 43 generator matrix 80 generator polynomial 80, 82 geodesic arc 11, 16 geodesic distance 11 geodesic metric 111 geodesic radius 95, 97, 111, 184 Goethals, J.M. 132, 145 Gohberg, I. 184 Golay, M.J.E. 84 Golay code 81, 85 Green, R. 102 Gregory, D. 10 GregoryNewton problem 10, 18, 21, 165 Gritzmann, P. 214 Groemer, H. 4, 19, 20, 128, 156 Gr¨ otschel, M. 142
Index Gruber, P.M. 20, 73, 178, 186 Gr¨ unbaum, B. 20 Guy, R.K. 200 Hadwiger, H. 2, 4, 20 Haj˘ os, G. 99 Halberg, C. 184 Hales, T.C. 14, 16, 18, 19 Hamming code 81, 84 Hamming distance 79, 82 Hardy, G.H. 100 Helmholtz, H. 43 Henk, M. 20, 110, 177, 185, 195, 200, 204, 214 Heppes, A. 162, 178, 185 Hermite, Ch. 43 Hermite’s constant 24, 40 Hermite’s reduction 40 Hilbert, D. 19, 44 Hilbert’s 18th problem 19 Hlawka, E. 4, 20, 47, 62, 99 hlocally averaged density 14 hneighbor 14 homeomorphic 169 homogeneous harmonic polynomial 127 Hoppe, R. 11, 21 Hornich, H. 183 Hornich number 185 Hornich’s problem 183, 189 Hortob´ agyi, I. 178 Horv´ ath, J. 163, 175, 177, 179, 181 Hoylman, D.J. 20 Hsiang, W.Y. 14, 16, 19 Hua, L.K. 99 Hurwitz, A. 43 icosahedron 10 inner product distribution 143, 146 integral lattice 139, 144, 146 isomorphic 169 isoperimetric inequality 200
239 Jacobi polynomial 125, 141 Jensen’s inequality 7, 9, 171 John, F. 186, 194 Jordan, C. 44, 102 Jordan measurable set 51, 156 Justesen code 82, 88 Kabatjanski, G.A. 125, 133, 136 Kanagasahapathy, P. 157 Kepler, J. 13 Kepler’s conjecture 13, 14, 16, 18, 19, 165 Kershner, R. 61 kfold packing 153, 155 Kirchhoﬀ, G.R. 43 Klein, F. 44, 102 Kleinschmidt, P. 214 Kneser, M. 140 Korkin, A.N. 33, 35, 38, 43 Kronecker, L. 43 Kronecker symbol 125 Kummer, E.E. 43 Kuperberg, W. 163 KZ reduced form 36, 38, 39 KZ reduction 40 Lagrange, J.L. 8, 19, 26, 43 Laplace operator 127 Larman, D.G. 20, 66, 74, 123 lattice 2, 9, 23, 36, 139, 153, 157, 183 lattice covering 55 lattice kissing number 2, 41 lattice packing 2, 7, 23, 42, 83 Leader, I. 187 Lebesgue measurable set 52 Lebesgue’s theorem 54 Leech, J. 11, 21, 82, 86, 88 Leech lattice 85, 140 Leichtweiß, K. 194, 200 Lekkerkerker, C.G. 20, 50, 178, 186 Lenstra, A.K. 40
240 Lenstra Jr, H.W. 40 Leven˘stein, V.I. 94, 125, 133, 136, 141 Levin, E. 184 Lie, S. 102 light ray 183, 188 Lindemann, F. 43 Lindsey II, J.H. 19 linear code 80, 83, 85, 88 Linhart, J. 163 Litsyn, S.N. 89 Littlewood, J.E. 100 Lloyd, S.P. 132 Lov´ asz, L. 40, 142 LSM reduced form 25, 27 LSM reduction 40 Macbeath, A.M. 50 Mahler, K. 19, 51, 62 Maly˘sev, A.V. 51, 62 ManiLevitska, P. 66 Markov, A.A. 44 Milman, V.D. 77 Milnor, J. 19 Minkowski, H. 2, 20, 25, 43, 47 MinkowskiHlawka theorem 44, 47, 50, 54, 84, 88 Minkowski inequality 74 Minkowski metric 44, 74 Minkowski sum 155 mixed volume 155 M¨ obius function 50 monomial 127 Mordell, L.J. 31 Muder, D.J. 19 Nemhauser, G.L. 142 Newton, I. 10, 21 Odlyzko, A.M. 132, 141 orthopoint 115 orthosimplex 114, 117, 119
Index Pachner, U. 214 pentagonal prism 18 perfect form 33, 35, 41 Petty, C.M. 123 P´ olya, G. 100 polyhedral convex cone 34 primitive point 50, 53 properly reduced form 51 quadratic form 19, 23, 25, 31, 34, 41, 47 quadratic residue code 81 quermassintegral 204 Rankin, R.A. 19, 94, 96, 99 ReedMuller code 82, 85, 88 ReedSolomon code 81 Riemann zeta function 45, 47 Riesz, F. 100 Rogers, C.A. 19, 47, 50, 54, 55, 58, 61, 103, 108, 122, 176, 178, 195 Rush, J.A. 84 Ry˘skov, S.S. 40, 45, 61, 175, 178, 181 Sanov, J.N. 51 saturated packing 14, 16, 17 sausage conjecture 199, 214 sausage catastrophe 214 Schaake, G. 99 Schl¨ aﬂi, L. 107 Schl¨ aﬂi’s function 107, 111, 121 Schmidt, W. 51, 54, 61 Schneider, T. 51, 62 Schrijver, A. 142 Sch¨ utte, K. 11, 21 score 16 Seeber, L.A. 19, 27 Seeber set 178, 181 Segre, B. 19 Seidel, J.J. 132, 145 Selling’s reduction 40
Index Shannon, C.E. 71 Shephard, G.C. 122 Sidelnikov, V.M. 94 Siegel, C.L. 51, 52, 62, 177 Siegel’s mean value formula 52, 54 simplical polytope 69 Sloane, N.J.A. 19, 82, 86, 88, 132, 137, 141, 142, 144, 146, 148, 178 Smith, H.J.S. 44 solid angle 58 Soltan, V. 189 spacecentered cubic lattice 166, 175 spherical code 127, 145 spherical conjecture 200 spherical harmonic 128, 145 spherical ldesign 145, 147 spherical polynomial 127 spherical simplex 107, 111, 121 stable form 24 star body 47 Steiner’s formula 155 Stirling’s formula 77, 110, 136 ˘ Stogrin, M.I. 40, 45 Straus, E. 184 sublattice 37 supersphere 74, 99 SwinnertonDyer, P. 72 Szeg¨ o, G. 125 Talata, I. 20, 77, 195 Tammela, P.P. 40, 45 Taylor’s theorem 96 Temesv´ ari, A.H. 163 thirteen spheres problem 10 Thue, A. 8, 19 tiling 6, 8, 51, 104, 154, 169 translative kissing number 2, 65, 72, 74, 111 translative packing 2 Tsfasman, M.A. 89
241 Venkov, B.A. 40, 45 Vet˘cinkin, N.M. 39 Voronoi, G.F. 34, 41, 43 Voronoi’s method 34 Waerden, B.L. van der 11, 21, 40 Watson, G.L. 42, 43, 45 Weber, H. 43 Weierstrass, C.T.W. 43 weight 79 Weil, A. 51, 62 Wills, J.M. 200, 204, 214, 215 Wirtinger, W. 62 Wolsey, L.A. 142 Wyner, J.M. 71 Yaglom, I.M. 137 Yakovlev, N.N. 162 Yudin, V.A. 95 Zassenhaus, H. 19 Ziegler, G. 21 Zolotarev, E.I. 33, 35, 38, 43 Zong, C. 20, 65, 67, 70, 72, 74, 186, 189, 194, 200, 204 Zucco, A. 217