MHF Lecture Notes SeriesKyushu University

21st Century COE ProgramDevelopment of Dynamic Mathematics with

High Functionality

The Proceedings of COE

Workshop on Sphere Packings

Eiichi Bannai

　　 　　 　

MHF Lecture Notes 2004-1

( Received March 4, 2005 )

Faculty of MathematicsKyushu UniversityFukuoka, JAPAN

Preface

The COE Workshop on Sphere Packings was held from November 1 toNovember 5, 2004, in Kyushu University, Fukuoka Japan. About 40 peopleparticipated in this workshop. The highlight of the workshop was the seriesof lectures of Professor Oleg Musin on the solution of the kissing numberproblem in dimension 4. We owe and greatly thank Professor Musin aswell as the other speakers and participants for the success of the workshop.We believe that exchanges of many ideas were made under the friendly andinformal atmosphere of the workshop. We hope that this workshop willcontribute to the future developments of mathematics. We also hope thatthis workshop will help to stimulate the research of discrete geometry inJapan.

This volume is the proceedings of this workshop. This is an unofficialpublication, like a preprint series. The manuscripts were submitted by allthe speakers (except for the speakers of short talks). The papers were notrefereed, and the contents of each paper rest on the responsibility of eachauthor(s). This is because we believe that the timeliness of the publicationis important in these unofficial publications. These proceedings will also beavailable on the web, as the first volume of a new lecture notes series of ourCOE program, i.e., MHF Lecture Note Series Vol 1 (2004). (See the followinghomepage of COE: http://www.math.kyushu-u.ac.jp/coe/index en.html)

This workshop is supported by Kyushu University 21st Century COEProgram: Development of Dynamic Mathematics with High Functionality.We greatly thank the program for the financial and moral supports to thisworkshop. We thank the staff members of the COE program and of our de-partment office for their help. In addition, we thank Dr. Makoto Tagami forpreparing these proceedings for publication, and Dr. Osamu Shimabukurofor preparing the home page of this workshop as well as making these pro-ceedings available on the web.

Eiichi BannaiKyushu University

January 31, 2005

Program of COE Workshop on Sphere Packings

Please note that on Monday afternoon and Thursday morning, the room isdifferent.

Monday Nov. 1

Morning Session (at International Hall)9:30 Opening and the introduction to this Workshop (E. Bannai)10:00-11:30 Oleg Musin: The kissing number in four dimensions, I11:30-12:00 (Discussion)

Afternoon Session (at Room 3109, Faculty of Science Building No. 3)13:30-14:20 Achill Schuermann: Computational Approaches to Lattice Packing and CoveringProblems,14:30-15:20 Hiroshi Maehara : Pearl numbers vs stick numbers for knots,15:30-16:20 Michel Deza: Isometric embeddings of Archimedean Wythoff polytopes intohypercubes and half-cubes,16:30-17:00 Elena Deza: Some generalizations of the divisor problem for Gaussian integers.

Tuesday Nov. 2

Morning Session (at International Hall)9:30-11:00 Oleg Musin: The kissing number in four dimensions, II11:00-12:00 (Discussion)

Afternoon Session (at International Hall)13:30-14:20 Makoto Tagami: On optimal sets in Musin’s paper ”The kissing number in fourdimensions”,14:30-15:20 Yoshiaki Itoh: One-dimensional random sequential packing and its extension tohigher dimesnsion,16:00-17:00 Martin Henk : Finite and Infinite Sphere Packings,

Wednesday Nov. 3

Morning Session (at International Hall)9:30-11:00 Oleg Musin: The kissing number in four dimensions, III11:00-12:00 (Discussion)

Afternoon Session (at International Hall)13:30-15:00 (two talks) Masaharu Tanemura: Problems of optimal configuration of pointson the sphere. andTeruhisa Sugimoto : Packing and Minkowski covering of congruent spherical caps on asphere,16:00-17:00 Tohru Ogawa : How far Mesh Models Simulate a Perfect Sphere?,

Thursday Nov. 4

Morning Session (at Room 3109 Faculty of Science Building No. 3)9:30-9:50 Kijung Kim: Association Schemes and Equitable Partition,9:50-10:10 Han Kang: On semiregular automorphisms of vertex transitive graphs,10:10-10:30 Masashi Shinohara: On (2k − 1)-point k-distance sets in S1.Afternoon: Excursion to Dazaifu.

Friday Nov. 5

Morning Session (at International Hall)10:00-11:30 Karoly Bezdek: On the packings of congruent balls in 3-space,

Afternoon Session (at International Hall)13:00-14:00 Florian Pfender : On Delsarte’s method,14:15-15:15 Hajime Tanaka: Schrijver bound for nonbinary codes.15:30. Closing.

The Proceedings of COE Workshop on Sphere Packings I Nov. 1st – Nov. 5th

Organizer Eiichi Bannai Contents 1. Oleg Musin AN EXTENSION OF DELSARTE'S METHOD. THE KISSING PROBLEM IN THREE AND FOUR DIMENSIONS.‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥１ 2. Eiichi Bannai and Makoto Tagami. On optimal sets in Musin’s paper “The kissing number in four dimensions.” ‥‥‥‥26 3. Károly Bezdek Sphere packings in 3-space. ‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥32 4. Elena Deza and Lidiya Varukhina Some generalizations of the divisor problem for Gaussian integers. ‥‥‥‥‥‥‥‥50 5. Michel Deza, Mathieu Dutour and Sergey Shpectorov Isometric embeddings of Archimedean Wythoff polytopes into hypercubes and half-cubes. ‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥55 6. Nikolai Dolbilin and Yoshiaki Itoh On random tilings and packings of space by cubes. ・‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥70 7. Martin Henk FINITE AND INFINITE LATTICE PACKINGS. ・‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥80 8. Hiroshi Maehara Pearl numbers vs stick numbers for knots. ‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥92 9. Tohru Ogawa How Far Mesh Models Can Simulate A Perfect Sphere? ‥‥‥‥‥‥‥‥‥‥‥‥‥101 10. Florian Pfender IMPROVED DELSARTE BOUNDS VIA EXTENSION OF THE FUNCTION SPACE(EXTENDED ABSTRACT). ‥‥‥‥‥‥‥・‥‥‥‥‥‥‥‥‥‥‥‥‥‥110 11. A. Schürmann and F. Vallentin METHODSIN THE LOCAL THEORY OF PACKINGS AND COVERING LATTICES‥‥112 12. T. Sugimoto and M. Tanemura Packings and Minkowski covering of congruent spherical caps on a sphere. ‥‥‥‥137

13. Hajime Tanaka Schrijver bound for nonbinary codes. ‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥167 14. Masaharu Tanemura Problems of optimal configuration of points on the sphere. ‥‥‥‥‥‥‥‥‥‥‥173

AN EXTENSION OF DELSARTE’S METHOD.

THE KISSING PROBLEM IN THREE AND

FOUR DIMENSIONS

Oleg R. Musin ∗

1 Introduction

The kissing number k(n) is the highest number of equal nonoverlapping spheresin Rn that can touch another sphere of the same size. In three dimensions thekissing number problem is asking how many white billiard balls can kiss (touch)a black ball.

The most symmetrical configuration, 12 billiard balls around another, is ifthe 12 balls are placed at positions corresponding to the vertices of a regularicosahedron concentric with the central ball. However, these 12 outer balls donot kiss each other and may all moved freely. So perhaps if you moved all ofthem to one side a 13th ball would possibly fit in?

This problem was the subject of a famous discussion between Isaac Newtonand David Gregory in 1694. (May 4, 1694; see interesting article [33] for detailsof this discussion.) It is commonly said that Newton believed the answer was 12balls, while Gregory thought that 13 might be possible. However, Bill Casselman[9] found some puzzling features in this story.

This problem is often called the thirteen spheres problem. R. Hoppe [19]thought he had solved the problem in 1874. But, Thomas Hales [18] in 1994published analysis of Hoppe’s mistake (see also [32]). Finally this problem wassolved by Schutte and van der Waerden in 1953 [31]. A subsequent two-pagessketch of an elegant proof was given by Leech [23] in 1956. No much doubtsthat Leech’s proof is correct, but there are gaps in his exposition, many involvedsophisticated spherical trigonometry. (Leech’s proof was presented in the firstedition of the well known book by Aigner & Ziegler [1], the authors removedthis chapter from the second edition because a complete proof to include somuch spherical trigonometry.) The thirteen spheres problem continues to be ofinterest, new proofs have been published in the last few years by Wu-Yi Hsiang[21], Karoly Boroczky [7], and Kurt Anstreicher [2].

∗Institute for Math. Study of Complex Systems, Moscow State University, Moscow, Russiaomusin@mail.ru

1

1

Note that k(4) 24. Indeed, the unit sphere in R4 centered at (0, 0, 0, 0)has 24 unit spheres around it, centered at the points (±√

2,±√2, 0, 0), with any

choice of signs and any ordering of the coordinates. The convex hull of these 24points yields a famous 4-dimensional regular polytope - the “24-cell”. Its facetsare 24 regular octahedra.

Coxeter proposed upper bounds on k(n) in 1963 [11]; for n = 4, 5, 6, 7, and8 these bounds were 26, 48, 85, 146, and 244, respectively. Coxeter’s boundsare based on the conjecture that equal size spherical caps on a sphere Sk canbe packed no denser than k+ 1 spherical caps on Sk that simultaneously touchone another. Boroczky proved this conjecture in 1978 [6].

The main progress in the kissing number problem in high dimensions wasin the end of 1970’s. Vladimir Levenshtein [24], and independently AndrewOdlyzko and Neil Sloane [27], [10, Chap.13] using Delsarte’s method in 1979proved that k(8) = 240, and k(24) = 196560. This proof is surprisingly short,clean, and technically easier than all proofs in three dimensions.

However, n = 8, 24 are the only dimensions in which this method gives aprecise result. For other dimensions (for instance, n = 3, 4) the upper boundsexceed the lower. In [27] the Delsarte method was applied in dimensions up to24 (see [10, Table 1.5]). For comparison with the values of Coxeter’s boundson k(n) for n = 4, 5, 6, 7, and 8 this method gives 25, 46, 82, 140, and 240,respectively. (For n = 3 Coxeter’s and Delsarte’s methods only gave k(3) 13[11, 27].) Kabatiansky and Levenshtein have found an asymptotic upper bound20.401n(1+o(1)) for k(n) in 1978 [22]. The lower bound 20.2075n(1+o(1)) was foundin [34].

Improvements in the upper bounds on kissing numbers (for n < 24) wererather weak during next years ([10, Preface to Third Edition] gives a brief reviewand references). Arestov and Babenko [3] proved that the bound k(4) 25cannot be improved using Delsarte’s method. Hsiang [20] claims a proof ofk(4) = 24. His work has not received yet a positive peer review.

If M unit spheres kiss the unit sphere in Rn, then the set of kissing pointsis an arrangement on the central sphere such that the (Euclidean) distancebetween any two points is at least 1. So the kissing number problem can bestated in other way: How many points can be placed on the surface of Sn−1 sothat the angular separation between any two points is at least 60?

This leads to an important generalization: a finite subset X of Sn−1 is calleda spherical z-code if for every pair (x, y) of X the scalar product x·y ≤ z. Spheri-cal codes have many applications. The main application outside mathematics isin the design of signals for data transmission and storage. There are interestingapplications to the numerical evaluation of n-dimensional integrals [10, Chap.3].

The Delsarte method (also known in coding theory as Delsarte’s linearprogramming method, Delsarte’s scheme, polynomial method) is described in[10, 22]. Let f(t) be a real polynomial such that f(t) ≤ 0 for t ∈ [−1, z], thecoefficients ck’s in the expansion of f(t) in terms of Gegenbauer polynomialsG

(n)k are nonnegative, and c0 = 1. Then the maximal number of points in a

spherical z-code in Sn−1 is bounded by f(1). Suitable coefficients ck’s can be

2

2

found by the linear programming method [10, Chapters 9,13].We found an extension of the Delsarte method in 2003 [25](see details in

[26]), that allowed to prove the bound k(4) < 25, i.e. k(4) = 24. This extensionyields also a proof k(3) < 13.

The first version of these proofs was relatively short, but used a numericalsolution of some nonconvex optimization problems. Later on [26] these calcula-tions have been reduced to calculations of roots of polynomials in one variable.(This is not a big problem now, all computer algebra systems such as Maple,Mathematica, and Matlab can find roots. Also these calculations can be inde-pendently verified. If you have approximate values all roots of a polynomial,then you can check the existence of these roots by simple computations.)

We present in this paper a new proof of the Newton-Gregory problem, anextension of Delsarte’s method, and a proof that k(4) = 24.

2 The thirteen spheres problem: a new proof

Let us recall the definition of Legendre polynomials Pk(t) by recurrence formula:

P0 = 1, P1 = t, P2 =32t2 − 1

2, . . . , Pk =

2k − 1k

t Pk−1 − k − 1k

Pk−2;

or equivalently Pk(t) =1

2k k!dk

dtk(t2 − 1)k (Rodrigues’ formula).

Lemma 1. Let X = x1, x2, . . . , xn be any finite subset of the unit sphere S2

in R3. By φi,j = dist(xi, xj) we denote the spherical (angular) distance betweenxi and xj . Then

n∑i=1

n∑j=1

Pk(cos(φi,j)) 0.

Let

f(t) =243180

t9 − 128720

t7 +18333400

t5 +34340

t4 − 8310t3 − 213

100t2 +

t

10− 1

200.

Lemma 2. Suppose X = x1, x2, . . . , xn ⊂ S2. Then

S(X) =n∑i=1

n∑j=1

f(cos(φi,j)) n2.

Lemma 3. Suppose X = x1, x2, . . . , xn is a subset of S2 such that the angularseparation φi,j between any two distinct points xi, xj is at least 60. Then

S(X) =n∑i=1

n∑j=1

f(cos(φi,j)) < 13n.

3

3

Theorem 1. k(3) = 12.

Proof. Suppose X is a kissing arrangement on S2 with n = k(3). Then X issatisfying the assumptions in Lemmas 2, 3. Therefore, n2 S(X) < 13n. Fromthis follows n < 13, i.e. n 12. From other side we have k(3) 12, thenn = k(3) = 12.

We need the only one fact from spherical trigonometry, namely the law ofcosines:

cosφ = cos θ1 cos θ2 + sin θ1 sin θ2 cosϕ,

where for spherical triangle ABC the angular lengths of its sides are θ1, θ2, φ andthe angle between AB,AC is ϕ (Fig. 1). If ϕ = 90, then cosφ = cos θ1 cos θ2(spherical Pythagorean theorem).

−1 −0.5 0 0.5 0.8

0

Fig. 2. The graph of the function f(t)

ϕ

A

B C

θ1 θ2

φ

Fig. 1

Proof of Lemma 1.

This lemma easily follows from Schoenberg’s theorem [29] for Gegenbauerpolynomials. Note that Pk = G

(3)k . For completeness we give a proof of Lemma

1 here. In this proof we are using original Schoenberg’s proof that based on theaddition theorem for Gegenbauer polynomials.1

The addition theorem for Legendre polynomials was discovered by Laplaceand Legendre in 1782-1785:

Pk(cos θ1 cos θ2 + sin θ1 sin θ2 cosϕ) =k∑

m=0

cm,k Pmk (cos θ1)Pmk (cos θ2) cosmϕ

= Pk(cos θ1)Pk(cos θ2) + 2k∑

m=1

(k −m)!(k +m)!

Pmk (cos θ1)Pmk (cos θ2) cosmϕ,

1Pfender and Ziegler[28] give a proof as a simple consequence of the addition theorem forspherical harmonics. This theorem is not so elementary. The addition theorem for Legendrepolynomials can be proven by elementary algebraic calculations.

4

4

wherePmk (t) = (1 − t2)

m2dm

dtmPk(t).

(See details in [8, 16].)

Proof. Let X = x1, . . . , xn ⊂ S2 and xi has spherical (polar) coordinates(θi, ϕi). Then from the law of cosines we have:

cosφi,j = cos θi cos θj + sin θi sin θj cosϕi,j , ϕi,j = ϕi − ϕj ,

which yields

∑i,j

Pk(cosφi,j) =∑i,j

k∑m=0

cm,kPmk (cos θi)Pmk (cos θj) cosmϕi,j

=∑m

cm,k∑i,j

um,ium,j cosmϕi,j , um,i = Pmk (cos θi).

Let us prove that for any real u1, . . . , un

∑i,j

uiuj cosmϕi,j 0.

Pick n vectors y1, . . . , yn in R2 with coordinates yi = (cosmϕi, sinmϕi). Ify = u1y1 + . . .+ unyn, then

y · y = ||y||2 =∑i,j

uiuj cosmϕi,j 0.

This inequality and the inequalities cm,k > 0 complete our proof.

Proof of Lemma 2.

Proof. The expansion of f in terms of Pk is

f =9∑

k=0

ckPk = P0 + 1.6P1 + 3.48P2 + 1.65P3 + 1.96P4 + 0.1P5 + 0.32P9.

We have c0 = 1, ck 0, k = 1, 2, . . . , 9. Using Lemma 1 we get

S(X) =9∑

k=0

ck

n∑i=1

n∑j=1

Pk(cos(φi,j)) n∑i=1

n∑j=1

c0P0 = n2.

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5

Proof of Lemma 3.

Proof. 1. The polynomial f(t) satisfies the following properties (see Fig.2):(i) f(t) is a monotone decreasing function on the interval [−1,−t0];(ii) f(t) < 0 for t ∈ (−t0, 1/2];where f(−t0) = 0, t0 ≈ 0.5907.

These properties hold because f(t) has the only one root −t0 on [−1, 1/2],and there are no zeros of the derivative f ′(t) (8th degree polynomial) on [−1,−t0].

Let Si(X) :=n∑j=1

f(cos(φi,j)), then S(X) =n∑i=1

Si(X). From this follows

if Si(X) < 13 for i = 1, 2, . . . , n, then S(X) < 13n.We obviously have φi,i = 0, so f(cosφi,i) = f(1). Note that our assumption

on X (φi,j 60, i = j) yields cosφi,j 1/2. Therefore, cosφi,j lies in theinterval [-1,1/2]. Since (ii), if cosφi,j ∈ [−t0, 1/2], then f(cosφi,j) 0. LetJ(i) := j : cosφi,j ∈ [−1,−t0). We obtain

Si(X) Ti(X) := f(1) +∑j∈J(i)

f(cosφi,j). (1)

Let θ0 = arccos t0, θ0 ≈ 53.794. Then j ∈ J(i) iff φi,j > 180 − θ0, i.e.θj < θ0, where θj = 180 − φi,j . In other words all xi,j , j ∈ J(i) lie inside thecircle of center e0 and radius θ0, where e0 = −xi is the antipodal point to xi.

2. Let us consider on S2 points e0, y1, . . . , ym such that

φi,j = dist(yi, yj) 60 for all i = j, dist(e0, yi) θ0 for 1 i m. (2)

Denote by µ the highest value of m such that the constraints in (2) define anon-empty set of points y1, . . . , ym.

Suppose 0 m µ and Y = y1, . . . , ym satisfies (2). Let

H(Y ) = H(y1, . . . , ym) := f(1)+f(− cosθ1)+. . .+f(− cosθm), θi = dist(e0, yi)

hm := maxY

H(Y ), hmax := max h0, h1, . . . , hµ.It is clear that Ti(X) hm, where m = |J(i)|. From (1) it follows that

Si(X) hm. Thus, if we prove that hmax < 13, then we prove Lemma 3.

3. Now we prove that µ 4.Suppose Y = y1, . . . , ym ⊂ S2 satisfies (2). If e0 is the North pole and yi haspolar coordinates (θi, ϕi), then from the law of cosines we have:

cosφi,j = cos θi cos θj + sin θi sin θj cos(ϕi − ϕj).

From (2) we have cosφi,j 1/2, then

cos(ϕi − ϕj) 1/2 − cos θi cos θjsin θi sin θj

. (3)

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6

Let Q(α) =1/2 − cosα cosβ

sinα sinβ, then Q′(α) =

2 cosβ − cosα2 sin2 α sinβ

.

From this follows, if 0 < α, β θ0, then cosβ > 1/2 (because θ0 < 60); sothen Q′(α) > 0, and Q(α) Q(θ0). Therefore,

1/2 − cos θi cos θjsin θi sin θj

1/2 − cos2 θ0sin2 θ0

=1/2 − t201 − t20

.

Combining this inequality and (3), we get

cos(ϕi − ϕj) 1/2 − t201 − t20

.

Note that arccos((1/2 − t20)/(1 − t20)) ≈ 76.582 > 72. Then m 4 becauseno more than four points can lie in an unit circle with the minimum angularseparation between any two points greater than 72.

4. Now we have to prove that hmax = max h0, h1, h2, h3, h4 < 13.We obviously have h0 = f(1) = 10.11 < 13.

From (i) follows that f(− cos θ) is a monotone decreasing function in θ on[0, θ0]. Then for m = 1 : H(y1) = f(1) + f(− cos θ1) attains its maximum atθ1 = 0,

h1 = f(1) + f(−1) = 12.88 < 13.

5. Let us consider for m = 2, 3, 4 an optimal arrangement e0, y1, . . . , ymin S2 that gives maximum of H(Y ) = hm. Note that for optimal arrangementpoints yk cannot be shifted towards e0 because in this case H(Y ) increases.

For m = 2 this yields: e0 ∈ y1y2, and dist(y1, y2) = 60. If e0 /∈ y1y2, thenwhole arc y1y2 can be shifted to e0. Also if dist(y1, y2) > 60, then y1 (and y2)can be shifted to e0.

For m = 3 we prove that ∆3 = y1y2y3 is a spherical regular triangle withedge length 60. As above, e0 ∈ ∆3, otherwise whole triangle can be shiftedto e0. Suppose dist(y1, yi) > 60, i = 2, 3, then dist(y1, e0) can be decreased.From this follows that for any yi at least one of the distances dist(yi, yj) isequal to 60. Therefore, at least two sides of ∆3 (say y1y2 and y1y3) have length60. Also dist(y2, y3) = 60, conversely y3 (or y2, if e0 ∈ y1y3) can be rotatedabout y1 by a small angle towards e0 (Fig.3).

When m = 4 first we prove that ∆4 = y1y2y3y4 is a convex quadrangle.Conversely, we may assume that y4 ∈ y1y2y3.

The great circle that is orthogonal to the arc e0y4 divides S2 into two hemi-spheres: H1 and H2. Suppose e0 ∈ H1, then at least one yi (say y3) belongs toH2 (Fig.4). So the angle ∠e0y4y3 greater than 90, then (again from the lawof cosines) dist(y3, e0) > dist(y3, y4). Thus,θ3 = dist(y3, e0) > dist(y3, y4) 60 > θ0 − a contradiction.

Arguing as for m = 3 it is easy to prove that ∆4 is a spherical equilateralquadrangle (rhomb) with edge length 60.

7

7

e0

y1

y3 y2

60 60

Fig. 3

y1 y2

y3

yc

e0θ1 θ2

θ360 60

Fig. 5

y1y2

y3

y4

e0

H2

H1

Fig. 4

6. Now we introduce the function F1(ψ),2 where ψ ∈ [60, 2θ0]:

F1(ψ) := maxψ/2θθ0

F1(θ, ψ), F1(θ, ψ) = f(− cos θ) + f(− cos(ψ − θ)).

So if dist(yi, yj) = ψ, then

f(− cos θi) + f(− cos θj) F1(ψ). (4)

Therefore,

H(y1, y2) h2 = f(1) + F1(60) ≈ 12.8749 < 13.

7. When m = 4, ∆4 is a spherical rhomb. Let d1 = dist(y1, y3), andd2 = dist(y2, y4), then cos(d1/2) cos(d2/2) = 1/2 (Pythagorean theorem, thediagonals of ∆4 are orthogonal). So if ρ(s) := 2 arccos[1/(2 cos(s/2))], thend1 = ρ(d2), d2 = ρ(d1), ρ(90) = 90.

Suppose d1 d2. Since θi θ0, d2 2θ0, thenρ(2θ0) d1 90 d2 2θ0.

Now we consider two cases:1) ρ(2θ0) d1 < 77, and 2) 77 d1 90.1) F1(ψ) is a monotone decreasing function in ψ. Then (4) implies

f(− cos θ1) + f(− cos θ3) F1(ρ(2θ0)), f(− cos θ2) + f(− cos θ4) < F1(ρ(77)),

so then

H(Y ) < f(1) + F1(ρ(2θ0)) + F1(ρ(77)) ≈ 12.9171 < 13.

2) In this case we have

H(Y ) f(1) + F1(77) + F1(90) ≈ 12.9182 < 13.

Thus, h4 < 13.

8. Our last step is to show that h3 < 13.3

2For given ψ, the value F1(ψ) can be find as the maximum of the 9th degree polynomial

Ω(s) = F1(θ, ψ), s = cos (θ − ψ/2), on the interval [cos(θ0 − ψ/2), 1].3More detailed analysis shows h3 ≈ 12.8721, h4 ≈ 12.4849.

8

8

Since ∆3 is a regular triangle, H(Y ) = f(1) + f(− cos θ1) + f(− cos θ2) +f(− cos θ3) is a symmetric function in θi, so we can consider only the caseθ1 θ2 θ3 θ0.In this case R0 θ3 θ0, where R0 = arccos

√2/3 ≈ 35.2644. (Note that

the circumradius of ∆3 equals R0.)Let yc is the center of ∆3. Denote by u the angle ∠e0y3yc. Then (see Fig.5)

cos θ1 = cos 60 cos θ3 + sin 60 sin θ3 cos (R0 − u),

cos θ2 = cos 60 cos θ3 + sin 60 sin θ3 cos (R0 + u),

where ∠y1y3yc = ∠y2y3yc = R0, 0 u u0 = arccos(cot θ3/√

3) −R0

(if u = u0, then θ2 = θ3).For fixed θ3 = ψ, H(y1, y2) becomes the polynomial of degree 9 in s = cosu.

Denote by F2(ψ) the maximum of this polynomial on the interval [cosu0, 1].Let

ψ1, . . . , ψ6 = R0, 38, 41, 44, 48, θ0.It is clear that F2(ψ) is a monotone increasing function in ψ on [R0, θ0]. Fromother side, f(− cosψ) is a monotone decreasing function in ψ. Therefore forθ3 ∈ [ψi, ψi+1] we have

H(Y ) = H(y1, y2) + f(− cos θ3) < wi := F2(ψi+1) + f(− cosψi).

Since,

w1, . . . , w5 ≈ 12.9425, 12.9648, 12.9508, 12.9606, 12.9519,

we get h3 < maxwi < 13.Thus, hm < 13 for all m as required.

3 Delsarte’s method

Let X = x1, x2, . . . , xM be any finite subset of the unit sphere Sn−1 ⊂ Rn,Sn−1 = x : x ∈ Rn, x ·x = ||x||2 = 1. From here on we will speak of x ∈ Sn−1

alternatively of points in Sn−1 or of vectors in Rn.By φij we denote the spherical (angular) distance between xi, xj . It is clear

that for any real numbers u1, u2, . . . , uM the relation

||∑

uixi||2 =∑i,j

cosφijuiuj ≥ 0

holds, or equivalently the Gram matrix T (X) is positive semidefinite, whereT (X) = (tij), tij = cosφij = xi · xj .

Schoenberg [29] extended this property to Gegenbauer (ultraspherical) poly-nomials G(n)

k of tij . He proved that if gij = G(n)k (tij), then the matrix (gij) is

positive semidefinite. Schoenberg proved also that the converse holds: if f(t) is

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9

a real polynomial and for any finite X ⊂ Sn−1 the matrix (f(tij)) is positivesemidefinite, then f is a sum of G(n)

k with nonnegative coefficients.Let us recall the definition of Gegenbauer polynomials. Suppose C(n)

k (t) bethe polynomials defined by the expansion

(1 − 2rt+ r2)1−n/2 =∞∑k=0

rkC(n)k (t).

Then the polynomials G(n)k (t) = C

(n)k (t)/C(n)

k (1) are called Gegenbauer or ul-traspherical polynomials. (So the normalization of G(n)

k is determined by thecondition G(n)

k (1) = 1.)Also the Gegenbauer polynomials G(n)

k can be defined by recurrence formula:

G(n)0 = 1, G

(n)1 = t, . . . , G

(n)k =

(2k + n− 4) tG(n)k−1 − (k − 1)G(n)

k−2

k + n− 3.

They are orthogonal on the interval [−1, 1] with respect to the weight func-tion ρ(t) = (1− t2)(n−3)/2 (see details in [8, 10, 16, 29]). In the case n = 3, G(n)

k

are Legendre polynomials Pk, and G(4)k are Chebyshev polynomials of the second

kind (but with a different normalization than usual, Uk(1) = 1),

G(4)k (t) = Uk(t) =

sin ((k + 1)φ)(k + 1) sinφ

, t = cosφ, k = 0, 1, 2, . . . .

For instance, U0 = 1, U1 = t, U2 = (4t2 − 1)/3, U3 = 2t3 − t,U4 = (16t4 − 12t2 + 1)/5, . . . , U9 = (256t9 − 512t7 + 336t5 − 80t3 + 5t)/5.

Let us now prove the bound of Delsarte’s method. If a matrix (gij) is positivesemidefinite, then for any real ui the inequality

∑gijuiuj 0 holds, and then

for ui = 1, we have∑i,j

gij ≥ 0. Therefore, for gij = G(n)k (tij), we obtain

M∑i=1

M∑j=1

G(n)k (tij) 0. (3.1)

Suppose

f(t) = c0G(n)0 (t) + . . .+ cdG

(n)d (t), where c0 0, . . . , cd 0. (3.2)

Let S(X) =∑i

∑j

f(tij). Using (3.1), we get

S(X) =d∑k=0

M∑i=1

M∑j=1

ckG(n)k (tij)

M∑i=1

M∑j=1

c0G(n)0 (tij) = c0M

2. (3.3)

Let X = x1, . . . , xM ⊂ Sn−1 be a spherical z-code, i.e. for all i = j,tij = cosφij = xi · xj z, i.e. tij ∈ [−1, z] (but tii = 1). Suppose f(t) 0 for

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10

t ∈ [−1, z], then S(X) = Mf(1)+ 2f(t12) + . . .+ 2f(tM−1M ) Mf(1). If wecombine this with (3.2), then for c0 > 0 we get

M f(1)c0

. (3.4)

The inequality (3.4) play a crucial role in the Delsarte method (see details in[3, 4, 5, 10, 14, 15, 22, 24, 27]). If z = 1/2 and c0 = 1, then (3.4) implies k(n) f(1). In [24, 27] Levenshtein, Odlyzko and Sloane have found the polynomialsf(t) such that f(1) = 240, when n = 8; and f(1) = 196560, when n = 24.Then k(8) 240, k(24) 196560. When n = 8, 24, there exist sphere packings(E8 and Leech lattices) with these kissing numbers. Thus k(8) = 240 andk(24) = 196560. When n = 4, a polynomial f of degree 9 with f(1) = 25.5585...was found in [27]. This implies 24 k(4) 25.

4 An extension of Delsarte’s method.

Let us now generalize the Delsarte bound M f(1)/c0.

Definition. Let f(t) be any polynomial on the interval [−1, 1]. Consider onSn−1 points y0, y1, . . . , ym such that

yi · yj z for all i = j, f(y0 · yi) 0 for 1 i m. (4.1)

Denote by µ = µ(n, z, f) the highest value of m such that the constraints in(4.1) define a non-empty set of points y0, . . . , ym.

Suppose 0 m µ. Let

H(Y ) = H(y0; y1, . . . , ym) := f(1) + f(y0 · y1) + . . .+ f(y0 · ym),

hm := maxY

H(Y ), hmax := max h0, h1, . . . , hµ.Remark. hmax depends on n, z, and f. Throughout this paper it is clear whatf, n, and z are; so we denote by hmax the value hmax(n, z, f).

Theorem 2. Suppose X = x1, . . . , xM ⊂ Sn−1 is a spherical z-code andf(t) = c0G

(n)0 (t) + . . .+ cdG

(n)d (t), where c0 > 0, c1 0, . . . , cd 0. Then

M hmaxc0

=1c0

maxh0, h1, . . . , hµ.

Proof. Since f satisfies (3.2), then (3.3) yields

S(X) c0M2.

Let J(i) := j : f(xi · xj) > 0, j = i, and X(i) = xj : j ∈ J(i). Then

Si(X) =M∑j=1

f(xi · xj) f(1) +∑j∈J(i)

f(xi · xj) = H(xi;X(i)) hmax,

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11

so then

S(X) =M∑i=1

Si(X) Mhmax.

We have c0M2 S(X) Mhmax, i.e. c0M hmax as required.

Note that h0 = f(1). If f(t) 0 for all t ∈ [−1, z], then for a z-code Xwe have hmax = h0 = f(1). Therefore, this theorem yields the Delsarte boundM f(1)/c0.

The problem of evaluating of hmax in general case looks even more compli-cated than the upper bound problem for spherical z-codes. It is not clear howto find µ? Here we consider this problem only for a very restrictive class offunctions f(t): f(t) < 0 for t ∈ [−t0, z], t0 > z 0.

Let us denote by A(k, ω) the maximal number of points in a spherical s-codeΩ ⊂ Sk−1 of minimal angle ω, cosω = s. (Note that A(n, 60) is the kissingnumber k(n).)

Theorem 3. Suppose Y = y1, . . . , ym is a spherical z-code in Sn−1, andpoints yi lie inside the sphere of center e0 and radius θ0, where t0 = cos θ0 z.Then

m A(n− 1, arccos

z − t201 − t20

).

Proof. We have φi,j = dist(yi, yj) δ = arccos z for i = j;θi = arccos(e0 · yi) θ0 for 1 i m; and θ0 δ.

Let Π be the projection of Y onto equator Sn−2 from pole e0. Denote by γi,jthe distances between points of Π in Sn−2. Then from the law of cosines andthe inequality cosφi,j z, we get

cos γi,j =cosφi,j − cos θi cos θj

sin θi sin θj z − cos θi cos θj

sin θi sin θj

Let Q(α) =z − cosα cosβ

sinα sinβ, then Q′(α) =

cosβ − z cosαsin2 α sinβ

.

From this follows, if 0 < α, β θ0, then cosβ z (because θ0 δ); so thenQ′(α) 0, and Q(α) Q(θ0). Therefore,

cos γi,j z − cos θi cos θjsin θi sin θj

z − cos2 θ0sin2 θ0

=z − t201 − t20

that completes our proof.

Corollary 1. Suppose f(t) < 0 for t ∈ [−t0, z], t0 z 0, then

µ(n, z, f) A(n− 1, arccos

z − t201 − t20

).

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12

Proof. The assumption on f yields f(y0 · yi) > 0 only if

θi = dist(e0, yi) < θ0 = arccos t0,

where e0 = −y0 is the antipodal point to y0. Therefore, this set of pointse0, y1, . . . , ym satisfies the assumptions in Theorem 3.

The next claim will be applied to prove that k(4) = 24.

Corollary 2. Suppose f(t) < 0 for t ∈ [−t0, 1/2], t0 0.6058, thenµ = µ(4, 1/2, f) 6.

Proof. Note that for t0 0.6058, arccos[(1/2 − t20)/(1 − t20)] > 77.87. SoCorollary 1 implies µ(4, 1/2, f) A(3, 77.87).

Denote by ϕk(M) the largest angular separation that can be attained ina spherical code on Sk−1 containing M points. In three dimensions the bestcodes and the values ϕ3(M) presently known for M 12 and M = 24 (see[12, 17, 30]). For instance, Schutte and van der Waerden [30] proved thatϕ3(5) = ϕ3(6) = 90 and ϕ3(7) ≈ 77.86954 (cosϕ3(7) = cot 40 cot 80).

Since 77.87 > ϕ3(7), then A(3, 77.87) < 7, i.e. µ 6.

Corollary 1 shows that if t0 is close enough to 1, then µ is small enough.Then one gets relatively small - dimensional optimization problems for compu-tation of numbers hm for small n. If additionally f(t) is a monotone decreasingfunction on [−1,−t0], then these problems can be reduced to low-dimensionaloptimization problems of a type that can be treated numerically.

5 Optimal sets for monotonic functions

In this section we consider f(t) that satisfies the monotonicity assumption:

f(t) is a monotone decreasing function on the interval [−1,−t0],f(t) < 0 for t ∈ [−t0, z], t0 > z 0. (∗)

Consider on Sn−1 points y0, y1, . . . , ym that satisfy (4.1). Denote by θk fork > 0 the distance between yk and e0, where e0 = −y0 is the antipodal point toy0. Then y0 · yk = − cos θk, and H(Y ) is represented in the form:

H(Y ) = f(1) + f(− cos θ1) + . . .+ f(− cos θm). (5.1)

A subset C of Sn−1 is called (spherical) convex if it contains, with every twononantipodal points, the small arc of the great circle containing them. If, inaddition, C does not contain antipodal points, then C is called strongly convex.The closure of a convex set is convex and is the intersection of closed hemispheres(see details in [13]). If a subset Z of Sn−1 lies in a hemisphere, then the convexhull of Z is well defined, and is the intersection of all convex sets containing Z.

Suppose f(t) satisfies (∗), then Qm = y1, . . . , ym lies in the hemisphere ofcenter e0. Denote by ∆m the convex hull of Qm in Sn−1, ∆m = convQm.

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13

Now we consider an optimal arrangement of Qm for H. Let δ = arccos z,φi,j = dist(yi, yj), N(Qm) = number of φi,j = δ (yi · yj = z).

Definition We say that Qm is optimal if H(Y ) = hm. If optimal Qm is notunique up to isometry, then we call Qm as optimal if it has maximal N(Qm).

The function f(t) is monotone decreasing on [−1,−t0]. By (5.1) it followsthat the function H(Y ) increases whenever θk decreases. This means that foran optimal Qm no yk ∈ Qm can be shifted towards e0.

That yieldse0 ∈ ∆m (5.2)

because in the converse case whole Qm can be shifted to e0.From this follows that for m = 1, e0 = y1. Thus

h1 = f(1) + f(−1).

It was proved in Section 2 that for m = 2 : dist(y1, y2) = δ, thus

h2 = f(1) + maxδ/2θθ0

(f(− cos θ) + f(− cos(δ − θ)), θ0 = arccos t0.

It was also proved that ∆3 is a spherical regular triangle with edge length δ.Using similar arguments it is not hard to prove that for n > 3, ∆4 is a sphericalregular tetrahedra with edge length δ. 4

Let ∆m, m n, is a spherical regular simplex with edge length δ, and

Ωm = y : y ∈ ∆m, y · yk t0, 1 k m.Note that Ωm is a convex set in Sn−1. Let

Hm(y) = f(1) + f(−y · y1) + . . .+ f(−y · ym).

Then hm is the maximum of Hm(y) on Ωm.

hm = maxy∈Λm

Hm(y), Ωm ⊂ ∆m ⊂ Sn−1, 2 m min(n, µ). (5.3)

When n > m any yk ∈ Qm is a vertex of ∆m. In other words, no yk that liesinside ∆m. In fact, that has been proved in Section 2 (see 5, Fig. 4).

In the first version of the paper [26] has been claimed thatfor optimal Qm with m > n, for any yk ∈ Qm there are at least n− 1 distinctpoints in Qm at the distance of δ from yk.However, Eiichi Bannai and Makoto Tagami found some gaps in our exposition.Most of them are related to “degenerated” configurations. In this paper we needonly the case n = 4, m = 5. For this case they verified each step of our proof,considered all “degenerated” configurations, and finally gave clean and detailedproof. I wish to thank Eiichi Bannai and Makoto Tagami for this work. Nowthis claim in general case can be considered only as conjecture.

4For m n, ∆m is a spherical regular simplex with edge length δ. In this paper we needjust cases m = 3, 4.

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14

6 An algorithm for computation suitable poly-nomials f(t)

In this section is presented an algorithm for computation “optimal” 5 polynomi-als f such that f(t) is a monotone decreasing function on the interval [−1,−t0],and f(t) < 0 for t ∈ [−t0, z], t0 > z 0. This algorithm based on ourknowledge about optimal arrangement of points yi for given m. Coefficients ckcan be found via discretization and linear programming; such method had beenemployed already by Odlyzko and Sloane [27] for the same purpose.

Let us have a polynomial f represented in the form f(t) = 1+d∑k=1

ckG(n)k (t).

We have the following constraints for f : (C1) ck 0, 1 k d;(C2) f(a) > f(b) for −1 a < b −t0; (C3) f(t) < 0 for −t0 t z.

When m n, hm = maxHm(y), y ∈ Λm. We do not know y where Hm

attains its maximum, so for evaluation of hm let us use yc − the center of ∆m.All vertices yk of ∆m are at the distance of Rm from yc, wherecosRm =

√(1 + (m− 1)z)/m.

When m = 2n− 2, ∆m presumably is a regular (n − 1)-dimensional cross-polytope. (It is not proven yet.) In this case cosRm =

√z.

Let In = 1, . . . , n⋃2n− 2, m ∈ In, bm = − cosRm, whenceHm(yc) = f(1) + mf(bm). If F0 is such that Hm(y) E = F0 + f(1), then(C4) f(bm) F0/m, m ∈ In. A polynomial f that satisfies (C1-C4) and givesthe minimal E (note that E = F0 + 1 + c1 + . . .+ cd = F0 + f(1) will become alower estimate of hmax) can be found by the following

Algorithm.

Input: n, z, t0, d, N.Output: c1, . . . , cd, F0, E.

First replace (C2) and (C3) by a finite set of inequalities at the pointsaj = −1 + εj, 0 j N, ε = (1 + z)/N :

Second use linear programming to find F0, c1, . . . , cd so as to minimize

E − 1 = F0 +d∑k=1

ck subject to the constraints

ck 0, 1 k d;d∑k=1

ckG(n)k (aj)

d∑k=1

ckG(n)k (aj+1), aj ∈ [−1,−t0];

1+d∑k=1

ckG(n)k (aj) 0, aj ∈ [−t0, z]; 1+

d∑k=1

ckG(n)k (bm) F0/m, m ∈ In.

Let us note again that E = maxm∈In

Hm(yc) hmax here, and that E = hmax

only if hmax = Hm0(yc) for some m0 ∈ In.

5Open problem: is it true that for given t0, d this algorithm defines f with minimal hmax?

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15

7 On calculations of hm for m n

Here we explain how to solve the optimization problem (5.3). Let ∆m ⊂ Sm−1 isa spherical regular simplex with edge length δ = arccos z; yi, i = 1, . . . ,m, arethe vertices of ∆m; ti = y · yi = cos θi t0 = cos θ0; t0 > z; f(t) is a monotonedecreasing function on the interval [−1,−t0]; hm is the maximum of Hm(y)subject to the constraints ti t0; Hm(y) = f(1)+ f(−y · y1)+ . . .+ f(−y · ym).

The first method.Hm(y) is a symmetric function in the variables θ1, . . . , θm. Then we can considerthis problem only on the domain Λ = y : θm . . . θ2 θ1. Note that Λ isa spherical simplex. Let us consider a barycentric triangulation of this simplexsuch that the diameter of any simplex σi of this triangulation is not exceed ε.

It is easy to prove that for any yk, y · yk attains its maximum on σi at somevertex of σi. Denote this vertex by yk,i. Let I = i : y1,i · y1 t0. So for i ∈ Iwe have

f(−yk,i · yk) = maxy∈σi

f(−y · yk),then

hm maxi∈I

m∑k=1

f(−yk,i · yk).

That yields a very simple method for calculation of hm. For f from Section9 this method gives h3 ≈ 24.8345, h4 ≈ 24.818.

The second method.For m n the values hm can be calculated another way. We are using here thatf(t) = f0+f1t+ . . .+fdtd is a polynomial. The first method is technically easierthen the second one. However, the second method doesn’t assume that f is amonotone decreasing function on [−1,−t0], and it can be applied to functionswithout monotonicity assumption.

Let us consider Hm(y) as the symmetric polynomial Fm(t1, . . . , tm) in thevariables ti = y · yi : Fm(t1, . . . , tm) = f(1) + f(−t1) + . . .+ f(−tm). Denote bysk = sk(t1, . . . , tm) the power sum tk1 + . . .+ tkm. Then

Fm(t1, . . . , tm) = Ψm(s1, . . . , sd) = f(1) +mf0 − f1 s1 + . . .+ (−1)dfd sd.

From the fact that ∆m is a spherical regular simplex follows

s2 = σ(s1) :=z

(m− 1)z + 1s21 + 1 − z. (7.1)

Any symmetric polynomial in m variables can be expressed as a polynomialof s1, . . . , sm. Therefore, in the case k > m the power sum sk is Rk(s1, . . . , sm).Combining this with (7.1), we get

Ψm(s1, σ(s1), s3, . . . , sd) = Φm(s1, s3, . . . , sm).

Therefore, we have

hm = max Φm(s1, s3, . . . , sm), (s1, s3, . . . , sm) ∈ Dm ⊂ Rm−1,

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16

where Dm is the domain in Rm−1 defined by the constraints ti t0 and (7.1).Let us show now how to determine Dm for m > 2. The equation (7.1) defines

the ellipsoid E : s2 = σ(s1) in space (t1, . . . , tm). Then s1 = t1 + . . . + tmattains its maximum on E at the point with t1 = t2 = . . . = tm, and s1 achievesits minimum on E

⋂ti t0 at the point with t2 = . . . = tm = t0. From thisfollows w1 s1 w2, where

w1 =

√(p− t20) (p− z2) + z t0

p+ (m− 1) t0, p =

1 + (m− 2) zm− 1

,

w2 =√m (m− 1) z +m.

The equation s1 = ω gives the hyperplane, and the equation s2 = σ(ω)gives the (m− 1)-sphere in space: (t1, . . . , tm). Denote by S(ω) the (m− 2)-sphere that is the intersection of these hyperplane and sphere. Let lk(ω) be theminimum of sk on S(ω)

⋂ti t0, and vk(ω) is its maximum. Now we have

hm = maxs1

maxs3

. . .maxsm

Φm(s1, s3, . . . , sm), where

w1 s1 w2, lk(s1) sk ≤ vk(s1), k = 3, . . . ,m.

For the polynomial f from Section 9 (and Section 2) we can give more detailsabout calculations of hm for m = 3, 4.

Let us consider the case m = 3 with d = 9. In this case Fω(s3) = Φ3(ω, s3)is a polynomial of degree 3 in the variable s3.

Lemma 4. Let f be a 9th degree polynomial f(t) = f0 + f1t+ . . .+ f9t9 such

that f9 > 0, f6 = f8 = 0, and f7 > −15f9/7. If F ′ω(s) ≤ 0 at s = l3(ω), then the

function Fω(s) achieves its maximum on the interval [l3(ω), v3(ω)] at s = l3(ω).

Proof. The expansion of s9 in terms of si1sj2sk3 , i+ 2j + 3k = 9, is

s9 =19s33 + s23(

23s31 + s2s1) + s3(

38s32 −

38s22s

21 −

78s2s

41 +

524s61) +R(s1, s2).

The coefficient of s23s1 in s7 equals 7/9. Thus

Fω(s) = −s3 f9/9 − s2 (f9 ω σ(ω) + 2f9 ω3/3 − 7f7 ω/9) + sR1(ω) +R0(ω).

Fω(s) is a cubic polynomial with negative coefficient of s3. Then Fω(s) is aconcave function for s > r, where r : F ′′

ω (r) = 0. Therefore, if r < l3(ω), thenFω(s) is a concave function on the interval [l3(ω), v3(ω)]. r < l3(ω) iff

B(ω) := 3l3(ω) + 6ω3 + 9ω σ(ω) > −7ωf7/f9.

This inequality holds for t0 < −z ≤ 0. Indeed,

ω w1 1 + 2z, σ (ω) ≥ 1, l3(ω) > 0;

so thenB(ω) > 15ω > −7ωf7/f9.

The inequality F ′ω(l3(ω)) 0 implies that Fω(s) is a decreasing function on the

interval [l3(ω), v3(ω)].

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The polynomial f from Section 9 satisfies the assumptions in this lemma.Then Φ3(ω, s) attains its maximum at the point s = l3(ω), i.e. at the point witht1 = t2 t3, or with t1 t2 ≥ t3 = t0. If t1 = t2 t3, then p(ω) = Φ3(ω, l3(ω))is a polynomial in ω. This polynomial is a decreasing function in the variable ωon the interval t3 t0. Therefore, p(ω) achieves its maximum on this intervalat the point with t3 = t0. The calculations show that for f from Section 9h3 = max p(ω) ≈ 24.8345, when θ3 = θ0, θ1 = θ2 ≈ 30.0715.

Corollary 3. Let f be the polynomial from Section 9, then h3 ≈ 24.8345.

Consider the function Fω(s3, s4) = Φ4(ω, s3, s4) on S(ω). Let qi ∈ S(ω) andq1 : t1 = t2 > t3 = t4, q2 : t1 = t2 = t3 > t4, and q3 : t1 > t2 = t3 = t4.

Lemma 5. Let f be a 9th degree polynomial f(t) =∑fit

i. If f9 > 0 andf6 = f8 = 0, then the function Fω(s3, s4) achieves its maximum on S(ω) withω > 1 at one of the points (s3(qi), s4(qi)), i = 1, 2, 3.

Proof. The expansion of s9 in terms of si1sj2sk3sl4 is

s9 =916s24s1 +

19s33−

13s23s

31 +

34s4s3s1 +

38s4s2s

31−

38s3s

22s

21−

124s3s

61 +R(s1, s2).

The coefficient of s23s1 in s7 equals 0. We have f6 = f8 = 0, thenFω(s3, s4) = −f9 s9 + . . . = −f9(s33/9 − s23 ω

3/3) + . . . Therefore,

F33 =∂2Fω(s3, s4)

∂2s3= −f9(2

3s3 − 2

3ω3) =

2f93

(ω3 − s3).

If Fω(s3, s4) has its maximum on S(ω) at the point x, and x is not a criticalpoint of s3 on S(ω), then F33 ≤ 0. From other side, for all ti ∈ [0, 1] ands1 = ω > 1 we have s3 ω < ω3, so then F33 > 0. The function s3 on S(ω)(up to permutation of labels) has critical points at qi, i = 1, 2, 3.

Corollary 4. Let f be the polynomial from Section 9, then h4 ≈ 24.818.

Proof. By direct calculations it can be shown thatFω(s3(q1), s4(q1)) > Fω(s3(qi), s4(qi)) for i = 2, 3. Then Lemma 5 impliesh4 = max p(ω), where p(ω) = Fω(s3(q1), s4(q1)) = Φ4(ω, s3(q1), s4(q1)).

The polynomial p(ω) attains its maximum h4 ≈ 24.818 at the point withθ1 = θ2 ≈ 30.2310, θ3 = θ4 ≈ 51.6765.

8 On calculations of h5 in four dimensions

Let us consider the case n = 4, m = 5. For simplicity here we consider only thecase z = 1/2. Then δ = 60 and θ0 = arccos t0 < 60.

Denote by Γ5 the graph of the edges of ∆5 with length 60 , where Q5 is anoptimal set. The degree of any vertex of Γ5 is not less than 3 (see Section 5).This implies that at least one vertex of Γ5 has degree 4. Indeed, if all verticesof Γ5 are of degree 3, then the sum of the degrees equals 15, i.e. is not an evennumber. There exists only one type of Γ5 with these conditions (Fig. 6).

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Fig. 6

αy5

y2

y4

y3

y1

For fixed dist(y2, y4) = α, Q5 is uniquely defined up to isometry. Therefore,we have the 1-parametric family ∆5(α) on S3. If dist(y3, y5) = β, then

2 cosα cosβ + cosα+ cosβ = 0 (8.1)

The equation (8.1) defines the function β = λ(α). Then α = λ(β), λ(90) = 90.For all i we have dist(yi, e0) θ0, then

dist(yi, yj) dist(yi, e0) + dist(yj , e0) 2θ0.

Suppose α β, then (8.1) and the inequality β 2θ0 yield

α0 α 90 β 2θ0, α0 := max60, λ(2θ0).Let

H5(y, α) = f(1) + f(−y · y1(α)) + . . .+ f(−y · y5(α)).

Then

h5 = maxy,α

H5(y, α), y ∈ S3, y ·yk(α) t0, 1 k 5, α0 α 90 (8.2)

We have four-dimensional optimization problem (8.2). Our first approachfor this problem was to apply numerical methods [25]. For the polynomial ffrom Section 9 this optimization problem was solved numerically by using theNelder-Mead simplex method: H5(y, α) achieves its maximum h5 ≈ 24.6856 atα = 60 and y with θ1 ≈ 42.1569, θ2 = θ4 ≈ 32.3025, θ3 = θ5 = θ0.(The similar approach for the case n = 4, m = 6 gives the 3-parametric family∆6(α, β, γ), and for f from Section 9: h6 ≈ 22.5205.)

Note that (8.2) is a nonconvex constrained optimization problem. In thiscase, the Nelder-Mead simplex method and other local improvements methodscannot guarantee finding a global optimum. It is possible (using estimations ofderivatives) to organize computational process in such way that it gives a globaloptimum. However, such a kind solutions are very hard to verify and somemathematicians do not accept such kind proofs. Fortunately, an estimation ofh5 can be reduced to discrete optimization problems.

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Let dist(y1, y) = ψ, and Φi,j(y, ψ) = f(−y · yi) + f(−y · yj). It is clearthat for fixed ψ, Φi,j(y, ψ) attains its maximum at some point that lies in thegreat 2-sphere that contains y1, yi, yj. Now we introduce the function F (ψ, γ).6

Suppose y1yiyj is a spherical triangle in S2 with dist(y1, yi) = dist(y1, yj) =60, dist(yi, yj) = γ, denote by F (ψ, γ) the maximum of Φi,j(y, ψ) on S2 subjectto the constraints y · yk t0, k = i, j. Then Φi,j(y, ψ) F (ψ, γ), so thenΦ2,4(y, ψ) F (ψ, α), Φ3,5(y, ψ) F (ψ, β). Thus

h5 f(1) + f(− cosψ) + F (ψ, α) + F (ψ, λ(α)). (8.3)

Let α0 < α1 < . . . < αk < αk+1 = 90. It is easy to see that F (ψ, γ) is amonotone decreasing function in γ. That implies for α ∈ [αi, αi+1] :F (ψ, α) F (ψ, αi), F (ψ, λ(α)) F (ψ, λ(αi+1)). Therefore, from (8.3) follows

h5 f(1) + f(− cosψ) + max0ik

F (ψ, αi) + F (ψ, λ(αi+1)). (8.4)

Note that (8.4) to reduce the dimension of the optimization problem (8.1)from 4 to 2. It is not too hard to solve this problem in general case. However,the polynomial f from Section 9 satisfies an additional assumptions that allowedto find a weak bound on h5 even more easier.

Let us briefly explain how to check the following assumptions for f :1) Φi,j(y, ψ) achieves its maximum at one of the ends of the arc ω(ψ, γ), whereω(ψ, γ) := y : y ∈ S2, dist(y1, y) = ψ, y · y t0, = i, j;2) F (ψ, γ) is a monotone increasing function in ψ.

For given γ (γ = dist(yi, yj)) and ψ the function Φi,j(y, ψ) becomes a poly-nomial p(s) of degree d on [s0, 1], where s = cosu, u = ∠yiy1yc, and yc isthe center of y1yiyj (see Section 2, 8). Then 1) holds iff p′(s) has no roots on(s0, 1), either if s : p′(s) = 0, then p′′(s) > 0.

Using 1) it is easy to check 2). For the polynomial f(t) from Section 9 ifγ > 62.41, then p(s) achieves its maximum at s = s0 (i.e. dist(yj , y) = θ0), soit is clear that 2) holds. From other side if γ < 69.34, then the arc ω(ψ, γ) liesinside the triangle y1yiyj , therefore F (ψ, γ) increases whenever ψ increases.

Note that 1) gives us the explicit expression for F (ψ, γ) = max(p(s0), p(1)).For fixed γ and ψ ψ from 2) follows F (ψ, γ) F (ψ, γ).

Denote by ψL(i), ψU(i) the lower and upper bounds on ψ that defined bythe constraints α ∈ [αi, αi+1], y · yq t0, q = 1, . . . , 5.Let ψL(i) = ψi,0 < ψi,1 < . . . < ψi, < ψi,+1 = ψU(i). Recall that f(− cosψ) is

a monotone decreasing function in ψ. Then 2) and (8.4) yield

h5 f(1) + max0ik

max0j

Ri,j, (8.5)

whereRi,j = f(− cosψi,j) + F (ψi,j+1, αi) + F (ψi,j+1, λ(αi+1)).

It is very easy to apply this method. Here we need just to calculate thematrix (Ri,j) and the maximal value of its entries gives the bound on h5. For

6F (ψ, 60) = F2(ψ) − f(1) (see Section 2, 8, Fig. 5).

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f from Section 9 and t0 ≈ 0.60794, θ0 = arccos t0 ≈ 52.5588, f(−t0) = 0, thismethod gives the bound h5 < 24.8434.7

Now we show how to find an upper bound on h6. Let e0, y1, . . . , y6 ∈ S3,H(y1, . . . , y6) = f(1) + f(− cos θ1) + . . . + f(− cos θ6), where θi = dist(e0, yi).Suppose θ1 θ2 . . . θ6. Now we prove that θ6 45. That can beproven as Corollary 2 (Section 4). Conversely, all θi < 45. In this caset0∗ = cos θ6 > 1/

√2, and ω = arccos [(1/2 − t20∗)/(1 − t20∗)] > 90. But if

u > 90, then A(3, ω) 4 (see [30, 17]) - a contradiction. (In fact we provedthat θ5 45 also.)

Let us consider two cases: (i) θ0 θ6 > 50 (ii) 50 θ6 45.(i) H(y1, . . . , y6) = H(y1, . . . , y5) + f(− cos θ6). We have

H(y1, . . . , y5) h5 < 24.8434, f(− cos θ6) < f(− cos 50) ≈ 0.0906,

then H(y1, . . . , y6) < 24.934.(ii) In this case all θi 50. Therefore, we can apply (8.5) for θ0 = 50. Thismethod gives h5(50) < 23.9181, then H(y1, . . . , y5) h5(50) < 23.9181, sothen

H(y1, . . . , y6) < 23.9181 + f(− cos 45) ≈ 23.9181 + 0.4533 = 24.3714.

Thush6 < max24.934, 24.3714 = 24.934.

9 k(4) = 24

For n = 4, z = cos 60 = 1/2 we apply this extension of Delsarte’s method with

f(t) = 53.76t9−107.52t7+70.56t5+16.384t4−9.832t3−4.128t2−0.434t−0.016.

The expansion of f in terms of Uk = G(4)k is

f = U0 + 2U1 + 6.12U2 + 3.484U3 + 5.12U4 + 1.05U9.

The polynomial f has two roots on [−1, 1]: t1 = −t0, t0 ≈ 0.60794, t2 = 1/2,f(t) 0 for t ∈ [−t0, 1/2], and f is a monotone decreasing function on theinterval [−1,−t0]. The last property holds because there are no zeros of thederivative f ′(t) on [−1,−t0]. Therefore, f satisfies (∗) for z = 1/2.

Remark. The polynomial f was found by using the algorithm in Section 6.This algorithm for n = 4, z = 1/2, d = 9, N = 2000, t0 = 0.6058 givesE ≈ 24.7895. For the polynomial f the coefficients ck were changed to “betterlooking” ones with E ≈ 24.8644.

7R achieves its maximum at α = 60, ψ ≈ 30.9344. Note that this bound exceeds thetight bound on h5 ≈ 24.6856 given by numerical methods.

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−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

0

1

2

3

4

5

6

−1

Fig. 7. The graph of the function f(t)

We have t0 > 0.6058. Then Corollary 2 gives µ 6. Consider all m 6.

h0 = f(1) = 18.774, h1 = f(1) + f(−1) = 24.48.

h2 = f(1) + max30θθ0

(f(− cos θ) + f(− cos(60 − θ)) ≈ 24.8644,

where θ0 = arccos t0 ≈ 52.5588.Note that h2 can be calculated by the same method as in Section 2. Here

h2 = f(1) + 2f(− cos30) also.In Sections 7, 8 have been shown that

h3 ≈ 24.8345, h4 ≈ 24.818, h5 < 24.8434, h6 < 24.934.

Theorem 4. k(4) = 24

Proof. Let X be a spherical 1/2-code in S3 with M = k(4) points. The poly-nomial f is such that hmax < 25, then combining this and Theorem 2, we getk(4) hmax < 25. Recall that k(4) 24. Consequently, k(4) = 24.

10 Concluding remarks

The algorithm in Section 6 can be applied to other dimensions and sphericalz-codes. If t0 = 1, then the algorithm gives the Delsarte method. E is anestimation of hmax in this algorithm.

Direct application of the method developed in this paper, presumably couldlead to some improvements in the upper bounds on kissing numbers in dimen-sions 9, 10, 16, 17, 18 given in [10, Table 1.5]. (“Presumably” because theequality hmax = E is not proven yet.)

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In 9 and 10 dimensions Table 1.5 gives:306 k(9) 380, 500 k(10) 595.The algorithm gives:n = 9 : deg f = 11, E = h1 = 366.7822, t0 = 0.54;n = 10 : deg f = 11, E = h1 = 570.5240, t0 = 0.586.For these dimensions there is a good chance to prove thatk(9) 366, k(10) 570.

From the equality k(3) = 12 follows ϕ3(13) < 60. The method givesϕ3(13) < 59.4 (deg f = 11). The lower bound on ϕ3(13) is 57.1367 [17].Therefore, we have 57.1367 ϕ3(13) < 59.4.

The method gives ϕ4(25) < 59.81, ϕ4(24) < 60.5. (This is a theorem thatcan be proven by the same method as Theorem 4.) That improve the bounds:

ϕ4(25) < 60.79, ϕ4(24) < 61.65 [24] (cf. [5]); ϕ4(24) < 61.47 [5];

ϕ4(25) < 60.5, ϕ4(24) < 61.41 [4].

Now in these cases we have

57.4988 < ϕ4(25) < 59.81, 60 ϕ4(24) < 60.5.

For all cases that were considered (z 0.6) this method gives better boundsthan Fejes Toth’s bounds for ϕ3(M) [17] and Coxeter’s bounds for all ϕn(M)[11]. However, for n = 5, 6, 7 direct use of this generalization of the Delsartemethod does not give better upper bounds on k(n) than the Delsarte method.It is an interesting problem to find better methods.

Acknowledgment. I wish to thank Eiichi Bannai, Ivan Dynnikov, DmitryLeshchiner, Sergei Ovchinnikov, Makoto Tagami and Gunter Ziegler for helpfuldiscussions and useful comments on this paper.

References

[1] M. Aigner and G.M. Ziegler, Proofs from THE BOOK, Springer, 1998 (firsted.) and 2002 (second ed.)

[2] K. Anstreicher, The thirteen spheres: A new proof, Discrete and Compu-tational Geometry, 31(2004), 613-625.

[3] V.V. Arestov and A.G. Babenko, On Delsarte scheme of estimating thecontact numbers, Trudy Mat. Inst. im. V.A.Steklova 219, 1997, 44-73;English translation, Proc. of the Steklov Inst. of Math. 219, 1997, 36-65.

[4] V.V. Arestov and A.G. Babenko, On kissing number in four dimensions,in Proc. Conf. memory of Paul Erdos, Budapest, Hingary, July 4-11, 1999,A.Sali, M.Simonovits and V.T.Sos (eds), J. Bolyai Math. Soc., Budapest,1999, 10-14.

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[5] P.G. Boyvalenkov, D.P. Danev and S.P. Bumova, Upper bounds on theminimum distance of spherical codes, IEEE Trans. Inform. Theory, 42(5),1996, 1576-1581.

[6] K. Boroczky, Packing of spheres in spaces of constant curvature, Acta Math.Acad. Sci. Hung. 32 (1978), 243-261.

[7] K. Boroczky, The Newton-Gregory problem revisited, Proc. Discrete Ge-ometry, Marcel Dekker, 2003, 103-110.

[8] B.C. Carlson, Special functions of applied mathematics, Academic Press,1977.

[9] B. Casselman, The difficulties of kissing in three dimensions, Notices Amer.Math. Soc., 51(2004), 884-885.

[10] J.H. Conway and N.J.A. Sloane, Sphere Packings, Lattices, and Groups,New York, Springer-Verlag, 1999 (Third Edition).

[11] H.S.M. Coxeter, An upper bound for the number of equal nonoverlappingspheres that can touch another of the same size, Proc. of Symp. in PureMath. AMS, 7 (1963), 53-71 = Chap. 9 of H.S.M. Coxeter, Twelve Geo-metric Essays, Southern Illinois Press, Carbondale Il, 1968.

[12] L. Danzer, Finite point-sets on S2 with minimum distance as large as pos-sible, Discr. Math., 60 (1986), 3-66.

[13] L. Danzer, B. Grunbaum, and V. Klee. Helly’s theorem and its relatives.Proc. Sympos. Pure Math., vol. 7, AMS, Providence, RI, 1963, pp. 101-180.

[14] Ph. Delsarte, Bounds for unrestricted codes by linear programming, PhilipsRes. Rep., 27, 1972, 272-289.

[15] Ph. Delsarte, J.M. Goethals and J.J. Seidel, Spherical codes and designs,Geom. Dedic., 6, 1977, 363-388.

[16] A. Erdelyi, editor, Higher Transcendental Function, McGraw-Hill, NY, 3vols, 1953, Vol. II, Chap. XI.

[17] L. Fejes Toth, Lagerungen in der Ebene, auf der Kugel und in Raum,Springer-Verlag, 1953; Russian translation, Moscow, 1958.

[18] T. Hales, The status of the Kepler conjecture, Mathematical Intelligencer16(1994), 47-58.

[19] R. Hoppe, Bemerkung der Redaction, Archiv Math. Physik (Grunet) 56(1874), 307-312.

[20] W.-Y. Hsiang, The geometry of spheres, in Differential Geometry (Shang-hai,1991), Word Scientific, River Edge, NJ, 1993, pp. 92-107.

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[21] W.-Y. Hsiang, Least Action Principle of Crystal Formation of Dense Pack-ing Type and Kepler’s Conjecture, World Scientific, 2001.

[22] G.A. Kabatiansky and V.I. Levenshtein, Bounds for packings on a sphereand in space, Problemy Peredachi informacii 14(1), 1978, 3-25; Englishtranslation, Problems of Information Transmission, 14(1), 1978, 1-17.

[23] J. Leech, The problem of the thirteen spheres, Math. Gazette 41 (1956),22-23.

[24] V.I. Levenshtein, On bounds for packing in n-dimensional Euclidean space,Sov. Math. Dokl. 20(2), 1979, 417-421.

[25] O.R. Musin, The problem of the twenty-five spheres, Russian Math. Sur-veys, 58(2003), 794-795.

[26] O.R. Musin, The kissing number in four dimensions, preprint, September2003, math. MG/0309430.

[27] A.M. Odlyzko and N.J.A. Sloane, New bounds on the number of unitspheres that can touch a unit sphere in n dimensions, J. of Combinato-rial Theory A26(1979), 210-214.

[28] F. Pfender and G.M. Ziegler, Kissing numbers, sphere packings, and someunexpected proofs, Notices Amer. Math. Soc., 51(2004), 873-883.

[29] I.J. Schoenberg, Positive definite functions on spheres, Duke Math. J., 9(1942), 96-107.

[30] K. Schutte and B.L. van der Waerden, Auf welcher Kugel haben 5,6,7,8oder 9 Punkte mit Mindestabstand 1 Platz? Math. Ann. 123 (1951), 96-124.

[31] K. Schutte and B.L. van der Waerden, Das Problem der dreizehn Kugeln,Math. Ann. 125 (1953), 325-334.

[32] G.G. Szpiro, Kepler’s conjecture, Wiley, 2002.

[33] G.G. Szpiro, Newton and the kissing problem,http://plus.maths.org/issue23/features/kissing/

[34] A.D. Wyner, Capabilities of bounded discrepancy decoding, Bell Sys. Tech.J. 44 (1965), 1061-1122.

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On optimal sets in Musin’s paper ”The kissing number in four

dimensions”

Eiichi Bannai and Makoto TagamiGraduate School of Mathematics,

Kyushu University

1 Introduction

The kissing number problem in four dimensions was solved by Oleg Musin[4]. There is a lemma whichplays an important role in the proof of four dimensional kissing number problem, that is to say, a lemmaabout optimal sets on the sphere (Lemma 1 in Musin[4]). But the proof of the lemma in that paperseems to be not clear enough. Therefore, in this manuscript, we shall show the lemma clearly by addingsome conditions (those conditions may be assumed for the Musin’s proof for the four dimensional kissingnumber). Finally we place the table implying the possiblity that we can improve the bounds of minimaldistance φ3(M) by using Musin’s method where we define φ(X) for the spherical codes X to be theminimal distance of distinct elements and we define φn(M) for the positive integer n and M to be themaximal value of φ(X) among codes X with M points on (n− 1)-dimensional sphere.

2 Optimal sets

In this section, we shall review on optimal sets introduced by Musin[4]. Fix z > 0 in an interval [−1, 1]and t0 in an interval [−1,−z) (in Musin’s proof of four dimensional kissing number, we take z = 0.5 andt0 = −0.60794). And we take a polynomial f satisfying the following conditions:

1. f is expanded as f =∑akG

(n)k (x) with a0 > 0 and ai ≥ 0 for any i = 0 where G(n)

k (x) is theGegenbauer polynomial of degree k in n dimensions.

2. f is nonpositive on the interval (t0, z] and f is a monotone decreasing function on the interval[−1, t0].

For a polynomial f like above, we consider a subset Y = y0, y1, . . . , ym satisfying the following condi-tions:

(i) (y0, yi) ∈ [−1, t0] for any i = 0,

(ii) (yi, yj) ∈ [−1, z] for any i = j.

For such a Y , we define

h(Y ) =m∑i=0

f(y0 · yi).

Since Y moves in a compact set and h is a continuous function, h(Y ) can take the maximal value underthe conditions (i) and (ii), we put hm to its value:

hm := maxh(Y ) : |Y | = m+ 1, Y satisfies the above conditions (i) and (ii).Then we define the optimality as follows. This definition follows Musin.

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Definition 2.1 (optimal sets). Let Y be a subset on Sn−1 with m+ 1 points satisfying the conditions(i) and (ii). Then we say that Y is an optimal set if h(Y ) = hm and Y has the maximal number of pairsyi, yj with (yi, yj) = z among subsets W satisfying |W | = m+ 1 and h(W ) = hm.

Below for a Y = y0, y1, . . . , ym satisfying conditions (i) and (ii), we consider a graph Γ = Γ(Y )associated to Y as follows: the vertex set V Γ = y1, . . . , ym, the edge set EΓ = yi, yj| (yi, yj) = z.We write x ∼ y if the x and y are adjacent. Otherwise, we write x ∼ y.

We want to calculate the value of hm in order to obtain an upper bound of z-code. Since f is amonotone decreasing function on the interval [−1, t0], when one of yi’s moves closer to y∗0 = −y0, h(Y )increases. This phenomenon makes us guess that the shape of an optimal set is almost decided. Musinconjectured the following (which is stated as Lemma 2.1 in the paper [4], but it is not still provedcompletely except for (i) in the lemma.):

Conjecture 1 (Musin). Let Y be an optimal set with m+ 1 points. Then,

(1) if m ≤ n, Γ is a complete graph.

(2) if m > n, any vertex of Γ has the degree at least n− 1.

In this proceeding, we shall show the following Lemma.

Lemma 2.1. Let Y be an optimal set with m + 1 points on Sn−1, z = 1/2, and (i) the case n = 3,m ≤ 4, t0 = −0.5907 or (ii) the case n = 4, m ≤ 5, t0 = −0.60794. Then the conjecture is correct.

Considering the cases of Lemma 2.1 are sufficient to complete the proof of the kissing number problemin dimension 4.

3 The proof of Lemma 2.1

In this section, we will show the proof for the case (ii) of Lemma 2.1, since case (i) of Lemma 2.1 issimilarly and easily treated. So we always assume that n = 4 and m ≤ 5 and t0 = −0.60794, unlessotherwise stated. arc(x, y) denotes the arc connecting x and y on the sphere.

Step 1. Suppose that x ∼ y (x, y ∈ V Γ) and m ≥ 3. Then y∗0 /∈arc(x, y).

Proof. Suppose y∗0 ∈arc(x, y). It is shown that if z ∈ V Γ and if (z, x) ≤ 1/2 and (z, y) ≤ 1/2, then(y∗0 , z) is less than 0.60794, a contradiction.

Step 2. For any x ∈ V Γ deg(x) ≥ 2.

Proof. Suppose x is of degree 0, then x is free with respect to the elements in V Γ other than x itself.So, x is moved closer to y∗0 , this contradicts the optimality of V Γ. Suppose x is of degree 1 and x ∼ y.Then since x is free with respect to the elements in V Γ other than x and y and since y∗0 /∈arc(x, y), x ismoved closer to y∗0 , this contradicts the optimality of V Γ.

Step 3. Suppose y1 is of degree 2 and y1 ∼ y2 and y1 ∼ y3. Then y∗0 ∈< y1, y2, y3 >, where< y1, y2, y3 > is the vector space over the real number field R generated by the 3 elements y1, y2, y3.

Proof. A geometric consideration proves the claim. Note that we consider in the 3-dimensional vectorsubspace orthogonal to y2− y3. The y1 moves freely on a circle of a fixed radius. Let x be the componentof y1 in the 2-dimensional subspace on which the circle lies. Then note that y1 is moved closer to y∗0 ifthe component of y∗0 in the 3-dimensional subspace does not lie on the 1-dimemsional subspace spannedby x.

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An Alternative Proof. Without loss of generality, we may assume that

y2 = (1, 0, 0, 0), y3 = (cos θ3, sin θ3, 0, 0), y1 = (1/2, (1 − cos θ3)/2 sin θ3, r cos θ1, r sin θ1).

Then y∗0 /∈< y1.y2, y3 > means that y∗0 = (∗, ∗, r∗ cos θ∗, r∗ sin θ∗) with r∗ = 0 and θ1 ≡ θ∗(mod 2π). Ifwe move θ1 closer to θ, then y1 gets closer to y∗0 , a contradiction to optimality of V Γ. QED.

Note that the above Step 1, Step 2 and Step 3 work not only for n = 4 but also for all n.

In what follows, we want to get a contradiction by assuming that at least one y ∈ V Γ is of degree 2and Y is optimal. Without loss of generality, we may assume that deg(y1) = 2.

Step 4. We first consider the case of

y1 ∼ y2, y1 ∼ y3, y2 ∼ y4, and y1 ∼ y4, y1 ∼ y5.

Then y4 ∼ y5.

Proof. If y4 ∼ y5, then as in the Proof of Step 3, we can move y4 closer to y∗0 , a contradiction.An alternative proof. Geometrically, y1, y4, y5 are in the circle of distance ψ from y2 and y3, and movefreely. So either by moving one of them makes the distance from y∗0 closer or make one of the distancesbetween y1, y4, y5 equal to ψ, a contradiction to the optimality. QED.

Step 5. We consider the case of

y1 ∼ y2, y1 ∼ y3, y2 ∼ y4, y4 ∼ y5and y1 ∼ y4, y1 ∼ y5.

We devide into the following two cases.

(1) y2 ∼ y5, and

(2) y2 ∼ y5.

We need the following Lemma. Here we put φ0 =Arccos(−t0).

Lemma. Let Y = y0, y1, . . . , ym ⊂ S2, (yi, yj) ≤ 1/2(i = j, i ≥ 1, j ≥ 1), (yi, y0) ≥ −t0. Then we getm ≤ 4.

Proof of Lemma. This comes from Lemma 2 in Musin’s paper. That is,

ρ(z, φ0) = 1.36242 · · ·> ϕ2(5) = 2π/5 = 1.256637 · · · .QED.

Suppose that Step 5 (1) holds. Then by Step 2 y3 ∼ y5. We devide into the following three subcases.

Subcase (1-a). y2 ∼ y3 and y3 ∼ y4.

Subcase (1-b). y2 ∼ y3 and y3 ∼ y4 or y2 ∼ y3 and y3 ∼ y4

Subcase (1-c). y2 ∼ y3 and y3 ∼ y4.

Proof of Subcase (1-a).By Step 3, we have y∗0 ∈< y1, y2, y3 > . By Step 1, y∗0 ∈< y1, y2 > and so y3 ∈< y1, y2, y

∗0 > . In

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particular < y1, y2, y3 >=< y1, y2, y∗0 > . In a similar way to the above, since y0∗ ∈< y1, y2, y4 >,

we have y4 ∈< y1, y2, y∗0 >=< y1, y2, y3 > . Similarly,we see that y5 ∈< y1, y2, y3 > . Therefore <

y∗0 , y1, y2, y3, y4, y5 >⊂ S2. By Lemma, we have a contradiction. QED.

Proof of Subcase (1-b). Without loss of generality, we assume that the former case y2 ∼ y3 and y3 ∼ y4holds. Since y∗0 ∈< y1, y2, y3 >, any three elements of y∗0 , y1, y2, y3 form a basis of < y1, y2, y3 > . Sincey∗0 ∈< y2, y4, y5 > and since y∗0 /∈ arc(y4, y5), we have y2 ∈< y∗0 , y4, y5 > . Similarly, since y∗0 ∈<y3, y4, y5 > and since y∗0 /∈ arc(y4, y5), we have y3 ∈< y∗0 , y4, y5 > . Therefore, y1 ∈< y∗0 , y4, y5 >, and soy4, y5 ∈< y1, y2, y3 > . That is, y∗0 , y1, y2, y3, y4, y5 ⊂ S2. By Lemma, we have a contradiction. QED.

The Subcase (1-c) will be treated later, i.e., in Subcase (2-b-2).

Now, we consider Case (2). We divide into the following subcases.

Subcase (2-a) y2 ∼ y3 and

Subcase (2-b) y2 ∼ y3.

Note that, since deg(y3) ≥ 2, Subcase (2-a) is divided into the following two subcases:

Subcase (2-a-1) y3 ∼ y4 and y3 ∼ y5.

Subcase (2-a-2) y3 ∼ y4 and y3 ∼ y5, (or y3 ∼ y4 and y3 ∼ y5.)

Proof of Subcase (2-a-1). As in the proof of Step 4, we can move y4 (or y5) either closer to y∗0 or oneof the distances between y∗0 and y4 and y5 becomes ψ. To be more precise, without loss of generality, wemay assume that

y2 = (1, 0, 0, 0), y3 = (cos θ3, sin θ3, 0, 0), y1 = (1/2, (1 − cos θ3)/2 sin θ3, r, 0).

with r ≥ 0. Then we have that y∗0 = (α, β, γ, 0) with γ ≥ 0, since otherwise y1 can get closer to y∗0 , acontradiction to the optimality. Now if γ > 0, it is possible to move either one of y4 and y5 closer to y1(hence to y∗0) (without reducing the distance between y4 and y5. (Note that y1, y4, y5 are in the circle ofradius r.) If γ = 0, we can move y4 and y5 together without changing the distance between them andwithout changing the distance of both of them to y∗0 ., Hence they can move up to the place where thedistance between one of them and y1 becomes ψ, a contradiction to the optimality again. QED.

Proof of Subcase (2-a-2). This case is equivalent to the Subcase (1-b), and was already taken care of.QED.

Subcase (2-b) is divided into the following three subcases:

Subcase (2-b-1) y3 ∼ y4 and y3 ∼ y5.

Subcase (2-b-2) y3 ∼ y4 and y3 ∼ y5, (or y3 ∼ y4 and y3 ∼ y5.) Note that this case is equivalent toSubcase (1-c).

Subcase (2-b-3) y3 ∼ y4 and y3 ∼ y5.

Proof of Subcase (2-b-1). Similar as in the proof of Subcase (2-a-1). QED.

Proof of Subcase (2-b-2) and Proof of Subcase (2-b-3).

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29

In these two cases, y1.y3, y4, y5 are of distance ψ from y2. Consider the projection of these pointsto the 3 dimensional orthogonal complements to y2. Then these points are on the 2-dimensional sphereS2 of radius

√3/2, which we denote also by y1, y3, y4, y5. Then these points on S2 form either 2 pairs of

edges ((y1, y3) and (y4, y5)) or a line (y1, y3, y5, y4) on S2, according as Subcase (2-b-2) or Subcase (2-b-3)holds. Here the edge means that the distance between them is possible minimum, namely of distanceψ in the original 3 dimensional unit sphere. Also for y∗0 we project it to the 3 dimensional orthogonalcomplements to y2. Then we normalize it so that it is on the sphere S2, and we denote the point by y∗0 .On the sphere S2 of radius

√3/2, if we move x closer to y∗0 , then it certainly approaches y∗0 in the original

3 dimensional unit sphere. Suppose that (y1, y3) and (y4, y5) are edges. Then y∗0 is on both arc(y1, y3)and arc(y4, y5) by the optimality. Then, y1, y4, y3, y5 forms a quadrilateral on S2 such that the diagonalsintersect at y∗0 . However, it is easy to see that there is no such quadrilateral with the length of bothdiagonals and the lengths of all 4 edges as desired, a contradiction. If the latter case of (y1, y3, y5, y4)form a graph of line, then y∗0 must lies on the intersection of arc(y1, y3) and arc(y4, y5), which is againimpossible by considering the quadrilateral y1y5y3y4 on S2. This completes the proof of Proof of Subcase(2-b-2) and Proof of Subcase (2-b-3).QED.

4 Table of the upper bounds for minimal distances of codes on

two-dimensional sphere

In this section, we shall apply the Musin’s method to Tammes’ problem and we shall present the tableshowing the possibility of improvement of upper bounds for the minimal distances (the squared Euclideandistance) of spherical codes with a constant number of points. First we review the notation.

Let X be a finite subset on Sn−1. Then we define

φ(X) := min|x− y|2|x, y ∈ X, x = y.

Also, for a fixed natural number M , we define

φn(M) := maxφ(X)|X ⊂ Sn−1, |X | = M.

In n = 3, the value of φ3(M) are known only for 2 ≤ M ≤ 12 and M = 24. So we want to investigatehow much we can improve the known upper bounds for φ3(M) (13 ≤ M ≤ 23) by Musin’s method. Wecalculated the possibility by using a computer soft Maple. The Algorithm used in the computation followsthe method in Musin[4]. We put the computation on the table below. In the table below, ρCK denotes theminimal distances of the best known codes according to Clare-Kepert[2], and ρL, ρFT and ρBDB denotesthe Levenshtein bounds, Fejes Toth bounds and Boyvalenkov-Danev-Bumova bounds respectively. Werefered to Ericson and Zinoviev[3] for the values of ρCK , ρL, ρFT and ρBDB. The ρM denotes thepossibility of improvement of upper bounds by using Musin’s algorithm to a polynomial of degree 11 andt0 chosen such that µ ≤ 4. To be more precise, we calculated the possibility ρM under the assumptionthat configulations of optimal sets are given by cofigulations respectively that in the case m = 2, y∗0 isat the center of a line, in the case m = 3, y1, y2, y3 is a triangle with a edge length (2 − 2 ∗ z)/2 andy∗0 is at the center of the triangle, and in the case m = 4, y1, y1, y3, y4 is a square with a edge length(2 − 2 ∗ z)/2 and y∗0 is at the center of the square.

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30

M ρCK ρL ρFT ρBDB ρM13 0.9147 1.0213 1.0279 1.0136 0.9914 0.8772 0.9520 0.9604 0.9447 0.9215 0.8148 0.8956 0.9011 0.8866 0.8816 0.7754 0.8494 0.8487 0.8397 0.840717 0.7421 0.8000 0.8020 0.7950 0.799218 0.7025 0.7577 0.7602 0.7529 0.75319 0.6470 0.7216 0.7225 0.7181 0.71620 0.6470 0.6907 0.6883 0.6880 0.68721 0.5996 0.6580 0.6572 0.6550 0.65222 0.5862 0.6286 0.6288 0.6254 0.6223 0.5538 0.6026 0.6028 0.5996 0.594

References

[1] P.G. Boyvalenkov, D.P. Danev and S.P. Bumova, Upper bounds on the minimum distance of sphericalcodes, IEEE Trans. on Inform. Theory, Vol. IT-42, pp. 1576-1581, September 1996.

[2] B.W. Clare and D.L. Kepert, the closest packing of equal circles on a sphere, Proc. R. Soc. Lond.,Vol. 405, pp329-344, 1986.

[3] T. Ericson and V. Zinoviev, Codes on Euclidean Spheres, North-Holland Mathematical Library, 2001.

[4] O. Musin, The kissing number in four dimensions. preprint, arxiv:math.MG/0309430.

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31

Sphere packings in 3-space

Karoly Bezdek∗

December 7, 2004

Abstract

In this paper we survey results on packings of congruent spheres in 3-dimensionalspaces of constant curvature. The topics discussed are as follows:

- Hadwiger numbers of convex bodies and kissing numbers of spheres;- Touching numbers of convex bodies;- Newton numbers of convex bodies;- One-sided Hadwiger and kissing numbers;- Contact graphs of finite packings and the combinatorial Kepler problem;- Isoperimetric problems for Voronoi cells, the strong dodecahedral conjecture and

the truncated octahedral conjecture;- The strong Kepler conjecture.Each topic is discussed in details along with some open problems. Four topics from

the above list are treated in spaces of dimension different from three as well.

1 Introduction

A family of (not necessarily infinitely many) non-overlapping congruent balls in 3-dimensionalspace of constant curvature is called a packing of congruent balls in the given 3-space whichis either Euclidean (E3) or spherical (S3) or hyperbolic (H3). The goal of this paper is tosurvey several of the most recent results on 3-dimensional sphere packings. On the one hand,this area seems to be one of the most active research areas of Discrete Geometry on the otherhand, it is one of the oldest research areas of mathematics ever studied. The topics discussedin separate sections of this paper are the following ones:

- Hadwiger numbers of convex bodies and kissing numbers of spheres;

∗The authors were partially supported by the Hung. Nat. Sci. Found (OTKA), grant no. T043556 andby a Natural Sciences and Engineering Research Council of Canada Discovery Grant.

1

32

- Touching numbers of convex bodies;- Newton numbers of convex bodies;- One-sided Hadwiger and kissing numbers;- Contact graphs of finite packings and the combinatorial Kepler problem;- Isoperimetric problems for Voronoi cells, the strong dodecahedral conjecture and the

truncated octahedral conjecture;- The strong Kepler conjecture.Each section outlines the state of the art of relevant research along with some open

research problems. Also, we feel important to mention that although in this paper emphasesare on the 3-dimensional case, four topics from the above list are treated in all dimensions.Last but not least the paper intends to complement the very recent papers of Casselman [21]and of Pfender and Ziegler [55] on similar topics.

2 Hadwiger numbers of convex bodies and kissing num-

bers of spheres

Let K be a convex body (i.e. a compact convex set with nonempty interior) in d-dimensionalEuclidean space Ed, d ≥ 2. Then the Hadwiger number H(K) of K is the largest number ofnon-overlapping translates of K that can all touch K. An elegant observation of Hadwiger[30] is the following.

Theorem 2.1 For every d-dimensional convex body K,

H(K) ≤ 3d − 1,

where equality holds if and only if K is an affine d-cube.

On the other hand, in another elegent paper Swinnerton-Dyer [60] proved the followinglower bound for Hadwiger numbers of convex bodies.

Theorem 2.2 For every d-dimensional (d ≥ 2)convex body K,

d2 + d ≤ H(K).

Actually, finding a better lower bound for Hadwiger numbers of d-dimensional convexbodies is a highly challanging open problem for all d ≥ 4. (It is not hard to see that the abovetheorem of Swinnerton-Dyer is sharp for dimensions 2 and 3.) The best lower bound knownin dimensions d ≥ 4 is due to Talata [62], who applying Dvoretzky’s theorem on sphericalsections of centrally symmetric convex bodies succeded to show the following inequality.

2

33

Theorem 2.3 There exists an absolute constant c > 0 such that

2cd ≤ H(K)

holds for every positive integer d and for every d-dimensional convex body K.

Now, if we look at convex bodies different from a Euclidean ball in dimensions largerthan 2, then our understanding of their Hadwiger numbers is very limited. Namely, we knowthe Hadwiger numbers of the following convex bodies different from a ball. The result fortetrahedra is due to Talata [63] and the rest was proved by Larman and Zong [41].

Theorem 2.4 The Hadwiger numbers of tetrahedra, octahedra and rhombic dodecahedra areall equal to 18.

In order to gain some more insight on Hadwiger numbers it is natural to pose the followingquestion.

Problem 2.5 For what integers k with 12 ≤ k ≤ 26 does there exist a 3-dimensional convexbody with Hadwiger number k? What is the Hadwiger number of a d−dimensional simplex(resp., crosspolytope) for d ≥ 4?

The second main problem in this section is fondly known as the kissing number problem.The kissing number τd is the maximum number of nonoverlapping d-dimensional balls ofequal size that can touch a congruent one in Ed. In three dimension this question was thesubject of a famous discussion between Isaac Newton and David Gregory in 1964. So, itis not surprising that the literature on the kissing number problem is ”huge”. Perhaps thebest source of information on this problem is the book [22] of Conway and Sloane. In whatfollows we give a short description of the present status of this problem.

τ2 = 6 is trivial. However, determining the value of τ3 is not a trivial issue. Actually thefirst complete and correct proof of τ3 = 12 was given by Schutte and van der Waerden [59]in 1953. The subsequent (two pages) often cited proof of Leech [42], which is impressivelyshort, contrary to the common belief does contain some gaps. It can be completed though,see for example, [47]. Further more recent proofs can be found in [18], [1] and in [53]. Noneof these are short proofs either and one may wonder whether there exists a proof of τ3 = 12in THE BOOK at all. (For more information on this see the very visual paper [21].) Thus,we have the following theorem.

Theorem 2.6 τ2 = 6 and τ3 = 12.

3

34

The race for finding out the kissing numbers of Euclidean balls of dimension larger than3 was always and is even today one of the most visible research projects of mathematics.Following the chronological ordering, here are the major inputs. Coxeter [23] conjectured

and Boroczky [17] proved the following theorem, where Fd(α) = 2dUd!ωd

is the Schlafli function

with U standing for the spherical volume of a regular spherical (d− 1)-dimensional simplexof dihedral angle 2α and with ωd denoting the surface volume of the d-dimensional unit ball.

Theorem 2.7 τd ≤ 2Fd−1(β)

Fd(β), where β = 1

2arcsec d.

It was another breakthrough when Delsarte’s linear programming method (for details seefor example [55]) was applied to the kissing number problem and so, when Kabatiansky andLevenshtein [40] succeded to improve the upper bound of the previous theorem for large das follows. The lower bound mentioned below was found by Wyner [64] several years earlier.

Theorem 2.8 20.2075d(1+o(1)) ≤ τd ≤ 20.401d(1+o(1)).

As the gap between the lower and upper bounds is exponential it was a great surprisewhen Levenshtein [42] and Odlyzko and Sloane [54] independently found the following exactvalues for τd.

Theorem 2.9 τ8 = 240 and τ24 = 196560.

In addition, Bannai and Sloane [2] were able to prove the following.

Theorem 2.10 There is a unique way (up to isometry) of arranging 240 (resp., 196560)nonoverlapping unit spheres in 8-dimensional (resp., 24-dimensional) Euclidean space suchthat they touch another unit sphere.

The latest surprise came when Musin [51], [52] extending Delsarte’s method found thekissing number of 4-dimensional Euclidean balls. Thus, we have

Theorem 2.11 τ4 = 24.

In connection with Musin’s result we believe in the following conjecture.

Conjecture 2.12 There is a unique way (up to isometry) of arranging 24 nonoverlappingunit spheres in 4-dimensional Euclidean space such that they touch another unit sphere.

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35

3 Touching numbers of convex bodies

The touching number t(K) of a convex body K in d-dimensional Euclidean space Ed is thelargest possible number of mutually touching translates of K lying in Ed. In a very elegantpaper Danzer and Grunbaum [24] have shown the following fundamental inequality.

Theorem 3.1 For an arbitrary convex body K of Ed

t(K) ≤ 2d

with equality if and only if K is an affine d-cube.

It is natural to ask for a non-trivial lower bound for t(K). Brass [19] as an applicationof Dvoretzky’s well-known theorem gave a partial answer for the existence of such a lowerbound.

Theorem 3.2 For each k there exists a d(k) such that for any convex body K of Ed withd ≥ d(k)

k ≤ t(K).

It is remarkable that the natural sounding conjecture of Petty [61] stated next is stillopen for all d ≥ 4.

Conjecture 3.3 For each convex body K of Ed, d ≥ 4

d + 1 ≤ t(K).

A generalization of the concept of touching numbers was introduced by K. Bezdek, M.Naszodi and B. Visy [9] as follows. The mth touching number (or the mth Petty number)t(m,K) of a convex body K of Ed is the largest cardinality of (possible overlapping) translatesof K in Ed such that among any m translates always there are two touching ones. Note thatt(2,K) = t(K). The following theorem proved by K. Bezdek, M. Naszodi and B. Visy [9]states some upper bounds for t(m,K).

Theorem 3.4 Let t(K) be an arbitrary convex body in Ed. Then

t(m,K) ≤ min(m− 1)4d,

(2d + m− 1

2d

)holds for all m ≥ 2, d ≥ 2. Also, we have the inequalities

t(3,K) ≤ 2 · 3d, t(m,K) ≤ (m− 1)[(m− 1)3d − (m− 2)]

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36

for all m ≥ 4, d ≥ 2. Moreover, if Bd (resp., Cd) denotes a d-dimensional ball (resp.,d-dimensional affine cube) of Ed, then

t(2,Bd) = d + 1, t(m,Bd) ≤ (m− 1)3d, t(m,Cd) = (m− 1)2d

hold for all m ≥ 2, d ≥ 2.

We cannot resist on raising the following question (for more details see [9]).

Problem 3.5 Prove or disprove that if K is an arbitrary convex body in Ed with d ≥ 2 andm > 2, then

(m− 1)(d + 1) ≤ t(m,K) ≤ (m− 1)2d.

4 Newton numbers of convex bodies

According to L. Fejes Toth [28] the Newton number N(K) of a convex body K in Ed isdefined as the largest number of congruent copies of K that can touch K without havinginterior points in common. (Note that unlike in the case of Hadwiger numbers here it is notnecessary at all to use translated copies of the given convex body in fact, often it is betterto use rotated or reflected ones.) For the special case when K is a ball we refer the reader toSection 2 of this paper. Here we focus on the case when K is different from a ball. Somewhatsurprisingly, in this case only planar results are known. Namely, Linhart [46] and Boroczky[16] determined the Newton numbers of regular convex polygons.

Theorem 4.1 If N(n) denotes the Newton number of a regular convex n-gon in E2, then

N(3) = 12, N(4) = 8 and N(n) = 6 for all n ≥ 5.

L. Fejes Toth [27] proved the following - in some cases quite sharp - upper bound for theNewton numbers of convex domains (i.e. compact convex sets with nonempty interior) inE2.

Theorem 4.2 A convex domain with diameter D and minimum width W cannot be touchedby more than [

(4 + 2π)D

W+ 2 +

W

D

]non-overlapping congruent copies of it.

Applying this theorem to convex domains of constant width we get the upper bound 8for their Newton numbers. This result was improved by Schopp [58] as follows.

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37

Theorem 4.3 The Newton number of any convex domain of constant width in E2 is at most7 and the Newton number of a Reuleaux triangle is exactly 7.

We close this section with a rather natural question, which to the best of our knowledgehas not been yet studied.

Problem 4.4 Prove or disprove that the Newton number of a d-dimensional (d ≥ 3) Eu-clidean cube is 3d − 1.

5 One-sided Hadwiger and kissing numbers

K. Bezdek and P. Brass [10] assigned to each convex body K in Ed a specific positive integercalled the one-sided Hadwiger number h(K) as follows: h(K) is the largest number of non-overlapping translates of K that touch K and that all lie in a closed supporting half-spaceof K. In [10], using the Brunn-Minkowski inequality, K. Bezdek and P. Brass proved thefollowing sharp upper bound for the one-sided Hadwiger numbers of convex bodies.

Theorem 5.1 If K is an arbitrary convex body in Ed, then

h(K) ≤ 2 · 3d−1 − 1.

Moreover, equality is attained if and only if K is a d-dimensional affine cube.

The notion of one-sided Hadwiger numbers was introduced to study the (discrete) geome-try of the so-called k+-neighbour packings, which are packings of translates of a given convexbody in Ed with the property that each packing element is touched by at least k others fromthe packing, where k is a given positive integer. As this area of discrete geometry has arather large literature we refer the interested reader to [10] for a brief survey on the relevantresults. Here, we emphasize the following corollary of the previous theorem proved also in[10].

Theorem 5.2 If K is an arbitrary convex body in Ed, then any k+-neighbour packing bytranslates of K with k ≥ 2 · 3d−1 must have a positive density in Ed. Moreover, there is a(2 · 3d−1 − 1)+-neighbour packing by translates of a d-dimensional affine cube with density 0in Ed.

It is obvious that the one-sided Hadwiger number of any circular disk in E2 is 4. However,the 3-dimensional analogue statement is harder to get. As it turns out the one-sided Hadwigernumber of the 3-dimensional Euclidean ball is 9. One of the shortest proofs of this fact wasfound by A. Bezdek and K. Bezdek [3]. Since here we are studying Euclidean balls theirone-sided Hadwiger numbers we simply call one-sided kissing numbers.

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38

Theorem 5.3 The one-sided kissing number of the 3-dimensional Euclidean ball is 9.

As we have mentioned before Musin [52] has just announced a proof of the long-standingconjecture that the kissing number of the 4-dimensional Euclidean ball is 24. Based on thatK. Bezdek [12] gave a proof of the following.

Theorem 5.4 The one-sided kissing number of the 4-dimensional Euclidean ball is either18 or 19.

The proof of the above theorem supports the following conjecture.

Conjecture 5.5 The one-sided kissing number of the 4-dimensional Euclidean ball is 18.

6 Contact graphs of finite packings and the combina-

torial Kepler problem

Let K be an arbitrary convex body in Ed. Then the contact graph of an arbitrary finitepacking by non-overlapping translates of K in Ed is the (simple) graph whose vertices corre-spond to the packing elements and whose two vertices are connected by an edge if and onlyif the corresponding two packing elements touch each other. One of the most basic questionson contact graphs is to find out the maximum number of edges that a contact graph of ntranslates of the given convex body K can have in Ed. Harborth [36] proved the followingremarkable result on the contact graphs of congruent circular disk packings in E2.

Theorem 6.1 The maximum number of touching pairs in a packing of n congruent circulardisks in E2 is precisely

b3n−√

12n− 3c.

In a very recent paper [20] Brass extended the above result to the ”unit circular diskpackings” of normed planes as follows.

Theorem 6.2 The maximum number of touching pairs in a packing of n translates of a con-vex domain K in E2 is b3n−

√12n− 3c, if K is not a parallelogram, and b4n−

√28n− 12c,

if K is a parallelogram.

The analogue question in the hyperbolic plane has been studied by Bowen in [13]. Weprefer to quote his result in the following geometric way.

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39

Theorem 6.3 Consider circle packings in the hyperbolic plane, by finitely many congruentcircles, which maximize the number of touching pairs for the given number of congruentcircles. Then such a packing must have all of its centers located on the vertices of a trian-gulation of the hyperbolic plane by congruent equilateral triangles, provided the diameter Dof the circles is such that an equilateral triangle in the hyperbolic plane of side length D haseach of its angles is equal to 2π

Nfor some N > 6.

It is not hard to see that one can extend the above result to S2 exactly in the way as theabove phrasing suggests. However, we get a more general approach if we do the following:Take n non-overlapping unit diameter balls in a convex position in E3, that is assume thatthere exists a 3-dimensional convex polyhedron whose vertices are center points moreover,each center point belongs to the boundary of that convex polyhedron, where n ≥ 4 is agiven integer. Obviously, the shortest distance among the center points is at least one. Thencount the unit distances showing up between pairs of center points but, count only thosepairs that generate a unit line segment on the boundary of the given 3-dimensional convexpolyhedron. Finally, maximize this number for the given n and label this maximum byc(n). The following theorem was found by D. Bezdek [4] who also pointed out its interestingrelation to protein folding as well as to Durer’s unsolved geometric problem on edge-unfoldingof convex polyhedra. He calls the convex polyhedra showing up in the theorem below ”higherorder deltahedra” mainly because they form an extension of ”deltahedra” classified earlierby Freudenthal and van der Waerden in [29].

Theorem 6.4 c(n) ≤ 3n − 6, where equality is attained for infinitely many n namely, forthose for which there exists a 3-dimensional convex polyhedron whose each face is an edge-to-edge union of some regular triangles of side length one such that the total number ofgenerating regular triangles on the boundary of the convex polyhedron is precisely 2n−4 witha total number of 3n− 6 sides of length one and with a total number of n vertices.

Now, we are ready to phrase the Combinatorial Kepler Problem. As its namesuggests this problem is strongly related to the Kepler Conjecture on the densest unit spherepackings in E3 (for more details see Section 7 of this paper).

Problem 6.5 For a given n find the largest number K(n) of touching pairs in a packing ofn congruent balls in E3.

This problem is quite open. The first part of the following theorem was proved by D.Bezdek [4] the second part by K. Bezdek [12].

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40

Theorem 6.6

(i) K(4) = 6, K(5) = 9, K(6) = 12 and K(7) = 15.

(ii) K(n) < 6n− 0.59n23 for all n ≥ 4.

We close this section with two upper bounds for the number of touching pairs in anarbitrary finite packing of translates of a convex body, proved by K. Bezdek in [8] . In orderto state these theorems in a short way we need a bit of notation. Let K be an arbitraryconvex body in Ed, d ≥ 3. Then let δ(K) denote the density of a densest packing of translates

of the convex body K in Ed, d ≥ 3. Moreover, let Iq(K) = (Svold−1(bdK))d

(Vold(K))d−1 be the isoperimetric

quotient of the convex body K, where Svold−1(bdK) denotes the (d−1)-dimensional surfacevolume of the boundary bdK of K and Vold(K) denotes the d-dimensional volume of K.Moreover, let B denote the closed d-dimensional ball of radius 1 centered at the origin inEd. Finally, let K0 = 1

2(K+(−K)) be the normalized (centrally symmetric) difference body

assigned to K with H(K0) (resp., h(K0)) standing for the Hadwiger number (resp., one-sidedHadwiger number) of K0.

Theorem 6.7 The number of touching pairs in an arbitrary packing of n > 1 translates ofthe convex body K in Ed, d ≥ 3 is at most

H(K0)

2· n− 1

2d · (δ(K0)(d−1)

d

·( Iq(B)

Iq(K0)

) 1d · n

(d−1)d − (H(K0)− h(K0)− 1).

Theorem 6.8 The number of touching pairs in an arbitrary packing of n > 1 translates ofthe convex body K in Ed, d ≥ 3 is at most

3d − 1

2· n− ω

1dd

2d+1· n

(d−1)d ,

where ωd = πd2

Γ( d2+1)

is the volume of a d-dimensional ball of radius 1 in Ed.

7 Isoperimetric problems for Voronoi cells - the strong

dodecahedral conjecture and the truncated octahe-

dral conjecture

Recall that a family of non-overlapping 3-dimensional balls of radii 1 in Euclidean 3-space,E3 is called a unit ball packing in E3. The density of the packing is the proportion of space

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41

covered by these unit balls. The sphere packing problem asks for the densest packing ofunit balls in E3. The conjecture that the density of any unit ball packing in E3 is at most

π√18

= 0.74078 . . . is often attributed to Kepler that he stated in 1611. The problem of

proving the Kepler conjecture appears as part of Hilbert’s 18th problem [37]. Using aningenious argument which works in any dimension, Rogers [57] obtained the upper bound0.77963 . . . for the density of unit ball packings in E3. This bound has been improved byLindsey [45], and Muder [49], [50] to 0.773055 . . .. Hsiang [38], [39] proposed an elaborateline of attack (along the ideas of L. Fejes Toth suggested 40 years earlier), but his claimthat he settled Kepler’s conjecture seems exaggerated. However, so far no one has foundany serious gap in the approach of Hales [31], [32], [33], [34], although no one has been ableto fully verify it either. This is not too surprising, given that the detailed argument isdescribed in several papers and relies on long computer aided calculations of more than 5000subproblems. Thus, we have the following remarkable theorem.

Theorem 7.1 The densest packing of unit balls in E3 has density π√18

, which is attained bythe ”cannonball packing”.

For several of the above mentioned papers Voronoi cells of unit ball packings play acentral role. Recall that the Voronoi cell of a unit ball in a packing of unit balls in E3 is theset of points that are not farther away from the center of the given ball than from any otherball’s center. As it is well-known, the Voronoi cells of a unit ball packing in E3 form a tilingof E3. One of the most attractive problems on Voronoi cells is the Dodecahedral Conjecturefirst phrased by L. Fejes Toth in [25]. According to this the volume of any Voronoi cell in apacking of unit balls in E3 is at least as large as the volume of a regular dodecahedron withinradius 1. Very recently Hales and McLaughlin [35] announced a solution to this problem:

Theorem 7.2 The volume of any Voronoi cell in a packing of unit balls in E3 is at least aslarge as the volume of a regular dodecahedron with inradius 1.

Now, we can make a step further and take a look of the following stronger version ofthe Dodecahedral Conjecture called the Strong Dodecahedral Conjecture. It was firstarticulated in [7].

Conjecture 7.3 The surface area of any Voronoi cell in a packing with unit balls in E3 isat least as large as 16.6508 . . . the surface area of a regular dodecahedron of inradius 1.

It is easy to see that if true, then the above conjecture implies the Dodecahedral Conjec-ture. The strongest inequality known towards the Strong Dodecahedral Conjecture is due toK. Bezdek and E. Daroczy-Kiss published in [11]. In order to phrase it properly we introduce

11

42

a bit of terminology. A face cone of a Voronoi cell in a packing with unit balls in E3 is theconvex hull of the face chosen and the center of the unit ball sitting in the given Voronoicell. The surface area density of a unit ball in a face cone is simply the spherical area of theregion of the unit sphere (centered at the apex of the face cone) that belongs to the face conedivided by the Euclidean area of the face. It should be clear from these definitions that ifwe have an upper bound for the surface area density in face cones of Voronoi cells, then thereciprocal of this upper bound times 4π (the surface area of a unit ball) is a lower boundfor the surface area of Voronoi cells. Now, we are ready to state the main theorem of [11].

Theorem 7.4 The surface area density of a unit ball in any face cone of a Voronoi cell inan arbitrary packing of unit balls of E3 is at most

−9π + 30 arccos(√

32

sin(

π5

))5 tan

(π5

) = 0.77836 . . . ,

and so the surface area of any Voronoi cell in a packing with unit balls in E3 is at least

20π tan(

π5

)−9π + 30 arccos

(√3

2sin

(π5

)) = 16.1445 . . . .

Moreover, the above upper bound 0.77836 . . . for the surface area density is best possible in thefollowing sense. The surface area density in the face cone of any n-sided face with n = 4, 5of a Voronoi cell in an arbitrary packing of unit balls of E3 is at most

3(2− n)π + 6n · arccos(√

32

sin(

πn

))n tan

(πn

)and equality is achieved when the face is a regular n-gon inscribed in a circle of radius

1√3·cos(π

n)and positioned such that it is tangent to the corresponding unit ball of the packing

at its center.

The Kelvin problem asks for the surface minimizing partition of E3 into cells of equalvolume. According to Lhuilier’s memoir [44] of 1781, the problem has been described as oneof the most difficult in geometry. The solution proposed by Kelvin is a natural generaliza-tion of the hexagonal honeycomb in E2. Take the Voronoi cells of the dual lattice givingthe densest sphere packing. This gives truncated octahedra, the Voronoi cells of the bodycentered cubic lattice. A small deformation of the faces produces a minimal surface, which

12

43

is Kelvin’s proposed solution. Just recently Phelan and Weaire [56] produced a remarkablecounter-example to the Kelvin conjecture. Their work indicates also that Kelvin’s originalquestion is even harder than it was expected. In fact, the following simplier and quite funda-mental question seems to be still open. One can regard this as the isoperimetric inequalityfor parallelohedra and one can call the conjecture below the Truncated Octahedral Con-jecture. (Recall that a parallelohedron is a 3-dimensional convex polyhedron that tiles E3

by translation.)

Conjecture 7.5 The surface area of any parallelohedron of volume 1 in E3 is at least aslarge as the surface area of the truncated octahedral Voronoi cell of the body-centered cubiclattice of volume 1 in E3.

8 The strong Kepler conjecture

In this section we propose a way to extend Kepler’s conjecture to finite packings of congruentballs in 3-space of constant curvature that is in Euclidean 3-space E3, in spherical 3-spaceS3 and in hyperbolic 3-space H3. The idea goes back to the theorems of L. Fejes Toth [26]in E2, J. Molnar [48] in S2 and K. Bezdek [5], [6] in H2 which in short, can be phrased asfollows:

Theorem 8.1 If at least two congruent circular disks are packed in a circular disk in theplane of constant curvature, then the packing density is always less than π√

12.

The hyperbolic case of this theorem proved by K. Bezdek in [5] (see also [6]) seemedquite unexpected because there are (infinite) packings of congruent circular disks in H2 inwhich the density of a circular disk in its respective Voronoi cell is significantly larger than

π√12

. Also, we note that the constant π√12

is best possible in the above theorem. Last wehave to mention that since the standard methods do not give a good definition of densityin H2 (in fact all of them fail to work as it was observed by Boroczky [15]) and since eventoday we know only a rather ”fancy” way of defining density in hyperbolic space (see thework of Bowen and Radin [14]) it seems important to study finite packings in boundedcontainers of the hyperbolic space where there is no complication with the proper definitionof density. All this supports the idea of the following conjecture that we call the StrongKepler Conjecture:

Conjecture 8.2 The density of at least two non-overlapping congruent balls in a ball of the3-space of constant curvature (having radius strictly less than π

2in the case of S3) is always

less than π√18

= 0.74078 . . . .

13

44

The following theorem proved by K. Bezdek [12] supports the above conjecture.

Theorem 8.3 The density of at least two non-overlapping congruent balls in a ball of the3-space of constant curvature (having radius strictly less than π

2in the case of S3) is always

less than Rogers’ upper bound for the density of packings of congruent balls in E3 that is lessthan 0.77963 . . . .

References

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[27] L. Fejes Toth, On the number of equal discs that can touch another of the same kind,Studia Sci. Math. Hungar. 2 (1967), 363–367.

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[31] T.C. Hales, Sphere packings 1, Discrete Comput. Geom. 17 (1997), 1–51.

[32] T.C. Hales, Sphere packings 2, Discrete Comput. Geom. 18 (1997), 135–149.

[33] T.C. Hales, Overview of the Kepler conjecture, Discrete Comput. Geom. (to appear),see arXiv: math. MG/9811071.

[34] T.C. Hales, A proof of the Kepler conjecture, Annals of Math. (to appear).

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[36] H. Harborth, Losung zu Problem 664A, Elem. Math. 29 (1974), 14–15.

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[43] V.I. Levenshtein, On bounds for packings in n-dimensional Euclidean space, Dokl. Akad.Nauk SSSR 245 (1979), 1299–1303.

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[46] J. Linhart, Die Newtonsche Zahl von regelmasigen Funfecken, Periodica Math. Hungar.4 (1973), 315–328.

[47] H. Maehara, Isoperimetric theorem for spherical polygons and the problem of 13 spheres,Ryukyu Math. J. 14 (2001), 41–57.

[48] J. Molnar, Ausfullung und Uberdeckung eines konvexen spharischen Gebietes durchKreise I, Publ. Math. Debrecen 2 (1952), 266–275.

[49] D.J. Muder, Putting the best face on a Voronoi polyhedron, Proc. London Math. Soc.3/56 (1988), 329–348.

[50] D.J. Muder, A new bound on the local density of sphere packings, Discrete Comput.Geom. 10 (1993), 351–375.

[51] O.R. Musin, The problem of the twenty-five spheres, Russian Math. Surveys 58 (2003),794–795.

[52] O.R. Musin, The kissing number in four dimensions, Preprint (2003), 1–22.

[53] O.R. Musin, The kissing number in three dimensions, Preprint (2004), 1–10.

[54] A.M. Odlyzko and N.J.A. Sloane, New bounds on the number of unit spheres that cantouch a unit sphere in n dimensions, J. Combin. Theory Ser. A 26 (1979), 210–214.

[55] F. Pfender and G.M. Ziegler, Kissing numbers, sphere packings and some unexpectedproofs, Notices of the AMS 51/8 (2004), 873–883.

[56] R. Phelan and D. Weaire, A counter-example to Kelvin’s conjecture on minimal surfaces,Philos. Mag. Lett. 69 (1994), 107–110.

[57] C.A. Rogers, The packing of equal spheres, J. London Math. Soc. 3/8 (1958), 609–620.

[58] J. Schopp, Uber die Newtonsche Zahl einer Scheibe konstanter Breite, Studia Sci. Math.Hungar. 5 (1970), 475–478.

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[59] K. Schutte and B.L. van der Waerden, Das Problem der dreizehn Kugeln, Math. Ann.125 (1953), 325–334.

[60] H.P.F. Swinnerton-Dyer, Extremal lattices of convex bodies, Proc. Cambridge Philos.Soc. 49 (1953), 161-162.

[61] C.M. Petty, Equilateral sets in Minkowski spaces, Proc. Amer. Math. Soc. 29 (1971),369-374.

[62] I. Talata, Exponential lower bound for translative kissing numbers of d-dimensionalconvex bodies, Discrete Comput. Geom. 19 (1998), 447–455.

[63] I. Talata, The translative kissing number of tetrahedra is 18, Discrete Comput. Geom.22 (1999), 231–293.

[64] A.D. Wyner, Capabilities of bounded discrepancy decoding, Bell Systems Tech. J. 54(1965), 1061–1122.

Karoly Bezdek, Department of Mathematics and Statistics, University of Calgary, 2500University Drive N.W., Calgary, AB Canada T2N 1N4

E-mail address: bezdek@math.ucalgary.ca

18

49

Some generalizations of the divisor problem forGaussian integers

Elena Deza and Lidiya VarukhinaMoscow State Pedagogical University

November 11, 2004

The problem of calculating the number of integer points in some special domains iswell-known problem in Number Theory and Discrete Mathematics. Classical examples arethe Gauss Circle problem of calculating the number of integer points in a circle and theDirichlet divisor problem of calculating the number of the positive integer points underan hyperbola.

In fact, the number of integer points under the hyperbola x1x2 = x is given by theformula

D2(x) =∑

x1x2≤x

1 =∑n≤x

∑x1x2=n

1 =∑n≤x

∑

d|n1 =

∑n≤x

τ(n),

whereτ(n) =

∑

d|n1 (d, n ∈ N)

is the divisor function, and we come to the problem of the research of asymptotic behaviorof the divisor function on the average.

The classical result in this domain was obtained by P. Dirichlet in 1849 ([Diri49]):

D2(x) =∑n≤x

τ(n) = x log x + (2γ − 1)x + O(√

x),

where γ = 0, 5772157... = limn→+∞(∑N

k=1 k−1 − log N) is the Euler’s constant.The function

τk(n) =∑

n1·...·nk=n

1 (n1, . . . , nk, n ∈ N)

gives the number of representations of the positive integer number n as a product of kpositive integer factors and, as τ(n) = τ2(n), is a generalization of the divisor function.

Some generalizations of Dirichlet’s method allow us to obtain an asymptotic formulafor mean value of τk(n):

Dk(x) =∑n≤x

τk(n) = xPk(log x) + O(x1− 1k logk−2 x),

1

50

where Pk is a polynomial of degree k − 1.The theory of the Riemann zeta-function, especially the estimates of zeta sums by

Vinogradov’s method, allow as to obtain a new asymptotic formula for Dk(x) ([Pant88]):

Dk(x) = xPk(x) + θx1− 1

13k2/3 (C log x)k,

where Pk is a polynomial of degree k−1, θ is a complex number, |θ| ≤ 1, 2 ≤ k << log56 x,

C > 0 - an absolute constant.The same method can be used to obtain new asymptotic formulas for two arithmetic

functions, similar of τk(n) ([Pant94]):

∑n≤x

τk1(n) · . . . · τkl(n) = xPm(log x) + θx

1− 1

13m2/3 (C log x)m,

where l ≥ 1, k1, . . . , kl ≥ 2, m = k1 · . . . · kl, Pm is a polynomial of degree m − 1, θ is a

complex number, |θ| ≤ 1, m << log56 x, C > 0 - an absolute constant;

∑n≤x

τk(n2) = xPm(log x) + θx

1− 1

13m2/3 (C log x)m,

where k ≥ 2, m = k(k+1)2

, Pm is a polynomial of degree m − 1, θ is a complex number,

|θ| ≤ 1, m << log56 x, C > 0 - an absolute constant.

A generalization of the above method gives new results for Dirichlet divisor problemin some number fields ([Pant01]), in particular, in the ring of Gaussian integers

Z[i] = α = a + bi|a, b ∈ Z.

The norm of a Gaussian integer α = a + bi is defined by N(α) = a2 + b2. It iswell-known, that N(α · β) = N(α) · N(β).

The arithmetic function

τGk (n) =

∑

N(α1·...·αk)=n

1 (α1, . . . , αk ∈ Z[i], n ∈ N)

gives the number of representations of positive integer number n as the norm of a productof k Gaussian numbers and, as for positive integer m its norm N(m) = m, is a general-ization of the function τk(n).

We obtained new asymptotic formula for mean value of τGk (n) ([Pant01]):

DGk (x) =

∑n≤x

τGk (n) = xPk(log x) + θx1− 1

15k−2/3

(C log x)2k,

where Pk is a polynomial of degree k−1, θ is a complex number, |θ| ≤ 1, 2 ≤ k << log56 x,

C > 0 - an absolute constant.Now we obtain the similar result in the ring of Gaussian integers for the two arithmetic

functions of general form.

2

51

Theorem 1.

∑n≤x

τGk1

(n) · . . . · τGkl

(n) = xPm(log x) + θx1− 1

15m2/3 (C log x)2m,

where l ≥ 1, k1, . . . , kl ≥ 2, m = k1 · . . . · kl, Pm is a polynomial of degree m − 1, θ is a

complex number, |θ| ≤ 1, m << log56 x, C > 0 - an absolute constant.

Theorem 2.

∑n≤x

τGk (n2) = xPm(log x) + θx

1− 1

15m2/3 (C log x)2m,

where k ≥ 2, m = k(k+1)2

, Pm is a polynomial of degree m − 1, θ is a complex number,

|θ| ≤ 1, m << log56 x, C > 0 - an absolute constant.

Proof.The Dedekind zeta-function of the ring of Gaussian integers has the form

ζZ[i](s) =∑

α∈Z[i]

1

Ns(α)=

∏

π∈Z[i]

(1− 1

Ns(π))−1, Res > 1,

where the product is over all primitive elements π of Gaussian integers.Rational prime number p ∈ P is a primitive element in Z[i], if p ≡ 3(mod4), and is a

product of two primitive elements from Z[i], p = π · π, if p ≡ 1(mod4). In the first caseN(p) = p2, that gives in the product above the factor (1− 1

p2s )−1 = (1− 1

ps )−1(1 + 1

ps )−1.

In the second case N(π) = N(π) = p, and we obtain the factor (1− 1ps )

−1(1− 1ps )

−1. Define

χ(p) = 1 for p ≡ 1(mod4) and χ(p) = −1 for p ≡ 3(mod4). Then

∏

π∈Z[i]

(1− 1

Ns(π))−1 =

∏p∈P

(1− 1

ps)−1

∏p∈P

(1− χ(p)

ps)−1.

As χ(p) is a Dirichlet character modulo 4, we obtain for the Dedekind zeta-functionζZ[i](s) the equality

ζZ[i] = ζ(s)L(s, χ),

where ζ(s) is the Riemann zeta-gunction (ζ(s) =∑+∞

n=11ns , Res > 1) and L(s, χ) is the

Dirtichlet L-function with character χ modulo 4 (L(s, χ) =∑+∞

n=1χ(n)ns , Res > 1).

As for Res > 1

ζkZ[i](s) =

∞∑n=1

1

ns

∑

N(α1·...·αk)=n

1 =∞∑

n=1

τGk (n)

ns,

we should study the behavior of adding functions DGk1...kl

(y) =∑

n≤y τGk1

(n) . . . τGkl

(n) of

the Dirichlet series ζk1...kl(s)Lk1...kl(s, χ) and DGk,2(y) =

∑n≤y τG

k (n2) of the Dirichlet series

ζk(k+1)/2(s)Lk(k+1)/2(s, χ).

3

52

1. Let m = k1 . . . kl. Easy to see ([Marg39]), that

|DGk1...kl

(y)| =∑n≤y

τ2m(n) ≤ y(log y + 2m− 1)2m−1

(2m− 1)!.

For Dirichlet series f(s) =∑+∞

n=1 ann−s (Res > 1) and its adding function Φ(ξ) =∑n≤ξ an, there is ([Kara72]) following integral formula:

∫ x

1

Φ(ξ)dξ =1

2πi

∫ b+iT

b−iT

f(s)xs+1

s(s + 1)ds + R(x).

Let b = 1+ 1log x

, T = x1−η, η = 1−(30m)−2/3. Then for f(s) = ζk1...kl(s)Lk1...kl(s, χ)

and Φ(y) = DGk1...kl

(y) ([Pant88])

R(x) = θx1+η(C log x)2m,

and∫ x

1

DGk1...kl

(y)dy =1

2πi

∫ b+iT

b−iT

ζk1...kl(s)Lk1...kl(s, χ)xs+1

s(s + 1)ds + θx1+η(C log x)2m.

Consider the contour Γ with vertices η ± iT , b± iT and the integral

J =1

2πi

∫

Γ

f(s)xs+1

s(s + 1)ds

on the contour Γ.

By the residue theorem, J = Ress=1f(s)xs+1

s(s+1)= x2Pm(log x). Therefore,

1

2πi

∫ b+iT

b−iT

f(s)xs+1

s(s + 1)ds = J − J1 − J2 − J3,

where J1, J2 and J3 are the integrals on upper, lower, and left sides of Γ.

By the estimate ([Pant88]) of the Riemann zeta-function in critical strip

ζ(σ + it) << |t|15,25(1−σ)3/2

log2/3 |t|, |t| ≥ 2,1

2≤ σ ≤ 1

and similar estimate ([Pant88]) of a Dirichlet L-function with Dirichlet charactermodulo 4

L(σ + it, χ) << |t|15,25(1−σ)3/2

log2/3 |t|, |t| ≥ 2,1

2≤ σ ≤ 1

we obtain, that max|J1|, |J2|, |J3| = θx1+η(C log x)2m, and∫ x

1

DGk1...kl

(y)dy = xPm(log x) + θx1+η(C log x)2m,

4

53

where Pm is some polynomial of degree m−1, |θ| ≤ 1, C > 0 - an absolute constant.

By asymptotic derivation ([Kara72]) we obtain now

DGk1...kl

(x) = xPm(log x) + θx1− 115

m−2/3

(C log x)2m,

where Pm is a polynomial of degree m − 1, θ is a complex number, |θ| ≤ 1, C > 0-an absolute constant.

2. Let m = k(k+1)2

. Then

|DGk,2(y)| ≤

∑n≤y

τ2m(n) ≤ (log y + 2m− 1)2m−1

(2m− 1)!.

Choosing η = 1 − (30m)−2/3, b = 1 − (log x)−1, T = x1−η, we obtain by similarargument, that

∫ x

1

DGk,2(y)dy =

1

2πi

∫ b+iT

b−iT

ζk(k+1)/2(s)Lk(k+1)/2(s, χ)xs+1

s(s + 1)ds + θx1+η(C log x)2m,

∫ x

1

DGk,2(y)dy = Pm(log x) + θx1+η(C log x)2m,

DGk,2(x) = xPm(log x) + θx1− 1

15m−2/3

(C log x)2m,

where Pm is a polynomial of degree m − 1, θ is a complex number, |θ| ≤ 1, C > 0-an absolute constant.

References

[Diri49] Dirichlet P. Uber die Bestimmung der mittleren Werte in der Zahlentheorie, Abh.Acad. Wiss., Berlin, Werke 2, ss. 49–66, 1849.

[Marg39] Mardganishvili K.K. An estimate of one arithmetic sum, Dokl. Acad. of ScienceSSSR, Vol. 22, Nr. 7, 1939.

[Kara72] Karatsuba A.A. Uniform estimate of error term in Dirichlet divisor problem,Izv. Acad. of Science SSSR, Math., Vol. 36, Nr. 3, 1972.

[Pant88] Panteleeva E. Diriclet divisor problem in Number fields, Math. Notes, Vol. 44,Nr. 3, 1988.

[Pant94] On mean values of some arithmetics functions, Math.notes, Vol. 55, Nr. 2, 1994.

[Pant01] Dirichlet divisor problem in the ring of Gaussian integers, Works of MSPU,2001.

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Isometric embeddings of Archimedean Wythoff

polytopes into hypercubes and half-cubes

Michel DEZALIGA, ENS, Paris and Institute of Statistical Mathematics, Tokyo

Mathieu DUTOUR∗

LIGA, ENS, Paris and Hebrew University, Jerusalem

Sergey SHPECTOROV†

Bowling Green State University, Bowling Green

December 9, 2004

Abstract

We study polytopes, obtained by the Wythoff construction from regular

polytopes, and the isometric embeddings of their skeletons or dual skeletons

into the hypercubes Hm and half-cubes 12Hm.

1 Wythoff kaleidoscope construction

A flag in a poset is an arbitrary completely ordered subset. We say that a connectedposet K is a d-dimensional complex (or, simply, a d-complex) if every maximal flagin K has size d + 1. In a d-complex K every element x can be uniquely assigneda number dim(x) ∈ 0, . . . , d, called the dimension of x, in such a way, that theminimal elements of K have dimension zero and dim(y) = dim(x) + 1 wheneverx < y and there is no z with x < z < y.

The elements of a complex K are called faces, or k-faces if the dimension ofthe face needs to be specified. Furthermore, 0-faces are called vertices and d-faces(maximal faces) are called facets. If we reverse the order on K then the resultingposet K∗ is again a d-complex, called the dual complex. Clearly, the vertices of K∗

are the facets of K and, more generally, the dimension function on K∗ is given bydim∗(x) = d− dim(x).

We will often use the customary geometric language. If x < y and dim(x) = k,we will say that x is a k-face of y.

∗Research financed by EC’s IHRP Programme, within the Research Training Network “Alge-

braic Combinatorics in Europe,” grant HPRN-CT-2001-00272.†Research partly supported by an NSA grant.

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A d-complex is a polytope if every submaximal flag (that is, a flag of size d) iscontained in exactly two maximal flags. In the polytopal case, 1-faces are callededges, because each of them has exactly two vertices. Starting from the next sectionwe will deal exclusively with polytopes. The skeleton of a polytope K is the graphformed by all vertices and edges of K.

For a flag F ⊂ K define its type as the set t(F ) = dim(x)|x ∈ F. Clearly, t(F )is a subset of ∆ = 0, . . . , d and, reversely, every subset of ∆ is the type of someflag.

Let Ω be the set of all nonempty subsets of ∆ and fix an arbitrary V ∈ Ω. Fortwo subsets U,U ′ ∈ Ω we say that U ′ blocks U (from V ) if for all u ∈ U and v ∈ Vthere is a u′ ∈ U ′, such that u ≤ u′ ≤ v or u ≥ u′ ≥ v. This defines a binary relationon Ω, which we will denote as U ′ ≤ U . We also write U ′ ∼ U if U ′ ≤ U and U ≤ U ′,and we write U ′ < U if U ′ ≤ U and U 6≤ U ′.

It is easy to see that ≤ is reflexive and transitive, which implies that ∼ is anequivalence relation. Let [U ] denote the equivalence class containing U . It will beconvenient for us to choose canonic representatives in equivalence classes. It can beshown that if U ∼ U ′ then U ∩U ′ ∼ U ∼ U ∪U ′. This yields that every equivalenceclass X contains a unique smallest (under inclusion) subsetm(X) and unique largestsubset M(X). If X = [U ] then m(X) and M(X) can be specified as follows: m(X)is the smallest subset of U that blocks U , while M(X) is the largest subset of ∆that is blocked by U . The subsets m(X) will be called the essential subsets of ∆(with respect to V ). Let E = E(V ) be the set of all essential subsets of ∆. Clearly,the above relation < is a partial order on E. Also, V ∈ E and V is the smallestelement of E with respect to <.

We are now ready to explain theWythoff construction. Naturally, our descriptionis equivalent to the one given in [Cox35] and [Cox73], that generalized the originalpaper [Wyt18]. (See also a relevant paper [Sch90].) Suppose K is a d-complex and let∆, Ω, V , ≤, and E be as above. TheWythoff complex (orWythoffian) K(V ) consistsof all flags F such that t(F ) ∈ E. For two such flags F and F ′, we have F ′ < F

whenever t(F ′) < t(F ) and F ′ is compatible with F (that is, F ∪F ′ is a flag). It canbe shown that K(V ) is again a d-complex and that dim(F ) = d + 1 − |M([t(F )])|.It can also be shown that if K is a d-polytope then K(V ) is again a d-polytope forall V . (For a concrete Euclidean realization of such polytopes, see [HE93].)

Since there are 2d+1 − 1 different subsets V , there are, in general, 2d+1 − 1different Wythoffians constructed from the same complex K. It is easy to see thatK(V ) = K∗(d − V ), where d − V = d − v|v ∈ V . This means that the dualcomplex does not produce new Wythoffians. Furthermore, in the case of self-dualcomplexes (that is, where K ∼= K∗), this reduces the number of potentially pairwise

nonisomorphic Wythoffians to 2d + 2dd−1

2e − 1.

Some of the Wythoffians are, in fact, familiar complexes. First of all, K(0) = Kand K(d) = K∗. Furthermore, K(1) is also known as the median complexMed(K) of K and the dual of K(∆) is known as the order complex of K (see [Sta97]).We will call K(∆) the flag complex of K. Thus, the order complex is the dual of theflag complex.

Since in this paper we are going to deal with the skeletons of K(V ) and K(V )∗

(in the polytopal case), we need to understand elements of K(V ) of types 0, 1, d−1,

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and d. Since V is the unique smallest essential subset, the vertices (0-faces) of K(V )are the flags of type V . For a flag F to be a 1-face of K(V ), U = t(F ) must havethe property that M([U ]) misses just one dimension k from ∆. Clearly, k must bein V . Now U = Uk can be readily computed. Namely, Uk is obtained from V byremoving k and including instead the neighbors of k (that is, k − 1 and/or k + 1).Thus, K(V ) has exactly |V | types of 1-faces. Turning to the facets (d-faces), we seethat, for F to be a facet of K(V ), we need that U = t(F ) be an essential subset ofsize one, such that M([U ]) = U . The latter condition can be restated as follows: Ushould block no other 1-element set. From this we easily obtain that the relevantsets U = k are those for which k = 0 (unless V = 0), k = d (unless V = d),or min(V ) < k < max(V ). Finally, if F is a d − 1-face then U = t(F ) is essentialof size one or two and M([U ]) is of size exactly two. We will not try to make herea general statement about all such subsets U . However, in the concrete situationsbelow, it will be easy to list them all.

2 Archimedean Wythoffians: d = 3, 4

In this section we start looking at particular examples of Wythoffians, namely, at theArchimedean Wythoffians. These polytopes come by the Wythoff construction fromthe regular convex polytopes. A complex (in particular, a polytope) is called regularif its group of symmetries acts transitively on the set of maximal flags. Convexpolytopes are the ones derived from convex hulls H of finite sets of points in Rd.(We assume that the initial set of points contains d + 1 points in general position;equivalently, the interior ofH is nonempty.) The faces of the polytope are the convexintersections of the boundary of H with proper affine subspaces of Rd. In particular,the polytope is d− 1-dimensional, rather than d-dimensional.

It is well-known that the regular convex polytopes of dimension d−1 ≥ 2 fall intothree infinite series: simplices αd, hyperoctahedra βd, and hypercubes γd; and fivesporadic examples: the icosahedron and dodecahedrom for d = 3, and the 24-cell,600-cell, and 120-cell for d = 4.

We are interested in the following

Main Question: Which Archimedean Wythoffians have skeleton graph or dualskeleton graph isometrically embeddable, for a suitable m, in the hypercube graphHm or half-cube graph

12Hm?

Recall ([AsDe80], [GrWi85], [DeSh00], [Shp93], [DeSh96] and books [DeLa97],[DGS04]) that a mapping φ from a graph Γ to a graph Γ′ is an isometric embeddingif dΓ′(φ(u), φ(v)) = dΓ(u, v) for all u, v ∈ Γ. For brevity, we will often shorten “iso-metric embedding” to just “embedding”. Notice that Hm is an isometric subgraphof 1

2H2m, which means that every graph isometrically embeddable in a hypercube is

also embeddable in a half-cube. There is also an intermediate class of graphs—thosethat are embeddable in a Johnson graph J(m,n). Below, when we state our resultson embeddability of the skeleton graphs Γ, we will indicate the smallest class in theabove hierarchy, containing Γ.

We remark that γd is dual to βd, which means that they produce the sameWythoffians. Thus, we can skip the case K = γd altogether. Similarly, we can skip

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the cases where K is the dodecahedron, since the latter is dual to the icosahedron,and the 120-cell, since it is dual to the 600-cell.

In the remainder of this section we state the results of a computer calculationcarried out in the computer algebra system GAP [GAP].

We start with the case d = 3. In this case K is 2-dimensional, that is, K is a map(and so, we switch to the notationM = K). It is easy to see thatM(V ) with V=0,0, 1, 0, 1, 2, 0, 2, 1, 2, 1, and 2 correspond, respectively, to the followingmaps: original map M, truncated M, truncated Med(M), Med(Med(M)), trun-cated M∗, Med(M) and M∗.

In Table 1 we give a complete answer to our Main Question in the case d = 3.The table lists all Archimedean Wythoffians and dual Wythoffians, whose skeletongraph is embeddable. The details of the embedding, such as the dimension of theembedding and whether or not it is equicut, are also provided. Recall that anembedding of a graph Γ is a hypercube is called equicut if each cut on Γ, producedby a coordinate of the hypercube, splits Γ in half. An embedding is called q-balancedif each coordinate cut on Γ has parts of sizes q and |Γ| − q. We will indicate in thetable whether the embedding is equicut, q-balanced, or neither. Finally, for brevity,we truncated in the table the word “truncated” to just “tr”.

Embeddable Wythoffian n embedding equicut?Tetrahedron= α3(0) = α3(2) 4 = J(4, 1); = 1

2H3 q = 1; yes

Octahedron= β3(0) = α3(1) 6 = J(4, 2) yesCube= β3(2) = β3(0)

∗ 8 = H3 yesIcosahedron= Ico(0) 12 1

2H6 yes

Dodecahedron= Ico(2) 20 12H10 yes

tr Cuboctahedron= β3(0, 1, 2) 48 H9 yestr Icosidodecahedron= Ico(0, 1, 2) 120 H15 yesRhombicuboctahedron= β3(0, 2) 24 J(10, 5) yes

Rhombicosidodecahedron= Ico(0, 2) 60 12H16 yes

(tr Tetrahedron)∗ = α3(0, 1)∗ = α3(1, 2)

∗ 8 12H7 no

(tr Icosahedron)∗ = Ico(0, 1)∗ 32 12H10 yes

(tr Cube)∗ = β3(1, 2)∗ 14 J(12, 6) no

(tr Dodecahedron)∗ = Ico(1, 2)∗ 32 12H26 no

(Cuboctahedron)∗ = β3(1)∗ = α3(0, 2)

∗ 14 H4 yes(Icosidodecahedron)∗ = Ico(1)∗ 32 H6 yes

tr Octahedron= β3(0, 1) = α3(0, 1, 2) 24 H6 yes

Table 1: Embeddable Archimedean Wythoffians for d = 3

A striking property of this table is that it contains all possible Wythoffians (allfive regular polytopes and 11 of the 13 Archimedean polytopes; missing are the SnubCube and Snub Dodecahedron, which are not Wythoffian). Furthermore, for eachof these polytopes, exactly one of the skeleton and the dual skeleton is embeddable.

This nice picture does not extend to the case d = 4, where far fewer embeddingsexist. Our Table 2 gives a complete answer to the Main Question.

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Embeddable Wythoffian n embedding equicut?α4 = α4(0) = α4(3) 5 = J(5, 1) q = 1γ4 = β4(3) = β4(0)

∗ 16 = H4 yesβ4 = β4(0) 8 = 1

2H4 yes

α4(0, 1, 2, 3) 120 H10 yesβ4(0, 1, 2, 3) 384 H16 yes

24− cell(0, 1, 2, 3) 1152 H20 yesβ4(0, 1, 2) = 24− cell(0, 1) = 24− cell(2, 3) 192 H12 yes

α4(0, 3)∗ 30 H5 yes

β4(0, 3) 64 12H12 yes

α4(1) = α4(2) 10 = J(5, 2) q = 4600− cell(0, 1, 2, 3) 14400 H60 yes

Table 2: Embeddable Archimedean Wythoffians for d = 4

Notice that the total number of Archimedean Wythoffians for d = 4 is 45. (Thisfact was stated in Conway [Con67] without a proof or a complete list.) Thus, Table2 indicates that the embeddable cases become more rare as d grows, and that, likely,there are only finitely many infinite series of embeddings. Furthermore, Tables 1and 2 lead us to a number of concrete conjectures about possible infinite seriesof embeddings. In the next section we resolve those conjectures in affirmative byconstructing the series and verifying the embedding properties.

We conclude this section with some further remarks about the embeddings inTables 1 and 2. The majority of these embeddings are unique. The only exceptionis the Tetrahedron α3, whose skeleton, the complete graph K4, has two isometricembeddings. We also checked that all the skeleton graphs for d = 3, that turn out tobe non-embeddable, violate, moreover, the so-called 5-gonal inequality (see [DGS04],[DeLa97]).

3 Infinite series of embeddings

Since in this section we are only interested in the infinite series of embeddings, werestrict ourselves to the cases K = αd and βd. These polytopes can be described incombinatorial terms as follows: The faces of αd are all proper nonempty subsets ofthe set 1, . . . , d+ 1. The order on αd is given by containment, and the dimensionof the face X is |X| − 1. Clearly, αd has

(d+1k+1

)faces of dimension k. The faces of

βd are the sets ±i1, . . . ,±ik, where the signs are arbitrary and i1, . . . , ik is anonempty subset of 1, . . . , d. Again, the order is defined by containment and thedimension of a face X is |X| − 1. Thus, βn has 2k+1

(d

k+1

)faces of dimension k.

Two infinite series of embeddable skeletons are well-known:

(1) The skeleton of αd(0) = αd(d − 1) is the complete graph Kd+1, whichcoincides with J(d+ 1, 1).

(2) The skeleton of βd(d − 1) = βd(0)∗ = γd coincides with the hypercube

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graph Hd.

The first of these embeddings can be generalized as follows.

Proposition 1 If k ∈ 0, . . . , d − 1 then the skeleton of αd(k) coincides withJ(d+ 1, k + 1).

Proof. We refer to the discussion of the vertices and edges at the end of Section 1.According to that discussion, the vertices of αk are the k-faces, that is, the subsetsof 1, . . . , d + 1 of size k + 1. Furthermore, the only types, leading to edges, arek − 1 (if k > 1) and k + 1. This means that two vertices are on an edge if and onlyif their symmetric difference, as sets, has size two. 2

The above result explains a number of entries in Tables 1 and 2. We now turnto the series showing up in line 14 of Table 1 and in line 8 of Table 2.

Proposition 2 The skeleton of αd(0, d − 1)∗ coincides with Hd+1 with two an-tipodal vertices removed. It is an isometric subgraph of Hd+1.

Proof. Again we refer to the discussion in Section 1. When V = 0, d− 1, everyone-element subset U of ∆ = 0, . . . , d − 1 has the property that M([U ]) = U .This means that all elements of αd(V ) (that is, all nonempty proper subsets of1, . . . , d+1) are vertices of αd(V )∗. This also means that the types correspondingto edges necessarily have size two. If U = a, b ⊂ ∆ and a < b then M([U ]) =k|a ≤ k ≤ b. Therefore, edges of αd(V )∗ are flags, whose type is of the formk, k + 1. Thus, we come to the following description of the skeleton Γ of αd(V )∗:Its vertices are all nonempty proper subsets of 1, . . . , d+1; two subsets are adjacentwhen one of them lies in the other and their sizes differ by one. This matches thewell-known definition of Hd+1 as the graph on the set of all subsets of 1, . . . , d+1.In fact, Γ is Hd+1 with two antipodal vertices (∅ and the entire 1, . . . , d + 1)removed. The last claim is clear. 2

We remark that the dual Wythoff polytope in this proposition is, in fact, thezonotopal Voronoi polytope of the root lattice Ad.

The following series of embeddings is “responsible” for line 16 of Table 1 and line4 of Table 2.

Proposition 3 The skeleton of αd(0, . . . , d−1) isometrically embeds into H(d+1

2 ).

Proof. Here we find that the vertices of the skeleton Γ are the maximal flags andthe edges are the submaximal flags. Hence, the vertices can be identified with the(d+ 1)! tuples (x1, . . . , xd+1), which are permutations of (1, . . . , d + 1). Two tuplesare adjacent if one is obtained from the other by switching two consecutive entries.

Let Θ be the set of all two-element subsets of 1, . . . , d+1 and let the hypercubeHm, where m =

(d+12

), be defined with Θ as its base set. (That is, the vertices of

Hm are all subsets of Θ.) We now define a mapping from Γ into Hm. For a tupleX and a two-element subset T = i, j ∈ Θ, i < j, we say that T is an inversion inX if j precedes i in X. We map the tuple X to the subset φ(X) of Θ consisting ofall inversions from X. When we interchange two consecutive entries in a tuple, we

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either add an inversion, or remove an inversion. This means that adjacent verticesof Γ are mapped to adjacent vertices of Hm. It remains to see that the distancein Γ between two vertices, say X and Y , coincides with the Hamming distancebetween φ(X) and φ(Y ). In fact, it suffices to show that the distance between X

and Y is no greater than the Hamming distance between φ(X) and φ(Y ). SupposeX = (x1, . . . , xd+1) and Y = (y1 . . . , yd+1). Let k be the first position where X andY disagree and let t be defined by xt = yk. Then, clearly, k < t. Consider the tupleX ′ obtained from X by switching the consecutive numbers xt−1 and xt. Then X

′ isadjacent to X and the Hamming distance between φ(X ′) and φ(X) is one less thanthe Hamming distance between φ(X) and φ(Y ). Now the claim follows by inductionon the Hamming distance. 2

Notice that the polytope αd(0, . . . , d − 1) is known as the permutahedron. Itis the zonotopal Voronoi polytope of the dual root lattice A∗

d.We now turn to the case K = βd. The following result, which is, in a sense,

similar to Proposition 3, originates from line 6 of Table 1 and line 5 of Table 2.

Proposition 4 The skeleton of βd(0, . . . , d− 1) isometrically embeds into Hd2.

Proof. Again, the vertices of the skeleton Γ of βd(0, . . . , d− 1) are the maximalflags, while the edges are the submaximal flags. Thus, the vertices can be identifiedwith the 2dd! tuples (ε1x1, . . . , εdxd), where the sign factors εi = ±1 are arbitraryand (x1, . . . , xd) is a permutation of 1, . . . , d. Two such tuples are adjacent ifone is obtained from the other by one of the two operations: (1) interchanging twoconsecutive entries; or (2) changing the sign of the last entry of the tuple.

As in the proof of Proposition 3, we introduce the concept of an inversionin a tuple X. Furthermore, to each possible inversion we will attach its type,which will be an ordered pair of (possibly equal) numbers from 1, . . . , d. Theinversions in a tuple X = (ε1x1, . . . , εdxd) come from the deviations of the “dou-bled” tuple D(X) = (ε1x1, . . . , εdxd,−εdxd, . . . ,−ε1x1) from the normal order N =(1, . . . , d,−d, . . . ,−1). Notice that there is a symmetry in the location in D(X) ofthe deviations from the normal order. Namely, if i < j and the i-th and j-th entriesof D(X) appear in N in the reverse order then also the same is true for the entriesin positions 2d+1− j and 2d+1− i. We will view two symmetric deviations as oneinversion. According to this convention, we obtain three kinds of inversions. First,there may be self-symmetric deviations. Such a deviation arises when j = 2d+1− i.Since i < j, we have that i ≤ d. Furthermore, the i-th entry is εixi and the j-thentry is −εixi. Thus, we have a deviation simply when εi = −1. We let such aninversion have type (xi, xi). Secondly, there may be deviations in D(X) where thepositions i and j are in the same half of D(X). Because of our symmetry conven-tion, we may assume that i, j ≤ d. Then this deviation and the one symmetric to ittogether define one inversion, and the type of that inversion will be (i, j). Thirdly,there may be (non-self-symmetric) deviations with i and j in the different halves ofD(X). Then, clearly, i ≤ d. Furthermore, because of the symmetry, we may assumethat i < j ′ = 2d + 1 − j. Such a deviation, together with the deviation symmetricto it, defines again one inversion, and the corresponding type will be (2d+ 1− j, i).

Let us now see how the inversions can be determined from X alone. The inver-sions of the first kind are easy to find—they correspond to the negative entries in

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X. Let 1 ≤ i < j ≤ d. There are eight possibilities depending on the signs of εi andεj, and on whether or not xi < xj. If εi = 1 = εj then there are no inversions for iand j if xi < xj, and there is an inversion of type (i, j) if xi > xj. If −εi = 1 = εj

then, in addition to the inversion of the first kind, (xi, xi), there are two inversionsof types (i, j) and (j, i) if xi < xj, and one inversion of type (i, j), otherwise. Ifεi = 1 = −εj then, in addition to the inversion of the first kind, (xj, xj), there isno further inversion if xi < xj, and one additional inversion of type (j, i), otherwise.Finally, if −εi = 1 = −εj then, in addition to the inversions of the first kind, (xi, xi)and (xj, xj), there are two inversions of types (i, j) and (j, i) if xi < xj, and oneinversion of type (j, i), otherwise.

Clearly, there are exactly d2 types. Let Θ be the set of types (the Cartesiansquare of 1, . . . , d). Let m = d2 and let our copy of Hm be defined with Θ as itsbase set. For a tuple X let φ(X) be the set of types of the inversions found in X.Since two different inversions in X never have the same type, we have that |φ(X)|is the number of inversions in X. Each time we interchange two consecutive entriesor change the sign of the last entry in X, we add or remove one inversion. Thismeans that φ maps adjacent vertices of the skeleton Γ to adjacent vertices of Γm. Itremains to show, as in the proof of the preceding proposition, that the distance in Γbetween X and Y does not exceed the Hamming distance between φ(X) and φ(Y ).

Let X = (ε1x1, . . . , εdxd) and Y = (δ1y1, . . . , δdyd). Let k be the first positionwhere X and Y disagree and let t be defined by xt = yk. Define a tuple X ′ asfollows. If the signs δk and εt agree then X ′ is obtained from X by interchangingthe consecutive entries εt−1xt−1 and εtxt. If the signs disagree and t 6= d then X ′ isobtained by interchanging the consecutive entries εtxt and εt+1xt+1. Finally, if thesigns disagree and t = d then X ′ is obtained from X by changing the sign of thelast entry. In each case, it is easy to check that X ′ is adjacent to X and that theHamming distance between φ(X ′) and φ(Y ) is one less than the Hamming distancebetween φ(X) and φ(Y ). So the claim follows by induction. 2

We remark that βd(0, . . . , d − 1) is not a Voronoi polytope of a lattice, sinceits number of vertices, 2dd! is greater than (d + 1)!. It is easy to check inductivelythat βd(0, . . . , d − 1 and all its faces are centrally symmetric, which means thatβd(0, . . . , d− 1) is a zonotope.

It is interesting to note that Propositions 3 and 4 together with line 7 of Table 1and lines 6 and 10 of Table 2 show that the skeleton of K(0, . . . , d−1 isometricallyembeds in a hypercube for every regular convex polytope K.

There is an easy connection between the embeddings from Propositions 3 and4. Namely, in the situation of Proposition 4, a tuple (ε1x1, . . . , εdxd) has inversionssolely of type (i, j) with i < j if and only if all signs in it are pluses. Thus theskeleton graph of αd−1 is a convex subgraph of the skeleton of βd defined by thecondition that φ(X) is contained in the specific subhypercube H(d

2)of Hd2 .

However, there is also another possibility. Instead of taking the intersection ofthe embedded isometric subgraph with a specific subhypercube, we can take theprojection. Suppose Γ is isometrically embedded in a hypercube Hm with the baseset Θ (so that |Θ| = m) and suppose Θ0 ⊂ Θ. Set m0 = |Θ0|. The projectionof Γ onto the subhypercube Hm0

, defined by Θ0, is the subgraph induced by allintersections of Θ0 with the vertices of Γ (which are subsets of Θ). The following

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fact is well-known.

Lemma 1 The projection of an isometric subgraph of Hm onto an arbitrary subhy-percube Hm0

is an isometric subgraph of Hm0. 2

As a consequence of this lemma, we can derive a new series of embeddings startingfrom the series from Proposition 4. This new series leaves trace in Tables 1 and 2in lines 16 and 7, respectively.

Proposition 5 The skeleton of βd(0, . . . , d−2) isometrically embeds into Hd(d−1).

Proof. From the discussion in Section 1 we derive the following description of theabove skeleton. The vertices (flags of type 0, . . . , d − 2) can be viewed as tuplesX = (ε1x1, . . . , εd−1xd−1, xd), where (x1, . . . , xd) is a permutation of 1, . . . , d andthe last entry in X does not carry a sign (rather than has a plus as its sign).The edges are the pairs X,Y , where Y is obtained from X by interchanging twoconsecutive entries. If we interchange the entries in positions d− 1 and d, then theold sign εd−1 is discarded and the new sign in position d− 1 is chosen arbitrarily.

It is clear from this description that our new skeleton graph is obtained fromthe skeleton from Proposition 4 by factorizing over the equivalence relation comingfrom ignoring the last sign in the tuple.

Let now Θ be the Cartesian square of 1, . . . , d, that is, Θ is the base setof the hypercube Hd2 from Proposition 4. Let Θ0 consists of all (i, j) with i 6=j. Then |Θ0| = d(d − 1). Let ψ be the composition of the embedding φ fromProposition 4 and the projection to the subhypercube defined by Θ0. To prove ourproposition, it suffices to show that ψ is exactly the factorization from the precedingparagraph. In other terms, it suffices to show that tuples X = (ε1x1, . . . , εdxd) andY = (δ1y1, . . . , δdyd) have the same set of inversions of the second and third kinds(that is, of types other than (i, i)) if and only if they agree in all but maybe the lastsign. Observe that the symmetric difference of φ(X) and φ(Y ) is fully contained inthe set of inversions of the first kind. Since φ is an isometric embedding every edgeon the shortest path from X to Y comes from switching the sign of the last entry.The claim now follows. 2

For d = 3 the polytope βd(0, . . . , d − 2 is the zonotopal Voronoi polytope ofthe lattice A∗

3. For d = 4 and higher it is not a Voronoi polytope of a lattice, sincethe number of vertices n = 2d−1d! > (d + 1)!. However, it can be checked directlythat βd(0, . . . , d− 2) is a zonotope for all d.

The examples in lines 8 and 9 of Tables 1 and 2, respectively, suggest that theskeleton of βd(0, d − 1) might be embeddable in a half-cube for all d. The nextproposition demonstrates that the actual situation is somewhat more complicated.We first need to recall some further concepts.

Suppose Γ is a graph and φ is a mapping from Γ to a hypercube Hm. Wesay that φ is an embedding with scale λ if for all vertices x, y ∈ Γ we have thatthe distance in Hm between φ(x) and φ(y) (the Hamming distance) coincides withλdΓ(x, y). Clearly, isometric embeddings in a hypercube are scale 1 embeddings,while isometric embeddings in a half-cube are scale 2 embeddings. A graph is an`1-graph if it has a scale λ embedding into a hypercube for some λ. In fact, due

9

63

to [AsDe80], any finite rational-valued metric embeds isometrically into some spacelk1 if and only if this metric is scale λ embeddable into Hm for some λ and m. See[AsDe80], [Shp93], [DeSh96], [DeLa97], [DGS04] for details on l1-embedding.

Proposition 6 The skeleton of βd(0, d− 1) is an `1-graph for all d. However, ifd > 4, it is not an isometric subgraph of a half-cube.

Proof. Let Γ be the above skeleton graph. The vertices of this graph can beidentified with all tuples of the form (k; ε1, . . . , εd), where 1 ≤ k ≤ d and the signsεi are arbitrary. Thus, there are d2d vertices. The edges of Γ arise in two ways: (1)We have that (i; ε1, . . . , εd) is adjacent to (j; ε1, . . . , εd) for all i 6= j. (2) We alsohave that (i; ε1, . . . , εd) is adjacent to (i; ε1, . . . , εj−1,−εj, εj+1 . . . , εd), again for alli 6= j.

Let Γ1 be the graph whose vertices are all tuples (ε1, . . . , εd), εi = ±1, andwhere two tuples are adjacent whenever they differ in just one entry. Let Γ2 be thegraph whose vertices are ±k, 1 ≤ k ≤ d, and where vertices s and t are adjacentwhenever |s| 6= |t|. It is clear that Γ1 is isomorphic to the hypercube Hd, while Γ2 isisomorphic to the hyperoctahedron graph Kd×2 (complete multipartite graph withd parts of size two; also known as the cocktail-party graph). Mapping the vertex(k; ε1, . . . , εd) of Γ to the ordered pair ((ε1, . . . , εd), εkk) defines an embedding φ ofΓ into the direct product graph Γ1 × Γ2. It is easy to see that this embedding isisometric. Since Γ projects surjectively onto both Γ1 and Γ2, we can now determinethe Graham-Winkler direct product graph for Γ (cf. [GrWi85]). Namely, that directproduct graph has d complete graphs of size two and the cocktail-party graph Γ2 asits factors. (We assume that d > 2.) It follows from [Shp93] that Γ has a scale λembedding in a hypercube if and only if every factor has. In our case, every factoris an `1-graph, hence Γ is an `1-graph, too. Furthermore, the cocktail-party graphKd×2 with d > 4 requires λ > 2, which proves the second claim of the proposition.2

The infinite series exhibited in this section explain a majority of the examplesfrom Tables 1 and 2, including, in fact, all examples from Table 2. This allows usto make the following conjecture.

Conjecture 1 If Γ is the skeleton of the Wythoffian K(V ) or of the dual WythoffianK(V )∗, where K is a regular convex polytope, and Γ is isometrically embeddable ina half-cube then Γ can be found either in Table 1, Table 2, or in one of the infiniteseries discussed in this section.

4 From the hypercubes to the cubic lattices?

The above conjecture shows one direction of possible further research. Anotherpossibility is extending the results of this paper to cover the case of infinite regularpolytopes, that is, regular partitions of the Euclidean and hyperbolic space. In thissection we briefly discuss what is known about the easier Euclidean case, and putforward some conjectures.

In the infinite case, instead of embedding the skeleton graphs up to scale intohypercubes Hm, we embed them into the m-dimensional cubic lattice Zm (including

10

64

m =∞) taken with its metric `1. Notice that this is a true generalization, because,due to [AsDe80], a finite metric that can be embedded into a cubic lattice, can alsobe embedded into a hypercube.

All regular partitions of Euclidean d-space (d finite) are known [Cox73]. Theyconsist of one infinite series δd = δ∗d, which is the partition into the regular d-dimensional cubes, two 2-dimensional ones, (36) (partition into regular triangles)and (63) = (36)∗ (partition into regular 6-gons), and two 4-dimensional ones, hδ4(partition into 4-dimensional hyperoctahedra) and hδ∗4 (partition 24-cells). Noticethat the latter two partitions are the Delaunay and Voronoi partitions associatedwith the lattice D4. In particular, below we use the notation V o(D4) in place of hδ∗4.

In the following table we give a complete list of Wythoffians of regular partitionsof the Euclidean plane. We use the classical notation for the vertex-transitive par-tition of the Euclidean plane; namely, each partition is identified by its type, listingclock-wise the gonalities of the faces containing a fixed vertex. In particular, theregular partitions of the Euclidean plane are (44) = δ2, (3

6), and (63) = (36)∗. Inthe second column we indicate the embedding. We put Zm for an embedding withscale one and 1

2Zm for an embedding with scale two.

Wythoffian embeddingδ2 = δ2(0) = δ2(1) = δ2(2) = δ2(0, 2) Z2

(36) = (36)(0) 12Z3

(63) = (36)(2) = (36)(0, 1) Z3(4.82) = δ2(0, 1) = δ2(1, 2 = δ2(0, 1, 2 Z4

(4.6.12) = (36)(0, 1, 2) Z6(3.4.6.4) = (36)(0, 2) 1

2Z3

(3.6.3.6)∗ = (36)(1)∗ Z3(3.122)∗ = (36)(1, 2)∗ 1

2Z∞

Table 3: Embeddable Wythoffian cases for plane partitions.

All Archimedean Withoffians or their dual, which are not mentioned in Table 3,are nonembeddable and, moreover, they do not satisfy the 5-gonal inequality.

In this table we separated the three regular plane partitions from theArchimedean(i.e., vertex- but not face-transitive) ones. Notice that, out of the eight Archimedeanpartitions, five are Wythoffians. Missing are partitions (32.4.3.4), (33.42) and (34.6).It turns out that for all regular and Archimedean plane partitions (and in particular,for all our Wythoffians) exactly one out of itself and its dual is embeddable. In thisrespect the situation here repeats the situation for the Archimedean polyhedra ford = 3, see Section 2 and Tables 9.1 and 4.1–4.2 in [DGS04].

We now turn to the next dimension, d = 3. Here we identify the Wythoffiansas particular partitions of the Euclidean 3-space in two ways. First, in column 2we give the number of that partition in the list of 28 regular and Archimedeanpartitions of the 3-space from [DGS04]. Secondly, we identify in column 3 the tilesof the partition. Here, as before, β3 and γ3 are the Octahedron and the Cube,respectively. Also, “Cbt” stands for the Cuboctahedron and “Rcbt” stands for

11

65

the Rhombicuboctahedron. Clearly, “tr” stands for “truncated” and Prism8 is theregular 8-gonal prism. In some cases we also indicate the chemical names of thecorresponding partitions. In column 4 we give the details of the embedding. If theparticular Wythoffian is nonembeddable, we put “non 5-gonal” in that column toindicate that it fails the 5-gonality inequality. The information in column 4 is takenfrom Table 10.1 from [DGS04].

Wythoffian no tiles embbeddingδ3 = δ3(0) = δ3(3) = δ3(0, 3) 1 γ3 Z3

δ3(1, 2) = V o(A∗3) 2 tr β3 Z6

δ3(0, 1, 2) = δ3(1, 2, 3)=zeolit Linde A 16 γ3, tr β3, tr Cbt Z9δ3(0, 1, 2, 3)=zeolit ρ 9 Prism8, tr Cbt Z9

δ3(1) = δ3(2) = De(J − complex) 8 β3, Cbt non 5-gonalδ3(0, 1) = δ3(2, 3)=boride CaB6 7 β3, tr γ3 non 5-gonal

δ3(0, 2) = δ3(1, 3) 18 γ3, Cbt, Rcbt non 5-gonalδ3(0, 1, 3) = δ3(0, 2, 3)=selenide Pd17Se15 23 γ3, Prism8, tr γ3, Rbct non 5-gonal

Table 4: Wythoffians of regular partitions of the 3-space.

As Table 4 indicates, only eight out of 28 regular and Archimedean partitions ofthe 3-space arise as the Wythoffians of the cubic partition δ3.

Finally, in Table 5 we collected some information about the dimensions d ≥ 4.

Wythoffian tiles embbeddingδd = δd(0) = δd(d) = δd(0, d) γd Zd

δd(0, 1)=tr δd βd, tr γd non 5-gonalV o(D4) = V o(D4)(0 24− cell non 5-gonalV o(D4)

∗ = V o(D4)(4) β4 non 5-gonalV o(D4)(1 =Med(V o(D4)) γ4, Med(24− cell) non 5-gonalV o(D4)(0, 1=tr V o(D4) γ4, tr 24− cell Z12

Table 5: Some Wythoffians of regular partitions of the d-space, d ≥ 4.

Notice that again, as in Table 2, few Wythoffians for d = 3 possess embeddings.This gives hope that there is only a small number of infinite series of embeddings inthe Euclidean case. One of the infinite series is shown in line 1 of Table 5. Basedon examples in lines 4 of Tables 3 and 4, we put forward the following conjecture.

Conjecture 2 The skeleton graph of the Wythoffian flag partition δd(0, . . . , d) isisometrically embeddable in Zd2.

Line 4 of Table 3 and line 3 of Table 4 suggest that the following may be true.

Conjecture 3 The skeleton graph of the Wythoffian δd(0, . . . , d−1) = δd(1, . . . , d)is also isometrically embeddable in Zd2.

12

66

By way of speculation, there may be a connection between these two conjectures,just like there is a connection between Theorems 4 and 5. One simple connectionbetween the data and conjectures from this section and the results from the preced-ing section is that the tiles of the embeddable partition must be embeddable, too.One can check that the tiles of the above two partitions are closely related to thepolyhedra from Theorems 4 and 5.

It appears (see line 4 of Table 3 and line 2 of Table 4) that δd(1, 2) may havean embedding for all d. In this case, however, we are reluctant to formulate anexact conjecture. Perhaps, the situation will be more clear when the case d = 4 iscompleted.

We have already pointed out that only eight out of 28 regular and Archimedeanpartitions of the Euclidean 3-space are Withoffians of δ3. This indicates that, maybe,we need to derive Wythoffians from a larger class of Euclidean partitions. Theobvious candidates are the Delaunay and Voronoi partitions of interesting Euclideanlattices, in particular, the root lattices. For the case of such lattices themselves (thatis, for V = 0 or d), see Chapter 11 of [DGS04]. The zonotopal embeddings ofV o(Ad) in Zd+1 and V o(A∗

d) in Z(d+1

2 ), shown in there, correspond to the zonotopal

embeddings of the corresponding tiles in Theorems 2 and 3.It is, of course, also very interesting to consider the Wythoffians of the regular

partitions of the hyperbolic d-space. In fact, in [DeSh00] (see also Chapter 3 of[DGS04]), the embeddability was decided for any regular tiling P of the d-sphere,Euclidean d-space, hyperbolic d-space or Coxeter’s regular hyperbolic honeycomb(with infinite or star-shaped cells or vertex figures). The large program will be togeneralize it for all Wythoffians of such general P .

If P is a regular d-polytope having a Coxeter group of automorphisms, then theskeleton of its order complex is the Cayley graph of this group on the Coxeter set ofits generators. This raises the question of considering the Wythoff construction forthe case of Coxeter groups.

The regular polygon Pn has the Coxeter group I2(n) as symmetry group. Its ordercomplex is a regular polygon P2n. A regular 2m-gon is a 2-dimensional zonotope;its skeleton ia an isometric subgraph of Hm.

So, the skeletons of order complexes of all regular polytopes are hypercube em-beddable (moreover, are skeletons of zonotopes). Hence, the natural question is:do the Cayley graph of any finite Coxeter group embeds in Hm? This concerns theremaining cases E6, E7, E8 and the serie Dn, which we expect to embed into Hn(n−1).

Note that in [BKLM04], the skeletons of order complexes of Platonic polyhedraare called cubic inflation. Their hypecube embeddings are obtained in a straightfor-ward way. Also in [NiRe03] an embedding of finitely generated Coxeter group intocube complexes is given. But this is a topological embedding, while our embeddingsare isometric.

References

[AsDe80] P. Assouad and M. Deza, Espaces metriques plongeables dans un hyper-cube: aspects combinatoires, Annals of Discrete Mathematics 8 (1980)

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197–210.

[BKLM04] B. Bresar, S. Klavzar, A. Lipovec and B. Mohar, Cubic inflations, mirrorgraphs, regular maps, and partial cubes, European Journal of Combina-torics 25 (2004) 55–64.

[Con67] J.H. Conway, Four-dimensional Archimedean polytopes, Proc. Collo-quium on Convexity, Copenhagen 1965, Kobenhavns Univ. Mat. Institut(1967) 38–39.

[Cox35] H.S.M. Coxeter, Wythoff’s construction for uniform polytopes, Proc. ofthe London Math. Society (2), 38 (1935) 327–339; reprinted in H.S.M.Coxeter, Twelve geometrical essays, Carbondale, Illinois, 1968.

[Cox73] H.S.M. Coxeter, Regular polytopes, Dover Publications, New York, 1973.

[DeDu04] M. Deza and M. Dutour, Zigzag structure of complexes, submitted to Bul-letin of South-East Asean Mathematical Society, Special Issue in memoryof K. Beidar. http://www.arxiv.org/abs/math.CO/0405279

[DGS04] M. Deza, V. Grishukhin, and M. Shtogrin, Scale-Isometric PolytopalGraphs in Hypercubes and Cubic Lattices, Imperial College Press andWorld Scientific, 2004.

[DeLa97] M. Deza and M. Laurent, Geometry of Cuts and Metrics, Springer–Verlag, Berlin, 1997.

[DeSh00] M.Deza and M.Shtrogin, Embedding the graphs of regular tilings andstar honeycombs into the graphs of cubical lattices, in Advanced Studyin Pure Mathematics 27, Arrangements - Tokyo (2000) 73–92.

[DeSh96] M.Deza and S.Shpectorov, Recognition of l1-graphs with complexityO(nm), or football in a hypercube, European Journal of Combinatorics17-2,3 (1996) 279–289.

[GAP] The GAP Group, GAP – Groups, Algorithms, and Programming, Version4.4; 2004 (http://www.gap-system.org).

[GrWi85] R.L. Graham and P.M. Winkler, On isometric embeddings of graphs,Trans. Amer. Math. Soc. 288(2) (1985) 527–536.

[HE93] Z. Har’El, Uniform solutions for uniform polyhedra, Geom. Dedicata 47

(1993), 57–119.

[NiRe03] G.A. Niblo, L.D. Reeves, Coxeter groups act on CAT(0) cube complexes,J. Group Theory 6-3 (2003) 399–413.

[Sch90] R. Scharlau, Geometrical realizations of shadow geometries, Proc. LondonMath. Soc. (3) 61 (1990) 615–656.

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[Shp93] S. Shpectorov, On scale embeddings of graphs into hypercubes, EuropeanJournal of Combinatorics, 14 (1993) 117–130.

[Sta97] R. Stanley, Enumerative combinatorics, Cambridge University Press,Cambridge, 1997.

[Wyt18] W.A. Wythoff, A relation between the polytopes of the C600-family,Koninklijke Akademie van Wetenschappen te Amsterdam, Proceedingsof the Section of Sciences, 20 (1918) 966–970.

15

69

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79

FINITE AND INFINITE LATTICE PACKINGS

MARTIN HENK

Abstract. In the first part we survey some results on finite lattice packingswith respect to the parameterised density. In particular, we show that theparameterised densities of densest finite lattice sphere packings tend to thedensity of a densest lattice sphere packing if the parameter ρ is not less

thanp

21/4. This result is based on a kind of algorithm which refines anypacking lattice such the outcome is a dense packing lattice as well as thincovering lattice. In the last part of the paper we study this algorithm andgive some computational results.

1. Introduction

Let Kn be the set of all convex bodies in the n-dimensional Euclidean space

Rn with non-empty interior, i.e., int(K) 6= ∅. A set C ⊂ Rn is called a packing

set of K ∈ Kn if for x1 6= x2 ∈ C,

int(x1 + K) ∩ int(x2 + K) = ∅.

The family of all packing sets of K is denoted P(K). In the case #C < ∞ the

density of the finite packing C + K is given by the quotient

#C · vol(K)

vol(conv C + K),

where conv denotes the convex hull and vol(·) denotes the volume (n-dimen-

sional Lebesgue measure). This notion of density of a finite packing was gener-

alised by the concept of parameterised densities introduced in [BHW94]. The

parametrised density of C + K w.r.t. the parameter ρ > 0 is defined by

δ(ρ, C, K) =#C · vol(K)

vol(conv C + ρ · K),

and for ρ ∈ R>0, m ∈ N, the density of a densest finite packing of K of

cardinality m w.r.t. the parameter ρ is given by

δ(ρ, m, K) = sup δ(ρ, C, K) : C ∈ P(K), #C = m .

Moreover we define

δ(ρ, K) = lim supm→∞

δ(ρ, m, K)

as the limit density of densest finite densities w.r.t. ρ. It is not hard to see that

(cf. [BHW94])

δ(ρ, K) ≥ δ(K) for all ρ > 0,

where δ(K) denotes the “classical” density of a densest (infinite) packing of

K. By looking at the denominator vol(convC + ρ K) it is also not hard to see1

80

2 MARTIN HENK

that for large parameters ρ, optimal finite packing sets C are full-dimensional.

Therefore we call

ρc(K) = infρ ∈ R>0 : δ(ρ, K) = δ(K)

the critical parameter of K. In [BHW94] it was shown that

δ(K) ≥1

n[ρc(K)]

1−n .

Hence any upper bound on ρc(K) leads to a lower bound on δ(K) and in

[BHW94] the following bounds were proven:

ρc(K) ≤

2, if K = −K,

n + 1, else.

Although these bounds give only the trivial lower bounds on δ(K) (which are

still the best known bounds) it shows that parameterised densities of finite

packings might also be useful for infinite packings. Here we want to study

similar relations for finite lattice packings. For more information on parame-

terised densities and packing problems we refer to [Bor04], [GW93], [FTK93],

[FT97], [Rog64], [Zon99] and [GL87], [EGH89]. In particular, we also recom-

mend the last two books for some basic facts on lattices which will be used in

the following.

2. Finite lattice packings

Let Ln be the set of all lattices Λ ⊂ Rn with detΛ 6= 0. Λ ∈ Ln is called a

packing lattice if Λ ∈ P(K). The set of all lattice packing sets of K ∈ Kn is

denoted by P∗(K), i.e.,

P∗(K) = C ∈ P(K) : ∃Λ ∈ Ln ∩ P(K) with C ⊂ Λ .

In analogy with the non-lattice case we define

δ∗(ρ, m, K) = sup δ(ρ, C, K) : C ∈ P∗(K), #C = m ,

δ∗(ρ, K) = lim supm→∞

δ∗(ρ, m, K),

ρ∗c(K) = infρ ∈ R>0 : δ∗(ρ, K) = δ∗(K),

where δ∗(K) denotes the density of a densest lattice packing of K, i.e.,

δ∗(K) = sup

vol(K)

det Λ: Λ ∈ Ln ∩ P(K)

.

Proposition 2.1. Let K ∈ Kn. Then

i) δ∗(ρ, K) ≥ δ∗(K) for all ρ > 0.

ii) δ∗(K) ≥ 1n[ρ∗c(K)]1−n.

iii) For the n-dimensional unit ball Bn we have

δ∗(Bn) ≥κn

2κn−1[ρ∗c(K)]1−n,

where κj denotes the j-dimensional volume of the j-dimensional unit

ball Bj .

81

FINITE AND INFINITE LATTICE PACKINGS 3

Proof. All the statements can be proven completely analogously to the non-

lattice case and therefore we just refer to [BHW94]. ¤So again we see that any upper bound on the critical lattice parameter ρ∗c(K)

yields a lower bound on δ∗(K). For 0-symmetric convex bodies K, i.e., K ∈ Kn

with K = −K, the classical Minkowski-Hlawka Theorem gives the following

the lower bound (cf. e.g. [GL87])

(2.1) δ∗(K) ≥ ζ(n) 21−n, n ≥ 2.

Here ζ(n) =∑∞

i=1(1/i)n denotes the ζ-function. The Minkowski-Hlawka bound

is still the best known lower bound and even for the n-dimensional ball no

substantial better lower bounds are known (see [Bal92]).

Unfortunately, we are not aware of a proof showing that also in the case of

lattice packings the critical lattice parameter of a 0-symmetric convex body is

at most 2. Maybe this also too much to hope for because by Proposition 2.1 ii)

we would get a lower bound on δ∗(K) of the same order of magnitude as the

Minkowksi-Hlawka bound, but via finite packings. Here we will show

Theorem 2.2. Let m ∈ N and K ∈ Kn. Then

δ∗(ρ, m, K) ≤ δ∗(K) for ρ ≥

3, if K = −K,√21/4, if K = Bn,

(3/2)(n + 1), else.

By Proposition 2.1 i) Theorem 2.2 implies

Corollary 2.3. Let K ∈ Kn. Then

ρ∗c(K) ≤

3, if K = −K,√21/4, if K = Bn,

(3/2)(n + 1), else.

The proof of Theorem 2.2 is based on the following lemma about lattices

which are dense packing and thin covering lattices simultaneously. To this end

we denote by Kn0 the family of all 0-symmetric convex bodies, i.e., all K ∈ Kn

with K = −K.

Lemma 2.4. Let K ∈ Kn0 and let Λ ∈ P∗(K). Then there exists a packing

lattice Λ ∈ P∗(K) with Λ ⊂ Λ and

Λ + 3 · K = Rn.

If K = Bn then the dilation factor 3 can be replaced by√

21/4.

In order to prove Lemma 2.4, as well as the Theorem 2.2, we need a well

known functional from Geometry of Numbers. For K ∈ Kn0 and Λ ∈ Ln

µ(K, Λ) = min µ ∈ R>0 : Λ + µ K = Rn

is called the covering radius (or inhomogeneous minimum) of Λ with respect to

K. In other words, µ(K, Λ) is the smallest positive number µ such that Λ+µ K

is a covering. This covering property immediately implies that

vol(µ(K, Λ) · K) ≥ det Λ

82

4 MARTIN HENK

and the next proposition generalises this fact to finitely many lattice translates

of µ(K, Λ) ·K.

Proposition 2.5. Let K ∈ Kn0 . Let Λ ∈ Ln and let A ⊂ Λ with #A < ∞.

Then

vol(A + µ(K, Λ) · K) ≥ #A · det Λ.

Proof. Let T be an elementary cell (half-open parallelepiped generated by a

basis) of our lattice and for abbreviation we write µ instead of µ(K, Λ). Since

Λ + µ K is a covering we can find lattice points b1, . . . , bm ∈ Λ such that

T ⊂ ∪mi=1(bi + µ K).

For each i let Si = T ∩ (bi + µ K). By the definition of an elementary cell we

conclude that for b 6= c ∈ Λ and 1 ≤ i 6= j ≤ m

(Si + b) ∩ (Sj + c) = ∅.

Otherwise, we have b − c ∈ Sj − Si ⊂ T − T and since the origin is the only

lattice point contained in T − T we obtain the contradiction b = c.

Hence with S = ∪mi=1(Si − bi) we have S ⊂ µ K, vol(S) = vol(T ) = det Λ and

(S + b)∩ (S + c) = ∅, for b 6= c ∈ Λ. Thus we finally get

vol(A + µ(K, Λ) · K) ≥ vol(A + S) = #A vol(S) = #A det Λ.

¤The proof of Theorem 2.2 is an easy consequence of Proposition 2.5 and Lemma

2.4.

Proof of Theorem 2.2. First let us assume that K ∈ Kn0 . Let m ∈ N, ρ > 0,

and let C∗ ∈ P∗(K), #C∗ = m such that

δ∗(ρ, m, K) = δ(ρ, C∗, K) =m vol(K)

vol(conv C∗ + ρK).

Furthermore, let Λ ∈ Ln ∩ P∗(K) with C∗ ⊂ Λ. Next we observe that every

packing lattice Λ ∈ Ln ∩P∗(K) with Λ ⊂ Λ contains a finite lattice packing C∗

whose parametric density is at least as large as δ(ρ, C∗, K). Therefore, in view

of Lemma 2.4 and by the definition of µ(K, Λ) we may assume

µ(K, Λ) <

3, K ∈ Kn

0 ,√21/4, K = Bn .

Hence with Proposition 2.5 we get for ρ ≥ µ(K, Λ)

(2.2) δ∗(ρ, m, K) =m vol(K)

vol(convC∗ + ρK)≤

m vol(K)

m detΛ≤ δ∗(K).

Next let K ∈ Kn be an arbitrary convex body and let DK = 12(K − K) ∈ Kn

0

be its difference body. It is well known and easy to see that C ∈ P∗(K) ⇔ C ∈P∗(DK). In particular we have

δ∗(K) =vol(K)

vol(DK)δ∗(DK).

83

FINITE AND INFINITE LATTICE PACKINGS 5

Now let ρ ≥ (3/2)(n + 1) and let C∗, Λ as above with respect to the body K.

Since DK ⊂ t + n+12 K for a suitable vector t ∈ Rn (see [BF34]) we get with

(2.2)

δ∗(ρ, m, K) =m vol(K)

vol(conv C∗ + ρK)≤

m vol(K)

vol(conv C∗ + ρ 2n+1DK)

≤vol(K)

vol(DK)δ∗(ρ 2/(n + 1), m, DK) ≤

vol(K)

vol(DK)δ∗(DK) = δ∗(K).

¤

In remains to prove Lemma 2.4 which will be done in the next section.

3. Lattice refinements

A well known problem in Geometry of Numbers is to find lattices which are

simultaneously dense packing and thin covering lattices of a given 0-symmetric

convex body (cf. [GL87], [EGH89]). In terms of the covering radius µ(K, Λ) we

can formulate the problem as follow:

Simultaneous packing and covering problem. For K ∈ Kn0 find a packing lattice

Λ ∈ Ln ∩ P∗(K) such that µ(K, Λ) is minimal.

Here we are dealing with the following restricted version of this problem:

Lattice refinement problem. For K ∈ Kn0 and a packing lattice Λ ∈ Ln ∩P∗(K)

find a packing lattice Λ ∈ Ln ∩P∗(K) such that Λ ⊂ Λ and µ(K, Λ) is minimal.

In other words, starting with an arbitrary packing lattice Λ of a centrally

symmetric convex body we are looking among all packing lattices containing Λ

for one whose covering radius is minimal. The existence of such a “minimal”

lattice is guaranteed by the selection theorem of Mahler (cf. [GL87]). In view

of the Minkowski-Hlawka bound (2.1) and since

δ∗(K) ≥vol(K)

det Λ≥ [µ(K, Λ)]−n

for any Λ ∈ Ln ∩P∗(K), we can not expect to solve one of these problems with

a covering radius strictly less than 2. In 1972, however, Butler [But72] proved

that there exists a packing lattice Λ ∈ Ln ∩ P∗(K) such hat

µ(K, Λ) = 2 + o(1), n → ∞.

The proof of Butler is based on a mean value argument and therefore his result

can not be applied to the refinement problem. For recent results on the si-

multaneous packing and covering problem we refer to Zong [Zon02b], [Zon02a]

and [Zon03]. But Rogers [Rog50] proved in 1950 that the ratio of the den-

sity of a thinnest lattice covering to the density of a densest lattice packing of

a 0-symmetric convex body is at most 3n−1 (see also [Ban90]), and his proof

immediately gives us the bound 3 in Lemma 2.4. In order to prove the bound√21/4 for the unit ball Bn we just need a slight extension of Rogers’ proof.

84

6 MARTIN HENK

First we note the that covering radius is the maximum distance between a

point in Rn and the lattice Λ. Here the distance is measured with respect to

the distance function fK of K ∈ Kn0 , i.e.,

(3.1) µ(K, Λ) = maxx∈Rn

min fK(x − a) : a ∈ Λ ,

where fK(x) = minµ ∈ R≥0 : x ∈ µ K.For the proof of Lemma 2.4 in the case K = Bn we will also need the following

simple proposition, where ‖ · ‖ denotes the Euclidean norm.

Proposition 3.1. Let u, v ∈ Rn, α1, α2, α3, β ∈ R≥0, β < 1 such that

i) ‖u + β v‖ ≥ α1, ii) ‖u + v‖ ≤ α2 and iii) ‖v‖ ≥ α3. Then

‖u‖2 ≥α21 − β α22 + α23(β − β2)

1 − β.

Proof. Taking the squares of the assumptions i) and ii) gives

‖u‖2 + 2β u · v + β2‖v‖2 ≥ α21,

‖u‖2 + 2u · v + ‖v‖2 ≤ α22.

Multiplying the last inequality by −β and then taking the sum of the two

inequalities leads, on account of assumption iii), to the desired lower bound on

‖u‖2. ¤

Proof of Lemma 2.4. First we consider the general case K ∈ Kn0 . Let Λ ∈

Ln ∩ P∗(K) and let Λ ∈ Ln ∩ P∗(K) be a lattice containing Λ and having

minimal determinant. We claim that

(3.2) µ(K, Λ) < 3.

Suppose the contrary, i.e., µ(K, Λ) ≥ 3. By (3.1) there exists a point y ∈ Rn

such that

(3.3) fK(y − a) ≥ µ(K, Λ), for all a ∈ Λ.

On the other hand, since µ(K, 13 Λ) = 13µ(K, Λ) we can also find a point b ∈ Λ

such that

(3.4) fK(y −1

3b) ≤

1

3µ(K, Λ).

Let Λ′ = Λ + Z13b be the lattice generated by Λ and 1

3b. Next we show that Λ′

is a still packing lattice of K, i.e., fK(a′) ≥ 2 for all a′ ∈ Λ′. Let a′ ∈ Λ′ \ Λ.

Then there exists an a ∈ Λ such that a′ = a − 13b and from (3.3) and (3.4) we

get

fK(a′) = fK(a−1

3b) ≥ fK(a − y) − fK(y −

1

3b) ≥

2

3µ(K, Λ) ≥ 2.

Hence we conclude that fK(a′) ≥ 2 for all a ∈ Λ′ \ 0 and therefore, Λ′ ∈

P?(K). Of course, Λ ⊂ Λ′ and since 13b /∈ Λ we also know det Λ′ = 1

3 det Λ. So

we get a contraditction to the choice of Λ.

85

FINITE AND INFINITE LATTICE PACKINGS 7

In order to establish the bound√

21/4 for K = Bn we just explore some

properties of the Euclidean norm. Again let Λ ∈ Ln ∩ P∗(Bn) be a lattice

containing Λ and having minimum determinant. Suppose that

(3.5) µ(Bn, Λ) >

√21

4

and for abbrevaition we write µ instead of µ(Bn, Λ). As in the previous case

let y ∈ Rn such that

(3.6) ‖y − a‖ ≥ µ, for all a ∈ Λ.

Now we consider two possible constructions. First assume that there exists a

b ∈ Λ such that

(3.7) ‖y −1

2b‖ ≤

√5

21µ.

Let Λ′ = Λ + Z12b. Then we have Λ ⊂ Λ′, det Λ′ = 1

2 det Λ, and next we show

that Λ′ is still a packing lattice of Bn . To this end it suffices to verify

(3.8) ‖1

2b − a‖ ≥ 2, for all a ∈ Λ.

With u = a − 12b, v = y − a, β = 1

2 we get from (3.6) and (3.7)

i) ‖u + βv‖ ≥µ

2= α1, ii) ‖u + v‖ ≤

√5

21µ = α2

and by (3.6) we also have iii) ‖v‖ ≥ µ(= α3). Applying Proposition 3.1

w.r.t. these values and taking into account (3.5) we get

‖a −1

2b‖2 = ‖u‖2 ≥ µ2 −

5

21µ2 ≥ 4.

Hence (3.8) is verified, i.e., Λ′ is a packing lattice of Bn. Since it contains Λ and

since it has a smaller determinant than Λ we get a contradiction to the choice

of Λ. Thus we have to assume that there does not exist a b ∈ Λ satisfying (3.7),

which implies that

(3.9) ‖2

3y −

1

3a‖ =

2

3‖y −

1

2a‖ ≥

2

3

√5

21µ, for all a ∈ Λ.

Now let b ∈ Λ such that

(3.10) ‖y −1

3b‖ ≤

1

3µ.

Again we set Λ′ = Λ + Z13b and we want to show that Λ′ is a packing lattice of

Bn. This time we set u = a − 13b, v = a − y, β = 2

3 , and from (3.9) and (3.10)

we get

i) ‖u + βv‖ ≥2

3

√5

21µ = α1, ii) ‖u + v‖ ≤

√1

3µ = α2

Together with iii) ‖v‖ ≥ µ(= α3) we may apply Proposition 3.1 and with (3.5)

we get

‖1

3b− a‖2 = ‖u‖2 ≥

4

9µ2

(15

21+ 1

)≥ 4.

86

8 MARTIN HENK

Hence Λ′ is a packing lattice of Bn and as before we end up with a contradiction.

Thus the assumption (3.5) is wrong and the Lemma 2.4 is proven. ¤

4. An algorithmic version and some computational results

The proof of Lemma 2.4 in the previous section suggests the following algo-

rithm in order to find a “good” packing and covering lattice. We will present

the algorithm only for the n-dimensional unit ball Bn.

Input: Λ ∈ Ln ∩ P∗(Bn).

Output: Λ ∈ Ln ∩ P∗(Bn) with Λ ⊂ Λ and µ(Bn , Λ) <√

21/4.

(1) Let Λ = Λ.

(2) Find y ∈ Rn with ‖y‖ = µ(Bn, Λ).

(3) If ‖y‖ <√

21/4 Stop.

(4) Find b ∈ Λ such thati) ‖y − 1

2b‖ ≤ 5

21‖y‖ or ii) ‖y − 1

3b‖ ≤ 1

3‖y‖.

(5) In case i) replace Λ with Λ + 1

2bZ and in case ii) with Λ + 1

3bZ.

(6) Goto (2).

The existence of a b as in (4) as well as the correctness of the algorithm is

guaranteed by the proof of Lemma 2.4. Of course, in this form the algorithm is

only applicable in small dimensions since in each step we have to find at least

one deep hole of the lattice Λ, i.e., a point y ∈ Rn with ‖y‖ = µ(Bn , Λ). In

other words we have to find a vertex of the so called Dirichlet-Voronoıcell of Λ,

which is a hard algorithmic problem in higher dimensions. Therefore we use a

more heuristic version of the algorithm above. To this end we denote by

V (b1, . . . , bn) =

n∑

i=1

zi bi : zi ∈ 0, 1

the vertices of the parallelepiped generated by n linearly points b1, . . . , bn ∈ Rn.

Algorithm 1

Input: Λ ∈ Ln ∩ P∗(Bn) with basis b1, . . . , bn.

Output: Λ ∈ Ln ∩ P∗(Bn) with Λ ⊂ Λ.

(1) Let Λ = Λ.(2) For each vertex v ∈ V ( 1

2b1, . . . ,

1

2bn) do

If Λ + vZ ∈ Ln ∩P∗(Bn) then

replace Λ with Λ + vZ and replace a suitable vector bi with v

such that b1, . . . , bi−1, v, bi+1, . . . , bn is a basis of Λ.Goto (2).

End If.End Do.

So in this variant we hope that the vertices of the parallelepiped generated by12b1, . . . ,

12bn are close to the deep holes of the lattice Λ. However there is no

guarantee that this really happens and thus we cannot say anything about the

quality of the final lattice Λ. Nevertheless we have implemented this algorithm

where we use the following rule for testing the 2n vertices of V (b1, . . . , bn) in

step (2). For k = 0, . . . , 2n − 1 let vk =∑n

i=0 zk,ibi where (zk,1, . . . , zk,n) is

87

FINITE AND INFINITE LATTICE PACKINGS 9

the binary representation of k, i.e., k =∑n−1

i=0 zk,i 2i, zk,i ∈ 0, 1. If Λ + vkZis packing lattice, which can be verified by any algorithm for computing the

shortest non-trivial vector in a lattice, then we replace that bj with vk which is

determined by zj,k = 1 but zl,k = 0 for all l > j.

In dimension 24 we have started Algorithm 1 with the integer lattice Λ = 2Z24

and after 24 rounds it stops with a lattice Λ whose basis vectors are the columns

of the following matrix:

2 0 0 1 0 1 1 0 0 1 1 0 1 0 07

2

5

2

3

2

3

26

3

27 6

11

2

0 2 0 1 0 1 0 1 0 1 0 1 0 1 07

2

3

2

5

2

3

26

3

26

11

27

0 0 2 1 0 0 1 1 0 0 1 1 0 0 17

2

3

2

3

2

5

26

3

2

11

27 6

0 0 0 1 0 0 0 0 0 0 0 0 0 0 01

2

1

2

1

2

1

21

1

2

3

2

3

2

3

2

0 0 0 0 2 1 1 1 0 0 0 0 1 1 17

2

3

2

3

2

3

2

11

2

5

27 7 7

0 0 0 0 0 1 0 0 0 0 0 0 0 0 01

2

1

2

1

2

1

2

3

2

1

21

3

2

3

2

0 0 0 0 0 0 1 0 0 0 0 0 0 0 01

2

1

2

1

2

1

2

3

2

1

2

3

21

3

2

0 0 0 0 0 0 0 1 0 0 0 0 0 0 01

2

1

2

1

2

1

2

3

2

1

2

3

2

3

21

0 0 0 0 0 0 0 0 2 1 1 1 1 1 17

21 1 1

7

21

7

2

7

2

7

2

0 0 0 0 0 0 0 0 0 1 0 0 0 0 01

20 0 0

1

20

1

20 0

0 0 0 0 0 0 0 0 0 0 1 0 0 0 01

20 0 0

1

20 0

1

20

0 0 0 0 0 0 0 0 0 0 0 1 0 0 01

20 0 0

1

20 0 0

1

2

0 0 0 0 0 0 0 0 0 0 0 0 1 0 01

20 0 0 0 0

1

2

1

20

0 0 0 0 0 0 0 0 0 0 0 0 0 1 01

20 0 0 0 0

1

20

1

2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 11

20 0 0 0 0 0

1

2

1

2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 01

20 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 01

20

1

2

1

20

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 01

20

1

20

1

2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 11

20 0

1

2

1

2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01

20 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 11

2

1

2

1

2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01

20 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01

20

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01

2

So we have det Λ24 = 1 and indeed Λ is the Leech lattice which is known as a

densest lattice packing in dimension 24 [CS93].

Taking the first k columns as a basis of a k-dimensional packing lattice Λk

we find for k ∈ 2, 4, 7, 8, 11, 12, 13, 16, 17, 20, 23

k 1 4 7 8 11 12 13 16 17 20 23

det Λk 2 8 16 16 32 32 32 16 16 8 2

The corresponding densities δ∗(Λk) coincide with the densities of the densest

known laminated lattices (see [CS93]), which are for k ∈ 4, 7, 8, 16, 17, 20, 23are the densest known lattice packings. On account of the deterministic char-

acter of Algorithm 1 the lattices Λk are also the lattices generated by the

algorithm in dimension k.

In each successful step of Algorithm 1 the determinant is halved. Thus

starting with a lattice Λ, Algorithm 1 will only create lattices of determinant

det Λ 2−m, m ∈ N. Hence in order to find e.g. the densest packing lattices D3

or D5 in dimension 3 or 5 we have to start with a different lattice.

88

10 MARTIN HENK

For n ≥ 3 let Λ = 2 Zn−3 ⊕ 2√

2Z3. Then Algorithm 1 produces the

densest known packing lattices in dimensions 3, 5, 9, 15, 19, 21 which are again

laminated lattices (see [CS93]).

n 3 5 9 15 19 21

det Λ 4√

2 8√

2 16√

2 16√

2 8√

2 4√

2.

As an example we just give the basis of the lattice Λ found by the algorithm in

dimension 21:

2 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1

2− 1

2

1

2− 1

20 0

0 2 0 1 0 1 0 1 0 1 0 1 0 1 0 1

2− 1

2

1

2− 1

20 0

0 0 2 1 0 0 1 1 0 0 1 1 0 0 11

2− 1

2− 1

2

1

20

1

2

0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 −1

2

1

2

1

2

1

2−1

1

2

0 0 0 0 2 1 1 1 0 0 0 0 1 1 11

2− 1

2− 1

20

1

20

0 0 0 0 0 1 0 0 0 0 0 0 0 0 01

2

1

2

1

20

1

2−1

0 0 0 0 0 0 1 0 0 0 0 0 0 0 01

2

1

2

1

20

1

2

1

2

0 0 0 0 0 0 0 1 0 0 0 0 0 0 01

2

1

2

1

2−1

1

2

1

2

0 0 0 0 0 0 0 0 2 1 1 1 1 1 11

2−1 −1

1

2− 1

2

1

2

0 0 0 0 0 0 0 0 0 1 0 0 0 0 01

20 0

1

2

1

2

1

2

0 0 0 0 0 0 0 0 0 0 1 0 0 0 01

20 0

1

2

1

20

0 0 0 0 0 0 0 0 0 0 0 1 0 0 01

20 0

1

2

1

20

0 0 0 0 0 0 0 0 0 0 0 0 1 0 01

20 0 0 0

1

2

0 0 0 0 0 0 0 0 0 0 0 0 0 1 01

20 0 0 0

1

2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 11

20 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −1

20 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −1 0 − 1

2− 1

2− 1

2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −1 − 1

2− 1

2− 1

2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01√

20 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01√

20

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01√

2

In dimensions 11,12 and 13 there exist lattices K11, K12, K13 which are denser

than the maximal laminated lattices (see [CS93]). In order to produce them we

write n as n = n′ + n′′ such that |n′− n′′| ≤ 1 and if n is odd that n′′ has to be

odd, too. With this notation let Λ = 2Zn′⊕2√

3Zn′′ . For these lattices we have

to be a bit more careful in selecting the vertices from V (b1, . . . , bn). Roughly

speaking, we always try to find vertices which have non-zero components in

both spaces, Rn′ and Rn′′ .

For instance, if n = 11 we start with the lattice Λ = 2Z6 ⊕ 2√

3Z5 and first

we look at the vertices 12(2e1+2

√3e7),

12(2e2+2

√3e8) and so on. If this is not

longer possible we try to improve the lattice as in Algorithm 1. In dimension

11 this strategy leads to the following packing lattice Λ of determinant 18√

3:

2 0 0 1 0 1 1 1

20 1 1

2

0 2 0 1 0 2 01

20 0 −1

0 0 2 1 0 0 0 1 1 1

2− 1

2

0 0 0 1 0 0 0 0 0 1

20

0 0 0 0 2 1 0 1 0 1 − 1

2

0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0√

3

√

3

20 0 −

√

3

2

0 0 0 0 0 0 0

√

3

20 0 0

0 0 0 0 0 0 0 0√

3

√

3

2−√

3

2

0 0 0 0 0 0 0 0 0

√

3

20

0 0 0 0 0 0 0 0 0 0 −√

3

2

89

FINITE AND INFINITE LATTICE PACKINGS 11

Doing the same in dimension 12 gives a packing lattice Λ of determinant 27:

2 0 0 1 0 1 1 1

20 1 1

2

1

2

0 2 0 1 0 1 0 1

20 0 −1 − 1

2

0 0 2 1 0 0 0 1 1 1

2− 1

20

0 0 0 1 0 0 0 0 0 1

20 0

0 0 0 0 2 1 0 1 0 1 − 1

2

1

2

0 0 0 0 0 1 0 0 0 0 0 − 1

2

0 0 0 0 0 0√

3

√

3

20 0 −

√

3

2−√

3

2

0 0 0 0 0 0 0

√

3

20 0 0 −

√

3

2

0 0 0 0 0 0 0 0√

3

√

3

2−√

3

20

0 0 0 0 0 0 0 0 0

√

3

20 0

0 0 0 0 0 0 0 0 0 0 −√

3

2−√

3

2

0 0 0 0 0 0 0 0 0 0 0 −√

3

2

Finally, in dimension 13 we need one additional step in our algorithm. Namely,

so far we have refined our lattices only by vectors from the lattice 12 Λ. Of

course, as in the proof of Lemma 2.4, we can also look for suitable vectors in

the lattice 13Λ. And this is needed (as final “refinement step”) in order to get

the following packing lattice in dimension 13 of determinant 18√

3:

2 0 0 1 0 1 11

20 1

1

2

1

21

0 2 0 1 0 1 01

20 0 −1 − 1

20

0 0 2 1 0 0 0 1 11

2−1

20 1

0 0 0 1 0 0 0 0 01

20 0 0

0 0 0 0 2 1 0 1 0 1 −1

2

1

20

0 0 0 0 0 1 0 0 0 0 0 − 1

20

0 0 0 0 0 0√

3

√

3

20 0 −

√

3

2−√

3

2

1√

3

0 0 0 0 0 0 0

√

3

20 0 0 −

√

3

20

0 0 0 0 0 0 0 0√

3

√

3

2−√

3

20 1

√

3

0 0 0 0 0 0 0 0 0

√

3

20 0 0

0 0 0 0 0 0 0 0 0 0 −√

3

2−√

3

20

0 0 0 0 0 0 0 0 0 0 0 −√

3

20

0 0 0 0 0 0 0 0 0 0 0 0 2√

3

So in these dimensions we get

n 11 12 13

det Λ 18√

3 27 18√

3

and the corresponding densities coincide with the best known densities of the

lattices K11, K12 and K13 (see [CS93]).

References

[Bal92] K. Ball, A lower bound for the optimal density of lattice packings, Internat. Math.Res. Notices (1992), no. 10, 217–221.

[Ban90] W. Banaszczyk, On the lattice packing–covering ratio of finite-dimensional normed

spaces, Colloq. Math. 59 (1990), no. 1, 31–33.[BF34] T. Bonnesen and W. Fenchel, Theorie der konvexen Korper, Berichtigter Reprint,

1974 ed., Springer, Berlin, 1934.[BHW94] U. Betke, M. Henk, and J. M. Wills, Finite and infinite packings, J. Reine Angew.

Math. 453 (1994), 165–191.[Bor04] K. Boroczky, Jr., Finite packing and covering, Cambridge University Press, 2004.[But72] G. J. Butler, Simultaneous packing and covering in euclidean space, Proc. London

Math. Soc. (3) 25 (1972), 721–735.

90

12 MARTIN HENK

[CS93] J. H. Conway and N. J. A. Sloane, Sphere packings, lattices and groups, second ed.,vol. 290, Springer-Verlag, New York, 1993.

[EGH89] P. Erdos, P. M. Gruber, and J. Hammer, Lattice points, Pitman Monographs andSurveys in Pure and Applied Mathematics, vol. 39, Longman Scientific & Technical,Harlow, Essex/Wiley, New York, 1989.

[FT97] G. Fejes Toth, Packing and covering, Handbook of discrete and computationalgeometry, CRC Press Ser. Discrete Math. Appl., CRC, Boca Raton, FL, 1997,pp. 19–41.

[FTK93] G. Fejes Toth and W. Kuperberg, Packing and covering with convex sets, Handbookof convex geometry, Vol. A, B, vol. B, North-Holland, Amsterdam, 1993, pp. 799–860.

[GL87] P. M. Gruber and C. G. Lekkerkerker, Geometry of numbers, second ed., vol. 37,North-Holland Publishing Co., Amsterdam, 1987.

[GW93] P. Gritzmann and J. M. Wills, Finite packing and covering, Handbook of convexgeometry, Vol. A, B (P.M. Gruber and J.M. Wills, eds.), vol. B, North-Holland,Amsterdam, 1993, pp. 861–897.

[Rog50] C. A. Rogers, A note on coverings and packings, J. London Math. Soc. 25 (1950),327–331.

[Rog64] , Packing and covering, Cambridge Tracts in Mathematics and MathematicalPhysics, No. 54, Cambridge University Press, New York, 1964.

[Zon99] Ch. Zong, Sphere packings, Springer-Verlag, New York, 1999.[Zon02a] , Simultaneous packing and covering in the Euclidean plane, Monatsh. Math.

134 (2002), no. 3, 247–255.[Zon02b] , Simultaneous packing and covering of centrally symmetric convex bodies,

Rend. Circ. Mat. Palermo (2) Suppl. (2002), no. 70, part II, 387–396, IV Interna-tional Conference in “Stochastic Geometry, Convex Bodies, Empirical Measures &Applications to Engineering Science”, Vol. II (Tropea, 2001).

[Zon03] , Simultaneous packing and covering in three-dimensional Euclidean space,J. London Math. Soc. (2) 67 (2003), no. 1, 29–40.

Martin Henk, Universitat Magdeburg, Institut fur Algebra und Geometrie,

Universitatsplatz 2, D-39106 Magdeburg, Germany

E-mail address: henk@math.uni-magdeburg.de

91

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2

100

How Far Mesh Models Can SimulateA Perfect Sphere?

Tohru OGAWAogawa-t@koalanet.ne.jp

University of Tsukuba (Emeritus Professor),1-1-1, Tennodai, Tsukuba, 305-8577

andNPO-ISTA (Interdisciplinary Institute of Science, Technology and Art),

2-106-B-102, Kitahara, Asaka, Saitama, 351-0036, Japan

AbstractThe author likes to introduce some important problems to mathemati-

cians from scientists’ side for future collaboration. He believes that mu-tual cooperation between mathematicians and scientist is necessary withthe difference of the both side keeping in mind.

1 IntroductionMany computer simulations are performed with mesh models for global environment

problems, where at least two types of conservation laws, of materials and of energy,are important. Most of them are based on discrete models in some sense. Here thevertical freedoms are ignored. The number of regular polyhedrons (Platonic solids)is only five kinds and that of semi-regular polyhedrons (Archimedean solids with twokinds of regular polygons) is sixteen including both of two pairs of optical isomers.The maximum number of the equivalent vertices is 120 even if those with three kindsof regular polygons are used (see Appendix). In these structures, all of the vertex pointsare equivalent and all of the edges are of the same length.

Laplace equation, the most fundamental equation in such problems, is reduced intodifference equation in mesh models. It is almost equivalent to such an N × N matrixexpressing the corresponding planer graph, where N is the number of points in thatall the diagonal elements is the number of lines meet at there and the off-diagonalelement is −1 if and only if the corresponding pair of points share a line. The matrixis nonnegative. In the system with high symmetry as mentioned above, it has constanteigenvector associate with the minimum eigenvalue 0.

The main purpose of this paper is to show the inevitable difference between sphere1

101

and such mesh models. Another purpose is to review some of authors works on re-lated topics, among which Ehrenfest process (a stochastic process), representation ofrotational operator and spherical harmonics.

2 Summary on Rotational Symmetry

Laplace equation is the most fundamental equation in physics. For the system withspherical symmetry, it is natural to use spherical coordinate (r, θ.φ). Then, after theseparation of the radial part, Legendre’s differential equation is obtained for the angularpart (θ.φ)

−h

1sin θ

∂∂θ

¡sin θ ∂∂θ

¢+ 1

sin2 θ∂2

∂ϕ2

iY (θ,ϕ) = λY (θ,ϕ) .

In terms of the angular momentum operators

Lx ≡ −i³y ∂∂z − z ∂∂y

´, Ly ≡ −i

¡z ∂∂x − x ∂

∂z

¢, Lx ≡ −i

³x ∂∂y − y ∂

∂x

´.

the operator in the left hand side of the above equation is rewritten as

L2 ≡ L2x + L2y + L2z = −h

1sin θ

∂∂θ

¡sin θ ∂∂θ

¢+ 1

sin2 θ∂2

∂ϕ2

iNow, we have½L2Ylm (θ.φ) = l (l + 1)Ylm (θ.φ) ,LzYlm (θ.φ) =mYlm (θ.φ)

where l = 0, 1, 2, ... is azimuthal quantum number andm = l, l−1, ...,− (l − 1,−l) ismagnetic quantum number in quantum mechanical scheme. The function is the spheri-cal harmonic function and related with Legendre polynomial as

Ylm (θ.φ) = (−1)(m+|m|)/2q

2l+12

l−|m|l+|m|Plm (cos θ) exp (imϕ) .

where Plm (t)is associated Legendre function

Plm (t) ≡¡1− t3¢|m|/2 dm

dtmPl (t) ,with Legendre function

Pl (t) ≡ 12ll!

dl

dtl

¡t2 − 1¢l .

It is the representation of rotational group in function space.

The corresponding expression in two-dimensional special unitary group is as fol-lows L+Ψm ≡

p(l−m) (l +m+ 1)Ψm+1,

L−Ψm ≡p(l +m) (l −m+ 1)Ψm−1,

LzΨm ≡mΨm

2

102

£L+¤m0,m ≡

p(l −m) (l +m+ 1)δm0,m+1,£

L−¤m0,m ≡

p(l +m) (l −m+ 1)δm0,m−1,

[Lz]m0,m ≡ mδm0,m

L+ = Lx + iLy, L− = Lx − iLyLx ≡ (L++L−)

2 , Ly ≡ (L+−L−)2i ,

Then the matrix element corresponds to Lx for l is given for 0 ≤ |m| ≤ l and 0 ≤|m0| ≤ l. by

[Lx]m0m =1

2

hp(l +m) (l−m+ 1)δm0,m−1 +

p(l−m) (l +m+ 1)δm0,m+1

iFor example, for l = 3 and for l = 4,

Sx =1

2

0√6 0 0 0 0 0√

6 0√10 0 0 0 0

0√10 0

√12 0 0 0

0 0√12 0

√12 0 0

0 0 0√12 0

√10 0

0 0 0 0√10 0

√6

0 0 0 0 0√6 0

(1)

Sx =1

2

0√8 0 0 0 0 0 0 0√

8 0√14 0 0 0 0 0 0

0√14 0

√18 0 0 0 0 0

0 0√18 0

√20 0 0 0 0

0 0 0√20 0

√20 0 0 0

0 0 0 0√20 0

√18 0 0

0 0 0 0 0√18 0

√14 0

0 0 0 0 0 0√14 0

√8

0 0 0 0 0 0 0√8 0

(2)

The values¡ √

10,√18,

√24,

√28,

√30¢

arrange in the off-diagonal el-ements for l = 5 in the similar way.

All of the relations up to here are well-known facts and shown as a guide for theauthor’s discovery in the following sections.

2.1 Rational representation of the rotation operators

The matrix elements of Lx generally contain square root and the related computationis rather boring. The eigenvector of Lx for large values of l is a good example of that.The author happened to realize that rational representation obtain by a kind of affinetransformation Tm0,m ≡ δm0,m/

p2lC/−m

£T−1

¤m0,m ≡

p2lC/−mδm0,m as

3

103

[Lx]m0m =£T−1LxT

¤m0,m = (l+m) δm0,m−1 + (l −m) δm0,m+1

The next two are the new explicit forms for corresponding cases, respectively in Eq.1and

Eq.2,

Sx =1

2

0 1 0 0 0 0 06 0 2 0 0 0 00 5 0 3 0 0 00 0 4 0 4 0 00 0 0 3 0 5 00 0 0 0 2 0 60 0 0 0 0 1 0

for l = 3, (3)

Sx =1

2

0 1 0 0 0 0 0 0 08 0 2 0 0 0 0 0 00 7 0 3 0 0 0 0 00 0 6 0 4 0 0 0 00 0 0 5 0 5 0 0 00 0 0 0 4 0 6 0 00 0 0 0 0 3 0 7 00 0 0 0 0 0 2 0 80 0 0 0 0 0 0 1 0

for l = 4. (4)

This expression enable matrix calculations extremely easier. Transform the matrix atstart and then perform the necessary calculations. After all, do the inverse transforma-tion to get back to the original representation. [1]

2.2 Ehrenfest ProcessThe last matrix is proportional to the transition matrix for a stochastic process called asEhrenfest process, introduced in connection with so-called H-theorem [2] The modelby Ehrenfest is as follows. Prepare two vessels A and B. N numbered balls (from 1 toN on it) are vessel A at the beginning. Now, a random number decides the ball to moveto the other vessel. The attention is only paid to the number of balls in the vessels, thetransition matrix U of the process is of (N + 1)× (N + 1),Um0,m = [(N −M) δM0,M+1 +MδM 0,M−1] /N

The explicit forms of the transition matrix for N = 6 is 1/3 of Eq.3 and that forN = 4 is 1/8 of Eq.4.

The N + 1eigenvalues locate periodically between 1 and -1 as 1 − 2k/N (k =0, 1, 2, ..., N ) . The fact suggests the similarity to rotation operators. We can takeanother view in which the position of all the balls is recorded. It corresponds to the ran-dom walk on the N-dimensional cube: a vertex to another through the edge connectingthem.

4

104

2.3 Episode concerning the motivation

The author happened to find the above mentioned fact. Allow him to mention anepisode here. In 1984, something interesting was found among the articles left byNobel laureate Sin-itiro Tomonaga in his house after his death in 1979. The article wasconsist of six sheets of papers and ten pieces of papers with a number on each of them(1, 2, 3,...,10). On each of six sheets of papers, many numbers were written by hand.The author luckily successfully guess that they are the record of hand-made simulationon Ehrenfest process performed by moving the numbered piece of paper by Tomonagahimself. This simulation was the original material of the very last part of the unfinishedmanuscript of his last book “What is Physics?” [3] published in 1982 in Japanese. Theauthor wrote a paper about this material and Ehrenfest process and some related topicsin 1985 [4]. He tried to finish the last part of the paper by a kind of a joke in a style ofrakugo (a kind of comical performance by one person on stage and Tomonaga liked itso much as he sometimes play by himself). The author tried to connect this material tospin because spin is the subject of one of the Tomonaga’s books “Spin wa meguru (Spinis moving around)” (1974, in Japanese and English translation in 1998 as “The Story ofSpin”) [5].

Then the description in this section is completely opposite to time sequence actuallyhappened. As a matter of fact, the started with Ehrenfest process and then a set of theeigenvalues and eigenfunctions. Rotation was later.

3 Spherical Harmonics as Function of Unit Vector.

Spherical coordinate (r, θ.φ) is naturally used for the system related with spherical sym-metry. But, the existence of the axis breaks the spherical symmetry. Is it inevitable?The study in this section is motivated with this question.

Here, spherical harmonics is formulate as a function of unit vector, without intro-ducing any special direction [6]. It is the orthonormal polynomial of unit vectors.

3.1 Definition and some useful relations

The product of two scalar functions, f (r) and g (r) is defined as(f, g) ≡ R

Ωf (r) g (r) dr =4π hf (r) g (r)iΩ

whereRΩ...dr and h..iΩ are respectively integration and average over possible di-

rections. The Cartesian coordinate is sometimes useful to estimate something.[l,m, n] ≡ ­xlymzn®

Ω= 1

4π

RΩ x

lymzndr (l,m, n ∈ I)It is zero if any of three integers is odd.

[2l, 2m, 2n] = 14π

R π0 sin θdθ

R 2π0 dϕ

£sin2l+2m θ cos2n θ sin2m ϕ cos2l ϕ

¤=(2l− 1)!! (2m− 1)!! (2n− 1)!!

(2l+ 2m+ 2n+ 1)!!

5

105

Some explicit forms are[0, 0, 0] = 1,[2, 0, 0] = 1/3,[4, 0, 0] = 1/5, [2, 2, 0] = 1/15,[6, 0, 0] = 1/7, [4, 2, 0] = 1/35, [2, 2, 2] = 1/105,[8, 0, 0] = 1/9, [6, 2, 0] = 1/63, [4, 4, 0] = 1/105, [4, 2, 2] = 1/315.

Some of the basic relations between them are­x2ly2mz2n

¡x2 + y2 + z2

¢®Ω=­x2ly2mz2n

®Ω,

[2l + 2, 2m, 2n] + [2l, 2m+ 2, 2n] + [2l, 2m, 2n+ 2] = [2l, 2m, 2n] ,

1 =D¡x2 + y2 + z2

¢kEΩ=Pl+m+n=k

k!l!m!n! [2l, 2m, 2n] .

The next include arbitrary constant unit vectors.h(ar) (br)iΩ = 1

3 (ab)h(ar) (br) (cr) (dr)iΩ = 15!! [(ab) (cd) + (ac) (bd) + (ad) (bc)]h(ar) (br) (cr) (dr) (er) (fr)iΩ = 1

7!!

P(ab) (cd) (ef) -type (15 terms)DY

(ar)EΩ= 1

(2k+1)!!

Q(aiaj) -type

( (2k − 1)!!terms of the products over all the possible pairing)

It is noted that the pairing in these formula can be regarded as a kind of Wick’stheorem which are used in field theory [7].

The explicit form of normalized spherical harmonics in this scheme up to 4 are listedbelow.l = 0 1/

√4π

l = 1p3/ (4π) (ar)

l = 2

r5/h3 + (ab)2

i h3 (ar) (br)− (ab)2

il = 3 N3 [5 (ar) (br) (cr)− (bc) (ar)− (ca) (br)− (ab) (cr)]l = 4 N4 [35 (ar) (br) (cr) (cr)− 5

P(ab) (cr) (dr) +

P(ab) (cd)]

The available space for case l = 4 is to small, then some parts are in truncated form.The term

P(ab) (cr) (dr) consists of six terms and that

P(ab) (cd) of three terms.

The part withP

means summation over the term with the similar form. N3andN4,though the detailed forms are ignored here, are the normalization constant dependingon the choice of constant unit vectors as in lower order cases..

Here, the whole function space are uniquely divided into (2l + 1)-dimensional sub-space each of which corresponds to l-th order polynomials. It is important to point outthat further divisions within the subspace specified by l are arbitrary. Though mag-netic quantum number m is conventionally adopted there mainly because of quantummechanical image, it is not necessary. One can choose freely. There are several ways,

putting much of geometrical characters. One of the principle is not using complexnumber but only real numbers. Properly chosen mutually orthogonal three unit vectors

6

106

a, b, c makes mutually orthogonal three real l = 1functions with chosen three direc-tions. Though the cases of the larger values of l are the more and more complicatedbecause the functions are generally tensors, it is possible to keep geometrical image. Itis the merit of the present scheme.

4 Structure analysis by spherical harmonics

For any set of unit vectors expressing point distribution on a sphere, analysis by spher-ical harmonics is standard since it corresponds to Fourier analysis with spherical sym-metry is imposed in advance.

Let a set of N unit vectors rk (k = 1, 2, ..., N) expressing the point distributionon a sphere. Then the distribution function f (r) is defined as

f (r) ≡ 1

N

NXk=1

δ (r− rk) (5)

It is better here to point out the special character of problem different from in linearspaces before going in the analysis with spherical harmonics. There is the properlydefined representative value, for example, average value and median etc., in usual sta-tistics. It is not true for the data are only on the spherical surface. The center ofgravity should be inside of the sphere. It can be even at the center if the distribution issymmetrical enough. Anyway, it lies out of definition range.

5 Spherical harmonic analysis and concluding remarks

The coefficient Clm of Ylm (r), defined in Eq.5 is obtained asf (r) =

PClmYlm (r)

Clm =1√4π

RΩf (r)Ylm (r) dr =

1√4πN

∞Pl=0

lPm=−l

Ylm (r) δ (r− rk)

=1√4πN

∞Pl=0

lPm=−l

Ylm (rk)

Ql ≡s4π |Cl|22l + 1

, |Cl|2 ≡lP

m=−l|Clm|2

The value |Cl|2 is independent from the choice of 2l+1 orthonormal functions among(2l + 1)-dimensional subspace specified by l. The higher the symmetry of the struc-ture, The values of more numbers of Qls are zero. In case of dodecahedral and icosa-hedral symmetry, non vanishingQls are only, 0, 6, 15, for up to l = 15.

7

107

Though it is well-known fact, it should be also pointed out that more than 5-folddegeneracy cannot appear except spherical symmetry itself.

AcknowledgmentThe author is indebted to Professor Eiichi Bannai and Kyushu University 21-th Cen-

tury COE program, for giving opportunity of presentation to the author.

Appendix A. Symmetrical point arrangements on a spherical face

There are altogether eighteen symmetrical arrangements on a spherical surface, whosevertices and edges are both congluent, though the faces can be at most three kinds.

No. qp’s (the number ofp-gons at a vertex)

Ppqp V E F Fp(the number of p− gons)

1 q3 = 3 3 4 6 4 F3 = 42 q4 = 3 3 8 12 6 F4 = 63 q3 = 4 4 6 12 8 F3 = 84 q5 = 3 3 20 30 12 F5 = 125 q3 = 5 5 12 30 20 F3 = 206 q3 = 1, q6 = 2 3 12 18 8 F3 = 4, F6 = 47 q3 = 1, q8 = 2 3 24 36 14 F3 = 8, F8 = 68 q4 = 1, q6 = 2 3 24 36 14 F4 = 6, F6 = 89 q3 = 1, q10 = 2 3 60 90 32 F3 = 20, F10 = 1210 q5 = 1, q6 = 2 3 24 90 32 F5 = 12, F5 = 2011 q4 = 1, q6 = 1, q8 = 1 3 48 72 26 F4 = 12, F6 = 8, F8 = 612 q4 = 1, q6 = 1, q10 = 1 3 120 180 62 F4 = 30, F6 = 20, F5 = 1213 q3 = 2, q4 = 2 4 12 24 14 F3 = 8, F4 = 614 q3 = 3, q4 = 1 4 24 72 26 F3 = 8, F4 = 1815 q3 = 2, q5 = 2 4 30 60 32 F3 = 20, F5 = 1216 q3 = 1, q4 = 2, q5 = 1 4 120 180 62 F3 = 20, F4 = 30, F5 = 1217 q3 = 4, q4 = 1 5 24 60 38 F3 = 32, F4 = 618 q3 = 4, q5 = 1 5 60 150 92 F3 = 80, F5 = 12

8

108

References

[1] T. Ogawa, "The Ehrenfest Process, a Rational Representation of Spin Operatorsand Explicit Form of Rotation Operators" in "Quasicrystals" Eds. T. Fujiwara andT. Ogawa, pp.20-26, (Springer, 1990).

[2] P. and T. Ehrenfest, Phys. Z.,8, 311 (1907). P. and T. Ehrenfest, Encyklop. d. mathWissensch. IV 2, II (1911).

[3] S. Tomonaga, "Buturigaku towa nandarouka? (What is Physics?)"

[4] T. Ogawa, "Tomonaga-sensei no Tesagyo Simulation (Handmade Simulation ByTomonaga)" (in Japanese) Suuri Kagaku (Mathematical Sciences) No. 259, 20-16,(1985).

[5] S. Tomonaga, "Spin wa meguru" (in Japanese) (Chuo Koron, 1974); "The Story ofSpin" (English Translation) (University of Chicago, 1998)

[6] T.Ogawa, "A New Aspect of Spherical Harmonics" Forma, 13, 51-62 (1998).

[7] G. C. Wick, Phys. Rev. 80, 268 (1950).

9

109

IMPROVED DELSARTE BOUNDS VIA EXTENSION OF THEFUNCTION SPACE

(EXTENDED ABSTRACT)

FLORIAN PFENDER

ABSTRACT. We present a new extension to Delsarte’s linear program-ming method. It is used to find new upper bounds for kissing numbersand spherical codes in several dimensions. In particular, Musin’s recentwork in dimensions three and four can be reformulated in this framework.

1. INTRODUCTION

A spherical (n, N, )-code is a set x1, . . . , xN of unit vectors in Rn

such that the pairwise angular distance between the vectors is at least .Often, one tries to find codes which maximize N or , respectively, if theother two values are fixed.

The kissing number problem asks for the maximum number k(n) of non-overlapping unit balls touching a central unit ball in n-space. This is aspecial case of maximizing a spherical code, just fix = π

3.

Most successful in the task of finding upper bounds for these problems isan approach via linear programming that was pioneered by Delsarte in theearly seventies in the context of binary codes and association schemes ([2],[3]). In 1977, Delsarte, Goethals and Seidel [4] adapted this approach to thecase of spherical codes, and up to some minor improvements, their methodstill yields the best known bounds.

Let X = (x1, . . . , xN) ∈ RnN be an (n, N, )-code, and let

M = (xij) = (〈xi, xj〉) = X>X ∈ RNN

be the matrix containing all the scalar products of the xi. Then

• xii = 1, while xij ≤ cos for i 6= j,• M is positive semi-definite, and• M has rank ≤ n.

We get the following theorem with a one-line proof.

Supported by the DFG Research Center MATHEON “Mathematics for key technologies”in Berlin.

1

110

2 FLORIAN PFENDER

Theorem 1 (Delsarte-Goethals-Seidel). Let M = (xij) be as above. Letc > 0 and let f : R → R be a function such that

(i)∑

i,j≤N f(xij) ≥ 0,(ii) f(x) ≤ −c for −1 ≤ x ≤ cos , and

(iii) f(1) ≤ 1 − c.

Then N ≤ 1/c.

Proof. Let g(x) = f(x) + c. Then

N2c ≤ N2c +∑

i,j≤N

f(xij) =∑

i,j≤N

g(xij) ≤∑

i≤N

g(xii) = N g(1) ≤ N.

To prove bounds on N with the help of this theorem, we need to find a

“good” function f (and it needs to work for every conceivable code).Start with a set S of functions, for which (i) is true for every (n, N, )-

code. Then, find a linear combination of these functions with non-negativecoefficients, such that (ii) and (iii) are true. If we discretize (iii), this canbe done via linear programming. Afterwards, minor inaccuracies stemmingfrom the discretization are dealt with.

So, the set S will be the center of our attention. In Section 2, we willtake a closer look at the set S which is classically used in this method. InSections 3 and 4 we explore how one could add further functions to thisset. In Section 5 we present a family of functions which can be added to S,giving improvements to some best known bounds. In particular, we lowerthe upper bounds for the kissing number in dimensions 10, 16, 17, 25 and26.

We will find some limits to this approach in Section 6, and in the finalsection we show how Musin’s recent work on the kissing numbers in threeand four dimensions can be reformulated in our frame work.

In this extended abstract, most proofs are omitted. We refer to the journalversion for full details [7].

2. THE CLASSIC APPROACH

To guarantee condition (i) in Theorem 1, one looks for a function f thatwill return a matrix (f(xij)) which is positive semi-definite for all sets ofunit vectors (xi). Reason for this restriction is that one knows a lot aboutthese functions, by the following theorem of Schoenberg about Gegenbauerpolynomials. These polynomials (also known as the spherical or the ultra-spherical polynomials) may be defined in a variety of ways. One compactdescription is that for any n ≥ 2 and k ≥ 0, G

(n)k (t) is a polynomial of

degree k, normalized such that G(n)k (1) = 1, and such that G

(n)0 (t) = 1,

111

IMPROVED DELSARTE BOUNDS 3

-1 -0.5 0.5 1

-1

-0.5

0.5

1

FIGURE 1. A plot of G(4)7

G(n)1 (t) = t, G

(n)2 (t) = nt21

n1, . . . are orthogonal with respect to the scalar

product

⟨

g(t), h(t)⟩

:=

∫ +1

1

g(t)h(t)(1 − t2)n3

2 dt

on the vector space R[t] of polynomials; this product arises naturally inintegration over Sn1.

Theorem 2 (Schoenberg [8]). If (xij) ∈ RNN is a positive semidefi-nite matrix of rank at most n with ones on the diagonal, then the matrix(

G(n)k (xij)

)

is positive semidefinite as well.

In fact, Schoenberg proved the reverse implication as well. Every poly-nomial with this property is a non-negative combination of said Gegenbauerpolynomials.

To obtain bounds on N , conditions (ii) and (iii) in Theorem 1 are dis-cretized, and a linear program will give an optimal (i.e., one with a maximalvalue for c) non-negative linear combination of Gegenbauer polynomialsfor some fixed maximum degree. The minor inaccuracies arising from thediscretization can be dealt with.

In dimension three and four, Delsarte’s method gives bounds of k(3) ≤13 and k(4) ≤ 25, and the method as is is proven to not be able to lowerthese bounds. The true values are 12 and 24, respectively, but the proofs aremuch more complicated. In most other dimensions, this method gives thebest known upper bound for the kissing number.

112

4 FLORIAN PFENDER

3. EXTENDING THE FUNCTION SPACE

Instead of limiting ourselves to Gegenbauer polynomials, let us considerthe bigger space of candidates for f given by condition (i) in Theorem 1.

Definition 1. Let PnN() be the set of all functions f : R → R, for which

(i) f(1) = 1,(ii)

∑

i,j≤N f(〈xi, xj〉) ≥ 0 for every (n, N, )-code x1, . . . xN.

Let

• PnN := Pn

N(0),• Pn() :=

⋂

N PnN (),

• PN () :=⋂

n PnN (),

• PN , Pn, P() and P accordingly.

Fact 3. The following statements are true.

(i) PnN () ⊆ Pn

N(′) for ≤ ′.(ii) Pn

N () ⊆ Pn1N ().

(iii) Let f ∈ Pn be piecewise continuous, then∫

Sn1 f(〈e1, x〉) dx ≥ 0.(iv) Let f ∈ Pn

N() and N < n, then f(x) ≥ − 1N

for − 1N

≤ x ≤cos .

(v) If f, g ∈ PnN(), and 0 ≤ c ≤ 1, then cf + (1 − c)g ∈ Pn

N().(vi) All functions with f(x) ≥ 0 for −1 ≤ x ≤ 1 are in P .

(vii) G(n)k ∈ Pn.

Proof. Statements (i), (ii), (v) and (vi) follow directly from the definition,(vii) follows from Theorem 2. To show (iii), observe that if N → ∞, andthe xi are uniformly distributed over Sn1, the sum in Definition 1 dividedby N approaches the integral. To see (iv), note that for − 1

k≤ x ≤ cos ,

there is a regular simplex on k + 1 points with pairwise angular distancearccos x ≥ inside Sn1.

By the following lemma we may often restrict our observations to func-tions with f(x) ≤ 1.

Lemma 4. For f ∈ PnN , the function g(x) = min(f(x), 1) ∈ Pn

N .

Proof. Suppose there are vectors x1, . . . , xN ∈ Sn1 such that

S(g) :=∑

i,j≤N

g(〈xi, xj〉) < 0.

Assume further that S(g) is minimal amongst all multisets of N vectorsfrom x1, . . . , xN. As f ∈ Pn

N , there are k, ` such that f(〈xk, x`〉) > 1and

∑

i≤N g(〈xk, xi〉) ≤∑

i≤N g(〈x`, xi〉).

113

IMPROVED DELSARTE BOUNDS 5

-1 -0.5 0.5 1

-1

-0.5

0.5

1

FIGURE 2. A plot of the function from Fact 5 for = π3

Consider the configuration x1, . . . , xN ∈ Sn1, where x` is replaced byxk. Then

∑

i,j≤N

g(〈xi, xj〉) =

∑

i,j≤N

g(〈xi, xj〉) − 2∑

i≤N

g(〈x`, xi〉) + 2∑

i≤N

g(〈xk, xi〉)

≤∑

i,j≤N

g(〈xi, xj〉).

Since S(g) is minimal, equality must hold and thus∑

i≤N g(〈xk, xi〉) =∑

i≤N g(〈x`, xi〉) whenever f(〈xk, x`〉) > 1.We may assume that x1, . . . , xN ∈ Sn1 were chosen in a way that

minNk + ` : f(〈xk, x`〉) > 1 ≤ N 2 is maximal amongst all multi-sets which achieve S(g). But x1, . . . , xN ∈ Sn1 has a higher value for thisquantity, a contradiction.

The next two facts show that Pn is larger than just convex combinationsof Gegenbauer functions and positive functions.

Fact 5. Let < π/2, let

f(x) =

−1 : x < − cos β

2

0 : − cos β

2≤ x ≤ cos

1 : cos < x

.

Then f ∈ P .

114

6 FLORIAN PFENDER

Fact 6. The function f from Fact 5 is not a convex combination of Gegen-bauer polynomials and positive functions.

4. RESTRICTING THE FUNCTION SPACE

It is often difficult to test for a function f : R → R if f ∈ PnN (). We

will focus on a smaller space to avoid some of the complications.

Definition 2. Let RnN+1() ⊆ Pn

N+1() be the space of functions f : R →R, such that

1 +N

∑

i=1

f(〈x0, xi〉) ≥ 0

for every set x0, x1, x2, . . . , xN ∈ Sn1 with 〈xi, xj〉 < cos for all 0 ≤i < j ≤ N .

Define R, Rn, etc. analogously to the definitions of P , Pn, etc.

So, instead of just requiring that the sum of all matrix entries in M =(xij) is non negative, we ask for all row sums to be non negative. Again,we get a number of statements similar to Fact 3.

Fact 7. The following statements are true.

(i) RnN () ⊆ Rn

N (′) for ≤ ′.(ii) Rn

N () ⊆ Rn1N ().

(iii) Let f ∈ RnN() and N < n, then f(x) ≥ − 1

Nfor − 1

N≤ x ≤

cos .(iv) If f, g ∈ Rn

N(), and 0 ≤ c ≤ 1, then cf + (1 − c)g ∈ RnN ().

(v) All functions with f(x) ≥ 0 for −1 ≤ x ≤ 1 are in R.(vi) Let f ∈ Rn

N and g ∈ Rn. Then f(x) ≥ − 1N1

and g(x) ≥ 0 for−1 ≤ x ≤ 1.

Proof. Statements (i)-(v) follow from the definition. For (vi), supposethat f(x) < − 1

N1for some x. Then pick a point x0 ∈ Sn1, and N − 1

points x1, x2, . . . , xN1 ∈ Sn1 with 〈x0, xi〉 = x, showing that f /∈ RnN .

The same argument works for the second part of the statement, just pick N

large enough. The last statement in Fact 7 shows that we have nothing to gain by adding

functions in RnN to the set S, we will have to look for functions in Rn

N () \Rn

N . The following lemma gives a lower bound for functions in R().

Lemma 8. Let f ∈ R(), let z = cos , and let

gα(x) :=

− 1N

: −√

z + 1zN

≤ x < −√

z + 1zN+1

for N ∈ N

0 : −√

z ≤ x < 11 : x = 1

.

115

IMPROVED DELSARTE BOUNDS 7

Then f(x) ≥ gα(x) for −1 ≤ x ≤ z.

For a plot of gπ3(x), see Figure 3 in Section 5.

Observe that gα /∈ R(). A simple example of vectors showing this is

x1 =

√1 − z

0√z

, x2 =

0√1 − z√

z

, x0 = −(1 + ε)x1 + x2

||(1 + ε)x1 + x2||∈ S2.

Here, 1 + gα(〈x0, x1〉) + gα(〈x0, x2〉) = 1 − 1 − 12

< 0.The following lemma enables us to reduce the vector combinations which

have to be tested when we are searching for a function f ∈ R().

Lemma 9. Let z = cos , let θ0 < −√

z, and let f : [−1, θ0] → R. Letn > k and let x0, x1, . . . , xk ∈ Sn1 be a set of k + 1 points such that

(i) 〈xi, xj〉 ≤ z for 1 ≤ i < j,(ii) 〈xi, x0〉 ≤ θ0 for i ≥ 1,

(iii)∑

f(〈x0, xi〉) is minimal with respect to (i)-(ii),(iv) 〈xi, xj〉 is pointwise maximal with respect to (i)-(iii).

Then the xi (i ≥ 1) form a regular simplex with 〈xi, xj〉 = z for i 6= j.

From this we get the following corollary, showing that Lemma 8 is sharpin some sense.

Corollary 10. Let z = cos , let k ∈ N, and let

hkα(x) :=

− 1k

: −1 ≤ x < −√

z + 1zk+1

0 : −√

z + 1zk+1

≤ x < 1

1 : x = 1

.

Then hkα ∈ R().

Proof. Suppose for the sake of contradiction that hkα /∈ R(). Then there

is a set of points x0, x1, . . . , xk+1 ∈ Sn1 for some n > k + 1 such that

〈xi, xj〉 ≤ z and 〈x0, xj〉 < −√

z + 1zk+1

for 0 ≤ i < j ≤ N . By Lemma 9,

we may assume that the points x1, . . . , xk+1 form a regular simplex with〈xi, xj〉 = z for i 6= j. But this is impossible, since then

max1≤i≤k+1

〈x, xi〉 ≥ −

√

z +1 − z

k + 1

for any x ∈ Sn1, a contradiction proving the corollary.

116

8 FLORIAN PFENDER

-0.9 -0.8 -0.7 -0.6 -0.5

-1

-0.8

-0.6

-0.4

-0.2

FIGURE 3. A plot of f π3

and gπ3

5. THE MAIN RESULT

One can add the functions from Corollary 10 to the set of Gegenbauerpolynomials in some dimension n, and this will give a larger set S. But inmost cases, this does not give an improvement to the known bounds on N

and . This is different for the following function. With the addition of thisfunction to the set S, numerous improvements are possible.

Lemma 11. Let 0 ≤ z = cos < 1 and

fα(x) =

zx2

1z: x < −

√z

0 : −√

z ≤ x < 11 : x = 1

.

Then fα ∈ R().

We can add fπ3

to the list of Gegenbauer polynomials in dimension n toget new bounds on the kissing numbers k(n) through linear programming.This yields the new bounds in Table 1, where the known bounds are takenfrom [1] (with the exception of the bound k(9) ≤ 379 from [11]). For othern ≤ 30, the bounds are not improved.

Similarly, new bounds for the minimal angular separation in sphericalcodes can be achieved. Some of them are shown in Table 2 (here, the lowerbounds are from [10]).

6. POSSIBLE IMPROVEMENTS

Let us consider how we might improve our bounds by further exploringRn(). Lemma 8 gives a lower bound for functions in R() using regular

117

IMPROVED DELSARTE BOUNDS 9

n lower bound Delsarte bound new upper bound9 306 380 379

10 500 595 59416 4320 8313 831217 5346 12218 1221025 196656 278363 27808226 196848 396974 396444

: 379 and 12215 with some extra inequalities

TABLE 1. New upper bounds for the kissing number

n N lower bound Dels. bound new upper bound3 13 57.13 60.42 60.353 14 55.67 58.09 58.003 15 53.65 56.13 56.103 24 43.69 44.45 44.434 9 80.67 85.60 83.654 10 80.40 82.19 80.734 11 76.67 79.46 78.734 22 60.13 63.41 64.384 23 60.00 62.36 62.304 24 60.00 60.50 60.385 11 82.36 87.30 85.395 12 81.14 84.94 83.155 13 79.20 82.92 81.545 14 78.46 81.20 80.305 15 78.46 79.73 79.30

TABLE 2. New upper bounds for in (n, N, )-codes

simplices as the extremal configurations. But in Sn1, there can be at most npoints x1, x2, . . . , xn such that 〈x0, xi〉 = 〈x0, xj〉 and 〈xi, xj〉 = z = cos .Thus, there may be functions f ∈ Rn() for which − 1

n≤ f(x) < gα(x) for

some x ≥ −√

z + 1zn+1

. The analysis gets much more complicated, though,

as one has to consider configurations which are not regular simplices.To keep the analysis managable, we will require that f(x) ≥ 0 for

x ≥ −z and that f is monotone increasing. This way, one only has toconsider configurations such that x1, . . . , xN are the vertices of a sphericalpolytope, and −x0 is inside this polytope. We get a new lower bound for

118

10 FLORIAN PFENDER

such functions:

f(x) ≥ gnα(x) :=

− 1N

: −√

z + 1zN

≤ x < −√

z + 1zN+1

, N < n

1n

: −√

z + 1zn

≤ x < −z

0 : −z ≤ x < 11 : x = 1

.

Observe that the function gnα /∈ Rn(). But we can use gn

π3

in our linear pro-gram to find out the limitations of this approach towards improvements ofthe kissing number bounds—lower bounds to the upper bounds achievablewith this method. The results up to dimension 16 are given in Table 3.

n lower bd. Dels. bd. new bd. bd. using gnπ3

3 12 13 13 124 24 25 25 246 72 82 82 807 126 140 140 1359 306 380 379 366

10 500 595 594 56111 582 915 915 87712 840 1416 1416 137013 1130 2233 2233 213214 1582 3492 3492 326815 2564 5430 5430 526216 4320 8313 8312 8159

TABLE 3. Lower bounds for upper bounds on k(n)

7. MUSIN REVISITED: k(3) = 12 AND k(4) = 24

In dimensions three and four, using f π3

only gives marginal improvementsto the bounds on the kissing numbers achieved with Delarte’s method—notenough to show that k(3) = 12 and k(4) = 24. Several proofs for k(3) = 12are known, the first one by Schutte and van der Waerden [9]. In dimensionfour, only recently a proof for k(4) = 24 was found by Musin [5]. The sametechniques also yield the arguably simplest proof for dimension three [6].

Our techniques give a new frame work in which to formulate Musin’sproofs. We will give a short sketch of this reformulation for dimension

119

IMPROVED DELSARTE BOUNDS 11

four. Similarly to g4π3

, define the function

g(x) :=

− 1N

: −√

N+12N

≤ x < −√

N+22N+2

, N < 4

14

: −√

58≤ x < −0.6058

0 : −0.6058 ≤ x < 11 : x = 1

.

Run the linear program using g, G40, G

41, . . . , G

49, finding a linear combina-

tion with non negative coefficients

f = c + c0g +9

∑

i=1

G4i ,

maximizing c while maintaining f(x) ≤ 0 for x ≤ 12

and f(1) ≤ 1. Thefunction f is not necessarily in P4(π

3) as g 6∈ R4(π

3). To avoid this problem,

let

h(x) =

min

−f(x)c0g(x)1c0

, 0

: x ≤ 12

1 : x > 12

.

If we can show that h ∈ R4(π3), then this will prove that

k(4) ≤1

c≈ 24.79 < 25,

as

f = c + c0h +

9∑

i=1

G4i

is a function with f(x) ≤ 0 for x ≤ 12

and f(1) ≤ 1.Comparing h to the ninth degree polynomial used in Musin’s proof (see [5]),

one sees that they are identical up to a constant factor on [−1,−0.6058].Since h(x) ≥ 0 for x ≥ −0.6058, Musin shows that one only needs to con-sider configurations x0, x1, . . . , xN with N ≤ 6. Musin’s further analysiscan be directly translated into a proof of h ∈ R4(π

3).

REFERENCES

1. J. H. Conway and N. J. A. Sloane, “Sphere Packings Lattices and Groups,”Grundlehren Math. Wiss., vol. 290, Springer-Verlag, New York, third ed., 1993.

2. P. Delsarte, Bounds for unrestricted codes, by linear programming, Philips Res. Rep.27 (1972), 272–289.

3. P. Delsarte, An algebraic approach to the association schemes of coding theory, PhilipsRes. Rep. Suppl. (1973), vi+97.

4. P. Delsarte, J. M. Goethals, and J. J. Seidel, Spherical codes and designs, Geom. Ded-icata 6 (1977), 363–388.

5. O. R. Musin, The kissing number in four dimensions, preprint.6. O. R. Musin, The kissing problem in three dimensions, preprint.

120

12 FLORIAN PFENDER

7. F. Pfender, Improved Delsarte bounds via extension of the function space, preprint.8. I. J. Schoenberg, Positive definitive functions on Spheres, Duke Math. J. 9 (1942),

96–107.9. K. Schutte and B. L. van der Waerden, Das Problem der dreizehn Kugeln, Math. Ann.

53 (1953), 325–334.10. N. J. A. Sloane, with the collaboration of R. H. Hardin, W. D. Smith

and others, Tables of Spherical Codes, published electronically atwww.research.att.com/˜njas/packings/.

11. M. A. Vsemirnov and M. G. Rzhevskii, An upper bound for the contact number indimension 9, Russian Math. Surveys 57 (2002), no. 5, 1015–1016.

INSTITUT FUR MATHEMATIK, MA 6-2, TECHNISCHE UNIVERSITAT BERLIN, STRASSE

DES 17. JUNI 136, 10623 BERLIN, GERMANY

E-mail address: fpfender@math.tu-berlin.de

121

METHODS IN THE LOCAL THEORY OF PACKING AND COVERINGLATTICES

A. SCHURMANN AND F. VALLENTIN

ABSTRACT. In this paper we are concerned with three lattice problems: the lattice packingproblem, the lattice covering problem and the lattice packing-covering problem. One wayto find optimal lattices for these problems is to enumerate all finitely many, locally optimallattices. For the lattice packing problem there are two classical algorithms going backto Minkowski and Voronoi. For the covering and for the packing-covering problem wepropose new algorithms.

Here we give a brief survey about these approaches. We report on some recent com-puter based computations where we were able to reproduce and partially extend the knownclassification of locally optimal lattices. Furthermore we found new record breaking cov-ering and packing-covering lattices. We describe several methods with examples to showthat a lattice is a locally optimal solution to one of the three problems.

1. INTRODUCTION

Classical problems in the geometry of numbers are the determination of most econom-ical lattice sphere packings and coverings of the Euclideand-spaceRd. A lattice L is afull rank, discrete subgroup ofRd. Thus there exist matricesA ∈ GLd(R) with L = AZd

which we callbasesof L.If Bd denotes the Euclidean unit ball, then the Minkowski sumL + αBd = v + αx :

v ∈ L,x ∈ Bd, α ∈ R>0, is a lattice packing if the translates ofαBd have mutuallydisjoint interiors and a lattice covering ifRd = L + αBd. The packing radiusλ(L) of alatticeL is given by

λ(L) = maxλ : L + λBd is a lattice packing,

and the covering radiusµ(L) by

µ(L) = minµ : L + µBd is a lattice covering.

For a latticeL we define itsdeterminantdet(L) = |det(A)|, which is independent ofthe chosen basis. We consider the following three “quality measures” ofL:

(1) thepacking densityδ(L) =λ(L)d

det(L)· κd,

(2) thecovering densityΘ(L) =µ(L)d

det(L)· κd,

(3) thepacking-covering constantγ(L) =µ(L)λ(L)

.

Hereκd = πd/2/Γ(d/2 + 1) denotes the volume of the unit ballBd. So the packingdensityδ(L) for instance gives the ratio of space covered by spheres in the lattice packingL + λ(L)Bd. Note that all three quantities are invariant with respect to a scalingαL of Lwith α 6= 0. For each of the quantities we consider the problem of finding extremal latticesattaining a maximum or minimum respectively.

Date: 16th December 2004.The second author was supported by the Edmund Landau Center for Research in Mathematical Analysis and

Related Areas, sponsored by the Minerva Foundation (Germany).

1

122

2 A. SCHURMANN AND F. VALLENTIN

Problem 1.1(Lattice Packing Problem). Ford ≥ 2, determineδd = maxL

δ(L) and lattices

L attaining it.

Problem 1.2 (Lattice Covering Problem). For d ≥ 2, determineΘd = minL

Θ(L) and

latticesL attaining it.

Problem 1.3(Lattice Packing-Covering Problem). For d ≥ 2, determineγd = minL

γ(L)

and latticesL attaining it.

Note that all three optima are attained. All three problems have in common that thereexist only finitely many local optima for everyd (see Section 3 for definitions). In thisarticle we want to review some of the major tools available to find such local extrema andto verify their local optimality. In Section 2 we briefly summarize known results of thethree problems and in Section 3 we give a short introduction to the connection of latticesand positive definite quadratic forms, which gives the framework in that the problems areusually dealt with. In Section 4 we describe two classical approaches by Minkowski andVoronoi to enumerate all local optima of the lattice packing problem. In Section 5 weare concerned with local optima of the lattice covering and the lattice packing-coveringproblem, which can be treated in parallel. Most of these techniques are described in greaterdetail in [SV04a] and [SV04b].

2. KNOWN RESULTS

2.1. The Lattice Packing Problem. The lattice packing problem, arising from the studyof positive definite quadratic forms, is the oldest and most popular of the three problemsand has been considered by many authors in the past. As shown in Table 1, the solution tothe problem was known for dimensiond ≤ 8 since 1934. For a description of the extremalroot latticesAd,Dd andEd and the history of the problem we refer the interested readerto the book [CS88b]. Recently, Cohn and Kumar [CK04] showed that the Leech latticeΛgives the unique densest lattice packing inR24. Furthermore they showed: The density ofany sphere packing (without restriction to lattices) inR24 cannot exceed the one given bythe Leech lattice by a factor of more than1 + 1.65 · 10−30.

d lattice density δd author(s)2 A2 0.9069 . . . Lagrange [Lag73], 17733 A3 = D3 0.7404 . . . Gauß [Gau40], 18404 D4 0.6168 . . . Korkine, Zolotareff [KZ73], 18735 D5 0.4652 . . . Korkine, Zolotareff [KZ77],18776 E6 0.3729 . . . Blichfeldt [Bli34], 19347 E7 0.2953 . . . Blichfeldt [Bli34], 19348 E8 0.2536 . . . Blichfeldt [Bli34], 1934

24 Λ 0.0019 . . . Cohn, Kumar [CK04], 2004

TABLE 1. Optimal lattice packings.

2.2. The Lattice Covering Problem. The lattice covering problem has only been solvedup to dimension5. Recently, we were able to verify the list of known results computation-ally. Even more, we found the complete list of222 local covering optima in dimension5.Table 2 invites to a question formulated by Ryshkov [Rys67], who asked for the lowestdimension in whichA∗

d gives not the thinnest lattice covering. In Section 5 we describe themethod used to find a latticeLc

6 ⊂ R6 with Θ(Lc6) = 2.4648 . . . < 2.5511 . . . = Θ(A∗

6).Thus the answer to Ryshkov’s question is6. It remains an open problem to prove that thelatticeLc

6 gives the best lattice covering in dimension6. We do not even know the exact

123

METHODS IN THE LOCAL THEORY OF PACKING AND COVERING LATTICES 3

coordinates of the lattice yet. Currently, a complete solution in dimensiond ≥ 6 seems outof reach without completely new methods.

d lattice densityΘd author(s)2 A∗

2 1.2091 . . . Kershner [Ker39], 19393 A∗

3 1.4635 . . . Bambah [Bam54], 19544 A∗

4 1.7655 . . . Delone, Ryshkov [DR63], 19635 A∗

5 2.1242 . . . Ryshkov, Baranovskii [RB75], 1975

TABLE 2. Optimal lattice coverings.

2.3. The Lattice Packing-Covering Problem. As in the case of the lattice covering prob-lem the lattice packing-covering problem has been solved only for dimensionsd ≤ 5.And, as in the covering case, we recently were able to verify these results computation-ally. Moreover, we found a new best known latticeLpc

6 in dimension6, having a slightlylower packing-covering constantγ(Lpc

6 ) = 1.4110 . . . than the previously best known oneγ(E∗6) =

√2.

One reason for studying the lattice packing-covering problem is the open question whetherthere exists a dimensiond with γd ≥ 2. If so, then anyd-dimensional lattice packing withspheres would leave space large enough for spheres of the same radius. This would inparticular prove that densest sphere packings in dimensiond are non-lattice packings. Thisphenomenon is likely to be true for large dimensions, but has not been verified for anyd sofar.

d lattice densityγd author(s)2 A∗

2 1.1547 . . . Ryshkov [Rys74], 19743 A∗

3 1.2909 . . . Ryshkov [Rys74], 19744 H4 1.3625 . . . Horvath [Hor82], 19805 H5 1.4494 . . . Horvath [Hor86], 1986

TABLE 3. Optimal lattice packing-coverings.

3. LATTICES AND POSITIVE QUADRATIC FORMS

It is sometimes convenient to switch from the language of lattices to the language ofpositive definite quadratic forms (PQFs from now on). In this section we give a dictionary.For further reading we refer to [CS88b] and [SV04a].

Given ad-dimensional latticeL = AZd with basisA we associate ad-dimensional PQFQ[x] = xtAtAx = xtGx, where theGram matrixG = AtA is symmetric and positivedefinite. We will carelessly identify quadratic forms with symmetric matrices by sayingQ = G andQ[x] = xtQx. The set of quadratic forms is a

(d+12

)-dimensional real vector

spaceSd, in which the set of PQFs forms an open, convex coneSd>0. The PQFQ depends

on the chosen basisA of L. For two arbitrary basesA andB of L there exists aU ∈ GLd(Z)with A = BU . Thus,GLd(Z) acts onSd

>0 by Q 7→ U tQU . A PQFQ can be associatedto different latticesL = AZd andL′ = A′Zd. In this case there exists an orthogonaltransformationO with A = OA′. Note that the packing and covering density, as well asthe packing-covering constant, are invariant with respect to orthogonal transformations.

Thedeterminant(or discriminant) of a PQFQ is defined bydet(Q). Thehomogeneousminimumλ(Q) and theinhomogeneous minimumµ(Q) are given by

λ(Q) = minv∈Zd\0

Q[v], µ(Q) = maxx∈Rd

minv∈Zd

Q[x− v].

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4 A. SCHURMANN AND F. VALLENTIN

If Q is associated toL, thendet(L) =√

det(Q), µ(L) =√

µ(Q), λ(L) =√

λ(Q)/2.Using this dictionary we define

δ(Q) =

√λ(Q)d

detQ· κd

2d, Θ(Q) =

õ(Q)d

detQ· κd and γ(Q) = 2

√µ(Q)λ(Q)

.

We say that a latticeL with associated PQFQ gives alocally optimal lattice packing,locally optimal lattice coveringor locally optimal lattice packing-covering, if there is aneighborhood ofQ in Sd

>0, so that we haveδ(Q) ≥ δ(Q′), Θ(Q) ≤ Θ(Q′) or γ(Q) ≤γ(Q′) respectively, for allQ′ in this neighborhood.

4. ON PACKING LATTICES

A PQFQ attaining a local maximum ofδ(Q) is calledextreme. A PQF attainingδd =maxQ∈Sd

>0δ(Q) is calledabsolutely extremeor critical. One can characterize an extreme

PQF using the geometry of its minimal vectors

Min(Q) = v ∈ Zd : Q[v] = λ(Q).Before we state the characterization in Theorem 4.1, we give some more definitions. APQFQ′ is calledperfectif it is uniquely determined by its minimal vectors, i.e.Q′ is theunique solution of the linear equationsQ[v] = λ(Q′), v ∈ Min(Q′). A PQFQ is calledeutacticif

Q−1 ∈ relint conevvt : v ∈ Min(Q) =

∑v∈Min(Q)

λvvvt : λv > 0,v ∈ Min(Q)

.

The eutaxy and the polyhedral coneconevvt : v ∈ Min(Q) also plays an importantrole in the lattice packing-covering problem. As a general reference on basic facts aboutpolyhedral cones, which are used throughout this article, we refer to the book of Ziegler[Zie97].

Theorem 4.1(Voronoi [Vor07]). A PQF is extreme if and only if it is perfect and eutactic.

This provides an easy way for proving that a given PQF is extreme: after finding theminimal vectors, one has to solve a system of linear equations to show its perfectness.Then, one has to solve a linear programming problem to verify its eutaxy. By scaling wecan normalize an extreme PQFQ so thatλ(Q) is rational. Then, sinceQ is perfect, thematrix entries ofQ are rational as well.

It turns out that there exist only finitely many pairwise non-equivalent perfect PQFsin Sd

>0. We want to describe two classical algorithms to attain all perfect forms of a givendimensiond. The first one goes back to Minkowski, the second one is due to Voronoi. Herewe only state definitions and main results. Additionally, we briefly sketch the computationswhich we were able to perform in low dimensions. We compare them with correspondingresults in the literature.

For history and further remarks we refer to [GL87], [RB79], [vdW56], [Mar03] and toreferences therein. Chapter§v of Gruber and Lekkerkerker’s book [GL87] gives a compre-hensive survey about history, results and literature of the reduction theory of PQFs. Thearticle [RB79] introduces to methods for studying the geometry of PQFs and contains manyproofs. Van der Waerden’s paper [vdW56] is a classic resource for Minkowski’s approach.The recent book [Mar03] of Martinet gives a contemporary view on Voronoi’s approachand on possible generalizations.

4.1. Minkowski’s Approach.

Definition 4.2. A PQFQ = (qij) ∈ Sd>0 is calledMinkowski reducedif

(i) Q[x] ≥ qii whenevergcd(xi, . . . ,xd) = 1, i = 1, . . . , d,(ii) qi,i+1 ≥ 0, i = 1, . . . , d− 1.

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METHODS IN THE LOCAL THEORY OF PACKING AND COVERING LATTICES 5

Every PQF is equivalent to a Minkowski reduced PQF. The following procedure, whichis nothing but an algorithmic interpretation of the definition, finds a Minkowski reducedPQF equivalent to a given PQFQ. Choose a minimal vectorv1 of Q. Then, choose amongall vectors inZd, which can complementv1 to a lattice basis ofZd, a vectorv2 for whichthe valueQ[v2] is minimal. Using this greedy strategy we get a basisV = (v1, . . . ,vd) ∈GLd(Z). Now we choose signs forvi so thatvt

i−1Qvi ≥ 0. Hence,V tQV is Minkowskireduced.

The set of Minkowski reduced PQFs forms an unbounded cone inSd≥0 which is defined

by the linear inequalities (i) and (ii). ByM we denote the cone defined by the linearinequalities (i) and byM+ the one which is defined by the linear inequalities (i) and (ii).Minkowski [Min05] showed thatM, and henceM+, is a polyhedral cone, i.e. that finitelymany inequalities (i) imply all others. He showed that every extreme PQF is equivalentto a PQF lying on a ray (a one-dimensional face) ofM+. Ryshkov [Rys70] proved thatevery perfect PQF is equivalent to a PQF lying on a ray ofM+. On the other hand Cohn,Lomakina and Ryshkov [CLR82] found a ray ofM+ ⊂ S5

≥0 which contains non-perfectPQFs.

Minkowski [Min87] gave a list of conditions implying all others in (i) up to dimension6.Tammela [Tam81] enlarged this list to dimension7. Besides thed − 1 inequalitiesq11 ≤. . . ≤ qdd, the linear conditions forM, d = 2, . . . , 7, are attained by plugging the valuesx = (x1, . . . , xd) ∈ Zd from Table 4 into (i), where the indicesi1, . . . , id run through allpermutations of1, . . . , d. If d < 7 one has to omit the columnsd+1, . . . , 7 and the rowswith more thand non-zero entries.

xi1 ±xi2 ±xi3 ±xi4 ±xi5 ±xi6 ±xi7

1 1 0 0 0 0 01 1 1 0 0 0 01 1 1 1 0 0 01 1 1 1 1 0 02 1 1 1 1 0 01 1 1 1 1 1 02 1 1 1 1 1 02 2 1 1 1 1 03 2 1 1 1 1 0 1

1 1 1 1 1 1 12 1 1 1 1 1 13 1 1 1 1 1 12 2 1 1 1 1 13 2 1 1 1 1 12 2 2 1 1 1 13 2 2 1 1 1 14 2 2 1 1 1 13 3 2 1 1 1 1 2

4 3 2 1 1 1 1 3

3 2 2 2 1 1 14 3 2 2 1 1 1

TABLE 4. Tammela’s list of linear inequalities definingM

We checked these conditions for redundancy using the softwarelrs of Avis [Avi04].Our computation shows that in row1 the entries withx1 6= 3 (already mentioned by Tam-mela in [Tam73]), in row2 the entries withx1 6= 3 or x2 6= 3, and in row3 the entries withx1 6= 4 orx2 6= 3 are redundant. The remaining conditions are all non-redundant and define

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6 A. SCHURMANN AND F. VALLENTIN

a facet ofM. In Table 5 we list the number of facets and rays as far as we were able to com-pute them withcdd [Fuk03]. We hereby confirm earlier results bya Minkowski [Min87]and b Barnes and Cohn [BC76]. We made the data available from the arXiv.org e-printarchive. To access it, download the source files for the paper arXiv:math.MG/0412320. Thefilesmink2.ine , . . . ,mink6.ine (due to its size the filemink7.ine is only availablefrom the authors) contain the facets ofM, the filesmink2.ext , . . . , mink6.ext con-tain the rays ofM, the filesminkp2.ine , . . . ,minkp6.ine contain the facets ofM+,and the filesminkp2.ext , . . . , minkp5.ext contain the rays ofM+. For the dataformat we chose the common convention (PolyhedraH-format for the*.ine -files andPolyhedraV -format for the*.ext -files) of the software packagescdd andlrs .

The computational bottlenecks of Minkowski’s approach are apparent. It is not easyto find a sufficiently small system of linear inequalities definingM (or ofM+). Even ifone has a minimal system of linear inequalities, then computing its rays is a very difficultcomputational problem in higher dimensions.

d(d+12

)# FacetsM # RaysM # FacetsM+ # RaysM+

2 3 3a 3a 3a 3a

3 6 12a 19b 9a 11b

4 10 39a 323b 26b 109b

5 15 200 15971 122 39796 21 1675 ? 1084 ?7 28 65684 ? ? ?

TABLE 5. Known number of facets and rays ofM andM+.

4.2. Voronoi’s Approach. Now we describe Voronoi’s algorithm [Vor07] for finding allperfect forms of a given dimension. Letm be a positive number. In the remaining of thissection we assume that every perfect formQ is scaled so thatλ(Q) = m. The set

Pm =Q ∈ Sd

>0 : Q[v] ≥ m for all v ∈ Zd \ 0

is a convex, locally finite polyhedral cone. Its boundary consists of the PQFs with homo-geneous minimumm. A PQFQ is perfect if and only if it is a vertex ofPm. The set ofperfect PQFs of a given dimensiond naturally carries a graph structure which we denoteas theVoronoi graph in dimensiond: Two perfect PQFsQ, Q′ are connected by an edge ifthe line segmentconvQ,Q′ (convex hull ofQ andQ′) is an edge ofPm. In this case wesay thatQ andQ′ areVoronoi neighbors. The group GLd(Z) acts onPm, on its verticesand on its edges byQ 7→ U tQU . Therefore, one can enumerate perfect PQFs by a graphtraversal algorithm which we shall now describe.

Voronoi’s first perfect formQ[x] =∑d

i=1 x2i +

∑i<j xixj , which is associated to

the root latticeAd, can serve as a starting point in any dimension. For implementing agraph traversal algorithm one has to find the Voronoi neighbors of a given perfect formQ.Consider the unbounded polyhedral cone

P(Q) = Q′ ∈ Sd : Q′[v] ≥ m,v ∈ Min(Q).

We compute the raysQ + R≥0Ri, i = 1, . . . , n of P(Q). TheRi turn out to be indefinitequadratic forms. So there arev ∈ Zd with Ri[v] < 0. Then, the Voronoi neighbors ofQ areQ + ρiRi whereρi is the smallest positive number so thatλ(Q + ρiRi) = m andMin(Q + ρiRi) 6⊆ Min(Q). It is possible to determineρi, for example with the followingprocedure:

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METHODS IN THE LOCAL THEORY OF PACKING AND COVERING LATTICES 7

(α, β)← (0, 1)while Q + βRi 6∈ Sd

>0 or λ(Q + βRi) = m doif Q + βRi 6∈ Sd

>0 then β ← α+β2

else(α, β)← (β, 2β)end if

end whilewhile Min(Q + αRi) ⊆ Min Q do

γ ← α+β2

if λ(Q + γRi) < m then β ← γelseα← γend if

end whileρi ← α

Perfect forms were classified up to dimension7. A list of all these forms is given in thepaper [CS88a] of Conway and Sloane. One can find an electronic version in the Catalogueof Lattices1 by Nebe and Sloane. On his homepage2 , Martinet reports that up to now,10916 pairwise inequivalent perfect forms are known in dimension8 and lists them.

We verified the results for dimensionsd ≤ 6 using the programslrs by Avis [Avi04],isom by Plesken and Souvignier [PS97] andshvec by Vallentin [Val99]. In Table 6 wegive the known classifications of perfect forms, extreme forms and absolute extreme formstogether with the references where the classifications were established.

d(d+12

)# perfect # extreme # critical

2 3 1a 1a 1a

3 6 1b 1b 1b

4 10 2c 2c 1c

5 15 3d 3d 1d

6 21 7f 6e 1f

7 28 33h 30h 1g

8 36 ≥ 10916i ? 1g

9 45 > 500, 000 ? ?

24 300 ? ? 1j

a Lagrange [Lag73]b Gauß [Gau40]c Korkine, Zolotareff [KZ73]d Korkine, Zolotareff [KZ77]e Hofreiter [Hof33]f Barnes [Bar57]g Vetchinkin [Vet82]h Jaquet-Chiffelle [JC93]i Laihem, Baril, Napias, Batut, Martinet (see http://www.math.u-bordeaux.fr/˜martinet)j Cohn, Kumar [CK04]

TABLE 6. Classifications of perfect, extreme and absolute extreme forms

The computational bottleneck of Voronoi’s approach is mainly the enumeration of allrays of the polyhedral coneP(Q) in case of a large setMin(Q) of minimal vectors. Mar-tinet writes in [Mar03], Ch. 7.11 : ”The existence ofE8 [...] makes hopeless any attempt to

1 http://www.research.att.com/˜njas/lattices/perfect.html2 http://www.math.u-bordeaux.fr/˜martinet

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8 A. SCHURMANN AND F. VALLENTIN

construct the Voronoi graph in dimension8”. Another problem is the combinatorial explo-sion, whend ≥ 9. We found more than500000 inequivalent perfect forms in dimension9and we strongly believe there exist millions of them.

Finally, we want to remind of Coxeter’sAd-hypothesis: Although finding perfect formswith maximal packing density is a very difficult problem, finding perfect forms with min-imal packing density might be very easy. In [Cox51] Coxeter formulates the followingconjecture:

Conjecture 4.3. (Coxeter’sAd-hypothesis) Voronoi’s first perfect form gives the minimalpacking density among all perfect forms of a given dimension.

Our computations support Coxeter’s conjecture. But on the contrary, Conway andSloane [CS88a] conjecture that it is false for sufficiently larged.

5. COVERING AND PACKING-COVERING LATTICES

The lattice covering problem and the lattice packing-covering problem can be treated inparallel. We describe below that both problems have only finitely many local optima whichcan be found by solving finitely many convex optimization problems. This is mainly dueto Voronoi’s theory of Delone subdivisons, which we briefly review. For a detailed accountwe refer to [SV04a].

With an implementation of the proposed algorithms we found all local optima in dimen-siond ≤ 5 and some new best known lattices in dimensiond ≥ 6.

For both problems, recognition of local optima is not as easy as for the lattice packingproblem. Due to the involved convexity we can give sufficient conditions for local optima,allowing to compute a certificate for the local optimality of a lattice. This is in particularapplicable, if the Delone subdivison is a triangulation which is the generic case (for defini-tions see below). Exemplarily we give a proof of the local packing-covering optimality ofthe latticesA∗

d.In some cases it is possible to attain good or even tight “local lower bounds” for the

lattice covering density and the packing-covering constant. This is demonstrated for thelocal packing-covering optimality of the Leech lattice. A similar proof of the local coveringoptimality of the Leech lattice is given in [SV04b].

5.1. Voronoi’s Theory of Delone Subdivisions.Let Q be a positive semidefinite qua-dratic form. A polyhedronP = convv1,v2, . . .with v1,v2, . . . ∈ Zd, is called aDelonepolyhedronof Q if there exists ac ∈ Rd and a real numberr ∈ R with Q[vi − c] = r2 forall i = 1, 2, . . ., andQ[v − c] > r2 for all otherv ∈ Zd \ v1,v2, . . .. The setDel(Q)of all Delone polyhedra is called theDelone subdivisionof Q. It is a periodic face-to-facetiling of Rd. ThereforeDel(Q) is completely determined by all Delone polytopes havinga vertex at the origin0. We call two Delone polyhedraL,L′ equivalentif there exists av ∈ Zd so thatL = v ± L′. Note moreover that the inhomogeneous minimumµ(Q) isat the same time the maximum squared circumradius of its Delone polyhedra. We say thatthe Delone subdivision of a positive semidefinite quadratic formQ′ is a refinementof theDelone subdivision ofQ, if every Delone polytope ofQ′ is contained in a Delone polytopeof Q.

By a theory of Voronoi [Vor08], the set of positive semidefinite quadratic forms with afixed Delone subdivisionD is an open (with respect to its affine hull) polyhedral cone inSd≥0. We refer to this set as thesecondary cone∆(D) of the subdivision. In the literature

the secondary cone is sometimes calledL-type domain of the subdivision. The topologicalclosure∆(D) of a secondary cone is a closed polyhedral cone. The relative interior of eachface inSd

>0 is the secondary cone of another Delone subdivision. If a face is contained inthe boundary of a second face, then the corresponding Delone subdivision of the first is atrue refinement of the second one.

The interior of faces of maximal dimension(d+12

)contain PQFs whose Delone subdivi-

sion is a triangulation, that is, it consists of simplices only. We refer to such a subdivision

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METHODS IN THE LOCAL THEORY OF PACKING AND COVERING LATTICES 9

as asimplicial Delone subdivision orDelone triangulation. As mentioned in Section 3,the groupGLd(Z) acts onSd

≥0. One of the key observations of Voronoi is that under thisgroup action there exist only finitely many inequivalent Delone subdivisions, respectivelysecondary cones.

Theorem 5.1(Voronoi [Vor08]). The topological closures of secondary cones of Delonetriangulations give a face-to-face tiling ofSd

≥0. The groupGLd(Z) acts on the tiling, andunder this group action there are only finitely many non-equivalent secondary cones.

Given a Delone triangulationD, the Delone triangulationsD′ with ∆(D′) sharing afacet with∆(D) are attained bybistellar operations(flips). These change a triangulationonly in certainrepartitioning polytopesassociated to the facet. By this operation it becomespossible to enumerate all Delone triangulations, and hence all Delone subdivisions in agiven dimension. For details we refer to [SV04a].

5.2. Obtaining Local Optima via Convex Optimization. For a fixed triangulationD, wecan formulate the lattice covering, as well as the lattice packing-covering problem in theframework ofDeterminant Maximization Problems. Following Vandenberghe, Boyd, andWu [VBW98] their general form is

(1)minimize ctx− log det G(x)subject to G(x) 0, F (x) 0.

Here, the optimization vector isx ∈ RD. The objective function contains a linear partgiven byc ∈ RD andG : RD → Rm×m, F : RD → Rn×n are both affine maps

G(x) = G0 + x1G1 + · · ·+ xDGD,F (x) = F0 + x1F1 + · · ·+ xDFD,

whereGi ∈ Rm×m, Fi ∈ Rn×n, i = 0, . . . , D, are symmetric matrices. The notationG(x) 0 andF (x) 0 gives the constraints “G(x) is positive definite” and “F (x) ispositive semidefinite”. Note that we are dealing with a so-calledsemidefinite programmingproblem, if G(x) is the identity matrix for allx ∈ RD.

For the lattice covering, as well as the lattice packing-covering problem, we can expressµ(Q) ≤ 1 as alinear matrix inequalityF (qij) 0 with optimization vectorQ = (qij). Tosee this, it is crucial to observe that an inner product(·, ·) defined by(x,y) = xtQy givesa linear expression in the parameters(qij) for any fixed choicex,y ∈ Zd (or Rd). Delone,Dolbilin, Ryshkov and Stogrin [DDRS70] showed

Proposition 5.2. Let L = conv0,v1, . . . ,vd ⊆ Rd be ad-dimensional simplex. ThenL’s circumradius is at most1 with respect to(·, ·) if and only if

BRL(Q) =

4 (v1,v1) (v2,v2) . . . (vd,vd)

(v1,v1) (v1,v1) (v1,v2) . . . (v1,vd)(v2,v2) (v2,v1) (v2,v2) . . . (v2,vd)

......

......

...(vd,vd) (vd,v1) (vd,v2) . . . (vd,vd)

0.

Since a block matrix is semidefinite if and only if the blocks are semidefinite, we con-clude

Proposition 5.3. Let Q = (qij) ∈ Sd>0 be a PQF. LetD be a Delone triangulation refining

Del(Q), and letL1, . . . , Ln be a representative system of all non-equivalentd-dimensionalDelone polytopes inD. Then

µ(Q) ≤ 1 ⇐⇒

BRL1(Q) 0 0 . . . 0

0 BRL2(Q) 0 . . . 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

0 0 0 . . . BRLn(Q)

0.

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10 A. SCHURMANN AND F. VALLENTIN

Thusµ(Q) ≤ 1 can be brought into one linear matrix inequality of typeF (qij) 0.We can moreover add linear constraints on the parametersqij by extendingF by a1 × 1block matrix for each linear inequality. In this way we can get one linear matrix inequalityfor the two constraintsµ(Q) ≤ 1 andQ ∈∆(D).

For a fixed Delone triangulationD, we can therefore determine the optimal solutions ofthe lattice covering problem of all PQFsQ for whichD is a refinement ofDel(Q). Recall

that the covering density of a PQFQ in d variables isΘ(Q) =√

µ(Q)d

det Q · κd. Scaling of

Q by a positive real numberα leavesΘ invariant. Thus we may maximizedet(Q) whileµ(Q) ≤ 1. For allQ ∈∆(D) this can be achieved by solving

minimize − log det(Q)subject to Q 0,

Q ∈∆(D), µ(Q) ≤ 1.

With an analogues specialization of problem (1), we are able to attain optimal solutionsof the lattice packing-covering problem among all PQFsQ for whichD is a refinementof Del(Q). Becauseγ(Q) = 2 ·

√µ(Q)/λ(Q), we have to maximizeλ(Q) while µ(Q) ≤

1. Maximizingλ(Q) is not as straightforward as maximizing the determinant, since we donot know which vector is the shortest. By a theorem of Voronoi [Vor08] we know thoughthat among the (at most2(2d − 1)) edges[0,v] = conv0,v ∈ D, there exist some withv ∈ Min(Q). Consequently, if we requireQ[v] ≤ Q[w] for all w with [0,w] ∈ D, weknow v ∈ Min(Q), respectivelyλ(Q) = Q[v]. So we can solve for each[0,v] ∈ D thesemidefinite programming problem

minimize −Q[v]subject to Q[v] ≤ Q[w], for all w with [0,w] ∈ D,

Q ∈∆(D), µ(Q) ≤ 1.

Note that in many cases the constraints have no feasible solutions, since in general notall of thev with [0,v] ∈ D are elements ofMin(Q).

So both, the lattice covering as well as the lattice packing-covering problem, are reducedto convex programming problems if restricted to the closure of a secondary cone. Conse-quently, there is at most one local minimum ofΘ, respectivelyγ, for each of the cones∆(D). This was first observed by Barnes and Dickson [BD67] in the covering case and byRyshkov [Rys74] for the packing-covering problem.

Proposition 5.4. LetD be a Delone triangulation. Then there exists a unique minimum of

(1) Θ(Q) : Q ∈ ∆(D) and the set of PQFs attaining it is equal to all positivemultiples of a single PQF.

(2) γ(Q) : Q ∈∆(D) and the set of PQFs attaining it is convex.

In case of a triangulationD, it follows thatQ ∈∆(D) is a locally optimal solution withrespect toΘ or γ if and only if it is an optimal solution within∆(D). In general we havethe trivial

Proposition 5.5. A PQFQ is a locally optimal solution with respect toΘ or γ, if and onlyif it is an optimal solution for all Delone triangulationsD with Q ∈∆(D).

5.3. Computational Results. By the foregoing propositions we know that the number oflocal optima is bounded from above by the number of pairwise inequivalent Delone tri-angulations inRd. Voronoi [Vor08] classified these triangulation in dimension2, 3 (onlyone each) and4 (three). By the work of Baranovskii and Ryshkov [BR73], [RB78] En-gel [Eng98], and Engel and Grishukhin [EG02] we know of exactly222 Delone trian-gulations in dimensiond = 5. Using lrs [Avi04] and an implementation (in C++) ofVoronoi’s algorithm for enumerating Delone triangulations, we were able to confirm theseresults [SV04a]. For dimensiond ≥ 6 we experience a combinatorial explosion, e.g. Engel

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METHODS IN THE LOCAL THEORY OF PACKING AND COVERING LATTICES 11

[Eng04] reports on more than2, 129, 120 pairwise inequivalent Delone triangulations ford = 6.

d(d+12

)# covering optima # packing-covering optima

2 3 1a 1b

3 6 1c 1b

4 10 3d 3e

5 15 222f 47f

a Kershner [Ker39]b Ryshkov [Rys74]c Bambah [Bam54]d Baranovski [Bar65], Dickson [Dic67]e Horvath [Hor82]f Schurmann, Vallentin [SV04a]

TABLE 7. Classifications of locally optimal covering and packing-covering lattices.

For each of the triangulations in dimensiond ≤ 5 we determined the local optima withrespect toΘ andγ using the software packageMAXDET3 of Wu, Vandenberghe, and Boydas a subroutine. By this we confirmed the known results for dimensionsd = 2, 3, 4 andextended them tod = 5 (see Table 7). Note that for the lattice covering problem there existsa local optimum for each triangulation, while this is not the case in higher dimensions (see[SV04b]) and for the lattice packing-covering problem.

Using our implementation we also found two lattices which currently give the bestknown covering and packing-covering in dimension6:

Theorem 5.6([SV04a]). In dimension6, there exits a latticeLc6 with Θ(Lc

6) = 2.4648 . . .and a latticeLpc

6 with γ(Lpc6 ) = 1.4110 . . ..

In [SV04b] we show that the root latticeE8 does not give a locally optimal lattice cov-ering, by constructing a refining triangulationD of Del(QE8) in whichΘ’s local optimumis not attained by the PQFQE8 . The PQF found in this way even beats the formerly bestknown valueΘ(A∗

8) by more than12%. By looking at a bistellar neighbor of the triangula-tionD, we found the currently best known covering lattice in dimension8.

Theorem 5.7([SV04b]). In dimension8, there exists a latticeLc8 with Θ(Lc

8) = 3.1423 . . ..

Looking at the results in dimensiond = 6, 8 it is interesting to observe that we found thenew covering lattices by looking at triangulations refining the Delone subdivisions of thelatticesE∗d. By looking at a corresponding refinement ofE∗7, we also found a new coveringrecord in dimension7. It remains to see if these results have a common explanation. . .

5.4. Sufficient Conditions for Local Optima. A disadvantage of finding local optima viaconvex programming is that solutions can only be approximated. But this is an inherentproblem: In contrast to the lattice packing problem, local optima to the other two problemscan in general not be represented by rational numbers. One has to use algebraic num-bers instead. In some cases it might be possible, e.g. with additional information on theautomorphism group, to attain exact coordinates from a first approximation (see [SV04a]for an example). In other cases we might have a conjectured optimal formQ′ and wantto compute a “certificate” verifying its local optimality. The following two propositionsgive such a criterion in terms of the gradientgL(Q′) = grad |BRL |(Q′) of the regularsurfaces|BRL(Q)| = 0 at Q′. Both are a consequence of the geometric fact that we

3http://www.stanford.edu/˜boyd/MAXDET.html

132

12 A. SCHURMANN AND F. VALLENTIN

have a local optimum atQ′ ∈ ∆(D) with µ(Q′) = 1 if and only if there exists a hyper-plane throughQ′, separating the convex setsQ ∈ Sd

>0 : L ∈ D, |BRL(Q)| ≥ 0 andQ ∈ Sd

>0 : det(Q) ≥ det(Q′), respectivelyQ ∈ Sd>0 : λ(Q) ≥ λ(Q′).

Proposition 5.8 (Barnes and Dickson [BD67]). Let D be a Delone triangulation. ThenQ ∈ ∆(D) with µ(Q) = 1 is a unique locally optimal solution to the lattice coveringproblem if and only if

Q−1 ∈ − conegL(Q) : L ∈ D with |BRL(Q)| = 0.

The corresponding result for the lattice packing-covering problem is

Proposition 5.9 ([SV04a]). Let D be a Delone triangulation. ThenQ ∈ ∆(D) withµ(Q) = 1 is a locally optimal solution to the lattice packing-covering problem if and onlyif

conevvt : v ∈ Min(Q) ∩ − conegL(Q) : L ∈ D with |BRL(Q)| = 0 6= ∅.

Combining these two propositions we get

Corollary 5.10. Let D be a Delone triangulation. ThenQ ∈ ∆(D) is a unique locallyoptimal solution to the lattice packing-covering problem, ifQ is eutactic and a locallyoptimal solution to the lattice covering problem.

Example 5.11. We can use Corollary 5.10 to show thatA∗d, d ≥ 2, with γ(A∗

d) =√

d+23

is locally optimal for the lattice packing-covering problem. Ryshkov [Rys74] gave anotherproof of this fact. The latticeA∗

d is known to give a locally optimal lattice covering (see[Gam62], [Gam63], [Ble62]). A quadratic formQA∗

dassociated withA∗

d is

QA∗d

=

d −1 · · · −1

−1...

......

......

... −1−1 · · · −1 d

with µ(QA∗

d) = d(d+2)

12 andλ(Q) = d. The setMin(QA∗d) contains exactly2(d + 1)

elements, namely the standard basis vectorse1, . . . ,ed, their negatives and±∑d

i=1 ei

(see [CS88b]). Thus in particular

(d + 1) ·Q−1A∗

d= QAd

=

2 1 · · · 1

1...

......

......

... 11 · · · 1 2

=∑

vvt:v∈Min(QA∗d)

vvt

Hence,QA∗d

is eutactic becauseQ−1A∗

d∈ relint conevvt : v ∈ Min(QA∗

d) and therefore

the assertion follows.

Propositions 5.8 and 5.9 assume thatD = Del(Q) is a Delone triangulation. If this isnot the case, the situation becomes more complicated, in particular for the lattice packing-covering problem.

For the covering problem we only have to add a condition on the set

VD =⋃

D′<DQ ∈∆(D′) : |BRL(Q)| ≥ 0 for all L ∈ D′.

whereD′ < D denotes thatD′ is a Delone triangulations refiningD. This set is a subsetof Q ∈ Sd

>0 : µ(Q) ≤ 1. We require thatVD is separable atQ, that is, there existsa supporting hyperplane ofVD throughQ. This is in particular the case, if there exists a

smallr > 0 such that(Q + rB(d+12 )) ∩ VD is convex.

Proposition 5.12. LetD be a Delone subdivision andQ ∈∆(D) with µ(Q) = 1. Then

133

METHODS IN THE LOCAL THEORY OF PACKING AND COVERING LATTICES 13

(1) Q is a locally optimal solution to the lattice covering problem, if and only ifVD isseparable atQ and

Q−1 ∈ − conegL(Q) : L ∈ D′ < D with |BRL(Q)| = 0.

(2) Q is a locally optimal solution to the lattice packing-covering problem, ifVD isseparable atQ and

conevvt : v ∈ Min(Q) ∩ − conegL(Q) : L ∈ D′ < D with |BRL(Q)| = 0 6= ∅.

In case of the lattice packing-covering problem the “only if” part is missing, becausewe can not exclude the case of a locally optimal solutionQ′with VD not being separable atQ′. This is due to the fact thatQ ∈ Sd

>0 : λ(Q) ≥ λ(Q′) is not smooth in contrast toQ ∈ Sd

>0 : det(Q) ≥ det(Q′).This phenomenon seems to happen to PQFs associated to the root latticeE8. This lattice

is known to give a globally optimal solution to the lattice packing problem, but not a locallyoptimal solution to the lattice covering problem (see [SV04b]). Nevertheless computationalexperiments support the

Conjecture 5.13.The root latticeE8 gives a locally optimal solution for the lattice packing-covering problem.

Zong [Zon02] even conjectured thatE8 gives the unique globally optimal solution to thelattice packing-covering problem in dimension8.

5.5. Local Optima via Local Lower Bounds. In [SV04a] we describe a way to attainlocal lower bounds for the covering density and the packing-covering constant due toRyshkov and Delone. A variant of this method is successfully used in [SV04b] to provethe local covering optimality of the Leech lattice. Here we describe a corresponding locallower bound for the lattice packing-covering problem. As an example we use it to provethe local packing-covering optimality of the Leech lattice directly.

Proposition 5.14. Let L1, . . . , Ln be a collection of Delone simplices of a PQFQ. Then

γ(Q) ≥ 2

√trace(F ·QF )(d + 1)λ(QF )

with the PQFF = 1n(d+1)

∑i

∑k 6=l vi,kvt

i,l and a PQFQF with

F ∈ conevvt : v ∈ Min(QF ).

One can prove this Proposition by doing obvious modifications to the proof of Propo-sition 10.6 in [SV04a]. As in Proposition 5.9 we use the following fact: A linear functionf(Q) = trace(F · Q), with a PQFF , has a minimum on the homogeneous minimumλsurfaceQ ∈ Sd

>0 : λ(Q) = λ at QF if and only if F ∈ conevvt : v ∈ Min(QF ). Inparticular, ifQ is eutactic andF = Q−1, then Proposition 5.14 is immediately applicablewith QF = Q.

Example 5.15. We use Proposition 5.14 to show that the Leech lattice is a locally optimalpacking-covering lattice.

Let us briefly review some necessary properties of the Leech latticeΛ. For furtherreading we refer to [CS88b]. An associated PQFQΛ has (up to congruences)23 differentDelone polytopes attaining the maximum squared circumradiusµ(QΛ) = 2. One of themis the Delone simplexL of typeA24.

Now we apply Proposition 5.14 to the orbit ofL under the automorphism groupCo0 =T ∈ GL24(Z) : T tQΛT = QΛ of QΛ. We getF = 1

25|Co0 |∑

T∈Co0

∑e eet, wheree

runs through all the edge vectors ofTL. In [SV04b] it was shown thatF = 52

22·3Q−1Λ . Due

134

14 A. SCHURMANN AND F. VALLENTIN

to the fact thatMin QΛ is aspherical2-designwith respect to the inner product given byQΛ, we know (see [SV04b] for details)∑

v∈Min(QΛ)

vvt =|Min(QΛ)|

dQ−1

Λ .

Thus,QΛ is eutactic and we may useQF = QΛ in Proposition 5.14. Withλ(QΛ) = 4 wederive

γ(Q) ≥ 2

√52

22·3 · 2425 · 4

=√

2 = γ(QΛ)

for all PQFsQ with Delone simplicesTL, T ∈ Co0, which proofs the assertion.

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MATHEMATICS DEPARTMENT, UNIVERSITY OF MAGDEBURG, 39106 MAGDEBURG, GERMANY

E-mail address: achill@math.uni-magdeburg.de

EINSTEIN INSTITUTE OFMATHEMATICS, THE HEBREW UNIVERSITY OF JERUSALEM, ISRAEL

E-mail address: vallenti@ma.tum.de

136

Submitted to Experimental Mathematics

Packing and Minkowski covering of congruent spherical caps ona sphereT. Sugimoto1 and M. Tanemura2

1The Institute of Statistical Mathematics,4-6-7 Minami-Azabu, Minato-ku, Tokyo 106-8569, Japan.sugimoto@ism.ac.jp

2The Institute of Statistical Mathematics,Department of Statistical Science,The Graduate University for Advanced Studies,4-6-7 Minami-Azabu, Minato-ku, Tokyo 106-8569, Japan.tanemura@ism.ac.jp

Abstract

Let Ci ( i = 1; : : : ; N ) be the i-th open spherical cap of angular radius r and let Mi be its centerunder the condition that none of the spherical caps contains the center of another one in its interior.We consider the upper bound, rN , (not the lower bound !) of r for the case where the wholespherical surface of a unit sphere is completely covered with N congruent open spherical caps underthe condition, sequentially for i = 2; : : : ; N ¡ 1 , that Mi is set on the perimeter of Ci¡1, and thateach area of the set ([i¡1º=1Cº) \ Ci becomes maximum. In this study, for N = 2; : : : ; 12, we foundout that the solutions of the above covering and the solutions of Tammes problem were strictlycorrespondent. Especially, we succeeded to obtain the exact value r10 for N = 10.

1 Introduction\How must N congruent non-overlapping spherical caps be packed on the surface of a unit sphere sothat the angular diameter of spherical caps will be as great as possible?" This packing problem is alsocalled the Tammes problem and mathematically proved solutions have been known for N = 1; : : : ; 12;and 24 [Danzer 63], [Fejes T¶oth 69], [Fejes T¶oth 72], [SchÄutte and van der Waerden 51], [Teshima andOgawa 00]. On the other hand, the problem \How must the covering of a unit sphere be formed by Ncongruent spherical caps so that the angular radius of the spherical caps will be as small as possible?"is also important. It can be considered that this problem is dual to the problem of packing of Tammes[Fejes T¶oth 69]. Among the problems of packing and covering on the spherical surface, the Tammesproblem is the most famous. However, the systematic method of attaining these solutions has not beengiven.

In this paper, we would like to think of the covering in connection with the packing problem. Namely,we consider the covering of the spherical caps such that none of them contains the center of another onein its interior. Such a set of centers is said to be a Minkowski set [Fejes T¶oth 99]. Hereafter, we call thecondition of Minkowski set of centers \Minkowski condition." In addition, the covering which satis¯esthe Minkowski condition is called \Minkowski covering." If angular radii of spherical caps which coverthe unit sphere under the Minkowski condition are concentrically reduced to half, the resulting sphericalcaps do not overlap. Then, our purpose in this paper is to obtain the upper bound (not the lower bound!) of angular radius of spherical caps which cover the unit sphere under the Minkowski condition.

Suppose we have N congruent open spherical caps with angular radius r on the surface S of the unitsphere and suppose that these spherical caps cover the whole spherical surface without any gap underthe Minkowski condition. Further we suppose that the spherical caps are put on S sequentially in themanner which is described just below. Let Ci be the i-th open spherical cap and let Mi be its center(i = 1; : : : ; N). Our problem is to calculate the upper bound of r for the sequential covering, such thatN congruent open spherical caps cover the whole spherical surface S under the condition that Mi is seton the perimeter of Ci¡1, and that each area of set ([i¡1

º=1Cº) \ Ci becomes maximum in sequence fori = 2; : : : ; N ¡1 [Sugimoto and Tanemura 01a], [Sugimoto and Tanemura 01b], [Sugimoto and Tanemura02], [Sugimoto and Tanemura 03]. In this paper, we calculate the upper bound of r for N = 2; : : : ; 12theoretically; the case N = 1 is self-evident. It is shown in this paper that the solutions of our problemare strictly correspondent to those of the Tammes problem for N = 2; : : : ; 12. Especially, the exact valuefor N = 10 is obtained (see (21)). Further, it should be said that our method is a systematic and adi®erent approach to the Tammes problem from the works by SchÄutte and van der Waerden [SchÄutteand van der Waerden 51], etc.

1

137

In Section 2, to solve our problem, we consider the properties of spherical caps under the Minkowskicondition. Then, we explain the procedure of our sequential covering and de¯ne the upper bounds rN andrN¡1. In Section 3, we give the solutions of our problem successively for N = 2; : : : ; 12. Our conclusionis summarized in Section 4.

2 Sequential CoveringThroughout this paper we assume that the center of the unit sphere is the origin O = (0; 0; 0). Hereafter,we represent the surface of this unit sphere by the symbol S. In the following, open spherical caps aresimply written as spherical caps unless otherwise stated. We de¯ne the geodesic arc between an arbitrarypair of points Ti and Tj on S as the inferior arc of the great circle determined by Ti and Tj . Then thespherical distance between Ti and Tj is de¯ned by the length of geodesic arc of this pair of points.

2.1 Relationship between the Kissing Number and the Angular Radius ofSpherical Caps

First of all, we consider how many centers of congruent spherical caps can be placed on the perimeterof a spherical cap under the Minkowski condition. This problem is related to the kissing number ofspherical caps. In the plane, one circle can contact simultaneously with six other congruent circles. Thenthe kissing number is always six in the plane. On the sphere, on the contrary, the kissing number of acircle (spherical cap) on the spherical surface changes with its angular radius. Its maximum value is ¯veas we will see below. Here, we de¯ne a \half-cap" as the spherical cap whose angular radius is r

2 andwhich is concentric with that of the original cap. When the kissing number is k, there can k half-capscontact with the central half-cap and there is no space for another half-cap to enter. At this instance,let us increase r until the peripheries of k half-caps contact with one another. Then, if the centers of twohalf-caps in contact are joined by a geodesic arc, there arise k spherical equilateral triangles of side-lengthr and of inner angle ¾ = 2¼

k around the central half-cap. Thus by applying the spherical cosine theoremto one of the spherical equilateral triangles, we have

r = cos¡1µ

cos¾1 ¡ cos ¾

¶. (1)

By substituting ¾ = 2¼k into (1), we can calculate the upper bound value of r for a given kissing

number k (> 1). Note that, for k = 1, it is clear that the upper bound value is ¼. For k ¸ 6, (1) cannothave a solution. Namely, for k = 6, we get at once r = cos¡1 1 = 0, and for k > 6, we get the inequalitycos ¾=(1 ¡ cos ¾) > 1 indicating that no real value solution exists for r. Therefore the maximum value ofthe kissing number of a spherical cap (circle on the spherical surface) is ¯ve. As a result, we obtain therelationship between k and r as is shown in Table 1. For example, from Table 1, the kissing number kis four when the angular diameter of a spherical cap is in the range

¡tan¡1 2; ¼=2

¤. Note that inclusion

of the upper bound in the range of r in Table 1 is correspondent to the Minkowski condition.Next, let us consider the packing with half-caps and the covering with corresponding spherical caps

(of angular radius r) where each half-cap is concentric with correspondent spherical caps. Then it isobserved that this covering satis¯es the Minkowski condition; namely, this is the Minkowski covering.For the covering that the centers of spherical caps are chosen on the perimeters of other spherical capsunder the Minkowski condition, we see from Table 1 that, for example, four spherical caps can be placedon the perimeter of a spherical cap when angular radius is in the range

¡tan¡1 2; ¼

2

¤. We note that the

discussions of this Subsection are valid for both of open and closed spherical caps.

2.2 Overlapping Area of Congruent Spherical CapsIn order to solve our problem, we consider the overlapping area of congruent spherical caps under theMinkowski condition.

Assume Ci and Cj be two congruent spherical caps, of angular radius r, which are mutually overlap-ping under the Minkowski condition, and let Aij = A(Ci \ Cj) be the overlapping area where A(X) isthe area of X. Further, when Ti and Tj are the points on S, let ds(Ti; Tj) denote the spherical distancebetween points Ti and Tj . Especially, we denote sij = ds(Mi; Mj) as the spherical distance betweencenters of Ci and Cj (r · sij · 2r). Then let hij be the spherical distance between cross points ofperimeters of Ci and Cj . By using the spherical cosine formula for a spherical right triangle, we have

2

138

Table 1: Relationship Between k and r.

k Range of r

1¡ 2¼

3 ; ¼¤

2¡¼ ¡ cos¡1

¡ 13

¢; 2¼

3

¤

3¡¼

2 ; ¼ ¡ cos¡1¡ 1

3

¢¤

4¡tan¡1 2; ¼

2

¤

5¡0; tan¡1 2

¤

hij = 2 cos¡1µ

cos rcos( sij

2 )

¶.

Then the overlapping area of Ci and Cj is given by

Aij = ¡2 cos¡1µ

cos hij ¡ cos2 rsin2 r

¶cos r ¡ 4 cos¡1

µ1 ¡ cos hij

tan r ¢ sin hij

¶+ 2¼. (2)

For the detailed derivation of (2), see Appendix A2. It is obvious that Aij is a continuous function of rand sij . Then, we get the following Lemma.

Lemma. If the range of angular radius r is 0 < r · 2¼3 , then the overlapping area Aij of set Ci \ Cj

is a monotone decreasing function of sij = ds(Mi; Mj) when r is ¯xed. Then, Aij is maximum whensij = r.

Proof: At ¯rst, we show that the range of r should be 0 < r · 2¼3 in order Aij to be a function of sij.

Under the Minkowski condition, if 2¼3 < r · ¼, the area of set Ci \ Cj is always constant 4¼ cos r and, if

r > ¼, two spherical caps cannot be put on the spherical surface S. Therefore, the range of angular radiusis limited to 0 < r · 2¼

3 . Let G(a) and G(b) be the set Ci \Cj for sij = a and b (a < b), respectively. Weassume that Ci and Cj contact with each other ¯rst. Let e be the geodesic arc between ¯xed centers ofthese spherical caps. Next, let Ci be ¯xed and let us move Cj by moving Mj along e toward Mi. Then, itis obvious G(a) ¾ G(b) holds during this movement. Therefore, for r ¯xed, Aij is a monotone decreasingfunction of sij . Thus, Aij is maximum when sij = r. ¤

On the contrary, Aij is a monotone increasing function of hij when r is ¯xed. Further, when sij = r, we¯nd that the area of set Ci [ Cj is minimum.

Let N be the number of spherical caps when the whole spherical surface is completely covered. And,let @Ci be the perimeter of Ci ( i = 1; : : : ; N ). Then we de¯ne:

Wi =½

Wi¡1 [ Ci

¯¯ max

Mi2@Ci¡1A(Wi¡1 \ Ci)

¾; i = 2; : : : ; N ¡ 1; W1 = C1. (3)

In other words, Wi is the union of Wi¡1 and Ci satisfying the condition that the area A(Wi¡1 \ Ci)is maximum with the restriction Mi 2 @Ci¡1. Hereafter, we call that \[i

º=1Cº is in an extreme state"when the set of spherical caps C1; : : : ; Ci possesses the property (3). We are always necessary to examinewhether [i

º=1Cº is in an extreme state in the sequential covering procedure of our problem mentionedin Section 1. For this purpose, we calculate the area A(Wi¡1 \ Ci) by using (2) and the area formula ofspherical triangle. Here, we consider W2 de¯ned in (3). From Lemma, the area A(C1 \ C2) will becomemaximum when M2, the center of C2, is put on @C1. Therefore, for i = 2, the set W2 = [2

º=1Cº is in anextreme state when s12 = r for 0 < r · 2¼

3 as shown in Fig. 1 (a).Next, we consider W3. In order to make the situation that [3

º=1Cº is in an extreme state (A(W2\C3)is maximum with the restriction M3 2 @C2), from the de¯nition of (3), we can assume N ¸ 4. Therefore,the whole spherical surface must be covered by four or more spherical caps under the Minkowski condition.Then, we present the following theorem.

3

139

Fig. 1: The sketch of W2 and W2 [ C3.

Theorem 1 If the range of angular radius r is 0 < r · ¼ ¡ cos¡1¡ 1

3

¢, then [3

º=1Cº is in an extremestate when s12 = s13 = s23 = r.

Proof: We ¯rst examine the range of r. From the consideration of kissing number in the foregoingSubsection, four or more spherical caps cannot be placed on S under the Minkowski condition whenr > ¼ ¡ cos¡1

¡ 13

¢. Therefore, the range of r should be 0 < r · ¼ ¡ cos¡1

¡ 13

¢.

From Lemma, [2º=1Cº is in an extreme state when s12 = r (see Fig. 1 (a)). Namely, at this time,

[2º=1Cº is identical to W2. Then, we de¯ne T1 and T2 as the two cross points of perimeters @C1 and @C2.

Now, from (3), the center M3 is set on the perimeter of C2 outside of C1 and we need to consider the areaA(W2 \ C3). At this time, s23 = r. Hence, from Lemma, the area A23 = A(C2 \ C3) is always ¯xed andmaximum for any M3 2 @C2. Here, as shown in Fig. 1 (b), let T3 be the ¯xed point on @C2 such thatds(T1; T3) = r and that it is outside C1. First, we put M3 at T3. Next, let us move M3 along @C2 towardT1. Then, as shown in Fig. 1 (c), we see the relation A(W2\C3) = A((C1\C3)\(C2)c)+A23. Therefore,under the condition M3 2 @C2, A(W2 \C3) is maximum when A((C1 \C3)\(C2)c) attains its maximum.Then, we seek for the position of M3 when A((C1 \ C3) \ (C2)c) is maximum. Let G(a) and G(b) bethe set (C1 \ C3) \ (C2)c for s13 = a and b (a < b), respectively. Then, it is obvious that G(a) ¾ G(b)holds. Therefore, for r ¯xed, A((C1 \ C3) \ (C2)c) is a monotone decreasing function of s13. Hence,A((C1 \ C3) \ (C2)c) is maximum when M3 is put at T1. Therefore, A(W2 \ C3) attains its maximumwhen M3 is selected on T1. Then, s13 = r holds. Thus, as shown in Fig. 1 (d), [3

º=1Cº is in an extremestate when s12 = s13 = s23 = r for 0 < r · ¼¡cos¡1

¡ 13

¢. ¤

2.3 Procedure of Sequential CoveringAs mentioned in Section 1, our problem is to calculate the upper bound of r for the sequential covering,such that N congruent open spherical caps cover the whole spherical surface under the condition thatMi is set on the perimeter @Ci¡1, and that each area A(Wi¡1 \ Ci) becomes maximum in sequence fori = 2; : : : ; N ¡ 1. Note that, although N spherical caps are needed in our problem, N ¡ 1 spherical capsare used in the sequential covering since we want to make the situation that [N¡1

º=1 Cº is in an extremestate. First, before beginning covering, the angular radius r of spherical caps is chosen su±ciently smallso that [N¡1

º=1 Cº cannot cover the whole spherical surface in the Minkowski covering. Note that, in theresult which will be obtained in the procedure below, the set WN¡1 does not cover the whole sphericalsurface. Algorithmically, the procedure of sequential covering is the followings:

STEP 1: The center M1 of the ¯rst spherical cap C1 is put at (x; y; z) = (0; 0;¡1). Then, from (3),W1 = C1 holds.

4

140

STEP 2: The center M2 of C2 is put at a certain point on the perimeter @C1. As a result, [2º=1Cº

is in an extreme state (A(W1 \ C2) is maximum with the restriction M2 2 @C1) since s12 = r. If(N ¡ 1) ¸ 3, go to the next step; otherwise the procedure ends.

STEP 3: The center M3 of C3 is put on one of the cross points of @C1 and @C2. Then [3º=1Cº is in

an extreme state since s12 = s13 = s23 = r. If (N ¡ 1) ¸ 4, put i = 4 and go to the next step; otherwisethe procedure ends.

STEP 4: First, the center Mi of Ci is placed at a certain point on the acceptable part of @Ci¡1which is outside the other spherical caps. Next, move Mi among the acceptable points on @Ci¡1 andcompute A(Wi¡1 \ Ci) for respective points of Mi. Here, check whether [i

º=1Cº covers the whole of Sfor all possible positions of Mi. If it is true, the angular radius r is reduced to half for instance, puti = 4, and return to the top of STEP4; otherwise Mi is ¯xed at the position where A(Wi¡1 \Ci) attainsits maximum. As a result, the set Wi is formed on S. Go to STEP. 5.

STEP 5: If i = N ¡ 1, the procedure ends; otherwise put i à i + 1 and go to STEP 4.

Therefore, our sequential covering satis¯es the condition that, sequentially for i = 2; : : : ; N ¡ 1, eachA(Wi¡1 \ Ci) is maximum with the restriction Mi 2 @Ci¡1 ([i

º=1Cº is in an extreme state ).

2.4 Upper Bounds rN and rN¡1We de¯ne rN as the upper bound of r which is mentioned at the top of Subsection 2.3. Next, wede¯ne another upper bound of r, rN¡1, such that the set [N¡1

º=1 Cº which contains WN¡2 cannot coverS under the Minkowski condition. Then rN¡1 should be equal to the spherical distance of the largestinterval in the uncovered region (WN¡2)c of S. It is because, when the angular radius r is equal torN¡1, the set [N¡1

º=1 Cº which contains WN¡2 can cover S except for a ¯nite number of points or a linesegment under our sequential covering. Therefore, [N¡1

º=1 Cº is in an extreme state if and only if at leastone endpoint of the interval, which has the above mentioned spherical distance rN¡1, comes on theperimeter @CN¡2. Further, when there are two or more uncovered points, the spherical distance of anypair of these uncovered points is less or equal to rN¡1 since the largest interval is assumed to be rN¡1.Then, we can put the center MN of CN at one of the uncoverd points. At this moment, we see that[N

º=1Cº which contains WN¡1 covers S without any gap. Then, we notice the fact that rN is equal torN¡1.

Therefore, for our problem, it is necessary to know the spherical distance of the largest interval inthe uncovered region (WN¡2)c. In our cases in Section 3, it becomes important to consider trianglesor quadrangles as the shape of (WN¡2)c in the ¯nal steps of sequential covering. For this purpose, weinvestigate the farthest pair of points in spherical triangle and quadrangle so that it is useful for laterconsiderations.

First, we consider the spherical triangle. Let T1, T2, and T3 be the points on S and let the sphericaltriangle T1T2T3 be such that the side T1T2 of the triangle T1T2T3 is an inferior arc of the great circledetermined by T1 and T2: namely the side T1T2 is the geodesic arc and satis¯es 0 < ds(T1; T2) · ¼.Next, we de¯ne the point H as the middle point of the geodesic arc T1T2. Then, we get the position ofH as follows:

H =T1 + T2

2 cos¡cos¡1 ((T1;T2))

±2¢ ;

where a bold symbols H, T1, and T2 are unit vectors from the origin O to the points H , T1, and T2 onthe unit sphere, respectively, and where (T1;T2) is the inner product of vectors T1 and T2. Then, weget the following theorem.

Theorem 2 If the spherical triangle T1T2T3 satis¯es the condition that ¼2 ¸ ds(T1; H) = ds(H;T2) =

ds(T1;T2)2 ¸ ds(H; T3) > 0, then the farthest pair of points inside the spherical triangle T1T2T3 is the pair

of T1 and T2.

Proof: Let C0 be the closed spherical cap with its center at H and with its angular radius as ds(T1;H).From the condition that ds(T1; H) ¸ ds(H; T3), T3 is inside C0. In order to prove Theorem 2, we needto consider three cases: (I) ds(T1; H) = ds(H; T3) = ¼

2 and T3 2 @C0; (II) ds(T1; H) = ds(H; T3) < ¼2

and T3 2 @C 0; (III) ds(T1; H) · ¼2 and T3 =2 @C0. First, for the case (I), let T1, T2, and H be (0; 1; 0),

(0; ¡1; 0) and (0; 0; 1), respectively. Namely, the geodesic arc T1T2 is the half of the great circle.Therefore, the perimeter of C0 is the equator of unit sphere. Then, all the sides of spherical triangleT1T2T3 are the geodesic arcs. Furthermore, the spherical triangle T1T2T3 is formed with two sectors

5

141

T1HT3 and T2T3H , and is contained in C0. Therefore, it is clear that the spherical distance ds(T1; T2) = ¼is the largest distance. Next, for the case (II), ds(T1; T3) is shorter than the length of arc T1T3 of perimeter@C0 and the geodesic arc T1T3 is inside of C 0. Therefore, the spherical triangle T1HT3 is contained in thesector T1HT3. Hence, it is entirely contained in C0. Similarly, we ¯nd that the spherical triangle T2T3His contained in C0. Thus, the closed spherical cap C0 entirely contains the spherical triangle T1T2T3. Theproof for the case (III) is the same procedure as in the case (II). Therefore, in the cases (II) and (III), thespherical distance ds(T1; T2) is the largest distance. Thus, the farthest pair of points in spherical triangleT1T2T3 is the pair of T1 and T2. ¤

Note that if the condition of Theorem 2 are not satis¯ed, the claim of Theorem 2 does not hold.For example, when the spherical triangle is an equilateral triangle of side lengths between ¼

2 and ¼, itsheight is larger than the sides: namely the farthest pair of points in spherical equilateral triangle is notthe pairs of spherical triangular vertices.

Next, we consider here the quadrangle. Let us given the spherical triangle T1T2T3 which satis¯es thecondition of Theorem 2. Then, we add the point T4 on S to the opposite side of T3 against the greatcircle which passes T1 and T2. By joining T4 with T1 and T2 by the geodesic arcs, respectively, we makethe spherical quadrangle T1T4T2T3 that is formed with two spherical triangles T1T2T3 and T1T4T2. Then,we have the following Corollary from Theorem 2.

Corollary. If the spherical quadrangle T1T4T2T3 satis¯es the following conditions8<:

ds(T1; T2) ¸ ds(T3; T4);¼2 ¸ ds(T1; H) = ds(H; T2) = ds(T1;T2)

2 ¸ ds(H; T3) > 0;ds(T1; H) ¸ ds(H; T4) > 0;

then the farthest pair of points in the spherical quadrangle T1T4T2T3 is the pair of T1 and T2.

Proof: Let C 0 be the closed spherical cap whose center is H and whose angular radius is ds(T1;H).From Theorem 2, the spherical triangle T1T2T3 is entirely contained in C0 and its farthest pair is thepair of T1 and T2. Similarly, the spherical triangle T1T4T2 is contained in C0 and its farthest pair isthe pair of T1 and T2. Therefore, the spherical quadrangle T1T4T2T3 is entirely contained in C0 whoseangular diameter is determined by T1 and T2. Thus, the pair of T1 and T2 is the farthest pair of pointsin spherical quadrangle T1T4T2T3. ¤

Then, we consider the upper bounds rN and ¹rN¡1 in an illustrative example. Fig. 2 shows thedistinction between two upper bounds rN and ¹rN¡1. In Fig. 2 (a), the shaded region is the coveredregion WN¡2 and the white region is the uncovered region (WN¡2)c. At this time, the shape of (WN¡2)c

is the quadrangle T1T4T2T3 on S. We note that the sides of uncovered region (WN¡2)c are not geodesicarcs but are perimeters of spherical caps. Here, we suppose that the spherical quadrilateral T1T4T2T3 (thequadrilateral enclosed by dotted-and-dashed segments in Fig. 2 (a)) satis¯es the condition of Corollary ofTheorem 2. Then, it is obvious that the quadrangle T1T4T2T3 (the quadrangle enclosed by solid segmentsin Fig. 2 (a)) on S is inside the spherical quadrilateral T1T4T2T3. Therefore, from Corollary of Theorem2, we ¯nd that the pair of T1 and T2 is the farthest pair of points in the spherical quadrilateral T1T4T2T3and the quadrangle T1T4T2T3 on S. Namely, the spherical distance of the largest interval in (WN¡2)c

is ds(T1; T2). Then, when the radii r of N ¡ 1 spherical caps are all equal to ds(T1; T2) and the centerMN¡1 is put on T1 or T2, the set [N¡1

º=1 Cº which contains WN¡2 can cover S except for a point (T2 orT1). Therefore, the upper bound, rN¡1, of r such that the set [N¡1

º=1 Cº which contains WN¡2 cannotcover S under the Minkowski condition is ds(T1; T2). At this time, as shown in Fig. 2 (b), if T2 2 @CN¡2is selected as MN¡1, the set [N¡1

º=1 Cº is in an extreme state where T1 is the unique uncoverd point.Therefore, when MN is put at T1 2 @CN¡1, the set [N

º=1Cº which contains WN¡1 covers S without anygap. Thus, as shown in Fig. 2 (c), our upper bound rN is equal to ds(T1; T2). Namely, in Fig. 2, we seethe fact that rN = rN¡1 = ds(T1; T2).

Here, we note the advantage of using the upper bound rN¡1. The value of rN¡1 is easier to calculatethan rN , since it is better to examine the extreme situation where the spherical surface S cannot becovered by N ¡ 1 spherical caps than the situation where S is covered by N caps. Then, at the laststage of the process of covering, we only need to observe the situation where a few uncovered regionsremain since our covering is sequential. Moreover, the covering of our problem is ¯nished in fact whenN ¡ 1 spherical caps cover S except for a ¯nite number of points or a line segment since open sphericalcaps are considered. In such a case, as shown in the cases below, the position of the center of the N -thspherical cap is almost unique. Although rN¡1 and rN are not necessarily coincident, the value of rN¡1will give a strong candidate for rN .

6

142

Fig. 2: Upper bounds rN and ¹rN¡1.

7

143

3 Results

3.1 N = 2; 3; 4; 5 and 6For cases of N = 2; : : : ; 6, we ¯nd that rN is equal to the upper bound of the range of r for k in Table 1.

For N = 2, r2 (i.e. rN for N = 2) is equal to ¼. It is because that the radius of the ¯rst sphericalcap C1 should be equal to ¹r1 = ¼ in order C1 to cover the whole S except for a point. From the STEP1 as described in Subsection 2.3, the center M1 of C1 is put at the south pole (0; 0; ¡1). In this case,the north pole (0; 0; 1) is open for ¹r1 = ¼. Then it is obvious that we can put M2, the center of thesecond spherical cap C2, at the north pole. At this time, S is covered without any gap by C1 and C2which satisfy the Minkowski condition. Thus we see r2 = ¹r1 = ¼. We note that this value ¼ is the upperbound of the range of r for k = 1 as shown in Table 1.

For N = 3, r3 should be equal to the upper bound of the range of r for k = 2. It is because that, whenr = 2¼

3 , the set [2º=1Cº under the condition M2 2 @C1 can cover S except for a point. First, let M1 be

the south pole as described above. Then, if M2 is put at the point¡sin 2¼

3 ; 0; ¡ cos 2¼3

¢=

³p3

2 ; 0; 12

´

on @C1, the unique uncovered point P of S will be³¡

p3

2 ; 0; 12

´. Then, [2

º=1Cº is in an extreme state.

Therefore, we get ¹r2 = 2¼3 , and we can put the center M3 of C3 at P . At this time, [3

º=1Cº whichcontains W2 covers the whole of S under the Minkowski condition. It is obvious that the position ofcenters is the trisection point of a great circle. Thus the correspondent angular radius r3 is equal to 2¼

3 .For N = 4, r4 should be equal to the upper bound of the range of r for k = 3. The reason is the

following. First, we can assume that M3 is put on one of the cross points of perimeters @C1 and @C2 underthe condition M2 2 @C1. It is because that, from Theorem 1 in Subsection 2.2, [3

º=1Cº is in an extremestate when s12 = s13 = s23 = r. If r is equal to the spherical distance between the cross points of @C1 and@C2, the spherical surface S except for a point is covered by the set W3. As before, let M1 be the southpole and let M2 be at (sin r; 0; ¡ cos r). At this time, let us assume that the angular radius r is equal tothe upper bound of the range r for k = 3 in Table 1. Thus, when r = ¼¡cos¡1

¡ 13

¢, M3 is selected as one

of the trisection point of @C1 and let it be³¡ 1

2 sin r; ¡p

32 sin r; ¡ cos r

´=

³¡

p2

3 ; ¡p

2p3; 1

3

´. Then, it

is easy to see that the point P =³¡ 1

2 sin r;p

32 sin r; ¡ cos r

´=

³¡

p2

3 ;p

2p3; 1

3

´is the unique uncovered

point on S. We note, at the same time, that the coordinates of M2 turns out to be³

2p

23 ; 0; 1

3

´.

Therefore, we get ¹r3 = ¼ ¡ cos¡1¡ 1

3

¢, and we can put M4 on that point P . As a result, the set [4

º=1Cº

which contains W3 can cover the whole of S when r = ¹r3 and we get ¯nally r4 = ¼ ¡ cos¡1¡ 1

3

¢. Then,

we ¯nd that the position of centers of these four spherical caps is in accord with the vertices of regulartetrahedron.

For N = 5, before deriving r5, ¹r4 = ¼2 is shown ¯rst. When r = ¼

2 , from Theorem 1, the centers M1,M2, and M3 are put , for example, at (0; 0; ¡1), (1; 0; 0), and (0; ¡1; 0), respectively, in order for[3

º=1Cº to be in an extreme state. Then, the uncovered region (W3)c is the spherical equilateral triangleof side-length ¼

2 and vertexes (0; 0; 1), (¡1; 0; 0), and (0; 1; 0). Therefore, from the discussion inSubsection 2.4, we ¯nd that the largest spherical distance in (W3)c is equal to ¼

2 ; namely ¹r4 = ¼2 . Next,

when M4 is put on the point (0; 0; 1) 2 @C3 or (¡1; 0; 0) 2 @C3 (in this paper, we choose (¡1; 0; 0)),the set [4

º=1Cº is in an extreme state. As a result, the set W4 can cover S except for a line segmentwhich is an inferior great circle connecting (0; 1; 0) and (0; 0; 1) and whose length is ¼

2 . Therefore,when M5 is put at any point on this line segment except for points (0; 1; 0) and (0; 0; 1), we ¯nd that[5

º=1Cº covers S without gap under the Minkowski condition. Thus, r5 = ¹r4 = ¼2 . Namely, r5 is equal

to the upper bound of the range of r for k = 4.For N = 6, we can assume that r6 is equal to the result of the case N = 5. First, similarly to the case

N = 5, we put the centers of C1, C2, C3, and C4 of radius r = ¼2 at (0; 0; ¡1), (1; 0; 0), (0; ¡1; 0), and

(¡1; 0; 0), respectively. If we put M5 on (0; 0; 1) or (0; 1; 0) (in the following, we assume (0; 1; 0)is chosen as M5), the set [5

º=1Cº is in an extreme state and the spherical surface S except for a point,namely the point (0; 0; 1), can be covered by the set W5. As a result, the set [6

º=1Cº containing W5covers the whole of S when we put M6 on (0; 0 1). Thus, our assumption r5 = r6 = ¼

2 is con¯rmed.Then, we ¯nd that the position of centers of these spherical caps for N = 6 is in accord with the verticesof regular octahedron. Therefore, if all spherical caps of our covering for N = 6 are replaced by half-caps(the spherical cap whose angular radius is r

2 and which is concentric with that of the original cap), allof those half-caps contact other four half-caps and there is no space for those half-caps to move.

8

144

3.2 N = 7From the considerations for the cases N · 6 and Subsection 2.1, we can assume that, for N = 7, theangular radius r should satisfy the inequalities tan¡1 2 < r < ¼

2 . The reason is the following. First, thecenter M1 of the ¯rst spherical cap C1 is set at the south pole (0; 0; ¡1) as before. Now, we remindthe cases of N · 6 in Subsection 3.1, in order to investigate the position of centers of spherical caps forN = 7. For N = 2, M2 is placed on the perimeter @C1. For N = 3, M2 and M3 are both on @C1. ForN = 4, the centers M2, M3, and M4, are placed on @C1. For N = 5 and 6, there are four centers of otherspherical caps on @C1. Then, from the these facts and from Subsection 2.1, we can consider two casesfor N = 7 as follows: four centers M2, M3, M4, and M5 are on @C1; and ¯ve centers M2, M3, M4, M5,and M6 are on @C1. For N = 7, we ¯nd that the second case is excluded because of the following reason.When ¯ve centers M2, M3, M4, M5, and M6 are placed on @C1 according to our sequential covering,the range of angular radius r must be 0 < r · tan¡1 2 from the consideration in Subsection 2.1. Here,we consider the case that r is equal to tan¡1 2 ¼ 1:10715. Then, it is obvious that the spherical distancebetween M1: (0; 0; ¡1) and any point of the set which is covered by the our six spherical caps is smallerthan 2 tan¡1 2, while the length of longitude line joining south and north poles is larger than 2 tan¡1 2.Thus, the union of our six spherical caps would leave a big open area (whose size is at least comparableto the area of spherical cap of angular radius ¼ ¡ 2 tan¡1 2) near the north pole of the unit sphere.Furthermore, the open area near the north pole would become still bigger for 0 < r < tan¡1 2. Hence,the set [6

º=1Cº cannot cover S at all. Thus, we should exclude the second case. Therefore, we haveto consider the ¯rst case that four centers M2, M3, M4, and M5 of spherical caps are placed on @C1.Then, below, we investigate the range of r under this condition. From the consideration in Subsection2.1, when r · ¼

2 , it is possible to put four spherical caps on the perimeter of a spherical cap. Especially,when the range of r is

¡tan¡1 2; ¼

2

¤, four spherical caps can be placed on the perimeter of a spherical cap

and, at the same time, ¯ve spherical caps cannot be placed on the perimeter of a spherical cap. Further,from the setup of our problem, we want to make r the biggest possible. Therefore, we consider that r islarger than tan¡1 2. On the other hand, if r ¸ ¼=2, we cannot cover S by seven spherical caps withoutbreaking the Minkowski condition due to the results of cases N · 6. Thus, r should be in the rangetan¡1 2 < r < ¼

2 . We note that the equality sign does not enter in these inequalities.Let (x; y; z) be the coordinates of cross points where the perimeters of Ca (the coordinates of the

center: (a1; a2; a3)) and Cb (the coordinates of the center: (b1; b2; b3)) intersect. By solving the simul-taneous equations

8<:

a1x + a2y + a3z = cos r ,b1x + b2y + b3z = cos r ,x2 + y2 + z2 = 1 ,

(4)

we will have the coordinates of the cross points. Thus, in the case where the centers of two sphericalcaps are put respectively at (0; 0; ¡1) and (sin r; 0; ¡ cos r), we get

8<:

¡z = cos r ,sin r ¢ x ¡ cos r ¢ z = cos r ,x2 + y2 + z2 = 1 .

(5)

Let the points K1 = (x1; y1; z1) and K2 = (x2; y2; z2) be the solutions of simultaneous equations(5). Solving (5), we obtain

(x1; y1; z1) =µ

¡cos r (cos r ¡ 1)sin r

;(cos r ¡ 1)

p2 cos r + 1

sin r; ¡ cos r

¶, (6)

(x2; y2; z2) =µ

¡cos r (cos r ¡ 1)sin r

; ¡ (cos r ¡ 1)p

2 cos r + 1sin r

; ¡ cos r¶

. (7)

Here, the center M1 of C1 is set at the south pole (0; 0; ¡1) as already mentioned. We put the centersM2 and M3 of C2 and C3 each, at K1 and (sin r; 0; ¡ cos r), respectively. Note that K1 and K2 are thepoints where the perimeters of C3 and C1 intersect. We also note here that the locations of centers ofthe second and the third spherical caps de¯ned as above are di®erent from the cases N = 4; 5; 6 becauseof the convenience of our computation. We will use this convention for all cases of N ¸ 7 hereafter.Then, taking into account Theorem 1 in Subsection 2.2, [3

º=1Cº is in an extreme state. At this time,similarly, we calculate the points where the perimeters of C2 and C1 intersect. Among these cross points,let K3 = (x3; y3; z3) be the point which is outside C3. Let K4 = (x4; y4; z4) be one of the cross points

9

145

Fig. 3: The curve of A(W3 \ C4) when M4 is moved on the arc K4K2 of C3. Here, the arc K4K2 is dividedinto 100 equal intervals and the area A(W3 \ C4) is calculated on 101 end points of the intervals. Note that thecurve of r ' 1:10715 corresponds to the case of r = tan¡1 2. The values of r of other curves are described in thetext.

Fig. 4: The curve of A((W4 [ C5)c) when M5 is moved on the arc K6K5 of C4. The similar computationmethod as in Fig. 3 is taken. See the legend in Fig. 3.

between @C2 and @C3 and let it not be the south pole (0; 0; ¡1). The explicit expressions of coordinatesof cross points K3 and K4 are shown in Appendix A1.

Let us place the center M4 at K4 and move it to K2 along the arc K4K2 of C3. We note that, duringthis movement, the distance s43 between M4 and M3 is equal to the angular radius r and the overlappingarea A43 of C4 and C3 is invariant while the area A(W3 \ C4) is variable. Then, we want to know theposition of M4 where the area A(W3 \ C4) is maximum. Therefore, we calculate the area A(W3 \ C4)against the moving point M4 numerically for several ¯xed values of r among tan¡1 2 · r < ¼

2 . In orderto use the result later, the computation for r = tan¡1 2 is also performed. The results are shown in Fig.3. In this ¯gure, the horizontal axis is the position of M4 on the arc K4K2 of C3 and the vertical axisis the area A(W3 \ C4). In this computation, the arc K4K2 is divided into 100 equal intervals and thearea A(W3 \C4) is calculated on 101 end points of the intervals. Hereafter, the similar computations areperformed for determination of centers of spherical caps (see Figs. 4, 7, 10, 13, and 16). As a result, forthe ¯ve values of r in Fig. 3, we ¯nd that this curve of A(W3 \ C4) is symmetrical at the center of thearc K4K2 (it is evident from the spherical symmetry) and A(W3 \ C4) is maximum at both ends. Thesame fact as above would hold for every values of r in the range tan¡1 2 · r < ¼

2 . Therefore, we expectthat [4

º=1Cº is in an extreme state if and only if M4 is put at K2 or K4 for tan¡1 2 · r < ¼2 . To make

sure, we shall check that these points K2 and K4 satisfy the condition that [4º=1Cº is in an extreme

state after obtaining the exact values of the angular radius r at the last paragraph in this Subsection.At the moment, we choose M4 on the point K2. Then, let K5 = (x5; y5; z5) be one of the cross pointsof perimeters @C4 and @C1, and let it be outside C3. Further, let K6 = (x6; y6; z6) be one of the crosspoints of @C4 and @C3, and let it not be the south pole (0; 0; ¡1). The explicit expressions of crosspoints K5 and K6 are given in Appendix A1.

10

146

Fig. 5: The kite K8K4K6K7 on unit sphere.

Next, the center M5 is put at a certain point on the arc K6K5 of C4. Then, we need to calculate thearea A(W4 \ C5) when M5 is moved on the arc K6K5 of C4. However, in order to simplify calculation,we pay attention to the area A((W4 [ C5)c). It is because, for i ¸ 2, we ¯nd the relation that the areaA(Wi¡1 \ Ci) is maximum with the restriction Mi 2 @Ci¡1 ([i

º=1Cº is in an extreme state ) is thesame as that the area A((Wi¡1 [ Ci)c) is maximum with the restriction Mi 2 @Ci¡1. Therefore, fori = 5, we calculate A((W4 [ C5)c) against the moving point M5 numerically for several ¯xed values ofr among tan¡1 2 · r < ¼

2 . Here, the computation is performed as in the determination of M4. Fig. 4shows the results. In this ¯gure, the horizontal axis is the position of M5 on the arc K6K5 of C4 andthe vertical axis is the area A((W4 [ C5)c). As a result, for the ¯ve values of r in Fig. 4, the curve ofA((W4 [ C5)c) is symmetrical at the center of the arc K6K5 (it is evident from the spherical symmetry)and A((W4 [ C5)c) is maximum when M5 is placed on K6 or K5. The same fact as above would holdfor every values of r in the range tan¡1 2 · r < ¼

2 . Therefore, we expect that [5º=1Cº is in an extreme

state if and only if M5 is put at K5 or K6 for tan¡1 2 · r < ¼2 . To make sure, we shall check whether

the point K5 and K6 such points after obtaining the exact values of the angular radius r at the lastparagraph in this Subsection like the case of M4. At the moment, we choose M5 on the point K5. Then,let K7 = (x7; y7; z7) be one of the cross points of the perimeters @C5 and @C4, and let it be outside ofC1. Similarly, let K8 = (x8; y8; z8) be one of the cross points of @C5 and @C2, and let it be outside ofC1. The exact coordinates of K7 and K8 are also given in Appendix A1.

At the time when C5 is put on the sphere, the shape of the uncovered region (W5)c becomes a kite

on the unit sphere (see Fig. 5). We note that the sides of the kite K8K4K6K7 are not geodesic arcs butare perimeters of spherical caps. Then, we see that there are four centers M2, M3, M4, and M5 on theperimeter @C1.

From the con¯guration of the vertices K4, K6, K7 and K8 of the kite K8K4K6K7, for the rangetan¡1 2 · r < ¼

2 , we ¯nd that there hold always the following relations of the spherical distance betweeneach vertices.

8>><>>:

ds (K8; K4) = ds (K8; K7) ;ds (K6; K4) = ds (K6; K7) ;ds (K6; K4) < ds (K8; K4) < ds (K6; K8) ;ds (K6; K4) < ds (K4; K7) < ds (K6; K8) :

(8)

Note that, as de¯ned in Subsection 2.2, ds(Ki; Kj) is the spherical distance between points Ki =(xi; yi; zi) and Kj = (xj ; yj ; zj); namely

ds(Ki; Kj) = cos¡1 (xi ¢ xj + yi ¢ yj + zi ¢ zj) : (9)

Refer to Appendix A1 for the explicit coordinates of K4, K6, K7, and K8. We produced the relation (8)by using mathematical software. Especially two inequalities in (8) are obtained numerically.

So far, we have arranged ¯ve spherical caps. Next, let us consider the sixth and seventh spheri-cal caps. As mentioned in Subsection 2.4, if we take the angular radius of spherical caps to be equalto the largest spherical distance in the kite K8K4K6K7, the sixth spherical cap C6 can cover the re-gion except for a point in the kite under the Minkowski condition. Therefore, we ¯nd that ¹r6 is thelargest spherical distance in the kite. It is obvious that the kite K8K4K6K7 on the sphere is inside

11

147

Fig. 6: (a) Our sequential covering for N = 7. (b) Our solution of Tammes problem for N = 7. Both viewpointsare (0; 0; 10). In this example, the coordinates of the centers are respectively (0; 0; ¡1), (0:16977; ¡0:96282;¡0:21014), (0:97767; 0; ¡0:21014), (0:16977; 0:96282; ¡0:21014), (¡0:91871; 0:33438; ¡0:21014), (¡0:55588;¡0:46644; 0:68806), and (0:39850; 0:33438; 0:85404).

the spherical quadrilateral K8K4K6K7 for ¯xed vertices K8, K4, K6, and K7 (see Fig. 5). From (8)and the coordinates of K8, K4, K6, and K7 in Appendix A1, we ¯nd numerically that four verticesof the spherical quadrilateral K8K4K6K7 satis¯es the relations that ds(K6; K8) > ds(K4; K7), and¼2 ¸ ds(K6; H) = ds(H; K8) = ds(K6;K8)

2 > ds(H; K4) = ds(H; K7) > 0 for tan¡1 2 · r < ¼2 . Note that

H is the middle point of the geodesic arc K6K8. Therefore, from Corollary of Theorem 2, the farthestpair of points in the spherical quadrilateral K8K4K6K7 is the pair of K6 and K8; namely ds (K6; K8)is the largest spherical distance in the kite K8K4K6K7 and is equal to ¹r6. Then, from (9) and thecoordinates of K6 and K8 in Appendix A1, we have

r = ds (K6; K8) = cos¡1µ

41 cos4 r ¡ 8 cos3 r ¡ 18 cos2 r + 19 cos4 r + 8 cos3 r ¡ 2 cos2 r + 1

¶. (10)

In addition, we note K8 2 @C5. Thus, [6º=1Cº is in an extreme state (the set [6

º=1Cº covers S exceptfor the point K6) if and only if M6 is put on the point K8. At this time, K6 is the unique uncoveredpoint. Equation (10) is solved against r by using mathematical software. As a result, the value of theupper bound for N = 7 is obtained

r7 = ¹r6 = cos¡1µ

1 ¡ 4p3

cosµ

7¼18

¶¶¼ 1:35908 rad. (11)

Therefore, when M6 and M7 are put at K8 and K6, respectively, then [7º=1Cº which contains W6

covers the whole of S (see Fig. 6 (a)). Namely, our sequential covering for N = 7 is completed.Here, we check whether the position of points K2 and K5 for M4 and M5 satisfy the condition that

[4º=1Cº and [5

º=1Cº are in an extreme state, respectively. For that purpose, when r is equal to the valueof (11), we examine the position where the area A(Wi¡1\Ci) is maximum with the restriction Mi 2 @Ci¡1(i = 4 and 5). First, about M4, we calculate numerically the area A(W3 \ C4) against the moving pointM4 for r = cos¡1

³1 ¡ 4p

3cos

¡ 7¼18

¢´. As mentioned above, we checked numerically that A(W3 \ C4) is

maximum with the restriction M4 2 @C3 ([4º=1Cº is in an extreme) if and only if M4 is put at K2 or

K4 for r = cos¡1³1 ¡ 4p

3cos

¡ 7¼18

¢´. The fact is indicated by the curve of r ' 1:10715 corresponding to

r = cos¡1³1 ¡ 4p

3cos

¡ 7¼18

¢´in Fig. 3. Next, about M5, we use the fact that A(W4\C5) is maximum with

the restriction M5 2 @C4 is the same as that A((W4[C5)c) is maximum with the restriction M5 2 @C4. Asmentioned above, we checked numerically that A((W4[C5)c) is maximum with the restriction M5 2 @C4

([5º=1Cº is in an extreme) if and only if M5 is put at K5 or K6 for r = cos¡1

³1 ¡ 4p

3cos

¡ 7¼18

¢´. The

result is graphically presented by the curve of r ' 1:10715 corresponding to r = cos¡1³1 ¡ 4p

3cos

¡ 7¼18

¢´

in Fig. 4. Thus, our choice for M4 and M5 is justi¯ed.

12

148

Fig. 7: The curve of A((W5 [ C6)c) when M6 is moved on the arc K7K8 of C5. The similar computationmethod as in Fig. 3 is taken. See the legend in Fig. 3.

3.3 N = 8It is expected that the solution r8 for N = 8 should not be larger than r7. Then, we assume r · r7.Further, at the moment, we assume tan¡1 2 · r. If our answer for N = 8 is not obtained by thisassumption tan¡1 2 · r, we will consider the range of r < tan¡1 2 next. First, as in the case of N = 7,the centers M1, M2, and M3 are placed at (0; 0; ¡1), K1, and (sin r; 0; ¡ cos r), respectively. Since,from Theorem 1 in Subsection 2.2, the set [3

º=1Cº is in an extreme state when the centers M1, M2,and M3 satisfy the relations s12 = s13 = s23 = r. Next, from the considerations for determinations ofM4 and M5 in Subesction 3.2, we can assume that the allocation of points K2 and K5 for M4 and M5respectively satisfy the condition that [4

º=1Cº and [5º=1Cº each are in an extreme state. To make sure,

we shall check that K2 and K5 are such points after obtaining the exact values of r8. Therefore, we usethe same positions of the ¯rst ¯ve spherical caps for the case N = 7. When the ¯fth spherical cap C5 isput on the sphere, in the same way as the foregoing Subsection, a quadrilateral kite K8K4K6K7 on thesphere might be formed as the uncovered region. Hence, the relation (8) holds.

Next, we search the position of M6 which satis¯es the condition that the area A(W5\C6) is maximumwith the restriction M6 2 @C5 ([6

º=1Cº is in an extreme state). From the restriction M6 2 @C5, M6is put at a certain on the arc K7K8 of C5 and, during M6 is moved on the arc K7K8 of C5, the areaA(W5 \ C6) is calculated. At this time, in order to simplify calculation, we use the relation that thearea A(W5 \ C6) is maximum with the restriction M6 2 @C5 is the same as that the area A((W5 [ C6)c)is maximum with the restriction M6 2 @C5. Hence, we calculate the area A((W5 [ C6)c) against themoving point M6 numerically for several ¯xed values of r among tan¡1 2 · r < r7. As a result, forthe four values of r in Fig. 7, we ¯nd that A((W5 [ C6)c) is maximum when M6 is put on K8. Fig.7 shows the graph of computation results. In this ¯gure, the horizontal axis is the position of M6 onthe arc K7K8 of C5 and the vertical axis is the area A((W5 [ C6)c). Then, we expect that [6

º=1Cº isin an extreme if and only if M6 is put at K8 in the range tan¡1 2 · r < r7. We shall check that K8 issuch point after obtaining the exact values of the angular radius r as in the cases of M4 and M5. At themoment, we choose M6 on the point K8.

Then, when M6 is put at K8, we notice that three cases are possible to consider for the relationbetween r and ds(K8; K4): r = ds(K8; K4); r < ds(K8; K4); and r > ds(K8; K4). First, we consider thecase r = ds(K8; K4). From (9) and the coordinates of K4 and K8 in Appendix A1, we have

r = ds(K8; K4) = cos¡1µ

16 cos4 r + cos3 r ¡ 9 cos2 r ¡ cos r + 19 cos3 r ¡ cos2 r ¡ cos r + 1

¶. (12)

Equation (12) is the quartic equation of cos r and can be solved by using the algebraic formula of Ferrari.As a result, we get the following value of the angular radius r.

r = cos¡1

á1

7+

2p

27

!¼ 1:30653 rad. (13)

Then, we ¯nd the following relation among ds(K8; K4), ds(K4; K7), and r by using mathematical soft-ware.

13

149

ds (K4; K7) = ds (K8; K4) = r. (14)

Therefore, from (8) and (14), we see that the spherical triangle K8K4K7 is equilateral and the uncov-ered region (W6)

c is the concave isosceles triangle K4K6K7 on S inside the spherical isosceles triangleK4K6K7. From the value of (13) and the coordinates of K4, K6, and K7 in Appendix A1, we ¯nd thatthe spherical isosceles triangle K4K6K7 satis¯es the relations ¼

2 ¸ ds(K4; H) = ds(H; K7) = ds(K4;K7)2 ¸

ds(H; K6) > 0. Note that H is the middle point of the geodesic arc K4K7. Therefore, from Theorem 2in Subsection 2.4, the largest spherical distance in (W6)

c is ds(K4; K7). Hence, [7º=1Cº is in an extreme

state (the set [7º=1Cº covers S except for a point) if and only if r = ds(K4;K7) and M7 is chosen on

one of the points K7 or K4. Here we put M7 on K7; namely the spherical surface S is covered by the set[7

º=1Cº except for the point K4. Then, from the relation (14), we ¯nd ¹r7 = cos¡1³¡1

7 + 2p

27

´. Thus,

from the fact K7 2 @C6, it is obvious that ¹r7 is the upper bound r8 for N = 8. Finally, M8 is uniquelydetermined to be the uncovered point K4, and then the whole of S is covered by [8

º=1Cº which containsW7 (see Fig. 8 (a)).

Next, we consider the other two cases r < ds(K8; K4) and r > ds(K8; K4) by way of precaution.Here, we examine the relations among ds(K8; K4), ds(K4; K7), and r, and ¯nd numerically that thefollowing relations hold by using mathematical software.

For tan¡1 2 · r < cos¡1

á1

7+

2p

27

!,

ds (K6; K4) · r < ds (K4; K7) < ds (K8; K4) . (15)

For cos¡1

á1

7+

2p

27

!< r <

¼2

, ds (K8; K4) < ds (K4; K7) < r. (16)

In the case (15), since the angular radius r is smaller than ds(K8; K4), the uncovered region (W6)c isreduced to the pentagon which is bounded by perimeters of spherical caps and ds(K4; K7) inside thispentagon is larger than r due to (8) and (15). As was described in Subsection 2.4, we expect thesituation that the set W7 covers S except for ¯nite points. Therefore, in the uncovered pentagon (W6)c,we take ¹r7 to be the spherical distance of the largest interval, such that at least one endpoint of theinterval comes on the perimeter @C6. However, from the relation (15), we cannot take r to be thelargest spherical disatance in (W6)c. Further, we ¯nd that an uncovered region leaves on S when M7 isput on the uncovered pentagon (W6)c according to our sequential covering. Refer to the considerationsof determination for M7 and the fact that the uncovered region leaves when W7 is formed on S fortan¡1 2 · r < cos¡1

³¡ 1

7 + 2p

27

´in the following Subsections. In the case (16), on the other hand, since

the angular radius r is larger than ds(K8; K4), the uncovered region (W6)c is reduced to the trianglewhich is bounded by perimeters of spherical caps and the spherical distance of the largest interval insidethis triangle is smaller than r due to (8) and (16). Then, the seventh spherical cap C7 covers theuncovered triangle (W6)c completely when the center M7 is put in this triangle. Namely, the center ofC8 cannot be placed on S under the Minkowski condition. Therefore, in two cases above, the range of ris not suitable for our upper bound.

Now, when r is equal to the value of (13), we check whether the positions of points K2, K5, and K8for M4, M5, and M6 satisfy the condition that [4

º=1Cº , [5º=1Cº , and [6

º=1Cº are in an extreme state,respectively. For i = 4; 5, and 6, as mentioned above, we checked numerically that the points K2, K5, andK8 are the position where the area A(Wi¡1 \Ci) (or A((Wi¡1 [Ci)c)) are maximum with the restrictionMi 2 @Ci¡1 ([i

º=1Cº is in an extreme state), respectively, when r is equal to cos¡1³¡ 1

7 + 2p

27

´. Then,

the facts are indicated by the curve of r ' 1:30653 corresponding to r = cos¡1³¡ 1

7 + 2p

27

´in Figs. 3,

4, and 7. Thus, for the case N = 8, our choice of M4, M5, and M6 is justi¯ed.At the beginning of this Subsection, we initially assumed that r should be in the range

¡tan¡1 2; r7

¤.

In fact, after the investigation, our upper bound r8 (the value of (13)) has fallen within the range¡tan¡1 2; r7

¤. However, one might suspect that the fact is due to the assumption. So, if r is in the range¡

0; tan¡1 2¤, we examine whether W7 is able to cover S except for ¯nite points. When r is assumed to be

equal to tan¡1 2 ¼ 1:10715, we see that the set W7 must leave an uncovered region on S. For its detail,refer to the consideration of Subsection 3.6 where r11 = r12 = tan¡1 2 is discussed. Therefore, from

14

150

Fig. 8: (a) Our sequential covering for N = 8. (b) Our solution of Tammes problem for N = 8. Both viewpointsare (0; 0; 10). In this example, the coordinates of the centers are respectively (0; 0; ¡1), (0:19992; ¡0:94435;¡0:26120), (0:96528; 0; ¡0:26120), (0:19992; 0:94435; ¡0:26120), (¡0:88248; 0:39116; ¡0:26120), (¡0:68256;¡0:55319; 0:47759), (¡0:28273; 0:55319; 0:78361), and (0:48264; ¡0:39116; 0:78361).

the setup of our problem, r = tan¡1 2 cannot be r8. Furthermore, for 0 < r < tan¡1 2, the uncoveredregion would become still bigger. Thus, our assumption that r8 is in the range

¡tan¡1 2; r7

¤is con¯rmed

(r = tan¡1 2 is just excluded from the above consideration) and cos¡1³¡1

7 + 2p

27

´is certainly a solution

for N = 8.

3.4 N = 9According to the considerations for N = 7 and 8, the angular radius for N = 9 should be smaller thanr8. First, at the moment, we expect that the inequalities tan¡1 2 · r < r8 hold like the case of N = 8.Then, the same positional relation of the ¯rst ¯ve spherical caps for the case of N = 7 is used again.From Theorem 1 in Subsection 2.2 and from the considerations for determinations of M4 and M5 inSubsection 3.2, we can consider that the set [5

º=1Cº is in an extreme state when the centers M1, M2,M3, M4, and M5 are placed at the points (0; 0; ¡1), K1, (sin r; 0; ¡ cos r), K2 and K5, respectively.To make sure, after obtaining the exact value of r9, we shall check that the allocations of K2 and K5 toM4 and M5, respectively, satisfy the condition that [4

º=1Cº and [5º=1Cº are in an extreme state. In the

same way as in Subsection 3.2, the shape of the set (W5)c is again the kite K8K4K6K7 on the sphere.Then, the relatios (8) and (15) hold. We should cover this kite K8K4K6K7 except for a point by the setC6 [ C7 [ C8 under the conditions M6 2 @C5, M7 2 @C6, and M8 2 @C7.

From the consideration for determination of M6 in Subsection 3.3, and from the condition that [6º=1Cº

is in an extreme state, we assume that the center M6 is placed on the point K8 as in the case of N = 8.At this time, the uncovered region (W6)c is reduced to the pentagon which is bounded by perimeters ofspherical caps (see Fig. 9). Here, let K9 be one of the cross points of the perimeters @C6 and @C2, andlet it be outside C1. Further, let K10 be one of the cross points of @C6 and @C5, and let it be outsideC1. Obviously the relation of ds(K8; K9) = ds(K8; K10) holds. Now, we want to place three centers ofC7, C8, and C9 in the uncovered pentagon K9K4K6K7K10 under the Minkowski condition. Then, weconsider an spherical equilateral triangle of side-length r = ds(K8; K9) in the pentagon K9K4K6K7K10since three centers M7, M8, and M9 could be arranged on the vertexes of the spherical equilateral triangleunder the Minkowski condition. Therefore, we assume that K9 and K10 satisfy

ds(K8;K9)(= ds(K8; K10)) = ds(K9; K10) = ds(K6; K9) = ds(K10; K6): (17)

As shown in Fig. 9, we see the spherical rhombus K8K9K6K10 which is formed with two sphericalequilateral triangles K8K9K10 and K9K6K10. If the spherical rhombus K8K9K6K10 which satis¯es (17)is possible to be formed, we can consider that the angular radius r9 (¹r8) is equal to a side-length (e.g.the spherical distance between the points K8 and K9) of the spherical rhombus K8K9K6K10 if and onlyif M7 is placed on the point K9 or K10. We shall show that this allocation of M7 is actually possible,after obtaining the solution r9. Then, due to ds(K9; K2) = 2r (the proof is given in Appendix A3), C7 is

15

151

Fig. 9: The sketch of the kite K8K4K6K7 and the spherical rhombus K8K9K6K10 in the kite K8K4K6K7.

in contact with C4 at the point K6 and the shape of the uncovered region should be a triangle K10K6K7when M7 is placed on the point K9. At this time, we will see later that the side-length of sphericalrhombus K8K9K6K10 is the largest spherical distance in the uncovered triangular region. Thus, if M8is chosen either on the point K6 or K10, the triangle is covered by C8 except for a point in the triangle.Note that we supposed that the points K8 and K9 are the positions such that [6

º=1Cº and [7º=1Cº are

in an extreme state, respectively. However, it is not checked yet that K8 and K9 are such positions inthis step. We shall check these points after obtaining the exact value of r9 and the coordinates of K9.

Now we examine whether the point K9 or K10 which satis¯es (17) is possible. To begin with, wecalculate the coordinates of the point K9. Since K9 is one of the cross points of the perimeters @C6and @C2, the coordinates of K9 can be obtained by using the simultaneous equations (4). Refer to thecoordinates of K8 = M6 in Appendix A1 and the equation (6) (the coordinates of K1 = M2). The exactcoordinates of K9 = (x9; y9; z9) are given in Appendix A1. From our consideration that r9 (¹r8) is equalto a side-length of spherical rhombus K8K9K6K10 which satis¯es (17), we will solve the equation

r = ds(K6; K9) = cos¡1µ

cos r (81 cos4 r ¡ 28 cos3 r ¡ 46 cos2 r ¡ 4 cos r + 5)(cos r + 1)(9 cos3 r ¡ cos2 r ¡ cos r + 1)

¶(18)

against r. By developing the equation (18), we obtain the quartic equation of cos r. Therefore, (18) canbe solved by using the algebraic formula of Ferrari as in the case of N = 8. As a result, the value of theangular radius r is obtained as

r9 = ¹r8 = cos¡1µ

13

¶¼ 1:23096 rad. (19)

From (19) and the coordinates of K6, K8, K9, and K10, we checked that the relation (17) strictlyholds; namely ds(K8; K9) = ds(K8;K10) = ds(K9; K10) = ds(K6;K9) = ds(K10; K6) = cos¡1

¡ 13

¢. Then,

K10 is obtained by using the fact that it is one of the cross points of the perimeters @C6 and @C5. Notethat the coordinates of the point K10 are omitted due to their long expressions. As we have noted, wecheck here whether the positions of the point K2, K5, K8, and K9 for M4, M5, M6, and M7 satisfy thecondition that [4

º=1Cº , [5º=1Cº , [6

º=1Cº , and [7º=1Cº are in an extreme state, respectively, by using

the value of (19). First, we checked numerically that A(Wi¡1 \ Ci) (or A((Wi¡1 [ Ci)c)) are maximumwhen Mi (i = 4; 5; and 6) are put at K2, K5, and K8, respectively. These facts are indicated by thecurve of r ' 1:23096 corresponding to r = cos¡1

¡ 13

¢in Figs. 3, 4, and 7. Therefore, we ¯nd numerically

that the positions of K2, K5, and K8 satisfy the condition that [4º=1Cº , [5

º=1Cº , and [6º=1Cº are in

an extreme state, respectively. Next, we examine the position of M7 where the area A(W6 \ C7) ismaximum. Then, in order to simplify calculation, we use the relation that the area A(W6 \ C7) ismaximum with the restriction M7 2 @C6 is the same as that the area A((W6 [ C7)c) is maximum withthe restriction M7 2 @C6. Then, we calculate the area A((W6 [ C7)c) against the moving point M7 onthe arc K10K9 of C6 numerically for several ¯xed values of r among tan¡1 2 · r < r8. As a result, thecurve of A((W6 [ C7)c) is symmetrical at the center of the arc K10K9 (it is evident from the shape ofuncovered region (W6)c) and A((W6 [C7)c) is maximum at both end for the three values of r in Fig. 10.The same fact as above would hold for every values of r in the range tan¡1 2 · r < r8. Fig. 10 illustratesthe result of computation for M7. In this ¯gure, the horizontal axis is the position of M7 on the arcK10K9 of C6 and the vertical axis is the area A((W6 [ C7)c). Then, the result is graphically presented

16

152

Fig. 10: The curve of A((W6 [ C7)c) when M7 is moved on the arc K10K9 of C6. The similar computationmethod as in Fig. 3 is taken. See the legend in Fig. 3.

Fig. 11: (a) Our sequential covering for N = 9. (b) Our solution of Tammes problem for N = 9. Bothviewpoints are (0; 0; 10). In this example, the coordinates of the centers are respectively (0; 0; ¡1), (0:23570;¡0:91287; ¡0:33333), (0:94281; 0; ¡0:33333), (0:23570; 0:91287; ¡0:33333), (¡0:82496; 0:45644; ¡0:33333),(¡0:78567; ¡0:60858; 0:11111), (0:15713; ¡0:60858; 0:77778), (0:58926; 0:45644; 0:66667), and (¡0:54997;0:30429; 0:77778).

by the curve of r ' 1:23096 corresponding to r = cos¡1¡ 1

3

¢in Fig.10. Therefore, for r = cos¡1

¡ 13

¢, we

check numerically that [7º=1Cº are in an extreme state if and only if M7 is put at K9 or K10.

As mentioned above, in this paper, we put M7 at K9. Then, the uncovered region (W7)c is the triangleK10K6K7 on S that satis¯es the relations ¼

2 ¸ ds(K10; H) = ds(H; K6) = ds(K10;K6)2 ¸ ds(H; K7) > 0.

We note that H is the middle point of the geodesic arc K10K6. From Theorem 2 in Subsection 2.4, thelargest spherical distance in (W7)c is ds(K10; K6). Namely, we ¯nd ¹r8 = cos¡1

¡ 13

¢. Hence, if M8 is put

at K10 or K6, the set [8º=1Cº which contains W7 covers the spherical surface S except for a point. Then,

since the facts K6 2 @C7 and K10 2 @C7, we ¯nd that [8º=1Cº is in an extreme state. In this paper, we

choose M8 on K6.Then, cos¡1

¡ 13

¢satis¯es the initial assumption tan¡1 2 · r < r8. However, one can suspect this

result is owing to the initial assumption. When r is the range¡0; tan¡1 2

¤, we check whether W8 is able

to cover S except for ¯nite points. If r is equal to tan¡1 2, our ¯rst to eighth spherical caps must leave anuncovered region on S because of the consideration of Subsection 3.6. Furthermore, for 0 < r < tan¡1 2,the uncovered region would become still bigger. Hence, our upper bound r9 for N = 9 does not exist inthe range

¡0; tan¡1 2

¤like the case of N = 8. Therefore, we note that tan¡1 2 < r < r8 is con¯rmed

(r = tan¡1 2 is just excluded from the above consideration).Finally, M9 is put on the unique uncovered point K10, and then [9

º=1Cº which contains W8 coversthe whole of S (see Fig. 11 (a)). Thus, our consideration that the angular radius r9 (¹r8) is equal to a

17

153

Fig. 12: The spherical equilateral triangle K8K9K10 and the spherical square K9K11K12K10 in the kiteK8K4K6K7 .

side-length of spherical rhombus K8K9K6K10 which satis¯es (17) is con¯rmed and cos¡1¡ 1

3

¢is certainly

a solution for N = 9.

3.5 N = 10According to the considerations and results for N · 9, the angular radius for N = 10 should be smallerthan r9 and should be larger than tan¡1 2. Therefore, at the moment, let us consider the case forN = 10 under the assumption tan¡1 2 · r · r9. Then, we use the same positional relation of the ¯rst¯ve spherical caps for the cases of N = 7; 8, and 9. It is because, from the considerations for computationsof M4 and M5 in Subsection 3.2, we can assume that the allocations of points K2 and K5 for M4 andM5, respectively, satisfy the condition that [4

º=1Cº and [5º=1Cº each are in an extreme state. To make

sure, we shall check that K2 and K5 are such points after obtaining the exact value of r10. Under theabove assumptions, we can see that the shape of the set (W5)c is the kite K8K4K6K7 on the sphere asin the cases of N = 7, 8, and 9. Then, the relations (8) and (15) hold.

Now, we want to placed ¯ve centers of C6, C7, C8, C9, and C10 in the uncovered kite K8K4K6K7under the Minkowski condition. In this kite, we need to consider ¯ve points which keep spherical distancer with each other. From the consideration in Subsection 3.3, we should put M6 on K8. Therefore, oneof the ¯ve points should be K8. Then, the four remaining points be K9, K10, K11, and K12, and let usassume points K9 2 @C2, K10 2 @C5, K11 2 @C3, and K12 2 @C4. In addition, we suppose K9, K10,K11, and K12 satisfy the relations

ds(K8; K9) = ds(K8; K10) = ds(K9; K10)= ds(K9;K11) = ds(K10; K12) = ds (K11;K12) : (20)

Then, we see that K8K9K10 is a spherical equilateral triangle and that K9K11K12K10 is a sphericalsquare (see Fig. 12). If such an arrangement of vertices K9, K10, K11, and K12 are actually possible, theupper bound r10 (r9) is considered to be equal to a side-length (e.g. the spherical distance between thepoints K9 and K11) of the spherical square K9K11K12K10. Then, if M6, M7, M8, and M9 are placedon the points K8, K9, K10, and K12, respectively, the surface S is covered by the set [9

º=1Cº except forthe point K11. That is, our sequential covering is realized. At this time, K9 is just a cross point of theperimeters @C6 and @C2. Similarly, K10, K11, and K12 are just cross points of @C6 and @C5, @C7 and@C3, and @C8 and @C4, respectively. Note that the points K8, K9, and K10 are the positions such that[6

º=1Cº , [7º=1Cº , and [8

º=1Cº are in an extreme state, respectively. However, it is not checked yet thatK8, K9, and K10 are such positions. We shall check these facts after obtaining the exact values of theangular radius r and the coordinates of K11 and K12.

From the above consideration that our r is equal to a side-length of spherical square K9K11K12K10,the equation r = ds (K9; K11), for instance, should be satisfy. Then, we get the exact value of r whichsatis¯es this equation. Therefore, to begin with, we calculate the coordinates of the point K9 and K11which satisfy (20). Since the point K9 is a cross point of the perimeters @C6 and @C2, the coordinatesof K9 are calculable by using (4), (6), and the coordinates of K8 in Appendix A1 like the case of N = 9.The result is given in Appendix A1. Next, we need to calculate the coordinates of K11 without using

18

154

the coordinates of K9. Otherwise, we cannot get an equation of r in a closed form. So we notice thespherical distance ` = ds(K8; K11). Then, by applying the spherical cosine theorem to the sphericalisosceles triangle K8K9K11 of legs ds(K8; K9) = ds (K9; K11) = r, cos ` may be written as follows:

cos ` =3 cos3 r + 2 cos2 r ¡ cos r ¡ 2

¡1 ¡ cos2 r

¢ pcos r + 2 cos2 r

(1 + cos r)2.

It is because the inner angle at K9 of the spherical isosceles triangle K8K9K11 is the sum of the interiorangles of spherical equilateral triangle K8K9K10 and spherical square K9K11K12K10. In this connection,the inner angle of spherical equilateral triangle of side-length r is

cos¡1µ

cos rcos r + 1

¶,

and the inner angle of spherical square of side-length r is

cos¡1µ

cos r ¡ 1cos r + 1

¶.

As a result, we can obtain the coordinates of K11 = (x11; y11; z11) by using simultaneous equationsds (K11; M3) = r, ds (K8; K11) = `, and x2

11 + y211 + z2

11 = 1. Here, M3 = (sin r; 0; ¡ cos r) and referto Appendix A1 for the coordinates of K8. The explicit coordinates of cross point K11 are shown inAppendix A1.

From (9) and the coordinates of K9 and K11 in Appendix A1, we get the equation of the followingtype

r = ds(K9(r); K11(r)):

Then, the equation is solved against r by using mathematical software. The form of the solution and itsvalue is obtained as

r10 = ¹r9 = tan¡1

î

pp¯ + °

±

!

¼ 1:1544798334192707378319618404230 ¢ ¢ ¢ rad ;

(21)

where

® = ¡858694 ¡ 51517p

3p

58 sin µ ¡ 156065p

58 cos µ + 1172528 cos2 µ

+ 317144p

58 cos3 µ + 829400p

3 sin µ cos µ + 40368p

3p

58 sin µ cos2 µ

¡ 457504 cos3 µ cos³µ +

¼3

´¡ 107648 cos4 µ cos

³µ ¡ ¼

3

´;

¯ = 2116782535 ¡ 160201402p

3p

58 sin µ + 567207238p

58 cos µ

¡ 290254620 cos2 µ ¡ 542674448p

58 cos3 µ ¡ 2435016552p

3 sin µ cos µ

¡ 2859682576 cos3 µ cos³µ +

¼3

´;

° = 31573 + 5074p

58 cos³µ +

¼3

´¡ 53476 cos µ cos

³µ ¡ ¼

3

´;

± = 1472499³95 + 8

p58 cos

³µ +

¼3

´¡ 116 cos µ cos

³µ ¡ ¼

3

´´;

and

µ =13

tan¡1

Ã21

p3p

229691

!.

By using (21), we ¯nd numerically at an arbitrary precision that the relation (20) is attained as weexpected. Note that we get also another solution r ¼ 1:192753 ¢ ¢ ¢ rad (the exact equation for this valueis omitted since it is complicated) in the range of tan¡1 2 · r · r9 when r = ds (K9; K11) is solvedagainst r. But, for r ¼ 1:192753 ¢ ¢ ¢ rad, we ¯nd ds (K11;K12) < r numerically. Namely, K11 and K12

19

155

Fig. 13: The curve of A((W7 [ C8)c) when M8 is moved on the arc K10K11 of C7. The similar computationmethod as in Fig. 3 is taken. See the legend in Fig. 3.

do not satisfy the relation (20). Therefore, when M9 is put at K12, C9 will cover K11. Thus, we canexclude this answer r ¼ 1:192753 ¢ ¢ ¢ rad as the solution of inadequacy.

As we have noted, we check here whether the allocations of the points K2, K5, K8, K9, and K10 forM4, M5, M6, M7, and M8, respectively, satisfy the condition that [4

º=1Cº , [5º=1Cº , [6

º=1Cº , [7º=1Cº ,

and [8º=1Cº are in an extreme state, each, by using the value of (21). First, from the considerations for

determinations of Mi mentioned above, we checked numerically that A(Wi¡1 \ Ci) (or A((Wi¡1 [ Ci)c))are maximum with the restriction Mi 2 @Ci¡1 ([i

º=1Cº are in an extreme state) when Mi (i = 4; 5; 6and 7) are put at K2, K5, K8, and K9, respectively. Then, these results are graphically presented bythe curve of r ' 1:15448 corresponding to the value of (21) in Figs. 3, 4, 7, and 10. Therefore, forN = 10, our choice of Mi (i = 4; 5; 6 and 7) is justi¯ed. Next, in order to ¯nd the optimal position ofM8, let us place M8 at K10 and move it to K11 along the arc K10K11 of C7. Then, we notice the factthat [8

º=1Cº is in an extreme state (the area A(W7 \ C8) is maximum with the restriction M8 2 @C7)is the same as that the area A((W7 [ C8)c) is maximum with the restriction M8 2 @C7. Therefore, wecalculate A((W7 [ C8)c) against the moving point M8 on the arc K10K11. When r is equal to (21), we¯nd numerically that A((W7 [ C8)c) is maximum with the restriction M8 2 @C7 if and only if M8 isput at K10. The curve of r ' 1:15448 in Fig. 13 presents this result. In Fig. 13, the horizontal axis isthe position of M8 on the arc K10K11 of C7 and the vertical axis is the area A((W7 [ C8)c). Hence, forN = 10, our choice for M8 is justi¯ed.

As mentioned above, when eight spherical caps which the angular radius is the value of (21) areplaced on on S according to our sequential covering, the uncovered region (W8)c becomes a quadrangleon S. Then, from Corollary of Theorem 2 in Subsection 2.4 and the relations for four vertices of (W8)c,we ¯nd that ds (K11; K12) is equal to the spherical distance of the largest interval in the uncovered region(W8)c. Therefore, r9 is equal to (21). In addition, we ¯nd that [9

º=1Cº is in an extreme state if andonly if M9 is put on the point K12 2 @C8. Therefore, we choose M9 on K12, and then K11 is the uniqueuncovered point on S.

For N = 10, we initially assumed that r should be in the range¡tan¡1 2; r9

¤. Then, as a result of

the investigation, our upper bound r10, (21), has fallen within the range¡tan¡1 2; r9

¤. However, one

might suspect that the fact is due to the assumption. So, if r is in the range¡0; tan¡1 2

¤, we examine

whether W9 is able to cover S except for ¯nite points. When r is assumed to be equal to tan¡1 2, we ¯ndthat the set W9 leave an uncovered region on S. For its detail, refer to the consideration of Subsection3.6. Furthermore, for 0 < r < tan¡1 2, the uncovered region would become still bigger. Hence, theupper bounds r10 cannot be in the range

¡0; tan¡1 2

¤as in the cases of N = 8 and 9. Thus, we note

that our initial assumption tan¡1 2 < r · r9 is also con¯rmed (r = tan¡1 2 is excluded from the aboveconsideration).

As a result of consideration above, the set W9 covers S except for K11. Therefore, when M10 is putat the point K11, the whole of S is covered by [10

º=1Cº which contains W9 (see Fig. 14 (a)). Thus,our consideration that our r is equal to the upper bound r10 (a side-length of the spherical squareK9K11K12K10 which satis¯es (20)) is con¯rmed and (21) is certainly the upper bound for N = 10.

20

156

Fig. 14: (a) Our sequential covering for N = 10. (b) Our solution of Tammes problem for N = 10.Both viewpoints are (0; 0; 10). In this example, the coordinates of the centers are respectively (0; 0; ¡1),(0:26335; ¡0:87585; ¡0:40439), (0:91458; 0; ¡0:40439), (0:26335; 0:87585; ¡0:40439), (¡0:76292; 0:50440;¡0:40439), (¡0:77575; ¡0:57681; ¡0:25593), (¡0:13883; ¡0:78326; 0:60599), (¡0:79006; 0:092588; 0:60599),(0:084546; 0:74290; 0:66405), and (0:735778; ¡0:13295; 0:66405).

3.6 N = 11 and 12It is expected that the solution r11 for N = 11 should not be larger than r10. Then, we assumer · r10. Further, at the moment, we assume tan¡1 2 · r. Then, we use the same con¯guration ofthe ¯rst ¯ve spherical caps of the case N = 10. Namely, when the ¯fth spherical cap C5 is put on thesphere, a quadrilateral kite K8K4K6K7 on the sphere might be formed as the uncovered region. Hence,the relations (8) and (15) hold for the spherical distance among each vertices and r. Here, from theconsideration for determination of M6 in Subsection 3.3, the center M6 should be placed again on thepoint K8. Then, let K9 be one of the cross points of the perimeters @C6 and @C2, and let it be outsideC1. Similarly, let K10 be one of the cross points of @C6 and @C5, and let it be outside C1. At this time,the uncovered region (W6)c is reduced to the pentagon K9K4K6K7K10 which is bounded by perimetersof spherical caps. To make sure, we shall check whether the allocations of point K2, K5, and K8 for M4,M5, and M6, respectively, satisfy the condition that [4

º=1Cº , [5º=1Cº , and [6

º=1Cº are in an extremestate, each, after obtaining the exact value of angular radius r.

Before considering the case of N = 11, we look back upon the cases of N = 9 and 10. For N = 9, inSubsection 3.4, we considered that two spherical equilateral triangles K8K9K10 and K9K6K10 on the kiteK8K4K6K7 on the sphere (see Fig. 9). At that time, we placed the centers of four caps on each verticesof two spherical equilateral triangles and obtained the upper bound r9 by using the relation that theangular radius r is equal to a side-length of the spherical equilateral triangle. For N = 10, in Subsection3.5, we considered the spherical equilateral triangle K8K9K10 and the spherical square K9K11K12K10(see Fig 12). Then, the centers of ¯ve caps are put on each vertices and the upper bound r10 is obtainedby using the relation that r is equal to a side-length of the spherical square K9K11K12K10.

Now, for N = 11, we assume the spherical equilateral triangle K8K9K10 and the spherical regularpentagon of side-length r on the kite K8K4K6K7, in order to place six more caps under the Minkowskicondition. Therefore, here, let us assume K9, K10, K11(2 @C2 or @C3), and K12(2 @C4 or @C5) satisfy

ds(K8; K9) = ds(K8; K10) = ds(K9; K10) = ds(K9; K11)= ds(K11; K6) = ds (K6; K12) = ds (K12; K10) . (22)

In (22), the relation ds(K8; K9) = ds(K8; K10) which is automatically satis¯es is included. Then, we cantake the angular radius r to be a side-length of the spherical regular pentagon K9K11K6K12K10 similarto the cases of N = 9 and 10 (see Fig. 15). At this time, for example, when Mi's (i = 7; : : : ; 10) areplaced on the points K9, K10, K12, and K6, respectively, it will be obvious that each [i

º=1Cº is in anextreme state subsequently for i = 7; : : : ; 10 and that K11 is a tentative uniquely uncovered point onS. Therefore, ¯nally, M11 can be placed at K11. Thus, we consider that the upper bound r11 (r10) isthe abouve r. However, it is not checked yet that the positions of K9, K10, K12, and K6 satisfy thecondition that [i

º=1Cº are in an extreme state for i = 7; : : : ; 10, respectively. We shall check this factafter obtaining the exact value of the angular radius r and the coordinates of K11 and K12.

21

157

Fig. 15: The kite K8K4K6K7, the spherical equilateral triangle K8K9K10, and the spherical regular pentagonK9K11K6K12K10.

Now, by applying the spherical cosine and sine theorems, the inner angle of spherical regular pentagonof side-length r is given by

cos¡1

Ã2 cos r ¡ 1 ¡

p5

2 (cos r + 1)

!.

Here, we pay attention to the spherical isosceles triangle K9K11K6 whose legs satisfy ds(K11; K9)= ds (K11; K6) = r. Then, by applying the spherical cosine theorem to this isosceles triangle K9K11K6,we have

ds (K6; K9) = cos¡1

Ãcos2 r +

(1 ¡ cos r)¡2 cos r ¡ 1 ¡

p5¢

2

!. (23)

Note that the left hand side of (23) is presented as the function of r by using (9) and the coordinatesof K6 and K9 in Appendix A1 (see the equation (18)). Equation (23) is solved against r by usingmathematical software. As a result, the value of r is obtained as

r = tan¡1 2 ¼ 1:10715 rad. (24)

Therefore, the side-length of spherical regular pentagon K9K11K6K12K10 which satis¯es the relation(22) is equal to tan¡1 2.

Here, from the relation (15), let us consider the special case r = ds(K6; K4). When the equationr = ds(K6; K4) is solved against r by using mathematical software, we get again the solution r = tan¡1 2.Therefore, r = ds(K6; K4) = ds(K6; K7) = tan¡1 2 from the relations (8). On the other hand, from thefact that the side-length of spherical regular pentagon K9K11K6K12K10 which satis¯es (22) is tan¡1 2,ds(K6; K11) = ds(K6; K12) = tan¡1 2. Thus, we ¯nd the result as follows:

K11 ´ K4 and K12 ´ K7

if and only if r = tan¡1 2. It follows that the spherical regular pentagon K9K11K6K12K10 which satis¯es(22) is identical with the spherical regular pentagon K9K4K6K7K10 of side-length tan¡1 2. We note thatthe vertex K8 of kite K8K4K6K7 is on the perimeter of the ¯rst spherical cap C1 then. In addition,from Subsection 2.1, we have shown that tan¡1 2 is equal to the upper bound of the range of angulardiameter for the kissing number k = 5. Therefore, as a result, we ¯nd that the spherical regular pentagonK9K4K6K7K10 satis¯es the relation as follows:

ds(K8; K9) = ds(K8; K10) = ds(K9; K10) = ds(K9;K4)= ds(K4; K6) = ds (K6; K7) = ds (K7; K10) = ds (P; K9)= ds (P; K4) = ds (P; K6) = ds (P; K7) = ds (P; K10) = tan¡1 2,

(25)

where P is the center point of spherical regular pentagon K9K4K6K7K10. From the above considerations,for tan¡1 2 · r · r10, we see that the maximal spherical regular pentagon of side-length r on theuncovered region (W6)

c is coincided with the spherical regular pentagon K9K4K6K7K10 which satis¯esthe relation (25).

Then, for N = 11, let us examine whether our sequential covering is actually possible when theangular radius r of caps is equal to tan¡1 2 ¼ 1:10715. First, we ¯nd numerically that the allocations

22

158

Fig. 16: The curve of A((W8 [ C9)c) when M9 is moved on the arc K12P of C8. The similar computationmethod as in Fig. 3 is taken. See the legend in Fig. 3.

of points K2, K5, K8, and K9 for M4, M5, M6, and M7, respectively, satisfy the condition that eachA((Wi¡1[Ci)c) (or A((Wi¡1[Ci)c)) is maximum with restriction Mi 2 @Ci¡1 ([i

º=1Cº are in an extremestate) for i = 4; 5; 6 and 7. As mentioned above, the results are indicated by the curve of r ' 1:10715 radcorresponding to r = tan¡1 2 in Figs. 3, 4, 7, and 10. Then, we see that K10 is the cross point of @C6,@C5, and @C7. Futhermore, K4 is the cross point of @C2, @C3, and @C7. Therefore, from the restrictionM8 2 @C7, M8 is put at a certain point on the arc K10K4 of C7. Then, we move M8 on the arc K10K4of C7 and search for the position of M8 where the area A(W7 \ C8) is maximum. For r = tan¡1 2, wechecked numerically that [8

º=1Cº are in an extreme state if and only if M8 is put at K10 or K4. Thefact is graphically presented by the curve of r ' 1:10715 in Fig. 13. Note that the arc K10K11 in Fig. 13turns out the arc K10K4 since K11 ´ K4 for r = tan¡1 2. Thus, in this paper, M8 is put at K10. Then,the uncovered region (W8)c is the quadrangle K4K6K7P on S. Next, for r = tan¡1 2, let us place M9 atK7 and move it to P along the arc K7P of C8, and then we search for the position of M9 where the areaA((W8 [ C9)c) is maximum. As a result, we ¯nd that [9

º=1Cº is in an extreme state when M9 is putat K7. The curve of r ' 1:01715 in Fig. 16 indicates this fact. In this ¯gure, the horizontal axis is theposition of M9 on the arc K7P of C8 and the vertical axis is the area A((W8 [ C9)c). Therefore, we putM9 at K7. Then, K6 and P are the cross points on @C9 and the uncovered triangle K4K6P on S is theequilateral triangle of side-length tan¡1 2. Hence, when M10 is put on the point K6 2 @C9, we see that[10

º=1Cº is in an extreme state automatically and that S is covered by the set W10 except for two pointsK4 (or K11) and P (the center point of spherical regular pentagon K9K4K6K7K10) from the relation(25). At this time, we see that P is the cross point of perimeters @C7, @C8, and @C9. Furthermore, asa result of calculating by using the coordinates of K9, K10, and K7 for r = tan¡1 2, P is just the northpole (0; 0; 1) certainly. Therefore, if M11 is put on K4 2 @C10, [11

º=1Cº is in an extreme state and P is aunique uncovered point on S. Then, in order to cover whole of S, we have to put one more cap at P . Inother words, when r = tan¡1 2, according to our sequential covering, we can put twelve caps on S underMinkowski covering. Therefore, M11 and M12 are put on K4 and P , respectively, then [12

º=1Cº whichcontains W11 covers the whole of S. Namely, we are able to see that ds (P;K4) = tan¡1 2 = r11. Thus,we consider that tan¡1 2 is the upper bound r12 for N = 12. As a result, we ¯nd that the positions ofcaps of our sequential covering for N = 12 correspond to the regular icosahedral vertices (see Fig. 17(a)). Therefore, if all spherical caps of our sequential covering for N = 12 are replaced by half-caps asmentioned in Subsection 2.1, all of those half-caps contact other ¯ve half-caps and there is no space forthose half-caps to move.

Then, how about r11? From the results of N = 10 and 12, it becomes obvious that our initialassumption r12 = tan¡1 2 · r · r10 is indispensable. At the time the center M6 is placed on the pointK8, we had to place ¯ve more caps of the angular radius r on the uncovered region (W6)

c under theMinkowski covering. Then, from the above considerations, we see that the maximal spherical regularpentagon of side-length r (r12 · r · r10) on (W6)

c is identical with the spherical regular pentagonK9K4K6K7K10 which satis¯es the relation (25). So, we conclude that r11 is equal to r12 = tan¡1 2and use the same con¯guration of the ¯rst eleven spherical caps of the case N = 12. However, whenr11 = tan¡1 2, our sequential covering for N = 11 covers S except for the point P . But, if only theeleventh cap C11 is replaced by a closed cap, our eleven caps can completely cover the whole of S under

23

159

Fig. 17: (a) Our sequential covering for N = 12. (b) Our solution of Tammes problem for N = 12. Bothviewpoints are (0; 0; 10). In this example, the coordinates of the centers are respectively (0; 0; ¡1), (0:27639;¡0:85065; ¡0:44721), (0:89443; 0; ¡0:44721), (0:27639; 0:85065; ¡0:44721), (¡0:72361; 0:52573; ¡0:44721),(¡0:72361; ¡0:52573; ¡0:44721), (¡0:27639; ¡0:85065; 0:44721), (¡0:89443; 0; 0:44721), (¡0:27639; 0:85065;0:44721), (0:72361; 0:52573; 0:44721), (0:72361; ¡0:52573; 0:44721), and (0; 0; 1).

the Minkowski condition. Thus, for N = 11, we need a special care as above.

4 ConclusionIn Section 3, we calculated the upper bound of r for N = 2; : : : ; 12, such that the set [N

º=1Cº whichcontains WN¡1 covers the whole spherical surface S (see Table. 2).

It was described, in Subsection 2.1, that the covering with spherical caps of angular radius r iscorrespondent with the packing with half-caps (see Subsection 2.1 for its de¯nition) according to ourmethod that the centers of spherical caps are chosen on the perimeters of other spherical caps under theMinkowski condition. Let us suppose the centers of N half-caps are placed on the positions of the centersof spherical caps Ci (i = 1; : : : ; N) which are considered in Section 3. At this time, we get the packingwith N congruent half-caps (see Figs. 6 (b), 8 (b), 11 (b), and 14 (b), 18). Then, for N = 2; : : : ; 12,we ¯nd that the upper bound of angular radius of our problem with N congruent spherical caps andthe value of angular diameter of Tammes problem with N congruent spherical caps are equivalent. Inaddition, we ¯nd that the location of centers of our problem is correspondent with that of the Tammesproblem for N = 2; : : : ; 12, respectively [Danzer 63], [Fejes T¶oth 72], [SchÄutte and van der Waerden 51].

Accordingly, we ¯nd that the fact that the results of our problem are coincident with those of Tammesproblem about N = 2; : : : ; 12 at least. Especially, we obtained in this paper the exact value of r10 forN = 10 (see (21)), whereas Danzer [Danzer 63] has obtained the approximate range [1:154479; 1:154480]of angular diameter for N = 10. Futher, SchÄutte, van der Waerden [SchÄutte and van der Waerden 51], andDanzer [Danzer 63] have solved the Tammes problem for N = 7; 8; 9; 10, and 11 through the considerationon irreducible graphs obtained by connecting those points, among N points, whose spherical distance isexactly the minimal distance. Then, after establishing the theorem which states that such irreduciblegraphs can only have triangles and quadrangles, SchÄutte, van der Waerden, and Danzer proved andobtained the minimal distance r for respective values of N = 7; 8; 9 and 10. Further, they need theindependent considerations for N = 7; 8; 9; 10, and 11, respectively [Danzer 63], [Fejes T¶oth 72], [SchÄutteand van der Waerden 51]. In contrast to this, we presented in this paper a systematic method which isdi®erent from the approach by SchÄutte, van der Waerden, and Danzer. Namely, as shown in Subsection2.4 and Section 3, our method is able to obtain a solution for N by using the results for the case N ¡ 1or N ¡ 2 successively. In addition, in this study, we have considered the packing problem from thestandpoint of sequential covering. The advantages of our approach are that we only need to observeuncovered region in the process of packing and that this uncovered region decreases step by step as thepacking proceeds. At least, in the cases of N · 12, the solutions of Tammes problem can be found byour method. However, we may say that our method has not necessarily given a mathematical rigorousproof about our result.

24

160

Table 2: Upper bound rN of our problem.

Number of Spherical Caps

N

Upper Bound of Angular Radius of Our Problem

rN [rad]

2 ¼ ¼ 3:14159

3 2¼3 ¼ 2:09440

4 ¼ ¡ cos¡1¡ 1

3

¢¼ 1:91063

5 ¼2 ¼ 1:57080

6 ¼2 ¼ 1:57080

7 cos¡1³1 ¡ 4p

3cos

¡ 7¼18

¢´¼ 1:35908

8 cos¡1³¡ 1

7 + 2p

27

´¼ 1:30653

9 cos¡1¡ 1

3

¢¼ 1:23096

10 (See Equation (21)) 1:15448

11 tan¡1 2 ¼ 1:10715

12 tan¡1 2 ¼ 1:10715

In this paper, we used the mathematical software, Maple, which is capable of manipulating complicatealgebraic expressions exactly and is also useful for numerical computations. Consequently, we were ableto calculate strict coordinates and solutions.

It is interesting to point out that our method presented in this paper will be also useful for providinga conjectured value of optimal angular radius for any N by using the already known value for N ¡ 1.This will be done as follows: (i) ¯rst we set the angular radius of a spherical cap by using the knownexact or approximate value for N ¡ 1, namely, we put r = rN¡1 where rN¡1 is the exact or approximatevalue of r for N ¡ 1; (ii) then we put N spherical caps with radius r through our method of sequentialcovering; (iii) in this stage, it is possible that the Minkowski condition will be broken especially when theN -th spherical cap is placed. If this is true, decrease the size of r so far as the Minkowski condition issatis¯ed and go to Step (ii). Otherwise, check if the whole of S except for a point or a line segment iscovered by N ¡ 1 spherical caps. If it holds, the procedure ends, else increase r by a prescribed value¢r then go to Step (ii). We will be able to estimate the value of an optimal angular radius for any N bythis procedure.

Here, we remark on the e±ciency of covering. Now, let us de¯ne the e±ciency of covering on sphericalsurface by (area of S)=(N £ (area of a cap)) = 2= (N £ (1 ¡ cos r)). Then, from the value of rN andfrom the positions of Ci ( i = 1; : : : ;N ), it appears that our solutions of N = 2; : : : ; 12 give the worste±cient covering of S with N congruent spherical caps under the Minkowski condition respectively. Atleast, from Lemma and Theorem in Subsection 2.2, it is obvious that our solutions r2 and r3 for N = 2and 3, respectively, give the worst e±cient covering under the Minkowski condition. Namely, our resultsr2 = ¼ and r3 = 2¼

3 respectively give the e±ciency of covering 12 and 4

9 . However, its proof for othervalues of N is still open and it is taken as a future subject.

AcknowledgementThe authors would like to thank Prof. L. Danzer, Univ. of Dortmund, and Prof. H. Maehara, Univ.

of Ryukyu, for their helpful comments.The research was partly supported by the Grant-in-Aid for Scienti¯c Research (the Grant-in-Aid for

25

161

Fig. 18: Our solutions of Tammes problem for N = 8; 9; 10 and 12. Note that the sphere is drawn with thewireframe.

26

162

JSPS Fellows) from the Ministry of Education, Culture, Sports, Science, and Technology (MEXT) ofJapan.

References[Danzer 63] L. Danzer. \Endliche Punktmengen auf der 2-SphÄare mit mÄoglichst gro¼en Minimalabstand." Uni-

versitÄat GÄottingen, 1963. (\Finite Point-Set on S2 with Minimum Distance as Large as Possible." DiscreteMathematics, 60 (1986), 3-66.)

[Fejes T¶oth 69] G. Fejes T¶oth. \KreisÄuberdeckungen der SphÄare." Studia Scientiarum Mathematicarum Hun-garica, 4 (1969), 225-247.

[Fejes T¶oth 72] L. Fejes T¶oth. \Lagerungen in der Ebene, auf der Kugel und im Raum." 2nd ed (1972), Springer-Verlag, Heidelberg.

[Fejes T¶oth 99] L. Fejes T¶oth. \Minkowski Circle Packings on the Sphere."Discrete & Computational Geometry,22 (1999), 161-166.

[SchÄutte and van der Waerden 51] K. SchÄutte. and B. L. van der Waerden. \Auf welcher Kugel haben 5, 6, 7,8, oder 9 Punkte mit Mindestabstand Eins Platz?." Mathematische Annalen, 123 (1951), 96-124.

[Sugimoto and Tanemura 01a] T. Sugimoto and M. Tanemura. \Random Sequential Covering of a Sphere withIdentical Spherical Caps." Forma, 16 (2001), 209-212.

[Sugimoto and Tanemura 01b] T. Sugimoto and M. Tanemura. \Covering of a Sphere with Congruent SphericalCaps under the Condition of Minkowski Set of Centers." Research Memorandum, ISM, 817 (2001).

[Sugimoto and Tanemura 02] T. Sugimoto and M. Tanemura. \Sphere Covering under the Minkowski Condi-tion." Research Memorandum, ISM, 857 (2002).

[Sugimoto and Tanemura 03] T. Sugimoto and M. Tanemura. \Packing of 10, 11, and 12 Congruent Caps on aSphere." Research Memorandum, ISM, 877 (2003).

[Teshima and Ogawa 00] Y. Teshima and T. Ogawa. \Dense Packing of Equal Circle on a Sphere by theMinimum-Zenith Method: Symmetrical Arrangement." Forma, 15 (2000), 347-364.

Appendices

A1. The Coordinates of the Point Ki (i = 3; 4; 5; 6; 7; 8; 9 and 11)In the following, r (tan¡1 2 · r < ¼

2 ) is the angular radius.K3 = (x3, y3, z3):

Ãsin r

¡cos2 r ¡ 2 cos r ¡ 1

¢

(cos r + 1)2; ¡2 cos r sin r

p2 cos r + 1

(cos r + 1)2; ¡ cos r

!.

K4 = (x4, y4, z4):Ã

2 cos r sin r (2 cos r + 1)(cos r + 1)2

; ¡2 cos r sin rp

2 cos r + 1(cos r + 1)2

; ¡4 cos2 r ¡ cos r ¡ 1cos r + 1

!.

K5 = (x5, y5, z5):Ã

sin r¡cos2 r ¡ 2 cos r ¡ 1

¢

(cos r + 1)2;

2 cos r sin rp

2 cos r + 1(cos r + 1)2

; ¡ cos r

!.

K6 = (x6, y6, z6):Ã

2 cos r sin r (2 cos r + 1)(cos r + 1)2

;2 cos r sin r

p2 cos r + 1

(cos r + 1)2; ¡4 cos2 r ¡ cos r ¡ 1

cos r + 1

!.

K7 = (x7, y7, z7):

27

163

Ã2 cos r sin r (2 cos r + 1) (cos r ¡ 1)

(cos r + 1)3;

2 cos r sin r (3 cos r + 1)p

2 cos r + 1(cos r + 1)3

;

¡4 cos2 r ¡ cos r ¡ 1cos r + 1

¶.

K8 = (x8, y8, z8):

µ2 cos r sin r (cos r ¡ 1) (2 cos r + 1)

9 cos3 r ¡ cos2 r ¡ cos r + 1;

2 cos r sin r (cos r ¡ 1)p

2 cos r + 19 cos3 r ¡ cos2 r ¡ cos r + 1

;

¡4 cos4 r ¡ cos3 r + 5 cos2 r + cos r ¡ 19 cos3 r ¡ cos2 r ¡ cos r + 1

¶.

K9 = (x9, y9, z9):

Ãsin r

¡cos3 r ¡ 5 cos2 r ¡ cos r + 1

¢

9 cos3 r ¡ cos2 r ¡ cos r + 1; ¡ 4 sin r cos2 r

p2 cos r + 1

9 cos3 r ¡ cos2 r ¡ cos r + 1;

¡cos r¡cos3 r + 11 cos2 r ¡ cos r ¡ 3

¢

9 cos3 r ¡ cos2 r ¡ cos r + 1

!.

K11 = (x11, y11, z11) are shown in the textual form of Maple due to their lengthy expressions. Notethat, for example, cos(r)2 expresses cos2 r.

x11 = 2¤(¡1=2¤(31¤cos(r)7+173¤cos(r)6+18¤sqrt(cos(r)¤(2¤cos(r)+1))¤cos(r)6+65¤cos(r)5+124¤sqrt(cos(r)¤(2¤cos(r)+1))¤cos(r)5¡9¤cos(r)4+2¤sqrt(cos(r)¤(2¤cos(r)+1))¤cos(r)4+cos(r)3¡32¤sqrt(cos(r) ¤ (2 ¤ cos(r)+1))¤ cos(r)3 ¡5 ¤ cos(r)2 +14¤ sqrt(cos(r) ¤ (2 ¤ cos(r)+1))¤ cos(r)2 ¡ cos(r)+4 ¤ sqrt(cos(r) ¤ (2 ¤ cos(r) + 1)) ¤ cos(r) + 1 ¡ 2 ¤ sqrt(cos(r) ¤ (2 ¤ cos(r) + 1)))=(sin(r) ¤ cos(r)3 ¤ (¡2 ¤cos(r)+cos(r)2 +2¤cos(r)3 ¡1))+1=2¤sqrt((31¤ cos(r)7 +173¤ cos(r)6 +18¤sqrt(cos(r)¤ (2¤cos(r)+1))¤ cos(r)6 +65¤ cos(r)5 +124¤sqrt(cos(r)¤ (2¤ cos(r)+1))¤cos(r)5 ¡9¤ cos(r)4 +2¤sqrt(cos(r)¤ (2¤cos(r)+1))¤cos(r)4 +cos(r)3 ¡32¤sqrt(cos(r)¤(2¤cos(r)+1))¤cos(r)3 ¡5¤cos(r)2 +14¤sqrt(cos(r)¤(2 ¤ cos(r) + 1)) ¤ cos(r)2 ¡ cos(r) + 4 ¤ sqrt(cos(r) ¤ (2 ¤ cos(r) + 1)) ¤ cos(r) + 1 ¡ 2 ¤ sqrt(cos(r) ¤ (2 ¤cos(r)+1)))2=(sin(r)2 ¤ cos(r)6 ¤ (¡2 ¤ cos(r)+ cos(r)2 +2 ¤ cos(r)3 ¡ 1)2)+ (39 ¤ cos(r)4 + cos(r)6 +1+22¤ cos(r)5 +12¤ cos(r)3 ¡9¤ cos(r)2 ¡2¤ cos(r))=cos(r)6=(1¡3¤ cos(r)2 +2¤ cos(r)3)¤ (1+4¤ cos(r)¡36¤cos(r)3 ¡4¤cos(r)2 +30¤cos(r)4 ¡52¤cos(r)6 +548¤cos(r)7 +124¤cos(r)5 ¡4¤(cos(r)¤(2¤cos(r)+1))(1=2)+1177¤cos(r)8 ¡52¤(cos(r)¤(2¤cos(r)+1))(1=2)¤cos(r)5 +52¤(cos(r)¤(2¤cos(r)+1))(1=2)¤cos(r)4 + 828 ¤ cos(r)7 ¤ (cos(r) ¤ (2 ¤ cos(r) + 1))(1=2) + 196 ¤ (cos(r) ¤ (2 ¤ cos(r) + 1))(1=2) ¤ cos(r)6 ¡12¤ (cos(r)¤ (2¤ cos(r)+1))(1=2)¤ cos(r)3 +12¤ (cos(r)¤ (2¤ cos(r)+1))(1=2)¤ cos(r)2 +4¤ (cos(r)¤ (2¤cos(r)+1))(1=2)¤cos(r))=(¡4¤cos(r)¡4¤cos(r)2 +2¤cos(r)3 ¡1+5¤cos(r)4 +2¤cos(r)5)))¤ cos(r)4 ¤(1¡3¤cos(r)2+2¤cos(r)3)=(39¤cos(r)4 +cos(r)6+1+22¤cos(r)5+12¤cos(r)3 ¡9¤cos(r)2 ¡2¤cos(r)),

y11 = 1=2¤((2¤cos(r)2+3¤cos(r)3¡cos(r)¡2¤(1¡cos(r)2)¤sqrt(2¤cos(r)2 +cos(r)))=((1+cos(r))2)¡(4¤cos(r)4 ¡cos(r)3 +5¤cos(r)2 +cos(r)¡1)=(9¤cos(r)3 ¡cos(r)2 ¡cos(r)+1))¤(9¤cos(r)3 ¡cos(r)2 ¡cos(r)+1)=(cos(r)¤sin(r)¤(¡1+cos(r))¤sqrt(2¤cos(r)+1))¡(2¤(2¤cos(r)+1)¤(¡1+cos(r))¤cos(r)¤sin(r)=(9 ¤ cos(r)3 ¡ cos(r)2 ¡ cos(r)+1)¡ (4 ¤ cos(r)4 ¡ cos(r)3 +5 ¤ cos(r)2 + cos(r)¡1)=(9 ¤ cos(r)3 ¡cos(r)2 ¡ cos(r)+1) ¤ tan(r)) ¤ (¡1=2 ¤ (31 ¤ cos(r)7 +173 ¤ cos(r)6 +18 ¤ (cos(r) ¤ (2 ¤ cos(r)+1))(1=2) ¤cos(r)6 +65¤cos(r)5 +124¤ (cos(r)¤ (2¤cos(r)+1))(1=2)¤cos(r)5 ¡9¤cos(r)4 +2¤(cos(r)¤ (2¤cos(r)+1))(1=2) ¤ cos(r)4 + cos(r)3 ¡ 32 ¤ (cos(r) ¤ (2 ¤ cos(r)+ 1))(1=2) ¤ cos(r)3 ¡ 5 ¤ cos(r)2 +14 ¤ (cos(r) ¤ (2 ¤cos(r)+1))(1=2)¤cos(r)2 ¡cos(r)+4¤(cos(r)¤(2¤cos(r)+1))(1=2)¤cos(r)+1¡2¤(cos(r)¤(2¤cos(r)+1))(1=2))=sin(r)=cos(r)3=(¡2¤cos(r)+cos(r)2 +2¤cos(r)3 ¡1)+1=2¤((31¤cos(r)7 +173¤cos(r)6 +18¤(cos(r) ¤ (2 ¤ cos(r)+ 1))(1=2) ¤ cos(r)6 +65 ¤ cos(r)5 +124 ¤ (cos(r) ¤ (2 ¤ cos(r)+1))(1=2) ¤ cos(r)5 ¡ 9 ¤cos(r)4 +2¤(cos(r)¤(2¤cos(r)+1))(1=2)¤cos(r)4+cos(r)3¡32¤(cos(r)¤(2¤cos(r)+1))(1=2)¤cos(r)3¡5 ¤ cos(r)2 + 14 ¤ (cos(r) ¤ (2 ¤ cos(r) + 1))(1=2) ¤ cos(r)2 ¡ cos(r) + 4 ¤ (cos(r) ¤ (2 ¤ cos(r) + 1))(1=2) ¤cos(r)+1¡2¤(cos(r)¤(2¤cos(r)+1))(1=2))2=sin(r)2=cos(r)6=(¡2¤cos(r)+cos(r)2 +2¤cos(r)3 ¡1)2 +(39 ¤ cos(r)4 + cos(r)6 +1+22 ¤ cos(r)5 +12 ¤ cos(r)3 ¡9 ¤ cos(r)2 ¡2 ¤ cos(r))=cos(r)6=(1¡3 ¤ cos(r)2 +2 ¤ cos(r)3) ¤ (1 + 4 ¤ cos(r) ¡ 36 ¤ cos(r)3 ¡ 4 ¤ cos(r)2 + 30 ¤ cos(r)4 ¡ 52 ¤ cos(r)6 + 548 ¤ cos(r)7 + 124 ¤

28

164

Fig. 19: Overlapping Area Aij

cos(r)5 ¡4¤ (cos(r)¤ (2¤ cos(r)+1))(1=2)+1177¤ cos(r)8 ¡52¤ (cos(r)¤ (2¤cos(r)+1))(1=2)¤ cos(r)5 +52¤(cos(r)¤(2¤cos(r)+1))(1=2)¤cos(r)4+828¤cos(r)7¤(cos(r)¤(2¤cos(r)+1))(1=2)+196¤(cos(r)¤(2¤cos(r)+1))(1=2)¤cos(r)6¡12¤(cos(r)¤(2¤cos(r)+1))(1=2)¤cos(r)3+12¤(cos(r)¤(2¤cos(r)+1))(1=2)¤cos(r)2 +4¤(cos(r)¤(2¤cos(r)+1))(1=2)¤cos(r))=(¡4¤cos(r)¡4¤cos(r)2 +2¤cos(r)3 ¡1+5¤cos(r)4 +2¤cos(r)5))(1=2))=(39¤cos(r)4+cos(r)6 +1+22¤cos(r)5+12¤cos(r)3 ¡9¤cos(r)2 ¡2¤cos(r))¤cos(r)3¤(1¡3¤cos(r)2 +2¤cos(r)3)=sin(r)=(¡1+cos(r))=(2¤cos(r)+1)(1=2)¤(9¤cos(r)3 ¡cos(r)2 ¡cos(r)+1),

z11 = ¡1+2 ¤ tan(r) ¤ (¡1=2 ¤ (31 ¤ cos(r)7 +173 ¤ cos(r)6 +18 ¤ sqrt(cos(r) ¤ (2 ¤ cos(r)+1)) ¤ cos(r)6 +65 ¤ cos(r)5 +124 ¤ sqrt(cos(r) ¤ (2 ¤ cos(r)+1)) ¤ cos(r)5 ¡9 ¤ cos(r)4 +2 ¤ sqrt(cos(r) ¤ (2 ¤ cos(r)+1)) ¤cos(r)4+cos(r)3¡32¤sqrt(cos(r)¤(2¤cos(r)+1))¤cos(r)3¡5¤cos(r)2+14¤sqrt(cos(r)¤(2¤cos(r)+1))¤cos(r)2 ¡cos(r)+4¤sqrt(cos(r)¤ (2¤cos(r)+1))¤cos(r)+1¡2¤sqrt(cos(r)¤ (2¤ cos(r)+1)))=(sin(r)¤cos(r)3 ¤(¡2¤cos(r)+cos(r)2 +2¤cos(r)3¡1))+1=2¤sqrt((31¤cos(r)7 +173¤cos(r)6+18¤sqrt(cos(r)¤(2 ¤ cos(r) + 1)) ¤ cos(r)6 + 65 ¤ cos(r)5 + 124 ¤ sqrt(cos(r) ¤ (2 ¤ cos(r) + 1)) ¤ cos(r)5 ¡ 9 ¤ cos(r)4 + 2 ¤sqrt(cos(r)¤(2¤cos(r)+1))¤cos(r)4+cos(r)3¡32¤sqrt(cos(r)¤(2¤cos(r)+1))¤cos(r)3¡5¤cos(r)2+14¤sqrt(cos(r)¤(2¤cos(r)+1))¤cos(r)2¡cos(r)+4¤sqrt(cos(r)¤(2¤cos(r)+1))¤cos(r)+1¡2¤sqrt(cos(r)¤(2¤cos(r)+1)))2=(sin(r)2¤cos(r)6¤(¡2¤cos(r)+cos(r)2+2¤cos(r)3¡1)2)+(39¤cos(r)4+cos(r)6+1+22¤ cos(r)5 +12¤ cos(r)3 ¡9¤ cos(r)2 ¡2¤ cos(r))=cos(r)6=(1¡3¤ cos(r)2 +2¤ cos(r)3)¤ (1+4¤ cos(r)¡36¤cos(r)3 ¡4¤cos(r)2 +30¤cos(r)4 ¡52¤cos(r)6 +548¤cos(r)7 +124¤cos(r)5 ¡4¤(cos(r)¤(2¤cos(r)+1))(1=2)+1177¤cos(r)8 ¡52¤(cos(r)¤(2¤cos(r)+1))(1=2)¤cos(r)5 +52¤(cos(r)¤(2¤cos(r)+1))(1=2)¤cos(r)4 + 828 ¤ cos(r)7 ¤ (cos(r) ¤ (2 ¤ cos(r) + 1))(1=2) + 196 ¤ (cos(r) ¤ (2 ¤ cos(r) + 1))(1=2) ¤ cos(r)6 ¡12¤ (cos(r)¤ (2¤ cos(r)+1))(1=2)¤ cos(r)3 +12¤ (cos(r)¤ (2¤ cos(r)+1))(1=2)¤ cos(r)2 +4¤ (cos(r)¤ (2¤cos(r)+1))(1=2)¤cos(r))=(¡4¤cos(r)¡4¤cos(r)2 +2¤cos(r)3 ¡1+5¤cos(r)4 +2¤cos(r)5)))¤ cos(r)4 ¤(1¡3¤cos(r)2+2¤cos(r)3)=(39¤cos(r)4 +cos(r)6+1+22¤cos(r)5+12¤cos(r)3 ¡9¤cos(r)2 ¡2¤cos(r)).

A2. The Overlapping Area of Two Congruent Spherical CapsWe give the overlapping area Aij of two congruent spherical caps Ci and Cj of angular radius r,

where their centers are Mi and Mj , respectively (see Fig. 19). We denote by sij the spherical distancebetween Mi and Mj and we assume r · sij · 2r. Then, let us assume G be the middle point of thegeodesic arc MiMj . Further, we de¯ne T1 and T2 as the two cross points of perimeters @Ci and @Cj.Thus, the spherical distance hij between T1 and T2 is expressed by using the spherical cosine formulaabout a spherical right triangle MiGT1 as follows:

hij = 2 cos¡1µ

cos rcos( sij

2 )

¶:

If the points Mi, T1 and T2 are mutually connected by geodesic arcs, there arises a spherical isoscelestriangle MiT1T2 on the unit sphere. Then, let ¸ and ¹ be the interior angles at vertices Mi and T1 (T2)of this triangle, respectively. By the spherical cosine theorem, we have

¸ = cos¡1µ

cos hij ¡ cos2 rsin2 r

¶, ¹ = cos¡1

µ1 ¡ cos hij

tan r ¢ sinhij

¶.

Then the area A1 of the spherical isosceles triangle MiT1T2 turns out

29

165

Fig. 20: The sketch of the kite K8K4K6K7 and three spherical equilateral triangles K8K9K10, K9K6K10, andK6M3K2 in the pentagon K8K4M3K2K7 on S

A1 = ¸ + 2¹ ¡ ¼ = cos¡1µ

cos hij ¡ cos2 rsin2 r

¶+ 2 cos¡1

µ1 ¡ cos hij

tan r ¢ sin hij

¶¡ ¼.

On the other hand, the area of a spherical cap of the angular radius r is jCj = 2¼ (1 ¡ cos r). Thereforethe area A2 of a sector with angle ¸ of the spherical cap is equal to jCj ¢ ¸

2¼ , namely

A2 = cos¡1µ

cos hij ¡ cos2 rsin2 r

¶¢ (1 ¡ cos r) .

Thus the overlapping area Aij of Ci and Cj is

Aij = 2 (A2 ¡ A1) = ¡2 cos¡1µ

cos hij ¡ cos2 rsin2 r

¶¢ cos r ¡ 4 cos¡1

µ1 ¡ cos hij

tan r ¢ sin hij

¶+ 2¼.

A3. The proof of ds(K9; K2) = 2r for N = 9For N = 9, the ¯rst ¯ve centers M1, M2, M3, M4, and M5 are placed at the points (0; 0; ¡1), K1,

(sin r; 0; ¡ cos r), K2 and K5, respectively, in order to satisfy the condition that the set [5º=1Cº is in

an extreme state. Then, the shape of the set (W5)c is the kite K8K4K6K7 on the sphere (see Fig. 19).We note that the sides of the kite K8K4K6K7 are not geodesic arcs but are perimeters of spherical caps.Next, we assume that K9 2 @C2 and K10 2 @C5 satisfy the relations (17).

In order to prove the relation ds(K9; K2) = 2r, we ¯rst note that K4; K6; K2 2 @C3 and K7; K6; M3 2@C4. Then, obviously the relations r = ds(K7; K2) = ds(K2; K6) = ds(K2; M3) = ds(K6; M3) =ds(M3; K4) hold. As shown in Fig. 20, we see the pentagon K8K4M3K2K7 on S contains the kiteK8K4K6K7 and three spherical equilateral triangles K8K9K10, K9K6K10, and K6M3K2. Note that ourpentagon K8K4M3K2K7 is not a spherical pentagon since the sides K8K4 and K8K7 are perimetersof spherical caps. Here, let G be the middle point of the geodesic arc M3K2. Next, let q denotethe great circle determined by K8 and G. Then, from the relations (8), (17), and r = ds(K7; K2) =ds(K2; K6) = ds(K2; M3) = ds(K6; M3) = ds(M3; K4), the pentagon K8K4M3K2K7 and three sphericalequilateral triangles K8K9K10, K9K6K10, and K6M3K2 are symmetrical by re°ection with respect toq. Accordingly, the great circle q is the mirror arc re°ecting K9 to K10, K4 to K7, and M3 to K2,respectively. In addition, we see that K6 is on q. Therefore, K9, K6, and K2 are on one great circle.Thus, ds(K9; K2) = ds(K9; K6) + ds(K6; K2) = r + r = 2r holds.

30

166

Schrijver bound for nonbinary codes

Hajime TanakaDivision of Mathematics

Graduate School of Information SciencesTohoku University, Sendai, Japanhtanaka@ims.is.tohoku.ac.jp

1 Introduction

In this article, we apply a method recently developed by Alexander Schrijver [6] forbinary codes to nonbinary block codes, and derive a new upper bound on the maximumsize A(n, d) of codes of length n and minimum distance at least d. An important processof this method is to block diagonalize the Terwilliger algebra [7]. In June, I succeeded inobtaining the appropriate block diagonalization following Dunkl [4], and this was what Iwas planning to talk at the workshop. However in October, just before the workshop, Ibecame aware that Professor Schrijver and one of his students, Dion Gijswijt, had obtainedthe same result in a much simpler way. They kindly invited me as one of the authors oftheir paper [5]. This article is just a summary, and we refer to [6, 5] for the details.

In this introduction, we briefly review Delsarte’s linear programming bound [2, 3], forcomparison with the new technique. In fact, as pointed out in [6], it turns out that theSchrijver bound is always better than the Delsarte bound.

We consider block codes over an alphabet F = 0, 1, . . . , q − 1 of size q ≥ 3. LetP = F n be the set of all n-tuples over F , always taken as column vectors. For u =(u1, u2, . . . , un)

T ∈ P, we denote its support and weight by supp(u) and wt(u), respectively:

supp(u) = r | 1 ≤ r ≤ n, ui = 0, wt(u) = | supp(u)|.

Let C(P,P) be the set of complex matrices with rows and columns indexed by theelements of P . For notational convenience, we assume that F has the structure of anabelian group (F, +) for which 0 is the identity. For 0 ≤ i ≤ n, define Ai ∈ C(P ,P) by

Ai(u, v) =

1 if ∂(u, v) = i,

0 otherwise,

where ∂(u, v) = wt(u − v) is the Hamming distance. Let

H = SpanCA0, A1, . . . , An(⊆ C(P ,P)).

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167

This is the Bose-Mesner algebra of the Hamming scheme H(n, q). Let E0, E1, . . . , En bethe basis of the primitive idempotents of H. Then the base change matrices are expressedin terms of the Krawtchouk polynomials Kk(x):

Ai =n∑j=0

Ki(j)Ej, 0 ≤ i ≤ n, and Ej =1

|P|

n∑i=0

Kj(i)Ai, 0 ≤ j ≤ n,

where

Kk(x) = Kk(x; n, q) =k∑r=0

(−1)r(q − 1)k−r(

x

r

)(n − x

k − r

).

For a code C ⊆ P (i.e., a subset of P), we define its inner distribution a = (a0, a1, . . . , an)by

ai =1

|C|(χC)TAiχC =

1

|C||(u, v) ∈ C × C | ∂(u, v) = i|, 0 ≤ i ≤ n,

where χC is the characteristic vector of C (taken as a column vector). Clearly a0 = 1 and|C| =

∑ni=0 ai. Suppose that C has minimum distance at least d, that is, a1 = a2 = · · · =

ad−1 = 0. Then, it follows that

1 +n∑i=d

aiKj(i) =|P||C|

(χC)TEjχC ≥ 0 for 1 ≤ j ≤ n. (1)

This simple observation lead to the use of the linear programming techniques:

maximize 1 +n∑i=d

ai

subject to

1 +

∑ni=d aiKj(i) ≥ 0, 1 ≤ j ≤ n,

ai ≥ 0, d ≤ i ≤ n.

The optimum value of this problem gives an upper bound for A(n, d). We note that fromthe equality

(nj

)(q − 1)jKi(j) =

(ni

)(q − 1)iKj(i), condition (1) is equivalent to

n∑i=0

ai(ni

)(q − 1)i

Ai =n∑i=0

n∑j=0

aiKi(j)(ni

)(q − 1)i

Ej =n∑j=0

1(nj

)(q − 1)j

(n∑i=0

aiKj(i)

)Ej ≥ 0,

where ≥ 0 means that it is positive semidefinite.Now, let G = Sq ≀Sn, the wreath product of Sq with Sn, where the symmetric groups

Sq and Sn act on F and 1, 2, . . . , n, respectively. The stabilizer of 0 = (0, 0, . . . , 0)T inG is H = Sq−1 ≀Sn, where Sq−1 acts on F\0 = 1, 2, . . . , q− 1. Let P : G → C(P ,P)be the permutation representation of G, that is,

Pg(u, v) = δu,gv, u, v ∈ P.

Then H is just the centralizer algebra (or the Hecke algebra) of P [1]:

H = M ∈ C(P ,P) |PgM = MPg for all g ∈ G.

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168

2 Schrijver’s method

Schrijver [6] considered the centralizer algebra A of the stabilizer H, instead of H (in thecase q = 2):

A = M ∈ C(P ,P) |PhM = MPh for all h ∈ H.

We remark that the centralizer algebra A agrees with the Terwilliger algebra T (0) withrespect to 0.

For (u, v) ∈ P × P , the orbit of H on P × P containing (u, v) can be determined bythe following five parameters:

s1(u, v) = |r | 1 ≤ r ≤ n, ur = vr = 0|,s2(u, v) = |r | 1 ≤ r ≤ n, ur = 0, vr = 0|,s3(u, v) = |r | 1 ≤ r ≤ n, ur = 0, vr = 0|,s4(u, v) = |r | 1 ≤ r ≤ n, ur = 0, vr = 0, ur = vr|,s5(u, v) = |r | 1 ≤ r ≤ n, ur = vr = 0|.

In particular, the number of the nonempty orbits of H on P ×P is given by(n+4

4

). When

q = 2 it is(n+3

3

)since s4 = 0. However we adopt a different parametrization. Namely

we put s(u, v) = s4(u, v) + s5(u, v) = | supp(u) ∩ supp(v)| and t(u, v) = s5(u, v). Then wehave

s1(u, v) = n − wt(u) − wt(v) + s(u, v),

s2(u, v) = wt(u) − s(u, v),

s3(u, v) = wt(v) − s(u, v),

s4(u, v) = s(u, v) − t(u, v),

s5(u, v) = t(u, v).

Therefore, all the H-orbits on P × P are of the following form:

Os,ti,j = (u, v) ∈ P × P | wt(u) = i, wt(v) = j, s(u, v) = s and t(u, v) = t.

Clearly Os,ti,j is nonempty if and only if max0, i + j − n ≤ s ≤ mini, j and 0 ≤ t ≤ s.

As in [6], for any nonnegative integers b, b1, b2, . . . , bm,(b

b1, b2, . . . , bm

)stands for the number of the m-tuples (B1, B2, . . . , Bm) of pairwise disjoint subsets of aset B of size b, with |Br| = br (1 ≤ r ≤ m). With this notation, we have

|Os,ti,j | =

(n

i − s, j − s, s − t, t

)(q − 1)i+j−s(q − 2)s−t.

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169

Define the matrix As,ti,j ∈ C(P ,P) by

As,ti,j(u, v) =

1 if (u, v) ∈ Os,t

i,j ,

0 otherwise.

Then the matrices As,ti,j with max0, i+ j−n ≤ s ≤ mini, j and 0 ≤ t ≤ s form a basis

for A. When q = 2, only s = t is possible. In particular,

dimA =

(

n + 4

4

)if q ≥ 3,(

n + 3

3

)if q = 2.

Let C ⊆ P be a code. Define Π, Π′ ⊆ G by

Π = g ∈ G |0 ∈ g(C), Π′ = G\Π,

and let

R =1

|Π|∑g∈Π

χg(C)(χg(C))T, R′ =

1

|Π′|∑g∈Π′

χg(C)(χg(C))T.

Clearly R and R′ are positive semidefinite real symmetric matrices in C(P,P), and infact R,R′ ∈ A since they commute with the action of H.

For max0, i + j − n ≤ s ≤ mini, j and 0 ≤ t ≤ s, let

as,ti,j = |(u, v, w) ∈ C3 | (u − w, v − w) ∈ Os,ti,j|,

and put

τ s,ti,j =1

|C||Os,ti,j |

as,ti,j .

Conventionally, we set τ s,ti,j = 0 for the other values of i, j, s, t ∈ 0, 1, . . . , n.

Proposition 2.1. We have

R =∑i,j,s,t

τ s,ti,j As,ti,j , and R′ =

|C|qn − |C|

∑i,j,s,t

(τ 0,0i+j−s−t,0 − τ s,ti,j )A

s,ti,j .

The following theorem gives the block diagonalization of A:

Theorem 2.2. For each 0 ≤ l ≤ n and l ≤ k ≤⌊n+l2

⌋, let

Nl,k = k, k + 1, . . . , n + l − k.

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170

Then the centralizer algebra A is isomorphic to ⊕nl=0 ⊕

⌊n+l2

⌋k=l C(Nl,k, Nl,k), and the compo-

nent in C(Nl,k, Nl,k) of the image of∑

i,j,s,t τs,ti,j A

s,ti,j ∈ A (τ s,ti,j ∈ C) is given by((

n − 2k + l

i − k

)− 12(

n − 2k + l

j − k

)− 12

(√

q − 1)i+j∑s,t

βsl,i,j,kγs,tl τ s,ti,j

)n+l−k

i,j=k

,

where

βsl,i,j,k =k−l∑r=0

(−1)r(

k − l

r

)(n − 2k + l

i − s − r, j − s − r, s − k + r

),

andγs,tl = (−1)l−t(q − 1)−s(q − 2)s−t−lKt(s − l; s, q − 1).

Thus the positive semidefiniteness of R and R′ is equivalent to that of the real sym-metric matrices (∑

s,t

βsl,i,j,kγs,tl τ s,ti,j

)n+l−k

i,j=k

≥ 0 (2)

and (∑s,t

βsl,i,j,kγs,tl

(τ 0,0i+j−s−t,0 − τ s,ti,j

))n+l−k

i,j=k

≥ 0 (3)

for 0 ≤ l ≤ n and l ≤ k ≤⌊n+l2

⌋.

Moreover

|C| =1

|C|

n∑i=0

a0,0i,0 =

n∑i=0

(n

i

)(q − 1)iτ 0,0

i,0 .

Now suppose that C has minimum distance at least d. Then we have

τ s,ti,j = 0 if i, j, i + j − s − t ∩ 1, 2, . . . , d − 1 = ∅. (4)

These observations suggest us to solve the following semidefinite programming prob-lem:

maximizen∑i=0

(n

i

)(q − 1)iτ 0,0

i,0

subject to the conditions (2), (3) and (4).

(There are other constraints. See [5].) We obtain an upper bound for A(n, d) by solvingthis problem.

Finally, note that∑i,j,s,t

τ s,ti,j As,ti,j +

∑i,j,s,t

(τ 0,0i+j−s−t,0 − τ s,ti,j )A

s,ti,j =

n∑i=0

ai(ni

)(q − 1)i

Ai,

where a = (a1, a2, . . . , an) is the inner distribution of C. Therefore Schrijver’s methodalways gives us a sharper bound than the Delsarte bound, noting that

(ni

)(q−1)iτ 0,0

i,0 = ai.

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171

References

[1] E. Bannai and T. Ito, Algebraic Combinatorics I, The Benjamin/Cummings Pub-lishing Co., Inc., Menlo Park, CA, 1984.

[2] P. Delsarte, An algebraic approach to the association schemes of coding theory,Philips Res. Rep. Suppl. 10 (1973).

[3] P. Delsarte and V. I. Levenshtein, Association schemes and coding theory, IEEETrans. Inform. Theory 44 (1998), 2477-2504.

[4] C. F. Dunkl, A Krawtchouk polynomial addition theorem and wreath products ofsymmetric groups, Indiana Univ. Math. J. 25 (1976), 335-358.

[5] D. Gijswijt, A. Schrijver and H. Tanaka, New upper bounds for nonbinary codesbased on the Terwilliger algebra and semidefinite programming, preprint, 2004.

[6] A. Schrijver, New code upper bounds from the Terwilliger algebra, preprint, 2004.

[7] P. Terwilliger, The subconstituent algebra of an association scheme I, J. AlgebraicCombin. 1 (1992), 363-388.

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172

Problems of Optimal Configurations on the Sphere

Masaharu TanemuraThe Inst. Statist. Math.

Tokyo, Japan

COE Workshop on Sphere PackingsNov.1-5, 2004, Fukuoka, Japan

173

1. Introduction

• “Distributing a given number of points on the sphere evenly as far as possible.”

• A new method of obtaining an optimal configuration of points is proposed under a different principle.

• Applications:1. Settlement of inner-core observatories on the earth2. Launching of communications satellites to cover the

whole surface of the earth3. Modeling of spherical biological tissues4. Optimal spherical triangulation for FEM5. etc.

174

2. Optimal Configurations on the Sphere: Preceding Researches

2.1 Mini-Max Configurations

221 ;,,, ∈= iNN xxxxX L

Surface of unit sphere: 2 Origin : O

N points:

NjijiXaxx

N

≡−<

||minmax

Tammes Problem (Tammes, 1930)

175

Sugimoto’s talk (N=10)

2.1.1 Mini-Max Configurations for Small N

Relationship between closest packing of spherical capsand mini-max configuration

⎟⎟⎠

⎞⎜⎝⎛ −=

2cos1

2NaND

Schütte-van der Waerden (1951), Danzer (1963), etc.

176

2.2 Minimum Interaction Energy Configurations

Interaction potential between jiijijji xxrrxx −≡),(:, φ

∑ ∑−

= +=

⇒=1

1 1)()(

N

i

N

ijijNN rXE φ minimum

mrr 1)( =φ ; m = 1 (Coulombic potential)

Erber-Hockney (1991)

177

3. Spherical Adjustment Methods on Sphere

221 )();(,),(),()( ∈= txtxtxtxtX iNN L

Configuration of points at time t :

Spherical Voronoi Tessellation

Vertices of Voronoi cell of ki AAAtx ,,,:)( 21 L

corresponding unit vectors : kbbb ,,, 21 L

∑∑

=

==+⇒k

j j

k

j jii

b

btxtx

1

1)1()(Movement: for i=1,2,…,N

)()0( TXX NN → : “Center of Gravity Method”

3.1 Center of Gravity Method

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(c)

Poisson pattern: N = 500)0(NX

)(TX N )(TX N

Center of Gravity Method Spherical Adjustment Method

1000=T

179

jA jA′

1+jA1−jA

3.2 Spherical Adjustment Method

Vertices of Voronoi cell of ki AAAtx ,,,:)( 21 L

corresponding unit vectors : kbbb ,,, 21 L

1111 +−+− ′∆=∆ jjjjjj AAAAAA ;Spherical equilateral triangle

Lexell’s Circle

∑∑

=

=

′

′=+⇒

k

j j

k

j jii

b

btxtx

1

1)1()(Movement:

“Spherical Adjustment Method”

180

3.3 An Application of Spherical Adjustment Method

Observation of Shell Network of Radiolarian (Aulonia hexagona)

(a) Result of Spherical Adjustment Method with initial Poisson pattern

(b)

181

3.4 Spherical Adjustment Method for Small N

• Spherical adjustment method is a new kind of method of obtaining optimal configuration of points different from mini-max method or minimum energy method.

• It is interesting to see the result of our method for relatively small N.

• We performed computer simulations with initial Poisson pattern for respective N.

• For several values of N, a global optimal configuration was attained by our method.

Criteria of the Regularity of Configuration of Points:

NN NNN λπ

≡⎟⎠

⎞⎜⎝

⎛ ⋅−

⋅−≥Λ −

6232cos)2(6 1

32;

cos1coscos)2(3 1 πδ ⋅

−=≡⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−≥∆ −

NNa

aaN NN

N

NN

Edge length of Voronoi network

Edge length of Delaunay network

182

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(c)

N = 12 N = 20

N = 32

“Fulleren C60”

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•••••••

time step

log(

mov

emen

t dis

t.)

0 100 200 300 400 500

(a)

-15

-10

-50 •

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time step

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0 100 200 300 400 500

(b)

-20

2

3.5 Temporal Behavior of Movement

Movement of points per time step from initial Poisson pattern.

N = 12 N = 1000

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3.6 Spherical Adjustment Method for Large N

• For a relatively small N, our method gives local optimal configurations (rarely, global optimal ones).

• As N increases, it is expected that the number of local optimal configurations grows exponentially.

• Then, for large N, a certain scheme of getting “good”configurations is necessary.

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(a) (b)

3.6.1 Pure Spherical Adjustment Method for Initial Poisson Pattern N = 2000

Voronoi tessellation Delaunay tessellation

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(a) (b)

3.6.2 Markov Chain Monte Carlo Method

Gibbs distribution for interaction potential : )(rφ

,)(exp);(1

1 1ZrXf

N

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⎜⎝

⎛⎟⎟⎠

⎞−= ∑ ∑

−

= +=

φβφ 2000;1)( 2 == Nrrφ

“Equilibrium pattern”

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(a) (b)

3.6.3 Hybrid Method with Initial MCMC Equilibrium Pattern

• Spherical Adjustment Method was applied for the equilibrium pattern obtained by MCMC method.

• Among three methods so far, hybrid method showed the highest score against several criteria of regularity.

189

4. Concluding Remarks

• A new method (“spherical adjustment method”) of obtaining optimal configuration of points on the sphere is proposed.

• Our method is able to give a “good” evenly spaced configuration of points for any N.

• For some values of small N, our method is possible to give a globally optimal solution.

• For relatively small values of N, our method gives a finite number of locallyoptimal solutions.

• For large values of N, a combination of MCMC method and our method works well.

190