The Cube-A Window to Convex and Discrete Geometry (Cambridge Tracts in Mathematics)

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CAMBRIDGE TRACTS IN MATHEMATICS General Editors B. BOLLOBÁS, W. FULTON, A. KATOK, F. KIRWAN, P. SARNAK, B. SIMON

168

The Cube: A Window to Convex and Discrete Geometry

THE CUBE: A WINDOW TO CONVEX AND DISCRETE GEOMETRY CHUANMING ZONG Peking University

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Contents

page vii

Preface

ix

Basic notation Introduction

1

1 1.1 1.2 1.3 1.4

Cross sections Introduction Good’s conjecture Hensley’s conjecture Additional remarks

5 5 8 16 28

2 2.1 2.2 2.3 2.4

Projections Introduction Lower bounds and upper bounds A symmetric formula Combinatorial shapes

30 30 32 39 42

3 3.1 3.2 3.3 3.4

Inscribed simplices Introduction Binary matrices Upper bounds Some particular cases

45 45 48 52 70

4 4.1 4.2 4.3 4.4

Triangulations An example Some special triangulations Smith’s lower bound Lower-dimensional cases

73 73 75 79 88 v

vi

Contents

5 5.1 5.2 5.3 5.4

0/1 polytopes Introduction 0/1 polytopes and coding theory Classification The number of facets

92 92 93 99 105

6 6.1 6.2 6.3 6.4

Minkowski’s conjecture Minkowski’s conjecture An algebraic version Hajós’ proof Other versions

111 111 113 118 125

7 7.1 7.2 7.3 7.4

Furtwängler’s conjecture Furtwängler’s conjecture A theorem of Furtwängler and Hajós Hajós’ counterexamples Robinson’s characterization

128 128 129 131 132

8 8.1 8.2 8.3 8.4

Keller’s conjecture Keller’s conjecture A theorem of Keller and Perron Corrádi and Szabó’s criterion Lagarias, Mackey, and Shor’s counterexamples

150 150 151 155 163

References

166

Index

173

Preface

What is the simplest object in an n-dimensional Euclidean space? The answer should be a unit cube or a unit ball. On the one hand, compared with others, they can be easily described and intuitively imagined; on the other hand, they are perfect in shape with respect to symmetry and regularity. However, in fact, neither of them is really simple. For example, to determine the density of the densest ball packing is one of the most challenging problems in mathematics. As for the unit cube, though no open problem is as famous as Kepler’s conjecture, there are many fascinating problems of no less importance. What is the most important object in the n-dimensional Euclidean space? The answer should be the unit cube and the unit ball as well. Since, among other things, the unit ball is key to understanding metrics and surface area, and the unit cube is key to understanding measure and volume, in addition, a highdimensional unit cube is rich in structure and geometry. Deep understanding of the unit cube is essential to understanding combinatorics and n-dimensional geometry. This book has two main purposes: to show what is known about the cube and to demonstrate how analysis, combinatorics, hyperbolic geometry, number theory, algebra, etc. can be applied to the study of the cube. The first two chapters discuss the area of a cross section and the area of a projection of the cube. The key problems in these directions are: What is the maximal
minimal area of a k-dimensional cross section of an n-dimensional unit cube? What is the maximal
minimal area of a k-dimensional projection of an n-dimensional unit cube? These problems are very natural and simple sounding. However, their solutions have not been completely discovered yet! On the other hand, it will be shown how deep analysis and linear algebra can deal with problems of these types. The next two chapters study the maximal simplex inscribed in the cube and the triangulation problem. For example: What is the maximum volume vii

viii

Preface

of a k-dimensional simplex that can be inscribed into an n-dimensional unit cube? What is the smallest number of simplices needed to triangulate an n-dimensional cube? These problems are very combinatorial in nature. The inscribed simplex problem in fact is closely related to Hadamard matrices. However, to have a deep understanding of these problems, we have to apply knowledge of number theory and even hyperbolic geometry! In addition, so far our knowledge about these problems is very limited. Then we discuss the 0/1 polytopes. Clearly 0/1 polytopes do provide nice models for finite geometry, geometric combinatorics, and optimization. However, what we want to emphasize here is the connection with coding theory. In fact, many problems about binary codes are problems about 0/1 polytopes! The last three chapters deal with Minkowski’s conjecture, Furtwängler’s conjecture, and Keller’s conjecture, respectively. The conjectures themselves are fascinating; moreover their algebraic solutions are just magic! According to S. K. Stein, “which in total are almost as startling as the metamorphosis of a caterpillar to a butterfly.” Is there any other fascinating mathematical problem solved in a more surprising way? I know not one. This work has been supported by the National Science Foundation of China, 973 Project, and a special grant from Peking University. Parts of it were written while I was a visitor at the Institut des Hautes Études Scientifiques and Mathematical Sciences Research Institute. I am very grateful to these institutions for their support. For the helpful comments, remarks, and suggestions, I am very grateful to R. Astley, I. Bárány, R. J. Gardner, P. M. Gruber, M. Henk, J. Liu, W. Orrick, B. Solomon, J. M. Wills, L. Yu, T. Zhou, and G. M. Ziegler.

Basic notation

En x o
x y x y x X convX int
X v
X s
X vk
X K Wi
K In In Bn n P V
P Z R Zn


n k 
n k

Euclidean space of n-dimensions. A point (or a vector) of E n with coordinates
x1  x2      xn ; or an element of a group. The origin of E n . The inner product of two vectors x and y. The Euclidean distance between two points x and y. The Euclidean norm of x. A set of points in E n . The convex hull of X. The interior of X. The volume of X. The surface area of X. The k-dimensional measure of X. An n-dimensional convex body. The ith quermassintegral of K. 
The n-dimensional unit cube x  xi  ≤ 21 . The n-dimensional unit cube x  0 ≤ xi ≤ 1. The n-dimensional unit ball centered at o. The volume of an n-dimensional unit ball. An n-dimensional polytope. The set of the vertices of P. The set of all integers. The set of all real numbers. An n-dimensional lattice. The n-dimensional lattice 
z1  z2      zn   zi ∈ Z. The maximal volume of a k-dimensional cross section of I n . The maximal volume of a k-dimensional projection of I n . ix

x
n k ∗
n k


n k n A
n s 
n
g g Zk Ok


G

Basic notation The maximal value of det
AA , where A is a k × n binary matrix. Especially, we abbreviate
n n to n . The maximal value of det
AA , where A is a k × n matrix with ±1 elements. Especially, we abbreviate ∗
n n to n∗ . The maximal volume of a k-dimensional simplex inscribed in I n . Especially, we abbreviate
n n to n . The minimal cardinality of a triangulation of I n . The maximal cardinality of an n-dimensional binary code with separation s. The number of the n-dimensional 0/1 polytopes reduced from I n . The cyclic group generated by g. The order of g. The additive cyclic group 0 1 
   k − 1 modulo k.  The√multiplicative cyclic group 1 e2i/k      e2
k−1i/k , where i = −1. The group ring generated by a group G.

Introduction

Let E n denote the n-dimensional Euclidean space over the real number field R and with an orthonormal basis e1  e2      en . Then each point x of E n can be uniquely expressed as x = x1 e1 + x2 e2 + · · · + xn en =
x1  x2      xn  where xi is known as the ith coordinate of x with respect to the basis. Let
u v denote the inner product of two vectors u and v and let x y denote the Euclidean distance between two points x and y, by which the Euclidean metric is defined. With the coordinates of the vectors and the points, we can write n 
u v = ui vi i=1

and x y =

 n 

1/2
xi − yi 

2



i=1

For convenience, we abbreviate x o to x. Let denote the angle between u and v, then we have
u v = u · v · cos  Clearly two vectors are orthogonal to each other if and only if their inner product is zero. Now let us introduce two particular objects in E n , namely   I n = x ∈ E n  xi  ≤ 21 for all i and

  Bn = x ∈ E n  x ≤ 1  1

2

Introduction

A subset of E n is called an n-dimensional unit cube if it is congruent to I n , and is called an n-dimensional unit ball if it is congruent to Bn . For convenience of the forthcoming usage, we introduce another particular unit cube   I n = x ∈ E n  0 ≤ xi ≤ 1 for all i  In fact, I n is a translate of I n with a translative vector v =
21  21      21 . In the n-dimensional Euclidean space, the volume v
X of a set X is its Lebesgue measure; that is  v
X = 
x dx En

where 
x is the characteristic function of the set. For the unit cube I n and the unit ball Bn , we have v
I n  = 1

(0.1)

and n

v
Bn  =

2  n 
2 + 1

(0.2)

where 
x is the gamma function. For convenience, we will abbreviate the volume of the n-dimensional unit ball to n . In fact,
01 is the foundation in defining the measure of a general set in E n . For two subsets C and D of E n , we define their Minkowski sum as   C + D = x + y  x ∈ C y ∈ D  Then the surface area s
X of the set X is defined by v
X + Bn  − v
X  →0 

s
X = lim

if the limit does exist. Intuitively speaking, the set X + Bn is nothing else but the result of putting a tight coat of thickness  on X. Applying this formula to I n and Bn , respectively, we can easily deduce s
I n  = 2n and s
Bn  = n · n  A subset K of E n is convex if x +
1 − y ∈ K

3

Introduction

whenever both x and y belong to K and 0 <  < 1. In addition, if it is also compact, we call it a convex body. For example, all balls, cubes, and simplices are convex bodies. Let x and y be two points of the unit ball Bn and let  be a number satisfying 0 <  < 1. Then, by the Cauchy–Schwarz inequality we get  1/2 n  2
xi +
1 − yi  x +
1 − y = i=1

 ≤

n  i=1

1/2 xi2

 +
1 − 

n 

1/2 yi2

i=1

≤ 1 Therefore, the unit ball is indeed convex. The convexity of the cubes and the simplices can be deduced by similar routine arguments. There is another important concept about convexity, which will be useful in this book; namely, the convex hull of a given set X, which is defined by 
  convX = i xi  xi ∈ X i ≥ 0 i = 1  In fact, by Carathéodory’s theorem, we can restrict each of the sum over only n + 1 terms. In particular, if cardX is a finite number, then convX is a convex polytope. For example, both a cube and a simplex are polytopes and they are the convex hulls of the sets of their vertices. Let H be a supporting hyperplane of an n-dimensional polytope P. We call F = P ∩ H a k-face of P if it is k-dimensional. In particular, an
n − 1-face is called a facet and a 0-face is a vertex. Let Z denote the ring of integers and let a1  a2      an be n linearly independent vectors in E n , then the set

n  = zi ai  zi ∈ Z i=1

is called an n-dimensional lattice and a1  a2      an  is called a basis of the lattice. It is easy to see that a lattice is a free abelian group under addition and there are many different bases for a given lattice. For example   Zn =
z1  z2      zn   zi ∈ Z is an n-dimensional lattice with a basis e1  e2      en . In addition, the set u1  u2      un  is a basis for Zn if and only if
uij  is a unimodular integral matrix, where ui =
ui1  ui2      uin .

4

Introduction

To end this brief introduction, let us have a close look at the unit cube. Clearly, an n-dimensional unit cube is a cylinder of height 1 over an
n − 1dimensional one. In other words I n = I 1 ⊕ I n−1 

(0.3)

Therefore, it is easy to see that every k-face (0 ≤ k ≤ n − 1) of I n is a k-dimensional unit cube. Let f
n k denote the number of the k-faces of I n . By (0.3) we get f
n k = 2f
n − 1 k + f
n − 1 k − 1

(0.4)

Then, by induction on n and the identity    n−1 n−1 n + =  k k−1 k it can be deduced from (0.4) that f
n k = 2n−k

 n  k

As a conclusion, when 0 ≤ k ≤ n − 1, the unit cube I n has exactly 2n−k k-faces, each of which is a k-dimensional unit cube.

n k

1 Cross sections

1.1 Introduction Let us start with the two-dimensional case. Let H 1 denote a straight line in E 2 passing through the origin o (a one-dimensional subspace of E 2 ) and let   
I 2 H 1  denote the length of I 2 H 1 . If, without loss of generality, H 1 intersects the boundary of I 2 at
21  y and
− 21  −y, then y ≤ 21 and   
I 2 H 1  = 1 + 4y2  Therefore, for every H 1 , we have

√  1 ≤ 
I 2 H 1  ≤ 2

(1.1)

where the lower bound is attained if and only if H 1 is an axis of E 2 and the upper bound is attained if and only if H 1 contains a pair of antipodal vertices of I 2 . √ In fact, the length of any segment contained in I 2 is at most 2. This statement can be deduced by the following argument. Let L be a segment contained in I 2 . Since I 2 is centrally symmetric, the segment L , which is symmetric to L, is also contained in I 2 . Then by convexity we get 1
L + L  2

⊂ I 2

Since L is parallel with L and 21
L + L  contains the origin o, by (1.1) we get √ 
L = 
21
L + L  ≤ 2 The three-dimensional case is more complicated and more interesting. First, we study the one-dimensional sections. Without loss of generality, we assume that H 1 intersects the boundary of I 3 at
21  y z and
− 21  −y −z. Then we have y ≤ 21 , z ≤ 21 , and   
I 3 H 1  = 1 + 4y2 + 4z2  5

6

Cross sections

Therefore, for every H 1 , we get √  1 ≤ 
I 3 H 1  ≤ 3

(1.2)

where the lower bound is attained if and only if H 1 is an axis of E 3 and the upper bound is attained if and only if H 1 contains a pair of antipodal vertices of I 3 . As with the two-dimensional case, by symmetry and convexity √ we can deduce that the length of any segment contained in I 3 is at most 3. Next, let us discuss the two-dimensional cross sections of I 3 . Let u be a point on the boundary of I 3 and let H 2 denote the two-dimensional subspace  x  x ∈ E 3 
x u = 0. Since I 3 has six facets and every edge of I 3 H 2 is an intersection of H 2 with one of the facets, by symmetry it follows that  I 3 H 2 is either a parallelogram or a hexagon.   Let v2
I 3 H 2  denote the area of I 3 H 2 and write   U1 = u  u3 = 21  u1 ≥ 0 u2 ≥ 0 u1 + u2 ≤ 21    U2 = u  u3 = 21  u1 ≤ 21  u2 ≤ 21  u1 + u2 > 21  and, for i = 1 2 3

 Fi = x ∈ E 3  xi = 21  xj  ≤

1 2

 for j = i 

 Now, we estimate v2
I 3 H 2  by considering two cases. Case 1. u ∈ U1 . Then the corresponding plane H 2 does not intersect the  relative interior of F3 and therefore I 3 H 2 is a parallelogram (see Figure 1.1).  By projecting I 3 H 2 on to F3 we get  052 + u21 + u22  v2
F3  v2
I 3 H 2  = 05

v1

U1

F3 u v4

v2

o

v3

Figure 1.1

7

1.1 Introduction

v2

v1 U2

F3 u v6

v3

o

v4

v5

Figure 1.2

Thus, by a routine argument we can deduce √  1 ≤ v2
I 3 H 2  ≤ 2 where the lower bound is attained if and only if u =
0 0 21  and the upper bound is attained if and only if u =
21  0 21  or u =
0 21  21 . Case 2. u ∈ U2 . Then the corresponding plane H 2 does intersect the relative  interior of every facet of I 3 and therefore I 3 H 2 is a hexagon (see Figure 1.2). Assume that H 2 does intersect the boundary of F3 at two points v1 and v2 . By a routine computation we can determine that v1 =
2u4u2 −1  − 21  21  and v2 = 1 
− 21  2u4u1 −1  21 . Since the projection of I 3 H 2 to F3 has area 1 −
21 + 2u4u2 −1  2


21 + 2u4u1 −1 , we get

1

2

 v2
I 3 H 2  =      1 2u2 − 1 1 2u1 − 1  1 +
2u1 2 +
2u2 2 1 − + + 2 4u1 2 4u2 For any fixed number c, it is easy to see that   1 2u2 − 1 1 2u1 − 1 1− =c + + 2 4u1 2 4u2 is a quadratic curve which is symmetric with respect to the straight line  u1 = u2 . Therefore, in this case v2
I 3 H 2  attains its minimum and maximum on the boundary of U2 or on the line u1 = u2 . By checking these subcases we get  √ 3  < v2
I 3 H 2  < 2 2 As a conclusion, by symmetry, for every H 2 we have √  1 ≤ v2
I 3 H 2  ≤ 2

(1.3)

8

Cross sections

where the lower bound can be attained if and only if u is in the direction of an axis and the upper bound can be attained if and only if H 2 contains two pairs of antipodal vertices of I 3 . √ In fact, the area of any planar section of I 3 is at most 2. Let P be such a section. Then the set P  , which is symmetric to P with respect to o, is also a planar section of I 3 . In addition, P  is parallel with P. Therefore, we have o ∈ 21
P + P   ⊂ I 3 and, by the Brunn–Minkowski inequality and (1.3) √ v2
P ≤ v2
21
P + P   ≤ 2 These examples are relatively simple, at least they are manageable by elementary methods. However, similar problems in higher dimensions are much more challenging and fascinating. Let H k denote a k-dimensional linear subspace of E n containing o and let vk
S denote the k-dimensional volume (measure) of a set S. The purpose of this chapter is to study the measure and  the structure of I n H k .

1.2 Good’s conjecture Based on the examples listed in the previous section, according to Hensley (1979), Anton Good made the following conjecture about the lower  bound for vk
I n H k . Good’s conjecture. Let k be an integer satisfying 1 ≤ k ≤ n − 1. For every k-dimensional subspace H k of E n , we have  vk
I n H k  ≥ 1 This conjecture is simple-sounding in nature. However, we have to use complicated analytic machinery to prove it. Let Bk denote the k-dimensional ball with radius 1 
k + 1 k r= 2 1 2 and centered at the origin of E k . Then by (0.2) we have 
k2 + 1 2 = 1 · k 
k2 + 1 2 k

vk
Bk  = k · r k =

Let 
S x denote the characteristic function of a subset S of E n . In 1979 J. D. Vaaler proved the following theorem.

1.2 Good’s conjecture

9

Theorem 1.1 (Vaaler, 1979). Suppose that n1  n2      nj are positive integers, n = n1 + n2 + · · · + nj , D = Bn1 ⊕ Bn2 ⊕ · · · ⊕ Bnj ⊂ E n , and A is a real k × n matrix with rank k. Then we have  1 (1.4) 
D xA dx ≥ AA − 2  Ek



where A is the transpose of A. As usual, X ⊕ Y means the Cartesian product of X and Y . Let us take n1 = n2 = · · · = nj = 1 (then j = n) and choose A in this theorem so that its rows form an orthonormal basis for H k in E n . Then D is nothing else but the unit cube I n and    
I n  xA dx = 
I n  y dy = vk
I n H k  Ek

Hk

On the other hand, by the assumption, we get AA  = 1 Therefore, Good’s conjecture follows as a corollary of Theorem 1.1. Corollary 1.1 (Vaaler, 1979). For every k-dimensional subspace H k of E n , we have  vk
I n H k  ≥ 1 For a deep generalization of this result we refer to Meyer and Pajor (1988). Vaaler’s theorem is very geometric. However, as one can imagine, its proof is very analytical. To introduce the proof, let us start with a couple of definitions. Definition 1.1. Let f
x be a nonnegative function defined in E n . If f
x1 +
1 − x2  ≥ f
x1  f
x2 1− holds for every pair of points x1 and x2 in E n and every  satisfying 0 <  < 1, then f
x is said to be logconcave. Let g
x be a concave function defined in E n and define f
x = eg
x . Since g
x1 +
1 − x2  ≥ g
x1  +
1 − g
x2  holds for every pair of points x1 and x2 in E n and every  with 0 <  < 1, we have f
x1 +
1 − x2  = eg
x1 +
1−x2  ≥ eg
x1  · e
1−g
x2  = f
x1  f
x2 1−  Thus f
x is logconcave in E n .

10

Cross sections

Definition 1.2. Let  be a probability measure on E n . If 
K1 +
1 − K2  ≥ 
K1  
K2 1− holds for every pair of open convex sets K1 and K2 in E n and every  with 0 <  < 1, then  is said to be logconcave in E n . As usual, we define the support of a probability measure  to be the smallest closed subset S of E n such that 
E n \ S = 0. In other words, the support of  is the set of all points x ∈ E n such that 
rBn + x > 0 for all r > 0. For convenience, it is denoted by supp
. Comparing with the logconcave functions, it seems more difficult to get examples for logconcave probability measures. In fact, as we can see from the following lemma, they are closely related. Lemma 1.1 (Borell, 1975 and Prékopa, 1973). Let  be a logconcave probability measure on E n and suppose that supp
 spans a k-dimensional subspace H k of E n . Then there is a logconcave probability density function f
x defined on H k such that d = fdvk , where vk is the k-dimensional Lebesgue measure on H k . On the other hand, for any logconcave probability density function f
x defined on a k-dimensional subspace H k in E n , the probability measure defined by d = fdvk is logconcave. The first part of this lemma was proved by C. Borell, and the second part was proved by A. Prékopa. Both this result and the next lemma are intuitively imaginable. However, like many results in measure theory, their proofs are very complicated. Proof (A sketch). Let x and y be two distinct points in H k and let  be a number satisfying 0 <  < 1. Since  is a logconcave probability measure with a density function f
x



 1−  Bk + x +
1 − y ≥  Bk + x  Bk + y holds for all sufficiently small . Therefore, we can deduce f
x +
1 − y ≥ f
x f
y1− which means that f
x is logconcave. To prove the second part, we start with a basic inequality (see Prékopa, 1971 for the details). If g1
x, g2
x, and g
x are nonnegative Borel measurable functions satisfying g
z = sup g1
x g2
y x+y=2z

11

1.2 Good’s conjecture then we have 



−



g
z dz ≥



21 

2

−

g1
x dx

 −

21

2

g2
y dy



Applying this inequality to the multivariable case g
z = sup g1
x g2
y x+y=2z

we get 

 −



g
z dz1 ≥ sup x2 + y2 = 2z2 ··· xn + yn = 2zn

 −

21 

2

g1
x dx1

 −

2

21

g2
y dy1

and, by induction  En

i m

g
z dz ≥



21 

2

g1
x dx

En

2

En

21

g2
y dy

(1.5)



If both g1
x and g2
x are logconcave, m = 2l , where l is a positive integer, = , mj = 1 − , and g
z =

sup

g1
x g2
y

x+
1−y=z

By the logconcavity and subsequent applications of (1.5), we get1   g
z dz = sup g1
x g2
y dz En

E n x+
1−y=z

≥ ≥



 En 



sup xw m

=z

g1
x dx

Finally, we define

g1
x = g2
x =

1

g1
xw 

w=1 1 

En

i 

1 i



m 

g2
xw 

w=i+1

 En

g2
x

1 1−

1 j

dz

1− dx 

f
x if x ∈ K1  0

otherwise,

f
x if x ∈ K2  0

otherwise,

Prékopa’s paper has an error in the corresponding formula of the last step of (1.6).

(1.6)

12

Cross sections

and

g
x =

f
x

if x ∈ K1 +
1 − K2 

0

otherwise.

Then, by the logconcavity and (1.6), we get   f
x dx = g
x dx K1 +
1−K2 En  ≥ sup g1
x g2
y1− dz E n x+
1−y=z

≥ =



 

En

g1
x dx  



1− En

1− f
x dx 

f
x dx K1

g2
x dx

K2




Therefore,  is logconcave. Definition 1.3. Let 1 and 2 be two probability measures on E n . If 1
C ≥ 2
C

holds for every centrally symmetric convex body C, then we say that 1 is more peaked than 2 . Let f1
x and f2
x be two probability densities on E n . We say that f1
x is more peaked than f2
x if f1
xdx is more peaked than f2
xdx. Intuitively speaking, if f1
x is more peaked than f2
x, it means that f1
x is larger than f2
x in the vicinity of the origin. Let 1 and 2 be two probability measures on E n1 and E n2 , respectively, and let 1 ⊗ 2 denote the product of 1 and 2 on E n1 +n2 . Clearly, 1 ⊗ 2 is a probability measure on E n1 +n2 . Lemma 1.2 (Kanter, 1977). Suppose that 11 , 12 , 21 , and 22 are logconcave probability measures such that 11 is more peaked than 21 on E n1 and 12 is more peaked than 22 on E n2 . Then 11 ⊗ 12 is more peaked than 21 ⊗ 22 on E n1 +n2 . Proof (A sketch). First, by Lemma 1.1 it is easy to see that both 11 ⊗ 12 and 21 ⊗ 22 are logconcave. Let Ln denote the family of continuous nonnegative logconcave functions f
x on E n satisfying f
−x = f
x and lim f
x = 0

x→

13

1.2 Good’s conjecture

By approximation, it can be shown (see Kanter, 1977 for the details2 ) that a probability measure  is more peaked than another probability measure  if and only if   f
x d ≥ f
x d (1.7) En

En

holds for all f
x ∈ Ln . Let g
x y be a logconcave function defined on E n1 +n2 , where x ∈ E n1 and y ∈ E n2 , and define  g
x y dx G
y = E n1

We proceed to show that G
y is logconcave on E n2 . To this end, for y1  y2 ∈ E n2 and 0 <  < 1, we define g1
x = g
x y1 

g2
x = g
x y2 

and g3
x = g
x y1 +
1 − y2  Since g
x y is logconcave, we have g3
x ≥ Therefore, by (1.6) we get G
y1 +
1 − y2  ≥

sup

x1 +
1−x2 =x

g1
x1  g2
x2 1−  

 E n1

g
x1  y1  dx1

(1.8) 1−

E n1

g
x2  y2  dx2

= G
y1  G
y2 1− 

(1.9)

which means that G
y is logconcave. Assume that d11 = p
xdx, where p
x is a logconcave function defined on E n1 . For any f
x y ∈ Ln1 +n2 , it follows by the previous statement and the assumption on 11 that  F
y = f
x y p
x dx ∈ Ln1 +n2  E n1

Then, by (1.7) we get   f
x y d11 d12 = F
y d12 E n1 +n2 E n2  ≥ F
y d22 E n2  = f
x y d11 d22  E n1 +n2

2

Kanter did not use the notation of peakedness, which was introduced by Birnbaum, (1948). Instead, he used its complementary relation, the symmetric dominance.

14

Cross sections

Thus, 11 ⊗ 12 is more peaked than 11 ⊗ 22 . Similarly, we can get that 11 ⊗ 22 is more peaked than 21 ⊗ 22 and finally 11 ⊗ 12 is more peaked than 21 ⊗ 22 .
It is easy to see that 
Bn  x is a logconcave probability density function on E n . On the other hand, since    2 2 e−r r n−1 dr e−x dx = nn n 0 E nn   −x n −1 = e x 2 dx n 2 2 0 n · 
n  = 2 n 2 = 1 (1.10) 
2 + 1 we can verify that e−x is a logconcave probability density function as well. 2

Lemma 1.3 (Vaaler, 1979). 
Bn  x is more peaked than e−x . 2

Proof. We express the points in polar coordinates x = rx , where r = x and x ∈ 
Bn . Then, for any centrally symmetric convex body C, we have     2 2 e−x dx = 
C rx e−r r n−1 drdx  (1.11) C


Bn  0



Let x be a fixed point on 
Bn . Since both C and Bn are convex, we have either 
C rx  ≤ 
Bn  rx 

(1.12)


C rx  ≥ 
Bn  rx 

(1.13)

or for all 0 ≤ r < . If
112 holds at x , then we have     2 
C rx  e−r r n−1 dr ≤ 
C rx  r n−1 dr 0 0   = 
C rx  
Bn  rx  r n−1 dr

(1.14)

0

If
113 holds at x , then we have    2 
C rx  e−r r n−1 dr ≤ 0



e−r r n−1 dr 2

0

 − 2 
n2 + 1 = n   
Bn  rx  r n−1 dr = 0   = 
C rx  
Bn  rx  r n−1 dr n

0

(1.15)

15

1.2 Good’s conjecture As a conclusion, by
111,
114, and
115 we get 

e−x dx ≤ 2

C

=








C rx  
Bn  rx  r n−1 drdx


Bn  0




Bn  x dx C

Thus, 
Bn  x is more peaked than e−x . Lemma 1.3 is proved. 2




Now we can prove Vaaler’s theorem. Proof of Theorem 1.1. If k = n, then the theorem is trivial. So, we assume that j = n − k is positive. Let E k denote the k-dimensional subspace of E n spanned by the rows of A and let E j denote the j-dimensional subspace of E n which is orthogonal to E k . Then we may write a point of E n as w =
u v, where u ∈ E k and v ∈ E j . Let B be a j × n matrix such that its j rows form an orthonormal basis in E j . Then we define an n × n matrix T=

A  B

For  > 0, we define   H =
u v ∈ E n  max vi  ≤ 2 1≤i≤j

and

  H = v ∈ E j  max vi  ≤ 2  1≤i≤j

Clearly H is a closed centrally symmetric subset of E n . By Lemma 1.2 and Lemma 1.3, we have   2 e−wT  dw ≤ 
D wT  dw H

Multiplying both sides of
116 by

(1.16)

H

1 j

and writing H = E k ⊕ H , we get

1  1   −uA+vB2 e dvdu ≤ j 
D uA + vB dvdu j  Ek H  Ek H By the orthogonality assumption, we get uA + vB2 = uA2 + vB2

(1.17)

16

Cross sections and therefore

1   −uA+vB2 e dvdu →0 j E k H    1  −vB2 2 e dv · e−uA du = lim j  k →0  H E  −uA2 e du = Ek   1 2 e−r r k−1 dr = kk AA − 2

lim

0

 − 21

= AA 

(1.18)



where the last step follows from (1.10). On the other hand, we have 1   lim j 
D uA + vB dvdu →0  E k H    1  = lim j 
D uA + vB dv du →0  Ek H =Ek 
D uA du

(1.19)

By
117,
118, and
119, we finally get  1 AA − 2 ≤ 
D uA du Ek

Theorem 1.1 is proved.




1.3 Hensley’s conjecture Before Vaaler’s complete proof for Good’s conjecture, Hensley (1979) did prove the special case k = n − 1; that is  vn−1
I n H n−1  ≥ 1 holds for every
n − 1-dimensional subspace H n−1 of E n . In the same paper, he also got an upper bound  vn−1
I n H n−1  ≤ 5 In addition, he made the following conjecture. Hensley’s conjecture. For every
n − 1-dimensional subspace H n−1 of E n , we have √  vn−1
I n H n−1  ≤ 2

1.3 Hensley’s conjecture It is easy to see that

17

√  v1
I n H 1  ≤ n

holds for every straight line H 1 . However, for 2 ≤ k ≤ n − 2, it turns out to be  difficult even to make a conjecture for the best upper bounds for vk
I n H k  Nevertheless, we can ask the following basic question.  What is the optimal upper bound for vk
I n H k ? To answer this question, K. Ball proved the following theorems. Theorem 1.2 (Ball, 1989). For every k-dimensional hyperplane H k in E n , we have

k  vk
I n H k  ≤ nk 2  Especially, if kn, the upper bound is best possible. Theorem 1.3 (Ball, 1989). For every k-dimensional hyperplane H k in E n , we have  n−k vk
I n H k  ≤ 2 2  Especially, if 2
n − k ≤ n, the upper bound is optimal. Remark 1.1. Hensley’s conjecture is proved by Theorem 13 as a special case k = n − 1. However, when k < n2 and k is not a divisor of n, neither Theorem 12 nor Theorem 13 can answer the basic question. Remark 1.2. When k is comparatively small, Theorem 12 is stronger than Theorem 13; when k is comparatively large, Theorem 13 is better. In addition, a routine computation shows that  n  k2 k

 n − k  k2 n−k = 1+ ≤e 2  k

To prove these theorems, like the lower bound case, we need some deep results from analysis. For a unit vector u, let u ⊗ u denote the orthogonal projection from E n to the one-dimensional subspace u   ∈ R and let In denote the identity operator on E n . In other words, writing as matrices ⎛ ⎞ u 1 u 1 u 1 u2 · · · u 1 u n ⎜ u2 u1 u2 u2 · · · u2 un ⎟ ⎜ ⎟ u⊗u = ⎜      ⎟ ⎝     ⎠ u n u1 un u2 · · · u n un

18

Cross sections

and

⎛

0 ··· 1 ···      0 0 ···

1 ⎜0 ⎜ In = ⎜  ⎝ 

⎞ 0 0⎟ ⎟  ⎟  ⎠ 1

To verify the matrix expression for u ⊗ u, we can write an arbitrary vector v as u + w, where w is orthogonal to u, and deduce from the expression that v ·
u ⊗ u = u. Clearly, we have tr
u ⊗ u =
u u = 1

(1.20)

where tr
A indicates the trace of A. Lemma 1.4 (Brascamp and Lieb, 1976). Let u1  u2      um be m unit vectors in E n and c1  c2      cm be m positive numbers satisfying m 

ci ui ⊗ u i = I n 

(1.21)

i=1

where m ≥ n and 0 < ci < 1. Then, for every set of nonnegative integrable functions f1
x f2
x     fm
x, we have ci   m m    fi

ui  xci dx ≤ fi
x dx  (1.22) E n i=1

i=1

−

where equality holds if fi
x are identical Gaussian densities. Proof (A sketch). Let g
x be an integrable function over
−  and define g ∗
x to be a positive even function, decreasing in 0 , and satisfying v1
x  g
x ≥  = v1
x  g ∗
x ≥  for all  ≥ 0. It can be shown (see Brascamp, Lieb, and Luttinger, 1974 for the details) that     gi
x dx = gi∗
x dx (1.23) −

and

  m E n i=1

gi

ui  x dx ≤

−

  m E n i=1

gi∗

ui  x dx

(1.24)

where gi
x are m arbitrary integrable functions. Similarly, let g
x be an integrable function in E d and define g ∗
x = h
x (the Schwarz symmetrization of g
x) to be a positive function, where h
r is decreasing in 0 , satisfying



 vd x  g
x ≥  = vd x  g ∗
x ≥ 

19

1.3 Hensley’s conjecture for all  ≥ 0. Then, as with the previous case, it can be shown that   gi
x dx = gi∗
x dx Ed

Ed

and 

m 

E nd i=1

gi

ui  x1     
ui  xd  dx1 · · · dxd

≤



m 

gi∗

ui  x1     
ui  xd  dx1 · · · dxd 

E nd i=1

(1.25)

Based on (1.23) and (1.24), to prove the lemma, we may assume that fi
xci are nonnegative even integrable functions decreasing in 0 . In addition, for convenience, we assume that fi
xci =

l 

(1.26)

ij j
x

j=1

where l is a positive integer, ij are nonnegative numbers, and j
x are characteristic functions of some intervals −hj  hj . The general case can be handled by approximation. For convenience, we abbreviate the left-hand side of (1.22) to J
f. Then, for any positive integer d, we have 

d m  

E nd i=1 j=1

fi

ui  xj ci dx1 · · · dxd = J
fd 

Let Fi
x denote the Schwarz symmetrization of by (1.25) that J
fd ≤



m  E nd i=1

!d

ci j=1 fi
xj  .

Then it follows

Fi

ui  x1     
ui  xd  dx1 · · · dxd 

(1.27)

In addition, by (1.26) and some combinatorial argument, it can be deduced that 


d+1l

Fi
x =

ij 
rj Bd  x

j=1

where ij and rj are suitable nonnegative numbers. For convenience, we write  p 1 f
xp = f
x p dx 

(1.28)

20

Cross sections

By (1.28) and Hölder’s inequality, we get 

fi
xci dci = Fi
xci ≥
d + 1−ci l




d+1l

ij 
rj Bd  xci 

j=1

where ci = 1 − ci . It follows by (1.20) and (1.21) that follows from (1.27), (1.28), and (1.29) that ⎛ ⎞d

m

i=1 ci

(1.29)

= n. Thus it

⎜ ⎟ J
f ⎜ ⎟ m ⎝! ⎠ c fi
x i ci i=1




d + 1
m−nl

1j1 · · · mjm

j1 jm

≤



j1 jm

1j1 · · · mjm

"

m ! E nd

m ! i=1

where yi =

ui  x1     
ui  xd  Defining i
r x = e

ci d 2

i=1


rji Bd  yi  dx1 · · · dxd  (1.30)


rji Bd  xci

2

1− x2



r



it can be shown that3 
rBd  x ≤ i
r x √ i
r xci ≤
3 dci 
rBd  xci 

(1.31)

and therefore by (1.21) " E nd

m !


rji Bd  yi  dx1 · · · dxd

i=1 m !

i=1


rji Bd  xci

√

≤
3 d √

"

m ! E nd

n

i=1

i
rji  yi  dx1 · · · dxd

m !

i=1

i
rji  xci

=
3 dn  Thus by (1.30) we get J
f m ! i=1

fi
xci ci

The lemma is proved. 3

 ≤ lim

d→

d

√
d + 1
m−nl
3 dn = 1


1/ci

To show (1.31), we can estimate i
r xci /
rBd  xci and applying Stirling’s formula.

21

1.3 Hensley’s conjecture

Proof of Theorem 1.2. Let e1 , e2      en  be an orthonormal basis of E n and let  denote the orthogonal projection on to H k . For convenience, we write cj = 
ej 2 

uj =

1 √ 
ej  cj

and assume cj = 0, without loss of generality. Clearly, u1  u2      un are unit vectors in H k satisfying n 

cj uj ⊗ u j =

j=1

Thus we have

n 


ej  ⊗ 
ej  = 

j=1

n 

cj uj ⊗ u j = I k 

(1.32)

j=1

where Ik is the identity on H k . Let fj
x denote the characteristic function of the interval − 2√1c  2√1c . j j Then we have    I n H k = x ∈ H k  
x ej  ≤ 21  1 ≤ j ≤ n   = x ∈ H k  
x uj  ≤ 2√1c  1 ≤ j ≤ n j

and, by Lemma 1.4   vk
I n H k  = ≤

n  H k j=1

n   j=1



=

n 

fj

uj  xcj dx cj



fj
x dx

−

− 21 cj cj



j=1

By (1.32), it is easy to see that n 

cj = k

j=1

Then it can be shown by routine analysis that n n    k   ci ci k n = cj cj ≥  n n j=1 j=1 Therefore, we have

k  vk
I n H k  ≤ nk 2 

22

Cross sections

On the other hand, we can prove that, if k is a divisor of n, there is a  k k-dimensional hyperplane H k such that I n H k is a cube of volume
nk  2 . Theorem 1.2 is proved.
To prove Theorem 1.3, besides Lemma 1.4, we need another basic result. Lemma 1.5 (Ball, 1986). If  ≥ 2, then  1   ## sin t ## 2  # # dt ≤  − t  with equality being attained if and only if  = 2. Proof. We divide the proof into two cases. √ Case 1.  ≥ 4. When t ≤ 6/ 5, we have  sin t  t2 t4
−t2 j = < 1− +  t 6 120 j=0
2j + 1! t2

e− 6 =

 
−t2 j j=0

j!

6j

> 1−

t2 t4 t6 + −  6 72 1296

t4 t6 t4 ≥ 0 − − 72 1296 120 and therefore 0≤

sin t t2 ≤ e− 6  t

√ Writing m = 6/ 5 and applying  ≥ 4, we obtain # # 1   ## sin t ## 1   − t2 2   − 6 dt + dt < e t dt  − # t #  −  m √ 6 2 =√ + 
 − 1m−1   √  3 2 2 2 ≤ + 3 2 3m   2 <   Case 2. 2 ≤  < 4. Writing = 2 − 1, we have 0 ≤ < 1. In this case, we proceed to show that  1+ 1   sin2 t 1 dt ≤ √  (1.33)  − t2 1+ with equality being attained if and only if = 0.

23

1.3 Hensley’s conjecture For convenience, we write

j = and

j 1   − t2  t2 e  1 − e−  dt  −

j  sin2 t 1   sin2 t j = 1− 2 dt  − t2 t

It can be easily checked that 0 = 1. Now we have  1+    sin2 t sin2 t sin2 t 1− 1− 2 = 2 t2 t t ⎛ j ⎞  2  !j−1  sin
i −

 t sin2 t ⎝ i=0 ⎠ 1+ 1− 2 = 2 j! t t j=1 Hence, by the monotone convergence theorem, we get  1+  !j−1  1   sin2 t i=0
i −  j  dt = 1 +  − t2 j! j=1

(1.34)

On the other hand, as with (1.34), we may write the right-hand side of (1.33) as 1 1   −
1+ t2 e  dt = √  − 1+  !j−1  i=0
i −  = 1+

j  j! j=1 ! Since j−1 i=0
i −  ≤ 0 to prove the lemma it is sufficient to show that j < j for j = 1 2     Computing the first six values of j and j up to two decimal places, we get the following table. j

j j

1

2

3

4

5

6

029 · · · 033 · · ·

016 · · · 022 · · ·

011 · · · 017 · · ·

008 · · · 015 · · ·

007 · · · 013 · · ·

005 · · · 012 · · ·

It is easy to check that, for j ≥ 7 √

1

1 1  − <  2j 2 
j + 1 2ej

24

Cross sections

Therefore to complete the proof it suffices to show that in this case 1

j ≤ √ 2j

(1.35)

1 1 j ≥   − 2ej 2 
j + 1

(1.36)

and

First, it is routine to show that, for 0 < x ≤ 1  log x ≥ 2
1 − x2  Hence we have

j = = = ≤ =

j 1   − t2  t2 e  1 − e−  dt  − j 2   −u2  2 1 − e−u du e √  0 1  1
1 − xj dx  √  0  log x 1  1
1 − xj−1 dx √ 2 0 1  √ 2j

which is (1.35). On the other hand, clearly we have j ≥

 1 j 2   sin2 t 1 − dt  1 t2 t2

√ It is easy to see that t12
1 − t12 j is increasing for 1 ≤ t ≤ j + 1 and decreasing √ √ for t ≥ j + 1. Hence, if j ≥ 6, we have j + 1 > 1 + 2 

√

j+1

1+ 2

and

  √j+1− 2 cos2 t  1 j 1 j sin2 t 1 − 2 dt ≥ 1 − 2 dt t2 t t2 t 1

    sin2 t cos2 t 1 j 1 j 1 − 1 − dt ≥ dt √ √ 2 t2 t2 t2 j+1 t j+1+ 2





1.3 Hensley’s conjecture

25

Therefore, we get   sin2 t  1 j 1 − dt t2 t2 1 √    1 1 1 j 1 j 1  j+1+ 2 cos2 t ≥ 1 − 2 dt − √ 1 − 2 dt 2 1 t2 t 2 j+1− 2 t2 t  j  j     1 1 1 1 1 1 1 − 2 dt − max 2 1 − 2 · cos2 t dt ≥ 2 1 t2 t 2 t t t 0 1 j 1 1   1− =
1 − u2 j du − 2 0 4
j + 1 j +1  j+1  −1 j 1 4  2j 1− = − 4j j +1 2
2j + 1 j   1  ≥ − 4 j + 1 4ej and finally 1 1  j ≥  − 2ej 2 
j + 1 which proves (1.36). The equality case is easy to check. The lemma is proved.
Proof of Theorem 1.3. Assume the assertion inductively for sections of I n−1 . Let H k denote the
n − k-dimensional subspace of E n which is perpendicular with H k . We consider two cases. √ Case 1. H k has a unit vector u with u1 > 1/ 2. Let T denote the orthogonal projection on to e1 , let T  denote the orthogonal projection on to u, and write   C = x ∈ E n  xj  ≤ 21  2 ≤ j ≤ n   In this case T
I n H k  is a subset of a k-dimensional section of the
n − 1dimensional unit cube T
C and therefore, by the inductive assumption, we get

  n−k−1 vk T
I n H k  ≤ 2 2  (1.37) Then u has an orthogonal basis a1  a2      an−1 , where a1 = T 
e1  and aj ∈ e1 , 2 ≤ j ≤ n − 1. It is easy to see that u1 a1  if j = 1, T
aj  = aj  if 2 ≤ j ≤ n − 1.

26

Cross sections

Thus, since H k ⊂ u by (1.37), we get   n−k vk
I n H k  ≤ u1  · vk
T
I n H k  ≤ 2 2  1

√ Case 2. H k has no unit vector with a coordinate larger than 1/ 2. Let  denote the orthogonal projection on to H k , and write j = 
ej  and uj = 1 
ej  for 1 ≤ j ≤ n. Then j

j ≤

j = 1 2     n

√1  2

If some j is zero, then the problem reduces to that for I n−1 . Thus, we assume that j > 0 for each j. For v ∈ H k , we define  f
v = vk
I n
H k + v In addition, let x1  x2      xn be independent random variables, each uniformly distributed on − 21  21 , with respect to some probability. Then the random vector x =
x1  x2      xn  ∈ E n induces Lebesgue measure on I n and f
v is the continuous probability density function of the random vector 
x. Let E
x denote the expectation of the random variable x and write  
w = ei
vw f
v dv Hk

√ Here the i over e is −1. Then, by some basic results in probability theory, we have



n  
w = E ei
w
x = E ei j=1 xj
w
ej 

n  = E ei j=1 xj j
wuj  =

n  sin 21 j
w uj  j=1

1


w uj  2 j



By the standard Fourier inversion formula  vk
I n H k  = f
o =

 1 
w dw
2n−k H k   n sin 21 j
w uj 

1 dw
2n−k H k j=1 21 j
w uj  n sin j
w uj  1   = n−k dw  H k j=1 j
w uj  # # n # sin j
w uj  # 1   # # ≤ n−k #
w u  # dw  H k j=1 j j n 1   = n−k j

w uj  dw  H k j=1

=

(1.38)

27

1.3 Hensley’s conjecture where

# # # sin
j t # # j
t = ##

j t #

(1.39)

It is easy to check that n 

2j uj ⊗ uj = In−k

j=1

and therefore n 

2j = n − k

j=1

By Lemma 1.4 and (1.38), we get  vk
I n H k  ≤

n  

1  n−k

j=1

 −

1 2

j
t j dt

2j 

(1.40)

Writing pj = −2 j ≥2 and applying Lemma 1.5 with (1.39) to (1.40), we get vk
I

# #p p1 j 1   ## sin
j t ## j dt H ≤ # t # j j=1  − # #p p1  n j  1   ## sin x ## j = dx # # x j=1 j  − 1  $ p j n n 1   2 1 ≤ = 2 2pj pj j=1 j j=1

n

n 

k

=2



n−k 2




The theorem is proved. For convenience, we write   


n k = max vk
I n H k  

where the maximum is over all k-dimensional hyperplanes. By Theorem 1.2 and Theorem 1.3, many values of
n i are known when n is relatively

28

Cross sections

small. Let us end this section by listing them up to n = 10 as the following table.

1 √


3 k 3


4 k √2


5 k √5


6 k √6


7 k √7


8 k 8


9 k √3


10 k 10 k

2 3 4 5 6 7 8 9 10 √ 2 √1 2 √1 2 ?? √2 2 √1 2 √1 3 8 √2 ?? ?? 8 √2 2 √1 4 √?? 4 8 √2 2 √1 27 ?? √4 8 √2 2 √1 ?? 5 ?? ?? 32 4 8 2 2 1

1.4 Additional remarks About the cross sections of the n-dimensional unit cube there is another natural problem:  For a k-dimensional hyperplane H k , what is the possible shape of I n H k ?  When n = 3 and k = 2, as was shown in Section 1.1, I 3 H 2 is either a hexagon or a parallelogram. In higher dimensions, however, the situation is much more complicated. Let fj
P denote the number of the j-dimensional faces of a polytope P. By an argument similar to the case n = 3 and k = 2, it can be easily deduced that   fn−2
I n H n−1  is either 2n − 2 or 2n. In addition, fn−2
I n H n−1  = 2n − 2  holds if and only if I n H n−1 is a parallelopiped. Let v be a vertex of I n and let H denote the
n − 1-dimensional hyperplane which is perpendicular to v and passing o. It is easy to see that half of the 2n facets of I n contain v and the other half contain −v. In addition, there is no facet of I n containing both v and −v. Let F be a facet of I n , we have √ (1.41) d
F = n − 1 where d
X denotes the diameter of X. On the other hand, if v ∈ F and  H F = ∅, then we have  √ √ n n √ = < n − 1 d
F ≤ 2 · 2 2

1.4 Additional remarks

29

which contradicts
141. Therefore H intersects the relative interior of every facet of I n and hence  fn−2
I n H = 2n  Especially, when n = 4, we can deduce that I 4 H is a three-dimensional  cross polytope. However, we do not know what I n H looks like when n is large.  When k  n, the shapes of I n H k can be very different from each other. It can be a k-dimensional parallelopiped; on the other hand, it can be very spherical, as it was shown by Dvoretzky’s well-known theorem (see Dvoretzky, 1961 or Zong, 1996). It was proved by Bárány and Lovász (1982) that almost every k-dimensional cross section of I n has at least 2k vertices. However, we do not know any good bound for the number of j-dimensional faces of a k-dimensional cross section of I n . Let E
n k j denote the expected number of j-dimensional faces of a random k-dimensional cross section of I n . Lonke (2000) recently proved the following results     2k kt2 k n E
n k 0 = 2 e− 2 n−k
tI n−k  dt k  0 where n−k is the
n − k-dimensional Gaussian probability measure, and E
n n − k n − j ∼


2nj−k
j − k!

for fixed 1 ≤ k < j as n → . As a consequence of the first formula, for n → , we can deduce
k−1 2k E
n k 0 ∼ √
 log n 2 k

and especially E
n n − 1 0 ∼

√ 2n n  

2 Projections

2.1 Introduction Let u be a unit vector in E 2 and let H be the one-dimensional subspace of E 2 % which is perpendicular to u. Let I 2 H denote the projection of I 2 on to H. It % is easy to see that, for any u, I 2 H is a segment. Without loss of generality, % by symmetry, to search for bounds for 
I 2 H we may assume that u =
cos  sin  with 0 ≤ ≤ 4  By elementary geometry (see Figure 2.1), we get √

 √

 % 
I 2 H = 2 · 22 · cos 2 − 4 − = 2 · sin 4 + and therefore

√ % 1 ≤ 
I 2 H ≤ 2


21

where the lower bound can be attained if and only if H is an axis of E 2 and the upper bound can be attained if and only if H contains two antipodal vertices of I 2 . In three-dimensional space, as with the cross-section case, the situation is more complicated. We start with an easy case. Let H 1 denote a onedimensional subspace of E 3 , defined by a unit vector u
 H 1 = u   ∈ R  % and let I 3 H 1 denote the orthogonal projection of I 3 on to H 1 . Clearly, for % any H 1 , the projection I 3 H 1 is a segment. Then it can be shown that   % 
I 3 H 1  = 2 max 
u x  x ∈ I 3   = 2 max u1 x1 + u2 x2 + u3 x3  xi ≤1/2

= u1  + u2  + u3  30

31

2.1 Introduction H u

o

θ

π–π–θ 2 4

Figure 2.1

Therefore, we have

√ % 1 ≤ 
I 3 H 1  ≤ 3


22

where the lower bound can be attained if and only if H 1 is an axis of E 3 and the upper bound can be attained if and only if H 1 contains a pair of antipodal vertices of I 3 . Let u be a unit vector in E 3 and let H 2 denote the two-dimensional subspace of E 3 which is perpendicular to u. In other words, u is a unit norm of H 2 . For % % a subset X of E 3 , let X H 2 denote its projection on to H 2 . Clearly, I 3 H 2 is a centrally symmetric polygon. For convenience, we write w1 =
1 0 0 w2 =
0 1 0 w3 =
0 0 1 and define   F1 = x ∈ I 3 
x w1  = 21    F2 = x ∈ I 3 
x w2  = 21  and

  F3 = x ∈ I 3 
x w3  = 21 

% By symmetry, to look for bounds for v2
I 3 H 2  we may assume that ui ≥ 0 (see Figure 2.2). Then we have & % & % % % I 3 H 2 =
F1 H 2 
F2 H 2 
F3 H 2  and therefore % % % % v2
I 3 H 2  = v2
F1 H 2  + v2
F2 H 2  + v2
F3 H 2  =
u w1  · v2
F1  +
u w2  · v2
F2  +
u w3  · v2
F3  = u1 + u2 + u3 

32

Projections

v1

v1′

v6

v2′

F1

o

H2

v6′

u

F3

v2

v5′ v5

F2 v3′ v3

v4 v4′

Figure 2.2

Since u21 + u22 + u23 = 1, as with (2.2) we get √ % 1 ≤ v2
I 3 H 2  ≤ 3


23

where the lower bound can be attained if and only if u is an axis of E 3 and the upper bound can be attained if and only if u is in the direction of a vertex % of I 3 . In addition, it is easy to see that I 3 H 2 is either a parallelogram or a hexagon. Comparing with the cross sections of I 3 , which were discussed in Section 1.1, it is easy to see that the cross sections and the projections are quite different. In fact, as we will see in higher dimensions, not only the results but also the proof methods are very different. While the key methods to deal with cross sections are analytic, the main ideas for projections are algebraic. In this chapter we will concentrate on the projections of I n .

2.2 Lower bounds and upper bounds % Let H k denote a k-dimensional subspace of E n and let I n H k denote the % orthogonal projection of I n on to H k . Clearly, I n H k is a k-dimensional centrally symmetric polytope. As with the cross-section case, it is natural to ask the following question: % What are the best lower bound and the best upper bound for vk
I n H k ? In 1986, G.D. Chakerian and P. Filliman did study this problem. For a lower bound, they proved the following result:

33

2.2 Lower bounds and upper bounds

Theorem 2.1 (Chakerian and Filliman, 1986). Whenever 1 ≤ k ≤ n − 1, for % any k-dimensional orthogonal projection I n H k of I n , we have % vk
I n H k  ≥ 1 where equality is attained if and only if H k is spanned by k axes of E n . Proof. It is easy to see that  % In Hk ⊆ In Hk and therefore

 % vk
I n H k  ≥ vk
I n H k 

Then Theorem 2.1 follows from Corollary 1.1.




Remark 2.1. It seems that Chakerian and Filliman were not aware of Vaaler’s  work on the lower bound of vk
I n H k . Their original proof did use a different method which will be mentioned in the next section. It is obvious that In ⊆ and therefore

√

n n B 2

% In Hk ⊆

√

n k B 2

% holds for every k-dimensional projection I n H k of I n . Thus, we have  √ k %
24 vk
I n H k  ≤ k 2n  Clearly this upper bound is not optimal, especially when k is comparatively large. To improve this upper bound, we will prove the following results in this section. Theorem 2.2 (Chakerian and Filliman, 1986). k  n k/2 %  vk
I n H k  ≤ k−1 k k−1 k Theorem 2.3 (Chakerian and Filliman, 1986). $ n! n% k vk
I H  ≤ 
n − k! k!


25


26

Remark 2.2. For any fixed k and relatively small n, among these three upper bounds
26 is the most efficient and
24 is the least efficient. On the

34

Projections

other hand, for sufficiently large n,
25 is the most efficient and
24 is the least efficient. Nevertheless, for any fixed k and sufficiently large n, the three upper bounds given by
24,
25, and
26 are essentially the same, all of the order nk/2 . Remark 2.3. Let c and d be integers defined by c = n/
k + 1 and d = n− c
k + 1. For i = 1 2     k + 1, let pi =
pi1  pi2      pin  be the vertex of I n with coordinates  1 if
i − 1c + 1 ≤ j ≤ ic, pij = 0 otherwise.  pk+1 are the vertices of a k-dimensional regular simplex S Then p1  p2     √ of edge length 2c. It is easy to see that S ⊆ I n and therefore there is a % corresponding I n H k which contains a translate of S. Thus we have % vk
I n H k  ≥ vk
S ∼ k
2ck/2 ∼ k nk/2  where both k and k are suitable constants depending only on k. It follows by this example that, though none of the three upper bounds given by
24,
25, and
26 is optimal in general, they do have the correct asymptotic order. To prove these theorems, we need some basic results in convex geometry. Let K and K  be convex sets in E n , it is known as Steiner’s formula that n   n Vi
K K   · ti
27 vn
K + tK   = i i=0 holds for all t ≥ 0, where Vi
K K   are constants only depending on K, K  , and i. Especially V0
K K   = vn
K and Vn
K K   = vn
K   When both K and K  are polytopes, the formula can be proved by a routine argument. The general case can be deduced by approximation. When K  = Bn , the number Wi
K = Vi
K Bn  is known as the ith quermassintegral of K. In particular, it is known that n · W1
K is the surface area of K, Wn
K = n , and  n · Wn−1
K = h
K u du
28 
Bn 

35

2.2 Lower bounds and upper bounds where


 h
K u = max
u x  x ∈ K 

Since h
K + K   u = h
K u + h
K   u holds for every u ∈ 
Bn , by (2.8) we get the following lemma. Lemma 2.1. For every pair of convex sets K and K  in E n , we have Wn−1
K + K   = Wn−1
K + Wn−1
K   For a convex set K ⊂ E n and a number  ∈ 0 1, we define K = K +
1 − Bn and 1

1

1

f
 = vn
K  n −  · vn
K n −
1 −  · vn
Bn  n  It follows by (2.7) that 

vn
K  = n · vn K + 1− Bn  n   n = Wi
K ·
1 − i n−i i i=0 and hence, by a routine computation   n−1 n−1 1 f 
0 = Wn
K− n Wn−1
K − Wn
K n W0
K n   n−1 1 − n−1 = n n Wn−1
K − n n vn
K n  It follows by the Brunn–Minkowski inequality that f
 is a concave function in 0 1. On the other hand, it is easy to see that f
0 = f
1 = 0 Therefore, we have f 
0 ≥ 0 and hence n−1

1

Wn−1
K − n n vn
K n ≥ 0 In other words, we have proved the following inequality. Lemma 2.2 (Urysohn’s inequality). For every n-dimensional convex set K, we have W n
K vn
K ≤ n−1n−1  n

36

Projections

Proof of Theorem 2.2. Let u1  u2      un  be an orthonormal set in E n and let o ui  denote the segment with endpoints o and ui . Then u1  u2      un generate a unit cube C given by

n n   C = o ui  = i u i  0 ≤  i ≤ 1  i=1

i=1

Let H k be the k-dimensional subspace of E n defined by   H k = x ∈ E n  xj = 0 j = k + 1     n 


29

and let ui denote the orthogonal projection of ui on to H k . For convenience, we identify H k with E k by disregarding the last n − k coordinates. So, if ui =
ui1  ui2      uin , then ui =
ui1  ui2      uik  0 0     0. Thus the projection of C on to H k is a zonotope in E k

n n  % k  C H = o ui  = i u i  0 ≤  i ≤ 1 
210 i=1

i=1

Since the n × n matrix ⎛

u11 u12 ⎜ u21 u22 ⎜ U =⎜   ⎝   un1 un2

⎞ · · · u1n · · · u2n ⎟ ⎟  ⎟    ⎠ · · · unn

is orthogonal, letting li denote the length of o ui , we have

n n n k     2 2 2 li = ui  = uij i=1

i=1

=

k n   j=1



i=1

u2ij =

i=1

j=1

k 

1 = k

(2.11)

j=1

It is well known and trivial that Wk−1
o ui  =

k−1 l k i

Therefore, by Lemma 2.1, we have n n    % Wk−1
C H k  = Wk−1
o ui  = k−1 l k i=1 i i=1


212

2.2 Lower bounds and upper bounds

37

Then, applying the Cauchy–Schwarz inequality to (2.12) and applying (2.11), we get 1/2   n n k−1 √  % k 2 Wk−1
C H  ≤ n li =k−1  k k i=1 Finally, by Lemma 2.2, we have

% W k
C H k  kk−1  n k/2 % vk
C H k  ≤ k−1 k−1 ≤ k−1  k k k


Theorem 2.2 is proved.

Let i1  i2      ik  be a subset of 1 2     n. Then the k vectors ui1  ui2      uik produce a parallelopiped

k k   Pi1 i2 ik = o uij  = j uij  0 ≤ j ≤ 1 j=1

j=1

with k-dimensional volume

' ' u i 1 ui 2 1 ' 1 ' ui 1 ui 2 2 ' 2 vk
Pi1 i2 ik  = '   '   ' 'u u ik 1 ik 2

' · · · ui1 k ' ' · · · ui2 k ' '  '    ' ' · · · uik k '

 It is obvious that, for a given set u1  u2      un , there are nk parallelopipeds of this kind. To prove Theorem 2.3, we need the following result about % the volume of a zonotope. For convenience, we consider C H k instead of % I n H k. Lemma 2.3 (Shephard, 1974). % vk
C H k  =



vk
Pi1 i2 ik 

i1 i2 ik 

where the summation is over all possible subsets of 1 2     n of cardinality k. Proof. We apply induction on n − k. If n − k = 0, there is nothing to prove. Assume that the statement is true for n = m − 1 ≥ k, we proceed to show it for n = m. Suppose that u1  u2      um−1 , and um are m orthonormal vectors, and write i (  ) Ci = o uj  j=1

38

Projections

Then we have C m = C m−1 + o um  and

% % C m H k = C m−1 H k + o um  


213

If um = o, then the lemma follows by (2.13) and the inductive assumption. If um = o and if H k−1 is the subspace of H k which is perpendicular to um , then it follows by (2.13) and the inductive assumption that % % % vk
C m H k  = vk
C m−1 H k  + vk
C m−1 H k−1 + o um    = vk
Pi1 i2 ik  + vk
Pi1 i2 ik−1 m  i1 i2 ik ∗

=



i1 i2 ik−1 

vk
Pi1 i2 ik 

i1 i2 ik 

where i1  i2      ik ∗ is over all possible subsets of 1 2     m − 1 of cardinality k, and i1  i2      ik−1  is over all possible subsets of 1 2     m − 1 of cardinality k − 1. The lemma is proved.
Proof of Theorem 2.3. Let C, ui , U , and Pi1 i2 ik be defined as above. Since U is an orthogonal matrix, we have U  U = UU  = In  Thus, writing

⎛

u11 u12 ⎜ u21 u22 ⎜ A=⎜   ⎝   un1 un2

⎞ · · · u1k · · · u2k ⎟ ⎟  ⎟     ⎠ · · · unk

we can deduce A A = Ik  By computing the determinants of ' ' ui 1 ' 1  ' ' ui2 1 '  ' i1 i2 ik  '  'u ik 1

in other words



both sides, it follows that '2 ui1 2 · · · ui1 k ' ' ui2 2 · · · ui2 k ' '     ' = 1    ' ' u ··· u ' ik 2

ik k

vk
Pi1 i2 ik 2 = 1

i1 i2 ik 

2.3 A symmetric formula

39

Then, by Lemma 2.3 and the Cauchy–Schwarz inequality, we get  2  % k 2 vk
C H  = vk
Pi1 i2 ik  i1 i2 ik 

    n v
P 2 ≤ k i i i  k i1 i2 ik 1 2 k  n =  k


Theorem 2.3 is proved.

2.3 A symmetric formula The projections of a unit cube have the following symmetric property. Theorem 2.4 (McMullen, 1984 and Chakerian and Filliman, 1986). If E k and E n−k are orthogonal complementary subspaces of E n , then % % vk
I n E k  = vn−k
I n E n−k  Remark 2.4. The analogous statement of this theorem for cross sections is not always true. For example, defining   E 1 = x ∈ E 3  x1 = x 2 = x 3 and

  E 2 = x ∈ E 3  x 1 + x 2 + x3 = 0 

it follows by
13 that   v1
I 3 E 1  = v2
I 3 E 2  Now let us introduce a basic result in linear algebra, from which we can easily deduce the above theorem. Lemma 2.4 (Jacobi’s identity). Let D be a subdeterminant of an n × n orthogonal matrix U and let D∗ denote its algebraic complement, then D∗  = D Proof. Assume that

 D=U

 i 1  i2      ik  j1  j 2      j k

40

Projections

where il and jl denote respectively the rows and columns of the entries of D in U , and write ∗  i 1  i2      ik ∗ D =U  j1  j 2      j k Let V denote the k × n submatrix of the n × n unit matrix In consisting of its j1  j2      jk rows and let W denote the n × k submatrix of In consisting of its i1  i2      ik columns. Since U is an orthogonal matrix, it is easy to see that U  = U −1 = In U −1 In  Therefore, we have



   i 1  i2      ik  j1  j2      j k D=U =U  j1  j 2      j k i1  i 2      i k = det
VU −1 W

(2.14)

Now we consider an
n + k ×
n + k matrix  U W Q=  V 0k×k By (2.14) we get ' ' ' U ' W ' ' det
Q = ' 0k×n −VU −1 W ' # # = #det
VU −1 W# = D On the other hand, by Laplace’s expansion formula, we get # ∗ #  # n + 1 n + 2     n + k ## # det
Q = #det
Ik  · Q # j2      jk j1  #  #  # i  i      ik ∗ # # = D∗   = ##U 1 2 j1  j 2      j k # The lemma follows by (2.15) and (2.16).

(2.15)

(2.16)


Proof of Theorem 2.4. As before, assume that u1  u2      un  is an orthonormal set in E n and the considered unit cube is

n n   C = o ui  = i u i  0 ≤  i ≤ 1  i=1

i=1

2.3 A symmetric formula

41

By Lemmas 2.3 and 2.4, we have

#  # # i1  i2      ik # # # vk
C E  = #U 1 2     k # i1 i2 ik  #  #  # i1  i2      ik ∗ # # # = #U 1 2     k # i1 i2 ik  #  # # #   j      j j 1 2 n−k #U # = # k + 1 k + 2     n # j1 j2 jn−k  % = vn−k
C E n−k  %

k






Theorem 2.4 is proved.

Remark 2.5. Comparing Theorems 22 and 23 with Theorems 12 and 13, we can see that our knowledge about projections is not as good as that about cross sections. Let 
n k denote the maximum area of a k-dimensional projection of I n . By Theorem 2.4 and Theorem 2.3, we have √ 
n 1 = 
n n − 1 = n
217 Besides this simple result, the only known values of 
n k are

 
n 2 = 
n n − 2 = cot 2n  Assume that C=

n ( 

o uj


218

)

j=1

and P2 =

n ( 

) o uj 

j=1

uj

where is the projection of uj on to a two-dimensional plane E 2 . By (2.11) we can deduce that n  uj 2 = 2 j=1

Let L denote the perimeter of P2 . Then we have  2 n n    2 L = 2 uj  ≤ 4n uj 2 = 8n j=1

j=1

and by the isoperimetric inequality for 2n-gons in E 2 (see Fejes Tóth, 1964),

 v2
P2  ≤ cot 2n 

42

Projections

where equality is attained if and only if P2 is a regular 2n-gon with edge √ length 2/n. On the other hand, we can construct such a projection. Let u1  u2      un be complex numbers defined by  uj = uj1 + iuj2 = 2/n · e
j−1i/n  √ where i = −1. For convenience, we may view them as points in E 2 . Then we have  2  2  uj1 + uj2 = uj 2 = 2  2  2   uj1 − uj2 + 2i uj1 uj2 = u2j = 0 and therefore

 

u2j1 =



u2j2 = 1

uj1 uj2 = 0

It is known in linear algebra that then we can construct an n × n unimodular matrix U =
ujl  Writing uj =
uj1  uj2      ujn , it can be verified that  C = uj is an n-dimensional unit cube and the projection of C on to E 2 is exactly the regular 2n-gon with vertices ±u1 , ±u2      ±un . As a conclusion, (2.18) is proved. Now, we end this section by listing the known values of 
n k up to n = 7 as the following table. k 
3 k 
4 k 
5 k 
6 k 
7 k

1 √ 3 2 √ √5 √6 7

2 √

3  cot 8   cot 10   cot 12   cot 14

3

4

5

6

2   cot 10 ?? ??

√

5  cot 12 ??

√

6  cot 14

√

7

2.4 Combinatorial shapes Defining

  H k = x ∈ E n  xi = 0 for i > k 

it is easy to see that the projection of I n on to H k is a k-dimensional unit cube. On the other hand, according to Dvoretzky (1963) and Larman and Mani (1975), roughly speaking, I n has a k-dimensional projection almost

2.4 Shapes

43

spherical when n is sufficiently large and k is relatively small. In addition, the projection of I n does change continuously while the image space changes. Based on the above observations, we can ask the following question. Problem 2.2. Determine the maximal
minimal number of the j-dimensional faces of a k-dimensional projection of I n . So far no exact result to this problem is known. However, we do know its answer from the probability point of view. Let G
n k denote the Grassmannian of k-dimensional linear subspaces of E with the usual topology. It is well known that there is a unique rotation invariant probability measure k on G
n k. Let P be an n-dimensional polytope % and let H k ∈ G
n k, then it is clear that P H k is a k-dimensional polytope. % For j ∈ 0 1     k − 1, let Fj
P H k  denote the family of the j-dimensional % k % % k faces of P H and let fj
P H  denote the cardinality of Fj
P H k . Clearly, % k fj
P H  is a random variable of integer values and its expectation is  % fj
P H k  dk  E
P k j = n

G
nk

Let F and G be two faces of P with F ⊆ G, let v be a relative interior point of F , and write
 H = 
x − v + v   ∈ R and x ∈ F and


 C = 
y − v + v   ≥ 0 and y ∈ G 

Clearly H is a hyperplane and C is a cone with v as its apex. Let C  denote the set of outward normals to hyperplanes supporting C at H and assume that G is d-dimensional, then we define 
F G =

vd

Bd + v ∩ C vd
Bd 


F G =

vd

Bd + v ∩ C    vd
Bd 

and

According to Grünbaum (1967) they are called the internal angle and the external angle of G at F , respectively. By the definition, we have 
F F  =
F F  = 1 For convenience, if F ⊆ G we write 
F G =
F G = 0

44

Projections

Based on the results in Grünbaum (1968), McMullen (1975), and Santaló (1952), Affentranger and Schneider (1992) deduced a general formula for E
P k j. Namely    E
P k j = 2 
F G ·
G P
219 s≥0 F ∈Fj
P G∈Fk−1−2s
P

It is observed by Böröczky and Henk (1999) that, for F ∈ Fj
I n  and G ∈ Fl
I n  with F ⊆ G, we have 
F G = 
F I l  = 2j−l 


F I n  = 2j−n   n n−j n  cardFj
I  = 2 j and the number of the l-dimensional faces containing a fixed j-face is Therefore, applying
219 to P = I n , we get the following result. Theorem 2.5. E
I n  k j = 2

  2nk−1 n−j n  ∼  j s≥0 k − 1 − 2s − j
k − 1 − j!j!

n−j  l−j

.

3 Inscribed simplices

3.1 Introduction What is the maximum area of a triangle contained in the unit square I2 ? Let T be such a triangle with vertices v1 , v2 , and v3 . It is easy to see that T is inscribed in I 2 ; in other words, all the three vertices are on the boundary of I 2 . If, without loss of generality, v1 is a relative interior point of an edge of I 2 , then, by moving v1 along the edge in a suitable direction until a vertex of I 2 , we can get a new triangle with at least the same area. Consequently, v1 , v2 , and v3 can be three of the four vertices of I 2 and therefore v2
T  = 21  In fact, by a similar idea, we can deduce the following result about the k-dimensional maximal simplices contained in an n-dimensional unit cube. Theorem 3.1. For any integer k with 1 ≤ k ≤ n, there is a k-dimensional maximal simplex contained in I n that all its vertices are vertices of I n . Proof. Let S be a k-dimensional simplex in the n-dimensional Euclidean space with vertices v1  v2      vk , and vk+1 , let v1∗  v1  be a segment containing v1 as a relative interior point, and let S ∗ and S  denote the corresponding simplices with v1 being replaced by v1∗ and v1 , respectively. First, let us show the following assertion. Assertion 3.1. Either vk
S ∗  or vk
S   is not smaller than vk
S. For convenience, we write H=



k+1 

i v i 

i=2

k+1  i=2

45

i = 0

46

Inscribed simplices

and

  H  = u ∈ E n 
u v = 0 for all v ∈ H 

In other words, H is a
k − 1-dimensional subspace parallel with the hyperplane determined by v2      vk , and vk+1 , and H  is the
n−k+1-dimensional subspace which is perpendicular with H. Let v
v2      vk+1  denote the
k − 1-dimensional measure of the simplex with vertices v2      vk , and vk+1 , and let p
x denote the projection of x on to H  . Then we have vk
S = k1 · p
v1 − v2  · v
v2      vk+1  vk
S ∗  = k1 · p
v1∗ − v2  · v
v2      vk+1  and vk
S   = k1 · p
v1 − v2  · v
v2      vk+1  Clearly p
v1 − v2  is contained in the segment p
v1∗ − v2  p
v1 − v2  and therefore one of the two ends is not a relative interior point of the
n − k + 1dimensional ball of radius p
v1 − v2  and center o in H  . Thus either p
v1∗ − v2  or p
v1 − v2  is not smaller than p
v1 − v2 , which implies the assertion. If S is a k-dimensional simplex inscribed in I n and v1 is a relative interior point of a d-dimensional face of I n , replacing v1 by some suitable boundary point of the face, then by Assertion 3.1 we can get a new inscribed simplex which is not smaller than S in volume. Therefore the theorem can be proved by repeating this process for every vertex of the starting simplex for at most n − 1 times.
By this theorem, to search for the maximum volume of a k-dimensional simplex inscribed in a unit cube, it is sufficient to deal with the vertex simplices, the ones whose vertices are vertices of the unit cube as well, of the corresponding dimensions. Now, as an example, we can easily answer the following question. Question 3.1. What is the maximum area (volume) of a triangle (tetrahedron) contained in a three-dimensional unit cube? Let us start with the triangle case (see Figure 3.1). By Theorem 3.1, we consider a vertex triangle T2√ . It is easy √ to see that an edge of T2 can√take only three possible lengths, 1, 2, and 3, and one of the three must be 2. Then, by comparing the distance from a vertex to such an edge, we can deduce that v2
T2  ≤

√ 3  2


31 √ where equality is attained if and only if T2 is a regular one of edge length 2.

47

3.1 Introduction v3

v4 v1

v2

E1 v8

v7

v6

v5

Figure 3.1 v3

v4 v1

v2

E1 v8

v7 v6

v5

Figure 3.2

Let T3 be a vertex tetrahedron of I 3 . Since every vertex triangle contains √ at least one √ edge of length 2, the tetrahedron T3 has two edges E1 and E2 of length 2. If E1 and E2 are co-planar (see Figure 3.2), by comparing the height of the possible tetrahedra with respect to the plane, we can deduce that v3
T3  ≤ 1 − 4 · 3!1 = 13 


32

√ where equality holds if and only if T3 is a regular one of edge length 2. If √ T3 does not have two co-planar edges of length 2, then the two vertices of T3 , which are not the ends of E1 , must belong to v1  v3  v6  v8 . In addition, one of them must be contained in the facet containing E1 and the other must be contained in the opposite facet. Then we get v3
T3  ≤ 16 


33

As a conclusion of (3.1), (3.2), and (3.3), we have proved the following result.

48

Inscribed simplices

Theorem 3.2. For every triangle T2 contained in I 3 , we have v2
T2  ≤

√ 3  2

where equality holds if and only if T2 is a regular triangle of edge length For every tetrahedron T3 contained in I 3 , we have v3
T3  ≤ 13  where equality holds if and only if T3 is a regular triangle of edge length

√

√

2.

2.

In this chapter we deal with problems of following types. Problem 3.1. What is the maximum volume of a k-dimensional simplex contained in the n-dimensional unit cube I n ? Problem 3.2. For which n and k does there exist a k-dimensional vertex simplex of I n which is both maximal and regular?

3.2 Binary matrices Let S  denote a k-dimensional simplex with vertices v0 =
0 0     0 v1 =
1 0     0     vk =
0 0     1 in the k-dimensional Euclidean space E k . Then it is well known that    vk
S   = dx = · · ·  dx1 · · · dxk = k!1  xi ≤1 xi ≥0

S

Let S be a k-dimensional simplex with vertices a0 =
0 0     0 a1 =
a11 , a12      a1n      ak =
ak1  ak2      akn  in the n-dimensional Euclidean space E n . For convenience, we assume that a1 , a2      ak are linearly independent; that is vk
S > 0 Let H denote the k-dimensional subspace spanned by a1 , a2      ak−1 , and ak , and let H  denote the
n − k-dimensional subspace which is perpendicular with H. In other words

k  H=  i ai   i ∈ R i=1

and

  H  = x ∈ E n 
x y = 0 for all y ∈ H 


34

49

3.2 Binary matrices Let ak+1 , ak+2      an  be a basis of H  such that  1 if i = j
ai  aj  = 0 otherwise


35

holds for k + 1 ≤ i ≤ j ≤ n. Then we define

n  ∗ S = S+ i a i  0 ≤  i ≤ 1  i=k+1

It is easy to see that vn
S ∗  = vk
S Let A denote the n × n matrix with entries aij , let Ak denote its k × n submatrix consisting of the first k rows, and let T denote the linear transformation determined by T
ai  = ei 

i = 1 2     n

where e1  e2      en  is an orthonormal basis of E n . Thus we have

n   ∗ T
S  = S +  i ei  0 ≤  i ≤ 1 i=k+1

and therefore vk
S = vn
S ∗  = =

1 k!

 S∗

dx = det
A



· det
A

T
S ∗ 

dx
36

By (3.4) and (3.5) it follows that  Ak Ak 0  AA = 0 In−k and therefore det
AA  = det
Ak Ak 


37

As a conclusion of (3.6) and (3.7) we have proved the following result. Theorem 3.3. vk
S =

1 k!



det
Ak Ak 

Let a0 =
0 0     0, a1 =
a11  a12      a1n      ak =
ak1  ak2      akn  be k + 1 vertices of I n , let S denote the simplex determined by them, and let A denote the k × n matrix with entries aij , where i = 1 2     k and j = 1 2     n. Clearly, for any fixed i and j, aij is either zero or one,

50

Inscribed simplices

and therefore A is a binary matrix. Then, by Theorem 3.3, Problem 3.1 can be restated as follows. Problem 3.1∗ . Determine the values of  
n k = max det
AA   where the maximum is over all k × n binary matrices. For any simplex S determined by k + 1 vertices of I n , by Theorem 3.3, there is an integer m between 0 and
n k satisfying √ vk
S = k!1 m
38 Let 
n k m denote the number of the simplices satisfying (3.8) and let 
n k m denote the number of the simplices satisfying (3.8) and taking the origin as one of their vertices. It is easy to see that 
n k m = 

2n 
n k m k+1 


nk


n k i =

i=0

and 

2n − 1  k




nk i=0

2n  k+1


n k i =


39

However, the following problem seems very hard. Problem 3.3. Determine or estimate 
n k m and 
n k m. Let S denote a simplex with n + 1 vertices a0 =
a01  a02      a0n , a1 =
a11  a12      a1n      an =
an1  an2      ann  in E n . We define a00 = a10 =    = an0 = 1 and let A denote the
n + 1 ×
n + 1 matrix with entries aij , where i j = 0 1     n It is easy to see that ' ' '1 a01 ··· a0n ' ' ' ' 0 a11 − a01 · · · a1n − a0n ' ' ' det
A = ' 
310 ' = n! · vn
S    '  '    ' ' '0 a −a ··· a −a ' n1 01 nn 0n On the other hand, when S is a vertex simplex of I n , multiplying aij (i = 0 1     n j = 1 2     n) by 2 we get an
n + 1 ×
n + 1 matrix with ±1 entries. Thus, by Theorem 3.1, the special case k = n of Problem 3.1 is equivalent to the following matrix problem.

51

3.2 Binary matrices

Problem 3.1 . What is the maximum determinant of an
n + 1 ×
n + 1 matrix with ±1 entries? Remark 3.1. When m = 0, it can be shown that the number of k × n binary matrices satisfying det
AA  = m is k!
n k m. Let 
n l denote the number of the n × n binary matrices satisfying det
A = l and let  ∗
n l denote the number of the n × n matrices with aij = ±1 and satisfying det
A = l. It is a challenging problem to estimate 
n l and  ∗
n l, especially 
n 0 and  ∗
n 0. For related references we refer to Kahn, Komlós, and Szemerédi (1995). Remark 3.2 (Williamson, 1946). Let n denote the maximum determinant of an n × n binary matrix and let n∗ denote the maximum determinant of an n × n matrix of ±1 entries. By
310 it is easy to see that ∗ n+1 = 2 n n 

Therefore, to determine the n-dimensional maximal vertex simplex of I n , we can deal with either n × n binary matrices or
n + 1 ×
n + 1 matrices of ±1 entries. An n × n matrix A is called a Hadamard matrix if its entries are either 1 or −1 and satisfying AA = nIn  Hadamard matrices have many important applications, especially in applied mathematics (see Agaian, 1985 for detailed information). Let A be an
n + 1 ×
n + 1 Hadamard matrix with entries aij , where both i and j run from 0 to n. If ai0 = 1 and ai =
ai1  ai2      ain  for i = 0 1     n, then by definition it follows that  n if i = j,
ai  aj  = −1 otherwise, and thus ai − aj  =




ai − aj  ai − aj  =

√ 2n + 2

holds for every pair of distinct indices i and j. Therefore, the simplex with vertices a0 , a1      an−1 , and an is regular. On the other hand, if I n contains an n-dimensional regular vertex simplex, then we can construct an
n + 1 ×
n + 1 Hadamard matrix. As a conclusion, we have proved the following result.

52

Inscribed simplices

Theorem 3.4 (Grigorév, 1982). There is an n-dimensional regular vertex simplex in I n if and only if there is an
n + 1 ×
n + 1 Hadamard matrix.

3.3 Upper bounds To approach Problem 3.1∗ , we will estimate the values of
n k in this section. Let Mkn denote the family of the k × n binary matrices and let m∗A denote the matrix obtained by concatenating a matrix A in m times. For convenience, let Jk denote the k × k matrix with identical entries 1 and let Ik denote the k × k unit matrix. Theorem 3.5 (Hudelson, Klee, and Larman, 1996 and Neubauer, Watkins, and Zeitlin, 1997). If A ∈ Mkn , then ⎧ ⎨
k+1k+1k nk if k ≡ 1 (mod 2),
4k  det
AA  ≤
k+2k nk ⎩ k if k ≡ 0 (mod 2). 4
k+1k−1 Proof. We will deduce this theorem by studying

k + 1Ik − Jk AA . It is easy to establish that det

k + 1Ik − Jk  =
k + 1k−1 and therefore det


k + 1Ik − Jk AA  =
k + 1k−1 det
AA 


311

Let tr
B denote the trace of an n × n matrix B; that is tr
B =

n 

bii 

i=1

and let 1
B 2
B     n
B denote the eigenvalues of B. It is known that det
B =

n 

i
B


312

i
B = tr
B


313

i=1

and

n  i=1

If B1 and B2 are n × n matrices and B2 is nonsingular, then the sets of the eigenvalues of B1 B2 and B2 B1 = B2 B1 B2 B2−1 are identical. By a suitable limit process it can be shown that, if B1 is an n×m matrix and B2 is an m×n matrix, then the set of the nonzero eigenvalues of B1 B2 is identical with that of B2 B1 .

53

3.3 Upper bounds

By (3.12), (3.13), and the arithmetic-geometric mean inequality, we have k

det


k + 1Ik − Jk AA  ≤ k1 · tr


k + 1Ik − Jk AA 

k = k1 · tr
A

k + 1Ik − Jk A  (3.14) Suppose the ith column c of A has r ones and k − r zeros. Then the
i i-entry of A

k + 1Ik − Jk A is c

k + 1Ik − Jk c =
k + 1c c − c Jk c =
k + 1r − r 2 = r
k + 1 − r If A has nr columns each having exactly r ones, then tr
A

k + 1Ik − Jk A =

k 

r
k + 1 − rnr 


315

r=1

Now we consider two cases. Case 1. k = 2l − 1. It is obvious that r
k + 1 − r ≤ l2 and equality holds if and only if r = l. Thus we have k 

k 

r
k + 1 − rnr ≤ l2

r=1

nr = l2 n


316

r=1

By (3.14), (3.15), and (3.16) we get  

det


k + 1Ik − Jk AA  ≤

l2 n k

k =


k + 12k nk
4kk

and, by (3.11) det
AA  ≤


k + 1k+1 nk 
4kk

Case 2. k = 2l. It is obvious that r
k + 1 − r ≤ l
l + 1 and equality holds if and only if r = l or r = l + 1. Thus we have k 

r
k + 1 − rnr ≤ l
l + 1

r=1

k 

nr = l
l + 1n

r=1

By (3.14), (3.15), and (3.17) we get  det


k + 1Ik − Jk AA  ≤

l
l + 1n k

k =


k + 2k nk 4k


317

54

Inscribed simplices

and, by (3.11) det
AA  ≤


k + 2k nk  4k
k + 1k−1


The theorem is proved.

By Theorem 3.3, one can restate this theorem in terms of inscribed simplex in I n as follows. Theorem 3.5∗ . For every k-dimensional simplex S contained in I n , we have

vk
S ≤

⎧ ⎨ ⎩

 1 k!2k 1 k!2k




k+1k+1 nk kk

if k ≡ 1
mod 2,


k+2k nk
k+1k−1

if k ≡ 0
mod 2.

Remark 3.3. The first general upper bound in this setting was achieved by Hudelson, Klee, and Larman (1996). It is known
see Fejes Tóth, 1964) that the maximal k-dimensional simplices S contained in an n-dimensional unit ball are regular. Based on this observation, they were able to prove 
k + 1k+1 nk 1 vk
S ≤  k!2k kk a maximal k-dimensional regular where equality holds if I n contains

 vertex

 simplex. Let Gj denote a k× kj matrix such that its columns are kj distinct

 binary vectors and each has exactly j ones. When k = 2l − 1, for n = m· kl and A = m∗Gl , we indeed have
n k = det
AA  =


k + 1k+1 nk 
4kk


318

When k = 2l, this upper bound was improved by Neubauer, Watkins,

 and Zeitlin (1997) into the listed form. In this case, by choosing n = m· 2l+1 and l A = m∗Gl  Gl+1 , we get
n k = det
AA  =


k + 2k nk  4k
k + 1k−1


319

Clearly, as a general problem, Problem 31 as well as Problem 31∗ is still far from solved. Corollary 3.1 (Neubauer, Watkins, and Zeitlin, 1997). For every fixed k, we have ⎧
k+1k+1 if k ≡ 1
mod 2,
n k ⎨
4kk = lim k k
k+2 n→ ⎩ n if k ≡ 0
mod 2. 4k
k+1k−1

3.3 Upper bounds

55

Proof. Let
n k denote the maximum volume of a k-dimensional simplex inscribed in the n-dimensional unit cube I n . It is easy to see that


n k <
n + 1 k Therefore, by Theorem 3.3, we have
n k <
n + 1 k For convenience, we write

k 

l c = k+1  l

=

if k = 2l − 1, if k = 2l,

⎧ ⎨
k+1k+1 k

if k = 2l − 1,

⎩

if k = 2l,


4k


k+2k 4k
k+1k−1

and n = mc + r where 0 ≤ r < c. By (3.18) and (3.19) we have
mc k = 
mck ≤
n k ≤

m + 1c k = 

m + 1ck  Then it follows that
mc k
n k

m + 1c k ≤ < nk nk nk and therefore

 

mc mc + r

k ≤


n k
m + 1c k <   nk mc + r


The corollary follows.

Now we discuss the most interesting and the most important case, k = n. Let us start with an upper bound discovered by Hadamard in 1893. Theorem 3.6 (Hadamard, 1893). For every n × n matrix A with ±1 entries, we have det
AA  ≤ nn  where equality holds if and only if A is a Hadamard matrix.

56

Inscribed simplices

Proof. It is easy to see that tr
AA  =

n n  

a2ij = n2 

i=1 j=1

Thus, by (3.12), (3.13), and the arithmetic-geometric mean inequality, we get det
AA  =

n 

i
AA  ≤

1 n

n

· tr
AA 

= nn 

i=1

where equality holds if and only if 1
AA  = 2
AA  = · · · = n
AA  = n then A is a Hadamard matrix. The theorem is proved.




Remark 3.4. For every positive integer k, Sylvester (1867) proved that  1 1 S= 1 −1 tensor with itself k times gives a 2k × 2k Hadamard matrix. Thus, the upper bound in Theorem 36 is tight for infinite values of n. However, it is not always tight. If there is an n × n Hadamard matrix, by Remark 32 we have n∗ = nn/2 = 2n−1 n−1  Thus, since both n−1 and n∗ are integers, n must be 1, 2 or a multiple of 4. This fact was discovered by Paley in 1933. On the other hand, up to now, we do not know if there is a 428 × 428 Hadamard matrix. There are hundreds of papers dealing with Hadamard matrices. However, the following conjecture is still open. Conjecture 3.1 (Paley, 1933). There is an n × n Hadamard matrix whenever n is a multiple of four. For convenience, we write

  ∗
n k = max det
AA  

where the maximum is over all k × n matrices with ±1 entries. Based on Theorem 3.6 and Remark 3.4 it is reasonable to believe that ∗
n n = 1 n→ nn However, as we will see from the following theorem, this is far from the truth. lim

57

3.3 Upper bounds Theorem 3.7 (Barba, 1933; Ehlich, 1964a, 1964b; and Wojtas, 1964). ⎧ n−1 if n ≡ 1 (mod 4), ⎨
2n − 1
n − 1 ∗ 2
n n ≤ 4
n − 1
n − 2n−2 if n ≡ 2 (mod 4), ⎩ 4·116 7 n
n − 3n−7 if n ≡ 3 (mod 4) and n ≥ 63. 77

The proof of this theorem, based on some basic results about symmetric matrices, is very complicated. We will introduce it case by case. Let us start with a couple of lemmas. Lemma 3.1. Let C be an n × n symmetric matrix satisfying  c11 ≥ c22 ≥ · · · ≥ cnn > 0 cin  ≥ cnn for i = 1 2     n and the submatrix C ∗ =
cij , i j = 1 2     n − 1, is positive definite, then det
C ≤ cnn

n−1 


cii − cnn 

i=1

Proof. The assertion is obvious when det
C ≤ 0. If det
C > 0, then C is positive definite. For convenience, we write c =
cn1  cn2      cnn−1  and D = C ∗ − c1 c c nn

It is easy to see that

 D 0  C∼ 0 cnn

Therefore D is positive definite det
D ≤

n−1  i=1

≤

dii =

 n−1 

cii −

i=1

2 cin cnn



n−1 


cii − cnn 

i=1

and finally det
C = cnn · det
D ≤ cnn

n−1 


cii − cnn 

i=1

The lemma is proved.




58

Inscribed simplices

Lemma 3.2. Let C be an n × n positive definite symmetric matrix satisfying  c11 ≥ c22 ≥ · · · ≥ cnn ≥ d cij  ≥ d whenever i = j. Then n−1 

det
C ≤
cnn +
n − 1d


cii − d

i=1

where equality can be attained. Proof. The assertion is obvious if d = 0. For convenience, we assume d > 0, write ⎞ ⎛ c11 c12 · · · c1k ⎜c21 c22 · · · c2k⎟ ⎟ ⎜ Ck = ⎜     ⎟ ⎝      ⎠ ck1 ck2 · · · ckk and let Dk denote the k × k matrix obtained by replacing ckk by d in Ck . Now we proceed to prove the lemma by induction. When n = 2 the assertion is clear 2 ≤
c11 − d
c22 + d det
C2  = c11 c22 − c12

If the lemma is true for n − 1, then by Lemma 3.1 we get det
Cn  =
cnn − d· det
Cn−1  + det
Dn  ≤
cnn − d
cn−1n−1 +
n − 2d

n−2 


cii − d + d

n−1 

i=1

≤
cnn +
n − 2d
cn−1n−1 − d

n−2  i=1

≤
cnn +
n − 1d


cii − d

i=1


cii − d + d

n−1 


cii − d

i=1

n−1 


cii − d

i=1

The equality case can be easily verified and the lemma is proved.




Proof of the first case of Theorem 3.7. Assume that C = AA , where A is an n × n matrix with ±1 entries. It is easy to see that det
C keeps invariant if we multiply a whole row or a whole column of A by −1. Therefore, since n is odd, by suitable modification we can assume that the number ri of the −1 entries in ith row of A is even. For convenience, we write
 ij = card l  ail = ajl = −1 

3.3 Upper bounds

59

Then we have c11 = c22 = · · · = cnn = n cij =

n 

ail ajl = n −
ri + rj − ij  + ij −
ri + rj − 2ij 

l=1

= n − 2
ri + rj  + 4ij ≡ n (mod 4)


320

and therefore cij  ≥ 1 holds for all indices i and j. Applying Lemma 3.2 with d = 1, we get det
C ≤
2n − 1
n − 1n−1  The first case of Theorem 3.7 is proved.




To prove the second case of Theorem 3.7, we need the following technical lemma. Lemma 3.3 (Ehlich, 1964a). Let 
C denote the number of the elements cij ≡ 0
mod 4 in C = AA , where A is an n × n matrix with ±1 entries. If n ≡ 2
mod 4, then 
C ≤ n2 /2 Proof. Let ri denote the number of the −1 entries in the ith row of A and write
 ij = card l  ail = ajl = −1  then we have cij =

n 

ail ajl = n − 2
ri + rj  + 4ij

l=1

and therefore cij ≡ 0
mod 2 for all indices i j = 1     n In addition, since there is no integer triple ri  rj  rl  satisfying ri + rj ≡ rj + rl ≡ ri + rl ≡ 1
mod 2 one of the three elements cij , cjl , and cil is 2
mod 4. By suitable modification, we may assume that
 s = max card l  cil ≡ 0
mod 4 i=1n

60

Inscribed simplices

and cs+11 ≡ cs+12 ≡ · · · cs+1s ≡ 0
mod 4 Then, based on the argument in previous paragraph, we have cij ≡ 2
mod 4 for all i j = 1 2     s. In addition, we may assume that
 t = max card l  cil ≡ 0
mod 4 i=1s

and css+1 ≡ css+2 ≡ · · · ≡ css+t ≡ 0
mod 4 Then we have cij ≡ 2
mod 4 for i j = s + 1 s + 2     s + t If s ≥ n/2, then we have t ≤ n − s and 
C ≤ 2st +
n − s2 −
n − s − t2 + t By determining the maximum of 2st − t2 + t at 0 ≤ t ≤ n − s, we get 
C ≤ 2s
n − s ≤ n2 /2 If s < n/2, then it follows by the assumption on s that 
C ≤ n · s < n2 /2 As a conclusion of the two cases, the lemma is proved.




Proof of the second case of Theorem 3.7. We define Ck as those in the proof of the first case and assume that
 sk = max card j  cij ≡ 0
mod 4 i=1k

and ck1 ≡ ck2 ≡ cksk ≡ 0
mod 4 in Ck . It is easy to see that sk < k. Then, by an argument similar to the proof of the previous lemma, we can deduce cij  ≥ 2

61

3.3 Upper bounds

for i j = 1 2     sk and for i = k, j = sk + 1     k. For convenience, we write ⎞ ⎛ csn +1sn +1 csn +1sn +2 · · · csn +1n ⎜cs +2s +1 cs +2s +2 · · · cs +2n⎟ n n n ⎟ ⎜ n n Cs∗n = ⎜    ⎟   ⎝     ⎠ cnsn +1 By considering two cases 3.1 and 3.2, we get

det
S0∗ 

cnsn +2 > 0 or

···

det
S0∗ 

2 ≤ 0 and applying Lemmas

# # # # det
C ≤
n − 2· det
Cn−1  + #det
Csn # · #det
Cs∗n # ≤
n − 2· det
Cn−1  + 2
n − 2n−2
n − 2 + 2sn 

Repeating this process inductively, we can deduce that  n  l n−2 det
C ≤
n − 2 · det
Cn−l  + 2
n − 2
n − 2l + 2 si i=n−l+1

holds for some l such that Cn−l contains no element of 0
mod 4. Then, by Lemmas 3.2 and 3.3, we get det
C ≤
n − 2n−1
n + 2
n − l − 1 + 2
n − 2n−2

n − 2l + n2 /2 = 4
n − 12
n − 2n−2 


The second case is proved.

In the rest of this section we will deal with the third case of Theorem 3.7. In this case, by (3.20) we have cij ≡ n ≡ 3
mod 4


321

for all indices i and j. Let F
m n denote the family of the m × m positive definite symmetric matrices Cm =
cij  satisfying cii = n

i = 1 2     m

and cij ≡ 3
mod 4 1 ≤ i j ≤ m and let Cm∗ denote a matrix in F
m n satisfying   det
Cm∗  = max det
Cm   Cm ∈ F
m n  By (3.21) it is easy to see that ∗
n n ≤ det
Cn∗ 


322

62

Inscribed simplices

Therefore, to prove the third case of Theorem 3.7 it is sufficient to show that 4 · 116
n − 3n−7 n7 77 if n ≡ 3
mod 4 and n ≥ 63. For this purpose, first let us study the matrices Cm∗ in detail. det
Cn∗  ≤

Lemma 3.4 (Ehlich, 1964b). If 2 ≤ m ≤ n, then ∗ det
Cm∗  >
n − 3· det
Cm−1 

Proof. When m = 2, we may take C2∗ =



n −1 −1 n

and therefore det
C2∗  = n2 − 1 >
n − 3· det
C1∗  Assume that the assertion is true for m < n, we proceed to show it for m + 1. If Cm∗ =
cij∗ , we define Cm+1 =
cij  by ⎧ ∗ ⎪ ⎪cij∗ if 1 ≤ i j ≤ m ⎨ if i ≤ m − 1 and j = m + 1 c cij = cji = im ⎪ 3 if i = m + 1 and j = m ⎪ ⎩ n if i = j = m + 1 Then we have Cm+1 ∈ Fm+1n and det
Cm+1  =
n − 3· det
Cm∗  +
n − 3· det
Dm  where Dm is different from Cm∗ only at the last diagonal element with dmm = 3. On the other hand, by the inductive assumption, we have ∗ det
Dm  ≥ det
Cm∗  −
n − 3· det
Cm−1  > 0

Therefore, we get ∗ det
Cm+1  ≥ det
Cm+1  >
n − 3· det
Cm∗ 




The lemma is proved.

Lemma 3.5 (Ehlich, 1964b). All the non-diagonal elements of Cm∗ =
cij∗  are either −1 or 3. Proof. For convenience, let Cm−1 and Cm−1 denote the complementary sub∗ ∗ matrices of cmm and cm−1m−1 , respectively. If the lemma is false, we may assume that ∗ ∗ cmm−1 = cm−1m =c>3

3.3 Upper bounds

63

and det
Cm−1  ≤ det
Cm−1 


323

det
Cm∗  =
n − 3· det
Cm−1  + det
Dm 


324

It is clear that

where Dm was defined in previous lemma. By Lemma 3.4, we have det
Dm  = det
Cm∗  −
n − 3· det
Cm−1  ∗  > 0 ≥ det
Cm∗  −
n − 3· det
Cm−1

Then Dm is positive definite and therefore 2

det
Dm  ≤
n − c3 · det
Dm−1  <
n − 3· det
Dm−1 


325

where Dm−1 =
dij  is the complementary submatrix of dm−1m−1 in Dm . Let Cm =
cij  be an m × m symmetric matrix defined by ⎧ ⎪ if 1 ≤ i j ≤ m − 1 and i = j, ⎨dij cij = dm−1j if i = m ⎪ ⎩ n if i = j. Then it follows by (3.23), (3.24), and (3.25) that det
Cm  =
n − 3· det
Cm−1  +
n − 3· det
Dm−1  ≥
n − 3· det
Cm−1  +
n − 3· det
Dm−1  > det
Cm∗  which contradicts the maximum assumption on det
Cm∗ . The lemma is proved.
To characterize Cm∗ further, let us introduce a couple of new terms first. We say a matrix Cm =
cij  ∈ F
m n has a block of length l if, after suitable modification (exchange rows and corresponding columns)  3 if k ≤ i j ≤ k + l − 1 and i = j cij = −1 if k ≤ i ≤ k + l − 1 and j < k or j > k + l − 1 holds for some suitable k. If Cm has r blocks of lengths l1  l2      lr satisfying r  i=1

then we call Cm a block matrix.

li = m

64

Inscribed simplices

Lemma 3.6 (Ehlich, 1964b). Cm∗ is a block matrix. Proof. If, on the contrary, Cm∗ is not a block matrix, then, by suitable modification, we may assume that 

∗ ∗ cmm−1 = cmm−2 =3 ∗ cm−1m−2 = −1

Then the lemma can be proved by repeating the proof argument of Lemma 3.5.
Lemma 3.7 (Ehlich, 1964b). If Cm ∈ F
m n is a block matrix with r blocks of lengths l1  l2      lr , respectively, then  det
Cm  =
n − 3

1−

m−r

r  i=1

li n − 3 + 4li



r 


n − 3 + 4li 

i=1

Proof. When m = 1 or 2, the lemma is obvious. Assume that the assertion is true for m ≤ k < n then we proceed to show it for m = k + 1 by considering two cases. Case 1. r = k + 1. In this case it is easy to deduce that  det
Ck+1  =
n + 1
n − k =
n + 1 k

k+1

k+1 1−  n+1

Case 2. r < k + 1. For convenience, we assume that lr ≥ 2. Subtracting the kth row from the
k + 1th row and the kth column from the
k + 1th column we get ⎧ 0 ⎪ ⎪ ⎨ 2
n − 3  cij = ⎪ 3−n ⎪ ⎩ cij

if if if if

i = k + 1, j < k or i < k, j = k + 1 i = j = k + 1, i = k + 1, j = k or i = k, j = k + 1, i j ≤ k

in the resulting matrix. Therefore, we have det
Ck+1  = 2
n − 3· det
Ck  −
n − 32 · det
Ck−1 


326

where Ck and Ck−1 denote the k × k and the
k − 1 ×
k − 1 principal submatrices of Ck+1 , respectively. Clearly both Ck and Ck−1 are block matrices. Now we consider two subcases.

65

3.3 Upper bounds Subcase 2.1. lr > 2. By (3.26) and the inductive assumption, we have .  lr − 1 k+1−r det
Ck+1  =
n − 3 2
n − 7 + 4lr  1 − n − 7 + 4lr   r−1  lr − 2 li −
n − 11 + 4lr  1 − − n − 11 + 4lr n − 3 + 4l i i=1 / r−1 r−1   li −
n − 3 + 4li  i=1 n − 3 + 4li i=1   r r   l i =
n − 3k+1−r 1 −
n − 3 + 4li  i=1 n − 3 + 4li i=1 Subcase 2.2. lr = 2. Similar to the previous subcase, we get .   r−1  1 li k+1−r det
Ck+1  =
n − 3 2
n + 1 1 − − n + 1 i=1 n − 3 + 4li  / r−1 r−1   li −
n − 3 1 −
n − 3 + 4li  n − 3 + 4l i i=1 i=1   r r   li k+1−r =
n − 3 1−
n − 3 + 4li  i=1 n − 3 + 4li i=1




As a conclusion of the two cases the lemma is proved. Writing



F
n = max
n − 3

n−r

1−

r  i=1

li n − 3 + 4li



r 


n − 3 + 4li  

i=1

where the maximum is over all sets of positive integers l1  l2      lr  satis fying ri=1 li = n, by (3.22) and Lemmas 3.5–3.7 we have n∗ = ∗
n n ≤ F
n


327

Therefore, to prove the third case of Theorem 3.7 it is sufficient to show that F
n ≤

4 · 116 7 n
n − 3n−7 77

if n ≥ 63. For this purpose, we define

 G
n r =
n − 3n−r
n − 3 + 4su
n + 1 + 4sv 1 −

us v
s + 1 −  n − 3 + 4s n + 1 + 4s

66

Inscribed simplices

where

(n) ⎧ ⎨s = r  v = n − rs ⎩ u = r − v


328

and introduce a couple of technical lemmas to estimate F
n. Lemma 3.8 (Ehlich, 1964b).

  F
n = max G
n r  1 ≤ r ≤ n 

Proof. Assume that



F
n =
n − 3

n−r

1−

r  i=1

li n − 3 + 4li



r 


n − 3 + 4li 

i=1

with smallest r. If r = 1, there is nothing to prove. If r > 1, without loss of generality, we assume that     l1 = min li  1 ≤ i ≤ r and l2 = max li  1 ≤ i ≤ r  We fix li for i ≥ 3 and assume l1 = x ≤ h/2 and l2 = h − x, where h = l1 + l2 is a constant. Then we get F
n = c·f
x, where c is a positive constant and  x h−x f
x =
n−3+4x
n−3+4
h−x 1− − − n − 3 + 4x n − 3 + 4
h − x with

=

r  i=3

li  n − 3 + 4li


329

By a routine computation, we get f 
x = 8
h − 2x
1 − 2  If 1 − 2 ≤ 0, both f
x and F
n will not decrease when we substitute l1 and l2 by 0 and h, respectively, which contradicts the assumption on r. If 1 −(2 ) > 0,(both) f
x and F
n will increase when we substitute l1 and l2 by h2 and h+1 , respectively. By repeating this process for at most a finite 2 number of times we get that u of the r integers l1  l2      lr  must be s, and v of them must be s + 1. In other words (n) ⎧ ⎨s = r  v = n − rs ⎩ u = r − v Then F
n can be expressed as some G
n r. Lemma 3.8 is proved.




3.3 Upper bounds

67

Lemma 3.9 (Ehlich, 1964b).     max G
n r  1 ≤ r ≤ n = max G
n r  1 ≤ r ≤ 7  Proof. Assume that r ∗ is the smallest r at which the maximum is attained. We define its corresponding f
x and as those in the proof of Lemma 3.8. To prove r ∗ ≤ 7, we proceed to show that 1 − 2 < 0 whenever r ≥ 8. First, let us observe two basic facts. Case 1. v = 0. If s <
n + 3/8 by (3.29) and (3.28), we get
n − 3 + 4s
1 − 2  = n − 3 + 4s − 2
u − 2s = −n − 3 + 8s < 0 Case 2. v > 0. If s <
n − 3/8 we have
n + 1 + 4s
1 − 2  ≤ n + 1 + 4s − 2
u − 1s − 2
v − 1
s + 1 = −n + 3 + 8s < 0 In both cases, we get 1 − 2 < 0 whenever s <
n − 3/8. On the other hand, it is easy to see that, if r ≥ 12 and n > 9, or if r ≥ 9 and n > 27, we have n n−3 s≤ <  8 r In addition, we can verify case by case that 1 − 2 < 0 when 9 ≤ r ≤ 11 and 11 ≤ n ≤ 27. Hence, whenever r ≥ 9 we have 1 − 2 < 0 Now we deal with the case r = 8. If n ≡ 3
mod 8, it follows by n = 8s + 3 that 4
s + 1 8s 1 − 2 = 1 − − 8s + 4
s + 1 8s + 4s 2 s+1 = 1− − < 0 3 3s + 1 If n ≡ 7
mod 8, it follows by n = 8s + 7 that 1 − 2 = 1 −

12
s + 1 3s + 3 = 1− < 0 8s + 8 + 4s 3s + 2

As a conclusion of these cases, the lemma is proved. Lemma 3.10 (Ehlich, 1964b). If n > 39, then     max G
n r  1 ≤ r ≤ 7 = max G
n 6 G
n 7 




68

Inscribed simplices

Proof. We define =


u − 1s v
s + 1 + n − 3 + 4s n + 1 + 4s

and G
n r x =
n − 3n−r−1
n − 3 + 4su−1
n + 1 + 4sv g
x where

 g
x =
n − 3 + 4x
n − 3 + 4s − 4x 1 −  −

s−x x −  n − 3 + 4x n − 3 + 4s − 4x

By a routine computation, we get G
n r = G
n r 0 = G
n r s In fact, both G
n r x and g
x attain their maxima at x = 0. Otherwise, we can easily deduce G
n r x > G
n r 0 for 0 < x < s and hence can construct a block matrix of larger determinant. Since g 
x = 8
s − 2x
1 − 2 we must have 1 − 2 < 0. However, if r ≤ 5 and n > 39, by (3.28) and the definition of  we have n−4 n+3 >  s≥ 5 6
n − 3 + 4s
1 − 2 ≥ −n − 3 + 6s > 0 and therefore 1 − 2 > 0 By this contradiction, the lemma is proved.




Proof of the third case of Theorem 3.7. By (3.27) and Lemmas 3.8–3.10, it is sufficient to show that G
n 6 ≤ G
n 7 ≤

4 · 116 7 n
n − 3n−7 77


330

for n ≥ 63 and n ≡ 3
mod 4. It is easy to see that 1−

n us v
s + 1 n ≤ 1− − ≤ 1−  n − 3 + 4s n − 3 + 4s n + 1 + 4s n + 1 + 4s

Therefore, we have
n − 3r−n G
n r ≥
n − 3 + 4su−1
n + 1 + 4sv
4s − 3


331

69

3.3 Upper bounds and for v > 0
n − 3r−n G
n r <
n − 3 + 4su
n + 1 + 4sv−1
4s + 1


332

When n ≥ 63, we have
11n − 41
11n − 17 ≤
11n − 212 and
11n − 17
4n − 45 ≥
11n − 33
4n − 40 Therefore, by considering different u v pairs in (3.31), it follows that 77
n − 37−n G
n 7 ≥
11n − 415
11n − 17
4n − 45 ≥ 4 · 116
n − 45
n − 10
n − 3 Similarly, it follows by (3.32) that 36
n − 36−n G
n 6 < 2
5n + 15
n − 1 < 2 · 55
n + 15
n − 1 Hence for n > 104 , we have

 5  G
n 7 2 · 36 · 116 5 9 1 − 1 − > G
n 6 55 · 77 n+1 n−1 > 100363 · 099955 · 0999 > 1000117

In addition, it can be verified by a computer that   max G
n 6 G
n 7 = G
n 7 when 63 ≤ n ≤ 104 and n ≡ 3
mod 4. Finally, by considering different u v pairs in (3.32), we get G
n 7 ≤

4 · 116 7 n
n − 3n−7  77

By (3.30) the third case of the theorem is proved.




Remark 3.5. It is natural to ask: Are the upper bounds in Theorem 37 tight? To answer this question, by some skilful constructions, Neubauer and Radcliffe (1997) were able to show that both the first two bounds can be attained at an infinite number of n. In the third case, we are not so lucky. Based on Theorem 3.3 and Remark 3.4, Theorem 3.6 and Theorem 3.7 together can be restated in terms of inscribed simplices in I n as follows.

70

Inscribed simplices

Theorem 3.7∗ . Let S denote an n-dimensional simplex contained in I n . Then ⎧ 1  n if n ≡ 0 (mod 4), ⎪ ⎪ n!2n
2n+ 1n ⎪ ⎪ 1 ⎪ n−1 ⎨
n−1!2n−1
n − 1 if n ≡ 1 (mod 4),  vn
S ≤ n−6 7
n−2
n+1 113 ⎪ ⎪ if n ≡ 2 (mod 4) and n ≥ 62, ⎪ n!2n−1 77 ⎪ ⎪  ⎩ 1
n + 1n+1 if n ≡ 3 (mod 4). n!2n We end this section with the following problem. Problem 3.4. Determine the value of

= lim inf n→

∗

∗
n n  nn

Is the sequence 
n n/n  n = 1 2    dense in   1? n

3.4 Some particular cases In this section we deal with the following problems. 1 What is the maximum area of a triangle inscribed in I n ? 2 What is the maximum volume of a tetrahedron inscribed in I n ? 3 For relatively small n, what do we know about the maximum volume of the n-dimensional simplex inscribed in I n ? Let
n k denote the maximum volume of a k-dimensional simplex inscribed in I n , especially let us abbreviate
n n to n . By Theorem 3.3 and the definition of
n k, we have 


n k = k!1
n k
333 Now we can determine
n 2 and
n 3 by studying binary matrices. Theorem 3.8 (Hudelson, Klee, and Larman, 1996, and Neubauer, Watkins, and Zeitlin, 1997). Write l = n/3 and j = n − 3l, then ⎧1√ 2 if j = 0 ⎪ ⎨ 2 √3l 1


n 2 = 2 3l2 + 2l if j = 1 ⎪ ⎩1√ 2 3l + 4l + 1 if j = 2 2 Proof. If A is a 2 × n binary matrix with l1 columns identical with
1 0 , l2 columns identical with
0 1 , and l3 columns identical with
1 1 , then

3.4 Some particular cases we have AA =

71

 l3 l 1 + l3 l3 l 2 + l3

and therefore det
AA  = f
l1  l2  l3  = l1 l2 + l1 l3 + l2 l3  Since l1 + l2 + l3 = n and li ≥ 0, assuming that l1 ≤ l2 ≤ l3  it is easy to show that f
l1  l2  l3  attains its maximum if and only if l1 , l2 , and l3 are balanced; that is, l3 − l1 ≤ 1. Consequently, l1 must be l. Then, by considering three cases, we get ⎧ 2 if j = 0 ⎨3l f
l1  l2  l3  ≤ 3l2 + 2l if j = 1 ⎩ 2 3l + 4l + 1 if j = 2 where the upper bounds can be attained in every case. By (3.33) the theorem is proved.
Theorem 3.9 (Hudelson, Klee, and Larman, 1996, and Neubauer, Watkins, and Zeitlin, 1997). Write l = n/3 and j = n − 3l, then 


n 3 = 13 l3−j
l + 1j  Proof. Let S = convo a1  a2  a3  be a tetrahedron inscribed in I n and let A =
aij  be the corresponding 3 × n binary matrix. If three of the four vertices of S belong to a facet of I n , then it follows by Theorem 3.8 that  v3
S ≤ 13
n − 1 2 < 13 l3−j
l + 1j  Therefore, if v3
S ≥

1 3



l3−j
l + 1j 

since o =
0 0     0 is a vertex of S, A has no column identical with
1 0 0 ,
0 1 0 ,
0 0 1 , or
1 1 1 . Assume that A has l1 columns identical with
1 1 0 , l2 columns identical with
1 0 1 , and l3 columns identical with
0 1 1 , then we have l 1 + l 2 + l3 = n and

⎞ ⎛ l1 l2 l 1 + l2 AA = ⎝ l1 l 1 + l3 l3 ⎠  l2 + l 3 l2 l3

72

Inscribed simplices

Therefore we get det
AA  = 4l1 l2 l3 ≤ 4l3−j
l + 1j  where equality holds if and only if l1 , l2 , and l3 are balanced; that is maxl1  l2  l3  ≤ minl1  l2  l3  + 1 Consequently, we get


n 3 =

1 3



l3−j
l + 1j 


The theorem is proved.

Remark 3.6. By studying D-optimal designs, Neubauer, Watkins, and Zeitlin (1998b, 2000) were able to determine
n 4,
n 5, and
n 6. For suffi´ ciently large n, Neubauer and Watkins (2002) obtained
n 7 and Abrego, Fern´andez–Merchant, Neubauer, and Watkins (2003) determined
n 11,


n 15,
n 19, and
n 23. Clearly, among the problems concerning inscribed simplices in I n , to determine the exact value of n for a particular n is both important and interesting. As mentioned in Remark 3.4 and Remark 3.5, we do know the value of n for an infinite number of n (see Agaian, 1985; Neubauer and Radcliffe, 1997). To end this chapter, let us list the known results about n , n∗ (defined in Remark 3.2), and n up to n = 11 as follows. n ∗ n+1 ∗ n = 2n+1 n

n = n!n Author

2 4 1 0.5 Williamson

7 4096 32 0.00063492 Sylvester

3 16 2 0.3333333 Sylvester

8 14336 56 0.0013888 Ehlich and Zeller

4 48 3 0.125 Ehlich

5 160 5 0.0416666 Ehlich

6 576 9 0.0125 Williamson

9 73728 144 0.0003968 Ehlich

10 327680 320 0.0000881 Ehlich

11 2985984 1458 0.0000365 Hadamard

4 Triangulations

4.1 An example Let v1  v2      v8 be the eight vertices of the unit box I 3 , as illustrated in Figure 4.1. The box can be divided into six tetrahedra T1  T2      T6 defined by T1 = convv1  v5  v6  v8 

T2 = convv1  v6  v7  v8 

T3 = convv1  v2  v6  v7 

T4 = convv1  v2  v3  v7 

T5 = convv1  v3  v4  v8 

T6 = convv1  v3  v7  v8 

and

Then we have v3
Ti  =

1 3!

for all i = 1 2     6. v3

v4

v2

v1

v8

v7

v5

v6

Figure 4.1

73

74

Triangulations

Can we do better? In other words, can we divide the box into a set of tetrahedra of smaller cardinality? Yes, in fact, we can divide I 3 into five tetrahedra T1  T2    , T5 defined by T1 = convv1  v2  v3  v6 

T2 = convv1  v3  v4  v8 

T3 = convv1  v5  v6  v8 

T4 = convv3  v6  v7  v8 

and T5 = convv1  v3  v6  v8  In this case, we can verify that v3
Ti  =

1 3!

for all i = 1 2 3, and 4, and v3
T5  = 3!2  Can we do still better? The answer is “no.” Assume that the unit box I 3 can be divided into  tetrahedra T1 , T2      T and its facets are divided into t corresponding triangles. Then, since I 3 has six facets, we have t ≥ 12


41

It is easy to see that each of the triangles belongs to one tetrahedron and each of the tetrahedra takes at most three of the triangles as its facets. Thus we have ≥

12 3

= 4


42

If equality holds in (4.2), then equality in (4.1) must be attained as well. Therefore, after suitable permutation, we have T1 = convv1  v2  v4  v5 

T2 = convv2  v3  v4  v7 

T3 = convv2  v5  v6  v7 

and

T4 = convv4  v5  v7  v8 

Then a fifth tetrahedron T5 = convv2  v4  v5  v7  must be on the list too and therefore  ≥ 5 As a conclusion, we have shown the following result. Theorem 4.1 (Mara, 1976). The unit box I 3 can be divided into 5 tetrahedra, but not fewer.

4.2 Some special triangulations

75

4.2 Some special triangulations Let V
P denote the set of the vertices of a polytope P. A finite set ℘ of n-dimensional polytopes is called an n-complex if the intersection of every pair of the polytopes is a common face of them. An n-complex  of simplices is called a triangulation4 of ℘ if 0 0 S= P P∈℘

S∈

for every P ∈ ℘ there is a subset ∗ ⊆  such that 0 S P= S∈∗

and, for every S ∈ ∗ V
S ⊆ V
P


43

Clearly, for a triangulation  of ℘, its cardinality is a natural measure of efficiency. We define   
℘ = min card
  

where the minimum is over all the triangulations  of ℘. To find efficient triangulations and to determine the value of 
℘ for general n-complexes is important, fascinating, and challenging. In this chapter, we only focus on the very special case ℘ = I n . For convenience, we abbreviate 
I n  to n . In this section, we will introduce several special triangulations for I n and therefore produce some corresponding upper bounds for n . Triangulation I. Assume that  = S1  S2      Sk  is a triangulation of I n . By Theorem 3.3, it follows that m vn
Si  = i n! holds for some suitable positive integer mi . Therefore, by comparing these volumes with the volume of I n , we have n ≤ n!


44

In fact, I n indeed has a triangulation of cardinality n!. The two-dimensional case is obvious. It is known that I n has n facets F1  F2      Fn opposite to a given vertex v, and all the facets are
n − 1-dimensional unit cubes. Assume that Sij  j = 1 2    
n − 1! is a triangulation for Fi , then 4

Triangulation is a particular type of decomposition, which will be introduced in Section 4.4.

76

Triangulations

conv
v ∪ Sij   i = 1 2     n j = 1 2    
n − 1! will be a triangulation of cardinality n! for I n . Triangulation II (Sallee, 1982a, 1982b, and 1984). If P = Fn ⊃ Fn−1 ⊃ · · · ⊃ F0 = ∅ is a sequence of faces of a polytope P and vn  vn−1      v0 is a sequence of vertices of P such that vi ∈ Fi and vi+1 ∈ Fi hold for 0 ≤ i ≤ n − 1, then convv0  v1      vn  is an n-dimensional simplex. Let V = v1  v2      vm  be an ordering of the vertices of an n-complex ℘. For each face F of P ∈ ℘, we define   i
F  = min i  vi ∈ F and v
F  = vi
F  Then each sequence of faces P = Fn ⊃ Fn−1 ⊃ · · · ⊃ F0 = ∅ satisfying v
Fi+1  ∈ Fi for 0 ≤ i ≤ n−1 has an associated simplex convv
Fn  v
Fn−1      v
F0 . Let 
℘ V  denote the set of all simplices generated by sequences of faces defined as above. Then we have the following lemma. Lemma 4.1 (Sallee, 1982a). Let ℘ be a complex with a vertex ordering V . Then the corresponding 
℘ V  is a triangulation of ℘. Proof. For n = 2 the assertion is obvious. Assuming that the statement is true for n = k, we proceed to show it for n = k + 1. Let ℘ be a
k + 1-complex and let ℘ denote its facet complex, the k-complex consisting of the facets of the polytopes of ℘. By the assumption 
℘  V  is a triangulation of ℘ , then, for each polytope P ∈ ℘, the k-complex of facets opposite v
P is triangulated by a subset ∗ of 
℘  V . Thus   & 
P = conv v
P S  S ∈ ∗ is a triangulation of P. On the other hand, each simplex in ∗ has an associated sequence of faces Fk ⊃ Fk−1 ⊃ · · · ⊃ F0 = ∅. Since v
P ∈ Fk , this sequence can be extended into a sequence P = Fk+1 ⊃ Fk ⊃ · · · ⊃ F0 = ∅. Thus 
℘ V  =

0


P

P∈℘

is a triangulation of ℘. The lemma is proved.




4.2 Some special triangulations

77

Based on this lemma, we can construct some relatively efficient triangulations for I n . For example, we define 
 P1
n k = x ∈ I n  xi ≤ k  
 xi ≥ k  P2
n k = x ∈ I n  and

It is easy to see that

  ℘
n k = P1
n k P2
n k  & I n = P1
n k P2
n k

In addition, since the two endpoints of an edge of I n are different only at one coordinate and the difference is one, we have V
Pi
n k ⊆ V
I n  Let V be an ordering of V
I n  satisfying the following conditions.    1 If vi < k and vi < ui , then v is after u.   2 If k < vi < ui , then v is before u. By Lemma 4.1, the complex 
℘
n k V  does provide a triangulation for ℘
n k and consequently for I n . For the special case k = n/2, based on some detailed combinatorial analysis, Sallee (1984) deduced that   n ≤ card 
℘
n n/2 V  ≤ o
1 · n!
45 Triangulation III (Haiman, 1991). For convenience, let S i denote an i-dimensional simplex. In 1988, Billera, Cushman, and Sanders proved the following result about the triangulations of a Cartesian product of simplices. Lemma 4.2. Every triangulation of S k ⊕ S l has exactly
k + l!/k! · l! simplices. Proof. Let S be a
k + l-dimensional vertex simplex of S k ⊕ S l . We proceed to show that v
S k! · l! = 
46 v
S k ⊕ S l 
k + l! Since the ratio v
S/v
S k ⊕ S l  keeps invariant under nonsingular affine transformations, without loss of generality we assume that S k = convo e1      ek  S l = convo ek+1      ek+l , and S = convo a1      ak+l . Then it is easy to see that S k ⊕ S l is a
k + l-dimensional polytope with vertices o, ei ,

78

Triangulations

and ej1 ⊕ ej2 , where 1 ≤ i ≤ k + l, 1 ≤ j1 ≤ k and k + 1 ≤ j2 ≤ k + l. Thus we have 1 v
S k ⊕ S l  = vk
S k  · vl
S l  =
47 k! · l! and v
S =

det
A 
k + l!


48

where A is the
k + l ×
k + l matrix with entries aij . If ai = ej holds for some i and j, then by induction we can show det
A = 1


49

In this case, (4.6) follows from (4.7), (4.8), and (4.9). If, on the contrary ai = ej1 ⊕ ej2


410

holds for all indices i, then we claim that det
A = 0 which contradicts the assumption that S is
k + l-dimensional. To see this, we may look at a
k l-bipartite graph G with k + l vertices v1  v2      vk+l such that vj1 and vj2 is connected if and only if ej1 ⊕ ej2 is a vertex of S. Since G is a bipartite graph with k + l vertices, it can be easily shown that each set of k + l edges contains an even cycle. Without loss of generality, assume that a1  a2      a2m  is an even cycle in a1  a2      ak+l , then by (4.10) we get
−11 a1 +
−12 a2 + · · · +
−12m a2m = 0 and hence det
A = 0


The lemma is proved.

Since I k+l = I k ⊕ I l , based on Lemma 4.2, Haiman (1991) discovered the following triangulation for I k+l . If k = S1k  S2k      Skk  is a triangulation for I k and l = S1l  S2l      Sl l  is a triangulation for I l , then   ℘ = Sik ⊕ Sjl  i = 1 2     k  j = 1 2     l is a
k + l-complex satisfying I k+l =

0 ij

Sik ⊕ Sjl 

79

4.3 Smith’s lower bound By Lemma 4.1, we can produce a triangulation 0  = ij ij

for ℘, where ij is a triangulation of Sik ⊕ Sjl . By Lemma 4.2, we have
k + l! · k · l  k! · l! Consequently, for any fixed k, we can deduce
nk! · 
n−1k · k  k n nk ≤ ·
nk! = nk ≤ k ·
nk!

n − 1k! · k! k! card =

where

 k =

k


411

k  k!

Especially, since k is known for some small k (see Section 4.4), we can deduce the following upper bound from (4.11). Theorem 4.2 (Haiman, 1991). When n is large n ≤ 0871n · n! Remark 4.1. Recently, Orden and Santos (2003) were able to reduce 0871 to 0816. So far this is the best-known upper bound for n .

4.3 Smith’s lower bound Let  = S1  S2      Sn  be a triangulation for I n . By Theorem 3.5∗ , we have vn
Si  ≤


n + 1 2n n!

n+1 2

and therefore n ≥

1 2n n!  ≥ maxvn
Si 
n + 1 n+1 2


412

This is the first lower bound for n . For the convenience of further comparison, we abbreviate the number on the right-hand side of (4.12) to !n . By Stirling’s formula, we get   2 2 n n !n ∼ n2  e e Let F be a facet of I n and let  be a triangulation of I n . We will say that a simplex S ∈  belongs to F if a facet of S is a subset of F . If 1 and 2

80

Triangulations

are the sets of simplices belonging to two adjacent facets of I n , respectively, then it is easy to see that  1
413 vn
S = n S∈1 and



1  n
n − 1

vn
S ≤

S∈1 ∩2


414

Let Fi  −Fi , i = 1 2     n, be the n distinct pairs of opposite facets of I and define   ℵi = S ∈   S belongs to Fi or −Fi n

and

k =

k 0

ℵi 

i=1

By a combinatorial argument and (4.14), it follows that 

vn
S ≤

S∈ k−1 ∩ℵk

4
k − 1  n
n − 1

Hence, by induction and applying (4.13) and (4.14), we get     vn
S = vn
S + vn
S − vn
S S∈ k

S∈ k−1

≥

k   i=1

and especially

S∈ℵk

2 4
i − 1 − n n
n − 1 



S∈ k−1 ∩ℵk

vn
S ≥ 21 

S∈ 
n+1/2

On the other hand, for all S ∈ 
n+1/2 , we have vn
S ≤ Thus we get n ≥ 21
!n + n !n−1  

1  n !n−1 

1 2

  + 41 e
n + 1 !n 


415

√ which improves (4.12) by an asymptotic factor 025 en. This lower bound was discovered by Sallee (1984).

4.3 Smith’s lower bound

81

Note that both (4.12) and (4.15) are based on volume estimation of the simplices, to get a better lower bound for n it is natural to introduce a weight function to the measure. In fact, this idea can be realized by applying some basic results in Hyperbolic Geometry. By such an approach, W.D. Smith was able to prove the following theorem. Theorem 4.3 (Smith, 2000). n ≥

1 2

3  n2 2

!n 

The n-dimensional hyperbolic space has several models. Its Poincaré disk model H n is defined as the open unit ball int
Bn  with the Riemannian metric ds2 =

n  4
dxi 2 
1 − x2 2 i=1

In this model the geodesics are circles orthogonal to the boundary of the unit ball and an n-dimensional h-simplex with vertices v0  v1      vn ∈ Bn is the closed subset of H n bounded by the n + 1 spheres, which contain all the vertices except one and which are orthogonal to the boundary of the unit ball. Let Hpn denote the projective model of the n-dimensional hyperbolic space; that is, the open ball int
Bn  with the Riemannian metric ds2 =

n   1 1 dxi2 + x x dx dx 2 2 2 1 − x i=1
1 − x  ij i j i j

and the volume form d" =
1 − x2 −

n+1 2

dx


416

In fact, we can obtain Hpn from H n by a map g1
x =

2 x 1 + x2

The advantage of the projective model is that the geodesics become straight lines in Bn . Therefore, if S is an h-simplex in H n with vertices on 
Bn , then g1
S is simply the Euclidean simplex with the same vertices. Let Hsn denote the half space model of the n-dimensional hyperbolic space; that is, the half space x ∈ Rn  xn > 0 with the Riemannian metric ds2 =

n 1 
dxi 2 xn2 i=1

and the volume form d" = xn−n dx


417

82

Triangulations

In fact, we can obtain Hsn from H n by a map   1 2      2x  1 − x 2x   2x g2
x = n−1 1 2 x − en 2 The geodesics in Hsn are half circles and half lines orthogonal to the plane P = x  xn = 0 Let S n be a maximal simplex contained in Hpn and let S n be a maximal regular simplex contained in Hpn , both with respect to ". It is easy to see that the vertices of both S n and S n are on the boundary of Hpn . Assume that S = convv0  v1      vn  is a simplex with vn = en and define   S  = conv g
v0  g
v1      g
vn−1   where g
x = g2
g1−1
x and x =
x1  x2      xn−1  Then, by (4.16) and (4.17), we have  n+1
418 "
S =
1 − x2 − 2 dx S

and "
S =

  S


1−x 2 1/2

t−n dt dx

1  n−1 =
1 − x 2 − 2 dx  n − 1 S


419

It is well known that the following actions do not change the measure of a set in Hsn : translations parallel to P, rotations leaving the xn -axis pointwise fixed, and multiplications by positive scalars. Therefore, we may assume that S  is inscribed in Bn−1 . Especially, we can see that S  will be regular if S is regular. For convenience, we write  
n z =
1 − xz dx Sn

Now let us introduce several basic results about "
S n  and "
S n . Lemma 4.3 (Haagerup and Munkholm, 1981). For n ≥ 2, we have n − 1 "
S n+1  1 ≤  ≤ n n2 "
S n  Proof. First of all it follows by (4.18) and (4.19) that  n+1 "
S n  =
1 − x2 − 2 dx Sn


420

83

4.3 Smith’s lower bound and n "
S n+1  =




1 − x2 − 2 dx n

Sn


421

Next we try to prove  n−1 n−1 "
S n  =
1 − x2 − 2 dx n Sn


422

Let us define a vector field V
x =
1 − x2 −

n−1 2

x

for x < 1. Then we have div
V
x =
1 − x2 −

n−1 2

+
n − 1
1 − x2 −

n+1 2

and, when x ∈ 
S n  and n is an outward normal of S n at x
V
x n = n1
1 − x2 −

n−1 2



Let F denote one of the n + 1 facets of S n . For x ∈ F , we let x denote the Euclidean distance between x and the center of gravity of F , and write r =
1 − n−2 1/2 . By Gauss’ divergence formula  
V
x n ds div
V
x dx = 
S n 

Sn

we get  +
n − 1 
n n+1 = 
n n−1 2 2 = = = =

1 n−1
1 − x2 − 2 ds n n 
S  n+1  2 n−1
r − x2 − 2 ds n F n+1  n−1
r 2 − r 2 x2 − 2 r n−1 dx n Sn−1 n+1  n−1
1 − x2 − 2 dx n Sn−1 n+1 
n − 1 n−1  2 n

Applying (4.21) and (4.20) to this formula, we get n+1 
n − 1 n−1  −
n − 1 
n n+1  2 2 n
n + 1
n − 1 = "
S n  −
n − 1 "
S n  n n−1 = "
S n  n


n n−1 = 2

84

Triangulations

which is (4.22). Finally, by (4.20), (4.21), and (4.22), we can deduce n−1 n

"
S n  ≤ n "
S n+1  ≤ "
S n 


which proves the lemma.

Lemma 4.4 (Haagerup and Munkholm, 1981). Let S be a simplex with all vertices on the boundary of Hpn , let r denote the Euclidean distance between its center of gravity and the origin, and let f
x be a concave function defined on
0 1. Then 1  1  f
1 − x2  dx ≤ f

1 − r 2 
1 − x2  dx n n vn
S S S vn
S  whenever both integrals converge. Proof. For convenience, let us abbreviate the left-hand side and the righthand side of the inequality to J1 and J2 , respectively. Let v0  v1      vn be the vertices of S and write 
 T n =
t0  t1      tn  ∈ Rn+1  ti ≥ 0 ti = 1  Then we have J1 =

 Tn

 ' '2  f 1 − ' ti vi ' d

where  is the normalized Lebesgue measure on T n . Since  is invariant under the transformation ti → t
i for any permutation  of 0 1     n, we have   ' '2  J1 = f 1 − ' t
i vi ' d Tn

Let E denote the formation of mean values over all such permutations . Then, by the concavity assumption, we have   ' '2  ' ' J1 = E f 1− t
i vi d Tn    ' '2  ≤ f E 1 − ' t
i vi ' d (4.23) Tn

Since

' '   ' t
i vi '2 = t
i t
j
vi  vj  + t2  i=j

E
t
i t
j  = =

 1 tt n
n + 1 k=l k l  

1 1 − ti2  n
n + 1

i

4.3 Smith’s lower bound

85

and ' '2  
vi  vj  = ' vi ' − vi 2 i=j

=
n + 12 r 2 −
n + 1 by (4.23) we get J1 ≤ =

 T

n

    2 r 2 −
n+1 1 − ti2 d f 1 − ti2 −
n+1n
n+1 f

n+1

Tn

n

 


1 − r 2  1 − ti2 d

(4.24)

On the other hand, we note that equality in (4.24) will hold if S is regular. Therefore, applying (4.24) to S n and to g
x = f

1 − r 2 x, we get J2 =

 f Tn

n+1 n

 


1 − r 2  1 − ti2 d


425


The lemma follows from (4.24) and (4.25).

Lemma 4.5 (Haagerup and Munkholm, 1981). For every simplex S in Hpn , we have

 "
S ≤ " S n  Proof (A sketch). The two-dimensional case is trivial. For n = 3, we can deduce the assertion by Lobachevsky’s volume formula (see Thurston, 1977). Assume that the assertion is true for some n ≥ 3, we proceed to show it for n + 1. Let S = convv0  v1      vn+1  be an
n + 1-dimensional simplex inscribed in Hpn+1 with vn+1 = en+1 and write S  = convg
v0  g
v1      g
vn  (g
x is defined just above (4.18)). Then we define cn =

n "
S n+1  "
S n 

and f
t = t− 2 − cn t− n

n+1 2



By Lemma 4.3, we get cn ≥

n−1 n
n + 2 > n
n + 1
n + 3

86

Triangulations

for n ≥ 3 and therefore f
t is strictly concave on
0 1. Then it follows by (4.18), (4.19), Lemma 4.4, and the inductive assumption that   n+1 n n "
S − cn "
S   =
1 − x 2 − 2 dx − cn
1 − x 2 − 2 dx   S S   2  = f
1 − x   dx S  ≤ f

1 − r 2 
1 − x2  dx Sn

=
1 − r 2 − 2 n "
S n+1  − cn
1 − r 2 −   n ≤
1 − r 2 − 2 n "
S n+1  − cn "
S n  n

n+1 2

"
S n 

= 0 Thus, we have n "
S ≤ cn "
S n  = n "
S n+1  which proves the lemma.




Remark 4.2. This result was first conjectured by W. Thurston (1977). Lemma 4.6 (Haagerup and Munkholm, 1981). n+1 √  n+1 2 n n "
S  ≤  n−1 n! Proof (A sketch). By a routine analytic argument (considering the second order derivative of log 
n z and applying the Cauchy–Schwarz inequality), it can be shown that 
n z, as a function of z, is logconvex5 on 0 n+1 . 2 Thus, we have  n−1  2 
n n−1   
n n+1 2 2  ≤ 
n 0 
n n−1  2 Since 
n 0 = vn
S n  by (4.20) and (4.22), we have   n−1 2  n  n+1 
n n+1  n 2 2 ≤ =  n−1 n n−1 vn
S  n − 1 
n 2  "
S n 

5

A function f
x is logconvex in a b if f
x1 +
1 − x2  ≤ f
x1  f
x2 1− holds whenever x1  x2 ∈ a b and 0 ≤  ≤ 1.

4.3 Smith’s lower bound

87

On the other hand, it is known that n+1 √  n+1 2 n n vn
S  =  n n! Therefore, we get  "
S n 

≤

n+1 n−1

√ n+1 2

n  n!


which proves the lemma.

Proof of Theorem 4.3. Let I1n denote a largest cube inscribed in Hpn and assume that  = S1  S2      Sn  is a triangulation of I1n . Then we have

 vn I1n =



2 √ n

n =

 n2 4 n

and "
I1n  =

n 

"
Si 


426

i=1

Next we observe that 1   2 1  xi dx

 x2 dx =  vn I1n I1n vn I1n I1n

√  3 vn I1n n 1 1 √ ·2· n ·n· 3 2 =

 vn I1n = 13  Therefore, for any convex function f
t, we can deduce   1  1  2 2

 f
x  dx ≥ f

 x dx = f
13  vn I1n I1n vn I1n I1n By choosing f
t =
1 − t− 2 , we get

  n+1  n+1 " I1n 1 − 2 3 2 = 

 ≥ 1− 3 2 vn I1n n+1

and therefore

 " I1n ≥

 n2  n+1 3 2 4  n 2


427


428

88

Triangulations

Then, applying (4.28), Lemma 4.5, and Lemma 4.6 to (4.26), we get

 4  n2 3  n+1 2 " I1n 2 n ≥  ≥ n  n+1 √ n n+1 2 " Sn n−1 n!  n 1 3 2 ≥ !n  2 2


The theorem is proved.

Remark 4.3. By a detailed computation, Smith (2000) was able to improve
427 into   n1 "
I1n  lim = 1261522510     n n→ v
I1 n

4.4 Lower-dimensional cases In 1976 Mara introduced a convenient way to enumerate the vertices of I n . If a =
a1  a2      an  is a vertex of I n , then we denote it by a number a=

n 

2i−1 ai 

i=1

Clearly the numbers 0 1     2n − 1 exactly represent the 2n vertices of I n . Then we denote the simplex with vertices a, b     p by a b     p. With this notation, he discovered a triangulation of I 4 with 16 simplices S1 = 0 1 2 4 8 S4 = 1 2 4 8 14 S7 = 1 4 5 7 13 S10 = 1 7 11 13 14 S13 = 2 4 6 7 14

S2 = 1 2 3 7 11 S5 = 1 2 7 11 14 S8 = 1 4 7 13 14 S11 = 1 8 9 11 13 S14 = 2 8 10 11 14

S3 = 1 2 4 7 14 S6 = 1 2 8 11 14 S9 = 1 4 8 13 14 S12 = 1 8 11 13 14 S15 = 4 8 12 13 14

and S16 = 7 11 13 14 15 Therefore, we have 4 ≤ 16


429

4 ≥ 16


430

On the other hand, we claim

4.4 Lower-dimensional cases

89

Let  = S1  S2      S4  be a triangulation for I 4 , let Fi1  Fi2 , i = 1 2 3, and 4, be the four pairs of opposite facets of I 4 , and let ij denote the subset of  which belong to Fij . By Theorem 3.7∗ , we can deduce mi 
431 4! where mi only takes three possible values 1, 2, and 3. Since ij induces a triangulation on Fij , by Theorem 4.1 we get v4
Si  =

cardij  = lij 


432

where lij only takes two possible values 5 and 6. In addition, when cardij  = 5 there is a simplex Sk ∈  satisfying the following conditions. 1 It belongs to one and only one facet Fij of I n . 2

v4
Sk  =

2 4!

=

1  12


433

Now let us consider three cases. Case 1. cardij  = 5 holds for all indices i and j. Let S1 , S2      S10 be the ten different simplices belonging to either F11 or F12 . In addition, for each facet Fij , i ≥ 2, by condition 1 we get an extra simplex. Therefore,  has at least six extra simplices and thus 4 ≥ 10 + 6 = 16 Case 2. cardi1  + cardi2  ≤ 11 holds for i = 1 2 3, and 4, and equality holds for i = 1. Let S1  S2      S11 be the eleven simplices which belong to either F11 or F12 . Then we have 11 

v4
Si  = 21 


434

i=1

In addition, by conditions 1 and 2 we have three extra simplices S12 , S13 , and S14 belonging to F21  F22 , F31  F32  and F41  F42 , respectively, and satisfying 14  v4
Si  = 41 
435 i=12

Then, by (4.34) and (4.35), we get 4  i=15

v4
Si  = 1 − 21 − 41 =

1 4

90

Triangulations

and therefore 4 ≥ 14 + 41 / 18 = 16 Case 3. card11  + card12  = 12 Let S1  S2      S12 be the 12 simplices which belong to either F11 or F12 . Then we have 12 

v4
Si  =

1 2

i=1

and therefore by (4.31) 4 ≥ 12 + 21 / 18 = 16 As a conclusion of these cases, we have proved (4.30) and therefore the following theorem. Theorem 4.4 (Mara, 1976; Cottle, 1982; Sallee, 1982a; and Lee, 1985). 4 = 16 Remark 4.4. Lee’s proof was based on some deep results about the f -vectors and h-vectors. Our proof here is based on Sallee’s arguments. Theorem 4.5 (Hughes, 1993; Hughes and Anderson, 1993 and 1996). 5 = 67

6 = 308

7 = 1493

So far these are the only known results about the exact values of n . As one can imagine, Theorem 4.5 was proved by complicated linear and integer programs, with computer aid. It is neither possible nor in our interest to discuss such long proofs in this book. We refer the interested readers to the original papers. Besides triangulations, let us introduce another interesting concept. Let

= S1  S2      Sk  be a set of simplices. If In =

k 0

Si

i=1

and

 int
Si  int
Sj  = ∅

holds for all distinct i and j, then we say that is a decomposition of I n . Similar to n , we define   n = min card  

4.4 Lower-dimensional cases

91

where the minimum is over all decompositions of I n . It was proved by Mara (1976), Sallee (1982a), and Hughes and Anderson (1996) that 3 = 3 = 5 4 = 4 = 16 and 5 = 5 = 67 respectively. In fact, the proof arguments for Theorem 4.1 and Theorem 4.4 can show the first two cases. So far these are the only known results about the exact values of n . There are many open problems about n and n . Here we only mention one of them. Problem 4.1. Does n = n hold for all n? For some polytopes, even in three-dimensional space, the answer to the similar problem is negative (see Below, Brehm, De Loera, and Richter-Gebert, 2000). Now let us end this chapter with the following remark. Remark 4.5. As it was pointed out by Mara
1976, to determine the minimal triangulation of I n is closely related with minimizing the number of pivot steps in simplicial algorithms for finding approximate fixed points. Besides its own interest, this is another motive for studying the cube triangulations.

5 0/1 polytopes

5.1 Introduction A 0/1 polytope is a convex hull of a subset of the vertices of the unit cube   I n = x ∈ E n  0 ≤ xi ≤ 1  In the planar case, the only 0/1 polygons are a point, a segment, a triangle, and a square. However, in higher dimensions the situation turns out to be extraordinarily complicated. For example, according to Aichholzer (2000), there are 1 226 525 different classes of 0/1 polytopes in five dimensions, with respect to 0/1 equivalence. Another example, let #
n m denote the average volume of the 0/1 polytopes in E n and with m vertices. It was shown by Dyer, Füredi, and McDiarmid (1992) that, let  be any positive number and √ write $ = 2/ e  lim #
n m =

n→

1 if m ≥
$ + n , 0 if m ≤
$ − n .

In fact, the higher-dimensional 0/1 polytopes are rich in structure. Besides their own geometric and combinatorial interest, they do provide intuitive models for coding theory, combinatorial optimization, etc. There are several fundamental problems about the geometry and the combinatorics of 0/1 polytopes. For example: Problem 5.1. Determine or estimate the number of the different classes
with respect to a fixed equivalence of all n-dimensional 0/1 polytopes. Problem 5.2. Determine or estimate the maximal number of the k-faces of an n-dimensional 0/1 polytope. 92

5.2 0/1 polytopes and coding theory

93

These natural problems are simple or even trivial when n is small. However, in higher dimensions they are indeed challenging and fascinating. So far our knowledge about them is very limited. Perhaps the next problem is not as natural as the previous ones. However, it is one of the key problems in coding theory. Problem 5.3. Given integers n and s. What is the maximal number A
n s such that there is an n-dimensional 0/1 polytope with A
n s vertices and √ the minimal distance between them is not smaller than s? Therefore deep study of 0/1 polytopes will provide better understanding of coding theory. In this chapter, we will introduce some basic results about 0/1 polytopes.

5.2 0/1 polytopes and coding theory Let F2 denote the binary field and let %n denote the set of the 2n vertices of I n ; that is   %n =
x1  x2      xn   xi ∈ F2  It is easy to see that, for any subset X of %n V
convX = X Thus, every subset X of %n can be the vertices of a 0/1 polytope, possibly not n-dimensional. In addition, for any two points x and y of %n  
x y = card i  xi = yi  On the other hand, if we treat %n as a linear space over F2 and define a Hamming metric   x yH = card i  xi = yi on it, then we get an n-dimensional binary Hamming space H2n . Clearly, we have  x y = x yH 
51 whenever both x and y belong to %n . Usually, a point c ∈ H2n is called a binary codeword, a subset C of H2n is called a binary code, and the minimum Hamming distance between distinct points in C is called the separation of C, denoted by s
C. In addition, the number of the nonzero coordinates of a codeword is called its weight.

94

0/1 polytopes

For convenience, a code of length n, size m, and separation s is called an
n m s-code. Then we can restate Problem 5.3 as follows. Problem 5.3∗ . Given n and s. What is the maximal number A
n s such that there is a code C in H2n with cardinality A
n s and separation s? Roughly speaking, an information transmission process can be described as follows. First, design a code C and encode the information into codewords. Second, transmit the codewords through a channel to a receiver. Since the channel may add errors, the received words (in H2n ) perhaps are not the sent ones. Third, design a decoder to eliminate the errors. In this step, a received word w will be replaced by a codeword c∈C satisfying   w cH = min w w H  w ∈ C  It is easy to imagine that, if s = s
C is relatively large, then the errors caused by the transmission can be eliminated more easily. On the other hand, if cardC is relatively large, then the code is more efficient. To measure the efficiency of a code, we define log2
cardC n and call it the information rate of the code C. Based on the above arguments, it is easy to see that Problem 53∗ is indeed a key problem in coding theory. $
C =

Let us start with some basic results about A
n s. First of all, it is obvious that A
n 1 = 2n and A
n n = 2 Second, if C is a binary
n m s-code with m = A
n s and for i = 0 and 1 define   Ci = c ∈ C  c 1 = i  then C0 will reduce to an
n − 1 m0  s-code with a suitable m0 and C1 will reduce to an
n − 1 m1  s-code with a suitable m1 . Since one of them has a cardinality not smaller than A
n s/2, we get A
n s ≤ 2 A
n − 1 s Third, if C is a binary
n m 2k − 1-code with m = A
n 2k − 1, by adding an overall parity check to each codeword, we can produce an
n + 1 m 2kcode. On the other hand, suppose that C is a binary
n + 1 m 2k-code with

5.2 0/1 polytopes and coding theory

95

m = A
n + 1 2k, by puncturing C in a position at which two codewords disagree, we get an
n m 2k − 1-code with m = A
n + 1 2k. Thus we have A
n 2k − 1 = A
n + 1 2k


52

Now we introduce one lower bound and three upper bounds for A
n s. Theorem 5.1 (The Gilbert–Varshamov bound). 2n A
n s ≥ s−1 n  i=0

i

Proof. Let C be an
n m s-code with m = A
n s. For every codeword c ∈ C, we define   Bc = w ∈ H2n  c wH < s and

  Bci = w ∈ H2n  c wH = i 

By the maximum assumption on m, it follows that 0 Bc  H2n =


53

c∈C

In addition, it is easy to see that cardBci  = and

cardBc  = card

s−1 0

 n i

Bci =

i=0

s−1   n  i i=0


54

Therefore, by (5.3) and (5.4), we get A
n s ≥

2n cardH2n  = s−1 n  cardBc  i=0 i


Theorem 51 is proved.

As a counterpart of this lower bound, we have the following upper bound. Theorem 5.2 (The Hamming bound). 2n A
n s ≤ s −1 n  i=0 

where s = 
s − 1/2.

i

96

0/1 polytopes

Proof. Let C be an
n m s-code with m = A
n s. Similar to the previous proof, for every codeword c ∈ C we define   Bc = w ∈ H2n  c wH < s  Since s
C = s and s = 
s − 1/2, it is easy to see that  Bc Bd = ∅


55

holds for every pair of distinct codewords c and d of C. In addition, similar to (5.4), we can deduce  −1  s n  cardBc  =  i i=0 Therefore, by (5.5) we obtain A
n s ≤

2n cardH2n  = s −1 n   cardBc  i=0 i


Theorem 52 is proved.

The Hamming bound is also known as the sphere packing bound. As a counterpart, Theorem 51 is also known as the sphere covering bound. The best-known result about A
n s, obtained by ideas of packing and covering, is the following theorem. Theorem 5.3 (The Elias bound). Assume that r is an integer satisfying both r ≤ n/2 and r 2 − nr + ns/2 > 0. Then A
n s ≤

ns 2r 2 − 2nr

+ ns

2n · r n  i=0

i

To prove this theorem we need two technical lemmas. Lemma 5.1. For each pair of subsets C1 and C2 of H2n there is a w ∈ H2n such that card
w + C1  ∩ C2  cardC2  ≥  cardC1  2n  Proof. Assume that card
x +C1  C2  as a function of x attends its maximum at x = w. Then we have     1    card
x + C1  C2 card
w + C1  C2 ≥ n 2 x∈H n 2


 1    = n card x + c1  ∩ c2  2 x∈H n c1 ∈C1 c2 ∈C2 2

97

5.2 0/1 polytopes and coding theory
 1    card x + c1  ∩ c2  n 2 c1 ∈C1 c2 ∈C2 x∈H n

=

2

1   = n 1 2 c1 ∈C1 c2 ∈C2 1 cardC1  · cardC2  2n

=




The lemma is proved.

Lemma 5.2. If the weights of the words of an
n m  s-code C have an upper bound r with r ≤ n/2, then ns m ≤ 2  2r − 2nr + ns Proof. Let us list the codewords of the code as rows of an m × n matrix and, for j = 0 and 1, let pij denote the number of occurrences of the symbol j in the ith column of the matrix. Then we have pi0 + pi1 = m and, by the weight assumption n 

pi0 ≥ m
n − r


56

i=1

For convenience, we denote the left-hand side of (5.6) by p. Therefore, we get 2  n n  1  p2 2 pi0 ≥ pi0 =  n i=1 n i=1 Based on these preparations it can be deduced that  cc ∈C

c c H =

1 n  

pij
m − pij 

i=1 j=0

= nm2 −

n  

2 2 + pi1 pi0



i=1

= nm2 −

n  

2 + m2 − 2m pi0 2pi0

i=1



2p2 ≤ nm − + nm2 − 2m p n 2p2 = 2m p −  n 2





98

0/1 polytopes

In addition, by (5.6) and the assumption r ≤ n/2, we get p ≥ m
n − r ≥ 21 nm and thus, by comparing the values of a quadratic form   2p2 r   2 ≤ 2m r 1 −  c c H ≤ 2m p − n n cc ∈C On the other hand, we have  c c H ≥ m
m − 1s


57


58

cc ∈C

It follows by (5.7) and (5.8) that

 r m
m − 1s ≤ 2m2 r 1 − n

and therefore m ≤

ns  2r 2 − 2nr + ns


The lemma is proved.

Proof of Theorem 5.3. Let C2 be a binary
n m s-code with m = A
n s and define   C1 = x ∈ H2n  o xH ≤ r  By Lemma 5.1, there is a suitable
n m  s-code

 C3 = C 1 + w C 2 satisfying 

m = cardC3  ≥

m

r n i=0

i



2n

Then applying Lemma 5.2 to C3 , we get   m ri=0 ni ns ≤ m ≤ 2 n 2 2r − 2nr + ns and thus m≤ The theorem is proved.

ns 2r 2 − 2nr

+ ns

2n · r n  i=0

i




As has been mentioned before, to determine or to estimate the values of A
n s is a key problem in coding theory. Besides the upper bounds

99

5.3 Classification

introduced above, there are still several others, such as the Plotkin bound, the Griesmer bound, the Johnson bound, the linear programming bound, etc. (see van Lint, 1982 and Pless, Huffman, and Brualdi, 1998). The linear programming bound, though it is hard to get an exact expression, perhaps is the most interesting one. Let Kj
x denote the Krawtchouk polynomial of degree j; that is   j  n−x i x  Kj
x =
−1 i j −i i=0 where

 x x
x − 1 · · ·
x − i + 1 =  i i!

Then the linear programming bound can be stated as follows. Theorem 5.4 (Delsarte, 1972 and, 1973). If s is even  n  al  a0 = 1 al = 0 for 1 ≤ l ≤ s or l is odd; A
n s ≤ max l=0  n  al ≥ 0 al Kj
l ≥ 0 for 0 ≤ j ≤ n  l=0

Clearly, we can apply (5.2) if s is odd. When n is relatively small, this method is powerful. However, when n is large, it is difficult for application. For technical reasons, we will not prove it here. To end this section we list some known values of A
n s in the following table. For more about A
n s we refer to Pless, Huffman, and Brualdi (1998), Sloane (1977) or van Lint (1982). n

5

6

7

8

9

10

11

12

13

14

15

A
n 3 A
n 5 A
n 7

4 2 −

8 2 −

16 2 2

20 4 2

40 6 2

72 12 2

144 24 4

256 32 4

512 64 8

1024 128 16

2048 256 32

5.3 Classification Let 
n denote the number of the n-dimensional 0/1 polytopes reduced from I n . It is easy to see that 2n  n  2 n < 22 
59 
n < i i=n+1

100

0/1 polytopes

On the other hand – note that any
n − 1-dimensional subset of a facet and a nonempty set of the opposite facet will produce an n-dimensional set – we can deduce n−1  2 2n−1 
n ≥ 
n − 1 · i i=1   n−1
510 = 22 − 1 · 
n − 1 and therefore 
n ≥

n  

i−1

22 − 1



i=1

=

n−1 

1 − 2−2

i=0

i

 n−1 

22

i

i=0 n

∼ c · 22  where c=


511

   1 i 1 − 2−2  2 i=0

Comparing (5.11) with (5.9), we note that these bounds are reasonably good, although they are possibly not optimal. There are several types of classification for 0/1 polytopes based on different equivalent relations; for example, the classifications based on combinatorial equivalence, affine equivalence, congruence, or 0/1 equivalence. Needless to say, the affine equivalence and the congruence are two fundamental concepts in Geometry and the combinatorial equivalence is basic to understanding polytopes. Only the 0/1 equivalence is a relatively new concept restricted to 0/1 polytopes. In this section, we will introduce some known results about these classifications. Let
P denote the face lattice of a polytope P; that is, the set of all faces of P partially ordered by inclusion. Two poytopes P1 and P2 are combinatorially equivalent if
P1 is isomorphic to
P2 . The combinatorial equivalence has two basic properties. 1 If P1 is equivalent with P2 and P2 is equivalent with P3 , then P1 is equivalent with P3 . 2 If  is an isomorphism from
P1 to
P2 , then dim
F  = dimF

5.3 Classification

101

holds for every face F of P1 and therefore     card F ∈
P1  dimF = k = card F ∈
P2  dimF = k holds for every k satisfying 0 ≤ k ≤ n In the two-dimensional case, there are only two classes of 0/1 polygons with respect to the combinatorial equivalence, one is represented by a triangle and the other is represented by a square. In three dimensions, the situation is much more complicated. By applying the second property, we can see from Figure 5.1 that there are exactly eight different classes of three-dimensional 0/1 polytopes with respect to the combinatorial equivalence.

Figure 5.1

102

0/1 polytopes

Two 0/1 polytopes P1 and P2 are 0/1 equivalent if one can be transformed into the other by a symmetry of the unit cube I n . It is known that any symmetry of I n can be represented as a product of transformations of the following two types. Type I. x −→ y, where

⎧ ⎨xi if k = j yk = xj if k = i ⎩ xk otherwise

holds for some index pair i j. Type II. x −→ y, where

 yk =

1 − xi if k = i xk otherwise

holds for some index i. Therefore the 0/1 equivalence has the following properties. 1 If P1 is equivalent with P2 and P2 is equivalent with P3 , then P1 is equivalent with P3 . 2 If P1 and P2 are two equivalent n-dimensional 0/1 polytopes with V
P1  = u1  u2      uk  and V
P2  = v1  v2      vk , then there is a permutation  of 1 2     n such that k 

uij =

i=1

k  i=1

vi
j or k −

k 

vi
j

i=1

holds for each index j. As an example, by a routine comparison based on Figure 5.1, we can conclude that there are 12 different classes of three-dimensional 0/1 polytopes with respect to the 0/1 equivalence. For n-dimensional 0/1 polytopes, we have the following relations between the 0/1 equivalence (E1 ), the congruence (E2 ), the affine equivalence (E3 ), and the combinatorial equivalence (E4 ). Theorem 5.5 (Ziegler, 2000). E1 ⇒ E2 ⇒ E3 ⇒ E4  This assertion is easy to show. Therefore, we will not give a proof here. However, the converse to any of the three implications is false. Clearly, the

103

5.3 Classification

first two tetrahedra of the last row in Figure 5.1 are affine equivalent but not congruent. To show the other cases, we have the following examples. Example 5.1. Let S1 be a five-dimensional simplex with vertices u1 =
0 0 0 0 0, u2 =
0 0 1 1 0, u3 =
0 1 0 1 0, u4 =
1 0 0 1 0, u5 =
0 1 1 0 0, and u6 =
0 1 1 0 1, and let S2 be a five-dimensional simplex with vertices v1 =
0 0 0 0 0, v2 =
0 0 1 1 0, v3 =
0 1 0 1 0, v4 =
0 1 1 0 0, v5 =
1 0 0 1 0, and v6 =
1 0 0 1 1. It is routine to verify that ui − uj  = vi − vj  holds for all index pairs i j. Thus S1 and S2 are congruent. However, by the second property of the 0/1 equivalence, we can easily deduce that S1 and S2 are not 0/1 equivalent. Example 5.2. Let P1 be a five-dimensional polytope with vertices u1 =
0 0 0 0 0, u2 =
1 0 0 0 0, u3 =
0 1 0 0 0, u4 =
0 0 1 0 0, u5 =
0 0 0 1 0, u6 =
0 0 0 0 1, and u7 =
1 1 1 1 1, and let P2 be a fivedimensional polytope with vertices v1 =
0 0 0 0 0, v2 =
1 1 0 0 0, v3 =
0 1 1 0 0, v4 =
0 0 1 1 0, v5 =
0 0 0 1 1, v6 =
1 0 0 0 1, and v7 =
1 1 1 1 1. In fact, both P1 and P2 are bipyramids over fourdimensional simplices. Therefore, they are combinatorially equivalent. However, since the main diagonals of P1 and P2 are divided by the simplices in the ratios 1  4 and 2  3, respectively, they are not affinely equivalent. Let 1
n, 2
n, 3
n, and 4
n denote the numbers of the different classes of the n-dimensional 0/1 polytopes with respect to the 0/1 equivalence, the congruence, the affine equivalence, and the combinatorial equivalence, respectively. It follows by Theorem 5.5 that 4
n ≤ 3
n ≤ 2
n ≤ 1
n ≤ 
n


512

For large n to determine the exact values of i
n or even 
n is a very hard job. So far, our knowledge of this kind is very limited. We list the known ones in the following table. n


n

1
n

2
n

3
n

4
n

2 3 4 5

5 151 60879 4292660729

2 12 347 1226525

2 12 347 ??

2 8 ?? ??

2 8 172 ??

104

0/1 polytopes

It follows by (5.9) and (5.12) that i
n < 22

n


513

holds for all i = 1 2 3, and 4. This upper bound is certainly not optimal, since, on the one hand, many of the 0/1 polytopes are lower-dimensional; on the other hand, many of the full-dimensional ones are equivalent. However, so far no essentially better upper bound for i
n is known. As a counterpart of
513, we have the following lower bound for i
n. Theorem 5.5 (Ziegler, 2000). When n ≥ 6, we have i
n ≥ 22

n−2

for all i = 1 2 3, and 4. Proof. By (5.12), to prove the theorem it is sufficient to show n−2

4
n ≥ 22 


514

Let Fi0 and Fi1 denote the facets of I n given by xi = 0 and xi = 1, respectively. For convenience, we will call Fn0 the bottom facet, call Fn1 the top facet, and call all the others vertical facets of I n . Let n denote the family of the 0/1 polytopes P reduced from I n and satisfying the following conditions. 1 It contains the whole bottom facet of I n . 2 It contains both en =
0 0     1 and e =
1 1     1. 3 It contains neither en + e1 =
1 0     1 nor e − e1 =
0 1     1. Clearly, all the polytopes contained in n are n-dimensional and  2n−1 −4 2n−1 − 4 n−1 cardn  = = 22 −4  i i=0


515

Assume that n can be divided into combinatorially equivalent classes 1  2      k , we claim that

516 card  j ≤ 2n−1 ·
n − 1! holds for all j = 1 2     k To prove (5.16) let us start with looking at the facial structure of an individual polytope P ∈ n . By condition 2, it follows that each vertical facet of I n induces a facet for P which is adjacent to the cubic facet Fn0 . Thus, its vertices v ∈ Fn0 are completely determined by its vertical facets. On the other hand, P has no other cubic facet except Fn0 . To see this, if H is a hyperplane which contains at least 2n−1 vertices of I n , then by symmetry −H will contain

105

5.4 The number of facets

at least 2n−1 vertices as well. Thus H contains either a whole facet of I n or the origin. By conditions 1, 2, and 3, it follows that P has no other cubic facet. Assume that P1 and P2 are two polytopes of n and they are combinatorially equivalent. If  is an isomorphism from
P1 to
P2 , then it follows from the above observations that  induces a symmetry of I n−1 , which is determined by 
u −→ v, where u v ∈ V
Fn0 . Since the order of the symmetry group of I n−1 is 2n−1 ·
n − 1!, then (5.16) is proved. By (5.15) and (5.16), it follows that 22 −4 22 · 22 =  2n−1 ·
n − 1! 2n+3 ·
n − 1! n−2

n−1

4
n ≥ Writing f
n = 22 f
n = f
n − 12

n−2

n−2


517

and g
n = 2n+3·
n−1!, it is easy to see that f
6 > g
6,

g
n = 2
n − 1 · g
n − 1 ≤ g
n − 12  and therefore f
n ≥ g
n provided n ≥ 6. Thus, when n ≥ 6, by (5.17), we get 4
n ≥ 22

n−2




The theorem is proved.

5.4 The number of facets In this section, we will discuss some known results about Problem 5.2. Let f
n k denote the maximal number of the k-dimensional faces of an n-dimensional 0/1 polytope, and especially abbreviate f
n n − 1 to f
n. The known exact values of f
n can be listed as follows (see Ziegler, 2000). n

2

3

4

5

f
n

4

8

16

40

It is interesting to notice that f
5 > 25 . Thus we may imagine that, when n is large, to determine or to estimate f
n is a very hard job.

106

0/1 polytopes

Let e1  e2      en  be a standard basis of E n , and write e = easy to see that 
Tn = conv e1  e − e1      en  e − en

n

i=1 ei .

It is

is centrally symmetric with respect to the center of I n and therefore it is an n-dimensional 0/1 cross polytope. By this example, we get f
n ≥ 2n 


518

Based on this simple observation, we can deduce the following result. Lemma 5.3 (Kortenkamp, Richter–Gebert, Sarangarajan, and Ziegler, 1997). For i = 1 and 2, if I ni has an ni -dimensional centrally symmetric 0/1 polytope with fi facets, then I n1 +n2 has an
n1 + n2 -dimensional centrally symmetric 0/1 polytope with f1 f2 facets. Writing n = n1 + n2 , it is routine to check that   Q1 = x ∈ I n  xn1 = xn1 +1 = · · · = xn is affinely equivalent with I n1   Q2 = x ∈ I n  x1 = · · · = xn1 = 1 − xn1 +1 is affinely equivalent with I n2 , and Q1 intersects Q2 at the center of I n . If Pi is an ni -dimensional centrally symmetric polytope with fi facets and satisfying V
Pi  ⊆ V
Qi , for i = 1 and 2, then it can be shown that convP1 ∩ P2  is an n-dimensional centrally symmetric 0/1 polytope with f1 f2 facets. Let f ∗
n denote the maximal number of the facets of an n-dimensional centrally symmetric 0/1 polytope. Then, by Lemma 5.3, we get f
n ≥ f ∗
n ≥ f ∗
n1  · f ∗
n2 


519

On the other hand, Christof and Reinelt (2001) discovered a 13-dimensional centrally symmetric 0/1 polytope with 17 464 356 facets and therefore f ∗
13 ≥ 17464356 > 3613 


520

Thus, by (5.19) and (5.20), Ziegler (2000) got the following lower bound for f
n. Theorem 5.6. When n is sufficiently large, we have f
n ≥ f ∗
n ≥ 36n 

5.4 The number of facets

107

Remark 5.1. Based on the number of the facets of certain ten-dimensional centrally symmetric 0/1 polytope, Kortenkamp, Richter-Gebert, Sarangarajan, and Ziegler
1997 were able to deduce f
n ≥ f ∗
n ≥ 276n for sufficiently large n, the first lower bound better than
518. The following lower bound was proved by a random method. Since its proof is too complicated to be introduced here, we only cite the statement. Theorem 5.7 (Bárány and Pór, 2001; Gatzouras, Giannopoulos, and Markoulakis, 2005). There is an absolute constant c such that  c n n/2 f
n ≥  log2 n Now, let us introduce an upper bound for f
n. Theorem 5.8 (Fleiner, Kaibel, and Rote, 2000). There is a positive number c such that f
n ≤ c ·
n − 2! Proof. First of all, let us make some observations. 1 The volume of any n-dimensional 0/1 polytope P is an integer multiple of 1/n!. In particular, v
P ≥ 1/n!. 2 Let P be an n-dimensional polytope with facets F1 , F2      Fp and let  denote a projection from E n to an
n − 1-dimensional hyperplane H. Then p  vn−1

Fi  = 2 · vn−1

P i=1

3 Let i denote the projection from E n to Hi = x  xi = 0 If a hyperplane H contains a subset Q with 0 < vn−1
Q < , then it has a normal vector n of the form   n = ± vn−1
1
Q ±vn−1
2
Q     ±vn−1
n
Q  Assume that P is an n-dimensional 0/1 polytope with facets F1  F2      Fp and abbreviate vn−1
i
Fj  to ij . Then it follows by Observations 1 and 3 that Fj has an integral normal vector of the form   nj =
n − 1! · ± 1j  ±2j      ±nj 

108

0/1 polytopes

Let  ·  denote the 1 norm; that is x =

n 

xi 

i=1

Then, by Observation 2 p 

nj  =
n − 1!

j=1

=
n − 1! ≤
n − 1! = 2 · n!

p n   j=1 i=1 p n  

ij ij

i=1 j=1 n 

2

i=1


521

If p ≤
n − 2!, then there is nothing to prove. If p >
n − 2!, then we proceed to show p ≤ c ·
n − 2!
522 For convenience, we write


 zi  ≤ r  Gr = z ∈ Zn  Jr = Gr \ Gr−1 

and &r =



x =

x∈Gr

r 

i · cardJi 

i=0

It is easy to see that cardGr  = v
Gr + I n  ≤ and Thus we have

2n
r + n2 n
2r + nn = n! n!


523



card Jr+1 ≥ card Jr  r   1 i · cardJi  +
r − i · cardJr−i  2 i=0 r  1 r ≥ cardJi  + cardJr−i  2 i=0 2 r r = cardJr  2 i=1 r ≥ cardGr  2

&r =

(5.24)

5.4 The number of facets Let k be a number satisfying 

card Gk ≤ p < card Gk+1 

109


525

By the assumption p >
n − 2!, (5.23), and (5.25), we get
n − 2!
− −1 8 4 2 > d · n2 

k>

(5.26)

where d is a suitable constant satisfying 0 < d < 1 Then, by (5.21), (5.25), (5.24), and (5.26), we get 2 · n! ≥

p 

  nj  ≥ &k + k p − cardGk 

j=1

  k cardGk  + k p − cardGk  2  k ≥ cardGk  + p − cardGk  2 d · n2 ≥ ·p 2

≥

and therefore p≤

4 ·
n − 2! d


Thus we have proved (5.22) and the theorem.

Remark 5.2. Before Theorem 58, based on Observation 1, Bárány
see Ziegler, 2000 proved f
n ≤ n! + 2n In fact, Fleiner, Kaibel, and Rote
2000 also obtained an upper bound for f
n k; that is f
n k ≤ c ·
2
k + 1

n
n−1 n+1

·
n − 2!

110

0/1 polytopes

As we can see, when k = 0 this bound is much worse than the trivial bound 2n , and when k = n − 1 it is not as good as Theorem 58. Remark 5.3. As a conclusion of Theorem 57 and Theorem 58, we get  c1 n n/2 ≤ f
n ≤ c2 ·
n − 2! log2 n where c1 and c2 are constants.

6 Minkowski’s conjecture

6.1 Minkowski’s conjecture Let C be an n-dimensional centrally convex body and let X be a discrete set in E n . If  0 En = C +x x∈X

and



int
C + x



 int
C + y = ∅

holds for all distinct points x and y of X, then we call C + X a tiling of E n and call C a tile. For example, both a regular hexagon and a square are two-dimensional tiles. In fact, up to linear transformations, they are the only tiles in E 2 . This fact was discovered by Fedorov in 1885. In 1908, it was proved by Voronoi (1908/1909) that there are exact five different types of three-dimensional tiles, the parallelotope (also known as parallelopiped), the hexagonal prism, the rhombic dodecahedron, the elongated dodecahedron, and the truncated octahedron (see Erdös, Gruber, and Hammer, 1989). ˇ According to Delone (1929) and Stogrin (1975), there are 52 different types of tiles in four-dimensional Euclidean space. In higher dimensions, as we can imagine, the situation becomes extremely complicated and our knowledge is very limited. Nevertheless, for every n ≥ 2, it is obvious that I n is a tile in E n . Let a1  a2      an be n linearly independent vectors in E n , then the set

n  zi ai  zi ∈ Z = i=1

is called an n-dimensional lattice and the set a1  a2      an  is called a basis for the lattice. For example, the set of all points of integer coordinates is a lattice. Clearly lattices are very regular discrete sets in E n , periodic and 111

112

Minkowski’s conjecture

I 2 + y′ I2 + y I2 + x

Figure 6.1

centrally symmetric. We will call K + a lattice tiling of E n if it is a tiling and if is a lattice6 . There are thousands of references about tiling and lattice. We refer the interested readers to Engel (1993), and Erdös, Gruber, and Hammer (1989), and Schulte (1993). Now let us observe a simple phenomenon. Assume that X is a discrete set and I 2 + X is a tiling of the two-dimensional Euclidean plane E 2 . If I 2 + x touches I 2 + y at its boundary, where x and y are two distinct points of X, then their intersection will be a whole edge or a part of an edge of I 2 + x. In the second case, since I 2 + X is a tiling of E 2 , as we can see from Figure 6.1, there is another square I 2 + y , which touches I 2 + x at a whole edge. Thus, we get the following conclusion. If I 2 + X is a tiling of E 2 , then there are two squares sharing a whole edge. Especially, if I 2 + is a lattice tiling, then we have either 
z 0  z ∈ Z ⊂ or 
0 z  z ∈ Z ⊂ . The three-dimensional case is much more complicated than the twodimensional one, but is still elementary in nature. Let I 3 + X be a tiling of E 3 and, without loss of generality, assume that o ∈ X. If I 3 + x touches I 3 at its vertex v, then v is either a relative interior point of a face of I 3 + x, or is a relative interior point of an edge of I 3 + x, or a vertex of I 3 + x. In the first case, by projecting all the cubes (except I 3 + x), which contain v to the hyperplane that contains the mentioned face of I 3 + x and repeating the argument for the two-dimensional case, it can be shown that one of them touches I 3 at a whole face. If none of the cubes touching I 3 at v =
21  21  21  shares a whole face with it, by a routine argument we can deduce that, up to permutations of coordinates, the tiling has three cubes I 3 +
1 0 1 , I 3 +
0 2  1, and I 3 +
3  1 0, where 0 < i < 1 holds for i = 1, 2, and 3. Then the vertex
21  2 − 21  21  of I 3 +
0 2  1 will be a relative interior point of a face of 6

It was proved by Venkov (see Zong, 1996) that there is a lattice tiling K + whenever K is a tile.

6.2 An algebraic version

113

I 3 +
1 0 1  and therefore one of the cubes touches I 3 +
0 2  1 at a whole face. Thus, we get the following conclusion. If I 3 + X is a tiling of E 3 , then two of the unit cubes share a whole face. Especially, if I 3 + is a lattice tiling, then we have either 
z 0 0  z ∈ Z ⊂ , or 
0 z 0  z ∈ Z ⊂ , or 
0 0 z  z ∈ Z ⊂ . For convenience, we call two n-dimensional unit cubes a twin if they share a whole facet. Based on the above observations, we can make several conjectures, which will be the subjects of the remaining chapters of this book. In this chapter, we will deal with the following one. Minkowski’s conjecture. Every lattice tiling I n + of E n has a twin.

6.2 An algebraic version Clearly a lattice is an abelian group under addition. This is why we can get an algebraic version for Minkowski’s conjecture. As usual, let G denote a group. For g ∈ G, let
g denote the cyclic group generated by g and let g denote its order. Let us start with a basic result about the structure of the abelian groups. Lemma 6.1. A finitely generated abelian group G is the direct sum of a finite number of cyclic groups. Proof. Let 0 denote the unit of G. Let n denote the minimal cardinality of the generators and define   m = min g1   where the minimum is over all the possible generators g1  g2      gn  of G. Based on induction, we assume that G =
g1  + G∗  where g1 satisfies g1  = m and G∗ is a direct sum of n − 1 cyclic groups, say
g2     
gn . If, on the contrary, G is not a direct sum of
g1  and G∗ , then z1 g1 + z2 g2 + · · · + zn gn = 0 holds for some suitable integers zi with 0 < z1 < g1 

(6.1)

If
z1  z2      zn  = k (the common divisor of z1  z2      zn ) and zi = kzi , then

114

Minkowski’s conjecture

I 2 + x4

I 2 + x3 I 2 + x2 I 2 + x1

Figure 6.2

we have
z1  z2      zn  = 1 It is well known in linear algebra that then we can construct a set of new generators g1  g2      gn  for G, where g1 = z1 g1 + z2 g2 + · · · + zn gn  Since kg1 = z1 g1 + z2 g2 + · · · + zn gn = 0 by the assumption on g1 , we get g1  ≤ k On the other hand, by (6.1) and the definition of k, we have k ≤ z1 < g1  By this contradiction, the lemma is proved.




Let I n + be a lattice tiling of E n . Two cubes I n + u and I n + v are called adjacent if u1 − v1  = 1 and ui − vi  < 1 for all indices i = 1. In other words, two cubes are adjacent if their intersection is an
n − 1-dimensional set, which is perpendicular to the first coordinate axis. In fact, the adjacency between cubes does induce an equivalent relation on them, which will be one of the key ideas to deduce the algebraic version of Minkowski’s conjecture. Let I n + x and I n + x be two cubes of the tiling. We say that they are equivalent to each other if there is a sequence of cubes I n + x1 , I n + x2      I n + xm , where x1 = x and xm = x , such that I n + xi is adjacent to I n + xi+1 for each i = 1 2     m − 1. As shown in Figure 6.2, both I 2 + x3 and I 2 + x4 are equivalent with I 2 + x1 . Clearly, the equivalence is well defined and similar relations can be defined with respect to the other axes. For convenience, we denote this equivalent relation by ∼.

6.2 An algebraic version

115

By this equivalence, the unit cubes in I n + are divided into different classes. It is easy to see that an individual cube cannot be shifted in the direction of the first axis, but the union of a whole class can be shifted in this direction. In other words, the union of the unit cubes which belong to the same class form a cylinder of infinite length. For convenience, we say a lattice is rational if all the coordinates of its points are rational. Otherwise, we say it is irrational. Based on this preparation, we can now prove the following lemma. Lemma 6.2 (Schmidt, 1933). If there is a lattice tiling I n + without a twin, then we can find a rational lattice tiling I n +  without a twin. Proof. Assume that I n + is a lattice tiling of E n , which has no twin, and is irrational. For convenience, we assume that the first coordinate of some lattice point is irrational and define   o = x ∈  x ∼ o and

   = x ∈  x1 is rational 

Clearly,  is a sublattice of and o ⊆  . Since has a basis, the quotient group /  is finitely generated. It follows by Lemma 6.1 that /  has a basis. In addition, since will be rational if l is rational for some integer l u =  holds for all u ∈  . Thus can be written as a direct sum of two sublattices =  ⊕   Let u1  u2      ur  be a basis for  and let ur+1  ur+2      un  be a basis for  . Then u1  u2      un  will be a basis for . In addition, the first coordinate ui1 of ui is irrational whenever i ≤ r and is rational whenever i ≥ r + 1. Now we define u + i e1 if i ≤ r (6.2) vi = i ui otherwise, where i are real numbers to be chosen in a moment, and define

n  zi vi  zi ∈ Z   = i=1

116

Minkowski’s conjecture

Clearly, we have n  i=1

zi v i =

n 

zi ui +

i=1

 r 

 zi i e1 

i=1

Thus, changing from I n + to I n +  , the relative position of the cubes, which belong to the same class u +  , does not change, but the whole class  shifts ri=1 zi i in the direction of e1 . This means that I n +  is also a lattice tiling of E n . Let % be a tiling lattice for I n and let 
% denote the length of the shortest vectors of % \ o. Since the interior of 2I n contains no other point of % but the origin o, we get 
% ≥ 1

(6.3)

where equality holds if and only if I n + % has a twin. It follows by a routine argument that, for a fixed lattice % with a basis a1  a2      an  and for any fixed number ' > 0, there is a corresponding number  > 0 such that 
%  − 
% < '

(6.4)

holds whenever % is a lattice with a basis b1  b2      bn  satisfying ai − bi  <  for all indices i. Now we take % = , ai = ui , and % =  . Since I n + has no twin, we get 
 = 1 + 2'

(6.5)

for some ' > 0 Then we have a corresponding number  such that
64 holds. Now we choose i = −ui1 + i for
62, where i are rational numbers satisfying i  =  i − ui1  <  Then the first coordinate of any point of  is rational and it follows by (6.4) and (6.5) that 
  > 
 − 
 − 
  ≥ 1 + ' > 1 Therefore, by (6.3), I n +  is a lattice tiling of E n which has no twin.

117

6.2 An algebraic version

The lemma can be proved by repeating this process at most n times, each time dealing with one coordinate.
Let be a rational lattice with a basis  q1n q11 q12 a1 =    p p p1n  11 12 q21 q22 q a2 =       2n  p2n p21 p22   qn1 qn2 q an =       nn  pn1 pn2 pnn where
qij  pij  = 1 holds for all indices i and j. Then we define7 ( ) di = p1i  p2i      pni  vi =

1 e di i

 P = x ∈ E n  0 ≤ xi ≤ and

=

n 

1 di

 

zi v i  z i ∈ Z 

i=1

It is easy to see that P +  is a lattice tiling of E n and is a sublattice of . Consequently, we get a quotient group G = / = A1 + A2 + · · · + An  where

  Ai = o vi  2vi     
di − 1vi

and vi is the quotient of vi . Now we proceed to deduce a connection between the group G and the two systems I n + and P + . If u ∈ , v ∈ , and    int
P + v I n + u = ∅ then we have P + v ⊆ I n + u 7

As usual, here a1  a2      an  denotes the least common multiple of a1  a2      an .

118

Minkowski’s conjecture

Therefore, if one parallelotope P + v belongs to two of the unit cubes, then the point v can be represented in two distinct ways v = u+

n 

zi vi 

(6.6)

i=1

where u ∈ and 0 ≤ zi ≤ di − 1. On the other hand, if there is a parallelotope P + v does not belong to any of the unit cubes, then the point v cannot be represented in the above form. As a conclusion, I n + is a tiling of E n if and only if any point v ∈  can be uniquely represented in the form of (6.6), therefore G = A1 ⊕ A2 ⊕ · · · ⊕ An  On the other hand, if I n + has a twin, say I n and I n + ei , then we have di vi = ei ∈ and thus Ai is a cyclic group. Thus, by Lemma 6.2, we get an algebraic version for Minkowski’s conjecture. For the convenience of its proof, we state it in multiplication form. Minkowski’s conjecture (Hajós, 1941). Let G be a finite abelian group with unit 1. If g1  g2      gn are elements of G and d1  d2      dn are positive integers such that each element g of G can be uniquely written in the form g=

n 

z

gi i 

0 ≤ zi ≤ di − 1

i=1 d

then gi i = 1 holds for some i with 1 ≤ i ≤ n.

6.3 Hajós’ proof Let G be an abelian group under multiplication and let 1 denote its unit. For convenience, we will call a set gm = 1 g     gm−1 , g ∈ G, a cyclic set. Clearly, a cyclic subgroup is a special cyclic set. Lemma 6.3 (Hajós, 1941). Any cyclic subset gm of G may be decomposed into a direct product of cyclic subsets of prime orders. Moreover, gm is a subgroup of G if and only if one of the cyclic subsets is a subgroup. Proof. Assume that m = p1 p2 · · · pk , where pi are primes. It is known that every integer l, 0 ≤ l < m, can be uniquely written as l = c0 + c1 p1 + c2 p1 p2 + · · · + ck−1 p1 p2 · · · pk−1 

119

6.3 Hajós’ proof

where ci are integers satisfying 0 ≤ ci ≤ pi+1 − 1. Therefore, gm can be decomposed as a direct product gm = gp1 ⊗ gp1 p2 ⊗ · · · ⊗ gp1 p2 ···pk−1 pk  If gm is a subgroup of G, then we have gm = gp1 p2 ···pk =
gp1 p2 ···pk−1 pk = 1 Therefore, gp1 p2 ···pk−1 pk is a subgroup. Now we proceed to show that, if gm =
b ⊗ H holds for some cyclic group
b and a set H, then gm is a cyclic group as well. Since 1 ∈ gm , we have 1 = b · h for some integer and some element h ∈ H. Then we get b = b +1 · h ∈ gm and therefore b = gs

(6.7)

for some positive integer s ≤ m − 1. On the other hand, it follows by gm =
b ⊗ H that b · c ∈ gm if c ∈ gm  Therefore, if l = rs + t and 0 ≤ t ≤ s − 1, by (6.7) we have gl = grs · gt = br · gt ∈ gm and finally gm =
g


The lemma is proved.

Now let us introduce a concept, group ring, which is important for Hajós’ proof. Let G = g1  g2      gn  be a finite abelian group under multiplication, then the group ring
G is the set  
r = z i gi  z i ∈ Z in which the addition is defined by    zi gi + zi gi =
zi + zi gi and the multiplication is defined by



     zi gi = zi gi ·



 gj gk =gi

 zj zk

gi 

120 where

Minkowski’s conjecture 

denotes a finite sum. For example, if G = 1 a with a = 2, then  


G = z1 1 + z2 a  zi ∈ Z 

in which
z1 1 + z2 a +
z1 1 + z2 a =
z1 + z1 1 +
z2 + z2 a and
z1 1 + z2 a ·
z1 1 + z2 a =
z1 z1 + z2 z2 1 +
z1 z2 + z1 z2 a Assume that ri =

li 

zij gij 

j=1

where zij = 0, then we define G
r1      rk  to be the subgroup of G generated



m by gij  i = 1     k j = 1     li . In addition, if p1 1 p2 2    pm is a standard factorization for G
r1      rk , then we define p
r1      rk  =

m 

i 

i=1

For convenience, for g ∈ G, we use g to denote either 1−g or 1+g +· · ·+gp−1 for some prime number p. Clearly, g ∈
G. Now we are ready to introduce another key lemma for Hajós’ proof. Lemma 6.4 (Hajós, 1941). Assume that in the group ring
G an equation r g1 · · · gk = 0 holds and none of the gi can be omitted without violating the equality. Then we have p
r g1      gk  < p
r + k

(6.8)

Proof. For convenience, we write r=

q 

zi hi 

i=1

where zi = 0 and hi ∈ G. When k = 1, we proceed to show g1 ∈ G
r and hence p
r g1  < p
r + 1 If g1 = 1 − g1 , then it follows by r · g1 = 0 that   q q   zi hi = zi hi · g1 i=1

i=1

121

6.3 Hajós’ proof and therefore h1 = hj · g1

holds for some suitable hj ∈ G. Then we have g1 = h1 hj−1 ∈ G
r If g1 = 1 + g1 + · · · + g1p−1 , then we get r
1 − g1p  = 0. By repeating the above arguments we get g1p ∈ G
r. On the other hand, by the assumption we get   r · g1 + · · · + g1p−1 = −r and therefore h1 = hj · g1s holds for some j and s with 1 ≤ s ≤ p − 1. Then we get g1s ∈ G
r. It is known in number theory that up + vs = 1 holds for two integers u and v. Thus we get g1 = g1up+vs =
g1p u ·
g1s v ∈ G
r As a conclusion of the two cases, (6.8) is true for k = 1. Inductively, assuming that k ≥ 2 and p
r g1 · · · gt  gt+1      gk  < p
r g1 · · · gt  + k − t

(6.9)

holds for all t = 1     k − 1, we proceed to prove (6.8). Assume that G1 and G2 are two subgroups of G such that G1 ⊂ G2 and K is a subset of G. Let Hi denote the subgroup generated by Gi  K; it is easy to see that the order of the quotient group H2 /H1 is a factor of the order of the quotient group G2 /G1 . Thus we have p
G2  K − p
G1  K ≤ p
G2  − p
G1  Applying this inequality to G1 = G
r g1 · · · gt  G2 = G
r g1      gt , and K = gt+1      gk , we get p
r g1      gk  − p
r g1 · · · gt  gt+1      gk  ≤ p
r g1      gt  − p
r g1 · · · gt  Then it follows by (6.9) that p
r g1      gk  − p
r g1      gt  ≤ p
r g1 · · · gt  gt+1      gk  − p
r g1 · · · gt  < k − t Now we consider two cases.

(6.10)

122

Minkowski’s conjecture

Case 1. p
gi  = 1 holds for some gi . Without loss of generality, we assume that i = 1. Applying (6.10) with t = 1, we get p
r g1      gk  < k − 1 + p
r g1  ≤ p
r + k and hence (6.8). Case 2. p
gi  ≥ 2 holds for all gi . In this case, we apply induction on f = p
g1  + · · · + p
gk . By multiplying gk with a suitable element u ∈
G, we can get a new element g0 = 1 − g0 such that 1 ≤ p
g0  = p
gk  − 1 To see this, we consider three subcases. For convenience, we assume that gk  = q l m, where q is a prime and
q m = 1. Subcase . gk = 1 − gk . Then we can choose u = 1 + gk + · · · + gkq−1 g0 = gkq  Subcase . gk = 1 + gk + · · · + gkp−1 and p is a factor of q l m. Then we can choose u = 1 − gk g0 = gkp  Subcase . gk = 1 + gk + · · · + gkp−1 and p is not a factor of q l m. Then we can choose ⎧   ⎨ u =
1 − g  1 + gp + · · · + g
q−1p k k k ⎩g = gqp  0

k

Now we get a new equation r g0 · · · gk = 0 where k ≤ k − 1 and 

k  i=0

p
gi 
0,

0

otherwise,

q
t

m−1 · · · gm−1

−sm−1 



which contradicts the assumption (6.12). By deleting as many factors as possible from (6.13), we get a new equation, without loss of generality, say gmpm · g1 p1 ⊗ · · · ⊗ gw pw = g1 p1 ⊗ · · · ⊗ gw pw with 1 ≤ w ≤ m − 1. For convenience, we write W = g1 p1 ⊗ · · · ⊗ gw pw and H = G
g1      gw 

(6.14)

124

Minkowski’s conjecture

For every h ∈ H, we have h = h1 · h2  where h1 ∈ W and h2 ∈ gw+1 pw+1 ⊗ · · · ⊗ gm pm  Since both h1 and h2 belong to H, we get h2 · W ⊆ H and therefore H can be divided into disjoint sets of the form h2 · W . As a consequence, we get cardW divides cardH; that is p1 · · · pw  cardH Let us define ℘=



(6.15)

g

g∈W

and p −1

gi = 1 + g i + · · · + g i i for i = 1 2     w. It is easy to see that ℘=

w 

gi 

i=1

On the other hand, by (6.14) we get gmpm g1 · · · gw = g1 · · · gw and thus g1 · · · gw
1 − gmpm  = 0 Apply Lemma 6.4 with r = 1, k = w + 1, and gw+1 = gmpm , we get p
g1      gw  gw+1  < w + 1 and thus p
g1      gw  ≤ w

(6.16)

By (6.15) and (6.16), we get G
g1      gw  = g1 p1 ⊗ · · · ⊗ gw pw  where w ≤ m − 1. The conjecture follows by the inductive assumption.




Now let us restate the proved conjecture as the following theorem. The Minkowski–Hajós theorem. Every lattice tiling I n + of E n has a twin.

6.4 Other versions

125

6.4 Other versions Let I n + be a lattice tiling of E n . By the Minkowski–Hajós theorem, without loss of generality, we may assume that
1 0     0 ∈ . Then we n have
k 0     0 ∈ for any integer k and ∪ k=−
I +
k 0     0 is a cylinder of infinite length. Let H denote the hyperplane x ∈ E n  x1 = 0, let  denote the projection of to H, and let I  denote the projection of the cylinder to H. It is clear that I  is an
n − 1-dimensional unit cube,  is an
n − 1-dimensional lattice, and I  +  is a tiling of H. By repeating this argument, one can easily deduce the following corollary. Corollary 6.1. If I n +Zn A is a lattice tiling of E n , after a suitable permutation of the axes, we can take ⎛

⎞ 1 0 ··· 0 ⎜ 21 1 · · · 0⎟ ⎜ ⎟ A = ⎜    ⎟     ⎝    ⎠

n1 n2 · · · 1 Let C be an n-dimensional centrally symmetric convex body and let = Zn A be a lattice with a basis a1  a2      an , where ai =
ai1  ai2      ain . According to Minkowski’s first theorem in geometry of numbers, if v
C ≥ 2n d
 = 2n det
A then C contains a lattice point u ∈ \ o As a special case C = 2I n  we can easily deduce that for any n × n real matrix A =
aij  with det
A = 1 there is a z ∈ Zn \ o satisfying ⎧ ⎪ ⎪ a11 z1 + a21 z2 + · · · + an1 zn  ≤ 1 ⎨ a12 z1 + a22 z2 + · · · + an2 zn  ≤ 1 ⎪ ··· ⎪ ⎩ a1n z1 + a2n z2 + · · · + ann zn  ≤ 1

(6.17)

This is a basic result in Diophantine approximation. Therefore it is both important and interesting to study the equality cases. In 1896 Minkowski discovered that, for any 2 × 2 real matrix A =
aij  with det
A = 1, there is a z ∈ Z2 \ o satisfying a11 z1 + a21 z2  < 1 a12 z1 + a22 z2  < 1

126

Minkowski’s conjecture

unless A has an integral column. In that book he also stated an n-dimensional analogue and promised a proof, which has never appeared. In fact, the n-dimensional analogue is nothing else but a Diophantine approximation version of Minkowski’s conjecture. If A =
aij  is an n × n real matrix with det
A = 1, then there is a z ∈ Zn \ o satisfying ⎧ a11 z1 + a21 z2 + · · · + an1 zn  < 1 ⎪ ⎪ ⎨ a12 z1 + a22 z2 + · · · + an2 zn  < 1 (6.18) ⎪ ··· ⎪ ⎩ a1n z1 + a2n z2 + · · · + ann zn  < 1 unless A has an integral column. In fact, this is the first version. The geometric version was formulated in Minkowski (1907) 11 years later. Next we will show their equivalence. It is easy to see that I n + is a packing in E n if and only if (6.18) has no nontrivial integer solution. In addition, since v
I n  v
I n  = = 1 d
 det
A I n + will be a tiling of E n if it is a packing. If the Diophantine approximation version is right, then A has an integral column satisfying
a1i  a2i      ani  = 1 whenever I n + Zn A is a lattice tiling of E n . Assume that z1  z2      zn are integers satisfying a1i z1 + a2i z2 + · · · + ani zn = 1 We define u=

n 

zj aj 

j=1

U=

0

 I n + zu 

z∈Z

and

  H = x ∈ E n  xi = 0 

Clearly, I n + consists of translates of U and the intersection of these translates with H is an
n − 1-dimensional lattice cube tiling of H. By an inductive argument, we can deduce the geometric version. On the other hand, the Diophantine approximation version follows directly from the Minkowski– Hajós theorem.

6.4 Other versions

127

In 1998, Kolountzakis stated another version of this conjecture. Let A =
aij  be an n × n real matrix with det
A = 1. If, for all z ∈ Zn \ o, some coordinate of the vector zA is a nonzero integer, then A has an integral column. We will not show their equivalence here. Instead, let us end this chapter with a conjecture about a general tile C. Just like the cube case, we call C + x and C + y a twin if they join at a whole facet. Conjecture 6.1. Every lattice tiling C + of E n has a twin.

7 Furtwängler’s conjecture

7.1 Furtwängler’s conjecture Let I n + X be a set of unit cubes. If every point of E n , which is not on the boundary of any cube, lies in exactly k cubes, then we say that I n + X is a k-fold tiling of E n . If at the same time X is a lattice, then we say that I n + X is a k-fold lattice tiling of E n . Especially, a tiling is a 1-fold tiling and a lattice tiling is a 1-fold lattice tiling of E n . By overlapping a given tiling exactly k times, we can get a k-fold tiling of E n . Clearly, the multiple tilings of this kind cannot reflect the nature of their complexity. The following example does show another class of multiple tilings. Example 7.1. Let I n + be a tiling of E n . Then, for any positive integer k, the system I n + k1 is a kn -fold lattice tiling of E n . This statement can be deduced by the following argument. First of all, is a lattice tiling of E n . Second, for any point p, the system k1 I n + a tiling of E n . Finally, there are kn points p1  p2      pkn such that  1 2  1  int I n + pi I n + pj = ∅ int k k

1 n I + k1 k p + k1 is

holds for every pair of distinct points pi and pj , and k  0 1 n

In

=

i=1

k

 I n + pi 

Therefore I n + k1 is a kn -fold lattice tiling of E n . Remark 7.1. In fact, by a similar argument, we can show that, for any ndimensional tile C, if C + is a lattice tiling of E n and k is a positive integer, then C + k1 is a kn -fold lattice tiling of E n . 128

7.2 A theorem of Furtwängler and Hajós

129

It is clear that the multiple lattice tiling is a very natural generalization for the lattice tiling. In 1936, even before Hajós’ proof for Minkowski’s conjecture, Furtwängler made the following generalization and did prove the n ≤ 3 cases. Furtwängler’s conjecture. In every multiple lattice tiling I n + there is a twin. A few years later, without knowledge of Furtwängler’s work, G. Hajós did study the same problem. First of all, with the very same argument as the lattice tiling case, he was able to restate the problem into the following algebraic version. Then he proved the n ≤ 3 cases and discovered counterexamples for the other cases. Furtwängler’s conjecture (Hajós, 1941). Let G be a finite abelian group and let A1  A2      An be cyclic sets of G. If every element g ∈ G can be expressed in exact k distinct ways as g = a1 a2 · · · an with ai ∈ Ai , then one of the n cyclic sets is a subgroup of G. In this chapter, we will discuss this conjecture, especially the works of G. denote the multiplicative Hajós and R.M. Robinson. For convenience, let Ok √ cyclic group 1 e2i/k      e2
k−1i/k , where i = −1. This notation will appear frequently in this chapter.

7.2 A theorem of Furtwängler and Hajós In this section, we will prove Furtwängler’s conjecture for n ≤ 3. The two proofs discovered respectively by Furtwängler and by Hajós are very different in nature. In order to have a systematic and coherent chapter, we choose the algebraic proof which belongs to Hajós. Theorem 7.1 (Furtwängler, 1936 and Hajós, 1941). Let k be a fixed positive integer, let G be a finite abelian group, and let A1 , A2 , and A3 be three cyclic sets of G. If every element g ∈ G can be expressed in exact k distinct ways as g = a 1 a 2 a3 with ai ∈ Ai , then one of the three cyclic sets is a subgroup of G.

130

Furtwängler’s conjecture

Proof. Let 1 be the unit of G and let
G denote the group ring generated by G. In addition, for a subset A of G, we write  1 A= a a∈A

Then the assumption of Theorem 7.1 can be restated as 1=A 11 · A 12 · A 13  kG

(7.1)

q −1

Assume that Ai = 1 gi      gi i , then we can deduce 1 ·
1 − gi  = k G 1−kG 1=0 kG and 1i ·
1 − gi  = 1 − giqi  A Therefore, multiplying both sides of (7.1) by
1 − g1 
1 − g2 
1 − g3 , we get

q  q  q  1 − g11 · 1 − g22 · 1 − g33 = 0 (7.2) q

For convenience, we write xi = gi i , then (7.2) can be rewritten as x1 + x2 + x3 + x1 x2 x3 = 1 + x1 x2 + x2 x3 + x1 x3 

(7.3)

If xi = 1 holds for one of the three indices, then the theorem is proved. Assume that xi = 1 for i = 1, 2, and 3, then by (7.3) we get ⎧ ⎪ ⎪x1 x2 x3 = 1 ⎪ ⎪ ⎨x x = x 1 2 3 (7.4) ⎪ x x = x ⎪ 1 3 2 ⎪ ⎪ ⎩ x2 x3 = x 1 and therefore x12 = x22 = x32 = 1

(7.5)

Multiplying the two sides of (7.1) by
1 − g2 
1 − g3 , we get  

11 · 1 − g2q2 · 1 − g3q3 = A 11 ·
1 − x2  ·
1 − x3  0=A and therefore by (7.4) 11 ·
x2 + x3  = A 11 ·
1 + x2 x3  = A 11 ·
1 + x1  A

(7.6)

Clearly every term on the right-hand side of (7.6) belongs to the cyclic group
g1 . Then every term on the left-hand side also belongs to
g1  and thus both x2 and x3 belong to
g1 .

7.3 Hajós’ counterexamples

131

If g1  is odd, then x2 = 1 has only one solution x = 1 in
g1 . Then by (7.5) we get x1 = x2 = x3 = 1 The theorem is true. If g1  = 2l is even, then x2 = 1 has two solutions x = 1 and x = g1l in
g1 . Then by (7.4) and (7.5) we get xi = 1 for one of the three indices. In this case, the theorem is also true. As a conclusion, the theorem is proved.
By a similar but much simpler argument, we can prove the case of two cyclic sets as well. Therefore, we get the following theorem. Theorem 7.1∗ . When n ≤ 3, every multiple lattice tiling I n + has a twin.

7.3 Hajós’ counterexamples Theorem 7.2 (Hajós, 1938 and 1941). Whenever n ≥ 4, Furtwängler’s conjecture is not true for some k. Proof. First, we claim that, if the conjecture is not true for some n and k, then it is false for n + 1 and 2k. Assume that 1=A 11 · A 12 · · · A 3n kG

(7.7)

holds with n cyclic sets Ai , none of which is a subgroup. If   q −1 A1 = 1 g1      g11  it is easy to see that An+1 = 1 g1  is a cyclic set but not a subgroup. Then by (7.7) we get 11 · A 12 · · · A4 1 1 A n+1 = k G ·
1 + g1  = 2k G which implies the claim. Second, we claim that the conjecture is false for n = 4 and k = 9. Let us take G = O12 ⊗ O2 and write g1 =
ei/6  1 and g2 =
1 ei . Clearly, g1  = 12 and g2  = 2. Then we define   A1 = 1 g1      g15    A2 = 1 g1 g2 
g1 g2 2 
g1 g2 3   2 

A3 = 1 g12 g2  g12 g2 

132

Furtwängler’s conjecture

and

 2 

 A4 = 1 g14 g2  g14 g2

It is easy to see that all A1 , A2 , A3 , and A4 are cyclic sets, and none of them is a subgroup of G. Then, by a routine computation, we can deduce 12 · A 13 · A 14 = 9 G 1 11 · A A As a conclusion, the theorem is proved.




Theorem 7.2 can be restated in the following geometric form. Theorem 7.2∗ . Whenever n ≥ 4, there is a multiple lattice tiling of E n , which has no twin. Geometrically speaking, the example constructed in the proof is corresponding to the nine-fold lattice tiling I 4 + of E 4 , where

1 1 is1 the  lat1 1 tice generated by a =
2 0 0 0 a =  −  0 0  a =    0 , and 3 1 2 3 2 6 4 3 

a4 = 21  41  0 13  Once we get an n-dimensional example, by considering layers we can easily extend it to
n + 1 dimensions. Remark 7.2. The geometric example was discovered by Hajós in 1938. Three years later he published the algebraic version in his famous paper (Hajós, 1941).

7.4 Robinson’s characterization To improve Theorem 7.2, in 1979 R.M. Robinson was able to determine all the integer pairs n k for which Furtwängler’s conjecture is false. Theorem 7.3 (Robinson, 1979). There is a k-fold lattice tiling I n + of E n , which has no twin if and only if 1 n = 4 and k is divisible by the square of an odd prime. 2 n = 5 and k = 3 or k ≥ 5. 3 n ≥ 6 and k ≥ 2. This theorem can be restated in an algebraic version as follows. Theorem 7.3∗ . The group ring equation 12 · · · A 3n 1=A 11 · A kG

(7.8)

has a solution, where Ai are cyclic subsets but not cyclic subgroups of G if and only if

7.4 Robinson’s characterization

133

1 n = 4 and k is divisible by the square of an odd prime. 2 n = 5 and k = 3 or k ≥ 5. 3 n ≥ 6 and k ≥ 2. To prove this theorem, as we can imagine, we need several technical lemmas. For convenience, we write q

h i = gi i and 1i ·
gi − 1 = giqi − 1 = hi − 1 Ai = A Then it follows by (7.8) that 12 · · · A 3n =
g1 − 1 A1 · A

n 

1−kG 1=0 1i = kg1 · G A

i=1

and, similarly



Ai ·

i∈I

 i∈I 

1i = 0 A

(7.9)

whenever I is a nonempty subset of 1 2     n and I  = 1 2     n \ I Especially, when I = 1 2     n, we have n 

Ai =

i=1

n 


hi − 1 = 0

(7.10)

i=1

We will call h1  h2      hn  a primitive solution to (7.10) if none of the factors can be omitted and will call it a solution of type
h1  h2      hn . For convenience, we assume that hi  are nondecreasing; that is h1  ≤ h2  ≤ · · · ≤ hn  Of course, every solution of (7.10) contains a primitive solution of some equation 
hi − 1 = 0 i∈I

where I is a subset of 1 2     n. In fact, Furtwängler’s conjecture predicts that hi − 1 = 0 holds at least for one of the n indices. Lemma 7.1 (Robinson, 1979). If n ≥ 3 and h1  h2      hn  is a primitive solution to
710, then we have hi  ≤ 2n−2 for all i. Especially, if both n and hi  are odd, then hi  ≤ 2n−3 

134

Furtwängler’s conjecture

Proof. Without loss of generality, we only consider hn . By a routine computation, we have n−1 

Ai =

i=1

n−1 


hi − 1 = U0 − V0 = U − V = 0

(7.11)

i=1

where we first multiply out, collect terms with same signs, and then cancel all the common terms. It is easy to see that both U0 and V0 have 2n−2 terms and not all of the terms are cancelled. By (7.10) we get
U − V  ·
hn − 1 = hn · U + V −
hn · V + U  = 0 and therefore hn · U + V = hn · V + U Since U and V have no common term, it follows that hn · U = U hn · V = V

(7.12)

This means that the multiplication by hn permutes the terms of U in cycles of length hn  and therefore hn  is a divisor of the number of the terms in U . Thus we have hn  ≤ U  ≤ U0  = 2n−2  where U  denotes the number of the terms of U . If n is odd, then n − 1 is even. It follows by (7.11) that  hi ∈ U 0 i∈I

if and only if



hi ∈ U 0 

i∈I

Thus the number of the terms in U is even. Since hn  is an odd factor of U , which cannot exceed 2n−2 , we get hn  ≤ 2n−3  The lemma is proved.




Remark 7.3. By repeating part of the argument of the previous proof, we can deduce that there is no primitive solution for
710 when n = 2. In other

7.4 Robinson’s characterization

135

words, Furtwängler’s conjecture is true in this case. It follows by
712 that hn belongs to the subgroup of G generated by h1  h2      hn−1 . In the next three lemmas we will determine the possible types of the primitive solutions to (7.10) for n = 3, 4, and 5. Lemma 7.2 (Robinson, 1979). When n = 3, equation
710 has only primitive solutions of
2 2 2 type. Proof. Let U and V be the elements of
G defined by (7.11), it can be deduced that U = h 1 h2 + 1 V = h 1 + h2  Then, by applying (7.12), we get ⎧ ⎪ h1 h2 h3 = 1 ⎪ ⎪ ⎪ ⎨h h = h 1 2

3

⎪ h1 h3 = h 2 ⎪ ⎪ ⎪ ⎩ h2 h3 = h 1 and therefore h12 = h22 = h32 = 1 On the other hand, such a solution can be realized in the group ring
G with G = O2 ⊗ O2 . Taking h1 =
ei  1, h2 =
1 ei , and h3 =
ei  ei  it can be verified that h1  h2  h3  is a primitive solution to (7.10). The lemma is proved.
Lemma 7.3 (Robinson, 1979). When n = 4, equation
710 has and only has primitive solutions of
3 3 3 3 type and
2 2 4 4 type. Proof. Assume that h1  h2  h3  h4  is a primitive solution to (7.10). It follows by A1 · A2 · A3 =
h1 − 1
h2 − 1
h3 − 1 = U0 − V0 that



U 0 = h 1 h 2 h 3 + h1 + h2 + h3 V0 = h1 h2 + h1 h3 + h2 h3 + 1

Then the only pairs of terms which might cancel out are among h1 h2 h3  1 h1  h2 h3 , h2  h1 h3 , and h3  h1 h2 . Now we consider three cases.

136

Furtwängler’s conjecture

Case 1. h4  = 2 Since h1  h2  h3  h4  is primitive, by the assumption that hi  are nondecreasing we have hi  = 2 for all i. Thus, if any pair of the terms are cancelled, then all the terms would be cancelled. So we have U = U0 and V = V0 . By (7.12), without loss of generality, we get h1 h2 h3 h4 = h1 and therefore h2 h3 h4 = 1 which leads to a solution of
2 2 2 type on h2 , h3 , and h4 . Hence there is no primitive solution in this case. Case 2. h4  = 3 In this case the multiplication by h4 to U will permute three terms cyclically. Hence exactly one of the four pairs of terms must be cancelled between U0 and V0 . Without loss of generality, we may suppose that h1 h2 h3 = 1 (if h1 = h2 h3 , we can take h1 = h1−1 and get h1 h2 h3 = 1) and, by (7.12) ⎧ ⎪ ⎪ ⎨h1 h4 = h2 h2 h4 = h 3 ⎪ ⎪ ⎩h h = h  3 4 1

Then we get h1 h4 · h1 = h2 · h3 h4  h12 = h2 h3  h13 = h1 h2 h3 = 1 and therefore h13 = h23 = h33 = h43 = 1 On the other hand, such a solution can be realized in the group ring
G with G = O3 ⊗ O3 . To see this, we define h1 =
e2i/3  1, h2 =
e2i/3  e2i/3  h3 =
e2i/3  e4i/3  and h4 =
1 e2i/3  It can be verified that h1  h2  h3  h4  is a primitive solution to (7.10). Thus, in this case (7.10) has and only has a primitive solution of
3 3 3 3 type. Case 3. h4  = 4 In this case there is no cancellation between U0 and V0 . Therefore, we have U = U0 and V = V0 . Then the multiplication by h4 must

7.4 Robinson’s characterization

137

permute the terms of U and of V cyclically. We may renumber the hi so that the terms of U occur in the order h1 h2 h3 , h1 , h3 , h2 . Then we get ⎧ ⎪ h2 h3 h4 = 1 ⎪ ⎪ ⎪ ⎨h h = h 1 4 3 ⎪ = h h h ⎪ 3 4 2 ⎪ ⎪ ⎩ h1 h3 = h 4 and therefore



h12 = h22 = 1 h34 = h44 = 1

On the other hand, such a solution can be realized in the group ring
G with G = O2 ⊗O4 . Let us define h1 =
ei  1, h2 =
ei  ei  h3 =
ei  ei/2  and h4 =
1 ei/2  It can be verified that h1  h2  h3  h4  is a primitive solution to (7.10). Hence, in this case (7.10) has and only has a primitive solution of
2 2 4 4 type. As a conclusion of these cases, the lemma is proved.
Lemma 7.4 (Robinson, 1979). If n = 5 and h1  h2      h5  is a primitive solution to
710, then hi  = 2 4, or 8. Proof. For convenience, in this proof we remove the assumption that hi  are nondecreasing and only consider h5 . Assume that A1 · A2 · A3 · A4 =
h1 − 1
h2 − 1
h3 − 1
h4 − 1 = U0 − V0 = U − V It was shown in the proof of Lemma 7.1 that h5  is a divisor of U  and U  is an even number less than or equal to 8. Thus the only doubtful case is U  = 6. In this case we need to show that we cannot have h5  = 3 or 6. If h5  = 6, clearly h1  h2  h3  h4  h52  is also a solution to (7.10). If it is primitive, we get a primitive solution with an element of order 3. If it is not primitive, then we can omit some factor
hi − 1 from (7.10) for some i < 5 and get a new equation. This can lead only to a primitive solution of type
3 3 3 3. Then the original solution h1  h2      h5  would have three elements of order 3. Thus, if h5  = 6 is possible, then h5  = 3 is also possible. So it is sufficient to exclude the latter. Assume that h5  = 3, then we have U0 = h1 h2 h3 h4 + h1 h2 + h1 h3 + h1 h4 + h2 h3 + h2 h4 + h3 h4 + 1 V0 = h1 h2 h3 + h1 h2 h4 + h1 h3 h4 + h2 h3 h4 + h1 + h2 + h3 + h4 

138

Furtwängler’s conjecture

If h1 h2 h3 h4 is cancelled by one of the first four terms of V0 , then one of the four elements h1 , h2 , h3 , and h4 will be 1, which contradicts the assumption. If h1 h2 h3 h4 is cancelled by one of the last four terms of V0 , then 1 will be cancelled by one of the first four terms of V0 . By similar arguments, we can conclude that the number of the cancelled terms of U0 is even and therefore U  = 6. By multiplying hi−1 to every term, we can deduce that 1 plays the same role as any other term in U0 . Therefore, we may assume that 1 is one of the two cancelled terms in U0 . Since it has to be cancelled by one of the four terms h1 h2 h3 , h1 h2 h4 , h1 h3 h4 , and h2 h3 h4 in V0 , the other term in U0 to be cancelled must be h1 h2 h3 h4 . Therefore, we get U = h1 h 2 + h 1 h 3 + h 1 h 4 + h 2 h 3 + h2 h 4 + h3 h 4  Since h5 U = U , the terms of U must fall into two three-term cycles and one of them must contain at least two terms involving h1 . We may suppose that the first cycle contains h1 h2 and h1 h3 in this order. Then we have h2 h5 = h3 and h2 = h3 h52 . The second cycle contains at least one of the two terms h2 h4 and h3 h4 , and hence also contains a term equal to the other. Exchanging two equal terms if necessary, we may suppose that both h2 h4 and h3 h4 occur in the second cycle. Without loss of generality, we assume that h1 h4 belongs to the first cycle. Then we get ⎧ ⎪ ⎪ ⎨h2 h5 = h3 (7.13) h3 h5 = h4 ⎪ ⎪ ⎩h h = h  4 5

2

Similarly, since h5 V = V , the terms of V must fall into two three-term cycles. Applying (7.13) to V and exchanging two equal terms if necessary, we may suppose that one of the two cycles is h2  h3  h4  and the other cycle contains the three complements h1 h2 h3 , h1 h2 h4 , and h1 h3 h4 . It follows that the terms deleted from V0 are h1 and h2 h3 h4 . They must have cancelled h1 h2 h3 h4 and 1 of U0 . Hence we have h2 h3 h4 = 1 However, this condition and (7.13) together imply that h2  h3  h4  h5  is a solution of
3 3 3 3 type. By this contradiction, the lemma is proved.
Let 
g be a homomorphic mapping from G to the multiplicative group of the complex number field. It is easy to see that 
g is an mth root of 1 for every g ∈ G, if G = m. In fact, this function can be extended from G to the group ring
G by defining

    zi gi = zi 
gi 

7.4 Robinson’s characterization

139

This map, known as a character on
G, has some important properties. For example, there are exactly m different characters. Let g1 = 1, g2      gm be the m elements of G and let 1 , 2      m be the m characters, where 1
g = 1 holds for all g ∈ G. Then we have m 

i
gj  = 0

i = 1

i
gj  = 0

j = 1

j=1

and

m  i=1

Let r=

m 

zi g i

i=1

be an element in
G. It can be deduced from the above properties that zi =

m 1 
r j
gi−1  m j=1 j

Especially, if j
r = 0 for all j, then we get zi = 0 for all i and therefore r = 0. If j
r = 0 for all j = 1, then we get zi = 1
r/m which is independent of i, and hence 1 r = kG for some integer k. Lemma 7.5 (Robinson, 1979). If h1  h2      hn  is a primitive solution to
710 and, for every i, hi  is a factor of hn , then 3n = 0 A1 · A2 · · · An−1 · A Proof. Based on the properties of the characters, it is sufficient to show that 3n  = 0, and there is some character  satisfying 
hi  = 1 for i < n, 
A qn 
hn  = 
gn  = 1 Since 3n  = 1 + 
gn  + · · · + 
gn qn −1 
A 
hn  − 1  = 
gn  − 1 3n  = 0 unless 
gn  = 1. Therefore it is sufficient to find a we have 
A character  satisfying  
hi  = 1 if i < n, (7.14) 
hn  = 
gn  = 1

140

Furtwängler’s conjecture

Let H denote the subgroup generated by h1  h2      hn . We claim that, if there is such a character whenever gn ∈ H, then we can extend it into the general case. Let g be an element of G such that hn ∈
g and let s be the smallest positive integer such that gs ∈ H. Since hn ∈
g and hn ∈ H, we must have hn ∈
gs . Based on the assumption, there is a character  satisfying 
hi  = 1 if i < n and 
gs  = 1. It is known in group theory that then we can extend the character from H to the group generated by H and g with 
g = 1. Now we show the gn ∈ H case. Since hi  is a factor of hn  for all i, we have hhn  = 1 for all h ∈ H. Since hn ∈
gn  and gnhn  = 1, we must have
hn  =
gn 

(7.15)

By the assumption that h1  h2      hn  is a primitive solution to (7.10), it follows that there is a character  on H satisfying 
hi  = 1 for all i < n and 
hn  = 1. Then by (7.15) we get 
gn  = 1 as well. As a conclusion we have shown (7.14). The lemma is proved.
Proof of Theorem 7.3, Case 1. First, we deal with the necessary part. Suppose that 11 · A 12 · A 13 · A 14 = k G 1 A (7.16) hi = 1 i = 1 2 3 4 holds for some positive integer k. Then we get A1 · A 2 · A3 · A4 = 0 14 = 0 A1 · A 2 · A3 · A and therefore, by lemma 7.5 A1 · A2 · A3 = 0 By the proof of Lemma 7.2 we get h1 h2 h3 = 1 h12 = h22 = h32 = 1

(7.17)

From these equations, we can deduce h2 ∈
g1  and h3 ∈
g1 . Otherwise, since x2 = 1 has only two solutions in
g1  and they are x = 1 and x = h1 , we can deduce h2 = h1 and therefore h3 = 1.

7.4 Robinson’s characterization

141

By (7.16) we also have 11 · A2 · A3 · A4 = 0 A from which we can deduce 11
h2 h3 h4 + h2 + h3 + h4  = A 11
h2 h3 + h2 h4 + h3 h4 + 1 A Then, applying (7.17), we can deduce 11
1 + h1 
h2 + h4  = A 11
1 + h1 
1 + h2 h4  A Now the right-hand side, when expanded, contains powers of g1 , hence the left-hand side must also. Since h2 ∈
g1  we must have h4 ∈
g1  Simi13 · A4 = 0, we get h4 ∈
g2  and 12 · A3 · A4 = 0 and A1 · A2 · A larly, by A1 · A h4 ∈
g3 . Now we claim that s = h4  is odd. If, on the contrary, s = 2t, then we have
h4t 2 = 1 and h4t ∈
g1  Since h4t = 1, we must have h4t = h1 . Similarly, h4t = h2 . Hence h1 = h2 , so that by (7.17) h3 = 1, which contradicts the hypothesis. Since h4 ∈
gi  for i = 1 2, and 3, we have s  gi . On the other hand, by (7.17), we have gi  = 2qi and therefore s  qi . For convenience, we write qi = sdi . Furthermore, all solutions of xs = 1 in
gi  must be powers of h4 , 2d therefore gi i ∈
h4  Writing   ∗ 1i = A 1i 1 + hi + · · · + hihi −1 A q hi −1

= 1 + gi + · · · + gi i



since hi  = 2 for i = 1 2, and 3 and h4  = s, it can be deduced by (7.16) that ∗

∗

∗

11 · A 12 · A 13 · A 14 = 23 k G 1 A and ∗

∗

∗

∗

11 · A 12 · A 13 · A 14 = 23 sk G 1 A

(7.18)

On the other hand, since qi = sdi , for i = 1 2, and 3, we have ∗

1i = 1 + gi + · · · + gi2sdi −1 A    2
s−1di 2d −1 2d 1 + gi i + · · · + g i  = 1 + gi + · · · + g i i

(7.19)

142

Furtwängler’s conjecture

It follows by the last paragraph that each term in the last factor is a power of ∗ ∗ 14 . Thus by (7.18) and (7.19) 14 = A h4 and hence of g4 . Clearly, we have g4j A we get 3    2d −1 1 ∗ 1 s3 1 + gi + · · · + g i i A4 = 23 sk G i=1

Since every term on the right-hand side has a coefficient 8sk, it follows that s3  8sk, or s2  8k. Since s is odd and s > 1, we conclude that k is divisible by the square of an odd prime. Now let us show the sufficient part. Assume that p is an odd prime and k is a positive integer. Let us take G = O2p ⊗ O4 and write e1 =
ei/p  1 and e2 =
1 ei/2 . Clearly, e1  = 2p and e2  = 4. We define 11 = 1 + e1 + · · · + e1kp−1 A   
k−1p  = 1 + e1 + · · · + e1p−1 1 + e1p + · · · + e1 

12 = 1 + e1 e2 + · · · + e1 e2 p−1  A 2 2 13 = 1 + e1 e2 + · · · +
e1 e2 2p−1  A and 14 = 1 + e2 e2 + e4 e2 + e6 e3  A 1 1 2 1 2 Then it can be verified that 12 · A 13 · A 14 = kp2 G 1 11 · A A hi = 1

i = 1 2 3 4

The first case of the theorem is proved.




Proof of Theorem 7.3, Case 2. First, we show the necessary part. Suppose that 12 · A 13 · A 14 · A 15 = k G 1 11 · A A (7.20) hi = 1 i = 1 2 3 4 5 holds with some positive integer k, then we get A1 · A2 · A3 · A4 · A 5 = 0 15 = 0 A1 · A2 · A3 · A4 · A and therefore, by Lemma 7.5 A1 · A2 · A3 · A4 = 0

7.4 Robinson’s characterization

143

Thus, by Lemmas 7.2, 7.3, and 7.4, we only need to deal with the following subcases. Subcase 2.1. A1 · A2 · A3 = 0 has a primitive solution of
2 2 2 type. First, we have h1 h2 h3 = 1 (7.21) h12 = h22 = h32 = 1 and 11 · A2 · A3 · A4 · A5 = 0 A

(7.22)

It follows by (7.21) that h2 h3 = h1 and h1 h2 = h3 . Therefore, by (7.22), we get 11
1 + h1 
h2 h4 h5 + h2 + h4 + h5  11
1 + h1 
1 + h2 h4 + h2 h5 + h4 h5  = A A (7.23) When expanded, there will be powers of g1 on the left-hand side. Therefore, there must also be powers of g1 on the right-hand side, and thus h2 , h4 , h5 or h2 h4 h5 must be a power of g1 . First of all, we cannot have h2 ∈
g1  since the only solutions for x2 = 1 in
g1  are x = 1 or h1 , and h2 = h1 would lead to h3 = 1. Thus h4 , h5 or h2 h4 h5 belongs to
g1 . In the latter case, we also 11
1 + h1  is cyclic have h1 h3 h4 h5 ∈
g1  and therefore h3 h4 h5 ∈
g1  Since A and is unchanged by multiplying by a power of g1 , we get 11
1 + h1  · h2 h4 h5  11
1 + h1  = A A 11
1 + h1  · h4 h5 = A 11
1 + h1  · h2 A and 11
1 + h1  · h−1 h−1  11
1 + h1  · h2 = A A 4 5 Then it can be deduced from (7.23) that 11
1 + h1 
h4 + h5  11
1 + h1 
h−1 + h−1  = A A 4 5 and hence 11
1 + h1 
h2 + h4 h5  11
1 + h1 
1 + h4 h−1  = A A 5 4

(7.24)

Since h2 ∈
g1 , we get h4 h5 ∈
g1  By comparing the two sides of (7.24), we get h42 ∈
g1 . Similarly, we find that h52 ∈
g1 .

144

Furtwängler’s conjecture

Summarizing these conclusions and similar conclusions drawn from the 12 · A3 · A4 · A5 = 0 and A1 · A2 · A 13 · A4 · A5 = 0, we have equations A1 · A ⎧ 2 2 ⎪ ⎪ ⎨h4 ∈
g1  or h5 ∈
g1  or h4  h5  h2 h4 h5  h3 h4 h5  ⊆
g1  (7.25) h4 ∈
g2  or h5 ∈
g2  or h42  h52  h1 h4 h5  h3 h4 h5  ⊆
g2  ⎪ ⎪ ⎩h ∈
g  or h ∈
g  or h2  h2  h h h  h h h  ⊆
g  4

3

5

3

4

1 4 5

5

2 4 5

3

Next we claim that it is impossible to have h42 = h52 = 1. If, on the contrary, these equations hold, we proceed to deduce a contradiction. Suppose that h4 or h5 belongs to some
gi  with i ≤ 3. We may assume that h4 ∈
g1 . It follows that h4 = h1 . We cannot also have h4 ∈
g2 , since this would lead to h4 = h2 , and hence h1 = h2 and h3 = 1. Similarly, we get h4 ∈
g3 . Since h1 h4 h5 = h5 , it follows by (7.25) that h5 ∈
g2  and h5 ∈
g3 , which leads to a contradiction in the similar way. Thus we must have h4 ∈
gi  and h5 ∈
gi  for i ≤ 3. Hence we must use the third alternative in each of the three relations in (7.25). Since
hi h4 h5 2 = 1, we have h i h4 h5 = 1

h i h4 h5 = h j 

or

where j ∈ 1 2 3 \ i. Then we can deduce h3 h4 h5 = h2 h4 h5 = 1 and hence h2 = h3 and h1 = 1, which contradicts the assumption. Since we cannot have h42 = h52 = 1, we may assume that h52 = 1. Then it follows by (7.25) that h4 ∈
gi 

or

h52 ∈
gi 

for i ≤ 3. Thus one of the two alternatives must hold for at least two indices. Suppose h4 ∈
g1  and h4 ∈
g2 , repeating the proof arguments in Case 1 about s, we can get s = h4  is odd and qi = sdi hold for some suitable integers di for i = 1 2, and 4. Then it follows from (7.20) that ∗

∗

∗

11 · A 12 · A 13 · A 14 · A 15 = 22 sk G 1 A 2    ∗ 2d −1 13 · A 14 · A 15 = 4sk G 1 s2 1 + gi + · · · + g i i ·A i=1

and therefore k is divisible by an odd prime. q Suppose h52 ∈
g1  and h52 ∈
g2 . We write q5 = 2q5  h5 = g55 = h52 , and 



15 = A 15
1 + h5  = 1 + g5 + · · · + gq5 −1 A 5

7.4 Robinson’s characterization

145

it follows by (7.20) that 

11 · A 12 · A 13 · A 14 · A 15 = 2k G 1 A Since h5 ∈
g1  and h5 ∈
g2 , exchanging the subscripts 4 and 5, it follows by the result proved above that k is divisible by an odd prime. As a conclusion, in this subcase k must be divisible by an odd prime. Subcase 2.2. A1 · A2 · A3 · A4 = 0 has a primitive solution of
3 3 3 3 type. First, by the proof of Lemma 7.3, we have ⎧ ⎪ h13 = h23 = h33 = h43 = 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ h1 h 2 h 3 = 1 (7.26) h1 h4 = h 2 ⎪ ⎪ ⎪ ⎪ h2 h4 = h 3 ⎪ ⎪ ⎪ ⎩h h = h 3 4

1

and 11 · A2 · A3 · A4 · A5 = 0 A Then it can be deduced that 11
1 + h1 + h2 h3 
1 + h2 h5  = A 11
1 + h1 + h2 h3 
h2 + h5  A and, therefore, multiplying both sides by h1



 11 1 + h1 + h2
1 + h2 h5  = A 11 1 + h1 + h2
h2 + h5  A 1 1 Hence we get h2 ∈
g1  or h5 ∈
g1 . Note the only solutions of x3 = 1 in
g1  are x = 1, h1 , and h12 . If h2 ∈
g1 , by (7.26), then either h2 = h1 and hence h4 = 1, or else h2 = h12 and hence h3 = 1. Therefore, we must have h5 ∈
g1 . Similarly, we also get h5 ∈
g2  and h5 ∈
g3 . 14 · A5 = 0 that In addition, it follows by A1 · A2 · A3 · A 14
h1 + h2 + h3 
h2 + h5  = A 14
h1 + h2 + h3 
1 + h2 h5  A Multiplying both sides by h1−1 , this yields



 14 1 + h4 + h2
h2 + h5  = A 14 1 + h4 + h2
1 + h2 h5  A 4 4 Thus we have either h2 ∈
g4  or h5 ∈
g4 . If h2 ∈
g4 , by (7.26), then either h2 = h4 and hence h1 = 1, or else h2 = h22 and hence h3 = 1. Thus we must have h5 ∈
g4 . So we have h5 ∈
gi  for all i ≤ 5. Now we write s = h5 . We proceed to show that s is not divisible by 3. If, on the contrary, s = 3t, then we have
h5t 3 = 1 and therefore h5t = hi or hi2

146

Furtwängler’s conjecture

for i ≤ 3. Thus two of the three elements h1 , h2 , and h3 would be equal and hence h4 = 1. Since h5 ∈
gi , we have s  gi  and gi   3qi . Thus we get s  qi for all i ≤ 4. Write qi = sdi . It is easy to see that all solutions of xs = 1 in
gi  are 3d powers of h5 , therefore gi i ∈
h5  for all i ≤ 4. Then it follows by (7.20) that ∗

∗

∗

∗

∗

12 · A 13 · A 14 · A 15 = 34 sk G 1 11 · A A which leads to s4

4  

3di −1

1 + gi + · · · + g i



∗

1 15 = 81sk G A

i=1

Thus we have s3  k. In other words, in this subcase the multiplicity k is divisible by the cube of a prime other than 3. Subcase 2.3. A1 · A2 · A3 · A4 = 0 has a primitive solution of
2 2 4 4 type. First, by the proof of Lemma 7.3, we have ⎧ ⎪ h12 = h22 = h34 = h44 = 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪h2 h3 h4 = 1 ⎨ (7.27) h1 h 4 = h 3 ⎪ ⎪ ⎪ ⎪ h3 h4 = h 2 ⎪ ⎪ ⎪ ⎩h h = h 1 3

4

and 11 · A2 · A3 · A4 · A5 = 0 A Then it can be deduced that 11
h3 + h4 
1 + h2 h5  11
h3 + h4 
h2 + h5  = A A and, therefore, multiplying both sides by h3−1 11
1 + h1 
h2 + h5  = A 11
1 + h1 
1 + h2 h5  A Hence we get h2 ∈
g1  or h5 ∈
g1 . If h2 ∈
g1 , then h2 = h1 . It follows by the last four equations of (7.27) and the proof of Lemma 7.2 that A2 ·A3 ·A4 = 0 has a primitive solution of type
2 2 2, which contradicts our assumption. Therefore, we must have h5 ∈
g1 . Similarly, we also get h5 ∈
g2 . 13 · A4 · A5 = 0 that In addition, it follows by A1 · A2 · A 13
1 + h3 
h1 + h2 + h5 + h1 h2 h5  13
1 + h3 
1 + h1 h2 + h1 h5 + h2 h5  = A A

7.4 Robinson’s characterization

147

Since h1 h4 · h2 = h3 · h3 h4 and hence h1 h2 = h32 , we can deduce from the above equation that  

13
1 + h3  1 + h2
1 + h2 h5  = A 13
1 + h3  1 + h2
h1 + h5  A 3 3 Then we have either h1 ∈
g3  or h5 ∈
g3 . If h1 ∈
g3 , since h1 h2 = h32 , then we also get h2 ∈
h3 . Since h12 = h22 = 1, we would find h1 = h2 , which is impossible. Thus we must have h5 ∈
g3  and similarly h5 ∈
g4 . So we have h5 ∈
gi  for all i ≤ 5. Now we write s = h5 . As in the other subcases, it can be deduced that s is odd, s  gi , and gi   qi hi . Since hi  = 2 or 4 for i ≤ 4, we get s  qi . Write qi = sdi . Since all solutions of xs = 1 in
gi  are powers of h5 , we get h d gi i i ∈
h5  for all i ≤ 4. Then it follows by (7.20) that ∗

∗

∗

∗

∗

11 · A 12 · A 13 · A 14 · A 15 = 22 · 42 sk G 1 A which leads to s4

4  

hi di −1

1 + gi + · · · + g i



∗

15 = 64sk G 1 A

i=1

Thus we have s3  k. In other words, in this subcase the multiplicity k is divisible by the cube of an odd prime. As a conclusion of these three subcases, the necessary part of Case 2 is proved. To prove the sufficient part, we have two constructions corresponding to the first two subcases. Construction 1. Assume that p is an odd prime and q is a positive integer. Let us take G = O2p ⊗ O4 and write e1 =
ei/p  1 and e2 =
1 ei/2 . Clearly, e1  = 2p and e2  = 4. We define 11 = 1 + e1 + · · · + e1pq−1 A    p
q−1 = 1 + e1 + · · · + e1p−1 1 + e1p + · · · + e1  

p−1 12 = 1 + e1 e2 + · · · + e1 e2 A  2 2 13 = 1 + e2  A 14 = 1 + e1p e2  A and 15 = 1 + e2 e2  A 1 2

148

Furtwängler’s conjecture

Then it can be verified that 11 · A 12 · A 13 · A 14 · A 15 = pq G 1 A hi = 1

i = 1 2     5

Construction 2. Assume that q is a positive integer. Let us take G = O18 ⊗O3 and write e1 =
ei/9  1 and e2 =
1 e2i/3 . Clearly, e1  = 18 and e2  = 3. We define 11 = 1 + e3 e2 + · · · +
e3 e2 2q−1 A 1 2 1 2

  = 1 + e13 e22 1 +
e13 e22 2 + · · · +
e13 e22 2
q−1  12 = 1 + e15 e2  A 1 2 13 = 1 + e9 e2  A 1 2

14 = 1 + e1 + · · · + e5  A 1 and

15 = 1 + e1 e2 + · · · +
e1 e2 8  A

Then it can be verified that 11 · A 12 · A 13 · A 14 · A 15 = 8q G 1 A hi = 1

i = 1 2     5

Let Z∗ denote the set of the positive integers and let Z denote the set of the positive integers, which is divisible by eight or by an odd prime. It is easy to see that 1 2 4 = Z∗ \ Z  Thus the sufficient part follows by the two constructions. The second case is proved.
Proof of Theorem 7.3, Case 3. The necessary part has been proved in Chapter 6. To show the sufficient part, we have the following construction for n = 6. The higher-dimensional case follows inductively. Assume that q is a positive integer. Let us take G = O6 ⊗ O6 ⊗ O3 and write e1 =
ei/3  1 1, e2 =
1 ei/3  1, and e3 =
1 1 e2i/3 . Clearly, e1  = 6, e2  = 6, and e3  = 3. Then we define 11 = 1 + e1 + · · · + e13q−1 A  

3
q−1  = 1 + e1 + e12 1 + e13 + · · · + e1 12 = 1 + e2 + e2  A 2 13 = 1 + e2 e3  A 1 2

14 = 1 + e3 e2  A 1 2 2

1 A5 = 1 + e1 e3 e3 + e1 e3 e3  2

2

7.4 Robinson’s characterization

149

and 16 = 1 + e4 e3 e3  A 1 2 It can be verified that

11 · A 12 · A 13 · A 14 · A 15 · A 16 = 2q G 1 A

hi = 1

i = 1 2     6

which means that Furtwängler’s conjecture is false when n = 6 and k is even. By Case 2, it follows that the conjecture is also false when n = 6 and k = 1 is odd. Hence the third case is proved.


8 Keller’s conjecture

8.1 Keller’s conjecture In 1930, ten years before Hajós’ proof, O. Keller generalized Minkowski’s conjecture from lattice tilings to translative tilings and made the following conjecture. Keller’s conjecture. Every translative tiling I n + X of E n has a twin. As was shown at the beginning of Chapter 6, the two- and three-dimensional cases of this conjecture can be proved by routine arguments. In 1940, Perron (1940a) extended this result to n ≤ 6 through elementary but complicated arguments. In fact, Keller himself did claim a proof for the n ≤ 6 cases in 1937. However, his sketch is hardly accepted as a proof. In 1949, based on an ingenious observation, G. Hajós was able to formulate Keller’s conjecture into the following algebraic version8 . Let G be an abelian group generated by g1 , g2      gn with gi  = 2qi . If G = q −1 H ·A1 · · · An is a factorization, where H = 2n and Ai = 1 gi      gi i , then 
−1    qi hi hj  hi  hj ∈ H gi  i ≤ n = ∅ In 1980s, while no essential progress was made in the positive direction, S. Szabó and K. Corrádi started to consider this problem in the negative direction. First, Szabó (1986) proved that, to search for a counterexample for Keller’s conjecture, it is sufficient to restrict qi = 2 for all i ≤ n in the algebraic version. Then Corrádi and Szabó (1990b) obtained a criterion in graph theory for such a counterexample, if it does exist. 8

In this chapter, we say G = A1 A2 · · · An (or G = A1 + A2 + · · · + An ) is a factorization if every element g ∈ G can be uniquely expressed as g = a1 a2 · · · an (or g = a1 + a2 + · · · + an ), where ai ∈ Ai .

150

8.2 A theorem of Keller and Perron

151

Consider the set z =
z1  z2      zn   zi ∈ Z4 , where Z4 = 0 1 2 3, and define a graph n on it in the following way. Two vertices z and z are adjacent if zi − zi ≡ 2
mod 4 for some i and zj = zj for some j, where j = i. It is easy to see that n has 4n vertices. A subgraph of n is called a clique if any two of its vertices are adjacent. The size of a clique is the cardinality of its vertices. For convenience, let n denote the maximal size of a clique in n . With these definitions, Corrádi and Szabó’s criterion can be stated as follows. If

n

= 2n , then there exists a counterexample for Keller’s conjecture in E n .

Based on this criterion, in 1992 J.C. Lagarias and P. Shor were able to construct such an example for n ≥ 10 and therefore got the first counterexample for Keller’s conjecture. In 2002, the dimensions were reduced to n ≥ 8 by J. Mackey. So far, Keller’s conjecture is open only for n = 7. Therefore, the known results about Keller’s conjecture can be concluded as follows. When n ≤ 6, Keller’s conjecture is true; whenever n ≥ 8 it is false.

8.2 A theorem of Keller and Perron In 1937, O.H. Keller sketched a proof for his conjecture for n ≤ 6. In 1940, O. Perron gave a complete proof for this result. Theorem 8.1 (Keller, 1937; Perron, 1940a). When n ≤ 6, every translative tiling I n + X of E n has a twin. As we can imagine, this theorem was proved by complicated case-bycase arguments. We will not reproduce the whole proof here. Instead, we will introduce some general results, which did play important roles in the proof. Then we will demonstrate how to apply them to deduce the two- and three-dimensional cases of the theorem. For convenience, for a discrete set X in E n , two fixed integers i and u with 0 ≤ i ≤ n, and a fixed fractional v, we define   Xi
u v = x ∈ X  xi  = u and xi  = v  where x and x denote the integer part and the fractional part of x, respectively. Lemma 8.1 (Keller, 1937). If I n + X is a tiling of E n and v is a fixed fractional number, then   I n + Xi
u v = I n + Xi
u  v +
u − u ei holds for any set of integers u, u , and i with 1 ≤ i ≤ n.

152

Keller’s conjecture

Proof. It is sufficient to show the i = 1 case. Let H t denote the hyperplane x ∈ E n  xn = t, and let H t denote the halfspace bounded by H t and containing
t + 1en . Then the system
t n   H
I + x 
int
I n  + x H t = ∅ is a tiling of H t . For convenience, we write this tiling as I n−1 + X t . Clearly the lemma is true when n = 1. Assume that it is true in
n − 1 dimensions, for all u u ∈ Z we have   I n−1 + X1t
u v = I n−1 + X1t
u  v +
u − u e1  On the other hand, it is easy to see that   I n−1 + X1t
u v =
I n + x H t  x ∈ X1
u v
int
I n  + x



 H t = ∅ 

Thus we get 

0

I n + X1
u v =

−