Discrete Analytic Convex Geometry Introduction · Discrete Analytic Convex Geometry Introduction Martin Henk Otto-von-Guericke-Universit at Magdeburg Winter semester 2012/13 & Summer

Discrete Analytic Convex Geometry

Introduction

Martin Henk

Otto-von-Guericke-Universitat MagdeburgWinter semester 2012/13 & Summer semester 2013

webpage

http://shamash.math.uni-magdeburg.de/gilgamesh

CONTENTS i

Contents

Preface iiWS 2012/13

0 Some basic and convex facts 1

1 Support and separate 7

2 Radon, Helly, Caratheodory and (a few) relatives 17

3 The multifaceted world of polytopes 21

4 The space of convex bodies 37

5 A glimpse on mixed volumes 41

6 A glimpse on geometry of numbers 57

SS 2013

7 Packing 67

8 Count and generate 83

Index 99

ii CONTENTS

Preface

The material presented here is stolen from different excellent sources:

• First of all: A manuscript of Ulrich Betke on convexity which is partiallybased on lecture notes given by Peter McMullen.

• The inspiring books by

– Alexander Barvinok, ”A course in Convexity”

– Gunter Ewald, ”Combinatorial Convexity and Algebraic Geometry”

– Peter M. Gruber, ”Convex and Discrete Geometry”

– Peter M. Gruber and Cerrit G. Lekkerkerker, ”Geometry of Num-bers”

– Jiri Matousek, ”Discrete Geometry”

– Rolf Schneider, ”Convex Geometry: The Brunn-Minkowski Theory”

– Gunter M. Ziegler, ”Lectures on polytopes”

• and some original papers

!! and they are part of lecture notes on ”Discrete and Convex Geometry”jointly written with Maria Hernandez Cifre but not finished yet.

Some basic and convex facts 1

0 Some basic and convex facts

0.1 Notation. Rn ={x = (x1, . . . , xn)ᵀ : xi ∈ R

}denotes the n-dimensional

Euclidean space equipped with the Euclidean inner product 〈x,y〉 =∑n

i=1 xi yi,x,y ∈ Rn, and the Euclidean norm |x| =

√〈x,x〉.

0.2 Definition [Linear, affine, positive and convex combination]. Let m ∈N and let xi ∈ Rn, λi ∈ R, 1 ≤ i ≤ m.

i)∑m

i=1 λi xi is called a linear combination of x1, . . . ,xm.

ii) If∑m

i=1 λi = 1 then∑m

i=1 λi xi is called an affine combination of x1,. . . ,xm.

iii) If λi ≥ 0 then∑m

i=1 λi xi is called a positive combination of x1, . . . ,xm.

iv) If λi ≥ 0 and∑m

i=1 λi = 1 then∑m

i=1 λi xi is called a convex combinationof x1, . . . ,xm.

v) Let X ⊆ Rn. x ∈ Rn is called linearly (affinely, positively, convexly)dependent of X, if x is a linear (affine, positive, convex) combinationof finitely many points of X, i.e., there exist x1, . . . ,xm ∈ X, m ∈ N,such that x is a linear (affine, positive, convex) combination of the pointsx1, . . . ,xm.

0.3 Definition [Linearly and affinely independent points]. x1, . . . ,xm ∈ Rnare called linearly (affinely) dependent, if one of the xi is linearly (affinely) de-pendent of {x1, . . . ,xm}\{xi}. Otherwise x1, . . . ,xm are called linearly (affinely)independent.

0.4 Proposition. Let x1, . . . ,xm ∈ Rn.

i) x1, . . . ,xm are affinely dependent if and only if(x1

1

), . . . ,

(xm1

)∈ Rn+1 are

linearly dependent.

ii) x1, . . . ,xm are affinely dependent if and only if there exist µi ∈ R, 1 ≤i ≤ m, with (µ1, . . . , µm) 6= (0, . . . , 0),

∑mi=1 µi = 0 and

∑mi=1 µi xi = 0.

iii) If m ≥ n+ 1 then x1, . . . ,xm are linearly dependent.

iv) If m ≥ n+ 2 then x1, . . . ,xm are affinely dependent.

Proof. By definition we have that x1, . . . ,xm are affinely dependent if andonly if there exists an xi, say, and scalars λj , 1 ≤ j 6= i ≤ m, such thatxi =

∑j 6=i λj xj and

∑j 6=i λj = 1. This can be reformulated as(

0

0

)= −

(xi1

)+∑j 6=i

λj

(xj1

)

which is equivalent to the linear dependency of the vectors(xj

1

). For ii) we just

observe that the equation above is just a reformulation of what is to show if we


set µi = −1 and µj = λj , j 6= i. Fianlly, iv) follows from i) and iii), which istrivial.

�

0.5 Definition [Linear subspace, affine subspace, cone and convex set].X ⊆ Rn is called

i) linear subspace (set) if it contains all x ∈ Rn which are linearly dependentof X,

ii) affine subspace (set) if it contains all x ∈ Rn which are affinely dependentof X,

iii) (convex) cone if it contains all x ∈ Rn which are positively dependent ofX,

iv) convex set if it contains all x ∈ Rn which are convexly dependent of X.

0.6 Notation. Cn = {K ⊆ Rn : K convex} denotes the set of all convex setsin Rn. The empty set ∅ is regarded as a convex, linear and affine set.

0.7 Theorem. K ⊆ Rn is convex if and only if

λx+ (1− λ)y ∈ K, for all x,y ∈ K and 0 ≤ λ ≤ 1.

Proof. Of course, if K is convex and x,y ∈ K then for any λ ∈ [0, 1] the pointλx+ (1− λ)y is convexly dependent of K and hence it must belong to K.

Conversely, let v ∈ Rn be convexly dependent of K. Then there existx1, . . . ,xm ∈ K and scalars λ1, . . . , λm ≥ 0 with

∑λi = 1 and v =

∑λi xi.

We show by induction on m that v ∈ K. The case m = 1 is trivial, and so letm ≥ 2 and λm < 1. Then

v =m∑i=1

λi xi = (1− λm)

(m−1∑i=1

λi1− λm

xi

)︸︷︷︸

x

+λm xm = (1− λm)x+ λm xm.

Sincem−1∑i=1

λi1− λm

=1

1− λm

m−1∑i=1

λi =1− λm1− λm

= 1,

we get by our inductive argumention x ∈ K and thus, by our assumption v ∈ K.�

0.8 Example. The closed n-dimensional ball Bn(a, ρ) ={x ∈ Rn : |x− a| ≤

ρ}

with centre a and radius ρ > 0 is convex. The boundary of Bn(a, ρ), i.e.,{x ∈ Rn : |x− a| = ρ

}is non-convex. In the case a = 0 and ρ = 1 the

ball Bn(0, 1) is abbreviated by Bn and is called n-dimensional unit ball. Itsboundary is denoted by Sn−1.


0.9 Corollary. Let Ki ∈ Cn, i ∈ I. Then⋂i∈I Ki ∈ Cn.

Proof. Let x,y ∈⋂i∈I Ki and let λ ∈ [0, 1]. Since Ki is a convex set for all

i ∈ I, we have λx + (1 − λ)y ∈ Ki for all i ∈ I and hence λx + (1 − λ)y ∈⋂i∈I Ki. By Theorem 0.7 we obatin the convexity of

⋂i∈I Ki. �

0.10 Definition [Linear, affine, positive and convex hull, dimension].Let X ⊆ Rn.

i) The linear hull linX of X is defined by

linX =⋂

L⊆Rn, L linear,X⊆L

L.

ii) The affine hull aff X of X is defined by

aff X =⋂

A⊆Rn, A affine,X⊆A

A.

iii) The positive (conic) hull posX of X is defined by

posX =⋂

C⊆Rn, C convex cone,X⊆C

C.

iv) The convex hull convX of X is defined by

convX =⋂

K⊆Rn, K convex,X⊆K

K.

v) The dimension dimX ofX is the dimension of its affine hull, i.e., dim aff X.

0.11 Theorem. Let X ⊆ Rn. Then

convX =

{m∑i=1

λi xi : m ∈ N,xi ∈ X,λi ≥ 0,m∑i=1

λi = 1

}.

Proof. Let M(X) bet the set on the right hand side of the equation above.

For the inclusion convX ⊆ M(X) we note that by definition convX iscontained in any convex set containing X, and so it suffices to show that M(X)is convex. So let x,y ∈ M(X) with x =

∑m1i=1 νi xi and y =

∑m2j=1 µj yj ,

xi,yj ∈ X, νi, µj ≥ 0, and∑m1

i=1 νi =∑m2

j=1 µj = 1. Hence for λ ∈ [0, 1] we find

λx+ (1− λ)y =

m1∑i=1

λνi xi +

m2∑j=1

(1− λ)µj yj ,


where the coefficients of the above combination satisfy λνi, (1 − λ)µj ≥ 0 and∑m1i=1 λνi +

∑m2j=1(1 − λ)µj = λ + (1 − λ) = 1. Hence λx + (1 − λ)y ∈ M(X)

which shows that M(X) is convex (cf. Theorem 0.7), and so convX ⊆M(X).

In order to verify the reverse inclusion M(X) ⊆ convX, we observe thateach x ∈ M(X) is convexly dependent of X, and hence x is contained in anyconvex set containing X, i.e., x ∈

⋂K∈Cn,X⊆K K = convX. �

0.12 Remark.

i) conv {x,y} ={λx+ (1− λ)y : λ ∈ [0, 1]

}.

ii) linX ={∑m

i=1 λixi : λi ∈ R, xi ∈ X, m ∈ N}

.

iii) aff X ={∑m

i=1 λixi : λi ∈ R,∑m

i=1 λi = 1, xi ∈ X, m ∈ N}

.

iv) posX ={∑m

i=1 λixi : λi ∈ R, λi ≥ 0, xi ∈ X, m ∈ N}

.

0.13 Definition [(Relative) interior point and (relative) boundary point].Let X ⊆ Rn.

i) x ∈ X is called an interior point of X if there exists a ρ > 0 such thatBn(x, ρ) ⊆ X. The set of all interior points of X is called the interior ofX and is denoted by intX.

ii) x ∈ Rn is called boundary point of X if for all ρ > 0, Bn(x, ρ) ∩X 6= ∅and Bn(x, ρ)∩ (Rn\X) 6= ∅. The set of all boundary points of X is calledthe boundary of X and is denoted by bdX.

iii) Let A = aff X. x ∈ X is called a relative interior point of X if thereexists a ρ > 0 such that Bn(x, ρ)∩A ⊆ X. The set of all relative interiorpoints is called the relative interior of X and is denoted by relintX.

iv) Let A = aff X. x ∈ A is called a relative boundary point of X if for allρ > 0, Bn(x, ρ)∩X 6= ∅ and Bn(x, ρ)∩ (A\X) 6= ∅. The set of all relativeboundary points of X is called relative boundary of X and is denoted byrelbdX.

0.14 Remark. Let X ⊆ Rn be closed. Then X = relintX ∪ relbdX.

0.15 Theorem. Let K ∈ Cn, x ∈ relintK and y ∈ K. Then (1− λ)x+ λy ∈relintK for all λ ∈ [0, 1).

Proof. Let A = aff K, x ∈ relintK, and for λ ∈ [0, 1) let zλ = (1− λ)x+ λy.Since x = z0 ∈ relintK there exists a ρ > 0 such that Bn(x, ρ)∩A ⊆ K. By thetheorem on intersecting lines it follows immediately that Bn

(zλ, (1−λ)ρ

)∩A ⊆

K, and hence zλ ∈ relintK. Or more explicitly: For a ∈ Bn(zλ, (1− λ)ρ

)∩ A

we have

(1− λ)ρ ≥ |zλ − a| =∣∣(1− λ)x+ λy − a

∣∣ =∣∣(1− λ)x− (a− λy)

∣∣ .


Since λ < 1 we may divide both sides by 1−λ and get |x− (a− λy)/(1− λ)| ≤ρ. Thus (a − λy)/(1 − λ) ∈ Bn(x, ρ) ∩ A ⊆ K and by the convexity of K wefinally find

a = (1− λ)a− λy1− λ

+ λy ∈ K.

�

0.16 Corollary. Let K ∈ Cn be closed. Let x ∈ relintK and y ∈ aff K \K.Then the segment conv {x,y} intersects relbdK in precisely one point.

Proof. K ∩ conv {x,y} is a convex, compact 1-dimensional set. Hence K ∩conv {x,y} = conv {x,y} for some y ∈ K. Obviously, y /∈ relintK and soy ∈ relbdK. By Theorem 0.15, y is the only point of conv {x,y} lying onrelbdK. �

0.17 Definition [Polytope and simplex]. Let X ⊂ Rn of finite cardinality,i.e., #X <∞.

i) convX is called a (convex) polytope.

ii) A polytope P ⊂ Rn of dimension k is called a k-polytope.

iii) If X is affinely independent and dimX = k then convX is called a k-simplex.

0.18 Notation. Pn = {P ⊂ Rn : P polytope} denotes the set of all polytopesin Rn.

0.19 Lemma. Let T = conv {x1, . . . ,xk+1} ⊂ Rn be a k-simplex, and letλi > 0, 1 ≤ i ≤ k + 1, with

∑λi = 1. Then

∑λi xi ∈ relintT .

Proof. See Exercise ??. �

0.20 Corollary. Let K ∈ Cn, K 6= ∅. Then relintK 6= ∅.

Proof. Let k = dimK ≥ 0. Then there exist x1, . . . ,xk+1 ∈ K affinely inde-pendent such that aff K = aff {x1, . . . ,xk+1}. Let Tk = conv {x1, . . . ,xk+1} ⊆K. From Lemma 0.19 we get relintTk 6= ∅, and hence relintK 6= ∅. �

0.21 Theorem. Let P = conv {x1, . . . ,xm} ∈ Pn. A point x ∈ Rn belongs torelintP if and only if x admits a representation as x =

∑mi=1 λixi with λi > 0,

1 ≤ i ≤ m, and∑m

i=1 λi = 1.

Proof. Let x ∈ relintP and let y =∑m

i=1(1/m)xi ∈ P . Since x ∈ relintPthere exists a z ∈ P such that x = λ z + (1 − λ)y with λ ∈ [0, 1). Letz =

∑mi=1 µixi, with µi ≥ 0, 1 ≤ i ≤ m, and

∑µi = 1. Then x =

∑mi=1

(λµi +

(1 − λ)/m)xi, where all the scalars λµi + (1 − λ)/m are positive and sum up

to 1.Next we assume that x has a representation as x =

∑mi=1 λixi with λi > 0,

1 ≤ i ≤ m, and∑λi = 1. Let k = dimP and without loss of generality let


x1, . . . ,xk+1 be affinely independent. Setting λ =∑k+1

i=1 λi, Lemma 0.19 showsthat

y =k+1∑i=1

λiλxi ∈ relint conv {x1, . . . ,xk+1} ⊆ relintP.

If λ = 1 then k + 1 = m and hence x = y ∈ relintP . If λ < 1 let z =1/(1 − λ)

∑mi=k+1 λi xi ∈ P and with Theorem 0.15 we also find in this case

x = λy + (1− λ)z ∈ relintP . �

0.22 Notation.

i) For two sets X,Y ⊆ Rn the vectorial addition

X + Y = {x+ y : x ∈ X, y ∈ Y }

is called the Minkowski 1 sum of X and Y . If X is just a singleton, i.e.,X = {x}, then we write x+ Y instead of {x}+ Y .

ii) For λ ∈ R and X ⊆ Rn we denote by λX the set

λX = {λx : x ∈ X} .

For instance, Bn(a, ρ) = a+ ρBn.

1Hermann Minkowski, 1864–1909

http://en.wikipedia.org/wiki/Hermann_Minkowski

Support and separate 7

1 Support and separate

1.1 Notation. Let a ∈ Rn, a 6= 0, and α ∈ R. The closed halfspacesH+(a, α),H−(a, α) ⊂ Rn are given by

H+(a, α) ={x ∈ Rn : 〈a,x〉 ≥ α

}, H−(a, α) =

{x ∈ Rn : 〈a,x〉 ≤ α

}.

The hyperplane H(a, α) is defined by

H(a, α) ={x ∈ Rn : 〈a,x〉 = α

}.

1.2 Definition [Supporting hyperplane]. LetX ⊂ Rn. A hyperplaneH(a, α) ⊂Rn is called supporting hyperplane of X if:

i) H(a, α) ∩X 6= ∅ and ii) X ⊆ H−(a, α).

a is called outer normal vector of X and if, in addition, |a| = 1 then it is calledouter unit normal vector of X.

1.3 Proposition. Let X ⊂ Rn and let H(a, α) be a supporting hyperplane ofX. Then

H(a, α) ∩ convX = conv(H(a, α) ∩X

).

Proof. Let x ∈ H(a, α) ∩ convX. Since x ∈ convX there exist xi ∈ X andλi > 0, i = 1, . . . ,m, with

∑mi=1 λi = 1 and x =

∑mi=1 λixi. Since x ∈ H(a, α)

we have 〈a,x〉 = α and from X ⊆ H−(a, α) we get 〈a,xi〉 ≤ α, 1 ≤ i ≤ m.Hence

α = 〈a,x〉 =m∑i=1

λi 〈a,xi〉 ≤m∑i=1

αλi = α.

Thus 〈a,xi〉 = α, 1 ≤ i ≤ m, and so xi ∈ H(a, α) ∩ X which implies x ∈conv

(H(a, α) ∩X

).

The reverse inclusion is trivial since conv(H(a, α)∩X

)⊆ conv

(H(a, α)

)∩

convX = H(a, α) ∩ convX. �

1.4 Remark. Let X ⊂ Rn be compact and a ∈ Rn\{0}. Then there exists asupporting hyperplane of X with outer normal vector a.

1.5 Definition [Nearest point map (or metric projection)]. Let K ∈ Cn beclosed. The map ΦK : Rn → K, where for x ∈ Rn the point ΦK(x) ∈ K isgiven by |x− ΦK(x)| = min

{|x− y| : y ∈ K

}is called the nearest point map

(metric projection) with respect to K.

1.6 Remark. The nearest point map is well-defined: Notice that since K isclosed, for all x ∈ Rn there exist yx ∈ K such that |x− yx| = min

{|x− y| :

y ∈ K}

, and yx is uniquely determined. In fact, if there exists y ∈ K, y 6= yx,with |x− y| = |x− yx| then we may assume that x−yx and x−y are linearlyindependent. Hence∣∣∣∣x− yx + y

2

∣∣∣∣ =

∣∣∣∣12(x− yx) +1

2(x− y)

∣∣∣∣ < 1

2|x− yx|+

1

2|x− y| = |x− yvx| .

Since K is convex, (yx + y)/2 ∈ K which contradicts the minimality of yx.


1.7 Theorem. Let K ∈ Cn be closed and let x ∈ Rn \K. Let a = x−ΦK(x)and α = 〈a,ΦK(x)〉. Then H(a, α) is a supporting hyperplane of K with outernormal vector a.

Proof. By the choice of a and α it ΦK(x) ∈ K ∩H(a, α) and hence it remainsto show K ⊆ H−(a, α), i.e., that 〈a,y〉 ≤ α for all y ∈ K. Suppose the oppositeand let y ∈ K with 〈a,y〉 > α. For λ ∈ [0, 1] we consider the distance of x toz(λ) = (1− λ)ΦK(x) + λy ∈ K

h(λ) = |x− z(λ)|2 = |x− ΦK(x) + λ(ΦK(x)− y)|2

= |x− ΦK(x)|2 + 2λ 〈x− ΦK(x),ΦK(x)− y〉+ λ2 |ΦK(x)− y|2

= |x− ΦK(x)|2 + 2λ 〈a,ΦK(x)− y〉+ λ2 |ΦK(x)− y|2 .

Since 〈a,ΦK(x)− y〉 < 0 there exists a positive λ∗ ∈ (0, 1] with h(λ∗) < h(0).This, however, contradicts the fact that h(0) is the minimal (squared) distancebetween x and K. �

1.8 Corollary. Let K ∈ Cn, K 6= Rn, be closed. Then

K =⋂

H(a,α) supportinghyperplane of K

H−(a, α),

i.e., K is the intersection of all its “supporting halfspaces”.

Proof. Clearly K ⊆⋂H(a,α)H

−(a, α). In order to prove the reverse inclusionwe take x 6∈ K, and let H(a, α) be the supporting hyperplane defined in The-orem 1.7. Then x ∈ H+(a, α) but x 6∈ H(a, α), i.e, x 6∈ H−(a, α) and hence xis not contained in the intersection on the right hand side. �

1.9 Corollary. Let X ⊂ Rn such that convX is closed and convX 6= Rn.Then

convX =⋂

X⊆H−(a,α)

H−(a, α),

i.e., convX is the intersection of all halfspaces containing X.

Proof. Since each halfspace containing X also contains convX, we certainlyhave that convX is contained in the intersection of halfspaces above. On theother hand, if x 6∈ convX, by Corollary 1.8 there exists a supporting hyperplaneH(a, α) of convX with x 6∈ H−(a, α). Since X ⊆ convX ⊆ H−(a, α), x isalso not contained in the intersection of the right hand side. �

1.10 Lemma [Busemann-Feller Lemma]. 2,3 Let K ∈ Cn be closed. Then

|ΦK(x)− ΦK(y)| ≤ |x− y|

for all x,y ∈ Rn, i.e., the nearest point map does not increase distances. Inparticular, it is a continuous map.

2Herbert Busemann, 1905–19943William Feller, 1906–1970

http://en.wikipedia.org/wiki/Herbert_Busemann

http://en.wikipedia.org/wiki/William_Feller


Proof. We suppose that ΦK(x) 6= ΦK(y) and let a = ΦK(x) − ΦK(y),αx = 〈a,ΦK(x)〉 and αy = 〈−a,ΦK(y)〉. It suffices to show that x ∈ H+(a, αx)and y ∈ H+(−a, αy), because then 〈a,x〉 ≥ αx and 〈−a, y〉 ≥ αy, which implies

〈a,x− y〉 ≥ αx + αy = 〈a,ΦK(x)− ΦK(y)〉 = |ΦK(x)− ΦK(y)|2 .

By the Cauchy-Schwarz inequality we conclude |x− y| ≥ |ΦK(x)− ΦK(y)|. Sowe assume the contrary and without loss of generality let x /∈ H+(a, αx), i.e.,〈a,x〉 < αx. Then the ray

R(x) ={

ΦK(x) + λ(x− ΦK(x)

): λ ≥ 0

}has to intersect the hyperplaneH(−a, αy) in a point x, say, and by the Pythagoreantheorem we obtain

|x− ΦK(y)| < |x− ΦK(x)| .On the other hand, by Exercise ?? we have ΦK(z) = ΦK(x) for all z ∈ R(x),and so we get the contradiction |x− ΦK(x)| = |x− ΦK(x)| > |x− ΦK(y)|. �

1.11 Theorem. Let K ∈ Cn be compact and let ρ > 0 such that K ⊂int (ρBn). The nearest point map on ρSn−1, i.e., ΦK : ρSn−1 → bdK issurjective.

Proof. Let x ∈ bdK. Then ΦK(x) = x and for i ∈ N let xi ∈ int (ρBn) suchthat xi 6∈ K and limi→∞ xi = x. By Lemma 1.10 we have

|x− ΦK(xi)| = |ΦK(x)− ΦK(xi)| ≤ |x− xi| .

and so limi→∞ΦK(xi) = x as well. By Exercise ??, the intersection point ziof the ray R(xi) =

{ΦK(xi) + λ

(xi − ΦK(xi)

): λ ≥ 0

}with ρSn−1 verifies

ΦK(zi) = ΦK(xi), and thus limi→∞ΦK(zi) = x. By the compactness of ρSn−1

we may asssume (after restricting to a convergent subsequence) that (zi) isconvergent, and so let limi→∞ zi = z ∈ ρSn−1. Finally, the continuity of thenearest map point gives

ΦK(z) = limi→∞

ΦK(zi) = x.

�

1.12 Corollary. Let K ∈ Cn be closed and let x ∈ relbdK. Then there existsa supporting hyperplane H(a, α) of K with x ∈ H(a, α).

Proof. Without loss of generality let dimK = n. Let x ∈ bdK and let γ > 0with x ∈ int (γBn). The convex set K = K ∩ (γBn) is compact and x ∈ bdK.Let ρ > γ be such that K ⊂ int (ρBn). By Theorem 1.11 we can find z ∈ ρSn−1

with x = ΦK(z). Then Theorem 1.7 ensures that the hyperplane H(a, α), witha = z − x and α = 〈a,x〉, supports K at x. Finally we have to check thatH(a, α) is also a supporting hyperplane of K at x.

By definition we have K ⊂ H−(a, α) and suppose that there exists y ∈ Kwith 〈a,y〉 > α. Since 〈a,x〉 = α it holds 〈a, (1− λ)x+ λy〉 > α for anyλ ∈ (0, 1]. Let λ > 0 be sufficiently small such that y = (1− λ)x+ λy ∈ γBn.Since y ∈ K we have y ∈ K with 〈a,y〉 > α, a contradiction. �


K1

K2

H(a, α)

H−(a, α) H+(a, α)

Figure 1: A strictly separating hyperplane of two compact convex sets

1.13 Theorem [Separation theorem]. Let K1,K2 ∈ Cn with K1 ∩ K2 = ∅.Then there exists a separating hyperplane H(a, α) of K1 and K2, i.e., K1 ⊆H+(a, α) and K2 ⊆ H−(a, α).

If K1 is closed and K2 is compact, then there exists even a strictly sepa-rating hyperplane H(a, α) of K1 and K2, i.e., K1 ⊂ intH+(a, α) and K2 ⊂intH−(a, α).

Proof. If both, K1 and K2 are compact the statement is certainly true bystandard compactness arguments (see also Exercise ??). If K1 is closed and K2

is compact we take the intersection K1 = K1 ∩ ρBn for ρ > 0 sufficiently largesuch that the distance between K2 and K1 equals the distance between K2 andK1. Hence we are back in the case of two compact sets, and we finally havejust to observe that a strictly separating hyperplane of K2 and K1 is also onefor K2 and K1 (see also the proof of Corollary 1.12).

Next we consider the case of arbitrary disjoint convex sets K1 and K2. Letx1 ∈ relintK1 and x2 ∈ relintK2, and for i ∈ N let

Kij =

[cl

((1− 1

i

)(Kj − xj)

)+ xj

]∩ (iBn), for j = 1, 2.

Clearly Kij ⊂ K

i+1j ⊂ Kj , for j = 1, 2 and any i ∈ N, and for every x ∈ relintKj

there exists an index ix such that x ∈ Kij for all i ≥ ix. Moreover, Ki

1 and Ki2

are compact convex sets with Ki1 ∩Ki

2 = ∅ for any i ∈ N. By Exercise ?? weknow that there exists a separating hyperplane H(ai, αi) of Ki

1 and Ki2 with

|ai| = 1, and thus

〈ai,x〉 ≤ αi for all x ∈ Ki1 and 〈ai,x〉 ≥ αi for all x ∈ Ki

2.

Since 〈ai,x1〉 ≤ αi ≤ 〈ai,x2〉 for i large and |ai| = 1, the sequence (αi)i∈N isbounded and hence

{(aiαi

): i ∈ N

}⊂ Rn+1 is also a bounded sequence. Without

loss of generality we can assume that this sequence is convergent and we writea = limi→∞ ai and α = limi→∞ αi. In order to prove that the hyperplaneH(a, α) separates K1 and K2, let x ∈ K1. If x ∈ relintK1 then there exists anindex ix such that 〈ai,x〉 ≤ αi for all i ≥ ix, which implies that 〈a,x〉 ≤ α. Forx ∈ relbdK1, we approach x by the points xλ = (1 − λ)x + λx1 ∈ relintK1,λ ∈ (0, 1]. By the previous discussion we have 〈a,xλ〉 ≤ α for all λ ∈ (0, 1] and


so also 〈a,x0〉 = 〈a,x〉 = α. Analogously we obtain 〈a,x〉 ≥ α for all x ∈ K2.�

1.14 Definition [Support function, breadth]. Let K ∈ Cn, K 6= ∅. Thefunction h(K, ·) : Rn → R given by

h(K,u) = sup{〈u,x〉 : x ∈ K

}is called support function of K. For u ∈ Sn−1 the breadth of K in the directionu is defined by h(K,u) + h(K,−u).

1.15 Proposition. Let K ∈ Cn be non-empty and compact. Then

K =⋂

u∈Sn−1

{x ∈ Rn : 〈u,x〉 ≤ h(K,u)

}.

Proof. By Corollary (1.8) it suffices to observe that the intersection is justtaken over all supporting hyperplanes of K. In fact, given a supporting hyper-plane H(a, α) of K we may assume a ∈ Sn−1 and since K ⊆ H−(a, α) butK ∩H(a, α) 6= ∅, we have α = maxx∈K 〈a,x〉 = h(K,a). �

1.16 Definition [Convex function]. Let K ∈ Cn. A function f : K → R iscalled convex if

f(λx+ (1− λ)y

)≤ λf(x) + (1− λ)f(y), for all x,y ∈ K, λ ∈ (0, 1).

f is called strictly convex when the above inequality holds as a strict inequalityif x 6= y. If −f is convex then f is called concave.

1.17 Proposition. Let f : K → R be a function differentiable on an openconvex set K ⊂ Rn.

i) f is convex if and only if

f(x) ≥ f(y) + 〈∇f(y),x− y〉 , for all x,y ∈ K. (1.17.1)

ii) Let f be twice differentiable. Then f is convex if and only if its Hessianis positive semi-definite for all points in K.

Proof. i) We suppose first that f is convex. Then for any x,y ∈ K and anyλ ∈ (0, 1) it holds f

(y+ λ(x− y)

)= f

((1− λ)y+ λx

)≤ (1− λ)f(y) + λf(x),

which implies

f(y + λ(x− y)

)− f(y)

λ− 〈∇f(y),x− y〉 ≤ f(x)− f(y)− 〈∇f(y),x− y〉 .

The left hand side approaches zero when λ→ 0, which proves (1.17.1).Conversely, if (1.17.1) holds then interchanging the roles of x and y and

adding both inequalities we get

〈∇f(x)−∇f(y),x− y〉 ≥ 0, for all x,y ∈ K. (1.17.2)


Let x,y ∈ K. We define the function g : [0, 1] → R given by g(λ) = f(y +

λ(x− y)), which is differentiable. Then for 0 ≤ λ0 < λ1 ≤ 1 we get

g′(λ1)−g′(λ0) =⟨∇f(y + λ1(x− y)

),x− y

⟩−⟨∇f(y + λ0(x− y)

),x− y

⟩≥ 0

by using (1.17.2). Thus g′ is an increasing function which implies that g is aconvex function, and so g(λ) ≤ λ g(1) + (1− λ) g(0) = λ f(x) + (1− λ) f(y).

ii) For x,y ∈ K we consider the same function gx,y(λ) as in the previous proof.Then f is convex if and only if all gx,y are convex for all x,y ∈ K, which isequivalent to g′′x,y ≥ 0. Hence, f is convex if and only if for all x,y ∈ K andλ ∈ [0, 1]

(x− y)ᵀ [(Hess f)(x+ λ(x− y))] (x− y) ≥ 0.

This is certainly fulfilled if the Hessian is positive semi-definite. Otherwise, fora given z in the open set K and a given vector v ∈ Rn we can find x,y ∈ Kand λ ∈ [0, 1] and µ ∈ R>0 such that v = µ(x − y), z = x + λ(x − y) and sovᵀ(Hess f)(z)v = µ2(x−y)ᵀ [(Hess f)(x+ λ(x− y))] (x−y) ≥ 0, which showsthat the Hessian is positive definite for every z ∈ K. �

1.18 Theorem [Jensen’s inequality]. 4 Let K ∈ Cn and let f : K → R beconvex. For all x1, . . . ,xm ∈ K and 0 ≤ λ1, . . . , λm with

∑mi=1 λi = 1 it holds

f

(m∑i=1

λixi

)≤

m∑i=1

λif(xi).

Proof. Induction on m. �

1.19 Remark. Let f : K → R be defined on a convex set K. The set epi f ={(

xxn+1

)∈ Rn+1 : x ∈ K, xn+1 ≥ f(x)} is called the epigraph of f . Then f is

convex if and only if its epigraph epi f is convex.

1.20 Theorem*. Let K ∈ Cn be open and let f : K → R be convex. Then fis continuous.

1.21 Theorem. Let K ∈ Cn be bounded and let K 6= ∅.

i) h(K, ·) is a convex function.

ii) h(K, ·) is positively homogeneous of degree 1, i.e., h(K,λu) = λh(K,u)for all λ ≥ 0 and u ∈ Rn.

iii) If h : Rn → R is a function satisfying i) and ii) then there exists a convexclosed and bounded set K ∈ Cn such that h(K,u) = h(u) for all u ∈ Rn.

4Johan Jensen, 1859–1925

http://en.wikipedia.org/wiki/Johan_Jensen_(mathematician)


Proof. Let u,v ∈ Rn and λ ∈ [0, 1]. Then

h(K,λu+ (1− λ)v

)= sup

{λ 〈u,x〉+ (1− λ) 〈v,x〉 : x ∈ K

}≤ sup

{λ 〈u,x〉 : x ∈ K

}+ sup

{(1− λ) 〈v,x〉 : x ∈ K

}= λh(K,u) + (1− λ)h(K,v),

which shows i), and the last step also gives ii).

For iii) and the given function h : Rn → R we consider

K =⋂

v∈Rn

{x ∈ Rn : 〈x,v〉 ≤ h(v)

},

which is a closed convex(cf. Corollary 0.9) and bounded(consider v = ei) set.If it is non-empty then clearly h(K,u) ≤ h(u) for all u ∈ Rn, and so it sufficesto prove that for every u ∈ Rn there exist an a ∈ K with 〈u,a〉 = h(u) andthus h(K,u) = h(u).

Since h satisfies i) and ii), its epigraph epih is a closed (cf. Theorem 1.20)convex cone in Rn+1. Let u ∈ Rn. Since

(u, h(u)

)∈ bd epih, we can find by

Corollary 1.12 a supporting hyperplane H((a, t), α

)of epih through

(u, h(u)

)such that epih ⊆ H−

((a, t), α

). Since epih is a cone containing 0, H

((a, t), α

)must pass through 0, hence α = 0. If t = 0 then (b, h(b))ᵀ /∈ H−

((a, t), 0

)for

〈b,a〉 > 0. If t > 0, then for some xn+1 > h(u) the point (u, xn+1)ᵀ wouldnot be contained in H−

((a, t), 0

). Hence t < 0 and so we may assume t = −1.

Then we can write epih ⊆ H−((a,−1), 0

), and since (v, h(v))ᵀ ∈ epih we have

〈a,v〉 ≤ h(v) for all v ∈ Rn, which shows a ∈ K. Moreover, by constructionwe have 〈a,u〉 = h(u) and so h(K,u) = 〈a,u〉 = h(u). �

1.22 Definition [Polar set]. Let X ⊆ Rn.

X? ={y ∈ Rn : 〈x,y〉 ≤ 1 for all x ∈ X

}is called the polar set of X.

1.23 Proposition.

i) X? is a convex and closed set and 0 ∈ X?.

ii) If X1 ⊆ X2 then X?2 ⊆ X?

1 .

iii) Let M be a regular n× n matrix. Then (MX)? = M−ᵀX?.

iv) Let Xi ⊆ Rn, i ∈ I. Then(⋃

i∈I Xi

)?=⋂i∈I X

?i .

v) X ⊆ (X?)?.

vi) Let X ⊂ Rn. Then X = X? if and only if X = Bn.


Proof. i) X? is an intersection of closed and convex sets and hence X? is closedand convex(cf. Corollary 0.9). Obviously, 0 ∈ X?.

ii) Let y ∈ X?2 . Then 〈x,y〉 ≤ 1 for all x ∈ X2. In particular, 〈x,y〉 ≤ 1 for all

x ∈ X1 which proves that y ∈ X?1 .

iii) Let y ∈ (MX)?. By definition, 〈Mx,y〉 ≤ 1 for all x ∈ X, i.e., 〈x,Mᵀy〉 ≤ 1for all x ∈ X. This condition is equivalent to Mᵀy ∈ X?, which says thaty ∈M−ᵀX?.

iv) Just notice that(⋃i∈I

Xi

)?=

{y ∈ Rn : 〈x,y〉 ≤ 1 for all x ∈

⋃i∈I

Xi

}=⋂i∈I

{y ∈ Rn : 〈x,y〉 ≤ 1 for all x ∈ Xi

}=⋂i∈I

X?i .

v) Let x ∈ X. By definition of polar body 〈y,x〉 ≤ 1 for all y ∈ X?, i.e.,x ∈ (X?)?.

vi) We suppose first thatX = Bn. ThenBn? =

{y ∈ Rn : 〈x,y〉 ≤ 1 for all x ∈

Bn}

. If y ∈ Bn, it is clear that 〈y,x〉 ≤ 1 for all x ∈ Bn, which shows Bn ⊆ Bn?.If y ∈ Bn? with |y| > 1 then y/ |y| ∈ Bn. Since 〈y/ |y| ,y〉 = |y| > 1 we gety 6∈ Bn?, a contradiction.

Conversely, let x ∈ X = X?. By the definition of the polar body we know〈x,x〉 ≤ 1, which shows X ⊆ Bn. Now part ii) implies X = X? ⊇ Bn

? = Bn,and thus X = Bn.

�

1.24 Proposition.

i) Let P = conv {x1, . . . ,xm} ⊂ Rn. Then

P ? ={y ∈ Rn : 〈xi,y〉 ≤ 1, 1 ≤ i ≤ m

}.

ii) Let P ={x ∈ Rn : 〈ai,x〉 ≤ 1, 1 ≤ i ≤ m

}with ai ∈ Rn. Then

P ? = conv {0,a1, . . . ,am}.

Proof. i) P ? ⊆{y ∈ Rn : 〈xi,y〉 ≤ 1, 1 ≤ i ≤ m

}holds by the definition

of polar body. So let y ∈ Rn with 〈xi,y〉 ≤ 1, 1 ≤ i ≤ m. For any x ∈ Pthere exist λ1, . . . , λm ≥ 0 with

∑mi=1 λi = 1 such that x =

∑mi=1 λixi. Then

〈x,y〉 =∑m

i=1 λi 〈xi,y〉 ≤∑m

i=1 λi = 1, i.e., y ∈ P ?.ii) Clearly P ? ⊇ conv {0,a1, . . . ,am}. So suppose there exists y ∈ P ? withy 6∈ conv {0,a1, . . . ,am}. Theorem 1.13 ensures the existence of a strictlyseparating hyperplane H(a, α) with 〈a,x〉 < α for all x ∈ conv {0,a1, . . . ,am}and 〈a,y〉 > α. Since α > 0 we have, in particular, 〈a/α,ai〉 < 1 for 1 ≤ i ≤m, which shows that a/α ∈ P . Since 〈a/α,y〉 > 1 we conclude y 6∈ P ?, acontradiction. �

1.25 Lemma. Let K ∈ Cn be closed with 0 ∈ K. Then (K?)? = K.


Proof. In view of Proposition 1.23, part v), it suffices to show (K?)? ⊆ K. Wesuppose there exists y ∈ (K?)? with y 6∈ K. Then by Theorem 1.13 we can finda strictly separating hyperplane H(a, α) such that 〈a,y〉 > α and 〈a,x〉 < αfor all x ∈ K. Since α > 0 we have 〈a/α,x〉 < 1 for all x ∈ K and so a/α ∈ K?.From 〈a/α,y〉 > 1 we get the contradiction y 6∈ (K?)?. �


Radon, Helly, Caratheodory and (a few) relatives 17

2 Radon, Helly, Caratheodory and (a few) relatives

2.1 Theorem [Radon]. 5 Let X ⊂ Rn. If #X ≥ n + 2 then there existX1, X2 ⊂ X with X1 ∩X2 = ∅ and convX1 ∩ convX2 6= ∅.

Proof. Since #X ≥ n + 2, X contains n + 2 affinely dependent pointsx1, . . . ,xn+2 ∈ X. Thus there exist λi ∈ R, i = 1, . . . , n+ 2, not all zero, with∑n+2

i=1 λi = 0 and∑n+2

i=1 λi xi = 0. Without loss of generality let λ1, . . . , λk > 0

and λk+1, . . . , λn+2 ≤ 0. Then∑k

i=1 λi xi =∑n+2

i=k+1(−λi)xi, and with λ =∑ki=1 λi =

∑n+2i=k+1(−λi) we may write

k∑i=1

λi

λxi =

n+2∑i=k+1

(−λiλ

)xi ∈ conv {x1, . . . ,xk}︸︷︷︸

X1

∩ conv {xk+1, . . . ,xn+2}︸︷︷︸X2

.

�

2.2 Theorem [Helly]. 6 Let K1, . . . ,Km ∈ Cn, m ≥ n+ 1, such that for each(n+ 1)-index set I ⊆ {1, . . . ,m} we have

⋂i∈I Ki 6= ∅. Then all sets Ki have a

point in common, i.e.,⋂mj=1Ki 6= ∅.

Proof. We use induction on m. The case m = n + 1 is certainly true byassumption, and for the proof of the induction step m − 1 → m we set Xj =⋂mi=1,i 6=jKi, for j = 1, . . . ,m. By induction hypothesis Xj 6= ∅, and let xj ∈ Xj ,

j = 1, . . . ,m. Hence xj ∈ Ki for all i 6= j.

Since m ≥ n + 2, we may apply Radon’s Theorem 2.1 to {x1, . . . ,xm}and hence, without loss of generality, we may suppose that there exists x ∈conv {x1, . . . ,xk}∩ conv {xk+1, . . . ,xm}, for a suitable index k. Now conv {x1,. . . ,xk} ⊆ Ki for i > k and conv {xk+1, . . . ,xm} ⊆ Ki for i ≤ k, and sox ∈

⋂mi=1Ki. �

2.3 Remark.

i) Without any further restrictions/assumptions Helly’s theorem is not truefor infinitely many convex sets Ki. For instance, let Ki = (0, 1

i ], i ∈ N.

ii) Helly’s theorem, however, can be easily generalised to infinitely manycompact (bounded and closed) convex sets.

2.4 Corollary. Let C ⊂ Cn be compact. Then there exists t ∈ Rn with

−C ⊆ t+ nC.

Proof. For c ∈ C we set Xc = {t ∈ Rn : −c ∈ t+nC}. Clearly, Xc = −c−nC,and so it is convex and compact. We are loking for a t ∈ ∩c∈CXc. According

5Johann Karl August Radon, 1887–19566Eduard Helly, 1884–1943

http://en.wikipedia.org/wiki/Johann_Radon

http://en.wikipedia.org/wiki/Eduard_Helly


to Helly’s Theorem 2.2 (cf. also Remark 2.3 ii)) it suffcies to show that forc1, . . . , cn+1 ∈ C

n+1⋂i=1

Xci 6= ∅.

For it we consider t = −∑n+1

i=1 ci. Then for j = 1, . . . , n+ 1,

−cj − t =n+1∑

i=1,i 6=jci = n

n+1∑i=1,i 6=j

1

nci

∈ nC,i.e., t ∈ Xci , 1 ≤ i ≤ n+ 1. �

2.5 Definition [Centerpoint]. For a finite point set X ⊂ Rn a point c ∈ Rnis called centerpoint if every closed halfspace containing c containes at leastb 1n+1#Xc points of X.

2.6 Theorem. Every finite set X ⊂ Rn has a centerpoint.

Proof. With

M = {U ⊆ X : #U > d(n/(n+ 1))#Xe} ,

we first notice that c is a centerpoint of X if and only if

c ∈⋂U∈M

convU.

For if, c is not a centerpoint if and only if there exists a halfspace H− containingc such that #(H−∩X) < b1/(n+1)Xc. Hence c is not contained in convU withU = (intH+ ∩ X) ∈ M. On the other hand, if c /∈ convU for some U ∈ M,then Theorem 1.13 gives a halfspace H− containing c and H− ∩X = X \ U .

Hence we have to show that the intersection⋂U∈M convU , consisting of

finitely many compact convex sets, is nonempty. Due to the cardinality of thesets U ∈ M the intersection of each n + 1 of theses sets is nonempty, and sothe intersection of their convex hulls is non-empty. Hence, Helly’s Theorem 2.2yields

⋂U∈M convU 6= ∅. �

2.7 Theorem [Caratheodory]. 7 Let X ⊂ Rn. Then

convX =

{n+1∑i=1

λi xi : λi ≥ 0,n+1∑i=1

λi = 1,xi ∈ X, i = 1, . . . , n+ 1

}.

Proof. Let x ∈ convX. By Theorem 0.11 there exist a minimal m ∈ N andx1, . . . ,xm ∈ X such that x =

∑mi=1 λi xi for certain λi > 0, i = 1, . . . ,m, with∑n

i=1 λi = 1.

We have to show m ≤ n + 1. So we assume that m ≥ n + 2. Then{x1, . . . ,xm} are affinely dependent and there exist µ1, . . . , µm ∈ R not all

7Constantin Caratheodory, 1873 - 1950

http://en.wikipedia.org/wiki/Constantin_Carath�odory

Radon, Helly, Caratheodory and (a few) relatives 19

of them zero with∑m

i=1 µi = 0 and∑m

i=1 µi xi = 0. Hence we can writex =

∑mi=1(λi − αµi)xi for any α ∈ R. Let the index k be chosen such that

λkµk

= min1≤i≤m

{λiµi

: µi > 0

}.

Then λi − (λk/µk)µi ≥ 0 for i = 1, . . . ,m,∑m

i=1,i 6=k(λi − (λk/µk)µi) = 1 and

x =

m∑i=1,i 6=k

(λi −

λkµkµi

)xi,

which contradicts the minimality of m. The reverse inclusion is certainly trueby Theorem 0.11. �

2.8 Remark. Let X ⊂ Rn. Then

convX =

{dimX+1∑i=1

λi xi : λi ≥ 0,

dimX+1∑i=1

λi = 1,xi ∈ X

}.

As a direct consequence of Caratheodory’s Theorem 2.7 we get

2.9 Corollary. A polytope is the union of simplices.

2.10 Corollary. The convex hull of a compact set is compact.

Proof. Let X be a compact set and we suppose that X ⊆ Bn(0, ρ). Then it isclear that also convX ⊆ Bn(0, ρ), which shows convX is bounded.

In order to see convX is closed let x = limi→∞ xi, with xi ∈ convX, i ∈ N.By Caratheodory’s Theorem 2.7 we can write xi =

∑n+1j=1 λij xij , with xij ∈ X

and λij ≥ 0,∑n+1

j=1 λij = 1.

Since 0 ≤ λij ≤ 1 and X is bounded we may assume that for each j ∈{1, . . . , n + 1} the limits limi→∞ λij = λj and limi→∞ xij = xj exist. By thecloseness of X we have xj ∈ X, and moreover, λj ≥ 0 and

∑n+1j=1 λj = 1. So we

know

x = limi→∞

xi = limi→∞

n+1∑j=1

λij xij =n+1∑j=1

λj xj ,

which proves that x ∈ convX. �

2.11 Theorem* [(strong) Fractional Helly theorem]. Let K1, . . . ,Km ∈ Cn,m ≥ n+ 1, and let α ∈ (0, 1] such that for at least α

(mn+1

)of the (n+ 1)-index

sets I ⊆ {1, . . . ,m} we have⋂i∈I Ki 6= ∅. Then there exists a point in common

of at least (1− (1− α)1/(n+1)) ·m sets Ki.

2.12 Theorem [Colorful Caratheodory theorem]. Let X1, . . . , Xn+1 ⊂ Rnsuch that 0 ∈ convXi, 1 ≤ i ≤ n+ 1. There exist xi ∈ Xi, 1 ≤ i ≤ n+ 1, suchthat 0 ∈ conv {x1, . . . ,xn+1}.


Proof. It suffices to prove the statement for finite sets Xi, because otherwise wemay replace Xi be any finite subset Xi containing 0 in its convex hull. For xi ∈Xi, 1 ≤ i ≤ n+1, we call {x1, . . . ,xn+1} a rainbow set and conv {x1, . . . ,xn+1}a rainbow simplex. Suppose there is no rainbow simplex containing 0, and letS = {s1, . . . , sn+1}, si ∈ Xi, be a rainbow set such that convS has minimaldistance to 0 attained at the point s ∈ S, say. Let H = {x ∈ Rn : 〈s,x〉 =〈s, s〉} be the hyperplane perpendicular to s and containing s. Then we mayassume that S ⊂ H+ = {x ∈ Rn : 〈s,x〉 ≥ 〈s, s〉} which does not contain 0.

Since convS ∩ H = conv (S ∩ H) (cf. Proposition 1.3) we know by Cara-theodory’s Theorem 2.7 that there exists an n-point set T ⊂ S ∩ H with s ∈conv T . Without loss of generality let s1 /∈ T . If X1 ⊂ H+ then 0 /∈ convX1,and so we may assume that there exists a z ∈ X1 with 〈s, z〉 < 〈s, s〉. The newrainbow set S′ = S \{s1}∪{z} contains the segment conv {s, z}, which, by thechoice of z, contains a point closer to 0 than s — a contradiction. �

2.13 Theorem* [Tverberg]. 8 Let X ⊆ Rn and let k ∈ N≥1. If #X ≥ (k −1)(n+1)+1, k ∈ N, then there exist k subsets X1, . . . , Xk ⊂ X with Xi∩Xj = ∅,i 6= j, but convX1 ∩ convX2 ∩ · · · ∩ convXk 6= ∅.

8Helge Arnulf Tverberg, 1935–

http://en.wikipedia.org/wiki/Helge_Tverberg

The multifaceted world of polytopes 21

3 The multifaceted world of polytopes

3.1 Definition [Polyhedron]. The intersection of finitely many closed halfs-paces is called a polyhedron.

3.2 Theorem [Minkowski, Weyl]. 9,10

i) A bounded polyhedron is a polytope.

ii) A polytope is a bounded polyhedron.

Proof. i) Let P = {x ∈ Rn : 〈ai,x〉 ≤ αi, 1 ≤ i ≤ m} be bounded. We proceedby induction on n. The case n = 1 is obvious. Hence, we may assume n ≥ 2,and let Fi = P ∩H(ai, αi), 1 ≤ i ≤ m. Fi is a bounded polyhedron in an affinespace of dimension at most n − 1 and by our inductive argument there existfinite sets Vi such that Fi = conv Vi, 1 ≤ i ≤ m. It suffices to show that

P = conv (V1 ∪ V2 ∪ · · · ∪ Vm) .

The inclusion “⊇” follows from Vi ⊂ P and the convexity of P . For the reverseinclusion let x ∈ Pand let l be a line passing through x. The intersection l∩Pis a non-empty compact convex set of dimension at most 1. Hence we can findy, z ∈ P such that l ∩ P = conv {y, z}. Since both, y and z, has to lie in theboundary of P we can find k and j with y ∈ P ∩ H(ak, αk) = Fk = conv Vkand z ∈ P ∩H(aj , αj) = Fj = conv Vj . So we have

x ∈ conv {y, z} ⊆ conv (Vk ∪ Vj) ⊆ conv (V1 ∪ V2 ∪ · · · ∪ Vm).

For the second statement ii) we apply polarity to i). Let P = conv {v1, . . . ,vm}, and here we may assume that dimP = n and 0 ∈ intP . By Propositions1.23 ii) and 1.24 i) we find that P ? is a bounded polyhedron. Applying i) to P ?

we can find points w1, . . . ,wl ∈ Rn such that P ? = conv {w1, . . . ,wl}. Nextwe consider (P ?)?, which can be written as (cf. Proposition 1.24 i))

(P ?)? = {x ∈ Rn : 〈wi,x〉 ≤ 1, 1 ≤ i ≤ l}.

By Lemma 1.25 we know P = (P ?)? and we are done. �

3.3 Notation [V-Polytope, H-Polytope]. A polytope given as the convex hullof finitely many points is called a V-polytope. If it is given as the boundedintersection of finitely many closed halfspaces, then it is called an H-polytope.

3.4 Corollary. Let P ∈ Pn.

i) Let A ∈ Rm×n and t ∈ Rm. Then AP + t is a polytope.

ii) Let U ⊂ Rn be an affine subspace. Then P ∩ U is a polytope.

9Hermann Minkowski, 1864–190910Hermann Klaus Hugo Weyl, 1885 – 1955

http://en.wikipedia.org/wiki/Hermann_Minkowski

http://en.wikipedia.org/wiki/Hermann_Weyl


Proof. For i) we may assume that P is a V-Polytope, i.e., P = conv {v1, . . . ,vm}for some points vi ∈ Rn. Hence AP + t = t + conv {Av1, . . . , Avm} =conv {Av1 + t, . . . , Avm + t} is again a polytope.

For ii) we regard P as an H-polytope, i.e., P = ∩mi=1H−(ai, αi) for suitable

hyperplanes H(ai, αi). Each affine subspace U ⊂ Rn can be written as inter-section of hyperplanes and thus as intersection of halfspaces. Hence let U =∩ki=1H

−(ui, µi), 1 ≤ i ≤ k, and so P ∩ U = ∩mi=1H−(ai, αi) ∩ ∩ki=1H

−(ui, µi).Since P is bounded, P∩U is a bounded polyhedron, i.e., a polytope by Theorem3.2 �

3.5 Definition [Faces]. Let K ∈ Cn be closed and let H be a supportinghyperplane of K. If j = dim(K ∩ H), then K ∩ H is called a j-face of K.Moreover, K itself is regarded as a (dimK)-face and the empty set ∅ as (−1)-face of K.

3.6 Notation [Vertices, edges, facets]. A 0-face of K ∈ Cn, K closed, iscalled an exposed point or in the case of a polytope a vertex, a 1-face of apolytope is called edge and a (dimK−1)-faceof K is called facet of K. K itselfand the empty set are called improper faces, whereas the remaining faces arecalled proper faces of K.

The set of all vertices of a polytope P is denoted by vertP .

3.7 Remark.

i) Let K ∈ Cn be closed. Every (relative) boundary point of K lies in asuitable j-face, 0 ≤ j ≤ dimK − 1 (cf. Corollary 1.12).

ii) Let K ∈ Cn, dimK = n. Let F be a facet of K and H a supportinghyperplane of K with F = K ∩H. Then H = aff F .

3.8 Proposition. Every face of a polytope is a polytope, and a polytope hasonly finitely many faces.

Proof. Let P = conv {v1, . . . ,vm} and let F = P ∩ H(a, α) be a face of F .By Proposition 1.3 we have P ∩H(a, α) = conv (H(a, α)∩{v1, . . . ,vm}) whichshows that F is a polytope.

In particular, we can write each face F of P as conv VF for a suitable subsetVF ⊆ {v1, . . . ,vm}. Since different faces have different affine hulls and sincethe affine hull of a j-face, j ∈ {0, . . . , n − 1}, is uniquely determined by j + 1affinely independent points we may bound the number of all proper faces by∑n−1

i=0

(mi+1

). �

3.9 Definition [f-vector]. For P ∈ Pn let fi(P ) be the number of i-faces ofP , −1 ≤ i ≤ dim P . Furthermore, let fi(P ) = 0 for dim P + 1 ≤ i ≤ n. Thevector f(P ) with entries fi(P ), −1 ≤ i ≤ n, is called the f -vector of P .

3.10 Remark.

i) Let Tn be an n-dimensional simplex. Then fi(Tn) =(n+1i+1

), i.e., any (i+1)

subset of the vertices are the vertices of an i-face.


ii) For any n-polytope P ∈ Pn we have∑n

i=−1 fi(P ) ≥ 2n+1 with equality ifand only if P is an n-simplex.

3.11 Lemma. Let P ∈ Pn.

i) v ∈ vertP can not be written as a convex combination of two other pointsof P , i.e., v /∈ conv

(P\{v}

).

ii) If P = convW then vertP ⊆W .

iii) P = conv (vertP ).

Proof. For i) let H(a, α) be a supporting plane of v. Then we have 〈a,v〉 = αand 〈a,x〉 < α for all x ∈ P \{v}. Hence we cannot write v = λx1 +(1−λ)x2

with xi ∈ P \ {v}, λ ∈ [0, 1].Statement ii) is a direct consequence of i).For iii) let W ⊂ Rn be a minimal (w.r.t. its cardinality) set with P =

convW . By ii) we know already vertP ⊆ W , and so for w ∈ W it remainsto show that w is a vertex of P . By the minimality of W we have w /∈conv (W \ {w}). By Theorem 1.13 there exists a strong separation hyperplaneH(a, α) of w and conv (W \ {w}), and let 〈a,w〉 > α and 〈a,w〉 < α for allw ∈ conv (W \ {w}). With α∗ = 〈a,w〉 this implies firstly P = convW ⊂H−(a, α∗) and secondly (cf. Proposition (1.3))

P ∩H(a, α∗) = conv (W ∩H(a, α∗)) = {w}.

Hence w ∈ vertP . �

3.12 Lemma. Let P ∈ Pn be an n-polytope with 0 ∈ intP . For a proper faceF of P let

F � = {y ∈ P ? : 〈x,y〉 = 1 for all x ∈ F} .

Then

i) F � is a face of P ?.

ii) F = (F �)�.

iii) If G is a face of P and F ⊆ G, then G� ⊆ F �.

iv) dimF � = n− 1− dimF .

Proof. First we fix some notation. Let vertP = {v1, . . . ,vm}. Then P =conv (vertP ) (Lemma 3.11) and P ? = {y ∈ Rn : 〈vi,y〉 ≤ 1, 1 ≤ i ≤ m}(Proposition 1.24). We may assume dimF = k, F = conv {v1, . . . ,vl}, l ∈{k+ 1, . . . ,m}, v1, . . . ,vk+1 are affinely independent. Moreover, let H(a, 1) bea supporting hyperplane of F , i.e., F = {x ∈ P : 〈a,x〉 = 1} and 〈a,x〉 ≤ 1for all x ∈ P . Hence a ∈ P ? and, in particular, a ∈ F � and 〈a,vi〉 < 1 forl + 1 ≤ i ≤ m.


For i) we observe that we may write F � = {y ∈ P ? : 〈vi,y〉 = 1, 1 ≤ i ≤ l}.Since 〈vi,y〉 ≤ 1 for all y ∈ P ? we conclude

F � = {y ∈ P ? : 〈v1 + v2 + · · ·+ vl,y〉 = l}

and P ? ⊂ {y ∈ Rn : 〈v1 + v2 + · · ·+ vl,y〉 ≤ l}. Hence F � is a face of P ?

which shows i).By definition we have (F �)� = {x ∈ P : 〈y,x〉 = 1 for all y ∈ F �} and so

vi ∈ (F �)�, 1 ≤ i ≤ l, which implies F ⊆ (F �)�. For the reverse inclusion werecall that a ∈ F � and so (F �)� ⊆ {x ∈ P : 〈a,x〉 = 1} = F .

iii) is obvious. For iv) let

U = {y ∈ Rn : 〈vi,y〉 = 0, 1 ≤ i ≤ l} = {y ∈ Rn : 〈vi,y〉 = 0, 1 ≤ i ≤ k + 1}.

Then we have F � = (a + U) ∩ P ?. Since the vectors v1, . . . ,vk+1 are, in fact,linearly independent, because otherwise 0 ∈ aff F contradicting 0 ∈ intP , wehave dimU = n− (k+ 1) and so dimF � ≤ n− (k+ 1). Now let z ∈ U arbitraryand for µ ∈ R we consider a+ µ z. Then we have

〈vi,a+ µz〉 =

{1, 1 ≤ i ≤ l,〈vi,a〉+ µ 〈vi, z〉 , i > l.

Since 〈vi,a〉 < 1 for i > l we can find and εz > 0 such that a+ µ z ∈ P ?, andthus in F � for all µ < |εz|. Since z ∈ U was arbitrary we conclude dimF � ≥n− (k + 1). �

3.13 Theorem. Let P ∈ Pn be an n-polytope with 0 ∈ int P . Then

fn−1−i(P?) = fi(P ), −1 ≤ i ≤ n.

Proof. For i ∈ {−1, n} there is nothing to prove. By Lemma 3.12 we canassociate to every i-face F injevtively a dual n− (i+ 1)-face F � of P ?. Hencefn−i−1(P ?) ≥ fi(P ). Applying the same argument to the faces of P ? yields theassertion. �

3.14 Theorem. Let P ∈ Pn be an n-polytope with facets F1, . . . , Fm and letH(ai, αi), 1 ≤ i ≤ m, be the supporting hyperplanes of Fi, 1 ≤ i ≤ m. Then

P =

m⋂i=1

H−(ai, αi).

Proof. We may assume that 0 ∈ intP . In view of Theorem 3.13, P ? has m ver-tices v1, . . . ,vm, say, and by Lemma 3.11 iii) we have P ? = conv {v1, . . . ,vm}.By Lemma 1.25 and Proposition 1.24 we also have

P = P ?? = {x ∈ Rn : 〈vi,x〉 ≤ 1, 1 ≤ i ≤ n} =m⋂i=1

H−(vi, 1),

and according to Lemma 3.12, {vi}� = {x ∈ P : 〈vi,x〉 = 1} = P ∩H(vi, 1) isan (n − 1)-face of P . Hence, after an appropriate renumbering we must haveH(ai, αi) = H(vi, 1), 1 ≤ i ≤ m. �


3.15 Theorem. Let P ∈ Pn be an n-polytope.

i) The boundary of P is the union of all its facets.

ii) A k-face is the intersection of (at least) (n− k) facets.

iii) An (n− 2)-face is contained in exactly two facets.

iv) If F,G are faces of P with F ⊆ G, then F is a face of G.

v) A face of P is also a face of a facet of P .

Proof. We may assume that 0 ∈ intP . According to Theorem 3.14, eachboundary point of P is contained in a supporting hyperplane of a facet, whichshows i).

For ii) let F be a k-face of P , let P ? = conv {v1, . . . ,vm} with vi vertex of P ?

and let F � = conv {v1, . . . ,vl}. By Lemma 3.12 iv) we have dimF � = n−(k+1)and so l ≥ n− k. Moreover, with Lemma 3.12 ii) we also have

F = (F �)� = {x ∈ P : 〈vi,x〉 = 1, 1 ≤ i ≤ l} =

l⋂i=1

{x ∈ P : 〈vi,x〉 = 1}.

Since vi is a vertex of P ?, {vi}� = {x ∈ P : 〈vi,x〉 = 1} is a facet of P , whichyields ii). In particular, an (n− 2)-face F of P can be written as

F =r⋂i=1

{x ∈ P : 〈vi,x〉 = 1}

for some r ≥ 2. By Lemma 3.12, F � = conv {v1, . . . ,vr} is a 1-face of P ?. Sincevi are vertices we conclude r = 2, because otherwise a vertex could be writtenas a convex combination of other vertices, contradicting Lemma 3.11 i).

For iv) let H(a, α) be a supporting hyperplane of F , i.e., F = P ∩H(a, α).Then F = G ∩ F = G ∩ P ∩H(a, α) = G ∩H(a, α), which shows that F is aface of G.

Finally, we observe that ii) shows that any face is contained in a facet andthus, by iv), it is a face of that facet. �

3.16 Theorem. Let P ∈ Pn be an n-polytope.

i) Let G be a face of P and let F be a face of G. Then F is a face of P .

ii) Let Fj be a j-face of P and let Fk be a k-face of P with Fj ⊂ Fk. Thereexist i-faces Fi of P , j < i < k, such that

Fj ⊂ Fj+1 ⊂ · · · ⊂ Fk−1 ⊂ Fk.


Proof. For i) let F and G be proper faces and let 0 ∈ F ⊂ G. Let H(aF , 0)and H(aG, 0) be supporting hyperplanes of F and G, respectively, i.e., F =H(aF , 0) ∩G, G = H(aG, 0) ∩ P and G ⊂ H(aF , 0)−, P ⊂ H(aG, 0)−.

For µ ≥ 0 we consider the Hyperplane H(aF + µaG, 0) and observe that

〈aF + µaG, x〉 = 〈aF , x〉+ µ 〈aG,x〉

{= 0, for x ∈ F,< 0, for x ∈ G \ F.

Since for all x ∈ P \G we have 〈aG,x〉 < 0 there exists a µ > 0 such that forall v ∈ (vertP ) \G we have 〈aF + µaG,v〉 < 0. Hence for x ∈ P which we canwrite as a convex combination x =

∑v∈(vertP )\G λv v+

∑v∈((vertP )∩G)\F λv v+∑

v∈(vertP )∩F λv v we find

〈aF + µaG,x〉 =∑

v∈(vertP )\G

λv 〈aF + µaG,v〉

+∑

v∈((vertP )∩G)\F

λv 〈aF + µaG,v〉+∑

v∈(vertP )∩F

λv 〈aF + µaG,v〉

≤ 0,

with equality if and only if x ∈ F . Hence F is a face of P .In order to verify ii) we may assume j ≤ k − 2 and we first note that

by Theorem 3.15 iv) Fj is a face of Fk. According to Theorem 3.15 v) Fj iscontained in a facet G, say, of Fk, i.e., G is a (k− 1) face of FK and hence it isa face of P by i). With Fk−1 = G we have shown Fj ⊂ Fk−1 ⊂ Fk. In the casej < k − 2 we can apply the same reasoning as above to the pair Fj , Fk−1, andobtain so recursively the desired chain of faces. �

3.17 Remark. Let v0 be a vertex of an n-polytope P and let {v1, . . . ,vr} beall adjacent vertices of v0, i.e., conv {v0,vi} is an edge of P . In other words,{v1, . . . ,vr} are the neighbours of v0. Then

i)P ⊂ v0 + pos {v1 − v0, . . . ,vr − v0}.

ii) Let c ∈ Rn with 〈c,v0〉 ≥ 〈c,vi〉, 1 ≤ i ≤ r. Then

max{〈c,x〉 : x ∈ P} = 〈c,v0〉 .

3.18 Theorem [Euler-Poincare formula]. 11 12 Let P ∈ Pn. Then

n∑i=−1

(−1)i fi(P ) = 0. (3.18.1)

In particular, in the 3-dimensional case, i.e., dimP = 3, it holds f0−f1+f2 = 2.

11Leonhard Euler,1707–178312Henri Poincare, 1854–1912

http://en.wikipedia.org/wiki/Leonhard_Euler

http://en.wikipedia.org/wiki/Henri_Poincar�


Proof. It suffices to consider n-dimensional polytopes and we proceed byinduction with respect to n. In the case n = 1 there is nothing to prove as it is−1 + f0(P )− 1 = 0. Let n ≥ 2, m = f0(P ) and let a ∈ Rn \ {0} be chosen suchthat for any α ∈ R the hyperplane H(a, α) contains at most one vertex P .

Let α1 < α2 < · · · < α2m−1 such that the “odd” hyperplanes H2 i−1 :=H(a, α2 i−1), i = 1, . . . ,m, contains a vertex of P . The “even” hyperplanesH2 i := H(a, α2 i), i = 1, . . . ,m− 1, then, do not contain vertices, and H1 andH2m−1 are supporting hyperplanes of P . Hence, for i = 2, . . . , 2m − 2 thepolytopes Pi := Hi ∩P are (n− 1)-dimensional polytopes. For a j-face F of P ,j ≥ 1, and a polytope Pi we set

ψ(F, Pi) =

{1, Pi ∩ relintF 6= ∅,0, Pi ∩ relintF = ∅.

Observe, since F 6⊂ Hi, ψ(F, Pi) = 1 implies that F ∩Hi is a (j − 1) face of P .The first index i1 and the last index i2 of a hyperplane Hi with Hi ∩F 6= ∅ hasto be odd, and ψ(F, Pi) = 1 if and only if i ∈ {i1 + 1, . . . , i2 − 1}. Hence,

2m−2∑i=2

(−1)iψ(F, Pi) = 1.

Summing over all j-faces yields

n−1∑j=1

(−1)j∑

F j-face

2m−2∑i=2

(−1)iψ(F, Pi) =

n−1∑j=1

(−1)jfj(P ).

Changing the order of summation on the left hand side gives

n−1∑j=1

(−1)jfj(P ) =

2m−2∑i=2

(−1)i

n−1∑j=1

(−1)j∑

F j-face

ψ(F, Pi)

, (3.18.2)

and next we evaluate the interior sum on the right hand side.For an even index i or for j ≥ 2 each (j − 1)-face F of such a Pi is the

intersection of a j-face F of P with Hi. For odd i and j = 1 a vertex of Pi iseither the unique vertex of P lying in Hi or the intersection of an edge of Pwith Hi. Hence for j ≥ 1 we have

∑F j-face

ψ(F, Pi) =

{f0(Pi)− 1, j = 1 and i odd,

fj−1(Pi), otherwise.

By our inductive argument we get

n−1∑j=1

(−1)j∑

F j-face

ψ(F, Pi)

=

{1 +

∑n−1j=1 (−1)jfj−1(Pi) = 1−

∑n−2j=0 (−1)jfj(Pi) = (−1)n−1, i odd,∑n−1

j=1 (−1)jfj−1(Pi) = −∑n−2

j=0 (−1)jfj(Pi) = (−1)n−1 − 1, i even.


Summing over i gives

2m−2∑i=2

(−1)i

n−1∑j=1

(−1)j∑

F j-face

ψ(F, Pi)

= (−1)n−1 − (m− 1)

= −(−1)−1f−1(P )− (−1)nfn(P )− (−1)0 f0(P ),

which proves the theorem by (3.18.2). �

3.19 Proposition. The Euler-Poincare formula is the only linear equation sat-isfied by the f -vector, i.e., let λi ∈ R, such that

∑ni=−1 λi fi(P ) = 0 for all

P ∈ Pn. Then there exists a constant γ ∈ R, such that λi = γ (−1)i.

Proof. Let∑n

i=−1 λi fi(P ) = 0 be a linear equation which holds for anyP ∈ Pn. Taking a k-simplex Tk, k ∈ {0, . . . , n}, we obtain (cf. Remark 3.10)

n∑i=−1

λi

(k + 1

i+ 1

)= 0, k = 0, . . . , n.

The (n + 1) × (n + 2) matrix A with coefficients ak,i =(k+1i+1

)has rang n + 1

and hence any solution λ ∈ Rn+2 of the homogeneous system Aλ = 0 must bea multiple of the coefficients given by the Euler-Poincare identity. �

3.20 Definition [Simple and simplicial polytopes]. Let P ∈ Pn.

i) P is called simplicial if all proper faces are simplices.

ii) P is called simple if every vertex is contained in exactly dimP manyfacets.

3.21 Lemma. Let P ∈ Pn be an n-polytope with 0 ∈ intP . The followingstatements are equivalent:

i) P is simplicial.

ii) All facets of P are simplices.

iii) P ? is simple.

iv) Every k-face of P ? is contained in exactly n−k facets for k = 0, . . . , n−1.

Proof. We recall that by polarity we have P = conv {v1, . . . ,vm} with vivertex of P if and only if P ? = {y ∈ Rn : 〈vi,y〉 ≤ 1, 1 ≤ i ≤ m} withfacets {vi}� = P ? ∩ H(vi, 1), 1 ≤ i ≤ m. Moreover, F = conv {vi1 , . . . ,vik}is an l-face of P with vertices vi1 , . . . ,vik if and only if F � = {y ∈ P ? : y ∈H(vij , 1), j = 1, . . . , k} is an n − l − 1-face of P ? contained only in the facets{vij}� of P ?, j = 1, . . . , k.”i)⇔iv)”: Let F � be a k-face of P ?. Then (F �)� is an n− k− 1 face of P , thusan n− k − 1 simplex with n− k vertices. Hence by the foregoing remark F � is


contained in exactly n − k facets. On the other hand let F be a k-face of P .Then F � is an n− k − 1 of P ? contained in exactly k + 1 facets, and thus thek-face F has exactly k + 1 vertices and it is a simplex.”ii)⇔iii)”: Same argumentation as before where ”⇒” is the case k = 0 and”⇐” is the case k = n− 1.”i)⇔ii)”: Follows from the fact that every proper face of a polytope is a face offacet (see Theorem 3.15 v)). �

3.22 Theorem. Let P ∈ Pn be a simple n-polytope. Then

i) Every vertex is contained in exactly(nk

)k-faces of P , k = 0, . . . , n− 1.

ii) The intersection of k facets containing a common vertex is an (n−k)-faceof P .

iii) Let v1, . . . ,vn be the neighbours of a vertex v0 of P . For each subset ofk neighbours vi1 , . . . ,vik there exists a unique k-face F of P containingv0,vi1 , . . . ,vik .

iv) A face of a simple polytope is simple.

v) Every j face of P is contained in exactly(n−jk−j)k faces of P .

Proof. Without loss of generality we may assume that 0 ∈ intP and by Lemma3.21 we know that P ? is simplicial.i) Let v be a vertex of P and let F be a k-face of P . Then we have {v} ⊆ Fif and only if F � ⊆ {v}� is a (n − k − 1)-face of P ?. Now {v}� is facet of thesimplicial polytope P ? and so it has exactly

(n

n−k)

many (n−k−1)-dimensionalfaces.ii) Let v be a vertex of P . Since P is simple, v is contained in n-facets Fi =P ∩H(ai, 1), 1 ≤ i ≤ n, and we want to show that ∩ki=1Fi is an n−k-face fo P .Since P ? is simplicial, {v}� = conv {a1, . . . ,an} is an (n − 1)-simplex and soconv {a1, . . . ,ak} is a (k−1)-face of P ?. Hence conv {a1, . . . ,ak}� is an (n−k)face of P given by {x ∈ P : 〈ai,x〉 = 1, i = 1, . . . , k} = ∩ki=1Fi.iii) Let {v0}� = conv {w1, . . . ,wn} be a facet of P ?. For the edges (1-faces)conv {v0,vi} the associated polar face is an (n− 2)-face and so let

conv {v0,vi}� = conv ({w1, . . . ,wn} \ {wi}), 1 ≤ i ≤ n.

Since P ? is simplicial,

U = conv ({w1, . . . ,wn} \ {wi1 , . . . ,wik}) = ∩kj=1conv ({w1, . . . ,wn} \ {wij})

is an (n − k − 1)-face of P ?. Hence U� is a k face containing the edgesconv {v0,vij}, j = 1, . . . , k. For any k-face G of P containing these edgeswith have by polarity that G� ⊆ ∩kj=1conv ({w1, . . . ,wn} \ {wij}) = U . SincedimG� = dimU = n−k−1 we have G = U�. Thus U� is uniquely determined.iv) Let F be a k-face of P , k ∈ {1, . . . , n − 1}, and let v be a vertex of F .We want to show that v is contained in exactly k-facets of F , i.e., (k− 1)-faces


of P contained in F . Now G ⊂ F is a (k − 1)-face containing v if and only ifG� is a (n − k)-face of P ? with F � ⊂ G� ⊂ {v}�. Now P ? is simplicial andso we may assume that {v}� = conv {w1, . . . ,wn} and F � = {w1, . . . ,wn−k}.Hence there are exactly k many k-faces G� with the required property,namelyF � ∪ {wj}, j = n− k + 1 dots, n.iv) left as an exercise. �

3.23 Theorem. Let P ∈ Pn be a simple n-polytope.

i) n f0(P ) = 2 f1(P ).

ii)∑n

k=0 fk(P ) ≤ 2nf0(P ).

iii) f0(P ) ≤ 2fdn/2e(P ).

Proof. For i) we note that every edge contains exactly 2 vertices and everyvertex is contained in exactly n edges.

By Theorem 3.22 i) every vertex is contained in exactly∑n

i=0

(nk

)= 2n faces

of P and each face has at least one vertex. This gives ii).For iii) we assume that P = conv {v1, . . . ,vm} with vertices vi and all

vertices have different last coordinate. For a fixed vertex v with its n neighborsv1, . . . ,vn, say, let

L(v) = {vi : 〈en,vi〉 < 〈en,v〉} and U(v) = {vi : 〈en,vi〉 > 〈en,v〉}.

Next we distinguish two cases depending on the cardinality of these sets.

a) #L(v) ≥ dn/2e. On account of Theorem 3.22 iii) each dn/2e subsetS ⊆ L(v) determines an unique dn/2e-face F of P containing the edgesconv {v,vi}, vi ∈ S. By Theorem 3.22 iv), F is simple and so conv {v,vi},vi ∈ S, are the only edges of F containing v. Thus, on account of Remark3.17, there exists an dn/2e face F of P with

〈en,v〉 = maxx∈F〈en,x〉 .

b) #U(v) ≥ dn/2e. In the same way we conclude that there exists an dn/2eface F of P with 〈en,v〉 = minx∈F 〈en,x〉.

Hence for each vertex v there exists an dn/2e face F of P such that v haseither the largest or smallest last coordinate among all points of F . Since eachdn/2e face contains a larest as well as a smallest vertex (with respect to thelast coordinate) we must have 2 fdn/2e(P ) ≥ f0(P ). �

3.24 Corollary. Let P be a simple n-polytope with m facets. Then

f0(P ) ≤ 2

(m

bn/2c

)and

n∑k=0

fk(P ) ≤ 2n+1

(m

bn/2c

).

Or equivalently: Let P be a simplicial n-polytope with m vertices. Then

fn−1(P ) ≤ 2

(m

bn/2c

)and

n∑k=0

fk(P ) ≤ 2n+1

(m

bn/2c

).


Proof. First we note that the statement for simplical polytopes follows by”polarity”. For any polytope with m facets we certainly have fk(P ) ≤

(mn−k),

k = 0, . . . , n− 1 (cf. Theorem 3.15 ii)), and, in particular, fdn/2e(P ) ≤(

mbn/2c

).

Hence with Theorem 3.23 iii) we obtain f0(P ) ≤ 2(

mbn/2c

)which also gives the

second inequality by Theorem 3.23 ii) �

3.25 Lemma*. Let P be an n-polytope.

i) There exists a simple n-polytope Q with the same number of facets as Pand fi(P ) ≤ fi(Q), 0 ≤ i ≤ n− 2.

ii) There exists a simplical n-polytope Q? with the same number of verticesas P and fi(P ) ≤ fi(Q?), 1 ≤ i ≤ n− 1.

3.26 Corollary. Let P be an n-polytope with m facets. Then

f0(P ) ≤ 2

(m

bn/2c

).

Or equivalently: Let P be an n-polytope with m vertices. Then

fn−1(P ) ≤ 2

(m

bn/2c

).

3.27 Definition [Cyclic polytopes]. The curve γ : R → Rn given by γ(t) =(t, t2, t3, . . . , tn)ᵀ is called moment curve. The convex hull of m points on themoment curve is called a cyclic polytope with m vertices and is denoted byC(n,m).

3.28 Proposition. Any n + 1 points on the moment curve are affinely inde-pendent. In particular, cyclic polytopes are simplicial polytopes.

Proof. Suppose γ(t1), . . . , γ(tn+1), ti 6= tj , are affinely dependent. Then theyare contained in a hyperplane H(a, a0), a 6= 0, say. Hence 〈a, γ(ti)〉 − a0 = 0,or in coordinates

n∑j=1

aj(ti)j − a0 = 0, 1 ≤ i ≤ n+ 1,

contradicting the fact that a polynomial has at most degree many roots.

�


3.29 Theorem* [McMullen’s Upper Bound Theorem, 1971]. 13 Let P be ann-polytope with m vertices. Then

fi(P ) ≤ fi(C(n,m)) =

{∑(n−1)/2j=0

i+2m−j

(m−jj+1

)(j+1i+1−j

), n odd,∑n/2

j=1mm−j

(m−jj

)(j

i+1−j), n even.

In particular,

fn−1(P ) ≤ fn−1(C(n,m)) =

{2(m−bn/2c−1

bn/2c), n odd,(m−bn/2c

bn/2c)

+(m−bn/2c−1bn/2c−1

), n even.

For fixed n the right hand sides are of order mbn/2c.

3.30 Theorem* [Barnette’s Lower Bound Theorem, 1973]. 14 Let P be asimplicial n-polytope with m vertices. P has at least as many i-faces as the socalled stacked polytopes P (n,m) with m vertices for which

fi(P (n,m)) =

{m(ni

)− i(n+1i+1

), 0 ≤ i ≤ n− 2,

n+ 1 + (m− (n+ 1))(n− 1), i = n− 1.

P (n, n+ 1) is an n-simplex, and for m ≥ n+ 2 an m-vertex stacked n-polytopeP (n,m) is the convex hull of an (m− 1)-vertex stacked polytope with an addi-tional point that is beyond exactly one facet.

3.31 Remark. For any n-polytope P ∈ Pn we have nf0(P ) ≤ 2 f1(P ) withequality iff P simple and n fn−1(P ) ≤ 2 fn−2(P ) with equality iff P simplicial.

3.32 Theorem [Steinitz, 1906]. 15 A non-negative integral vector (f0, f1, f2)is the f -vector of a 3-polytope if and only if i) f0 − f1 + f2 = 2, ii) 3 f0 ≤ 2 f1,and iii) 3 f2 ≤ 2f1.

Proof. Equation i) is the Euler-Poincare formula (3.18.1) for polytopes in R3,ii) describes the fact that every vertex is contained in at least 3 edges and eachedge has exactly two vertices, and iii) is just the polar version of ii).

13Peter McMullen, born 194214David W. Barnette15Ernst Steinitz, 1871 – 1928

http://de.wikipedia.org/wiki/Peter_McMullen

http://en.wikipedia.org/wiki/Ernst_Steinitz


For the sufficiency part we have to construct a polytope with f0 verticesand f2 facets, where the non-negative integers f0, f2 satisfy

2 f0 − f2 − 4, 2f2 − f0 − 4 ≥ 0.

These inequalities are obtained by using i) in ii) and iii). Since their differenceis divisible by 3 they have the same reminder r, say, on division by 3. Thusthere exist c, s ∈ N such that 2 f0 − f2 − 4 = 3 c+ r and 2 f2 − f0 − 4 = 3 s+ r.Hence we have

f0 = (4 + r) + 2 c+ s and f2 = (4 + r) + c+ 2 s. (3.32.1)

Let P0 be a pyramid whose basis is a r+ 3-gon. Then f0(P0) = f2(P0) = r+ 4,and each facet containing the top vertex is a triangle, and each vertex of thebasis is contained in exactly three edges, which we call a simplex vertex. Nowwe apply two operations to P0, namely cutting of simple vertices and stackingover triangular faces.

If we cut off a simple vertex, then we obtain a polytope P1, say, withf0(P1) = f0(P0) + 2 and f2(P1) = f2(P0) + 1. Moreover, the three new verticesof P1 form a triangular face and all of them are simple. Hence, if we continuethis process of cutting of simple vertices (in a proper way) we obtain after csteps a polytope Pc with f0(Pc) = (4 + r) + 2 c and f2(Pc) = (4 + r) + c.

On the other hand, if we replace a triangular face of Pc by a suitable simplexthen we obtain a polytope Pc+1 with f0(Pc+1) = f0(Pc) + 1 and f2(Pc+1) =f2(P0)+2. Repeating this stacking process of triangular faces s-times, we finallyarrive at a polytope Pc+s with f0(Pc+s) and f2(Pc+s) as desired in (3.32.1). �

3.33 Conjecture [Kalai, 1989]. 16 Let P ∈ Pn be a 0-symmetric n-polytope.Then

n∑i=0

fi(P ) ≥ 3n.

Here we have equality, for instance, for the cube Cn and its polar, the cross-poyltope C?n, or, more generally, for the class of Hanner-polytopes. In 2007 theconjecture has been verified for all n ≤ 4 (see http://front.math.ucdavis.

edu/0708.3661).

3.34 Theorem [Figiel, Lindenstrauss, Milman, 1977]. 171819 Let P ∈ Pn bea 0-symmetric n-polytope, i.e., P = −P . Then

ln(f0(P )) ln(fn−1(P )) ≥ 1

16n.

Proof. For the proof we need two facts from convexity. First, there exits anA ∈ GL(n,R) such that Bn ⊂ AP ⊂

√nBn . Hence, in our combinatorial

setting we may assume that

Bn ⊂ P ⊂√nBn. (3.34.1)

16Gil Kalai, born 195517Tadeusz Figiel18Joram Lindenstrauss, 1936–201219Vitali Milman, born 1939

http://front.math.ucdavis.edu/0708.3661


http://en.wikipedia.org/wiki/Gil_Kalai

http://en.wikipedia.org/wiki/Joram_Lindenstrauss

http://en.wikipedia.org/wiki/Vitali_Milman


The second fact concerns spherical caps C(v, ε), which for v ∈ Sn−1 and ε ∈[0, 1] are defined by C(v, ε) = {w ∈ Sn−1 : 〈v,w〉 ≥ ε}. If µn−1(·) denotes theHaar probability measure on Sn−1 then µn−1(C(v, ε)) ∈ [0, 1] measures howmuch of Sn−1 is covered by C(v, ε). Here we need

µn−1(C(v, ε)) ≤ e−nε2

2 . (3.34.2)

For the proof we set m1 = f0(P ) and m2 = fn−1(P ), and we also need a V-and an H-representation of P

P = conv {vi : 1 ≤ i ≤ m1} = {x ∈ Rn : 〈ai,x〉 ≤ 1, 1 ≤ i ≤ m2}.

Moreover, let εi = 2√

ln(mi)/√n, i = 1, 2.

Let Vε1 = ∪m1i=1C(vi/ |vi| , ε1) and Fε2 = ∪m2

i=1C(ai/ |ai| , ε2). Then by(3.34.2) and the choice of εi we have µn−1 (Vε1) , µn−1 (Fε2) ≤ 1/4, e.g.,

µn−1 (Vε1) ≤m1∑i=1

µn−1 (C(vi/ |vi| , ε1)) ≤ m1 e−nε212 =

1

m1≤ 1

4.

Thus we can find a c ∈ Sn−1 \ (Vε1 ∪ Fε2). Since c ∈ Sn−1 \ Vε1 we get with(3.34.1)

maxx∈P| 〈c,x〉 | = max

1≤i≤m1

| 〈c,vi〉 | ≤√n max

1≤i≤m1

| 〈c,vi/ |vi|〉 | <√n ε1 = 2

√ln(m1).

(3.34.3)In order to get a lower bound on maxx∈P 〈c,x〉 we observe that Bn ⊂ P implies|ai| ≤ 1, and since c ∈ Sn−1 \ Fε2 we obtain for 1 ≤ i ≤ m2

| 〈c,ai〉 | ≤ | 〈c,ai/ |ai|〉 | ≤ ε2.

Hence (1/ε2)c ∈ P which gives

maxx∈P| 〈c,x〉 | ≥ 1

ε2=

√n

2√

ln(m2).

Together with (3.34.3) the assertion is proved. �

3.35 Definition [Graph, combinatorial diameter]. Let P ⊂ Rn be a polyhe-dron.

i) The distances δP (v,w) between two vertices v,w ∈ P (or in G(P )) is theminimum length of an ”edge” path connecting v and w in G(P ).

ii) δ(P ) = max{δP (v,w) : v,w ∈ vertP} is called the (combinatorial) di-ameter of P .

3.36 Example. δ(Tn) = 1 = (n + 1) − n, δ(Cn) = n = 2n − n and δ(C?n) =2 ≤ 2n − n.


3.37 Definition. For integers n,m let

∆(n,m) = max {δ(P ) : P ⊂ Rn polyhedron, dimP = n and fn−1(P ) = m} .

3.38 Remark. In 1957 Hirsch 20conjectured ∆(n,m) ≤ m − n. It is knownthat

i) the conjecture is true if n ≤ 3 or m ≤ n + 5. For unbounded polyhedrathe conjecture is false, namely, for m ≥ 2n it is ∆(n,m) ≥ m−n+ bn/4c.(Klee&Walkup, 1961/1965),

ii) ∆(n,m) ≤ m 2n−3, (Barnette, 1969; Larman, 1970),

iii) Disproof of the Hirsch conjecture for polytopes by Francisco Santos, 2010,see http://front.math.ucdavis.edu/1006.2814

3.39 Theorem [Kalai, 1992; Kalai&Kleitman, 1992]. 21

∆(n,m) ≤ mlogn+2.

Proof. First we will establish the recurrence

∆(n,m) ≤ ∆(n− 1,m− 1) + 2∆(n, bm/2c) + 2 (3.39.1)

Let P be an n-dimensional polytope with m facets. For an edge-path ω of thepolytope let F (ω) be the set of facets of P which are incident with one of thevertices of F (ω), i.e., all facets which are visited on the path ω. The length ofa path ω is denoted |ω|.

For a vertex w of P let

kw = max

p : #

⋃ω path starting in w, |ω|≤p

F(ω)

≤ bm/2c .

Now let v,u be two vertices of P attaining the diameter of P . By the definitionof kv (and ku) we have #(∪ω starting in v,|ω|=kv+1F(ω)) > m/2 and hence thereexists a facet F of P which can be reached from v by a path of length at mostkv + 1 and from u by a path of length at most ku + 1. Thus we conclude

δ(P ) ≤ ∆(n− 1,m− 1) + kv + ku + 2,

20Warren M. Hirsch21Daniel J. Kleitman, born 1934


http://en.wikipedia.org/wiki/Hirsch_conjecture

http://en.wikipedia.org/wiki/Daniel_Kleitman


and it remains to show kv, ku ≤ ∆(n, bm/2c).To this end let Q be given by all facet defining inequalities corresponding

to facets of ∪ω starting in v,|ω|≤kvF(ω). Then we have P ⊂ Q and Q has at mostq ≤ bm/2c facets. Let w be a vertex of P with δ(v,w) = kv. Then v,w areverties of Q as well, and let ωQ be a shortest edge-path in Q joining v,w. Nextwe claim that

|ωQ| = kv. (3.39.2)

By definition we have |ωQ| ≤ kv and so suppose that |ωQ| < kv. Then ωQuses an edge which is not an edge of P . Let eQ be the first such edge on ωQ.Then this edge must be intersected by one of the facet defining hyperplanes ofP which are not in ∪ω starting in v,|ω|≤kvF(ω). Hence this facet can be reachedby a path in P of length ≤ kv which contradicts the choice of kv.

This shows (3.39.2) and so kv = |ωQ| ≤ ∆(n, q) ≤ ∆(n, bm/2c).Finally, in order to get the desired bound ∆(n,m) ≤ mlogn+2 from the

recursion (3.39.1) we refer to Lecture 3 of Gunter Ziegler’s book. �

http://www.amazon.de/Lectures-Polytopes-Graduate-Texts-Mathematics/dp/038794365X

The space of convex bodies 37

4 The space of convex bodies

4.1 Notation. In the following let Kn be the set of all convex bodies in Rn,i.e., the set of all non-empty convex compact sets in Rn.

4.2 Remark. Let Ki ∈ Kn and λi ∈ R, i = 1, . . . ,m. Then∑m

i=1 λiKi ∈ Kn.

4.3 Definition [Outer parallel body, Hausdorff distance]. 22

i) Let M ⊂ Rn be a non-empty compact set and ρ ∈ R≥0. The set M +ρBnis called the outer parallel body of M at distance ρ.

ii) Let M1, M2 ⊂ Rn be non-empty compact subsets. The Hausdorff distanceof M1 and M2 is defined as

δ(M1,M2) = min{ρ ≥ 0 : M1 ⊆M2 + ρBn and M2 ⊆M1 + ρBn}.

4.4 Remark. The Hausdorff-distance δ(·, ·) defines a metric on the space ofnon-empty compact subsets of Rn.

4.5 Definition [Convergent sequences of convex bodies]. A sequence of con-vex bodies Ki ∈ Kn, i ∈ N, is called convergent (to K ∈ Kn) iff there exists aK ∈ Kn such that limi→∞ δ(Ki,K) = 0; in this case we also write Ki → K orlimi→∞Ki = K.

4.6 Lemma. Let Ki ∈ Kn, i ∈ N, with Ki+1 ⊆ Ki and let K =⋂∞i=1Ki. Then

K ∈ Kn and Ki → K.

Proof. First we show that K 6= ∅. To this end let xi ∈ Ki, i ∈ N. SinceKi ⊆ K1 for all i we may assume that the sequence (xi)i∈N converges to a pointx? ∈ K1, say. Since for any j ∈ N the subsequence (xi)i≥j is contained in Kj

we also have x? ∈ Kj . Hence x? ∈ K.

In order to verify Ki → K we have to show that for any ρ > 0 thereexists an iρ ∈ N such that δ(Ki,K) ≤ ρ for i ≥ iρ. Obviously, the inclusionK ⊂ Ki + ρBn holds for any ρ > 0. So suppose that for a given ρ > 0 theredoes not exist an iρ with Ki ⊂ K+ ρBn for i ≥ iρ. Then Ki 6⊂ K+ ρBn for alli ∈ N, and therefore there exit points xi ∈ Ki \ (K + ρBn). As in the first partwe can assume that xi → x? ∈ K. Hence, |xi − x?| ≤ ρ for i sufficiently large,and so xi ∈ x? + ρBn ⊆ K + ρBn, which contradicts the choice of xi. �

4.7 Lemma. Let Ki ∈ Kn, i ∈ N, be a Cauchy23-sequence of convex bodies,i.e., ∀ε > 0 ∃mε ∈ N such that δ(Ki,Kj) < ε, ∀i, j ≥ mε. Then there exists aK ∈ Kn with Ki → K. In other words, the space Kn is complete.

22Felix Hausdorff, 1868–194223Augustin-Louis Cauchy, 1789-1857

http://en.wikipedia.org/wiki/Felix_Hausdorff

http://en.wikipedia.org/wiki/Augustin-Louis_Cauchy


Proof. Let Lj = cl conv (∪i≥jKi). Since (Ki)i∈N is a Cauchy-sequence, the setcl ∪i≥j Ki is bounded and hence compact. Thus Lj ∈ Kn. Since Lj+1 ⊆ Lj wemay apply Lemma 4.6, and get Lj → K = ∩j≥1Lj . In view of the definition ofLj we conclude that for any ρ > 0 and j ≥ jρ, say,

Kj ⊆ Lj ⊂ K + ρBn.

On the other hand, since (Ki)i∈N is a Cauchy-sequence we have Ki ⊂ Kj +ρBnfor all i ≥ j ≥ mρ, and so

K ⊆ Lj ⊂ Kj + ρBn.

Thus δ(Kj ,K) ≤ ρ for all j ≥ max{jρ,mρ}, which means Kj → K. �

4.8 Definition [Bounded sequences]. A sequence of convex bodies Ki ∈ Kn,i ∈ N, is called bounded, if there exists an γ ∈ R>0 such that Ki ⊆ γ Bn for alli ∈ N.

4.9 Theorem [Selection theorem of Blaschke, 1916]. 24 A bounded sequenceof convex bodies Ki ∈ Kn, i ∈ N, contains a convergent subsequence.

Proof. On account of Lemma 4.7 it suffices to find a Cauchy-subsequencewithin the sequence (Ki)i∈N. After a suitable scaling and translation we mayassume (Ki)i∈N ⊂ [0, 1]n. For j ∈ N, the cube [0, 1]n can be subdivided into

2j n cubes of size [0, 2−j ]n, which we denote by W(j)k , k = 1, . . . , 2j n. Now we

construct for each j ∈ N a subsequence (K(j)i )i∈N in the following recursive way:

Let K(0)i = Ki, i ∈ N. For K

(j)i , j ≥ 1, we consider for each K

(j−1)i the cubes

W(j)k having a non-empty intersection with K

(j−1)i , i.e.,

U(j)i =

{W

(j)k : W

(j)k ∩K

(j−1)i 6= ∅

}.

Since there are only finitely many possible sets U(j)i , but infinitely many convex

bodies K(j−1)i , there exists a infinite family of convex bodies K

(j−1)i having the

same set U(j)l , say. This family forms our new sequence K

(j)i , i ∈ N. Since all

bodies K(j)i intersect the same cubes W

(j)k , and since the maximum distance

between two points in W(j)k is

√n 2−j we conclude

δ(K(j)r ,K(j)

s ) ≤√n2−j ,

for all r, s ∈ N. Finally, we consider the diagonal sequence of these sequences,

i.e., (K(i)i )i∈N. For m ≥ j we have K

(m)m ∈ (K

(j)i )i∈N and so

δ(K(m)m ,K

(j)j ) ≤

√n 2−j ,

which shows that (K(i)i )i∈N is a Cauchy-sequence. �

24Wilhelm Blaschke, 1885–1962

http://en.wikipedia.org/wiki/Wilhelm_Blaschke

The space of convex bodies 39

4.10 Theorem. Let K ∈ Kn and ρ > 0. Then there exists a polytope P ∈ Pnsuch that P ⊆ K ⊆ P + ρBn, in particular we have δ(K,P ) ≤ ρ.

Proof. Since ∪x∈K (x+ ρ int (Bn)) is an open covering of the compact set K,there exist finitely many x1, . . . , xm ∈ K, say, such that

K ⊆m⋃i=1

(xi + ρBn) ⊆ conv {x1, . . . , xm}︸︷︷︸P

+ρBn.

Since P ⊆ K the assertion is proved. �

4.11 Corollary. Let K ∈ Kn with 0 ∈ relintK and let λ > 1. Then thereexists a P ∈ Pn such that P ⊂ K ⊂ λP .

Proof. Without loss of generality we may assume dimK = n, and let r > 0such that r Bn ⊆ K. By Theorem 4.10 we can find for any ρ = r (λ − 1)/λ apolytope P such that P ⊆ K ⊆ P +ρBn. Since ρ < r, the last inclusion implies(r − ρ)Bn ⊆ P and thus

P ⊆ K ⊆ P +ρ

r − ρ(r − ρ)Bn ⊆

r

r − ρP = λP.

�


A glimpse on mixed volumes 41

5 A glimpse on mixed volumes

5.1 Definition [Volume, characteristic function]. Let M ⊆ Rn. The func-tion χM : Rn → {0, 1} given by

χM (x) =

{1, x ∈M,

0, x 6∈M,

is called characteristic function of M . If M is bounded and χM is Riemann25-integrable then

vol (M) =

∫RnχM (x) dx =

∫M

1 dx

is called the (n-dimensional) volume (Jordan 26-measure) of M . M is also calledRiemann or Jordan-measurable.

5.2 Remark.

i) The volume of a (rectangular) box B = [0, α1]× [0, α2]×· · ·× [0, αn] ⊂ Rnis given by vol (B) =

∏ni=1 αi.

ii) Compact convex sets are Jordan-measurable. If M ⊂ Rn is Jordan-measurable then vol (M) = limk→∞

1kn #

(M ∩ 1

kZn).

iii) Let M,M1,M2 ⊂ Rn, M1 ⊆M2, be Riemann-integrable. Then

a) vol (AM + t) = | detA|vol (M) for any A ∈ Rn×n with detA 6= 0,and t ∈ Rn.

b) vol (λM) = |λ|nvol (M) for any λ ∈ R.

c) vol (M1) ≤ vol (M2).

d) If dimM ≤ n− 1 then volM = 0.

5.3 Notation. Let K ⊂ Rn be compact and contained in a j-dimensionalaffine plane A. The j-dimensional volume of K with respect to A is denotedby vol j(K), i.e., vol j(K) =

∫A χK(x)dx, where here dx is the j-dimensional

volume element with respect to the space A.

5.4 Lemma [Cavalieri’s principle]. 27 Let K ⊂ Rn be compact and for t ∈ Rlet Kt = K ∩ {x ∈ Rn : xn = t}. Then

vol (K) =

∫R

vol n−1(Kt) dt.

25Bernhard Riemann, 1826–186626Camille Jordan, 1838–192227Bonaventura Cavalieri, 1598–1647

http://en.wikipedia.org/wiki/Bernhard_Riemann

http://en.wikipedia.org/wiki/Camille_Jordan

http://en.wikipedia.org/wiki/Bonaventura_Cavalieri


Proof. By Fubini’s theorem we may write

vol (K) =

∫R

(∫Rn−1

χK((x, t)ᵀ)dx

)dt =

∫R

(∫{x∈Rn:xn=t}

χKt(x)dx

)dt

=

∫R

vol n−1(Kt) dt.

�

5.5 Lemma [Pyramid, Prism]. Let Q ⊂ Rn be an (n − 1)-dimensional com-pact convex set.

i) Let P = conv {Q, s} be a pyramid, and let h be the distance between theapex s and aff Q. Then

vol (K) =h

nvol n−1(Q).

ii) Let P = conv {x+Q,y +Q} be a prism (i.e., conv {x,y} is not parallelto aff Q), and let h be the distance between x+aff Q and y+aff Q. Then

vol (P ) = h vol n−1(Q).

Proof. We may assume that Q ⊂ {x ∈ Rn : xn = 0}, s = (s, h)ᵀ with h > 0and s ∈ Rn−1. For µ ∈ R let Pµ = P ∩ {x ∈ Rn : xn = µ}. Clearly Pµ = ∅ forµ 6∈ [0, h], and for µ ∈ [0, h] it holds

Pµ =(

1− µ

h

)Q+

µ

hs,

With Lemma 5.4 we get

vol (P ) =

∫ h

0vol n−1(Pµ) dµ =

∫ h

0vol n−1

((1− µ

h

)Q+

µ

hs

)dµ

=

∫ h

0vol n−1

((1− µ

h

)Q

)dµ = vol n−1(Q)

∫ h

0

(1− µ

h

)n−1dµ

= −vol n−1(Q)h

n

(1− µ

h

)n]h0

=h

nvol n−1(Q).

Statement ii) for prims can be proven analogously, where here Pµ is always atranslate of Q or empty. �

5.6 Proposition.

i) If intK 6= ∅ then vol (K) > 0.


ii) Let K1,K2 ⊂ Rn be compact. Then

vol (K1 ∪K2) = vol (K1) + vol (K2)− vol (K1 ∩K2).

iii) Let K,M ∈ Kn with 〈x,y〉 = 0 for all x ∈ K, y ∈ M and let k = dimKand m = dimM . Then dim(K +M) = k +m and

vol k+m(K +M) = vol k(K)volm(M).

iv) Let W be an n-cube of edge length λ. Then vol (W ) = λn and, in partic-ular, vol

([−1, 1]n

)= 2n.

v) Let T =conv {v0,v1, . . . ,vn} be an n-simplex. Then

vol (T ) =1

n!

∣∣det(v1 − v0, . . . ,vn − v0)∣∣.

Proof.i) If intK 6= ∅ then there exist ε > 0 and t ∈ Rn such that t + [0, ε] ⊂ K.Hence, vol (K) ≥ vol (t+ [0, ε]) = εn.iii) It is a direct consequence of the corresponding identity for the characteristicfunctions χK1∪K2

= χK1+ χK2

− χK1∩K2.

iv) We may assume that n = k +m and that

K ⊂{

(x1, . . . , xn)ᵀ ∈ Rn : xk+1 = · · · = xn = 0}

M ⊂{

(x1, . . . , xn)ᵀ ∈ Rn : x1 = · · · = xk = 0}.

Then K + M ={

(x′,x′′)ᵀ : (x′,0)ᵀ ∈ K,x′ ∈ Rk, (0,x′′)ᵀ ∈ M,x′′ ∈ Rn−k}

and dim(K +M) = n.Now for x = (x1, . . . , xn)ᵀ we write x′ = (x1, . . . , xk)

ᵀ and x′′ = (xk+1, . . . , xn)ᵀ.Then χK+M (x) = χK((x′,0)ᵀ)χM ((0,x′′)ᵀ). By Fubini’s theorem we get

vol (K +M) =

∫RnχK+M (x) dx

=

∫Rn−k

(∫RkχK((x′,0)ᵀ)χM ((0,x′′)ᵀ) dx′

)dx′′

=

∫Rm

χM ((0,x′′)ᵀ)

(∫RkχK((x′,0)ᵀ) dx′

)dx′′

= vol k(K)volm(M).

iv) Follows by the volume formula for boxes.v) Let A = (v1 − v0, . . . ,vn − v0) and let T = conv {0, e1, . . . , en}. ThenT = AT + v0. Hence vol (T ) = | detA|vol (T ), and since T may be regardedas a pyramid with basis conv {0, e1, . . . , en−1} and apex en, we recursively findby Lemma 5.5

vol (T ) =1

n!.

Hence vol (T ) = |detA|/n!. �


5.7 Lemma. Let P ∈ Pn with 0 ∈ intP and let F1, . . . , Fm be the facets of Pwith outer normal vectors ui ∈ Rn, |ui| = 1. Then

vol (P ) =1

n

m∑i=1

vol n−1(Fi)h(P,ui).

Proof. Let Vi be the pyramids over the facets with apex 0, i.e., Vi =conv {0, Fi}, 1 ≤ i ≤ m. Then P =

⋃ni=1 Vi. Moreover, since Vi ∩ Vj =

conv {Fi ∩ Fj ,0} for 1 ≤ i, j ≤ m, we have dim(Vi ∩ Vj) ≤ n − 1 for i 6= j.Hence, in view Proposition 5.6 ii) we get vol (P ) =

∑mi=1 vol (Vi). The distance

of the apex 0 to the basis Fi of the pyramid Vi is h(P,ui) and with Lemma 5.5we get

vol (P ) =m∑i=1

vol (Vi) =m∑i=1

h(P,ui)

nvol n−1(Fi).

�

5.8 Corollary. Let P ∈ Pn, dimP = n, and let F1, . . . , Fm be the facets of Pwith outer normal vectors ui ∈ Rn, |ui| = 1. Then

m∑i=1

vol n−1(Fi)ui = 0.

Proof. We may assume that 0 ∈ intP . Let t =∑m

i=1 vol n−1(Fi)ui and letλ > 0 such that 0 ∈ int (λ t+ P ). By Lemma 5.7 we get

vol (P ) = vol (λ, t+ P ) =1

n

m∑i=1

vol n−1(Fi)h(λ t+ P,ui)

=1

n

m∑i=1

vol n−1(Fi)h(P,ui) +1

n

m∑i=1

vol n−1(Fi) 〈ui, λ t〉

= vol (P ) +λ

n〈t, t〉 .

Hence t = 0. �

5.9 Corollary. Let P ∈ Pn, dimP = n, and let F1, . . . , Fm be the facets of Pwith outer normal vectors ui ∈ Rn, |ui| = 1. Then

vol (P ) =1

n

m∑i=1

vol n−1(Fi)h(P,ui).

Proof. Let t ∈ Rn such that 0 ∈ int (t + P ). As in the proof of Corollary 5.8we find that

vol (P ) = vol (t+ P ) =1

n

m∑i=1

vol n−1(Fi)h(P, ui) +1

n

⟨m∑i=1

vol n−1(Fi)ui, t

⟩.

According to Corollary 5.8 the last summand is 0. �


5.10 Theorem* [Minkowski]. Let u1, . . . ,um ∈ Rn, |ui| = 1, such that 0 ∈int conv {u1, . . . ,um}. Let f1, . . . , fm ∈ R>0 be positive numbers such that∑m

i=1 fi ui = 0. Then there exists an unique (up to translations) n-polytopeP ∈ Pn with m facets F1, . . . , Fm such that vol n−1(Fi) = fi and u1, . . . ,umare the outer unit normal vectors of the facets.

5.11 Definition [Valuations on Kn]. A function f : Kn → R is called

i) monotonous, if K1 ⊆ K2 implies f(K1) ≤ f(K2),

ii) continuous, if Ki → K implies f(Ki)→ f(K),

iii) homogeneous of degree r if for λ > 0 it holds f(λK) = λrf(K),

iv) translation invariant, if for t ∈ Rn it holds f(t+K) = f(K),

v) rotation invariant, if for any rotation g : Rn → Rn it holds f(g(K)

)=

f(K),

vi) rigid motion invariant, if it is translation and rotation invariant,

vii) additive, if for K1,K2 ∈ Kn with K1 ∪K2 ∈ Kn it holds

f(K1 ∪K2) = f(K1) + f(K2)− f(K1 ∩K2),

viii) simple, if f(K) = 0 for dimK ≤ n− 1.

An additive functional is also called a valuation.

5.12 Lemma. vol : Kn → R≥0 is continuous.

Proof. Let (Ki)i∈N ⊂ Kn with limi→∞Ki = K, i.e., for each ρ > 0 there existsan iρ with

Ki ⊆ K + ρBn and K ⊆ Ki + ρBn for i ≥ iρ.We have to show limi→∞ vol (Ki) = vol (K). First we treat the case whendimK ≤ n − 1, and w.l.o.g. let K be contained in the hyperplane H = {x ∈Rn : xn = 0}. Then we can find an (n − 1)-dimensional polytope Q ⊂ H, say,such that (K + Bn) ∩ H ⊂ Q. For a positive number ρ let Pρ be the prismgiven by Pρ = conv {Q + ρ en, Q − ρ en}. The for ρ ≤ 1 and i ≥ iρ we haveKi ⊆ K + ρBn ⊂ Pρ, and thus (cf. Lemma 5.5 ii))

0 ≤ vol (Ki) ≤ vol (Pρ) = 2 ρ vol n−1(Q).

Hence limi→∞ vol (Ki) = vol (K). In the case dimK = n we may assume thatr Bn ⊆ K for some positive number r, and by Exercise ?? we obtain for i ≥ iρ(

1− ρ

r

)n≤ vol (Ki)

vol (K)≤(

1 +ρ

r

)n.

Thus vol (Ki)→ vol (K). �

5.13 Theorem. vol : Kn → R≥0 is a monotonous, continuous, rigid motioninvariant, simple valuation, which is homogeneous of degree n.

5.14 Theorem* [Hadwiger’s characterization of the volume]. 28 Let φ : Kn →28Hugo Hadwiger, 1908–1981

http://en.wikipedia.org/wiki/Hugo_Hadwiger


R be a continuous, rigid motion invariant and simple valuation. Then there ex-ists a constant c ∈ R such that φ(K) = c vol (K).

5.15 Definition [Homothetic bodies]. Two convex bodiesK,L ∈ Kn are calledhomothetic iff there exist µ ≥ 0 and t ∈ Rn such that K = t+ µL.

5.16 Theorem [Brunn-Minkowski’s Theorem]. 29 Let K,L ⊂ Rn be com-pact, and let 0 ≤ λ ≤ 1. Then

vol(λK + (1− λ)L

)1/n ≥ λ vol (K)1/n + (1− λ)vol (L)1/n, (5.16.1)

i.e., the n-th root of the volume is a concave function.Moreover, if K and L are convex then equality holds if and only if K and

L are homothetic or K and L lie in parallel hyperplanes.

Proof. [(without the equality case)] By the homogeneity of the volume theinequality is equivalent to

vol (K + L)1/n ≥ vol (K)1/n + vol (L)1/n. (5.16.2)

First we prove this inequality for the family of polyboxes, i.e., the union of afinite number of boxes with edges parallel to the coordinate axes and disjointinteriors, which we then use in order to approximate K and L.

Let P and Q be two polyboxes consisting of mP and mQ boxes, respectively.We proceed by induction on mP +mQ.

For mP + mQ = 2 we may assume P = a + ([0, α1] × · · · × [0, αn]), i.e., Phas edge length αi > 0 in direction ei, and let Q = b + ([0, β1] × · · · × [0, βn])for suitable a, b ∈ Rn. Then P +Q = a+ b+ ([0, α1 + β1]× · · · × [0, αn + βn]),and for (5.16.2) in this particular case we have to show

vol (P +Q)1/n =

(n∏i=1

(αi + βi)

)1/n

≥

(n∏i=1

αi

)1/n

+

(n∏i=1

βi

)1/n

= vol (P )1/n + vol (Q)1/n,

which is equivalent to(n∏i=1

αiαi + βi

)1/n

+

(n∏i=1

βiαi + βi

)1/n

≤ 1.

By the arithmetic-geometric mean inequality this is true.So let mP + mQ > 2 and we assume mp ≥ 2. We translate the boxes of P

such that two boxes P1, P2 of P lie on different sides of a coordinate hyperplaneH, i.e., P1 ⊂ H+ and P2 ⊂ H− and we may assume H = {x ∈ Rn : xn = 0}.Let P+ = P ∩ H+, P− = P ∩ H−. Then both, P+ and P− have less thanmp boxes. Next we translate the polybox Q along the direction en such that

29Hermann Brunn, 1862–1939

http://en.wikipedia.org/wiki/Hermann_Brunn


vol (Q+)/vol (Q) = vol (P+)/vol (P ), where Q+ = Q ∩ H+. Then with Q− =Q∩H− we also have vol (P )/vol (Q) = vol (P+)/vol (Q+) = vol (P−)/vol (Q−).

P+ +Q+ and P− +Q− have no common interior points and both of themhave less than mP +mQ boxes. Hence in view of our inductive argumentationwe get

vol (P +Q) = vol ((P +Q) ∩H+) + vol ((P +Q) ∩H−)

≥ vol (P+ +Q+) + vol (P− +Q−)

≥(

vol (P+)1/n + vol (Q+)1/n)n

+(

vol (P−)1/n + vol (Q−)1/n)n

= vol (P+)

(1 +

vol (Q+)1/n

vol (P+)1/n

)n+ vol (P−)

(1 +

vol (Q−)1/n

vol (P−)1/n

)n

= vol (P+)

(1 +

vol (Q)1/n

vol (P )1/n

)n+ vol (P−)

(1 +

vol (Q)1/n

vol (P )1/n

)n

= vol (P )

(1 +

vol (Q)1/n

vol (P )1/n

)n=(

vol (P )1/n + vol (Q)1/n)n.

Now given two Jordan measurable compact sets K and L we can approxi-mate their volume by the volume of two sequences of polyboxes Pm ⊂ K andQm ⊂ L, m ∈ N consisting fo small cubes, i.e., limm→∞ vol (Pm) = vol (K) andlimm→∞ vol (Qm) = vol (L). Since Pm +Qm ⊆ K + L we get

vol (K + L)1/n ≥ lim supm→∞

vol (Pm +Qm)1/n

≥ limm→∞

vol (Pm)1/n + limm→∞

vol (Qm)1/n

= vol (K)1/n + vol (L)1/n.

�

5.17 Corollary. Let K ∈ Kn with K = −K. For a k-dimensinal linear sub-space Lk ⊂ Rn, k ∈ {1, . . . , n−1}, and x ∈ Rn let v(Lk,x) = vol k(K∩(x+Lk)).Then

v(Lk,0) ≥ v(Lk,x)

for all x ∈ Rn, i.e., the volume maximal section of a o-symmetric convex bodywith a plane contains the origin.

Proof. LetK(x) = K∩(x+Lk). SinceK = −K it holdsK(x) = −K(−x), andso v(Lk,x) = v(Lk,−x). By the convexity we have (1/2)K(x)+(1/2)K(−x) ⊆K(0) and together with Theorem 5.16 we get

v(Lk,0) = vol k(K(0)) ≥ vol k

(1

2K(x) +

1

2K(−x)

)≥(

1

2vol k(K(x))

1k +

1

2vol k(K(−x))

1k

)k= vol k(K(x)) = v(Lk,x).

�


5.18 Proposition [Multiplicative version of the B-M inequality (5.16.1)].Let K,L ⊂ Rn be compact. Inequality (5.16.1) is equivalent to

vol(λK + (1− λ)L

)≥ vol (K)λ vol (L)1−λ for all λ ∈ [0, 1]. (5.18.1)

Proof. By the Brunn-Minkowski inequality (5.16.1) and the weighted arithmetic-geometric mean inequality we get for λ ∈ [0, 1]

vol(λK+(1−λ)L

)1/n ≥ λvol (K)1/n+(1−λ)vol (L)1/n ≥ vol (K)λ/nvol (L)(1−λ)/n.

For the reverse implication we may assume that vol (K), vol (L) 6= 0 and we set

λ =vol (K)1/n

vol (K)1/n + vol (L)1/n∈ (0, 1).

Then vol (K)/λn = vol (L)/(1 − λ)n = (vol (K)1/n + vol (L)1/n)n and with(5.18.1) we get

vol (K + L) = vol(λK

λ+ (1− λ)

L

1− λ)≥ vol

(K

λ

)λvol

(L

1− λ

)1−λ

=

(vol (K)

λn

)λ( vol (L)

(1− λ)n

)(1−λ)

=

(vol (K)

λn

)λ(vol (K)

λn

)(1−λ)

=

(vol (K)

λn

)= (vol (K)1/n + vol (L)1/n)n.

�

5.19 Proposition. LetKi ∈ Kn, λi ∈ R≥0, 1 ≤ i ≤ r, and letK =∑r

i=1 λiKi.Then h(K, ·) =

∑ri=1 λi h(Ki, ·) and for u ∈ Rn we have

K ∩H(u, h(K,u)

)=

r∑i=1

λi

[Ki ∩H

(u, h(Ki,u)

)].

Proof. Let u ∈ Rn \ {0}. Since Ki are compact 1 ≤ i ≤ r, there exist xi ∈Ki such that h(Ki,u) = 〈xi,u〉. Hence

∑ri=1 λi h(Ki,u) = 〈

∑ri=1 λixi,u〉 ≤

h(K,u). On the other hand, any x ∈ K can be written as x =∑r

i=1 λixi withxi ∈ Ki and so 〈x,u〉 =

∑ri=1 λi 〈xi,u〉 ≤

∑ri=1 λi h(Ki,u). Hence h(K,u) ≤∑r

i=1 λi h(Ki,u).For the second property we notice that x =

∑ri=1 λixi ∈ (K∩H

(u, h(K,u)

))

with xi ∈ Ki, if and only if∑r

i=1 λi 〈xi,u〉 = h(K,u) =∑r

i=1 λih(xi,u),which is equivalent to 〈xi,u〉 = h(Ki,u) for all i = 1, . . . , r, or in other wordsxi ∈ Ki ∩H

(u, h(Ki,u)

)for all i = 1, . . . , r. �

5.20 Theorem [Mixed volumes]. Let r ∈ N, λi ∈ R≥0 and Ki ∈ Kn, 1 ≤ i ≤r. There are non-negative coefficients V(Ki1 , . . . ,Kin), called mixed volumes,which are symmetric in the indices, such that

vol

(r∑i=1

λiKi

)=

r∑i1,...,in=1

λi1 · . . . · λinV(Ki1 , . . . ,Kin),


i.e., vol(∑

i λiKi

)is a homogeneous polynomial of degree n in the variables

λ1, . . . , λr. Moreover, V(Ki1 , . . . ,Kin) depends only on the bodies Ki1 , . . . ,Kin .

Proof. First we will prove the theorem for polytopes Pi ∈ Pn, 1 ≤ i ≤ r, byinduction on the dimension.

If n = 1 then Pi = [αi, βi] with αi, βi ∈ R, 1 ≤ i ≤ r. Hence∑r

i=1 λiPi =[∑ri=1 λ1αi,

∑ri=1 λ1βi

], and we get vol

(∑ri=1 λiPi

)=∑r

i=1 λivol (Pi).So let n ≥ 2, and we assume for a moment that 0 ∈ Pi, 1 ≤ i ≤ r, and

dimPr = n, λr > 0. Then P =∑r

i=1 λiPi is an n-dimensional polytope and wedefine

U(P ) ={u ∈ Sn−1 : dim

(P ∩H

(u, h(P,u)

))= n− 1

},

i.e., the set of outer normal vectors to the facets of P . For u ∈ U(P ) we writeFu(P ) = P ∩H

(u, h(P,u)

)and Fu(Pi) = Pi ∩H

(u, h(Pi,u). By Proposition

5.19 we get

h(P,u) =r∑i=1

λih(Pi,u) and Fu(P ) =r∑i=1

λiFu(Pi).

All Fu(Pi), 1 ≤ i ≤ r, lie in parallel hyperplanes and since the volume is invari-ant under translations we may apply our induction hypothesis to

∑ri=1 λiFu(Pi)

and obtain

vol n−1

(Fu(P )

)= vol n−1

(r∑i=1

λiFu(Pi)

)

=r∑

i1,...,in−1=1

λi1 · . . . · λin−1V(Fu(Pi1), . . . , Fu(Pin−1)

).

Together with Corollary 5.9 we get

vol (P ) =∑

u∈U(P )

h(P,u)

nvol n−1

(Fu(P )

)

=∑

u∈U(P )

(r∑

in=1

λinh(Pin ,u)

n

) r∑i1,...,in−1=1

λi1 · . . . · λin−1V(Fu(Pi1), . . . , Fu(Pin−1)

)=

r∑i1,...,in=1

λi1 · . . . · λinV(Pi1 , . . . , Pin),

where

V(Pi1 , . . . , Pin) =∑

u∈U(P )

h(Pin ,u)

nV(Fu(Pi1), . . . , Fu(Pin−1)

).

Since 0 ∈ Pi we have h(Pi,u) ≥ 0 and by our induction hypothesis we con-clude V(Pi1 , . . . , Pin) ≥ 0. This polynomial can also be rewritten as one withsymmetric non-negative coeeficients V(Pi1 , . . . , Pin) and so we know

vol

(r∑i=1

λiPi

)=

r∑i1,...,in=1

λi1 · . . . · λinV(Pi1 , . . . , Pin). (5.20.1)


Since the left hand side – and thus the right hand side – is invariant withrespect to translations of the Pi, (5.20.1) holds also if 0 /∈ Pi. Since the leftand right hand side are continuous in λr the equation also holds for λr = 0.If dimPi ≤ n − 1, 1 ≤ i ≤ r, then we choose an arbitrary n-polytope Pr+1,apply (5.20.1) to

∑r+1i=1 λiPr+1 and finally set λr+1 = 0 on both sides of (5.20.1).

Hence we have proven the theorem for arbitrary polytopes P1, . . . , Pr.Now let Ki ∈ Kn, 1 ≤ i ≤ r, and let P ji ∈ Pn with limj→∞ P

ji = Ki. For any

arbitrary (but fixed) λi ∈ R≥0 we have λiPji → λiKi and hence

∑ri=1 λi P

ji →∑r

i=1 λiKi. Therefore,

vol

(r∑i=1

λiKi

)= lim

j→∞vol

(r∑i=1

λiPji

)

= limj→∞

r∑i1,...,in=1

λi1 · . . . · λinV(P ji1 , . . . , Pjin

).

Observe that the coefficients V(P ji1 , . . . , Pjin

) are the same for any choice of

(fixed) scalars λ ≥ 0. V(P ji1 , . . . , Pjin

) is non-negative and by setting λi = 1,1 ≤ i ≤ r, we see that they are all bounded by vol (

∑iKi). Hence each sequence

(V(P ji1 , . . . , Pjin

))j∈N contains a convergent subsequence and so we may assume

that limj→∞V(P ji1 , . . . , Pjin

) = V(Ki1 , . . . ,Kin), say. Thus

vol

(r∑i=1

λiKi

)=

r∑i1,...,in=1

λi1 · . . . · λinV(Ki1 , . . . ,Kin) (5.20.2)

for any arbitrary (but fixed) λi ≥ 0, 1 ≤ i ≤ r. Finally we observe thatV(Ki1 , . . . ,Kin) are non-negative and by choosing the proper subsequences wemay also assume that these numbers V(Ki1 , . . . ,Kin) are symmetric in theindices.

Now let ji ∈ {1, . . . , r} for 1 ≤ i ≤ n and let J = {j1, . . . , jn}. Settingλi = 0 for i /∈ J in (5.20.2) gives

vol

(∑i∈J

λiKi

)=

r∑i1,...,in=1,ij∈J

λi1 · . . . · λinV(Ki1 , . . . ,Kin)

which shows that V(Kj1 , . . . ,Kjn) depends only on Kj1 , . . . ,Kjn . �

5.21 Notation. LetK1, . . . ,Kr ∈ Kn and let ki ∈ {0, . . . , n} with n =∑r

i=1 ki.

i) Instead of V(Ki1 , . . . ,Kin) we also write V(K1, k1; . . . ;Kr, kr), where theki are the multiplicities of the body Ki in the sequence (Ki1 , . . . ,Kin).

ii) Let(

nk1,...,kr

)be the multinomial coefficient, i.e.,

(n

k1,...,kr

)= n!

k1!···kr! .

5.22 Remark. With the notation above we may write

vol

(r∑i=1

λiKi

)=

∑k1+···+kr=n

(n

k1, . . . , kr

)λk11 · . . . · λ

krr V(K1, k1; . . . ;Kr, kr).


In particular, for r = 2 we get

vol (λ1K1 + λ2K2) =

n∑i=0

(n

i

)λi1λ

n−i2 V(K1, i;K2, n− i).

5.23 Theorem. Let K1, . . . ,Kn ∈ Kn.

i) V (Km, . . . ,Km) = vol (Km).

ii) Let A ∈ Rn×n, detA 6= 0, and let ti ∈ Rn, 1 ≤ i ≤ r. Then V(t1 +AK1, t2 +AK2, . . . , tn +AKn) = | detA|V(K1, . . . ,Kn).

iii) Mixed volumes are continuous functions on (Kn)n.

iv) Mixed volumes are linear in each argument, i.e., V(λK+µL,K2, . . . ,Kn) =λV(K,K2, . . . ,Kn) + µV(L,K2, . . . ,Kn) for µ, λ ∈ R≥0, K,L ∈ Kn.

v) Mixed volumes are valuations in each argument, i.e., for let K,L,K ∪L ∈Kn it holds

V(K ∪ L,K2, . . . ,Kn) =

V(K,K2 . . . ,Kn) + V(L,K2, . . . ,Kn)−V(K ∩ L,K2, . . . ,Kn).

Proof. For i) we just set λm = 1 and λi = 0 for i 6= m in the mixed volumeformula in Theorem 5.20. For ii) we apply Theorem 5.20 to the bodies ti+AKi

and use Remark 5.2 iii)

n∑i1,...,in=1

λi1 · . . . · λinV(ti1 +AKi1 , . . . , tin +AKin)

= vol

(n∑i=1

λi(ti +AKi)

)= vol

(n∑i=1

λiti +A

( n∑i=1

λiKi

))

= | detA| vol

(r∑i=1

λiKi

)=

n∑i1,...,in=1

λi1 · . . . · λin | detA|V(Ki1 , . . . ,Kin).

Comparing the coefficients of the two multivariate polynomials on the left andright hand side gives the resuls. The other properies are left as an exercise.

�

5.24 Lemma. Let K1,K2 ∈ Kn and k1 > dimK1. Then

V(K1, k1;K2, n− k1) = 0.

Proof. Let m = dimK1 < n, and we assume that

K1 ⊂ Lm = {x ∈ Rn : xm+1 = · · · = xn = 0}.


We write L⊥m = {x ∈ Rn : x1 = · · · = xm = 0}. Let K2 and K2 be, respectively,the orthogonal projections of K2 onto Lm and L⊥m. Then K2 ⊆ K2 + K2 andso

n∑i=0

(n

i

)λiV(K1, i;K2, n− i) = vol (λK1 +K2) ≤ vol (λK1 + K2 +K2)

= volm(λK1 + K2) · vol n−m(K2)

=m∑i=0

(m

i

)λiV(K1, i; K2,m− i)vol n−m(K2)

by Remark 5.6 iv), where V(K1, i; K2,m − i) represents the m-dimensionalmixed volume in Lm. Hence the coefficients on the left hand side of the termsλi for i > m have to be zero. �

5.25 Lemma. Let P ∈ Pn, dimP ≥ n − 1, and let u1, . . . ,um ∈ Sn−1 bethe unit outer normal vectors of the (n − 1)-faces Fi of P , 1 ≤ i ≤ m. LetG1, . . . , Gp be the (n− 2)-faces of P and let K ∈ Kn. For λ ≥ 0 it holds

vol (P ) + λm∑i=1

h(K,ui)vol n−1(Fi) +

p∑i=1

vol(Gi + 2λD(K)Bn

)≥ vol (P + λK) ≥ vol (P ) + λ

m∑i=1

h(K,ui)vol n−1(Fi),

where D(K) = max{|x− y| : x,y ∈ K

}denotes the diameter of K.

Proof. We may assume 0 ∈ relintK (cf. Corollary 5.8) and λ > 0 is fixed butarbitrary. Let v1, . . . ,vm ∈ K with h(K,ui) = 〈ui,vi〉 ≥ 0 and let Si(λ) =conv {Fi, λvi + Fi}. Si(λ) is a prism with basis Fi and height λh(K,ui) =λ 〈ui,vi〉, and so

vol(Si(λ)

)= λvol n−1(Fi)h(K,ui).

Next we observe that two of these prisms have no interior points in common.Otherwise, there exist yi ∈ relintFi, yj ∈ relintFj and µi, µj ∈ [0, 1] withyi +µiλvi = yj +µjλvj . Hence 〈ui,yi〉+µiλ 〈ui,vi〉 =

⟨ui,yj

⟩+µjλ 〈ui,vj〉.

Since 〈ui,yi〉 >⟨ui,yj

⟩and 〈ui,vi〉 ≥ 〈ui,vj〉 we get µi < µj . Analogously,

taking the inner product with uj we obtain µj < µi, a contradiction. SinceP ∪ (

⋃mi=1 Si(λ)) ⊆ P + λK and no interior points of P lie in one the prisms

Si(λ), we get the lower bound

vol (P + λK) ≥ vol (P ) + λ

m∑i=1

vol n−1(Fi)h(K,ui).

In order to prove the upper bound we estimate the volume of (P + λK) \[P ∪

(⋃mi=1 Si(λ)

)]. To this end let x ∈ P + λK but x 6∈ P ∪

(⋃mi=1 Si(λ)

).

Then we may write x = xi + λv with v ∈ K, xi is contained in an (n− 1)-faceFi, say, and 〈ui,v〉 > 0.


In the case xi ∈ relbdFi there exists an (n − 2)-face Gi of P containg xiand so

x ∈ Gi + λK ⊆ Gi + λD(K)Bn. (5.25.1)

Next let xi ∈ relintFi. The ray {xi + µv : µ ≥ 0} intersects bdSi(λ) ina point y = xi + µv with µ > 0. Since x 6∈ Si(λ) we have µ < λ and sincey ∈ bdSi(λ) there exist yi ∈ Fi and µi ∈ [0, 1] with y = yi + µi λvi. Thus

x = xi + λv = y + (λ− µ)v = yi + µiλvi + (λ− µ)v.

If µi = 1 then

〈ui,x〉 = 〈ui,yi〉+ λ 〈ui,vi〉+ (λ− µ) 〈ui,v〉 > h(P,ui) + h(λK,ui),

which contradicts x ∈ P + λK. Hence, y is not contained in the “upper”boundary of Si(λ). If µi = 0 then yi = xi + µv and we get the contradiction0 = 〈ui,yi − xi〉 = µ 〈ui,v〉. Hence, y is also not contained in the “lower”boundary of Si(λ) which implies that yi ∈ relbdFi. Thus there exists an(n− 2)-face Gj of P with yi ∈ Gj and so

x ∈ Gj + 2λD(K)Bn.

Together with (5.25.1) we have shown (P+λK)\[P ∪

(⋃mi=1 Si(λ)

)]⊂⋃pi=1

(Gi+

2λD(K)Bn)

which yields the required upper bound. �

5.26 Theorem. Let P ∈ Pn, dimP ≥ n−1, and let u1, . . . ,um ∈ Sn−1 be theunit outer normal vectors of the (n− 1)-faces Fi of P , 1 ≤ i ≤ m. For K ∈ Knwe have

V(P, n− 1;K, 1) =1

n

m∑i=1

h(K,ui)vol n−1(Fi).

Proof. For λ > 0 we have

vol (P + λK) = vol (P ) + nλV(P, n− 1;K, 1) +

n−2∑i=0

(n

i

)λn−iV(P, i;K,n− i),

and hence

nV(P, n− 1;K, 1) = limλ→0

vol (P + λK)− vol (P )

λ.

Plugging in the bounds of Lemma 5.25 for the numerator gives

m∑i=1

h(K,ui)vol n−1(Fi) ≤ nV(P, n− 1;K, 1)

≤m∑i=1

h(K,ui)vol n−1(Fi) + limλ→0

∑pi=1 vol

(Gi + 2λD(K)Bn

)λ

.


Since dimGi = n− 2 we get in view of Lemma 5.24

p∑i=1

vol(Gi + 2λD(K)Bn

)=

p∑i=1

n∑j=0

(n

j

)(2λD(K)

)n−jV(Gi, j;Bn, n− j)

=

p∑i=1

n−2∑j=0

(n

j

)(2λD(K)

)n−jV(Gi, j;Bn, n− j)

and so limλ→0

∑pi=1 vol

(Gi+2λD(K)Bn

)λ = 0. �

5.27 Corollary. Let K ∈ Kn and u ∈ Sn−1. Then

V(K,n− 1; conv {−u,u}, 1

)=

2

nvol n−1(K|u⊥i ),

where K|u⊥ denotes the orthogonal projection of K onto the hyperplane u⊥ =H(u, 0).

Proof. Let dimK ≥ n−1 and first we show it for a polytope P ∈ Pn with outerunit normal vectors v1, . . . ,vm ∈ Sn−1 of the (n − 1)-faces of Fi, 1 ≤ i ≤ m.By Theorem 5.26 we get for u ∈ Sn−1

V(P, n− 1; conv {−u,u}, 1

)=

1

n

m∑i=1

h(conv {−u,u},vi

)vol n−1(Fi)

=1

n

m∑i=1

|〈u,vi〉|vol n−1(Fi) =1

n

m∑i=1

vol n−1(Fi|u⊥)

=2

nvol n−1(P |u⊥);

thus, V(K,n − 1; conv {−u,u}, 1

)= (2/n)vol n−1(K|u⊥) for any convex body

K ∈ Kn by the usual approximation argument. �

5.28 Theorem*. Let Ki ∈ Kn, i = 1 . . . , r, K1 ⊆ K1 ∈ Kn, and ij ∈{1, . . . , r}, j = 2, . . . , n. Then

V(K1,Ki2 , . . . ,Kin) ≤ V(K1,Ki2 , . . . ,Kin).

5.29 Definition [Surface area]. Let K ∈ Kn.

F(K) = nV(K,n− 1;Bn, 1)

is called the surface area of K.

5.30 Corollary. Let P ∈ Pn, dimP ≥ n − 1, and let u1, . . . ,um ∈ Sn−1 bethe unit outer normal vectors of the (n− 1)-faces Fi of P , 1 ≤ i ≤ m. Then

F(P ) =m∑i=1

vol n−1(Fi).


Proof. By Theorem 5.26 we have

F(P ) = nV(P, n− 1;Bn, 1) =m∑i=1

h(Bn, ui)vol n−1(Fi) =m∑i=1

vol n−1(Fi).

�

5.31 Proposition.

i) F : Kn → R≥0 is a monotonous, continuous, rigid motions invariant valu-ation, which is homogeneous of degree n− 1.

ii) F(K) = 0 if (and only if) dimK ≤ n− 2.

iii) F(K) = limλ→0vol (K+λBn)−vol (K)

λ .

iv) F(Bn) = n vol (Bn).

Proof. i) is a consequence of the general properties of mixed volumes (cf.Theorem 5.28, Theorem 5.23 ii), iii), v)). The (n−1)-homogeneity follows fromproperty iv) of Theorem 5.23.

For (the one direction of) ii) see Lemma 5.24, and iii) and iv) are immediateconsequences of the definition of the surface area. �

5.32 Theorem [Minkowski’s inequalities]. Let K,L ∈ Kn with dimK = n.Then

i) V(K,n− 1;L, 1)n ≥ vol (K)n−1vol (L) and equality holds if and only if Kand L are homothetic.

ii) V(K,n− 1;L, 1)2 ≥ vol (K) V(K,n− 2;L, 2).

Proof. For λ ∈ [0, 1] let Kλ = (1− λ)K + λL. Then

vol (Kλ) =n∑i=0

(n

i

)(1− λ)iλn−iV(K, i;L, n− i). (5.32.1)

We consider the function

f(λ) = vol (Kλ)1/n − (1− λ)vol (K)1/n − λvol (L)1/n.

By Brunn-Minkowski’s theorem 5.16 f(λ) is the sum of a concave function andtwo linear functions; hence it is concave and by Theorem 5.16 we also knowthat it is non-neagtive. Moreover, since dimK = n we have vol (Kλ) > 0for λ ∈ [0, 1) which shows by (5.32.1) that f is differentiable on [0, 1). Theconcavity of f together with f(0) = f(1) = 0 gives f ′(0) ≥ 0 with f ′(0) = 0 ifand only if f(λ) = 0 for all λ ∈ [0, 1]. In view of (5.32.1) we get

0 ≤ f ′(0)

=1

nvol (K)(1/n)−1

[vol (Kλ)

]′0

+ vol (K)1/n − vol (L)1/n

=1

nvol (K)(1/n)−1

[nV(K,n− 1;L, 1)− nvol (K)

]+ vol (K)1/n − vol (L)1/n

= vol (K)−(n−1)/nV(K,n− 1;L, 1)− vol (L)1/n.


Hence, V(K,n − 1;L, 1) ≥ vol (K)(n−1)/nvol (L)1/n, with equality if and onlyif f(λ) = 0 for all λ ∈ [0, 1], i.e., if and only if K and L are homothetic(cf. Theorem 5.16). This shows i).

Since f is concave we also have f ′′(λ) ≤ 0, and so by using again (5.32.1)

0 ≥ f ′′(0)

= −(n− 1)vol (K)−(2n−1)/n[V(K,n− 1;L, 1)2 − vol (K)V(K,n− 2;L, 2)

].

On account of vol (K) > 0 we get ii). �

5.33 Corollary [Isoperimetric inequality]. Let K ∈ Kn with dimK = n.Then

F(K)n

vol (K)n−1≥ F(Bn)n

vol (Bn)n−1= nnvol (Bn),

and equality holds if and only if K is an n-dimensional ball.

Proof. Minkowski’s first inequality (Theorem 5.32 i)) gives

F(K)n = nnV(K,n− 1;Bn, 1)n ≥ nnvol (K)n−1vol (Bn),

with equality if and only if K and Bn are homothetic. �

A glimpse on geometry of numbers 57

6 A glimpse on geometry of numbers

6.1 Definition [Lattice]. Let b1, . . . , bn ∈ Rn be linearly independent. Theset

Λ = {z1b1 + z2b2 + · · ·+ znbn : zi ∈ Z, 1 ≤ i ≤ n}

is called lattice. The set of generating vectors {b1, . . . , bn} or the matrix B =(b1, . . . , bn) with columns bi is called basis(matrix) of Λ. An element b ∈ Λ iscalled lattice point of Λ. The set of all lattices in Rn is denoted by Ln.

6.2 Remark.

i) The unit vectors e1, . . . , en ∈ Rn form a basis of the integral lattice (stan-dard lattice) Zn = {z ∈ Rn : zi ∈ Z}.

ii) Let B = (b1, . . . , bn) be a basis of Λ. Then Λ = B Zn and, in particular,Λ is a subgroup of Rn, i.e., b− b ∈ Λ for all b, b ∈ Λ.

iii) Λ =

(25 6416 41

)Z2 = Z2.

6.3 Definition [Unimodular matrix]. An integral matrix U ∈ Zn×n is calledunimodular iff |detU | = 1. The group of all unimodular matrices is denotedby GL(n,Z).

6.4 Proposition. GL(n,Z) = {U ∈ Rn×n : UZn = Zn}.

Proof. U ∈ GL(n,Z) if and only if U,U−1 ∈ Zn×n which is equivalent toU Zn ⊆ Zn and U−1 Zn ⊆ Zn. Since the last inclusion is equivalent to Zn ⊆U Zn we are done. �

6.5 Lemma. Let Λ = BZn ∈ Ln. A = (a1, . . . ,an) is a basis of Λ if and onlyif there exists a U ∈ GL(n,Z) such that A = B U .

Proof. A is a basis of Λ if and only if AZn = Λ = B Zn which is equivalent toB−1A ∈ GL(n,Z) by Proposition 6.4. �

6.6 Definition [Determinant, fundamental cell]. Let Λ ∈ Ln with basis B =(b1, . . . , bn).

i) det Λ = |detB| is called determinant of Λ.

ii) PB = {ρ1b1 + · · · + ρnbn : 0 ≤ ρi < 1, 1 ≤ i ≤ n} = B [0, 1)n is calledfundamental cell or fundamental parallelepiped of Λ (w.r.t. the basis B).

6.7 Remark.

i) det Λ is independent of the choice of the basis (cf. Lemma 6.5).

ii) det Λ = vol (PB) and det(µΛ) = |µ|n det Λ, µ ∈ R.


iii) det Λ ≤ |b1| |b2| · . . . · |bn|, with equality if and only if the vectors bi arepairwise orthogonal (Hadamard inequality).

iv) PB∩Λ = {0}. Since (PB−PB) = B (−1, 1)n we even have (PB−PB)∩Λ ={0}.

6.8 Notation. Let a1, . . . ,an∈ Rn be linearly independent and letA=(a1, . . . ,an). Let x ∈ Rn with x =

∑ni=1 ρiai, ρi ∈ R. Then we write bxcA =

∑ibρicai.

In particular, bxcA ∈ AZn and x− bxcA ∈ PA.

6.9 Proposition. Let Λ = BZn ∈ Ln. Then

Rn =⋃·

b∈Λ (b+ PB) ,

i.e., Rn is the pairwise disjunct union of the lattice translates b+ PB.

Proof. Each x ∈ Rn can be decomposed as x = (x− bxcB) + bxcB. The firstsummand is in PB and second is a lattice point of Λ. To show that the union isdisjunct we observe that the intersection of two lattice translates b+PB, b+PBis non-empty, if and only if b − b ∈ (PB − PB) ∩ Λ. By Remark 6.7 iv) this isequivalent to b = b. �

6.10 Definition [Discrete set]. A set S ⊂ Rn is called discrete iff there existsan ε > 0 such that |s1 − s2| ≥ ε for all s1, s2 ∈ S, s1 6= s2.

6.11 Theorem. A subset S ⊂ Rn is a lattice if and only if S a discrete sub-group of Rn containing n linearly independent points.

Proof. Obviously, a lattice is a subgroup of Rn containing n linearly independentpoints. LetB be a basis of the lattice and let ε be the minimum of the continuousfunction |B x| on Sn−1. Then for z ∈ Zn \ {0} we have |Bz| ≥ ε |z|, whichshows the discreteness of the lattice.

For the other direction let s1, . . . , sn ∈ S be n linearly independent points.By an inductive construction we show that there exist b1, . . . , bn ∈ S such thatfor 1 ≤ k ≤ n

lin {s1, . . . , sk} ∩ S = (b1, . . . , bk) Zk. (6.11.1)

The case k = n of (6.11.1) implies the assertion. For k = 1 let b1 6= 0 be theshortest vector in conv {0, s1} ∩S. Since S is discrete such a choice is possible,and since S is a subgroup we certainly have b1 Z1 ⊆ lin {s1} ∩ S. Now lets ∈ lin {s1}∩S and let λ ∈ R such that s = λ b1. Then s−bλc b1 = (λ−bλc)b1

and by the minimality of b1 we must have λ = bλc ∈ Z. Hence (6.11.1) holdsfor k = 1.

Let us assume that we have already found b1, . . . , bk satisfying (6.11.1).Next we consider the (k + 1)-dimensional parallelepiped Pk = {

∑ki=1 αi bi +

αk+1sk+1 : 0 ≤ αi ≤ 1}. Let bk+1 ∈ Pk ∩ S having minimum positive distanceto lin {b1, . . . , bk}, i.e.,

bk+1 =k∑i=1

αi bi + αk+1sk+1,


and αk+1 > 0 is minimal among all points in Pk ∩ S. Obviously, in view of ourinductive construction we have lin {b1 . . . , bk+1} = lin {s1 . . . , sk+1} and by thegroup properties of S we also know (b1, . . . , bk+1)Zk+1 ⊆ lin {s1, . . . , sk+1}∩S.So let s ∈ lin {s1, . . . , sk+1} ∩ S given by s =

∑k+1i=1 βi bi. Then we have

S 3s−k+1∑i=1

bβic bi =k+1∑i=1

(βi − bβic) bi =k∑i=1

(βi − bβic) bi + (βk+1 − bβk+1c) bk+1

=k∑i=1

[(βi − bβic) + αi(βk+1 − bβk+1c)]︸︷︷︸µi

bi + αk+1(βk+1 − bβk+1c)︸︷︷︸µk+1

sk+1.

For abbreviation we denote by µi all the scalars in front of the vectors. Thenwe have 0 ≤ µk+1 < αk+1 and

s−k+1∑i=1

bβic bi −k∑i=1

bµic bi =k∑i=1

(µi − bµic) bi + µk+1 sk+1 ∈ S ∩ Pk.

By the choice of αk+1 we must have µk+1 = 0 and so βk+1 = bβk+1c ∈ Z. Thus

s− βk+1 bk+1 =k∑i=1

βi bi ∈ lin {s1, . . . , sk} ∩ S,

and (6.11.1) implies the integrality of βi for 1 ≤ i ≤ k. �

6.12 Corollary. Let a1, . . . ,an ∈ Λ ∈ Ln linearly independent. Then thereexists a basis b1, . . . , bn of Λ such that

ak ∈ (b1, · · · , bk)Zk, 1 ≤ k ≤ n.

Proof. Follows from (6.11.1) in the proof of Theorem 6.11 by setting si = ai,1 ≤ i ≤ n, and S = Λ. �

6.13 Definition [Index of a sublattice]. Let Λ ∈ Ln and let a1, . . . ,an ∈ Λbe linearly independent. Λ0 = (a1, . . . ,an)Zn is called a sublattice with basis{a1, . . . , an}. The number of cosets of the subgroup Λ0 with respect to Λ, i.e.,the index of Λ0 in Λ is denoted by |Λ : Λ0|.

6.14 Lemma. Let Λ0 ⊆ Λ ∈ Ln be a sublattice of Λ. Then

i) |Λ : Λ0| = #(PA ∩ Λ) for any basis A of Λ0.

ii) |Λ : Λ0| = det Λ0/ det Λ.

Proof. For i) it is to show that Λ =⋃·

c∈PA∩Λ (c+ Λ0). For if, let b ∈ Λ. Thenwe have bbcA ∈ Λ0 ⊆ Λ and so (b− bbcA) ∈ PA ∩ Λ. Thus

b = (b− bbcA) + bbcA ∈ (PA ∩ Λ) + Λ0.


If there exist b1, b2 ∈ PA ∩Λ such that (b1 + Λ0)∩ (b2 + Λ0) 6= ∅ then b1− b2 ∈(PA − PA) ∩ Λ0 = {0} (cf. Remark 6.7). Hence b1 = b2 and i) is shown.

For ii) we observe that

mPA =⋃·

0≤mi<m (m1a1 + · · ·+mnan + PA) ,

where mi, m ∈ N. Moreover, for every a ∈ Λ we have #((a + PA) ∩ Λ

)=

#(PA ∩ Λ) and in view of i) we find

#(mPA ∩ Λ

)= mn #(PA ∩ Λ) = mn |Λ : Λ0|.

Finally, since PA is Riemann-integrable we may write (cf. Remark 5.2)

det Λ0 = vol (PA) = limm→∞

#

(PA ∩

1

mΛ

)det Λ

mn= det Λ lim

m→∞

# (mPA ∩ Λ)

mn

= det Λ |Λ : Λ0|.

�

6.15 Corollary. Let z1, . . . ,zn ∈ Zn be linearly independent. Then

#((z1, . . . ,zn)[0, 1)n ∩ Zn

)= |det(z1, . . . ,zn)|.

Proof. We just apply Lemma 6.14 ii) with Λ = Zn and Λ0 being the latticegenerated by z1, . . . ,zn. �

6.16 Remark. Let Λ0 = AZn ⊆ Λ ∈ Ln be a sublattice of Λ. Then

A is basis of Λ⇔ |Λ : Λ0| = 1⇔ Λ ∩ PA = {0}⇔ Λ ∩A [0, 1]n = A {0, 1}n.

6.17 Lemma. Let Λ ∈ L2 and let a1,a2 ∈ Λ be linearly independent. Then

a1,a2 basis of Λ⇔ conv {0,a1,a2} ∩ Λ = {0,a1,a2}.

Proof. If a1,a2 are a basis then every point of Λ has an unique representationas an integral linear combination of a1 and a2 and so conv {0,a1,a2} ∩ Λ ={0,a1,a2}. For the reverse implication let b ∈ PA = (a1,a2)[0, 1)2∩Λ. In viewof Remark 6.16 we have to show b = 0, which, by assumption, is certainly trueif b ∈ conv {0,a1,a2}. So let b 6∈ conv {0,a1,a2}. Then b = ρ1a1 + ρ2a2 with0 < ρ1, ρ2 < 1 and ρ1 + ρ2 > 1. So we have (1− ρ1) + (1− ρ2) < 1 and thus

(a1 + a2)− b = (1− ρ1)a1 + (1− ρ2)a2 ∈ conv {0,a1,a2} ∩ Λ.

Hence b = a1 + a2 contradicting the choice of b ∈ PA. �


6.18 Remark. An analogous statement to Lemma 6.17 does not exist in di-mension ≥ 3. For n ≥ 3 and m ∈ N let b(m) = (1, . . . , 1,m)ᵀ ∈ Rn andTn(m) = conv {0, e1, . . . , en−1, b(m)}. Then

Tn(m) ∩ Zn = {0, e1, . . . , en−1, b(m)},

but the determinant of the lattice with basis {e1, . . . , en−1, b(m)} is m. Tn(m)are called Reeve simplices.

6.19 Lemma. Let u1, . . . ,um ∈ Zn and let ki ∈ N, ki ≥ 1, 1 ≤ i ≤ m. Theset

Λ = {z ∈ Zn : 〈ui, z〉 ≡ 0 mod ki, 1 ≤ i ≤ m}

is a lattice with det Λ ≤ k1k2 · . . . · km.

Proof. By definition Λ is a discrete subgroup of Rn, actually of Zn. Since then linearly independent vectors (k1 · · · km)ei, 1 ≤ i ≤ n, belong to Λ, Theorem6.11 shows that Λ is a lattice. By Lemma (6.14) we have det Λ = |Zn : Λ| andso it suffices to bound the number of cosets of Zn w.r.t. Λ. To this end letΦ : Zn → Zm be given by Φ(z)i = 〈ui, z〉 mod ki ∈ {0, . . . , ki − 1}, 1 ≤ i ≤ m.Then #Φ(Zn) = k1 · . . . · km and it remains to observe that z, z belongs todifferent cosets of Zn w.r.t. Λ if and only if (z − z) /∈ Λ, i.e., if and only ifΦ(z) 6= Φ(z). �

6.20 Lemma. Let X ⊂ Rn be a bounded measurable set.

i) If (z1 +X)∩ (z2 +X) = ∅, for all z1, z2 ∈ Zn, z1 6= z2, then vol (X) ≤ 1.

ii) If Zn +X = Rn then vol (X) ≥ 1.

Proof. Let P = [0, 1)n be the fundamental cell of Zn, and let M = {z ∈ Zn :(z + P ) ∩X 6= ∅}. By Proposition 6.9) we have Rn =

⋃·z∈Zn(z + P ) = Zn + P

and sovol (X) = vol ((Zn + P ) ∩X) =

∑z∈M

vol ((z + P ) ∩X) .

Now (z + P ) ∩X = (P ∩X − z)) + z and so

vol (X) =∑z∈M

vol (P ∩ (X − z)) .

In the first case we have [P ∩ (X − z1)] ∩ [P ∩ (X − z2)] = ∅ for z1 6= z2 ∈ Zn,and thus ∑

z∈Mvol (P ∩ (X − z)) ≤ vol (P ) = 1.

In the second case we observe that M = {z ∈ Zn : P ∩ (X − z) 6= ∅} and onaccount of Zn +X = Rn we must have ∪z∈M (P ∩ (X − z)) = P . Hence∑

z∈Mvol (P ∩ (X − z)) ≥ vol (P ) = 1.

�


6.21 Corollary [Blichfeldt]. 30 Let X ⊂ Rn with vol (X) > 1. Then thereexist x1,x2 ∈ X such that x1 − x2 ∈ Zn \ {0}, i.e., X −X ∩ Zn \ {0} 6= ∅.

Proof. With out loss of generality we may assume that X is bounded. ByLemma 6.20 i) there exist z1, z2 ∈ Zn, z1 6= z2, such that (z1 +X)∩(z2 +X) 6=∅. Hence Zn \ {0} 3 z1 − z2 ∈ X −X. �

6.22 Notation. Let Kn0 = {K ∈ Kn : K = −K} be the set of all 0-symmetricconvex bodies.

6.23 Theorem [Minkowski, 1896]. Let K ∈ Kn0 with vol (K) ≥ 2n. Then

K ∩ Zn \ {0} 6= ∅,

i.e., a 0-symmetric convex body of volume at least 2n contains a non-triviallattice point.

Proof. First we assume vol (K) > 2n. Then we have vol (12K) > 1 and by

Corollary 6.21 we get (12K −

12K) ∩ Zn \ {0} 6= ∅. By the symmetry we have

−12K = 1

2K and thus K = 12K −

12K.

Now let vol (K) = 2n and suppose K ∩Zn = {0}. Since K is compact thereexists a λ > 1 with λK ∩ Zn = {0}. However vol (λK) > 2n and thus we get acontradiction to the previous case. �

6.24 Corollary. Let Λ ∈ Ln and K ∈ Kn0 with vol (K) ≥ 2n det Λ. Then

K ∩ Λ \ {0} 6= ∅.

Proof. Let B be a basis of Λ. Then K ∩ Λ \ {0} 6= ∅ if and only if B−1(K ∩Λ \ {0}) 6= ∅ which is equivalent to B−1K ∩ Zn \ {0} 6= ∅. By assumption wehave vol (B−1K) = vol (K)/det Λ ≥ 2n, and so the corollary is an immediateconsequence of Theorem 6.23. �

6.25 Theorem [Theorem on linear forms]. Let ai ∈ Rn, 1 ≤ i ≤ n, be lin-early independent, and let τi ∈ R>0, 1 ≤ i ≤ n, such that τ1τ2 · . . . · τn ≥|det(a1, . . . ,an)|. Then there exists a z ∈ Zn \ {0} with

| 〈ai, z〉 | ≤ τi, 1 ≤ i ≤ n.

Proof. Let P = {x ∈ Rn : | 〈ai,x〉 | ≤ τi, 1 ≤ i ≤ n}. In order to calculatethe volume of that 0-symmetric parallelepiped we observe that P = A−1{x ∈Rn : |xi| ≤ τi, 1 ≤ i ≤ n} where A is the matrix with columns ai. Thus

vol (P ) = 2nτ1 · . . . · τn|detA|

≥ 2n.

The assertion follows from Theorem 6.23. �30Hans Frederick Blichfeldt, 1873–1945

http://en.wikipedia.org/wiki/Hans_Frederick_Blichfeldt


6.26 Theorem [Dirichlet]. 31 Let α1, . . . , αn ∈ R and let 0 < ε < 1. Thenthere exist p1, . . . , pn, q ∈ Z with 1 ≤ q ≤ ε−n such that∣∣∣∣αi − pi

q

∣∣∣∣ < ε

q, 1 ≤ i ≤ n.

Proof. Let ai = −ei + αien+1, 1 ≤ i ≤ n, and let an+1 = en+1. Then|det(a1, . . . ,an+1)| = 1 and by Theorem 6.25 there exists for every τ > 0 anintegral point z = (p1, . . . , pn, q)

ᵀ ∈ Zn+1 \ {0} (depending on τ) satisfying thesystem of linear forms

| 〈ai, z〉 | = |αi q − pi| ≤ τ−1/n, 1 ≤ i ≤ n, and | 〈an+1, z〉 | = |q| ≤ τ.

Now we choose a τ > ε−n such that bτc ≤ ε−n and get

| 〈ai, z〉 | = |αi q − pi| < ε, 1 ≤ i ≤ n, and | 〈an+1, z〉 | = |q| ≤ bτc ≤ ε−n.

Finally, we observe that q 6= 0, because otherwise pi = 0 for i = 1, . . . , n sinceε < 1 and so z = 0. Thus we may assume q ≥ 1. �

6.27 Proposition. Let p be prime. Then there exist a, b ∈ N with

a2 + b2 + 1 ≡ 0 mod p.

Proof. For p = 2 the statement is certainly true. So let p be odd. For0 ≤ a ≤ 1

2(p − 1) the numbers a2 belong to pairwise distinct residue classesmod p, because

a2 ≡ a2 mod p⇔ (a− a)(a+ a) ≡ 0 mod p.

The same is true if we look at the residue classes of −b2−1 for 0 ≤ b ≤ 12(p−1).

Since there are precisely p different residue classes mod p we can find integers0 ≤ a, b ≤ 1

2(p− 1) such that a2 and −(b2 + 1) belong to the same residue classmodp which proves the proposition. �

6.28 Theorem [Lagrange]. 32 Every positive number m ∈ N can be writtenas the sum of four integer squares, i.e., there exist m1,m2,m3,m4 ∈ N suchthat

m = (m1)2 + (m2)2 + (m3)2 + (m4)2.

Proof. First we observe that it suffices to prove the theorem for integers mwhich are not divisible by a square other than 1. Hence let m = p1 · . . . ·pk withdistinct primes pi, 1 ≤ i ≤ k. According to Proposition 6.27 we choose ai, bi,1 ≤ i ≤ k, such that

(ai)2 + (bi)

2 + 1 ≡ 0 mod pi, 1 ≤ i ≤ k. (6.28.1)

31Gustav Lejeune Dirichlet, 1805–185932Joseph-Louis Lagrange, 1736–1813

http://en.wikipedia.org/wiki/Johann_Peter_Gustav_Lejeune_Dirichlet

http://en.wikipedia.org/wiki/Joseph_Louis_Lagrange


With this notation we set

Λ ={z ∈ Z4 : z1 ≡ (aiz3 + biz4) mod pi, z2 ≡ (biz3 − aiz4) mod pi, 1 ≤ i ≤ k

}.

By Lemma 6.19 we know that Λ ⊂ R4 is a lattice with det Λ ≤ (p1)2 ·. . .·(pk)2 =m2. Thus we get

vol (√

2mB4) = (2m)2vol (B4) = (2m)2π2

2> 24m2 ≥ 24 det Λ.

Hence by Corollary 6.24 there exists a z ∈ Λ \ {0} ∩ intB4, i.e.,

0 < (z1)2 + (z2)2 + (z3)2 + (z4)2 < 2m.

In order to prove the theorem it suffices to show that m is a divisor of the sum(z1)2 + (z2)2 + (z3)2 + (z4)2.

By the choice of Λ we get for 1 ≤ i ≤ k

(z1)2 + (z2)2 + (z3)2 + (z4)2

≡((aiz3 + biz4)2 + (biz3 − aiz4)2 + (z3)2 + (z4)2

)mod pi

≡((z3)2((ai)

2 + (bi)2 + 1) + (z4)2((ai)

2 + (bi)2 + 1)

)mod pi

≡ 0 mod pi,

where the last relation is a consequence of (6.28.1). Thus all the distinct pi aredivisors of (z1)2 + (z2)2 + (z3)2 + (z4)2 and so m is a divisor of this sum. �

6.29 Theorem. Let k ∈ N and let X ⊂ Rn be a Jordan measurable set withvol (X) > k. Then there exist x1, . . . ,xk+1 ∈ X such that xi − xj ∈ Zn \ {0},i 6= j.

Proof. Since we have a Jordan-measurable set we know

k < vol (X) = limm→∞

#

(X ∩ 1

mZn)

1

mn.

Thus there exists an m ∈ N such that #(X ∩ 1

mZn)> kmn. Since there are

mn cosets of the sublattice Zn with respect to the lattice 1mZn there exist (at

least) (k + 1) different x1, . . . ,xk+1 ∈ X ∩ 1mZn belonging to the same coset

and thus xi − xj ∈ Zn \ {0}, i 6= j. �

6.30 Corollary. Let Λ ∈ Ln and let K ∈ Kn0 with vol (K) ≥ k 2n det Λ. Then

# (K ∩ Λ) ≥ 2k + 1.

Proof. Without loss of generality let Λ = Zn and vol (K) > k 2n (cf. the proofsof Corollary 6.24 and Theorem 6.23). By Theorem 6.29 there exist (k + 1)different points x1, . . . ,xk+1 ∈ 1

2K with xi − xj ∈ Zn. Let us assume thatx1 is one with maximal Euclidean length among these (k + 1) points and letzi = xi+1−x1, 1 ≤ i ≤ k. Then we have zi 6= zj , i 6= j, and zi ∈ K ∩Zn \{0}.Furthermore, by the choice of x1 all these points satisfy 〈x1, zi〉 < 0 whichimplies that the 2k points ±zi, 1 ≤ i ≤ k, are pairwise distinct. Together withthe point 0 this gives the desired lower bound. �


6.31 Definition [Successive minima]. Let K ∈ Kn0 , Λ ∈ Ln. For 1 ≤ i ≤ n,

λi(K,Λ) = min {λ > 0 : dim (λK ∩ Λ) ≥ i}

is called the i-th successive minimum of K w.r.t. Λ.

6.32 Remark.

i) λi(K,Λ) ≥ λi−1(K,Λ), 2 ≤ i ≤ n.

ii) λi(K,Λ) = λi(AK,AΛ), A ∈ GL(n,R), i.e., A ∈ Rn×n with detA 6= 0.

iii) λi(µK,Λ) = 1µλi(K,Λ) = λi(K,

1µΛ), µ ∈ R, µ 6= 0.

iv) intK ∩ Λ \ {0} = ∅ ⇔ λ1(K,Λ) ≥ 1.

v) λ1(Bn,Λ) = min{|b| : b ∈ Λ \ {0}

}.

6.33 Proposition. Let K ∈ Kn0 , Λ ∈ Ln. Let a1, . . . ,an ∈ Λ be linearlyindependent such that ai ∈ λi(K,Λ)K, 1 ≤ i ≤ n. Then

int (λi(K,Λ)K) ∩ Λ ⊆ lin {a1, . . . ,ai−1}, 1 ≤ i ≤ n,

where we set lin ∅ = {0}.

Proof. We set λ0(K,Λ) = 0, a0 = 0, and let 0 ≤ j < i be the maximal indexwith λj(K,Λ) < λi(K,Λ). Then each lattice point in λK with λj(K,Λ) ≤ λ <λi(K,Λ) must be linearly dependent on {a0, . . . ,aj}. Hence

int(λi(K,Λ)K

)∩ Λ ⊂ lin {a0, . . . ,aj}.

�

6.34 Theorem [Minkowski’s first theorem on successive minima]. Let K∈Kn0 and Λ ∈ Ln. Then

λ1(K,Λ)nvol (K) ≤ 2n det Λ.

Proof. By the definition of λ1(K,Λ) we have int (λ1(K,Λ)K) ∩ Λ \ {0} = ∅.Hence by Corollary 6.24 we get vol (λ1(K,Λ)K) ≤ 2n det Λ. �

6.35 Theorem* [Minkowski’s second theorem on successive minima].LetK∈ Kn0 and Λ ∈ Ln. Then

2n

n!det Λ ≤ λ1(K,Λ)λ2(K,Λ) · . . . · λn(K,Λ) vol (K) ≤ 2n det Λ.


Proof. Without loss of generality let Λ = Zn. For convenience we writeλi = λi(K,Zn), and let z1, . . . ,zn be n linearly independent lattice points with

zi ∈ λiK ∩ Zn, 1 ≤ i ≤ n. (6.35.1)

Here we only prove the easy lower bound, for which we just observe that(6.35.1) implies conv {± 1

λizi : 1 ≤ i ≤ n} ⊆ K. Thus

vol (K) ≥ vol

(conv

{± 1

λizi : 1 ≤ i ≤ n

})=

2n

n!

∣∣∣∣det

(1

λ1z1, . . . ,

1

λnzn

)∣∣∣∣=≥ 2n

n!

1

λ1 · . . . · λn|det (z1, . . . ,zn)| ≥ 2n

n!

1

λ1 · . . . · λn.

�

6.36 Theorem. Let K ∈ Kn0 and Λ ∈ Ln. Then

#(K ∩ Λ) ≤(⌊

2

λ1(K,Λ)+ 1

⌋)n.

In particular: Let K ∈ Kn0 with intK ∩ Λ = {0}. Then #(K ∩ Λ) ≤ 3n.

Proof. Without loss of generality let Λ = Zn and we write λ1 instead ofλ1(K,Zn). Let k = b2/λ1 + 1c. Since there are kn different cosets of Zn withrespect to the sublattice kZn it suffices to show that two different lattice pointsz, z ∈ K ∩Zn belong to two different cosets. Suppose there exist z, z ∈ K ∩Znsuch that (z − z) ∈ kZn. Then we find

Zn 3 1

k(z − z) =

2

k

(1

2z − 1

2z

)∈ 2

kK ⊂ int (λ1K) ,

since 2/k < λ1. Thus, by definition of λ1 we must have z = z. �

6.37 Conjecture. Let K ∈ Kn0 and Λ ∈ Ln. Then

#(K ∩ Λ) ≤n∏i=1

⌊2

λi(K,Λ)+ 1

⌋.

6.38 Remark. Conjecture 6.37 would imply the upper bound in Minkowski’ssecond theorem 6.35, because

vol (K) = limm→∞

(1

m

)n#

(K ∩ 1

mZn)

= limm→∞

(1

m

)n#(mK ∩ Zn)

≤ limm→∞

(1

m

)n n∏i=1

(2

λi(mK)+ 1

)=

(1

m

)nlimm→∞

n∏i=1

(2m

λi(K)+ 1

)

= limm→∞

n∏i=1

(2

λi(K)+

1

m

)=

n∏i=1

2

λi(K).

Packing 67

7 Packing

7.1 Definition [Packing sets]. A subset D ⊂ Rn is called a packing set ofK ∈ Kn if for all x, vy ∈ D, x 6= y,

(x+ intK) ∩ (y + intK) = ∅.

The family of all packing sets of K is denoted by P(K).

7.2 Definition [Density of a Packing]. Let K ∈ Kn and D ∈ P(K).

δ(K,D) = lim supλ→∞

vol (K)#{x ∈ D : x+K ⊂ λCn

}vol(λCn

)is called the density of the packing D+K (with respect to the gauge body Cn =[−1, 1]n). Here a gauge body is an arbitrary convex body G with 0 ∈ intG.

7.3 Remark. This definition of the density depends on the chosen gauge body.For instance, let D = {z ∈ Zn : z ≥ 0} ∈ P([0, 1]n). According to theDefinition 7.2 we get δ

([0, 1]n, D

)= 1

2n . Now let Hn = Cn ∩ {x ∈ Rn :|x1− x2 + x3 + · · ·+ xn| ≤ n− 1}. Then [0, 1]n ⊂ Hn, vol (Hn) = 2n− 2/n! andchanging in Definition 7.2 the gauge body Cn to Hn gives

lim supλ→∞

vol ([0, 1]n)#{x ∈ D : x+ [0, 1]n ⊂ λHn}vol (λHn)

=1

2n − 2n!

.

7.4 Theorem. Let K ∈ Kn. The supremum sup{δ(K,D) : D ∈ P(K)

}is

independent of the chosen gauge body, and there exists a packing set DK ∈P(K) such that sup

{δ(K,D) : D ∈ P(K)

}= δ(K,DK).

The proof of Theorem 7.4 is split into several lemmas, which are of interestin their own. To this end we need the following notation.

7.5 Notation. Let K ∈ Kn. For G ∈ Kn with 0 ∈ intG and for λ > 0, wewrite

Φ(λ,G) = max{

#D : D ∈ P(K) and D +K ⊆ λG}

and

Ψ(λ,G) =vol (K) Φ(λ,G)

vol (λG).

7.6 Lemma. Let K ∈ Kn and G ∈ Kn with 0 ∈ intG. Then limλ→∞Ψ(λ,G)exists and is independent of G.

Proof. First we show that in the case G = Cn the limit exists. For short wewrite Φ(λ) = Φ(λ,Cn), Ψ(λ) = Ψ(λ,Cn), and for a given λ > 0 we denote byD(λ) ∈ P(K) a maximal packing set, i.e., #D(λ) = Φ(λ) and D(λ)+K ⊂ λCn.Furthermore, let γ ≥ 1 be such that K −K ⊂ γ Cn.

Now let λ > γ and q ∈ N. The cube q λCn can be filled by qn non-overlapping copies of λCn, and let ti ∈ Rn, 1 ≤ i ≤ qn, be the centers of thesescopies, i.e.,

qλCn = {t1, . . . , tqn}+ λCn.

68 Packing

In particular, we see thatΦ(qλ) ≥ qn Φ(λ). (7.6.1)

Next we want to find an upper bound on Φ(qλ) in terms of Φ(λ). To thisend we set D1(qλ) =

{x ∈ D(qλ) : ∃ ti s. t. x + K ⊂ ti + λCn

}and let

D2(qλ) = D(qλ) \D1(qλ). Obviously,

Φ(qλ) = #D1(qλ) + #D2(qλ) ≤ qnΦ(λ) + #D2(qλ). (7.6.2)

On the other hand, since D(qλ) + K ⊂ qλCn and since K − K ⊂ γ Cn weconclude for the translates by points in D2(qλ) (cf. Exercise ??)

D2(qλ) +K ⊂qn⋃i=1

(ti + λCn

)\(ti + (λ− γ)Cn

).

Thus we have

vol (K) #D2(q λ) ≤ qn[λn−(λ− γ)n

]vol(Cn)≤ qnλn−1vol

(Cn)

c,

where c is a constant depending only on n and γ. Together with (7.6.2) and(7.6.1) we get

0 ≤ Φ(qλ)− qnΦ(λ) ≤qnλn−1c vol

(Cn)

vol (K).

Multiplying by vol (K)/vol(q λCn

)yields

0 ≤ Ψ(qλ)−Ψ(λ) ≤ c

λ. (7.6.3)

Now let p1, p2 ∈ Q with p1 ≥ p2 > γ and let m1,m2 ∈ N with m1 p1 = m2 p2.Then we get from (7.6.3)∣∣Ψ(p1)−Ψ(p2)

∣∣ ≤ ∣∣Ψ(p1)−Ψ(m1 p1)∣∣+∣∣Ψ(p2)−Ψ(m2 p2)

∣∣ ≤ 2c

p2. (7.6.4)

This shows that Ψ(λ) forms a Cauchy-sequence on the rational numbers. SinceΦ(Λ) is a right-continuous function, for every λ ∈ R there exists a rationalnumber pλ ∈ Q with

0 ≤ pλ − λ ≤1

2and Φ(λ) = Φ(pλ).

So we have

0 ≤ Ψ(λ)−Ψ(pλ) =vol (K) Φ(λ)

vol (pλCn)

((pλλ

)n− 1)≤(pλλ

)n− 1 ≤ c

λ,

where c is a constant only depending on n. Hence for µ, λ ∈ R, µ ≥ λ, we findtogether with (7.6.4)∣∣Ψ(µ)−Ψ(λ)

∣∣ ≤ ∣∣Ψ(pµ)−Ψ(pλ)∣∣+ ∣∣Ψ(µ)−Ψ(pµ)

∣∣+ ∣∣Ψ(λ)−Ψ(pλ)∣∣ ≤ 2

(c + c

)λ

,

which finally shows that Ψ(λ) is a Cauchy-sequence. Hence limλ→∞Ψ(λ) existsand it remains to show its independency from the gauge body. This is left asan exercise for the reader. �

Packing 69

7.7 Lemma. Let K ∈ Kn and G ∈ Kn with 0 ∈ intG. Then there existsDK ∈ P(K) such that

lim supλ→∞

vol (K) #{x ∈ DK : x+K ⊂ λG

}vol (λG)

= limλ→∞

Ψ(λ,G).

Proof. By definition of Ψ(λ,G) it suffices to show that the left hand sideis not less than the limit on the right. Let K − K ⊂ γ G and for λ > 0 letD(λ) ∈ P(K) be a packing set of maximal cardinality such thatD(λ)+K ⊂ λG.In particular, we have #D(λ) = Φ(λ,G), and for m ∈ N we define

Dm ={x ∈ D(2m

2) : (x+K) ∩ 2(m−1)2 G = ∅

}and setDK =

⋃m∈NDm. Since, Dm ⊂ D(2m

2) and (Dm+K)∩(Dm−1+K) = ∅,

the set DK is a packing set for K. Moreover, for x ∈ D(2m2) \ Dm we have

(x+K) ∩ 2(m−1)2 G 6= ∅ and so

0 ≤ Φ(2m2, G)−#Dm ≤

vol(

(2(m−1)2 + γ)G)

vol (K).

Hence we obtain

#{x ∈ DK : x+K ⊂ 2m

2G}

vol (K)

vol (2m2G)

≥ #Dm vol (K)

vol (2m2G)

≥ Φ(2m2, G) vol (K)

vol (2m2G)−

vol(

(2(m−1)2 + γ)G)

vol (2m2G)

≥ Ψ(2m2, G)−

(1 + γ

2m

)n.

(7.7.1)

By Lemma 7.6 we know limλ→∞Ψ(λ,G) = limm→∞Ψ(2m2, G), and so (7.7.1)

implies the assertion. �Proof. [Proof of Theorem 7.4] Let K ∈ Kn. For G ∈ Kn with 0 ∈ intG andD ∈ P(K) we set

δG(K,D) = lim supλ→∞

vol (K) #{x ∈ D : x+K ⊆ λG}vol (λG)

.

Obviously we have δG(K,D) ≤ limλ→∞Ψ(λ,G) and on account of Lemma 7.7and Lemma 7.6 we get

sup{δG(K,D) : D ∈ P(K)

}= max

{δG(K,D) : D ∈ P(K)

}= lim

λ→∞Ψ(λ,G) = lim

λ→∞Ψ(λ,Cn).

�

70 Packing

7.8 Definition [Density of a Densest Packing]. Let K ∈ Kn.

δ(K) = sup{δ(K,D) : D ∈ P(K)

}is called the density of a densest packing of K and a set DK ∈ P(K) withδ(K) = δ(K,DK) is called a densest packing set of K.

7.9 Proposition. Let K ∈ Kn. Then

i) 0 < δ(K) ≤ 1.

ii) δ(t+AK) = δ(K) for all A ∈ GL(n,R) and t ∈ Rn.

iii) Let K ∈ Kn and D ∈ P(K). Then

δ(K,D) = lim supλ→∞

vol (K) #(D ∩ λCn

)vol(λCn

) .

iv) P(K) = P(

12(K −K)

)and for D ∈ P(K) we have

δ(K,D) = δ(

12(K −K), D

) vol (K)

vol(

12(K −K)

) ,and consequently

δ(K) = δ(

12(K −K)

) vol (K)

vol(

12(K −K)

) .v) Let K ∈ Kn0 . Then D ∈ P(K) if and only if |x− y|K ≥ 2 for all x,y ∈ D,x 6= y. Here |x|K = min{λ ≥ 0 : x ∈ λK} is the norm induced by K.

Proof. Items i), ii) follow immediately from the definition. For iii) let γ > 0with K,K −K ⊂ γ Cn, and for a given λ > γ let

m1(λ) = #{x ∈ D : x+K ⊂ λCn

}and m2(λ) = #

(D ∩ λCn

).

If x ∈ D with x+K ⊂ λCn, but x /∈ λCn then x+K ⊂ λCn \ (λ− γ)Cn andso

vol (K)m1(λ) ≤ vol (K)m2(λ) + [λn − (λ− γ)n]vol(Cn)

≤ vol (K)m2(λ) + λn−1vol(Cn)

c,

where c is a constant only depending on γ and n. On the other hand, ifx ∈ D ∩ λCn but x+K 6⊂ λCn then x+K ⊂ (λ+ γ)Cn \ (λ− γ)Cn. Hence

vol (K)m2(λ) ≤ vol (K)m1(λ) +[(λ+ γ)n − (λ− γ)n

]vol(Cn)

≤ vol (K)m1(λ) + λn−1vol(Cn)

c,

for another constant c. So we obtain

− c

λ≤

vol (K) #{x ∈ D : x+K ⊂ λCn

}vol(λCn

) −vol (K) #

(D ∩ λCn

)vol(λCn

) ≤ c

λ,

Packing 71

which shows iii). For iv) we note that

(x+ intK) ∩ (y + intK) 6= ∅

⇔ x− y ∈ intK − intK = int (K −K) =1

2int (K −K)− 1

2int (K −K)

⇔(x+

1

2int (K −K)

)∩(y +

1

2int (K −K)

)6= ∅

and thus P(K) = P(

12(K −K)

). Here we have used the fact that int (K+L) =

int (K)+int (L) for convex bodiesK,L (cf. Exercise []). v) is just a reformulationof the packing property of an o-symmetric convex body. �

7.10 Lemma. Let S ⊂ Rn be a bounded and measurable set with vol (S) > 0and let D ∈ P(K). Then there exist v,w ∈ Rn such that

vol (K) #((w + S) ∩D

)vol (S)

≤ δ(K,D) ≤vol (K) #

((v + S) ∩D

)vol (S)

.

Proof. We just prove the upper bound, the lower bound can be done analo-gously. Let γ > 0 such that S ⊂ γ Cn and let ε(λ) ∈ R with ε(λ) → 0 as λtends infinity and (cf. Proposition 7.9 iv))

ε(λ) +δ(K,D)

vol (K)=

#(D ∩ λCn

)vol(λCn

) . (7.10.1)

Let x ∈ D ∩ λCn. Since {v ∈ Rn : x ∈ v + S} = x− S ⊂ (λ+ γ)Cn we get∫(λ+γ)Cn

# ((v + S) ∩D) dv ≥ vol (S) #(D ∩ λCn

).

Hence there exist vλ ∈ Rn such that

# ((vλ + S) ∩D) ≥ vol (S)#(D ∩ λCn

)vol((λ+ γ)Cn

)= vol (S)

#(D ∩ λCn

)vol(λCn

) (λ

λ+ γ

)n,

and with (7.10.1) we obtain

# ((vλ + S) ∩D) ≥(δ(K,D)vol (S)

vol (K)+ ε(λ) vol (S)

)(1− γ

λ+ γ

)n≥ δ(K,D)vol (S)

vol (K)+ ρ(λ),

for suitable numbers ρ(λ) satisfying limλ→∞ ρ(λ) = 0. Since the left hand sideis an integer we can find λ ∈ R>0 such that #

((vλ + S) ∩

(D ∩ λ Cn

))≥

δ(K,D)vol (S)/vol (K), which gives the upper bound. �

72 Packing

7.11 Remark. Let K ∈ Kn.

R(K) = min{R > 0 : ∃x ∈ Rn with K ⊆ x+RBn}

is called the circumradius of K. The uniquely determined point tc ∈ Rn withK ⊆ tc + R(K)Bn is called circumcenter of K. There exist k + 1 affinelyindependent points x0, . . . ,xk ∈ bdK ∩ bd

(tc + R(K)Bn

)and λi > 0, 0 ≤ i ≤

k, with∑k

i=0 λi = 1 such that tc =∑k

i=0 λi xi (cf. Exercise []), and these pointsare extreme points of K.

7.12 Theorem. Let K ∈ Kn0 . Then

i) δ(K) ≥ 2−n,

ii) δ(Bn) ≤ (n+ 1)√

2−n

.

Proof. i) Let Ds ∈ P(K) be a saturated packing, i.e., for every x ∈ Rn wehave (x + K) ∩ (Ds + K) 6= ∅. Thus (x + 2K) ∩Ds 6= ∅ and now we use thelower bound of Lemma 7.10 with respect to the set S = 2K and the packingset Ds.

ii) For r ∈ R>0 let f(r, n) = max{

#(D ∩ int (r Bn)

): D ∈ P(Bn)

}. In view

of the upper bound given in Lemma 7.10, applied to S =√

2 intBn, it sufficesto show

f(√

2, n)≤ n+ 1. (7.12.1)

Let l = f(√

2, n)

and D ∈ P(Bn) be a packing set attaining this bound, i.e.,

there exists xi ∈ int(√

2Bn)∩ D, 1 ≤ i ≤ l, and let D = {x1, . . . ,xl}. Let

R <√

2 be the circumradius of convD and without loss of generality we mayassume D ⊂ RBn, i.e., the circumcenter of convD is 0. Due to Remark 7.11,among the l points there exists k + 1 affinely independent points x0, . . . ,xkwith k ∈ {1, . . . , n}, say, and λi > 0, 0 ≤ i ≤ k, such that

k∑i=0

λi xi = 0. (7.12.2)

We further have for 1 ≤ i 6= j ≤ l,

4 ≤ |xi − xj |2 ≤ 2 R2 − 2 〈xi,xj〉 < 4− 2 〈xi,xj〉 .

In particular, for any xj ∈ D \ {x1, . . . ,xk+1} we would have 〈xi,xj〉 < 0, 0 ≤i ≤ k, which is impossible on account of (7.12.2). Hence D = {x1, . . . ,xk+1}and so f(

√2, n) ≤ n+ 1. On the other hand, for instance, the circumradius of

a regular simplex of edge length 2 is equal to√

2√n/(n+ 1), and so we also

know f(√

2, n) ≥ n+ 1. �

7.13 Remark. Chronologically, the first upper bound is due to Blichfeldt 33

(1928), who proved

δ(Bn) ≤ n+ 2

2

√2−n.

33Hans Frederick Blichfeldt, 1873–1945

http://en.wikipedia.org/wiki/Hans_Frederick_Blichfeldt

Packing 73

This was slightly improved by Rogers34 (1958) by a factor of 2/e, roughlyspeaking. In 1973/74 Sidelnikov35 showed that

δ(Bn) ≤ 2−(0.509+o(1))n, n large.

Subsequently, this bound was improved by Levenshtein36 (1975), and Kaba-tiansky37 and Levenshtein (1978) to

δ(Bn) ≤ 2−(0.599+o(1))n, n large,

which is still the best known bound.

7.14 Theorem. Let K ∈ Kn and Λ ∈ Ln ∩ P(K). Then

δ(K,Λ) =vol (K)

det Λ.

Proof. Let PB be a fundamental cell of Λ. By Proposition 6.9, Rn is thepairwise disjoint union of the translates b+PB, b ∈ Λ. Hence #((x+PB)∩Λ) ≤1 for all x ∈ Rn, whereas the property that Rn is covered by all translates isequivalent to the lower bound #((x + PB) ∩ Λ) ≥ 1 for all x ∈ Rn. Hence#((x + PB) ∩ Λ) = 1 for all x ∈ Rn. Together with vol (PB) = det Λ theidentity follows from Lemma 7.10.

�

7.15 Definition [Density of a densest Lattice Packing]. For K ∈ Kn the setPL(K) = Ln ∩ P(K) is called the family of all packing lattices of K. ForΛ ∈ PL(K) the arrangement Λ +K is called a lattice packing of K and

δL(K) = sup{δ(K,Λ) : Λ ∈ PL(K)

}is called the density of a densest lattice packing of K.

7.16 Definition [Critical determinant and admissible lattices]. LetK ∈ Kn0 .A lattice Λ is called admissible for K (or K-admissible) if intK ∩ Λ = {0}.

∆(K) = inf {det Λ : Λ admissible for K}

is called the critical determinant of K.

7.17 Remark. Let K ∈ Kn0 and Λ ∈ Ln. Then

i) Λ is a packing lattice iff λ1(K,Λ) ≥ 2, and Λ is admissible iff λ1(K,Λ) ≥ 1.

ii) Λ is a admissible iff 2 Λ is a packing lattice.

34Claude Ambrose Rogers, 1920–200535Vladimir Michilovich Sidelnikov, 1940–200836Vladimir Iosifovich Levenshtein, 193537Grigory A. Kabatiansky, 1949

http://en.wikipedia.org/wiki/Claude_Ambrose_Rogers

http://www.mathnet.ru/php/person.phtml?option_lang=eng&personid=17348

http://en.wikipedia.org/wiki/Vladimir_Levenshtein

http://dynkincollection.library.cornell.edu/biographies/856

74 Packing

iii) 1λ1(K,Λ) Λ is admissible for K, and 2

λ1(K,Λ) Λ is a packing lattice.

7.18 Proposition* [Critical lattice]. For K ∈ Kn0 there exists a K-admissiblelattice ΛK with det ΛK = ∆(K). Such a lattice will be called a critical latticeof K.

7.19 Proposition. Let K ∈ Kn. Then

δL(K) =vol (K)

∆(K −K).

Proof. By Proposition 7.9 v) and the definition of admissible lattices (cf. Defi-nition 7.16) the family of packing sets PL(K) = PL

(12(K −K)

)coincides with

the set of all admissible lattices for K−K. From Corollary 7.14 and Proposition7.18 we get the desired identity. �

7.20 Proposition. Let K ∈ Kn0 and Λ ∈ Ln. Then

vol (K)

2n≤ ∆(K) ≤ det Λ

λ1(K,Λ)n

Proof. By Minkowski’s theorem 6.34 we get λ1(K,ΛK)nvol (K) ≤ 2n∆(K).Since λ1(K,ΛK) ≥ 1 we get the lower bound. For the upper bound we justnotice that by Remark 7.17 iii) we have ∆(K) ≤ det( 1

λ1(K,Λ)Λ). �

7.21 Conjecture [Davenport]. 38 Let K ∈ Kn0 and Λ ∈ Ln. Then

∆(K) ≤ det Λn∏i=1

1

λi(K,Λ).

7.22 Remark.

i) 0 < δL(K) ≤ δ(K) ≤ 1.

ii) δL(AK + t) = δL(K) for all A ∈ GL(n,R) and t ∈ Rn.

iii) For K ∈ Kn0 we have δL(K) = 2−nvol (K)/∆(K).

7.23 Theorem [Minkowski-Hlawka, 1943]. 39 Let S ⊂ Rn be a boundedJordan-measurable set with vol (S) < 1. Then there exists a lattice Λ ∈ Lnwith

det Λ = 1 and S ∩ Λ \ {0} = ∅.

38Harold Davenport, 1907 –196939Edmund Hlawka, 1916–2009

http://en.wikipedia.org/wiki/Harold_Davenport

http://de.wikipedia.org/wiki/Edmund_Hlawka

Packing 75

Proof. Since S is bounded and Jordan-measurable of vol (S) < 1, there existsa prime p such that

i)1

pn−1#

(S ∩ 1

p(n−1)/nZn)< 1,

ii) S ⊂{x ∈ Rn : |xi| < p1/n, 1 ≤ i ≤ n

}.

(7.23.1)

If we suppose that we can find pn−1 sublattices of Zn of determinant pn−1,such that {0} is the only common point in each two of them, then (7.23.1) i)would immediately imply the assertion. However, although we can not findthose sublattices, we can find sublattices whose common points different from0 are ”far away”, which is good by (7.23.1) ii).

Let Up = {u ∈ Zn : u1 = 1, 0 ≤ ui < p, i = 2, . . . , n}. For u ∈ Up let Λ(u)be the lattice with basis u, p e2, . . . , p en. Obviously, we have det Λ(u) = pn−1

and there are pn−1 sublattices of that type. Since u1 = 1 we observe that

z ∈ Λ(u)⇔ zi ≡ (z1 ui) mod p, i = 2, . . . , n, (7.23.2)

and next we claim for u 6= u ∈ Up

Λ(u) ∩ Λ(u) ⊂ {0} ∪ {x ∈ Rn : ∃xi with |xi| ≥ p}. (7.23.3)

To see this let z ∈ Λ(u)∩Λ(u). By (7.23.2) we have z1(ui− ui) ≡ 0 mod p andsince −(p − 1) ≤ ui − ui ≤ p − 1, we conclude z1 ≡ 0 mod p. If z1 6= 0 we aredone, and if z1 = 0 we get from (7.23.2) that zi ≡ 0 mod p, i = 2, . . . , n. Thus,either z = 0 or at least one coordinate is not less than p in absolute value.

In view of (7.23.1) ii) the inclusion (7.23.3) shows that the pn−1 sets

S ∩ 1

p(n−1)/nΛ(u) \ {0}, u ∈ Up,

are pairwise disjoint, and so (cf. (7.23.1) i))∑u∈Up

#

(S ∩ 1

p(n−1)/nΛ(u) \ {0}

)≤ #

(S ∩ 1

p(n−1)/nZn)< pn−1.

Thus, at least one of the lattices 1p(n−1)/nΛ(u) has the desired properties. �

7.24 Remark. Theorem 7.23 remains true for Jordan-measurable, unbounded,closed sets.

7.25 Corollary. Let S ⊂ Rn be a bounded Jordan-measurable set with S =−S and with vol (S) < 2. Then there exists a lattice Λ ∈ Ln with

det Λ = 1 and S ∩ Λ \ {0} = ∅.

Proof. Apply Theorem 7.23 to the set S = {x ∈ S : x1 > 0}. �

76 Packing

7.26 Corollary. Let K ∈ Kn0 . Then

δL(K) ≥ 2−(n−1)

[⇔ ∆(K) ≤ vol (K)

2

].

Proof. Without loss of generality let vol (K) = 2− ε for some ε > 0. Corollary7.25 shows that there exist a lattice Λε, with det Λε = 1, which is admissiblefor K and thus 2Λε ∈ PL(K). Hence we get

δL(K) ≥ vol (K)

det(2Λε)= 2−(n−1) − ε

2n.

Since this is true for every ε > 0 the statement is shown. �

7.27 Theorem*. Let S ⊂ Rn, n ≥ 2, be a bounded ray set (i.e., if x ∈ Sthen λx ∈ S for all λ ∈ [0, 1]) with vol (S) < ζ(n). Then there exists a latticeΛ ∈ Ln with det Λ = 1 and S ∩ Λ \ {0} = ∅.Here ζ(n) =

∑∞k=1 k

−n denotes the (Riemann) zeta function (ζ-function).

7.28 Theorem* [K. Ball, 1992]. 40 δL(Bn) ≥ (n− 1)2−(n−1)ζ(n).

7.29 Lemma [Lagrange, 1773]. 41 Let Λ ∈ L2 be a planar lattice. Thereexists a basis b1, b2 of Λ such that

det Λ ≥√

3

2|b1| |b2| .

Proof. Let b1, b2 be a basis of Λ, and let b2 be the orthogonal projection of b2

onto the line perpendicular to lin (b1), i.e.,

b2 = b2 −〈b1, b2〉|b1|2

b1.

Then det Λ = |b1|∣∣b2

∣∣. Replacing b2 by b2 + z b1 with z ∈ Z does not change

b2 and for any z ∈ Z the vectors b1, b2 + z b1 form a basis of Λ as well. So wemay choose a z ∈ Z such that for b2 = b2 + z b1∣∣∣∣∣∣

⟨b1, b2

⟩|b1|2

∣∣∣∣∣∣ =

∣∣∣∣〈b1, b2〉|b1|2

+ z

∣∣∣∣ ≤ 1

2.

Hence we get

∣∣b2

∣∣2 =∣∣∣b2

∣∣∣2 −⟨b1, b2

⟩2

|b1|2≥∣∣∣b2

∣∣∣2 − 1

4|b1|2 ≥

3

4

∣∣∣b2

∣∣∣2 ,since b1 was a shortest vector. Combing this bound with det Λ = |b1|

∣∣b2

∣∣ givesthe desired inequality. �

40Keith M. Ball, 196041Joseph-Louis Lagrange, 1736–1813

http://keithmball.wordpress.com

http://en.wikipedia.org/wiki/Joseph_Louis_Lagrange

Packing 77

7.30 Corollary [Lagrange, 1773].

δL(B2) =π

2√

3

[⇔ ∆(B2) =

√3

2

]

Proof. Let Λ be a packing lattice of the unit circle B2 with basis b1, b2.According to Theorem 7.14 and Lemma 7.29 we have

δ(B2,Λ) =vol (B2)

det Λ≤ 2√

3

π

|b1| |b2|≤ π

2√

3.

On the other hand, the hexagonal lattice Λhex with basis (2, 0)ᵀ and (1,√

3)ᵀ isa packing lattice of B2 with δ(B2,Λhex) = π/(2

√3). �

7.31 Theorem* [Thue, 1890, 1910]. 42 δ(B2) = δL(B2).

7.32 Lemma [Gauß, 1840]. 43 Let Λ ∈ L3. Then there exists a basis b1, b2, b3

of Λ such that

det Λ ≥ 1√2|b1| |b2| |b3| .

Proof. Let b1, b2, b3 be a basis of Λ such that b1 is a shortest lattice vector,b2 is a shortest vector among all vectors c ∈ Λ which can be together with b1

extended to a basis of Λ. Finally, let b3 be a shortest lattice vector such thatb1, b2, b3 is a basis of Λ. Then we have

|bi ± bj |2 ≥ |bi|2 , 1 ≤ i 6= j ≤ 3,

and thus

2|βi,j | ≤ βj,j , 1 ≤ i 6= j ≤ 3, (7.32.1)

with βi,j = 〈bi, bj〉, 1 ≤ i ≤ j ≤ 3. Then

(det Λ)2 = β1,1β2,2β3,3 − β1,1β22,3 − β2,2β

21,3 − β3,3β

21,2 + 2β1,2β1,3β2,3.

We can assume that either βi,j ≥ 0 or βi,j ≤ 0 for 1 ≤ i < j ≤ 3.

I. βi,j ≥ 0, 1 ≤ i < j ≤ 3.

2(det Λ)2 = β1,1β2,2β3,3

+ β1,1β2,3(β2,2 − 2β2,3) + β2,2β1,3(β3,3 − 2β1,3)

+ β3,3β1,2(β1,1 − 2β1,2) + β2,3(β1,1 − 2β1,3)(β2,2 − 2β1,2)

+ β1,3(β3,3 − 2β2,3)(β2,2 − 2β1,2) + β1,2(β3,3 − 2β2,3)(β1,1 − 2β1,3)

+ (β3,3 − 2β2,3)(β1,1 − 2β1,3)(β2,2 − 2β1,2)

≥ β1,1β2,2β3,3,

42Axel Thue, 1863–192243Carl Friedrich Gauss, 1777–1855

http://en.wikipedia.org/wiki/Axel_Thue

http://en.wikipedia.org/wiki/Carl_Friedrich_Gauss

78 Packing

where the last inequality follows from βi,j ≥ 0, 1 ≤ i 6= j ≤ 3, and (7.32.1).

II. βi,j ≤ 0, 1 ≤ i < j ≤ 3. Here we need one more inequality, namely|b1 + b2 + b3|2 ≥ |bi|2, 1 ≤ i ≤ 3, which may be rewritten as

αi,j = βi,i + βj,j + 2β1,2 + 2β2,3 + 2β1,3 ≥ 0, 1 ≤ i < j ≤ 3. (7.32.2)

Furthermore, let γi,j = 2βi,j + βj,j . By (7.32.1) we have

γi,j ≥ 0, 1 ≤ i, j ≤ 3. (7.32.3)

8(det Λ)2 = 4β1,1β2,2β3,3

− 2β1,1β2,3(γ2,3 + α2,3)− 2β2,2β1,3(γ1,3 + α1,3)

− 4β3,3β1,2α1,2

+ (β3,3 + γ1,3)γ2,3γ1,2 + (β3,3 + γ2,3)γ1,3γ1,2

≥ 4β1,1β2,2β3,3,

where the last inequality is an immediate consequence of βi,j ≤ 0, 1 ≤ i 6= j ≤ 3,cf. (7.32.2) and (7.32.3). �

7.33 Corollary [Gauss].

δL(B3) =π√18

[⇔ ∆(B3) =

1√2

].

Proof. Analogously to the proof of Corollary 7.30 we argue via Lemma 7.32that for any packing lattice Λ of B3 we have

δ(B3,Λ) ≥ π√18.

On the other hand let Λfcc = {z ∈ Z3 : z1 + z2 + z3 ≡ 0 mod 2}. Thendet Λfcc = 2 and λ1(B3,Λfcc) =

√2. Hence

√2Λfcc is a packing lattice of B3

with determinant 2√

23, given the density δ(B3,Λfcc) = π√

18. Λfcc is called the

face-centered-cubic lattice. �

7.34 Theorem* [Hales, 1998/2005, Proof of the “Kepler-conjecture”]. 44

δ(B3) = δL(B3) =π√18.

44Thomas Callister Hales, 1958

http://en.wikipedia.org/wiki/Thomas_Callister_Hales

Packing 79

7.35 Theorem* [Korkin-Zolotarev, 1872/73; Blichfeldt, 1934]. 45 46

δL(B4) =π2

16

[⇔ ∆(B4) =

1

2

]δL(B5) =

π2

15√

2

[⇔ ∆(B5) =

1

2√

2

],

δL(B6) =π3

48√

3

[⇔ ∆(B6) =

√3

8

],

δL(B7) =π3

105

[⇔ ∆(B7) =

1

8

],

δL(B8) =π4

384

[⇔ ∆(B8) =

1

16

].

7.36 Theorem* [Cohn&Kumar, 2004]. The so called Leech lattice ΛLeech isthe optimal packing lattice in dimension 24, and it is

δL(B24) =π12

12!

[⇔ ∆(B24) =

1

2n

].

7.37 Theorem [Swinnerton-Deyer, 1953]. 47 Let K ∈ Kn0 and let ΛK ∈ Lnbe a critical lattice of K. Then

#(K ∩ ΛK \ {0}

)≥ n(n+ 1).

Proof. Without loss of generality let ΛK = Zn, and let {±a1, . . . ,±ak} =K ∩Zn \ {0}. For T ∈ Rn×n with entries tl,m and for ρ ∈ R let Iρ,T = In + ρ T .In the following we show that we can choose parameters tl,m and ρ such thatthe corresponding lattice Λρ,T = Iρ,TZn is still admissible for K, but det Λρ,T <det ΛK = 1.

Let Hi ={x ∈ Rn : 〈ui, x〉 = 1

}be a supporting hyperplane of K at the

point ai and let ai,ρ,T = Iρ,T ai = ai + ρ Tai. Then we have

ai,ρ,T ∈ Hi ⇔ 〈ui, Iρ,T ai〉 = 1⇔ 〈ui, T ai〉 = 0⇔n∑

l,m=1

cl,m,i tl,m = 0,

where cl,m,i are certain numbers, depending on ai,ui. Since k < n (n+ 1)/2 wecan find non-trivial scalars tl,m ∈ R, say, such that

i) tl,m = tm,l, ii) ai,ρ,T ∈ Hi, 1 ≤ i ≤ k, and ρ ∈ R. (7.37.1)

Next we argue that we can find a ρ ∈ R such that Λρ,T is a K-admissible latticefor all |ρ| ≤ ρ. First we notice that for sufficiently small ρ, Λρ,T ∈ Ln, i.e.,

45Yegor Ivanovich Zolotarev, 1847–187846Aleksandr Korkin, 1837–190847Henry Peter Francis Swinnerton-Dyer, 1927

http://en.wikipedia.org/wiki/Yegor_Ivanovich_Zolotarev

http://en.wikipedia.org/wiki/Korkin

http://en.wikipedia.org/wiki/Peter_Swinnerton-Dyer

80 Packing

det Λρ,T > 0. Next, suppose that there exists a sequence ρi → 0 such thatthere exists uρi,T ∈ Λρi,T \ {0} ∩ intK for i ∈ N. Since Λρi,T → ΛK we mayassume that uρi,T → a1, say. We also know that a1,ρi,T

→ a1 and so we havea1,ρi,T

− uρi,T → 0, which implies a1,ρi,T= uρi,T for sufficiently large i. Since

a1,ρi,T∈ H1 we have uρi,T 6∈ intK, a contradiction.

Now since Λρ,T is K-admissible for all |ρ| ≤ ρ we must have det Λρ,T ≥det ΛK and so det(In + ρT ) ≥ 1 for all |ρ| ≤ ρ. Expressing det(In + ρT ) as thesum over all permutations leads to

det(In + ρT ) =

n∏i=1

(1 + ρ ti,i)− ρ2∑

1≤i<j≤nti,jtj,i

∏k/∈{i,j}

(1 + ρtk,k)

+ ρ3 γ3 + · · ·+ ρnγn,

where the first two terms reflect the identity permutation and the transposi-tions, and γi are certain constants depending on T . Hence we get

det(In + ρT ) = 1 + τ1 ρ+ τ2 ρ2 + ρ3γ3 + · · ·+ ρnγn,

with

τ1 =n∑j=1

tj,j , τ2 =∑

1≤i<j≤n(ti,i tj,j − ti,j tj,i),

and some other constants γi. Since det(In + ρT ) ≥ 1 for all |ρ| ≤ ρ we first getτ1 = 0 and then τ2 ≥ 0, which gives by (7.37.1) i)

0 ≤ 2τ2 − (τ1)2 = 2∑i<j

ti,i tj,j − 2∑i<j

(ti,j)2 −

n∑i=1

(ti,i)2 − 2

∑i<j

ti,i tj,j

= −n∑i=1

(ti,i)2 − 2

∑i<j

(ti,j)2.

Thus we have T = 0, which contradicts the choice of T . �

7.38 Theorem. Let K ∈ K20.

i) Let b1, b2 ∈ bdK such that b2 − b1 ∈ bdK. Then the lattice Λ =(b1, b2)Z2 is admissible for K.

ii) Let Λ ∈ L2 be a critical lattice of K. Then there exists a basis b1, b2 ofΛ such that b1, b2, b2 − b1 ∈ bdK.

Proof. For i) we first notice that, since b1, b2, b2 − b1 ∈ bdK then

intK ∩{z1 b1 + z2 b2 : z2 ∈ {0,±1}, z1 ∈ Z

}= {0}.

Now assume that there exists b = z1 b1 + z2 b2 ∈ intK and without loss ofgenerality let z2 ≥ 2. Further let ε > 0 such that b ± εb1 ∈ intK and let

Packing 81

P = conv {±b1, b± εb1}. Then vol 1

(P ∩ (z2 b2 + lin {b1})

)≥ 2ε |b1|, and since

vol 1

(P ∩ lin {b1}

)= 2 |b1| we conclude

vol 1

(P ∩ (b2 + lin {b1})

)> |b1|

This shows, however, that either b2 or b2 − b1 belong to the interior of K.For ii) we observe that on account of det Λ = ∆(K) we can find two linearly

independent points b1, b2 ∈ Λ ∩ bdK. On account of Lemma 6.17 we mayassume that b1, b2 build a basis of Λ. Now let α1, α2 ∈ R≥0 be maximal suchthat b2 − α1 b1 ∈ bdK and b2 + α2 b1 ∈ bdK.

If α1 ≥ 1 then we must have b2 − b1 ∈ bdK and we are done. Similarly, ifα2 ≥ 1 we have b2 + b1 ∈ bdK and the basis b1, b1 + b2 of Λ has the requiredproperty. So we may assume αi < 1. If α1, α2 > 0 then ±b2 + lin {b1} aresupporting lines of K, and from this we conclude that K∩Λ\{0} = {±b1,±b2},which contradicts Theorem 7.37. Hence we may assume that α2 = 0 and sinceα1 < 1 we can find λ ∈ (0, 1) such that vol 1

(K∩ (λ b2 +lin {b1})

)= |b1|. Let u

and v be the corresponding points in the boundary, such that v = u−b1. By i)we know that the lattice with basis b1,u is admissible for K, but

∣∣det(b1,u)∣∣ =∣∣det(b1, λ b2)

∣∣ < ∆(K), which contradicts the definition of ∆(K). �

7.39 Corollary. Let K ∈ K20 and HK be an affinely regular hexagon of mini-

mal volume with vertices on bdK. Then

δL(K) =3

4

vol (K)

vol (HK)

[⇔ ∆(K) =

1

3vol (HK)

].

Proof. Let ±v1,±v2,±v3 be the vertices of HK on the boundary of K.Since HK is affinely regular we may assume that v3 = v2 − v1. Hence, byTheorem 7.38 i), the lattice Λ(HK) with basis v1,v2 is admissible for K and itis det Λ(HK) = vol (HK)/3. Thus ∆(K) ≤ vol (HK)/3.

On the other hand, by Theorem 7.38 ii), each critical lattice ΛK gives riseto such an affinely regular hexagon H(ΛK) with vertices on the boundary andvol (H(ΛK)) = 3∆(K). Thus we also have ∆(K) ≥ vol (H(ΛK))/3. �

7.40 Theorem* [Fejes Toth, 1950; Rogers, 1951]. Let K ∈ K2. Then

δ(K) = δL(K).

82 Packing

Count and generate 83

8 Count and generate

8.1 Definition [Lattice polytope]. A polytope P = conv {v1, . . . ,vm} ⊂ Rnis called a lattice polytope if vi ∈ Zn, 1 ≤ i ≤ m. The set of all lattice polytopesis denoted by PnZ .

8.2 Notation.

i) For S ⊂ Rn we denote by G(S) the lattice point enumerator, i.e., G(S) =#(S ∩ Zn).

ii) For integers m,n we denote by(x+m

n

)=

1

n!

n−1∏i=0

(x+m− i)

the polynomial of degree n with roots i−m, i = 0, . . . , n− 1, and leadingcoefficient 1/n!. In particular, the polynomials

(x+n−in

), i = 0, . . . , n, form

a basis of the space of all polynomials of degree at most n.

8.3 Lemma. Let T = conv {0,v1, . . . ,vn} ∈ PnZ be a lattice simplex, i.e.,v1, . . . ,vn ∈ Zn are linearly independent, and for 0 ≤ i ≤ n let

Ui =

n∑j=1

λjvj ∈ Zn : 0 ≤ λj < 1, i− 1 <n∑j=1

λj ≤ i

Then for all k ∈ N, k ≥ 1, we have

G(k T ) =n∑i=0

#Ui ·(k + n− i

n

).

In particualr, G(k T ) is a polynomial of degree n in k whose coefficients areintegral multiples of n!.

Proof. For m ∈ Z let

Qm =

{n∑i=1

qi vi : qi ∈ N and

n∑i=1

qi ≤ m

},

where we setQm = ∅ ifm < 0. Furthermore, let U = {∑n

i=1 λivi ∈ Zn : 0 ≤ λi < 1}be the half open fundamental cell of the lattice generated by v1, . . . ,vn. ThenU is the disjoint union of the Ui’s, and each integral point z ∈ kT ∩ Zn has aunique representation as

z = uz +wz (8.3.1)

with uz ∈ Ui and wz ∈ Qk−i for a suitable i ∈ {0, . . . , n}. For if, let z =∑ni=1 µi vi ∈ kT ∩ Zn. Then 0 ≤ µi ≤ k,

∑ni=1 µi ≤ k and we may write

z =

n∑i=1

(µi − bµic)vi +

n∑i=1

bµicvi.


The first sum on the right hand side is in U , and thus in some Uj , say, and onaccount of 0 ≤ µi ≤ k,

∑ni=1 µi ≤ k we also get that the second sum is in Qk−j .

Regarding the uniqueness we observe that u+w = u+w for u ∈ Ui, u ∈ Ui,w ∈ Qm, w ∈ Qm implies that

u− u ∈ (U − U) ∩ (Qm −Qm).

By Remark 6.7 iv), the intersection on the right hand side contains only 0.Hence the representation (8) is unique. Since we obviously have Ui + Qk−i ⊂k T ∩ Zn we have shown

k T ∩ Zn =⋃· ni=0 (Ui +Qk−i) , (8.3.2)

and #(Ui +Qk−i) = #Ui #Qk−i. It finally remains to verify that

#Qm =

(n+m

n

),

which follows easily, e.g., by induction on m,n. �

8.4 Remark. The statement of the lemma above is also true for lower dimen-sional simplices and arbitrary lattices.

8.5 Lemma [Inclusion-Exclusion Formula]. Let Ai ⊆ Rn, 1 ≤ i ≤ m, withcharacteristic functions χ(Ai). Then

χ(A1 ∪A2 ∪ · · · ∪Am) =∑

I⊆{1,...,m}I 6=∅

(−1)#I−1χ

⋂j∈I

Aj

.

Proof. First we observe that χ(A) · χ(B) = χ(A ∩ B) for any two subsetsA,B ⊆ Rn. Hence the right hand side can be rewritten as

1−n∏i=1

(1− χ(Ai)), (8.5.1)

where 1 is the 1-constant function. Now the function in (8.5.1) takes the value1 exactly for those x ∈ Rn for which one of the functions 1 − χ(Ai) takes thevalue 0, i.e., if and only if x ∈ A1 ∪A2 ∪ · · · ∪Am. �

8.6 Definition [Triangulation]. A triangulation of a convex n-polytope P isa finite collection T of n-simplices such that

i) P is the union of all simplices in T .

ii) For any two simplices τ1, τ2 ∈ T their intersection τ1∩τ2 is a face commonto both τ1 and τ2.


8.7 Theorem. Every n-polytope P can be triangulated such that the verticesof any simplex in the triangulation are vertices of P .

Proof. Let V = {v1, . . . ,vm} be the set of vertices of P . We want to findnon-negative numbers η1, . . . , ηm such that for any choice of n+ 1 affinely inde-pendent points vi1 , . . . ,vin+1 ∈ V the unique hyperplane containing the liftedpoints (vij , ηij )

ᵀ, 1 ≤ j ≤ n+ 1, contains no other lifted points. Then all facesof the convex hull of this lifted configuration, i.e., conv {(vij , ηij )ᵀ : 1 ≤ i ≤ m}are simplices and the projected faces of the lower convex hull (i.e., factes, whoseouter unit normal vector has a negative coordinate) gives the desired regulartriangulation. Now for a given set vi1 , . . . ,vin+1 ∈ V this is equivalent to saythat

det

1 1 . . . 1 1vi1 vi2 . . . vin+1 vkηi1 ηi2 . . . ηin+1 ηk

(8.7.1)

is non zero for all k ∈ {1, . . . ,m} \ {i1, . . . , in+1}. Evaluating this determinantwith respect to the last row shows that the determinant is zero if and onlyif (ηi1 , ηi2 , . . . , ηin+1 , ηk)

ᵀ satisfies a non-trivial linear equation. Thus, exceptpoints lying in a hyperplane of the form 〈w, (η1, . . . , ηm)ᵀ〉 = 0 for a certainw ∈ Rm yielding a non-zero determinant in (8.7.1).

Hence for almost any choice of (η1, . . . , ηm)ᵀ ∈ Rm≥0 all the determinants oftype (8.7.1) for any choice of affinely independent points vi1 , . . . ,vin+1 ∈ V andvk ∈ V \ {vi1 , . . . ,vin+1} are non zero. �

8.8 Theorem [Ehrhart, 1967]. 48 Let P ∈ PnZ . Then there exist numbersGi(P ) ∈ Q, 0 ≤ i ≤ n, depending only on P , such that for all k ∈ N≥1

G(k P ) =n∑i=0

Gi(P ) ki.

The right hand side is called Ehrhart-polynomial.

Proof. Without loss of generality let dimP = n. On account of Theorem 8.7there exists a triangulation T = {τ1, . . . , τm} of P where the vertices of each τiare vertices of P . Then each τi as well as the intersection of two τi’s are m-dimensional lattice simplices or empty. Hence with Lemma 8.5 and Lemma 8.3we get

G(k P ) = G

(m⋃i=1

k τi

)=

∑I⊆{1,...,m}

I 6=∅

(−1)#I−1G

k ⋂j∈I

τj

=∑

I⊆{1,...,m}I 6=∅

(−1)#I−1

dim(⋂j∈I τj)∑

i=0

Gi

⋂j∈I

τj

ki,

which also implies Gi(P ) ∈ Q. �

48Eugene Ehrhart, 1906– 2000

http://en.wikipedia.org/wiki/Eug�ne_Ehrhart


8.9 Proposition. Let P ∈ PnZ .

i) Gn(P ) = vol (P ).

ii) Gi : PnZ → R is

a) homogeneous of degree i,

b) invariant with respect to unimodular transformations, i.e., for anyU ∈ GL(n,Z), t ∈ Zn, it is Gi(t+ UP ) = G(P ),

iii) additive, i.e., for P,Q ∈ PnZ with P ∪Q,P ∩Q ∈ PnZ it is Gi(P ∪Q) =Gi(P ) + Gi(Q)−Gi(Q ∩ P ).

iii) Gi(P ) are independent of the dimension of the space in which P is em-bedded, i.e., let P ∈ PnZ and let P = conv {(v, 0)ᵀ : v ∈ P} ∈ Pn+1

Z . ThenGi(P ) = Gi(P ), i = 0, . . . , n.

Proof. By the Riemann integrability of the characteristic function of P we getin view of Theorem 8.8

vol (P ) = limm→∞

#(P ∩ 1mZn)

mn= lim

m→∞

G(mP )

mn

= limm→∞

n∑i=0

Gi(P )mi−n = Gn(P ).

For ii) we observe that for k,m ∈ N≥1, U ∈ GL(n,Z), t ∈ Zn and P,Q ∈ PnZn∑i=0

(Gi(P )mi

)ki = G((km)P ) = G(k(mP )) =

n∑i=0

(Gi(mP )) ki,

n∑i=0

Gi(t+ U P )ki = G(k(t+ U P )) = G(k P ) =

n∑i=0

Gi(P )ki,

n∑i=0

(Gi(P ) + Gi(Q)) ki = G(k P ) + G(k Q) = G(k(P ∪Q)) + G(k(P ∩Q))

=

n∑i=0

(Gi(P ∪Q) + Gi(P ∩Q)) ki,

where for the last we also assume P ∪Q,P ∩Q ∈ PnZ . Comparing the coefficientsin all three equations shows the required properties of Gi(P ).

Obviously, G(k P ) = G(k P ) for k ∈ N≥1, and this shows iii). �

8.10 Theorem* [Betke& Kneser, 1985]. 49 50 Every additive and unimodu-lar invariant functional on the space of all lattice polytopes is a linear combi-nation of the n+ 1 functionals Gi(·).

49Martin Kneser, 1928 – 200450Ulrich Betke, 1948 – 2008

http://en.wikipedia.org/wiki/Martin_Kneser


8.11 Remark. Some of the coefficients Gi(P ) might be negative. One familyof standard examples in this context are the so called Reeve-simplices: letRm = conv

{0, e1, e2, (1, 1,m)ᵀ

}∈ P3

Z for m ∈ N. The only lattice pointscontained in Rm are the four vertices, the volume, however, can be arbitrarilylarge. Hence some Gi(Rm) must be negative for large m. More precisely, it isG3(Rm) = m/6, G2(Rm) = 1, G1(Rm) = (12−m)/6 and G0(Rm) = 1.

8.12 Corollary. Let P ∈ PnZ . Then there exist numbers ai(P ) ∈ Q, 0 ≤ i ≤ n,such that for all k ∈ N≥1

G(k P ) =n∑i=0

ai(P )

(k + n− i

n

).

8.13 Example.

i) Let Tn = conv {0, e1, e2, . . . , en}. Then

#(k Tn ∩ Zn) =

(n+ k

n

),

and so we have ai(Tn) = 0 for 1 ≤ i ≤ n, and a0(Tn) = 1. The Gi(Tn) are– up to ±1 – Stirling numbers of the first kind.

ii) Let Cn = [−1, 1]n. Then G(k Cn) = (2 k+ 1)n and so Gi(Cn,Zn) = 2i(ni

),

0 ≤ i ≤ n. Here the ai(Cn) are some combinatorial numbers, the so calledEulerian numbers.

iii) Let C?n = conv {±ei : 1 ≤ i ≤ n}. Then

G(k C?n) =

n∑i=0

(n

i

)(k + n− i

n

),

and so ai(C?n) =

(ni

), i = 0, . . . , n.

8.14 Definition [Generating function]. For S ⊆ Rn the function (formal powerseries)

γ(S;x) =∑

m∈S∩Znxm

is called the generating function or the integer-point transform of S. Here xm

denotes the monomial xm11 xm2

2 · . . . · xmnn .

8.15 Example. Let S1 = [0,∞) and S2 = (−∞, 2]. Then

γ(S1;x) =∞∑i=0

xi =1

1− x, |x| < 1,

γ(S2;x) = x2 + x+−∞∑i=0

xi = x2 + x+1

1− 1/x, |x| > 1.


For P = [0, 2] = S1 ∩ S2 we observe that

γ(P ;x) = 1 + x+ x2 =1

1− x+ x2 + x+

1

1− 1/x= γ(S1;x) + γ(S2;x).

8.16 Definition [Simplicial, rational, pointed cones]. Let v1, . . . ,vk ∈ Rn.Then C = pos {v1, . . . ,vk} is called a polyhedral cone generated by v1, . . . ,vk.If vi ∈ Zn, then C is called a rational cone. C is called a polyhedral pointedcone, if 0 is a vertex of C. If v1, . . . ,vk are linearly independent, then C iscalled a simplicial cone.

8.17 Lemma. Let C = pos {v1, . . . ,vk} be a rational simplical cone, W ={x ∈ Cn : |xvi | < 1 : 1 ≤ i ≤ k}, and let t ∈ Rn.

i) γ(t+C;x) converges absolutely and uniformly on all compact subsets ofW to the rational function

γ(t+ C;x) =γ(t+ U ;x)

(1− xv1) · (1− xv2) · . . . · (1− xvk),

where U = {∑k

i=1 αi vi : 0 ≤ αi < 1}.

ii) γ(t+ relintC;x) converges absolutely and uniformly on all compact sub-sets of W to the rational function

γ(t+ relintC;x) =γ(t+ U ;x)

(1− xv1) · (1− xv2) · . . . · (1− xvk),

where U = {∑k

i=1 αi vi : 0 < αi ≤ 1}.

iii) If (t+ relbdC)∩Zn = ∅, then U or U may be replaced by any set U withrelintU ⊆ U ⊆ clU .

Proof. First we note that the given set W is non-empty. To this end let C? ={y ∈ Rn : 〈x,y〉 ≤ 0, for all x ∈ C} be the polar cone. Since C is a pointed conewe have dimC? = n, and for y ∈ intC?, s ∈ Cn let ey+i s ∈ Cn be the vectorwith entries eyj+i sj . Then |

(ey+i s

)vk | = |e〈vk,y〉| |ei〈vk,s〉| < |ei〈vk,s〉| ≤ 1, thus{ey+i s : y ∈ intC?, s ∈ Cn

}⊆W.

Now let CN = {∑k

i=1 qi vi : qi ∈ N}. For i) we once again exploit the factthat any vector z ∈ (t + C) ∩ Zn can be uniquely written as t + u + h withu ∈ U and h ∈ CN. For if, let z = t+

∑ki=1 ρi vi, ρi ≥ 0, then

z = t+k∑i=1

(ρi − bρic)vi +k∑i=1

bρicvi = t+ u+ h


with u ∈ U and h ∈ CN, and the uniqueness follows from Remark 6.7 iv).Hence

γ(t+ C;x) =∑

m∈(t+C)∩Znxm =

∑g∈t+U,h∈CN

xg xh

=∑

g∈t+Uxg

∞∑q1,...,qk=0

xq1 v1+···+qk vk

=

∑g∈t+U

xg

k∏i=1

∞∑qi=0

(xvi)qi

=∑

g∈t+Uxg

k∏i=1

1

1− xvi

= γ(t+ U ;x)

k∏i=1

1

1− xvi,

where the last equation means absolutely convergence for x ∈ W . Hence wealso have uniformly convergence on compact subsets of W .

In order to get a unique representation of a point z ∈ (t + relintC) ∩ Zn,i.e., z = t+

∑ki=1 ρi vi ∈ Zn with ρi > 0 (cf. Theorem 0.21), we write

z = t+

∑ρi /∈N

(ρi − bρic)vi +∑

ρi∈N≥1

vi

+

∑ρi /∈N

bρicvi +∑

ρi∈N≥1

(ρi − 1)vi

= t+ u+ h,

with u ∈ U and h ∈ CN, and as before such a representation is unique. It justremains to note that also t+ U + CN ⊆ (t+ relintC) ∩ Zn (cf. Theorem 0.21),and we can proceed as before.

For iii) we observe that there are no lattice points z = t +∑k

i=1 αi vi,0 ≤ αi ≤ 1, in the cone with some αi = 0 or some αi = 1. Of course, αi = 0implies z ∈ t+relbdC and αi = 1 would imply t+

∑ki=1(1−αi)vi ∈ t+relbdC.

Hence the sets U and U in i) and ii) may be replaced by any set U with therequired property. �

8.18 Lemma. Every polyhedral pointed cone C = pos {v1, . . . ,vk}, vi ∈ Rn,can be triangulated into (dimC)-dimensional simplicial cones generated byv1, . . . ,vk, i.e., there exists a collection S = {σ1, . . . , σm} of (dimC)-dimensionalsimplicial cones σi such that C = ∪mi=1σi, and for any σ1, σ2 ∈ S, σ1 ∩ σ2 is theface common to both σ1 and σ2.

Proof. Without loss of generality let dimC = n. Since 0 is a vertex C,there exists an a ∈ Rn with 〈a,vi〉 > 0 for all vi, 1 ≤ i ≤ k. Let P =conv {vi/ 〈a,vi〉 : 1 ≤ i ≤ k}. Then C = posP and each triangulation T ={τ1, · · · , τm} of P into (dimC)-dimensional simplices τi yields a triangulationS = {σ1, . . . , σm} of C into simplicial cones σi = pos τi. Hence the assertionfollows from Theorem 8.7. �

8.19 Corollary. Let C ⊂ Rn be a rational pointed cone, and let t ∈ Rn. Thenthere exists an open subset W ⊂ Cn such that γ(t+C;x) converges absolutelyand uniformly to a rational function on all compact subsets of W .


Proof. Let C = pos {v1, . . . ,vm}, vi ∈ Zn, and let S = {σ1, . . . , σl} be atriangulation of the cone C via its generators v1, . . . ,vm (cf. Lemma 8.18).Applying the Inclusion-Exclusion formula (Lemma 8.5) gives as formal powerseries the identity

γ(t+ C;x) =∑

I⊆{1,...,l}I 6=∅

(−1)#I−1γ

t+⋂j∈I

σj ;x

.

Each ∩j∈Iσj is a rational simplicial cone and by Lemma 8.17 we know thatγ (t+ ∩j∈Iσj ;x) converges absolutely and uniformly to a rational function onall compact subsets of W = {x ∈ Cn : |xvi | < 1 : 1 ≤ i ≤ m}. As in the proofof Lemma 8.17 it can be argued that W 6= ∅. �

8.20 Lemma. Let C ⊂ Rn be a rational pointed n-dimensional cone. Letv ∈ Rn, and let S = {σ1, . . . , σl} be a triangulation of C. Then there exists ans ∈ Rn such that

i) (v + int (C))∩Zn = (s+ C)∩Zn and ii) bd (±s+σi)∩Zn = ∅, 1 ≤ i ≤ l.

In particular, for such an s it holds (−v + C) ∩ Zn = (−s+ C) ∩ Zn.

Proof. Let C = {x ∈ Rn : 〈ai,x〉 ≤ 0, 1 ≤ i ≤ m} with ai ∈ Zn, and we mayassume gcd(ai) = 1. Then v + C = {x ∈ Rn : 〈ai,x〉 ≤ 〈ai,v〉 , 1 ≤ i ≤ m},and the points s satisfying i) are given by the points for which {z ∈ Zn :〈ai, z〉 < 〈ai,v〉} = {z ∈ Zn : 〈ai, z〉 ≤ 〈ai, s〉}, 1 ≤ i ≤ m. Hence the set ofall possible points s fulfilling i) is given by

S ={x ∈ Rn : d〈ai,v〉 − 1e ≤ 〈ai,x〉 < d〈ai,v〉e, 1 ≤ i ≤ m

}.

In particular, for such an s we also have (−v + C) ∩ Zn = (−s + C) ∩ Zn.S is a full dimensional set; on the other hand, ii) imposes only finitely manyconditions on s and so ii) is satisfied for almost all points in S, in particular,for any vector s ∈ S whose entries together with the number 1 are linearlyindependent over Q. �

8.21 Theorem [Stanley’s Reciprocity Theorem]. Let C be a rational pointedcone and let v ∈ Rn. Then

γ(v + C;x) = (−1)dimCγ(−v + int (C);x−1), 51

where x−1 = ( 1x1, . . . , 1

xn).

Proof. Let dimC = n, and let {σ1, . . . , σm} be a triangulation of C, and let−s be satisfying Lemma 8.20 with respect to the vector −v. Then we have

γ(−v + intC;x) = γ(−s+ C;x) =m∑i=1

γ(−s+ σi;x) =m∑i=1

γ(−s+ intσi;x)

γ(v + C;x) = γ(s+ C;x) =m∑i=1

γ(s+ σi;x),

51Here the equality between the generating functions is meant as equality between therational functions represented by these functions (cf. Lemma 8.17).


and it suffices to show the assertion for a cone s+C, say, where C is a rationalsimplicial cone generated by the vectors v1, . . . ,vn ∈ Zn. Let U = {

∑ni=1 αi vi :

0 ≤ αi < 1} and U = {∑n

i=1 αi vi : 0 < αi ≤ 1}. With this notion we haves+ U = s+ (v1 + · · ·+ vn)− U and thus

∑z∈(s+U)∩Zn

xz =∑

z∈(−(−s+U)+(v1+···+vn))∩Znxz =

∑w∈(−s+U)∩Zn

x−w xv1 ·. . .·xvn ,

and so

γ(s+ U ;x) = γ(−s+ U ;x−1)xv1 · . . . · xvn .

Finally, with Lemma 8.17 we get

γ(−s+ intC;x−1) =γ(−s+ U ;x−1)

(1− x−v1) · . . . · (1− x−vn)

=γ(−s+ U ;x−1)xv1 · . . . · xvn

(xv1 − 1) · . . . · (xvn − 1)

= (−1)nγ(s+ U ;x)

(1− xv1) · . . . · (1− xvn)= (−1)n γ(s+ C;x).

�

8.22 Definition [Ehrhart series]. For bounded S ⊂ Rn and t ∈ R the formalseries

L(S; t) = 1 +∑k≥1

G(k S) tk

is called the Ehrhart series of S.

8.23 Example. Let Q ⊂ Rn−1×{0} be an (n−1)-dimensional lattice polytope.

i) If P is the pyramid P = conv {Q, en} then

L(P ; t) =1

1− tL(Q; t), |t| < 1.

In particular, we obtain for the standard simplex: L(Tn; t) = 1(1−t)n+1 .

To see this we note that for such a pyramid P and an integer k the sectionof k P with a hyperplane {x ∈ Rn : xn = l}, 0 ≤ l ≤ k − 1, is just len +(k−l)Q. So G(k P ) = 1+

∑k−1l=0 G((k−l)Q) = 1+G(k Q)+

∑k−1l=1 G(l Q),


and we obtain for |t| < 1

L(P ; t) = 1 +∑k≥1

G(k P ) tk

= 1 +∑k≥1

tk +∑k≥1

G(k Q) tk +∑k≥1

k−1∑l=1

G(l Q) tk

= L(Q; t) +t

1− t+∑k≥1

k−1∑l=1

G(l Q) tk

= L(Q; t) +t

1− t+∑l≥1

G(l Q)∑k≥l+1

tk

= L(Q; t) +t

1− t+∑l≥1

G(l Q)tl+1

1− t

= L(Q; t) +t

1− tL(Q; t) =

1

1− tL(Q; t).

In the case Tn we observe that L(Tn; t) = 1 +∑

k≥1(k+ 1) tk =∑

k≥0(k+

1) tk = 1/(1− t)2, and together with the previous recursion for pyramidswe get L(Tn; t) = 1/(1− t)n+1.

ii) If P is the bypyramid P = conv {Q,± en} then

L(P ; t) =1 + t

1− tL(Q; t), |t| < 1.

In particular, we obtain for the crosspolytope: L(C?n; t) = (1+t)n

(1−t)n+1 .

This can be easily calculated in the same way as before, or we just observethat G(k P ) = 2 G(k P )−G(k Q), where P is the pyramid conv {Q, en}.Hence by the previous example we find for |t| < 1

L(P ; t) = 1 +∑k≥1

G(k P ) = 2 (1 +∑k≥1

G(k P )tk)− (1 +∑k≥1

G(k Q)tk)

= 2 L(P ; t)− L(Q; t) = 21

1− tL(Q; t)− L(Q; t)

=1 + t

1− tL(Q; t).

Finally, we note that L(C?1 ; t) = (1 + t)/(1− t)2.

8.24 Proposition. Let P ∈ Pn and let P = conv {(x1

): x ∈ P} be its canoni-

cal embedding into Rn+1. Then L(P ; t) = γ(pos P ; (1, t)), where 1 ∈ Rn is theall 1-vector.

Proof. Obviously, a point (y1, . . . , yn, yn+1)ᵀ belongs to pos P iff (y1, . . . , yn)ᵀ ∈yn+1 P . Thus

γ(pos P ; (x, t)) = 1 +∑

(z,k)ᵀ∈pos P∩Zn+1

k≥1

xz tk = 1 +∑k≥1

tk∑

z∈k P∩Znxz,


and the right hand side evaluates to L(P ; t) for x = 1. �

8.25 Lemma. Let f : N→ R and ci ∈ R, 0 ≤ i ≤ n. Then

∑k≥0

f(k) tk =

∑ni=0 ci t

i

(1− t)n+1

for all |t| < 1 if and only if f(k) =∑n

i=0 ci(k+n−in

)for all k ∈ N.

Proof. By taking the n+ 1-st power (or the n-the derivative) of the geometricseries

∑k≥0 t

k = 1/(1− t), |t| < 1, we see that

1

(1− t)n+1=∑k≥0

(n+ k

n

)tk, (8.25.1)

and so∑ni=0 ci t

i

(1− t)n+1=

n∑i=0

ci ti∑k≥0

(n+ k

n

)tk =

n∑i=0

∑k≥i

ci

(n+ k − i

n

)tk

=∑k≥0

min{k,n}∑i=0

ci

(n+ k − i

n

) tk =∑k≥0

(n∑i=0

ci

(n+ k − i

n

))tk.

(8.25.2)

Comparing the coefficients of∑

k≥0 f(k) tk with the powers series on the righthand side finishes the proof. �

8.26 Corollary. Let P ∈ PnZ , and let ai(P ) ∈ Q, 0 ≤ i ≤ n, given by Corollary8.12. Then

L(P ; t) =

∑ni=0 ai(P ) ti

(1− t)n+1, |t| < 1,

and, in particular, a0(P ) = 1 = G0(P ).

Proof. If P is an l-dimensional lattice simplex T , say, then we know by Lemma8.3 that G(k T ) =

∑li=0 ai(T )

(l+k−il

), k ≥ 1, and a0(T ) = 1. Hence, rewriting

as a polynomial in the basis(n+k−in

)gives G(k T ) =

∑ni=0 ai(T )

(n+k−in

), k ≥ 1,

with a0(T ) = 1. Setting f(k) = G(k T ), k ≥ 1, and f(0) = 1 we may writef(k) =

∑ni=0 ai(T )

(n+k−in

), k ≥ 0, and

L(T ; t) = 1 +∑k≥1

G(k T ) tk =∑k≥0

f(k) tk.

Thus, by Lemma 8.25, the statement is proved for lattice simplices.

Now, let {τ1, . . . , τm} be a triangulation of P , and let τ i, P be the cor-responding canonical embeddings. By Proposition 8.24 and the Inclusion-


Exclusion principle we may write

L(P ; t) = γ(P ; (1, t)) =∑

I⊆{1,...,m}I 6=∅

(−1)#I−1γ

⋂j∈I

τj ; (1, t)

=∑

I⊆{1,...,m}I 6=∅

(−1)#I−1L

⋂j∈I

τj ; t

.

Since the lemma is verified for lattice simplices, we conclude that there existnumbers ci ∈ Z such that

L(P ; t) =

∑ni=0 ci t

i

(1− t)n+1, |t| < 1.

Hence by Lemma 8.25 we get G(k P ) =∑n

i=0 ci(n+k−in

), and in view of Corol-

lary 8.12 it is ci = ai(P ), 0 ≤ i ≤ n.a0(P ) is the constant coefficient in the polynomial representation of G(k P )

via the binomial basis and with respect to the monomial basis G0(P ) is the con-stant coefficient. Hence a0(P ) = G0(P ) and from the identity is the statementwe get a0(P ) = L(P ; 0) = 1. �

8.27 Theorem [Stanley’s Non-Negativity Theorem]. Let P ∈ PnZ , dimP =n. Then ai(P ) ∈ N≥0, 0 ≤ i ≤ n.

Proof. Let P be the canonical embedding of P into Rn+1, and let {σ1, . . . , σm}be a triangulation of posP into simplicial cones generated by the generators ofposP . According to Lemma 8.20 we choose an −s ∈ Rn+1 with respect to thecone posP , i.e., posP ∩ Zn+1 = (s+ posP ) ∩ Zn+1, and thus −s ∈ posP , and{s+ σ1, . . . , s+ σm} is a triangulation of s+ posP such that no lattice pointsare contained in the boundaries of s+ σi. Hence we have

L(P ; t) = γ(posP ; (1, t)) = γ(s+ posP ; (1, t)) =

m∑i=1

γ(s+ σi; (1, t)). (8.27.1)

Now let σi = pos {(v(i)

11

), . . . ,

(v(i)n+11

), where v

(i)1 , . . . ,v

(i)n+1 are vertices of P , and

let

Ui =

n∑j=1

αj

(v

(i)j

1

): 0 ≤ αj < 1

.

Then by Lemma 8.17 i) we can continue (8.27.1) for |t| < 1 as

L(P ; t) =m∑i=1

γ(s+ σi; (1, t)) =m∑i=1

γ(s+ Ui; (1, t))

(1− t)n+1

=1

(1− t)n+1

m∑i=1

∑z∈(s+Ui)∩Zn+1

tzn+1 .

(8.27.2)


So we have written L(P ; t) as a (Laurent) polynomial with non-negative coef-ficients. Comparing (8.27.2) with the representation in Corollary 8.26 shows

ai(P ) = # {z ∈ ∪mi=1(s+ Ui) : zn+1 = i} , 0 ≤ i ≤ n. (8.27.3)

�

8.28 Theorem [Stanley’s Monotonicity Theorem]. Let P,Q ∈ PnZ , dimP =dimQ = n with P ⊆ Q. Then ai(P,Zn) ≤ ai(Q,Zn), 0 ≤ i ≤ n.

Proof. We start as in the proof of Theorem 8.27 with a triangulation ofthe cone posP by the generators of posP . Then we extend this triangulationto a triangulation of the cone posQ, where Q is the canonical embedding ofQ into Rn+1. Next, with respect to the cone posQ we choose a translationvector s as in Lemma 8.20 and by the interpretation of ai(Q) as the number of”level i lattice points of a triangulation” (see (8.27.3)) we have ai(P ) ≤ ai(Q),0 ≤ i ≤ n. �

8.29 Proposition. The functionals ai : PnZ → R, 0 ≤ i ≤ n, are invariantwith respect to unimodular transformations, monotonous and non-negative onP ∈ PnZ with dimP = n.

8.30 Lemma. Let p : Z → R be a polynomial, and let g+(t) =∑

k≥1 p(k) tk

and g−(t) =∑

k≤0 p(k) tk. Then both series evaluate to rational functions forcertain (and usually different) values of t, and these rational functions coincide,i.e., g+(t) + g−(t) = 0 (wherever they are defined).

Proof. Let p be a polynomial of degree n, and let p(k) =∑n

i=0 ci(k+n−in

)for

some ci ∈ R. By Lemma 8.25 we have for |t| < 1

g+(t) =

∑ni=0 ci t

i

(1− t)n+1− p(0). (8.30.1)

In order to evaluate g−(t) we proceed as in the proof of Lemma 8.25: First wenote that substituting t by 1/t in (8.25.1) leads for |t| > 1 to

− 1

(1− t)n+1=

∑k≤−(n+1)

(n+ k

n

)tk,

which then yields analogously to (8.25.2)

−∑n

i=0 ci ti

(1− t)n+1=∑k≤−1

p(k) tk = g−(t)− p(0).

Together with (8.30.1) this shows the assertion. �

8.31 Theorem [Ehrhart-Macdonald Reciprocity]. Let P ∈ PnZ , dimP = n.Then

G(int k P ) = (−1)nn∑i=0

Gi(P )(−k)i.


Proof. For k ∈ Z let p(k) =∑n

i=0 Gi(P )ki. We have to show that G(int k P ) =(−1)n p(−k) for k ∈ N. To this end let Lint (P ; t) =

∑k≥1 G(int k P ) tk be the

corresponding Erhart series for the interior lattice points, and, as before, let Pbe the canonical embedding of P into Rn+1. As in Proposition 8.24 we have

Lint (P ; t) = γ(int posP ; (1, t)).

Applying the ”Reciprocity Theorem” 8.21 to the corresponding rational func-tions (cf. Lemma 8.17)∑

k≥1

G(int k P ) tk = Lint (P ; t)

= γ(int posP ; (1, t)) = (−1)n+1γ(posP ; (1, t−1))

= (−1)n+1L(P ; t−1) = (−1)n+1∑k≥0

G(k P ) t−k

= (−1)n+1∑k≤0

p(−k) tk = (−1)n∑k≥1

p(−k) tk,

where the last identity follows from Lemma 8.30. Comparing the coefficientsgives G(int k P ) = (−1)n p(−k). �

8.32 Theorem. Let P ∈ PnZ , dimP = n. Then for i 6= n mod 2

Gi(P ) =1

2

n−1∑j=i

(−1)i+j∑

F is j−face of P

Gi(F ).

In particular,

Gn−1(P ) =1

2

∑F is facet of P

vol n−1(F )

det(aff F ∩ Zn).

Proof. For k ∈ Z and a face F of P let p(F ; k) =∑dimF

i=0 Gi(F ) ki, and for 0 ≤j ≤ n letHj(k) =

∑F is j−face of P p(F ; k). Since G(k P ) =

∑F is face of P G(int k F )

we get from the Reciprocity-Theorem 8.31

G(k P ) = p(P ; k) =∑

F face of P

(−1)dimF p(F ;−k)

=n∑j=0

(−1)j∑

F is j−face of P

p(F ;−k) =n∑j=0

(−1)jHj(−k).

Hence

G(k P )−G(int k P ) = p(P ; k)− (−1)np(P ;−k) =

n−1∑j=0

(−1)jHj(−k).

On the other hand we have by Theorem 8.31

G(k P )−G(int k P ) =

n∑i=0

Gi(P )(1− (−1)n+i

)ki = 2

∑i 6=n mod 2

Gi(P ) ki,


and so

∑i 6=n mod 2

Gi(P ) ki =1

2

n−1∑j=0

(−1)jHj(−k) =1

2

n−1∑j=0

(−1)j∑

F is j−face of P

p(F ;−k)

=1

2

n−1∑j=0

∑F is j−face of P

j∑i=0

(−1)i+jGi(F ) ki.

Comparing the coefficients on the left and on the right hand side gives

Gi(P ) =1

2

n−1∑j=i

(−1)i+j∑

Fj−face of P

Gi(F ).

Finally, we note that for a facet F of P by Proposition 8.9 i) gives Gn−1(F ) =vol n−1(F )/det(aff F ∩ Zn), and so we get the formula for Gn−1(P ). �

8.33 Proposition. Let P ∈ PnZ . Then

G0(P ) = a0(P ) = 1,

Gn(P ) =1

n!(a0(P ) + · · · an(P )) = vol (P ),

a1(P ) = G(P )− (n+ 1),

an(P ) = G(intP ).

Proof. For the constant coefficients see Corollary 8.26. Comparing the leadingcoefficient in the two polynomials

∑ni=0 Gi(P )xi and

∑ni=0 ai(P )

(x+n−in

)yields

the relations for Gn(P ) (cf. Proposition 8.9 i)). Comparing the polynomials atx = −1 gives

n∑i=0

Gi(P ) (−1)i =

n∑i=0

ai(P )

(−1 + n− i

n

)= (−1)nan(P ),

which shows iii) by Theorem 8.31. Finally, evaluating at x = 1 gives G(P ) =a0(P ) (n+ 1) + a1(P ) = (n+ 1) + a1(P ). �


INDEX 99

Index

λi(K,Λ), 65

C(n,m), 31

F �, 23

H(a, α), 7

H+(a, α), H−(a, α), 7

Sn−1, 2

Cn, 2

GL(n,Z), 57

Kn, 37

G(·), 83

Ln, 57

Pn, 5

PnZ , 83

Rn, 1

aff X, 3

Bn, 2

Bn(a, ρ), 2

bdX, 4

Kn0 , 62

χM , 41

convX, 3

dimX, 3

|x|, 1

epi f , 12

intX, 4

bxcA, 58

linX, 3

H-polytope, 21

V-polytope, 21

posX, 3

relbdX, 4

relintX, 4

vertK, 22

vol (K), 41

vol j(K), 41

xᵀy, 1

ζ-function, 76

f -vector, 22

Cavalieri

Bonaventura, 41

adjacent vertex, 26

admissible lattice, 73

affine

combination, 1

hull, 3

subspace, 2

affinely

dependent, 1

independent, 1

ball, 2

Ball, Keith M., 76

Barnette’s Lower Bound Theorem, 32

Barnette, David, 32

Betke, Ulrich, 86

Blaschke, Wilhelm, 38

Blichfeldt, 62, 72, 79

boundary, 4

point, 4

breadth, 11

Brunn

Hermann, 46

Brunn-Minkowski

Theorem of, 46

Busemann, Herbert, 8

Caratheodory, Constantin, 18

Cauchy

Augustin-Loius, 37

Cavalieri’s principle, 41

characteristic function, 41

circumradius, 72

Cohn&Kumar, 79

Combinatorial diameter, 34

compact set, 17

concave function, 11

cone, 2

pointed, 88

polyhedral, 88

rational, 88

simplicial, 88

convergent sequences of convex bodies,37

convex

combination, 1

function, 11

100 INDEX

hull, 3set, 2

convex bodies, 37convex body

0-symmetric, 62convexly dependent, 1critical determinant, 73critical lattice, 74cyclic polytope, 31

Davenport, 74density of a packing, 67density of a densest lattice packing of

K, 73density of a densest packing, 70diameter, 52dimension, 3Dirichlet, 63discrete set, 58

edge, 22Ehrhart, 85Ehrhart polynomial, 85Ehrhart series, 91Ehrhart, Eugene, 85Ehrhart-Polynomial, 85epigraph, 12Euclidean

inner product, 1norm, 1space, 1

Euler, Leonhard, 26Euler-Poincare formula, 26exposed point, 22

face-centered-cubic lattice, 78faces, 22facet, 22family of convex sets, 2family of polytopes, 5fcc-lattice, 78Feller, William, 8Figiel, Tadeusz, 33function

additive, 45continuous, 45homogeneous of degree r, 45monotonous, 45

rigid motion invariant, 45rotation invariant, 45simple, 45translation invariant, 45

fundamental cell, 57fundamental parallelepiped, 57

gauge body, 67Gauss, 77Gauß, 78generating function, 87Graph, 34

HadwigerHugo, 45volume characterization, 46

Hales, 78halfspace, 7Hausdorff distance, 37Hausdorff, Felix, 37Helly, Eduard, 17Hirsch conjecture, 35Hirsch, Warren M., 35Hlawka, 74homothetic bodies, 46hyperplane, 7

Separating, 10Supporting, 7

improper faces, 22Inclusion-Exclusion Formula, 84index of a sublattice, 59inequality

Isoperimetric, 56Minkowski, 55of Jensen, 12

integral lattice, 57interior, 4

point, 4

Jensen’s inequality, 12Jensen, Johan, 12Jordan

Camille, 41Jordan measurable, 41

Kabatiansky, 73Kalai

INDEX 101

Gil, 35Kalai, Gil, 33Kepler conjecture, 78Kleitman, Daniel J., 35Kneser, Martin, 86Korkin, 79

Lagrange, 63, 76, 77lattice

basispoint, 57

determinant, 57integral, 57sublattice, 59

lattice packing, 73lattice point enumerator, 83lattice polytope, 83Leech lattice, 79Lemma

of Busemann-Feller, 8Levenshtein, 73Lindenstrauss, Joram, 33linear

combination, 1hull, 3subspace, 2

linearlydependent, 1independent, 1

McMullen’s Upper Bound Theorem, 32McMullen, Peter, 32metric projection, 7Milman, Vitali, 33Minkowski, 45, 62, 74

first theorem on successive minima,65

second theorem on successive min-ima, 65

Minkowski sum, 6Minkowski’s inequalities, 55Minkowski, Hermann, 6, 21Minkowski-Hlawka, 74mixed volumes, 48moment curve, 31multinomial coefficient, 50

nearest point map, 7

neighbour, 26neighbours, 26

outer normal vector, 7outer parallel body, 37outer unit normal vector, 7

packingdensity, 67saturated, 72set, 67

packing set, 67Poincare, Henri, 26polar set, 13polyhedron, 21polytope

k-polytope, 5positive

combination, 1hull, 3

positively dependent, 1prism, 42proper faces, 22pyramid, 42

Radon, Johann, 17Reeve simplices, 61Reeve-simplices, 87relative

boundary, 4boundary point, 4interior, 4interior point, 4

RiemannBenhard, 41

Rogers, 73

separating hyperplane, 10Sidelnikov, 73simple polytopes, 28simplex

k-simplex, 5simplicial polytopes, 28stacked polytopes, 32Steinitz’s theorem, 32Steinitz, Ernst, 32strictly convex function, 11sublattice, 59

102 INDEX

successive minima, 65support function, 11supporting hyperplane, 7surface area, 54Swinnerton-Deyer, 79

Theoremof Caratheodory, 18of Helly, 17of Radon, 17

theoremMinkowski-Hlawka, 74of Brunn-Minkowski, 46of Separation, 10

ThueAxel, 77

Triangulationpolytope, 84

Tverberg, Helge, 20

unimodular matrix, 57unit ball, 2

valuation, 45vertex, 22volume, 41

j-dimensional, 41mixed, 48polytope, 44prism, 42pyramid, 42simplex, 43

Weyl, Hermann, 21

Zolotarev, 79

Discrete Analytic Convex Geometry Introduction · Discrete Analytic Convex Geometry Introduction Martin Henk Otto-von-Guericke-Universit at Magdeburg Winter semester 2012/13 & Summer

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