Extremal Problems for Finite Sets (Student Mathematical Library) 1470440393, 9781470440398

Table of contents :
Cover
Title page
Contents
Notation
Chapter 1. Introduction
Chapter 2. Operations on sets and set systems
Chapter 3. Theorems on traces
Chapter 4. The Erdős–Ko–Rado Theorem via shifting
Chapter 5. Katona’s circle
Chapter 6. The Kruskal–Katona Theorem
Chapter 7. Kleitman Theorem for no ? pairwise disjoint sets
Chapter 8. The Hilton–Milner Theorem
Chapter 9. The Erdős matching conjecture
Chapter 10. The Ahlswede–Khachatrian Theorem
Chapter 11. Pushing-pulling method
Chapter 12. Uniform measure versus product measure
Chapter 13. Kleitman’s correlation inequality
Chapter 14. ?-Cross union families
Chapter 15. Random walk method
Chapter 16. ?-systems
Chapter 17. Exponent of a (10,{0,1,3,6})-system
Chapter 18. The Deza–Erdős–Frankl Theorem
Chapter 19. Füredi’s structure theorem
Chapter 20. Rödl’s packing theorem
Chapter 21. Upper bounds using multilinear polynomials
Chapter 22. Application to discrete geometry
Chapter 23. Upper bounds using inclusion matrices
Chapter 24. Some algebraic constructions for ?-systems
Chapter 25. Oddtown and eventown problems
Chapter 26. Tensor product method
Chapter 27. The ratio bound
Chapter 28. Measures of cross independent sets
Chapter 29. Application of semidefinite programming
Chapter 30. A cross intersection problem with measures
Chapter 31. Capsets and sunflowers
Chapter 32. Challenging open problems
Bibliography
Index
Back Cover

Citation preview

S T U D E N T M AT H E M AT I C A L L I B R A RY Volume 86

Extremal Problems for Finite Sets Peter Frankl Norihide Tokushige

Extremal Problems for Finite Sets

S T U D E N T M AT H E M AT I C A L L I B R A RY Volume 86

Extremal Problems for Finite Sets Peter Frankl Norihide Tokushige

Editorial Board Satyan L. Devadoss Erica Flapan

John Stillwell (Chair) Serge Tabachnikov

2010 Mathematics Subject Classification. Primary 05-01, 05D05. For additional information and updates on this book, visit www.ams.org/bookpages/stml-86 Library of Congress Cataloging-in-Publication Data Names: Frankl, P. (Peter), 1953- author. | Tokushige, Norihide, 1963- author. Title: Extremal problems for finite sets / Peter Frankl, Norihide Tokushige. Description: Providence, Rhode Island : American Mathematical Society, [2018] | Series: Student mathematical library ; volume 86 | Includes bibliographical references and index. Identifiers: LCCN 2017061643 | ISBN 9781470440398 (alk. paper) Subjects: LCSH: Extremal problems (Mathematics)| Set theory. | AMS: Combinatorics – Instructional exposition (textbooks, tutorial papers, etc.). msc | Combinatorics – Extremal combinatorics – Extremal set theory. msc Classification: LCC QA295 .F855 2018 | DDC 511/.6–dc23 LC record available at https://lccn.loc.gov/2017061643 Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy select pages for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for permission to reuse portions of AMS publication content are handled by the Copyright Clearance Center. For more information, please visit www.ams.org/ publications/pubpermissions. Send requests for translation rights and licensed reprints to reprint-permission @ams.org.

c 2018 by the authors. All rights reserved. Printed in the United States of America. ∞ The paper used in this book is acid-free and falls within the guidelines 

established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ 10 9 8 7 6 5 4 3 2 1

23 22 21 20 19 18

Contents

Notation

vii

Chapter 1. Introduction

1

Chapter 2. Operations on sets and set systems

5

Chapter 3. Theorems on traces

13

Chapter 4. The Erd˝os–Ko–Rado Theorem via shifting

19

Chapter 5. Katona’s circle

23

Chapter 6. The Kruskal–Katona Theorem

31

Chapter 7. Kleitman Theorem for no s pairwise disjoint sets

37

Chapter 8. The Hilton–Milner Theorem

43

Chapter 9. The Erd˝os matching conjecture

47

Chapter 10. The Ahlswede–Khachatrian Theorem

53

Chapter 11. Pushing-pulling method

61

Chapter 12. Uniform measure versus product measure

69

Chapter 13. Kleitman’s correlation inequality

77 v

vi Chapter 14.

Contents r-Cross union families

83

Chapter 15. Random walk method

87

Chapter 16.

95

L-systems

Chapter 17. Exponent of a (10, {0, 1, 3, 6})-system

103

Chapter 18. The Deza–Erd˝ os–Frankl Theorem

109

Chapter 19. F¨ uredi’s structure theorem

115

Chapter 20. R¨odl’s packing theorem

121

Chapter 21. Upper bounds using multilinear polynomials

127

Chapter 22. Application to discrete geometry

137

Chapter 23. Upper bounds using inclusion matrices

141

Chapter 24. Some algebraic constructions for L-systems

149

Chapter 25. Oddtown and eventown problems

155

Chapter 26. Tensor product method

161

Chapter 27. The ratio bound

175

Chapter 28. Measures of cross independent sets

181

Chapter 29. Application of semidefinite programming

189

Chapter 30. A cross intersection problem with measures

195

Chapter 31. Capsets and sunflowers

201

Chapter 32. Challenging open problems

211

Bibliography

217

Index

223

Notation

• Let R be the set of real numbers. • Let Q be the set of rational numbers. • Let Z be the set of integers. • Let Z>0 = {1, 2, 3, . . .} be the set of positive integers. • Let N = {0, 1, 2, . . .} be the set of nonnegative integers. • Let Z/mZ be the ring of congruence classes modulo m. • Let Fq denote the q-element field. • Let Rm×n denote the set of m × n real matrices. • Let Fq [x] denote the set of polynomials in variable x with coefficients in Fq . • For a positive integer n, let [n] = {1, 2, . . . , n}. • For a finite set A, the size (cardinality) of A is denoted by |A| or #A. • For functions f, g : R → R we write – f (x) = o(g(x)) if limx→∞ |f (x)/g(x)| = 0, – f (x) = O(g(x)) if |f (x)| < Cg(x) for all x > x0 , where x0 and C are some fixed constants, – f (x) = Ω(g(x)) if g(x) = O(f (x)), – f (x) = Θ(g(x)) if f (x) = O(g(x)) and f (x) = Ω(g(x)), – f (n)  g(n) if limn→∞ f (n)/g(n) = 0, vii

viii

Notation – f (n) ∼ g(n) if limn→∞ f (n)/g(n) = 1. • (Kronecker’s delta) δij = 1 if i = j and δij = 0 if i = j. • For integers a and b we write a|b to mean that a divides b. • (Disjoint union) We write A  B for A ∪ B with A ∩ B = ∅. • For a family F ⊂ 2X and x ∈ X let

(0.1)

F(x) = {F \ {x} : x ∈ F ∈ F},

(0.2)

F(¯ x) = {F : x ∈ F ∈ F}. • We say that two families F, G ⊂ 2X are isomorphic if G is obtained from F by renaming vertices, that is, if there is a permutation π on X such that G = {{π(x) : x ∈ F } : F ∈ F}. In this case we write F ∼ = G. • For a family F ⊂ 2X , the complement family F c is defined by F c = {X \ F : F ∈ F}. • For a family F ⊂ 2X and x ∈ X, the degree of x in F is defined by degF (x) = #{F ∈ F : x ∈ F ∈ F}. If degF (x) = d for all x ∈ X, then F is called d-regular . • A vector x is usually considered to be a column vector, and xi denotes the ith entry. • For a matrix A, the transposed matrix is denoted by AT . The identity matrix and the all-ones matrix are denoted by I and J, respectively. If we need to specify the size, then In is the n × n identity matrix. Similarly we write 1 and 1n for the all-ones vector. • An n-vertex graph G is a graph with n vertices. The adjacency matrix A = (ai,j ) of G is an n × n matrix defined by ai,j = 1 if i and j are adjacent in G and ai,j = 0 otherwise. • Let V be a vector space over a field F and let U ⊂ V be a finite subset. Let span U denote the subspace spanned by  U , that is, span U = { u∈U αu u : αu ∈ F}.

Chapter 1

Introduction

Apart from numbers, sets are the most fundamental notion in mathematics. Arguably the simplest sets are finite sets, that is, sets with a finite number of elements. The present book deals with combinatorial, mostly extremal problems concerning systems of subsets of a given finite set. Let us introduce the basic notation first. For a set X let |X| denote its size, that is, the number of its elements. If |X| = n, then X is called an n-element set. For a positive integer n let [n] = {1, 2, . . . , n} denote the standard n-element set. The power set 2[n] consists of the 2n subsets (including [n] and the   denote the collection empty set ∅) of [n]. For 0 ≤ k ≤ n let [n] k of all k-element subsets of [n]. A subset F of 2[n] is called a family of subsets. We can think of F ⊂ 2X as a hypergraph. In this case the vertex set is X and the edge set is F itself. In this sense  we is sometimes call a member of F an edge of F. A family F ⊂ [n] k called k-uniform (or a k-uniform hypergraph). To explain the topics in this book, probably it is best to state and prove Sperner’s Theorem, the first result in this area. We need some simple notions. If F0  F1  · · ·  Fl then (F0 , . . . , Fl ) is called a chain of length l. If for a family F no F, G ∈ F satisfy F  G, F is called an antichain. The reason is 1

2

1. Introduction

  is an antichain for every that F contains no chains. Obviously, [n] k   fixed k, 0 ≤ k ≤ n. Among these very specific antichains, [n] is the n 2  [n]   [n]  largest for n even. For n odd, n−1 and n+1 are the largest (note 2

2

that they have the same size). In 1928 Sperner [103] proved that for all n these are the largest antichains. Theorem 1.1 (Sperner’s Theorem). Let F ⊂ 2[n] be an antichain. Then   n , |F| ≤  n2   [n]    for n even and F = n−1 or and in the case of equality, F = [n] n 2 2  [n]  F = n+1 for n odd. 2

Proof. Let us suppose that F ⊂ 2[n] is an antichain of maximum possible size, that is, |F| is as large as possible. Let p = max{|F | : F ∈ F} and q = min{|F | : F ∈ F}. We want to show that p ≤ n+1 and q ≥ n−1 2 2 . If we succeed, the proof for n even is complete. Indeed the above inequalities imply   p = q = n2 . Thus F ⊂ [n] n . If n is odd, the two inequalities imply 2  [n]   [n]  F ⊂ n−1 ∪ n+1 and we have some more work to do. 2

2

Let us prove p ≤ and define

n+1 2

by an indirect argument. Suppose p >

n+1 2

G = {G ∈ F : |G| = p}. Define the immediate shadow H of G by H = {H : |H| = p − 1 and H ⊂ G for some G ∈ G}. Claim 1.2. We have the following. (i) Let F  = (F \ G) ∪ H. Then F  is an antichain. (ii) |F  | > |F|. Once we prove the claim, the proof of p ≤ n+1 2 is complete because  |F | > |F| contradicts the maximal choice of the antichain F.

1. Introduction

3

Proof of Claim. The facts that H ⊂

 [n]  p−1

and

max{|F | : F ∈ F \ G} ≤ p − 1 imply that H is an antichain, and H ⊂ F for H ∈ H and F ∈ F \ G. We need to show that H ⊃ F is impossible as well. Here we use that for H ∈ H there is some G ∈ G with H ⊂ G. Consequently H ⊃ F would imply G ⊃ F thereby contradicting the assumption that F = (F \ G) ∪ G is an antichain. This proves (i). To prove (ii) we note that |F  | − |F| = |H| − |G|. Thus we need to prove |H| > |G|. We do it by a simple double counting argument. Let M be the number of pairs (G, H) with G ∈ G and H ∈ H. Since every p-element subset G has p subsets of size p − 1, M = p|G|. On the other hand, for every (p − 1)-element subset H ⊂ [n] there are   with H ⊂ G . Some of n − (p − 1) = n + 1 − p choices of G ∈ [n] p these G might lie outside of G. Thus M ≤ (n − p + 1)|H|. We infer that (1.1)

p|G| ≤ (n + 1 − p)|H|.

If p > n+1 2 , then n+1−p < the proof of the claim.

n+1 2

and |G| < |H| follows. This completes 

The argument showing q ≥ n−1 2 is basically the same. However, one can circumvent it by taking the family of the complements: F c = {[n] \ F : F ∈ F}. It is easy to check that |F c | = |F| and that F c is an antichain, too. for all F ∈ F. EquivBy the above argument, n − |F | ≤ n+1 2 n+1 n−1 via (1.1) alently, |F | ≥ n − 2 = 2 . In the odd case, p = n+1 2 still implies |G| ≤ |H|. To avoid contradiction, |G| = |H| must hold. This in turn implies M = (n − p + 1)|H|, that is, all p-element subsets   G ∈ [n] p containing H ∈ H must be in F. We do not want to go into   details here, but by a connectivity argument this implies G = [n] p   and consequently F = [n] .  p We will give an alternative proof of Theorem 1.1 in Chapter 5.

4

1. Introduction

Sperner’s Theorem served as the starting point for a lot of research both inside and outside of extremal set theory. There are several books dealing with related results; see, for example [30]. In the present book we deal mostly with problems of a slightly different nature. Let us give an example from the famous paper [36] written by Erd˝ os, Ko, and Rado. We need one definition: A family F ⊂ 2[n] is called intersecting if F ∩ F  = ∅ for all F, F  ∈ F. Theorem 1.3. Suppose that F ⊂ 2[n] is an intersecting family. Then (i) and (ii) hold. (i) |F| ≤ 2n−1 .     and n ≥ 2k, then |F| ≤ n−1 (ii) If F ⊂ [n] k k−1 . The easiest examples of intersecting families are those for which one fixed element is contained in all members of the family. These show that both (i) and (ii) are best possible. Let us reproduce the easy proof of (i). For A ⊂ [n] let Ac = [n]\A. If F ∈ F, then F ∩ F c = ∅ implies F c ∈ F. Thus F ∩ F c = ∅, where F c = {F c : F ∈ F}. Since |F| = |F c | and F  F c ⊂ 2[n] , we have 2|F| = |F| + |F c | = |F  F c | ≤ |2[n] | = 2n , implying (i). The proof of (ii) is much harder, so we postpone it to Chapter 4.

Chapter 2

Operations on sets and set systems

In this chapter we consider some natural and useful operations on sets and families of subsets (also called set systems). We start by recalling some notation from the previous chapter. Let X be a finite set, and let 2X denote the power X  set of X, that is, X 2 = {A : A ⊂ X}. For a positive integer k let k denote the family   of all k-element subsets. A family F ⊂ X k is called k-uniform (or simply uniform). Sometimes a family F ⊂ 2X is called non-uniform 1 . For two subsets A and B of X, there are four very natural operations: A ∩ B, A ∪ B, A \ B = {x ∈ A : x ∈ B}, and the symmetric difference AB = (A \ B) ∪ (B \ A). By defining the distance d(A, B) of two sets as d(A, B) = |AB|, 2X becomes a metric space. Exercise 2.1. Show that d(A, B) = 0 if and only if A = B. Verify the triangle inequality d(A, C) ≤ d(A, B) + d(B, C). For A ⊂ X the set X \ A is called the complement of A, and if X is fixed it is denoted simply by Ac . Note that A ∩ Ac = ∅ and A ∪ Ac = X. We will often use the following identity: |A ∪ B| + |A ∩ B| = |A| + |B|, 1 Exactly speaking, a non-uniform family just means a family that is not necessarily uniform; it can be uniform.

5

6

2. Operations on sets and set systems

where A, B ⊂ X. For a family F ⊂ 2X the complement family is defined by F c = {F c : F ∈ F}. Now we introduce two operations on families of sets that will turn out to be very useful in establishing various upper bounds. Let F ⊂ 2X and let x ∈ X be a fixed element. The operation squash at an element x is defined by Sx (F) = {Fx : F ∈ F}, where

 Fx =

F \ {x} if x ∈ F ∈ F and F \ {x} ∈ F, F

otherwise.

Let us explain the operation squash at x in words as well. All the members F of F with x ∈ F are left unchanged. For each F with x ∈ F ∈ F we check whether F \ {x} is in F or not (so Fx depends on F). In the first case, in order to avoid duplication, we leave it unchanged. In the latter case we replace F by F \ {x}. For example, if F = {{x, y, z}, {x, y}, {y, z}, {z}},

(2.1) then

Sx (F) = {{x, y, z}, {y}, {y, z}, {z}}. By definition we have |Sx (F)| = |F|. The following trivial observation is essential in proving that certain properties are not altered by squashing. Observation 2.2. Let F ∈ Sx (F) and x ∈ F . Then both F and F \ {x} are members of F ∩ Sx (F). Exercise 2.3. Let F ⊂ 2X and x ∈ X. Recall from (0.1) and (0.2) that F(x) = {F \ {x} : x ∈ F ∈ F}, F(¯ x) = {F : x ∈ F ∈ F}. Prove that x) and (Sx (F))(¯ x) = F(x) ∪ F(¯ x). (Sx (F))(x) = F(x) ∩ F(¯

2. Operations on sets and set systems

7

Let us define the diameter diam(F) of F by diam(F) = max |AB|. A,B∈F

Proposition 2.4 ([84]). diam(Sx (F)) ≤ diam(F). Proof. Consider the squash of F at x. Take two arbitrary sets A, B ∈ F. As long as Ax Bx ⊂ AB there is no problem. The only other possibility is Ax Bx = (AB)  {x}. In this case we have x ∈ A ∩ B. By symmetry, let x ∈ Ax and x ∈ Bx . By Observation 2.2, A \ {x} is in F as well. Now Ax Bx =  (A \ {x})B implies diam(Sx (F)) ≤ diam(F). Let us remark that applying the operation squash at different elements x, y ∈ X will produce a family independent of the order of operations, that is, Sx (Sy (F)) = Sy (Sx (F)). Consequently, applying the operation squash at all x ∈ X just once will produce a family F˜ satisfying the property that (2.2)

if x ∈ F ∈ F˜ then F \ {x} ∈ F˜ for all x ∈ X.

We call F˜ a squashed family obtained from F. For the example (2.1), it follows that F˜ = Sz (Sy (Sx (F))) = {{x}, ∅, {y}, {z}}. We say that a family F is hereditary (or a downset) if G ⊂ F ∈ F implies G ∈ F. In other words, F is hereditary if it is a union of power sets. Proposition 2.5. A squashed family F˜ is hereditary. Proof. Let G ⊂ F ∈ F˜ . We show that G ∈ F˜ by induction on |F \G|. For |F \ G| = 1, the statement is simply (2.2). For the induction step choose H with G  H  F ; then F ∈ F and the induction hypothesis yield H ∈ F. Applying the induction hypothesis to the pair (G, H) yields G ∈ F as desired.  F|T

For a family F ⊂ 2X and a subset T ⊂ X, we define the trace by F|T = {F ∩ T : F ∈ F}.

Let us remark that F|T is a usual family, that is, a subset of 2T . Consequently, |F|T | ≤ 2|T | .

8

2. Operations on sets and set systems

Proposition 2.6 ([48]). Let F ⊂ 2X and x ∈ X. For every T ⊂ X one has |Sx (F)|T | ≤ |F|T |.

(2.3)

Proof. If x ∈ T then the operation squash does not affect the trace on T . Let T = {x}  T  and T0 ⊂ T  . If {x} ∪ T0 ∈ Sx (F), then {T0 , {x} ∪ T0 } ⊂ Sx (F) ∩ F by Observation 2.2. If {x} ∪ T0 ∈ Sx (F) and T0 ∈ Sx (F), then T0 ∈ F or {x} ∪ T0 ∈ F by the definition of the squash operation. Therefore we always have |Sx (F)|T ∩ {T0 , {x} ∪ T0 }| ≤ |F|T ∩ {T0 , {x} ∪ T0 }|. Summing this over all 2|T |−1 choices of T0 ⊂ T  gives (2.3).



The second operation on families of sets is called shifting. For convenience let X = {1, 2, . . . , n} (or [n] for short). For 1 ≤ i = j ≤ n we are going to define the (i, j)-shift Si,j for families F ⊂ 2[n] . The formal definition that goes back to Erd˝ os, Ko, and Rado is as follows: Si,j (F) = {si,j (F ) : F ∈ F}, where



si,j (F ) =

(F \ {j}) ∪ {i} if i ∈ F, j ∈ F and (F \ {j}) ∪ {i} ∈ F, F

otherwise.

In words, we only try to make changes in members of F if they do not contain i but contain j. In this case we check whether the subset obtained by replacing j with i is already in F. Only if not do we then replace the subset with the new one. Exercise 2.7. Verify the following. (i) |Si,j (F)| = |F|. (ii) |si,j (F )| = |F |, so F ⊂

[n] k

implies Si,j (F) ⊂

[n] k .

Exercise 2.8. Let F(a, ¯b) = {F \ {a} : F ∈ F, F ∩ {a, b} = {a}} for a, b ∈ [n]. Show that Si,j (F)(i, ¯j) = F(i, ¯j) ∪ F(j, ¯i) and Si,j (F)(j, ¯i) = F(i, ¯j) ∩ F(j, ¯i). ∈ We will present two propositions that show the importance of shifting: the first one says that shifting preserves the intersecting

2. Operations on sets and set systems

9

property, and the second one claims that shifting does not increase the size of the shadow. To make the statements precise we need two definitions. Let t be a positive integer. We say that a family F ⊂ 2X is t-intersecting if |F ∩ F  | ≥ t for all F, F  ∈ F. We simply say that a family is intersecting if it is 1-intersecting. For a positive integer p with 1 ≤ p ≤ n, define the p-shadow σp (F) by   σp (F) = {P ∈ [n] p : ∃ F ∈ F, P ⊂ F }. Proposition 2.9. If F ⊂ 2X is t-intersecting, then Si,j (F) is tintersecting as well. Proof. Suppose for contradiction that there exist F, G ∈ F with |si,j (F ) ∩ si,j (G)| < t ≤ |F ∩ G|. In this case, by symmetry, we may assume that si,j (F ) = F but si,j (G) = G and, moreover, that F ∩ {i, j} = G ∩ {i, j} = {j}. When performing the (i, j)-shift on F , why did not we replace j with i? The only possible reason is that the set F  := (F \{j})∪{i} is already in F. However, |F  ∩G| = |si,j (F )∩si,j (G)|, contradicting the t-intersecting property of F.  Proposition 2.10. σp (Si,j (F)) ⊂ Si,j (σp (F)). Proof. Let P ∈ σp (Si,j (F)). If |P ∩ {i, j}| = 0 or 2, then we can find F ∈ F such that P ⊂ F , and it follows that P ∈ σp (F) and P ∈ Si,j (σp (F)). So we may assume that |P ∩ {i, j}| = 1. Set R = P \ {i, j}. Since P ∈ σp (Si,j (F)), there is Q ∈ Si,j (F) such that P ⊂ Q. First suppose that P = R  {i}. If Q ∈ F then P ∈ σp (F) and P ∈ Si,j (σp (F)). If Q ∈ F then F := (Q \ {i}) ∪ {j} ∈ F and R  {j} ⊂ F . Thus R  {j} ∈ σp (F) and P = R  {i} ∈ Si,j (σp (F)). Next suppose that P = R  {j}. Then Q ∈ F and P ∈ σp (F). If i ∈ Q then R  {i} ∈ σp (F). If i ∈ Q then (Q \ {j}) ∪ {i} ∈ F and  R  {i} ∈ σp (F) again. Thus si,j (P ) = P and P ∈ Si,j (σa (F)). We say that a family G ⊂ 2X is s-union if |G ∪ G | ≤ |X| − s for all G, G ∈ G. We say that a family is union if it is 1-union. Exercise 2.11. Let s be a positive integer with s ≤ |X|.

10

2. Operations on sets and set systems (i) Show that G ⊂ 2X is s-union if and only if the complement family G c is s-intersecting. (ii) Show that if G ⊂ 2X is s-union then Si,j (G) is s-union.

Exercise 2.12. Show that if a family G ⊂ 2X is union, then |G| ≤ 2|X|−1 . (Hint: Use (i) of Theorem 1.3 and (i) of Exercise 2.11.) If G := si,j (F ) = F then us the following.

 a∈F

a−

 a∈G

a = j − i. This gives

Observation 2.13. If 1 ≤ i < j ≤ n and Si,j (F) = F ⊂ 2X , then     a> a. F ∈F a∈F

G∈Si,j (F ) a∈G

Since X is finite, the above observation guarantees that after a finite number of repeated applications of the (i, j)-shift for various 1 ≤ ˜ = i < j ≤ n, eventually we end up with a family F˜ satisfying Si,j (F) ˜ F for all 1 ≤ i < j ≤ n. Such a family is called shifted. From Proposition 2.9 (resp. Exercise 2.11) it is clear that in establishing upper bounds for the size of t-intersecting (resp. s-union) families we can restrict ourselves to shifted families. By Proposition 2.10, the same is true when trying to prove lower bounds on the size of the p-shadow. Finally we establish the following useful bound concerning shadows. Theorem 2.14 (Kruskal–Katona Theorem (integer version)). Let 0 ≤ p < k ≤ x be integers. Suppose that F  is a family of k-element x x subsets with |F| ≥ k . Then |σp (F)| ≥ p . The following proof is taken from [51]. Proof. Let p < k be fixed. We prove the statement by induction on x. The initial step x = k is trivial. We suppose x ≥ k + 1, that is, x − 1 ≥ k. By Proposition 2.10 we may suppose that F is shifted. First we consider the case p = k − 1. Recall the definitions of F(x) and F(¯ x) from (0.1) and (0.2). 1)) ⊂ F(1). Claim 2.15. σk−1 (F(¯

2. Operations on sets and set systems

11

Proof. Let G ∈ σk−1 (F(¯ 1)). Then G∪{j} ∈ F for some j ∈ {1}G. By shiftedness, G ∪ {1} ∈ F, that is, G ∈ F(1).  x−1 Claim 2.16. |F(1)| ≥ k−1 .   Proof. Suppose the contrary, that is, |F(1)| < x−1 k−1 . We infer that       x x−1 x−1 ¯ |F(1)| = |F| − |F(1)| > − = . k k−1 k So we may apply the induction hypothesis to F(¯1). Then we obtain x−1 ¯ > |F(1)|. This contradicts Claim 2.15.  |σk−1 (F(1))| ≥ k−1

We note that σk−1 (F) ⊃ F(1)  {{1}  G : G ∈ σk−2 (F(1))}. In fact, clearly σk−1 (F) contains F(1), which consists of subsets not containing 1. If G ∈ σk−2 (F(1)), then there is H ∈ F(1) such that G ⊂ H, and {1}  G ⊂ {1}  H ∈ F, yielding that {1}  G is in σk−1 (F). Thus we get |σk−1 (F)| ≥ |F(1)| + |σk−2 (F(1))|. Applying the induction with Claim 2.16 to F(1), we infer   hypothesis that |σk−2 (F(1))| ≥ x−1 . This proves k−2       x−1 x−1 x + = . |σk−1 (F)| ≥ k−1 k−2 k−1 This completes the proof for the case p = k − 1. Next we prove the general case p < k by induction on i := k − p, where we have already proved the case i = 1.  Suppose that the x statement is true for the case i, that is, |F| ≥ k implies |σk−i (F)| ≥  x  k−i . Since σk−(i+1) (F) = σ(k−i)−1 (σk−i (F)), we can apply the  induction hypothesis to the case k = k−i and p = k −1 (that is, i = x x 1); we get |σk−(i+1) (F))| = |σk −1 (σk (F))| ≥ k −1 = k−(i+1) .    Remark 2.17. One can define xk for all real numbers x by   x x(x − 1) · · · (x − k + 1) . = k k!     Note that x is a polynomial of degree k. Since the equality xk − x−1 x−1k − k−1 = 0 is true for all integer values of x with x ≥ k, it is k

12

2. Operations on sets and set systems

an identity. That is, it is true for all real values of x. By the same proof one can show that Theorem 2.14 holds for all real x with x ≥ k. This is called the Lov´ asz version of the Kruskal–Katona Theorem.   If F ⊂ X k , then the (k − 1)-shadow σk−1 (F) is sometimes just called the shadow (or immediate shadow) of F and denoted by σ(F). We will return to the problem of estimating σ(F) in Chapter 6 and refine the Lov´ asz version.

Chapter 3

Theorems on traces

In the preceding chapter we worked hard to introduce some operations on families and prove some of their properties. In this chapter our efforts will be rewarded by some almost trivial proofs of classical results. Let us start with the following important result that was proved independently by three sets of authors: Perles and Shelah, Sauer, and Vapnik and Chervonensky. The names of the third set of authors are ordered according to the Russian alphabet. [n] Theorem 3.1. Let n > k ≥ 0 be integers.  [n]  If F ⊂ 2 satisfies k n |F| > i=0 i , then there exists a T ∈ k+1 such that F|T = 2T , where F|T = {F ∩ T : F ∈ F} denotes the trace of F on T .

Proof. Suppose, to the contrary, that F ⊂ 2[n] with F|T = 2T for X  . In view of Propositions 2.5 and 2.6, after repeatedly all T ∈ k+1 applying squashing we obtain a hereditary family G ⊂ 2[n] with |G| =  X |F| and still satisfying |G|T | < 2|T | for all T ∈ k+1 . We claim that |G| ≤ k for all G ∈ G. Indeed, if |G| ≥ k + 1 for some G ∈ G, then by the hereditary G we have G|H = 2H , a property 2G ⊂ G, and thus for all H ∈ k+1 contradiction. Consequently, we have proved that |G| ≤ k for all G ∈ 13

14

3. Theorems on traces

k   G and |G| ≤ i=0 ni . But this contradicts the fact that |G| = |F| k    and our assumption that |F| > i=0 ni . Let us introduce the arrow notation (n, m) → (a, b) with the following meaning: For every F ⊂ 2[n] with |F| = m there exists T ⊂ [n] with |T | = a such that |F|T | ≥ b. Again, by Proposition 2.6, in order to establish the veracity of an arrow relation, it is sufficient to check it for hereditary families. Theorem 3.2. The following hold. (i) (n, m) → (n − 1, m) for m ≤ n. (ii) (n, m) → (n − 1, m − 1) for m ≤ 1 + n + (iii) (n, m) → (3, 7) for m > 1 + n +

n−1 2 .

n2 4 .

Proof. (i) If G ⊂ 2[n] is hereditary, then ∅ ∈ G. If, moreover, m = |G| ≤ n, then not all the 1-element subsets are in G, that is, {x} ∈ G for some x ∈ [n]. Then, by the hereditary property, G ∈ G for all G containing x. Therefore G ⊂ 2[n]\{x} , implying |G|[n]\{x} | = |G|. (ii) If {x} ∈ G for some x ∈ [n], the above proof works. Thus we may assume that ∅ and all 1-element sets are in G. Since m = |G| ≤ n−1 1 + n + n−1 2 and there are at most  2  more subsets in G, there is an element y ∈ [n] which is not contained in any 2-element member of G. By the hereditary property, the only member of G containing y is {y}. Consequently, G|[n]\{y} = G \ {y}, proving (ii). (iii) If |G| = 3 for some G ∈ G, then |G|G | = 8 and we have nothing   [n] [n] to prove. Consequently, we may assume that G ⊂ [n] 0  1  2 .   (2) Let G (i) = G ∩ [n] for i = 0, 1, 2. Then G is a graph on n vertices. i 2 (0) (1) The number of edges is at least m−|G |−|G | ≥ m−1−n > n4 by the assumption. By Exercise 3.4 there are three edges {x, y}, {x, z}, and {y, z} in it (forming Now, letting T = {x, y, z}, it      a triangle). follows that G|T = T0  T1  T2 , and thus |G|T | = 1 + 3 + 3 = 7.  Remark 3.3. In the above theorem (i) and (ii) were originally proved by Bondy and Bollob´as, respectively. As to (iii), it was conjectured by Lov´ asz and proved by Frankl; see [48].

3. Theorems on traces

15

Exercise 3.4. Let G be a graph on n vertices with more than n2 /4 edges. Show that G contains a triangle. For any F ⊂ 2X and Y ⊂ X, the arrow relation (|X|, |F|) → (|Y |, |F|Y | + 1) is not true. We apply this fact to the following construction. Construction 3.5. Let n be of the form n = dq, where d and q are positive integers. Let X = X1  · · ·  Xq be a partition with |Xi | = d, 1 ≤ i ≤ q. Define a hereditary family F := F(d, q) = 2X1 ∪ · · · ∪ 2Xq . Note that 2Xi ∩ 2Xj = {∅} and |F| = 1 + (2d − 1) nd . If Y = X \ {x} for some x ∈ X, then F|Y = F \ {F ∈ F : x ∈ F }. So it follows that  X  |F|Y | = |F| − 2d−1 for all Y ⊂ n−1 . Thus (dq, 1 + (2d − 1) nd ) → (dq − 1, 1 + (2d − 1) nd − 2d−1 + 1) is not true. The next result shows that F(d, q) is an extremal example. Recall the definitions of F(x) and F(¯ x) from (0.1) and (0.2). Theorem 3.6. Let n and d be fixed positive integers, with n > d. Let |X| = n and let F ⊂ 2X be a hereditary family satisfying |F| ≤ d 1 + 2 d−1 n. Then one of the following holds. (i) |F(x)| < 2d−1 for some x ∈ X. In this case |F|Y | > |F| − 2d−1 for Y = X \ {x}. (ii) F is isomorphic to F(d, q) for some q (in particular,  X d . divides n). In this case |F|Y | = |F|−2d−1 for all Y ∈ n−1 Proof. The proof is based on the following lemma of some independent interest. Lemma 3.7. Let H be a hereditary family, with |H| = 2t for some  t+1 1 ≥ 2 t+1−1 and H attains the minimum if and t ≥ 0; then H∈H |H|+1 only if H = 2T for some t-element set T . Let us postpone the proof of the lemma and prove the theorem first. The following identity is easily established by reversing the order

16

3. Theorems on traces

of summation: |F \ {∅}| =

    1 = |F |

F ∈F x∈F

x∈X

H∈F (x)

 1 . |H| + 1

Supposing that (i) does not hold, we have |F(x)| ≥ 2d−1 for all x ∈ X. d By the lemma the quantity in the bracket is at least 2 d−1 , yielding d |F| ≥ 1 + 2 d−1 n. Since the assumption of Theorem 3.6 is |F| ≤ d 1 + 2 d−1 n, we have equality. Therefore, we must have equality all the d−1 T (x) with way, that is, |F(x)|  X  = 2 , and by the lemma F(x) = 2 some T (x) ∈ d−1 . We infer that every x ∈ X is contained in a unique d-element set {x}  T (x). Consequently, these d-element sets partition X, that is, F is isomorphic to F(d, q), where q = nd is the number of these d-element sets. That is, (ii) holds.  Proof of Lemma 3.7. Let us use the hereditary property of H to prove a simple inequality. Here hi denotes the number of i-element t sets in H. Then hi = 0 for i > t and i=0 hi = |H|. Since |H| = 2t    we have that ti=0 hi = ti=0 ti . We will show that (3.1)

r  i=0

hi ≥

r    t i=0

i

 for allr with t ≥ r ≥ 0. Suppose, to the contrary, that ri=0 hi < r t for r = t, there must i=0 i for some 0 ≤ r < t. Since (3.1) holds  exist some s with r < s ≤ t such that hs > st . Then we can choose     G ⊂ H ∩ Xs with |G| = st . By Theorem 2.14 |σi (G)| ≥ ti for 0 ≤ i < s, and  by the hereditary property it follows that σi (G) ⊂ H. Thus hi ≥ ti for 0 ≤ i < s. Since r < s, it then follows that r t r i=0 hi ≥ i=0 i . This contradicts our indirect assumption and thereby proves (3.1). We label the members of H in size-increasing order, that is, let t H = {H1 , H2 , . . . , H2t }, where |H1 | ≤ |H2 | ≤  · · · ≤ |H2 |. We partition H into H0  H1  · · ·  Ht with |Hi | = ti by choosing members starting from H1 according to the above order, that is, H0 = {H1 }, H1 = {H2 , . . . , Ht+1 }, H2 = {Ht+2 , . . . , H(t )+t+1 }, and so on. In 2

view of (3.1), it follows that if H ∈ Hi then |H| ≤ i. Thus, for

3. Theorems on traces

17

0 ≤ i ≤ t, we have that     1  t t+1 1 ≥ = /(i + 1) = /(t + 1). (3.2) i i+1 |H| + 1 i+1 H∈Hi

H∈Hi

Consequently,  H∈H

  t t    t+1 1 1 1 = ≥ |H| + 1 |H| + 1 t+1 i+1 i=0 i=0 H∈Hi

=

 t+1  2t+1 − 1 1  t+1 . = j t + 1 j=1 t+1

In the case of equality, in particular, equality must hold at i = t in (3.2). That is, Ht consists of a t-element set T . Then H = 2T by the hereditary property.  Let us close this chapter with the following open problem. Problem 3.8. For each n and s, determine or estimate the maximum value m = m(n, s) such that (n, m) → (n − 1, m − s). It follows from (i) and (ii) of Theorem 3.2 that m(n, 0) = n and m(n, 1) = (3n + 1)/2. In view of Theorem 3.6, m(n, 2d−1 ) ≥ 1 + (2d − 1)n/d. The fact that m(n, 2) = 2n is not hard to prove. However, m(n, s) is unknown in general. Exercise 3.9. Show that m(n, 2) = 2n as follows. (i) Define a hereditary family F ⊂ 2[n] by   [n] F = {∅}   {{1, 2}, {2, 3}, . . . , {n − 1, n}, {n, 1}}, 1 and verify that (n, 2n + 1) → (n − 1, 2n − 1) is not true. (ii) Let F ⊂ 2[n] be a hereditary family with |F| = 2n. Show that there is some x ∈ [n] such that degF (x) ≤ 2, where degF (x) := {F ∈ F : x ∈ F }, and deduce that (n, 2n) → (n − 1, 2n − 2). (Hint: If degF (x) ≥ 3 for all x, then it follows from the hereditary property that F contains ∅, all the 1-element subsets, and at least n 2-element subsets.)

Chapter 4

The Erd˝ os–Ko–Rado Theorem via shifting

One of the oldest results in extremal set theory, a theorem that still influences a good part of the ongoing research in this field, is the Erd˝os–Ko–Rado Theorem. It was proved in 1938 during the time that all three authors were visiting England. However, as Erd˝ os told us, there was little interest in this type of combinatorial investigation at that time. This is the reason why the paper appeared only in 1961. Theorem 4.1 (Erd˝ os–Ko–Rado Theorem). Let n and k be positive   is intersecting, that is, integers with n ≥ 2k. Suppose that F ⊂ [n] k F ∩ F  = ∅ for all F, F  ∈ F. Then (4.1)

  n−1 . |F| ≤ k−1

Proof. We apply double induction on n and k. Clearly the statement holds for k = 1 and all n ≥ 2. The other 2k−1base case is n = 2k, which we 2k settle first. If n = 2k then k = 2 k−1 . That is, (4.1) is equivalent   c to saying that 2|F| ≤ 2k k . Consider F = {[n] \ F : F ∈ F}. Then   c c F contains no two F c ⊂ [2k] k , |F | = |F| and F ∩ F = ∅ since  c and |F  F c | = 2|F|, complementary sets. Thus F  F ⊂ [2k] k implying the desired inequality. 19

20

4. The Erd˝ os–Ko–Rado Theorem via shifting

Now let k > 1 be fixed. Let us suppose that the statement holds for all (n , k ) with n ≥ 2k and k < k, and for all (n , k) with n > n ≥ 2k. We will consider the case (n, k). In view of Proposition 2.9, we may assume that F is shifted. Recall the definitions of F(n) and F(¯ n) from (0.1) and (0.2). Claim 4.2. Both F(¯ n) and F(n) are intersecting. Proof. Since F(¯ n) ⊂ F, the statement is trivial for F(¯ n). Suppose indirectly that G, H ∈ F(n) with G ∩ H = ∅. Since |G| + |H| = 2(k − 1) < n − 2, there exists i ∈ [n − 1] such that i ∈ G ∪ H. As F is shifted, the set H ∪ {i} is also in F. However, (H ∪ {i}) ∩ (G ∪ {n}) = H ∩ G = ∅, a contradiction. This completes the proof of the claim.



  and Applying the induction hypothesis to both F(¯ n) ⊂ [n−1] k [n−1] F(n) ⊂ k−1 gives us       (n − 1) − 1 (n − 1) − 1 n−1 |F| = |F(¯ n)| + |F(n)| ≤ + = , k−1 (k − 1) − 1 k−1 which completes the proof of the theorem.



Recall from Exercise 2.12 that if F ⊂ 2[n] is 1-union then |F| ≤ 2n−1 . To extend this result to s-union families let us define  {K ⊂ [n] : |K| ≤ p} if q = 2p, (4.2) K(n, q) = {K ⊂ [n] : |K ∩ [2, n]| ≤ p} if q = 2p + 1. Exercise 4.3. Let s = n − q. Show that K(n, q) is s-union, that is, |K ∪ K  | ≤ q for all K, K  ∈ K(n, q). Exercise 4.4. Verify that the size |K(n,   q)| can be represented as   if q = 2p + 1. Verify also (i) pi=0 ni if q = 2p and (ii) 2 pi=0 n−1 i that |K(n, n − 1)| = 2n−1 . Theorem 4.5 (Katona union theorem). Let n > s > 0 and q = n−s, and suppose that G ⊂ 2[n] is s-union, that is, (4.3)

|G ∪ G | ≤ q

for all G, G ∈ G. Then either (i) or (ii) holds.

4. The Erd˝ os–Ko–Rado Theorem via shifting (i) q = 2p and |G| ≤

p i=0

(ii) q = 2p + 1 and |G| ≤ 2

n i

p

.

i=0

21

n−1 . i

The following simple proof is due to Wang [114]. Proof. Apply induction on n. The initial case is n = s + 1. In this case, q = 1 and the only maximal s-union family is {∅, {i}} for some i ∈ [n]. Thus (ii) holds for p = 0. So suppose n ≥ s + 2, that is, q ≥ 2. Using (ii) of Exercise 2.11, assume that G is shifted. As usual let G(n) = {G \ {n} : n ∈ G ∈ G} and G(¯ n) = {G ∈ G : n ∈ G}. Since G(¯ n) ⊂ G, it follows that |F ∪ F  | ≤ q for F, F  ∈ G(n). Claim 4.6. For F, F  ∈ G(n), |F ∪ F  | ≤ q − 2. Proof. Since F ∪ {n} and F  ∪ {n} are in G, (4.3) implies |F ∪ F  | ≤ q − 1. Suppose indirectly that equality holds, that is, |F ∪ F  | = q − 1. Since (n − 1) − (q − 1) = s ≥ 1, we can find j ∈ [n − 1] \ (F ∪ F  ). By the shiftedness of G, the set F  ∪ {j} is also in G. However, |(F ∪ {n}) ∪ (F  ∪ {j})| = |F ∪ F  | + 2 = q + 1, a contradiction.  Let us apply the induction hypothesis to G(¯ n) and G(n), both in 2[n−1] . Adding up the two inequalities, we obtain the desired upper bound in both cases (i) and (ii). The reason that this simple argument works is the fact that q and q − 2 have the same parity. Let us do the detailed calculation for case (ii). Note that we have q = 2p + 1 for G(¯ n) while q − 2 = 2(p − 1) + 1 for G(n). Thus we get     p  p    (n − 1) − 1 n−1 n−2 =2 +2 , |G(¯ n)| ≤ 2 i 0 i i=0 i=1   p−1  p    (n − 1) − 1 n−2 =2 . |G(n)| ≤ 2 i i−1 i=0 i=1  n−2 n−1  + i−1 = i , we finally obtain Noting that n−2 i  p   n−1 |G| = |G(¯ n)| + |G(n)| = 2 . i i=0 The calculation for case (i) is slightly simpler and we leave it to the reader. 

22

4. The Erd˝ os–Ko–Rado Theorem via shifting

It is easy to show that (ii) implies the Erd˝os–Ko–Rado Theorem.  [n]  Let n ≥ 2(p + 1) and q = 2p + 1 < n, and let F ⊂ p+1 be an arbitrary intersecting family. Define G = F ∪ {G ⊂ [n] : |G| ≤ p}. We show that G is (n − q)-union. In fact, if G, G ∈ F then |G ∪ G | = |G| + |G | − |G ∩ G | ≤ 2(p + 1) − 1 = q, and if G ∈ G and G ∈ G \ F then |G ∪ G | ≤ |G| + |G | ≤ q. Now it follows from (ii) of Theorem 4.5 that |G| ≤ |K(n, q)|. Both G and K(n, q) contain  subsets of size at most p of [n]. In excess K(n, n−1q)  all (p+1)-element sets containing 1. Thus |F| ≤ contains the n−1 p p follows. Since AB ⊂ A∪B for all pairs of subsets, the following theorem is formally stronger than the Katona union theorem. Theorem 4.7 (Kleitman diameter theorem). Suppose that n > d > 0 are integers and F ⊂ 2[n] is a family of diameter at most d, that is, |F F  | ≤ d for all F, F  ∈ F. Then according to the parity of d either (i) or (ii) holds. p   (i) d = 2p and |F| ≤ i=0 ni .  p  (ii) d = 2p + 1 and |F| ≤ 2 i=0 n−1 . i Proof. In view of Proposition 2.4, squashing will not increase the diameter, max |F F  |. Thus, by Proposition 2.5 we obtain a hereditary family G ⊂ 2[n] such that |G| = |F| and diam(G) ≤ d. Claim 4.8. G is (n − d)-union, that is, |G ∪ G | ≤ d for G, G ∈ G. Indeed, by the hereditary property for G, G ∈ G, the set G = G \ G is also in G. Now G ∪ G = GG implies the claim. Now we get the desired inequalities from Theorem 4.5. 

Chapter 5

Katona’s circle

Trying to solve [n]extremal problems by looking straight at the whole of [n] 2 or even k might be too difficult or very complicated. Katona’s ingenious idea was to look at such problems on a much smaller family of subsets, that is, the circle. Let us define this circle. For an arbitrary ordering of the n elements of [n] on a circle, we consider only subsets that form an arc on this circle, that is, the elements of this set are consecutive. More formally, let (a1 , . . . , an ) be a circular permutation of 1, 2, . . . , n. By circular we mean that we consider the element a1 to be the consecutive element of an and we do not distinguish the n possible permutations (a1 , . . . , an ), (a2 , . . . , an , a1 ), (a3 , . . . , an , a1 , a2 ), . . . , (an , a1 , a2 , . . . , an−1 ) that yield the same circular arrangement. There are (n − 1)! circular permutations and each of them contains n subsets of size i, 1 ≤ i < n, forming an arc of the circle. For a circular permutation π of [n], let C(π) denote the family of all arcs of π. So |C(π)| = n2 − n because C(π) consists of n arcs of size i for each 1 ≤ i < n. We will always imagine that the n elements are arranged on the circle in clockwise order. This will allow us to speak of the first and last elements of an arc. Let us illustrate the use of the circle by giving an alternate proof of the Erd˝os–Ko–Rado Theorem.

23

24

5. Katona’s circle

Theorem 5.1 (Erd˝os–Ko–Rado Theorem on the circle). Let n ≥ 2k be positive integers and fix a circular permutation of [n]. Let E = {E1 , . . . , Em } be a collection of circular arcs of k elements such that Ei ∩ Ej = ∅ for any i, j. Then m ≤ k. Proof. Without loss of generality let A = (a1 , . . . , ak ) be a member of E. Since E consists of subsets of size k, no circular arc of size k (except for A itself) contains A completely. By the intersection property all the remaining arcs Ei have either their first or their last element in A. The candidates for the first element except for A are a2 , . . . , ak . For 1 ≤ i ≤ k − 1, we denote the arc starting at ai+1 by Si and the arc terminating at ai by Ti . Note that 2k ≤ n implies that Si ∩ Ti = ∅. Thus for each 1 ≤ i ≤ k − 1 at most one of the two subsets Si and Ti is in E. Together with A this gives us m ≤ (k − 1) + 1 = k.  Katona deduced the Erd˝ os–Ko–Rado Theorem from the circle version by simple double counting. Let n ≥ 2k. Note that a fixed k-element subset is an arc in exactly k!(n − k)! circular permutations.   Let F ⊂ [n] k be an intersecting family. Let us count, in two different ways, the number N of pairs (F, π), where F ∈ F appears as an arc of a circular permutation π, that is, F ∈ C(π). The following picture shows an example for the case n = 4, k = 2, and F = 1, 2, 1, 3, 2, 3. In this example the members of F are indicated by thick arcs, and we see that N = 12. (1234)

(1243)

1 {1, 2} 4

1 2

3

3

4

3

2

4

3

2

1 4

2

1 3

3

4 3

2

1 3

(1432)

1 3

4

1 2

(1423)

1 3

2

1 2

(1342)

1 2

4

1 {1, 3} 4

(1324)

1 4

2

4

3

4

2

4

2

3

1

1

1

1

1

1

{2, 3} 4

2 3

3

2 4

4

3 2

2

3 4

3

4 2

2

4 3

On the one hand, each F ∈ F appears in k!(n − k)! circular permutations, implying N = |F|k!(n − k)!. On the other hand, there

5. Katona’s circle

25

are (n − 1)! circular permutations and each of them contains at most k arcs of F by Theorem 5.1. Thus N ≤ (n − 1)!k. Therefore we get |F|k!(n − k)! ≤ (n − 1)!k, or equivalently

  n−1 |F| ≤ . k−1

Probably, a simpler way is by comparing F and the family  

[n] A1 = A ∈ :1∈A k for each circular permutation π. Since |C(π) ∩ A1 | = k for all π, we always have |C(π) ∩ F| ≤ |C(π) ∩ A1 |. This implies



 n−1 |F| ≤ |A1 | = k−1 because each k-element subset of [n] occurs with the same multiplicity k!(n − k)!. This completes an alternative proof of Theorem 4.1. 

Let us next prove the following important theorem concerning antichains. Recall that a family S ⊂ 2[n] is an antichain if S contains no two members S and S  such that S  S  . Theorem 5.2 (Yamamoto’s inequality1 [112]). Suppose that S ⊂ 2[n] is an antichain. Then  1  n  ≤ 1. (5.1) S∈S

|S|

Noting that ∅ ∈ S (resp. [n] ∈ S) implies S = {∅} (resp. S = {[n]}), in proving (5.1) we may assume that ∅, [n] ∈ S. Then we shall consider arcs on a circular permutation π again, that is, we look at S ∩ C(π). Proposition 5.3. Suppose that S0 ⊂ C(π) is an antichain. Then |S0 | ≤ n, and moreover, equality holds only if all members of S0 have the same size. 1 This inequality has been rediscovered several times by several authors, including Lubell [90] and Meshalkin [93], and it is also called the LYM inequality.

26

5. Katona’s circle

Proof. Note that if the last element of two arcs is the same, then the longer one contains the shorter one. Consequently, for the antichain S0 there can be at most one member of S0 for every last element. This proves |S0 | ≤ n. Should we have equality, there must be exactly one member of S0 for every last element. If not all are of the same size, then we can find two consecutive last elements, say ai and ai+1 , such that the arc in S0 corresponding to ai+1 is longer. However, this means that this arc contains the shorter arc ending in ai , a contradiction.  Proof of Theorem 5.2. We count the number N of pairs (S, π), where S ∈ S appears as an arc on a circular permutation π. Since each S ∈ S appears in |S|!(n − |S|)! circular permutations, it follows  that N = S∈S |S|!(n − |S|)!. On the other hand, there are (n − 1)! circular permutations and each of them contains at most n arcs of S by Proposition 5.3. This yields N ≤ (n − 1)!n = n!. Thus  |S|!(n − |S|)! ≤ n!, S∈S



and dividing both sides by n! gives (5.1).

Let us next deduce Sperner’s Theorem (Theorem 1.1) from Yamamoto’s inequality. For convenience we restate it here.   Theorem. Let S ⊂ 2[n] be an antichain. Then |S| ≤  nn  . More2     over, equality holds if and only if S = [n] for n even and S = [n] n n 2 2  [n]  or  n for n odd. 2

  Proof. Noting that nk attains its maximum (as a function of k) for n+1 k = n2 in the even case, and for k = n−1 2 , 2 in the odd case, (5.1) implies  1  1 |S|  n =  n ≤  n  ≤ 1, n 2

S∈S

n 2

that is,

S∈S

 |S| ≤

n

n   |S|



.  n2  Moreover, in the case of equality one must have |S| = n2 for all S ∈ S if n is even. If n is odd, we can only deduce that |S| =  n2  or

5. Katona’s circle

27

|S| =  n2  for all S ∈ S. However, by Proposition 5.3, |S| = only possible if all S ∈ S have the same size. That is, S =   . S = [n] n 2





n n  [n]2  n 2

is or 

Our next target is an old theorem of Erd˝ os. A family S ⊂ 2[n] is called a t-antichain if S contains no t + 1 members S0 , S1 , . . . , St forming a chain, that is, S0  S1  · · ·  St .     Let k1 , k2 , . . . , kt be such that kn1 , . . . , knt are the largest t binomial coefficients. Note that for n−t odd, {k1 . . . , kt } = [ n−t+1 , n+t−1 ], 2 2 n−t n+t n−t and for n−t even, {k1 . . . , kt } is either [ 2 , 2 −1] or [ 2 +1, n+t 2 ]. Proposition 5.4. Suppose that S ⊂ 2[n] is a t-antichain and π is a circular permutation of [n]. Then we have (i) |S ∩ C(π)| ≤ tn, (ii)

 S∈S∩C(π)

 1 n ≤ . |S|!(n − |S|)! k !(n − ki )! i i=1 t

Proof. The arcs with identical last element form a chain. This proves that in S ∩ C(π) there can be at most t members with the same last element, proving (i). n 1 /n! and thus the t = |S| To prove (ii) we note that |S|!(n−|S|)! largest values are for {k1 , . . . , kt }. By (i), the left-hand side (LHS) of (ii) has at most tn terms. Each value of |S| can occur at most n times. These facts imply (ii).  Theorem 5.5 (The Erd˝os t-antichain theorem). Let n ≥ t, and let S ⊂ 2[n] be a t-antichain. Then we have t    n |S| ≤ . k i i=1 Proof. Let us sum (ii) of Proposition 5.4 over all (n − 1)! circular permutations. Since each S ∈ S is counted |S|!(n − |S|)! times, it follows that t t     n n = , |S| ≤ (n − 1)! k k !(n − k )! i i i i=1 i=1

28

5. Katona’s circle 

as needed.

Next we prove a very simple butuseful  statement about the shad[n] ows. Let k ≥ 2. For a family S ⊂ k of arcs on a circular permutation π, let σ π (S) denote the family of arcs of size k − 1 that are contained in some member of S. If S consists of all n arcs of size k, then |σ π (S)| = n.   be a family of arcs on Proposition 5.6. Let k ≥ 2 and let S ⊂ [n] k a circular permutation π. Unless S consists of all n arcs of size k, it follows that |σ π (S)| ≥ |S| + 1. Proof. To every S ∈ S associate the arc of size k − 1 obtained by deleting the last element of S. This shows that |σ π (S)| ≥ |S|. Let |S| < n. Then there are two consecutive arcs, say (a1 , . . . , ak ) and (a2 , . . . , ak+1 ), such that the first is in S but the second is not. Then the arc (a2 , . . . , ak ) of size k − 1 is in σ π (S) but has not been counted  yet. This proves |σ π (S)| ≥ |S| + 1. The next statement is about several — not necessarily distinct — families. Such statements have proved useful in dealing with problems concerning one family as well as being interesting in their own right. Let t ≥ 2 and let F1 , . . . , Fr ⊂ 2[n] . We say that F1 , . . . , Fr are r-cross union if F1 ∪ · · · ∪ Fr = [n] for all choices of Fi ∈ Fi , i = 1, . . . , r. Proposition 5.7. Let 1 ≤ ki < n for 1 ≤ i ≤ r. Let Fi be a family of arcs of size ki on a circular permutation π = (a1 , . . . , an ). Suppose that k1 + · · · + kr ≥ n and F1 , . . . , Fr are r-cross union. Then we have r  (5.2) (|Fi | + ki ) ≤ rn. i=1

Proof. We use induction on k1 + · · · + kr . First we prove the base case k1 + · · · + kr = n. The idea is very simple. For every 1 ≤ j ≤ n (j) (j) (j) consider the r arcs F1 , . . . , Fr , where the first element of F1 is (j) (j) aj , the first element of F2 is aj+k1 , . . . , and the first element of Fr is aj+k1 +k2 +···+kr−1 . Then these r arcs partition the whole circle.

5. Katona’s circle

29

Therefore not all r can be members of the corresponding families. (j) That is, there is some i such that Fi ∈ Fi . For the n choices of j we get altogether at least n missing sets, yielding |F1 | + |F2 | + · · · + |Fr | ≤ rn − n = rn − (k1 + · · · + kr ), which is equivalent to (5.2). Now suppose that k1 + · · · + kr > n and we are in the induction step. Suppose that we can choose i, 1 ≤ i ≤ r, such that ki ≥ 2 and |Fi | < n. Then we replace Fi with σ π (Fi ) and leave the other families unchanged. In view of Proposition 5.6 we have that |σ π (Fi )| + ki − 1 ≥ |Fi | + ki . Consequently, (5.2) follows from the induction hypothesis. The only remaining case is where after reordering the families we have k1 = · · · = ks = 1 and |Fs+1 | = · · · = |Fr | = n for some s ≥ 0. We may further assume that |Fs | < n. If ks+1 + · · · + kr ≥ n, then |Fs+1 | = · · · = |Fr | = n implies that Fs+1 , . . . , Fr are not (r − s)cross union already, and F1 , . . . , Fr cannot be r-cross union. Thus we may assume that ks+1 + · · · + kr ≤ n − 1 and so s ≥ 1. Then without loss of generality |F1 | < n and therefore |F1 |+k1 ≤ n. Now consider the (r − 1)-cross union families F2 , . . . , Fr . Recall that we have assumed k1 + · · · + kr ≥ n + 1 and k1 = 1. Thus k2 + · · · + kr ≥ (n + 1) − k1 = n and we may apply the induction r hypothesis and obtain i=2 (|Fi |+ki ) ≤ (r−1)n. Adding |F1 |+1 ≤ n, the inequality (5.2) follows.  We say that a family F ⊂ 2[n] is r-wise union if F1 ∪· · ·∪Fr = [n] for all F1 , . . . , Fr ∈ F.   Theorem 5.8 ([43]). Let 0 < k < n, r ≥ 2 and kr ≥ n. If F ⊂ [n] k n−1 is r-wise union, then |F| ≤ k . Proof. Let π be a circular permutation. Applying Proposition 5.7 to F1 = F2 = · · · = Fr := F ∩ C(π), we get |F ∩ C(π)| ≤ n − k. Adding up this inequality over all (n − 1)! cyclic permutations gives n−1  k!(n − k)!|F| ≤ (n − k)(n − 1)!, or equivalently |F| ≤ k .

30

5. Katona’s circle

Let us note that the above theorem  best possible. Indeed, Y more   is if and only if F = for careful analysis shows that |F| = n−1 k k  [n]  some Y ∈ n−1 , except for the case r = 2 and n = 2k; cf. [50].

Chapter 6

The Kruskal–Katona Theorem

An important classical theorem, which has found many applications outside extremal set theory, is the Kruskal–Katona Theorem. It answers the following question. Given positive integers k and m, what is the minimum number of (k−1)-element sets that are contained in some members of a collection of m sets of size k?   If m is of the form ka for some integer a, a ≥ k, then weknow  a . the answer already. In Theorem 2.14 we have shown that it is k−1 [a] Moreover, k shows that one cannot do better. To define the best construction let us introduce the colexicographic (colex for short) ordering of subsets. Let A and B be finite subsets of the set of positive integers Z>0 . We say that A is smaller than B if either A ⊂ B or max(A \ B) < max(B \ A). For example, for A = {2, 7.10} and B = {1, 2, 11}, we have A \ B = {7, 10} and B \ A = {1, 11}. The maximal element of the former is 10, and of the latter is 11. Therefore A precedes B in the colex order.

31

32

6. The Kruskal–Katona Theorem

Say we want to take the first 11 3-element subsets in the colex order. They are as follows: {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 5}, {1, 3, 5}, {2, 3, 5}, {1, 4, 5}, {2, 4, 5}, {3, 4, 5}, {1, 2, 6}. Let us look at the 2-element subsets contained inat least one of these 3-element sets. First we have all members of [5] 2 . Then we have the two sets {1, 6} and {2, 6} containing 6. In general, for fixed k, we can    sort Nk in the colex order. By an initial segment of Nk we mean the   first m subsets of Nk for some m ∈ N. The reader is invited to verify the following. Proposition 6.1. If F is an initial segment of the k-element subsets, then its shadow, the collection of (k − 1)-element subsets that are contained in at least one member of F, forms an initial segment among the (k − 1)-sets in colex order as well. Let us use the notation σ(F) for the immediate shadow : σ(F) = {G : ∃ F ∈ F, G ⊂ F, |F \ G| = 1}. Also let A(k, m) denote the collection of the first m subsets in in the colex order. The above example presents A(3, 11).

Z>0  k

6.2 (Kruskal–Katona Theorem). For every family F ⊂ Theorem  Z>0 with |F| = m, one has k |σ(F)| ≥ |σ(A(k, m))|. Before proceeding to the proof let us look at the colex order more closely. Since each set containing the element n + 1 is preceded by all subsets of [n], in finding the members of A(k, m) it is useful to define ak as the unique integer satisfying     ak ak + 1 ≤m< . k k   m) and the rest of A(k, m) is of the Then we know that [n] k ⊂ A(k,  [ak ]  form G ∪ {ak + 1} with G ∈ k−1 . Now these (k − 1)-element sets

6. The Kruskal–Katona Theorem

33

  G should form A(k − 1, m − akk ). Proceeding in this way, we obtain the so-called k-cascade representation of m:       ak ak−1 al (6.1) m= + + ··· + k k−1 l with ak > ak−1 > · · · > al ≥ l ≥ 1. Claim 6.3. If (6.1) is the k-cascade representation of m, then       ak ak−1 al (6.2) |σ(A(k, m))| = + + ··· + . k−1 k−2 l−1   Proof. We prove this by induction on k. If k = 1 then m = a11 = a1 and A(1, m) = {{1},  {2}, . . . , {m}}. Then σ(A(1, m)) = {∅} and |σ(A(1, m))| = 1 = a01 , as required. So (6.2) holds for k = 1 and all m.     Let k > 1. If m = akk then A(k, m) = [akk ] and σ(A(k, m)) =  [ak ]  ak    < m < akk+1 . As we saw k k−1 . Thus (6.2) holds. Now let above,       . A(k, m) = [akk ] ∪ G ∪ {ak + 1} : G ∈ A k − 1, m − akk Consequently,    ak    + σ A k − 1, m − akk . |σ(A(k, m))| = k−1    k−1    Since m− akk = ak−1 +· · ·+ all is the (k−1)-cascade representation ak  for m − k , we can use the induction hypothesis to deduce that    

    al

σ A k − 1, m − ak = ak−1 + · · · + , k k−1 l−1 and (6.2) follows.



Let us recall from Proposition 2.10 that it is sufficient to prove the Kruskal–Katona Theorem for shifted families. The following simple proof is due to Frankl. Proof of Theorem 6.2. We use double induction on k and m. The case of k = 1 and the case of m = 1 are trivial. In fact, if k = 1 then σ(F) = σ(A(1, m)) = {∅} for all m, and if m = 1 then |σ(F)| =

34

6. The Kruskal–Katona Theorem

|σ(A(k, 1))| = k for all k. We proceed with the induction step, distinguishing some cases according to the k-cascade representation of m. The two main observations that follow from the shiftedness of F are the following: (a) σ(F(¯ 1)) ⊂ F(1), (b) |σ(F)| = |F(1)| + |σ(F(1))|, where F(1) = {F \ {1} : 1 ∈ F ∈ F} and F(¯1) = {F ∈ F : 1 ∈ F }. Statement (a) is immediate from shiftedness: if G ⊂ F ∈ F(¯1),

with |G| = k − 1, then G ∪ {1} is in F as well. As for statement (b), we note that σ(F) consists of two types of (k − 1)-element subsets: those not containing 1 and those containing 1. So let σ(F) = G¯1  G1 , where G¯1 = {G ∈ σ(F) : 1 ∈ G} and G1 = {G ∈ σ(F) : 1 ∈ G}. We will show that |G¯1 | = |F(1)| and |G1 | = |σ(F(1))|. If G ∈ G¯1 then there is i such that G∪{i} ∈ F, and by shiftedness {1} ∪ G ∈ F as well. This shows that G ∈ F(1), that is, G¯1 ⊂ F(1). Of course G ∈ F(1) implies G ∈ σ(F), and F(1) ⊂ G¯1 . Consequently G¯1 = F(1) and |G¯1 | = |F(1)|. Next let G1 = {G \ {1} : G ∈ G1 }. If G ∈ G1 then G ⊂ G ⊂ F for some 1 ∈ G ∈ G1 and F ∈ F. This means that G ∈ σ(F(1)), and G1 ⊂ σ(F(1)). If G ∈ σ(F(1)) then {1}  G ∈ G1 . This gives us G ∈ G1 and σ(F(1)) ⊂ G1 . Therefore G1 = σ(F(1)), and |G1 | = |G1 | = |σ(F(1))|. This completes the proof of (b). Let us state the inequality that we need to prove in a formal way. If F is a family of k-element subsets, |F| = m, and m is of the form (6.1), then       ak ak−1 al (6.3) |σ(F)| ≥ + + ··· + . k−1 k−2 l−1 The general idea is to show that       ak − 1 ak−1 − 1 al − 1 (6.4) |F(1)| ≥ + + ···+ k−1 k−2 l−1

and use the induction hypothesis to infer that       ak − 1 ak−1 − 1 al − 1 (6.5) |σ(F(1))| ≥ + + ···+ . k−2 k−3 l−2

6. The Kruskal–Katona Theorem

35

In view of (b), adding (6.4) and (6.5) yields (6.3). However, there can be a problem,  is, if l = 1 then the right-hand side (RHS) of (6.4)  that ends with al0−1 , which is not permitted in a cascade representation.  l −1 However, the last term on the RHS of (6.5) is a−1 = 0. Thus it is sufficient to use     ak − 1 al+1 − 1 |F(1)| ≤ + ···+ k−1 l   −1 . to obtain the inequality (6.5) ending with the last term al+1 l−1 Now we have to prove (6.4). The idea is very simple. If (6.4) fails, then by using |F| = |F(1)| + |F(¯1)| with (6.1) and the negation of (6.4), it follows that     al − 1 ak − 1 ¯ + ···+ . (6.6) |F(1)| > k l If the RHS is a k-cascade, then (b) and the induction hypothesis imply (6.4), a contradiction. How can the RHS of (6.6) fail to be a k-cascade? Only if at − 1 < t for some k ≥ t ≥ l. In this case we have at = t because ai ≥ i for i = k, k − 1, . . . , l by (6.1). We choose t to be maximal, that is, ai > i for i = k, k − 1, and  . . ., t + 1l−1 , . . . , aj = j for j = t, t − 1, . . . , l. In (6.6) the bad terms t−1 t l are all 0. What saves us is the fact that the inequality (6.6) is strict. Consequently, we have      ak − 1 at+1 − 1 t |F(¯ 1)| ≥ + ··· + + . k t+1 t Applying the induction hypothesis to F(¯1) now gives us       ak − 1 at+1 − 1 t ¯ |σ(F(1))| ≥ + ··· + + . k−1 t t−1   t  t t−1 = t + t−1 + · · · + 11 , it follows that In view of (a) and t−1          ak − 1 at+1 − 1 t t−1 1 |F(1)| ≥ +···+ + + +···+ , k−1 t t t−1 1 which means that (6.4) holds, a contradiction.



For a fixed k the behavior of the function fk (m) = |σ(A(k, m))| − m is rather complicated. Indeed one can show that after a suitable

36

6. The Kruskal–Katona Theorem

normalization fk converges uniformly to the Takagi function (as k → ∞), which is continuous but nowhere differentiable; see [59].

Chapter 7

Kleitman Theorem for no s pairwise disjoint sets

In this chapter we present a very short proof of an important classical result of Kleitman. Let s ≥ 2 be a fixed integer. A family F ⊂ 2[n] is said to be s-dependent if it contains no s pairwise disjoint members. Note that 2-dependent simply means intersecting. For an intersecting family F we have |F| ≤ 2n−1 , because for any subset A ⊂ [n] only one of A and [n] \ A can be a member of F. There are many ways to achieve equality; in particular, for n + 1 even one can take all subsets of [n] of size at least n+1 2 , or for n even one can take all subsets of [n] having at least n/2 elements in [n − 1]. These constructions are easy to generalize for s ≥ 3. Example 7.1. Let m and s be positive integers with s ≥ 2. Define (i) K(ms − 1, s) = {K ⊂ [ms − 1] : |K| ≥ m}, (ii) K(ms, s) = {K ⊂ [ms] : |K ∩ [ms − 1]| ≥ m}. Note that both K(ms − 1, s) and K(ms, s) are s-dependent, and (7.1)

2|K(ms − 1, s)| = |K(ms, s)|.

Let k(n, s) denote the maximum size |F| over all s-dependent families F ⊂ 2[n] ; for example, k(n, 2) = 2n−1 . We call a family 37

38

7. Kleitman Theorem for no s pairwise disjoint sets

F ⊂ 2[n] an upset if F ∈ F and F ⊂ F  imply F  ∈ F. Since larger subsets are less likely to be pairwise disjoint, in dealing with k(n, s) we can restrict our attention to upsets. Theorem 7.2 (Kleitman [85]). We have the following. (i) k(ms − 1, s) = |K(ms − 1, s)|. (ii) k(ms, s) = |K(ms, s)|. Proof. The proof given here is taken from [54]. Let us remark that (ii) implies (i) in the above theorem. To see this, we first notice that if F ⊂ 2[n] is s-dependent then G = {G ⊂ [n + 1] : G ∩ [n] ∈ F} is s-dependent as well, and |G| = 2|F|. This shows that k(n + 1, s) ≥ 2k(n, s). Thus if we assume (ii), then by using (7.1) we get 2|K(ms − 1, s)| = |K(ms, s)| = k(ms, s) ≥ 2k(ms − 1, s) ≥ 2|K(ms − 1, s)|, and (i) follows. We are going to prove (ii). Let n = ms and let F ⊂ 2[n] be an s-dependent family. We may assume that F is an upset and, in particular, ∅ ∈ F. We define a family H consisting of fewer than sn subsets. For 1 ≤ i ≤ s let Bi = [(i − 1)m + 1, im]. So [n] = B1  · · ·  Bs is a (l) partition. Moreover, for 1 ≤ l ≤ m − 1 let Ai be the first l elements (l) in Bi , that is, Ai = [(i − 1)m + 1, (i − 1)m + l]. Finally, for each (l) Ai and every i = i, 1 ≤ i ≤ s, we take the (2m − l)-element subset (2m−l) (l) (l) (2m−l) Ci,i = (Bi  Bi ) \ Ai . Note that Ai  Ci,i = B i  B i . (l)

(2m−l)

Let H be formed by these Bi , Ai , and Ci,i , plus the empty set, altogether s+s(m−1)+s(s−1)(m−1)+1 = s(1+s(m−1))+1 < sn subsets. We will use H as a window through which we look at F. Let z(l) denote the number of l-element subsets in H \ F. If F = K(n, s), then z(0) = 1, z(l) = s for 1 ≤ l ≤ m − 1, z(m) = 1, and z(l) = 0 for m + 1 ≤ l < 2m. The following easy claim is the most important ingredient of our proof.

7. Kleitman Theorem for no s pairwise disjoint sets

39

Claim 7.3. One of the following holds. (a) z(l) ≥ 2 for all 1 ≤ l ≤ m, or (b) z(m) = 1 and z(l) + z(2m − l) ≥ s for all 1 ≤ l ≤ m − 1. Proof. Since F is s-dependent, not all of B1 , . . . , Bs can be members of F. This shows that z(m) ≥ 1. Note that z(m) ≥ 2 implies (a). Indeed, if Bp and Bq (1 ≤ p < q ≤ s) are missing from F, then each (l) (l) of Ap , Aq ∈ F for 1 ≤ l < m because F is an upset. Thus it is sufficient to consider the case z(m) = 1. (l)

For definiteness suppose that Bs ∈ F. Then As ∈ F for all 1 ≤ l < m. Now we claim that for 1 ≤ i < s and 1 ≤ l < m at least (l) (2m−l) is missing from F. Suppose to the contrary one of Ai and Ci,s (l)

(2m−l)

that both of them are in F. We have Ai  Ci,s = Bi  Bs .   Together with the s − 2 subsets Bi ∈ F for i = i, 1 ≤ i < s, we would get s pairwise disjoint members of F, a contradiction. Thus we (l) (2m−l) }∩(H\F)| ≥ 1 for 1 ≤ i < s and 1 ≤ l < m. This, have |{Ai , Ci,s (l)

together with As ∈ F, yields z(l) + z(2m − l) ≥ s for 1 ≤ l < m.  In the above claim, if case (b) happens then we get (7.2) z(m)

  m−1     m−1   n n n n + (z(l) + z(2m − l)) ≥ +s . m l m l l=1

l=1

m   If (a) happens, then the LHS of (7.2) is at least 2 l=1 nl , and (7.2) holds without equality; see Exercise 7.4 below. Let y(l) denote the number of l-element subsets in 2[n] \ F. Then n 2m−1 2 − |F| = l=0 y(l) ≥ l=0 y(l). On the other hand, n

 2 −|K(n, s)| = 2 − n

n

      m−1 n  n n−1 n n−1 + = + . m l m−1 l l=m+1

l=0

Therefore, to prove |F| ≤ |K(n, t)| it suffices to show that (7.3)

2m−1  l=0

 y(l) ≥

 m−1  n n−1 + . m−1 l l=0

40

7. Kleitman Theorem for no s pairwise disjoint sets

We will derive (7.3) from (7.2) by an averaging manipulation. To this end we need some notation for permutations. Let Sn denote the set of permutations on [n], so |Sn | = n!. For a permutation π ∈ Sn let π(F) = {{π(x) : x ∈ F } : F ∈ F}, and let z (l) denote the number of l-element subsets in H \ π(F). Since π(F) is s-dependent as well, the inequality (7.2) still holds if we replace z with z π . That is, for every π ∈ Sn it follows that   m−1     m−1   n n n n π π π + (z (l) + z (2m − l)) ≥ +s z (m) . m l m l π

l=1

l=1

We average the above inequality over all π ∈ Sn . Of course the RHS is unchanged. For the LHS let G ∈ 2[n] \ F with |G| = m. How  many times does G appear in π∈Sn (H \ π(F))? It is m!(n − m)!s because for fixed i there are m!(n − m)! permutations that satisfy {π(x) : x ∈ G} = Bi , and there are s choices for i. This implies that  π π∈Sn z (m) = m!(n − m)! s y(m), or equivalently   n 1  π z (m) = s y(m). m n! π∈Sn

n  1 π For the same reason, we have n! π∈Sn z (l) l = s y(l) for l ≤ m. On the other hand, if |G| = 2m − l with l < m, then we need to count (2m−l) the number of subsets sent to Ci,i , and there are s(s − 1) choices  for i and i . Thus it follows that   n 1  π z (2m − l) = s(s − 1)y(2m − l), 2m − l n! π∈Sn

and

n   n 1  π z (2m − l) = s(s − 1)  nl  y(2m − l) ≤ s y(2m − l). l n! 2m−l π∈Sn

For the last inequality see Exercise 7.5. Consequently, by averaging, we obtain   m−1 m−1   n n (s y(l) + s y(2m − l)) ≥ +s . s y(m) + m l l=1

l=1

7. Kleitman Theorem for no s pairwise disjoint sets

41

 n−1  n /s = m−1 , we get Dividing both sides by s and noting that m     2m−1 m−1   n n−1 y(l) ≥ + . m−1 l l=1 l=1   Adding y(0) = 1 = n0 to both sides, we finally have (7.3). This completes the proof of the theorem.  Exercise 7.4. Let l, s, m,  n  integers. First verify  and n be positive . Use this inequality to that if 1 ≤ l ≤ m then nl > (s − 1) l−1 prove that   m   m−1   n n n 2 > +s . l m l l=1

l=1

Exercise 7.5. Let l, s, m, and n bepositive integers with n = ms,   n ≥ (s − 1) nl . (Hint: We have l < m, and s ≥ 2. Show that 2m−l equality for the case s = 2. Let s ≥ 3. The inequality is equivalent to (n − l) · · · (n − 2m + l + 1) ≥ (s − 1)(2m − l) · · · (l + 1). Show that n−l > (s−1)(l+1) and n−l−i ≥ 2m−l−i+1 for 1 ≤ i ≤ 2(m−l)+1.)

Chapter 8

The Hilton–Milner Theorem

  For n ≥ 2k the Erd˝ os–Ko–Rado Theorem states that |F| ≤ n−1 k−1   for all intersecting families F ⊂ [n] k . Taking all k-element subsets of [n] containing a fixed vertex i ∈ [n] shows that this bound is best  possible. In this example F ∈F F = ∅. Such families are called  trivial. If F ∈F F = ∅ then F is called non-trivial. Theorem 8.1 (Hilton–Milner Theorem [74]). Let k ≥ 2 and n ≥ 2k.   is a non-trivial intersecting family, then If F ⊂ [n] k     n−1 n−k−1 (8.1) |F| ≤ − + 1. k−1 k−1   For n = 2k the RHS is n−1 k−1 ; that is, (8.1) follows from the Erd˝os–Ko–Rado   Theorem. However, for n > 2k the RHS is strictly less than n−1 k−1 . Thus the Hilton–Milner Theorem shows in a strict sense that only trivial intersecting families can attain the bound in the Erd˝ os–Ko–Rado Theorem. The bound (8.1) is best possible. To see this, let H = G  {[2, k + 1]}, where

G=

 F ∈



[n] k

: 1 ∈ G, G ∩ [2, k + 1] = ∅ . 43

44

8. The Hilton–Milner Theorem

Then H is a non-trivial intersecting family, and its size is given by the RHS of (8.1). Following Frankl [53], we show Theorem 8.1 in two steps. First, we prove the statement for shifted families. To this end we construct an injection from F \ [2, k + 1] to G. Second, we prove that any nontrivial intersecting family can be transformed into a shifted non-trivial intersecting family of the same size.   be a shifted Now let us start with the first step. Let F ⊂ [n] k (not necessarily non-trivial) intersecting family. For F ∈ F let lF = max{i : |F ∩ [2i]| ≥ i}. Let us verify that lF is well-defined. Let F = {a1 , a2 , . . . , ak } ∈ F with 1 ≤ a1 < a2 < · · · < ak ≤ n. If ai ≤ 2i for some i, then we are done. Otherwise we have ai > 2i for all i. Then, since F is shifted, we may assume that F = {2, 4, . . . , 2k} ∈ F. Again, by the shiftedness, F  = {1, 3, . . . , 2k − 1} is also in F, but F and F  do not intersect. This contradiction justifies the definition of lF . For F ∈ F we define F  = F [2lF ], where the RHS denotes the symmetric difference of F and [2lF ]. Exercise 8.2. For F ∈ F and lF < l , show that |F ∩ [2lF ]| = |F  ∩ [2lF ]| = lF and |F ∩ [2l ]| = |F  ∩ [2l ]| < l . For F ∈ F(¯1) show that 1 ∈ F  and lF ≥ 2. Define a map φ : F(¯ 1) →

[2,n] k−1

by

φ(F ) = F  \ {1}. For example, if F = [2, k + 1] then lF = k and φ(F ) = [k + 2, 2k]. Claim 8.3. The map φ satisfies the following. (i) φ(F(¯ 1)) ∩ F(1) = ∅. (ii) φ is injective. (iii) If φ(F ) ∩ [2, k + 1] = ∅ then F = [2, k + 1]. Proof. (i) Suppose the contrary, that is, φ(F ) ∈ F(1) for some F ∈ F(¯ 1). Set l = lF and let F \ [2l] = {bl+1 , bl+2 , . . . , bk } with bl+1
n0 (k, t) then M (n, k, t) = n−t k−t . In this chapter we present a result by Ahlswede and Khachatrian which tells us M (n, k, t) for all n ≥ k ≥ t. Frankl introduced the following t-intersecting families:  

[n] : |F ∩ [t + 2r]| ≥ t + r F (k) (n, t, r) := F ∈ k for r ≥ 0. Without loss of generality we may assume that n ≥ t + 2r, that is, r ≤ n−t 2 . He conjectured in [44] that (10.1)

M (n, k, t) =

max

0≤r≤ n−t 2

|F (k) (n, t, r)|.

Exercise 10.1. Let n, k, and t be fixed, and let Fr = F (k) (n, r, t). Show that |Fr | − |Fr+1 | is positive, zero, or negative if and only if   t−1 n − (k − t + 1) 2 + r+1 53

54

10. The Ahlswede–Khachatrian Theorem

is positive, zero, or negative, respectively. For example, |F (5) (9, 3, 1)| = |F (5) (9, 3, 2)| = 21 as shown int he picture below. (Hint: Compare |Fr \ Fr+1 | and |Fr+1 \ Fr |.) •·

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F (5) (9, 3, 2)

F (5) (9, 3, 1)

The above exercise shows that max |F r

(k)

(n, t, r)| = |F

(k)

  n−t (n, t, 0)| = k−t

if and only if n ≥ (t + 1)(k − t + 1). So the result by Erd˝ os, Ko, and Rado confirms (10.1) for n > n0 (k, t). Also, the case n = 4m and k = 2m of the conjecture already appeared in the paper [36], and was popularized by Erd˝ os for a number of years (e.g. [33]). Frankl [45] proved (10.1) (for t ≥ 15) for the exact range of n, that is, n ≥ (t + 1)(k − t + 1), by using shifting and then counting the lattice

10. The Ahlswede–Khachatrian Theorem

55

paths corresponding to subsets in a shifted family; see Chapter 15. Then Wilson [115] gave a completely different proof for the same   result (for all t) by studying the spectra of a graph on [n] k reflecting the t-intersecting property; see Section 27.3. The case of small n was more difficult. Frankl and F¨ uredi [56] proved (10.1) for the cases  n > (k − t + 1)c t/ log t, where c is some absolute constant. Then it was Ahlswede and Khachatrian who finally established (10.1) in general. They gave two proofs, one based on the method of generating sets [5] and the other using the so-called pushing-pulling method [7]. Both approaches are purely combinatorial and may be considered ‘dual’ to each other in some sense. Theorem 10.2 (Alhswede–Khachatrian Theorem). Let n ≥ k ≥ t ≥ 1 and 0 ≤ r ≤ n−t 2 be integers. (i) If (10.2)

    t−1 r−1 (k − t + 1) 2 + < n < (k − t + 1) 2 + , r+1 r

then1 M (n, k, t) = |F (k) (n, t, r)|. Moreover, F (k) (n, t, r) is the unique extremal configuration up to isomorphism2 . (ii) If (10.3)

  t−1 (k − t + 1) 2 + = n, r+1

then M (n, k, t) = |F (k) (n, t, r)| = |F (k) (n, t, r + 1)|. If moreover t ≥ 2, then F (k) (n, t, r) and F (k) (n, t, r + 1) are the only extremal configurations up to isomorphism. We will present the second proof from [7]. Here we show a consequence of the pushing-pulling method, which will play a key role in the proof. To state the result we need some definitions. For integers 1 ≤ i, j ≤ n with i = j and a subset F ⊂ [n], let Fi,j denote the 1 We regard r−1 as being ∞ for r = 0 in (10.2), that is, we put no restriction on r the upper bound for n in this case.   2 is t-intersecting and |F | = M (n, k, t), then F ∼ That is, if F ⊂ [n] = k (k) F (n, t, r).

56

10. The Ahlswede–Khachatrian Theorem

subset obtained by renaming i and j to each other in F ; more precisely, Fi,j = F {i, j} if |F ∩ {i, j}| = 1, and Fi,j = F otherwise. We say that Fi,j is obtained by (i, j)-exchange from F . For a family of subsets F ⊂ 2[n] let Fi,j = {Fi,j : F ∈ F}. We say that F is exchange stable on [h] if Fi,j = F for all 1 ≤ i, j ≤ h (i = j). Note that in this case if {F ∩ [h] : F ∈ F} contains an aelement subset, then it contains all a-element subsets of [h]. Suppose that Fi,j = F for some i, j. Then there exists a unique h such that F is exchange stable on [h] but not on [h + 1]. This number h is called a head size (or homogeneous part size) of F and denoted by h(F). The following easy fact is useful. Claim 10.3. Let F ⊂ 2[n] be exchange stable on [s]. If A, B ∈ F satisfy |A ∩ [s]| = a and |B ∩ [s]| = b, then there are A , B  ∈ F such that |A ∩ B  ∩ [s]| ≤ max{0, a + b − s}. Proof. Since F is exchange stable on [s], we may choose A , B  from A, B by changing A ∩ [s], B ∩ [s], respectively, so that A ∩ [s] = [a], B  ∩ [s] = [s − b + 1, n]. If a < s − b + 1 then A ∩ B  ∩ [s] = ∅. Otherwise |A ∩ B  ∩ [s]| = |A ∩ [s]| + |B  ∩ [s]| − |[s]| = a + b − s.  The following is the key lemma for the proof of Theorem 10.2. We will prove the lemma in the next chapter. Lemma 10.4. Let n, k, t and r be integers with k ≥ t ≥ 2, 0 ≤ r ≤ [n] n−t be a shifted t-intersecting family 2 , and n > 2k − t. Let F ⊂ k with |F| = M (n, k, t). If   t−1 n < (k − t + 1) 2 + , r then h(F) ≥ t + 2r. (We put no restriction on the upper bound on n if r = 0.) Having this lemma, we can prove Theorem 10.2 relatively easily. We just recall a simple fact about a complement family. For a family of subsets F ⊂ 2[n] let F c denote the complement family, that is, F c = {[n] \ F : F ∈ F}.

10. The Ahlswede–Khachatrian Theorem

57

  [n] c It is readily verified that if F ⊂ [n] k is t-intersecting, then F ⊂ k is t -intersecting, where k = n − k and t = n − 2k + t. Moreover, if F is shifted, then F c is also shifted but in the opposite direction, that is, if F ∈ F c , i ∈ F , j ∈ F , and 1 ≤ i < j ≤ n, then (F \ {i}) ∪ {j} ∈ F c . Proof of Theorem 10.2 for shifted families. Let n ≥ k ≥ t ≥ 1. In Chapter 4 we dealt with  case t = 1 already, so we may [n]the assume that t ≥ 2. Let F ⊂ k be a shifted t-intersecting family of maximum size. Recall that shifting preserves the size and the tintersecting property of the family. We will show that F is actually F (k) (n, t, r) or F (k) (n, t, r + 1). Let k = n − k, t = n − 2k + t, and r  = k − t − r. For F ∈ F let us define its head part F h and tail part F t by F h = F ∩ [t + 2r] and F t = F \ F h . Case (i). Assume (10.2). Then, by Lemma 10.4, the head size of F is such that h(F) ≥ t + 2r, and F is exchange stable on [t + 2r]. Next we will verify that the complement family has the corresponding exchange property. By direct computation we see that (10.2) can be rewritten as     t − 1 r − 1     . (k − t + 1) 2 +  < n < (k − t + 1) 2 + r +1 r    Recall that the complement family F c ⊂ [n] k is t -intersecting and reverse shifted. So, again by Lemma 10.4, it follows that h(F c ) ≥ t + 2r  = n − (t + 2r). Since the direction of shifting for F c is the opposite, the head of F c starts from n, so F c is (and of course also F is) exchange stable on [t + 2r + 1, n]. If |F h | ≥ t + r for all F ∈ F, then F ⊂ F (k) (n, t, r), and by the maximality, F = F (k) (n, t, r). So we may assume that |F h | ≤ t + r − 1 for some F ∈ F. By the shiftedness we may assume that |F h | = t + r − 1. Since F is exchange stable on [t + 2r], all (t + r − 1)-element subsets of [t + 2r] are contained in {F ∩ [t + 2r] : F ∈ F}. Thus by Claim 10.3 we can find F, G ∈ F such that |F h | = |Gh | = t + r − 1 and |F h ∩ Gh | = |F ∩ G ∩ [t + 2r]| ≤ 2(t + r − 1) − (t + 2r) = t − 2. We also have |F t | = |Gt | = k − (t + r − 1) and |F t | + |Gt | − |[t + 2r + 1, n]| = 2k − t − n + 2 ≤ 1.

58

10. The Ahlswede–Khachatrian Theorem

Since F is exchange stable on [t+2r+1, n], we apply Claim 10.3 again, and we may assume that |F t ∩ Gt | = |F ∩ G ∩ [t + 2r + 1, n]| ≤ 1. Therefore we have |F ∩ G| = |F h ∩ Gh | + |F t ∩ Gt | ≤ (t − 2) + 1 < t, which contradicts the t-intersecting property.   Case (ii). Assume (10.3). Then we have n < (k − t + 1) 2 + t−1 r and h := h(F) ≥ t + 2r by Lemma 10.4. Since (10.3) is equivalent to   t − 1   n = (k − t + 1) 2 + , r we also get n < (k − t + 1)(2 +

t −1 r  −1 )

and

h := h(F c ) ≥ t + 2(r  − 1) = n − (t + 2r + 2). Thus F is exchange stable both on [t + 2r] and on [t + 2r + 3, n]. If |F h | ≥ t + r for all F ∈ F, then F ⊂ F (k) (n, t, r), and by the maximality we have F = F (k) (n, t, r). So we may assume that |F h | ≤ t + r − 1 for some F ∈ F. By the shiftedness we may assume that |F h | = t + r − 1. Moreover, since F is exchange stable on [t + 2r], we may assume that F ∩ [t + 2r + 2] = [r + 2, t + 2r + 2]. If |G∩[t+2r +2]| ≥ t+r +1 for all G ∈ F, then F ⊂ F (k) (n, t, r + 1), and by the maximality we have F = F (k) (n, t, r + 1). So we may assume that |G ∩ [t + 2r + 2]| ≤ t + r for some G ∈ F. By the shiftedness we may assume that G ∩ [t + 2r + 2] = [t + r]. Then we have |F ∩G∩[t+2r +2]| = t−1. Let R = [n]\[t+2r +2]. As for the remaining part, we have |F ∩ R| = k − (t + r + 1) and |G ∩ R| = k − (t + r). Since F is exchange stable on R, we can use Claim 10.3. Then we may assume that F ∩ G ∩ R = ∅ because |F ∩ R| + |G ∩ R| − |R| = 2k − t − n + 1 ≤ 0. Thus |F ∩ G| = t − 1, which contradicts the t-intersecting property.  We have proved Theorem 10.2 for shifted families (assuming the key lemma which will be shown in the next chapter). To complete the proof we need to show that if a family can be made to be one of the extremal configurations after applying shifting operations, then it was isomorphic to the extremal configuration from the beginning. In fact the following holds.

10. The Ahlswede–Khachatrian Theorem

59

  Lemma 10.5. Let t ≥ 2, let A ⊂ [n] k be a t-intersecting family, and let F = F (k) (n, t, r). If Si,j (A) = F, then A ∼ = F. Proof. If i, j ∈ [t + 2r] or i, j ∈ [t + 2r], then A = F and we are done. So without loss of generality we may assume that i = t + 2r and j = n. Define two disjoint subfamilies B and C of A by B = {A ∈ A : i ∈ A, j ∈ A, Ai,j ∈ A}, C = {A ∈ A : i ∈ A, j ∈ A, Ai,j ∈ A}, where Ai,j is obtained by (i, j)-exchange from A. Since Si,j (A) = F, we have |A ∩ [t + 2r − 1]| ≥ t + r − 1 for all A ∈ B  C. Claim 10.6. If A ∈ B  C then |A ∩ [t + 2r − 1]| = t + r − 1. Proof. Suppose, to the contrary, that |A ∩ [t + 2r − 1]| ≥ t + r for some A ∈ B  C. Then A = Ai,j , and both A and Ai,j are members of F, but only one of them is a member of Si,j (A). This contradicts  the fact that Si,j (A) = F. Note that Si,j (A \ B) = A \ B. Thus if B = ∅, then A = F. Note also that Si,j ((A \ C)i,j ) = (A \ C)i,j . Thus if C = ∅, then Ai,j = F. So we may assume that B = ∅ and C = ∅. Let   : |H ∩ [t + 2r − 1]| = t + r − 1} H = {H ∈ [n]\{i,j} k−2 Then |H| = |B  C|. Moreover, for every H ∈ H we have H ∪ {j} ∈ B or H ∪ {i} ∈ C (but not both). This defines a bijection φ : B  C → H. Now we define a graph G on V (G) = H, by {H, K} ∈ E(G) if and only if |H ∩ K ∩ [t + 2r − 1]| = t − 1 and H ∩ K ∩ [t + 2r, n] = ∅. Then G is isomorphic to a connected graph obtained by a direct product of two Kneser graphs (see the exercises below). Therefore, for arbitrary B ∈ B and C ∈ C there is a path from φ(B) to φ(C) in G, and on this path there is an edge {φ(B  ), φ(C  )} where B  ∈ B and C  ∈ C. But then |B  ∩C  | = |φ(B  )∩φ(C  )| = t−1, which contradicts the fact that A is t-intersecting.  [n] Exercise 10.7. A Kneser graph G = (V, E) is defined by V = k and {x, y} ∈ E if and only if x ∩ y = ∅ for x, y ∈ V . Prove that G is connected provided n ≥ 2k + 1. (Hint: Let vi = [i, i + k − 1] ∈ V . There is a path from v1 to v2 . In fact {v1 , vk+2 }, {v2 , vk+2 } ∈ E. So

60

10. The Ahlswede–Khachatrian Theorem

there is also a path from v2 to v3 , and thus from v1 to v3 . Continue this process.) Exercise 10.8. For i = 1, 2 let Gi = (Vi , Ei ) be a Kneser graph,    [t+2r,n]  and V2 = (k−2)−(t+r−1) . Define a graph where V1 = [t+2r−1] r    G = (V1 × V2 , E ) by {(u1 , u2 ), (v1 , v2 )} ∈ E if {u1 , v1 } ∈ E1 and {u2 , v2 } ∈ E2 . Show that G is connected and isomorphic to the graph G defined in the proof of Lemma 10.5. (Hint: For G1 consider the complement.) Exercise 10.9. Verify that we have completed the proof of Theorem 10.2 using Lemma 10.5 (assuming Lemma 10.4). Exercise 10.10. Define F(n, t, r) = {F ⊂ [n] : |F ∩ [t + 2r]| ≥ t + r}. Let t ≥ 2 and let A ⊂ 2[n] be a t-intersecting family. Show that if Si,j (A) = F(n, t, r), then A ∼ = F. (Hint: This easily follows from Lemma 10.5. See Lemma 6 of [87] for an extension.)

Chapter 11

Pushing-pulling method

This chapter is a continuation of the previous chapter. Here we use the pushing-pulling method to analyze the structure of intersecting families. Then we prove Lemma 10.4, which is the key to the proof of the Ahlswede–Khachatrian Theorem.

11.1. Structure of shifted families Let F ⊂ 2[n] be a family with head size h = h(F). For F ∈ F we define its head part F h and tail part F t by F h = F ∩ [h] and F t = F \ [h], respectively. For F we also define its head part and tail part by F h = {F h : F ∈ F},

F t = {F t : F ∈ F},

respectively. We define a subfamily F − by F − = {F ∈ F : Fl,h+1 ∈ F for some l ∈ [h]}, where Fl,h+1 is obtained by (l, h+1)-exchange from F . We understand F − = ∅ if h = n. By definition F − = ∅ if h < n. We partition F − into h parts according to the size in its head part: F− =

h 

Fi− ,

i=1

61

62

11. Pushing-pulling method

where Fi− = {F ∈ F − : |F ∩ [h]| = i}. Finally we define another family Fi+ by Fi+ = {Fj,h+1 : j ∈ [h], F ∈ Fi− , Fj,h+1 ∈ F}. We note that Fi− is a subfamily of F while Fi+ is disjoint from F. Roughly speaking the pushing-pulling method gives a way to modify F to get a “better” family F˜ by choosing appropriate i1 and i2 such that F˜ = (F \ Fi−1 )  Fi+2 . For this to make sense we will need to assume that these families satisfy some additional properties. Recall that a family of subsets F ⊂ 2[n] is shifted if (F \ {j})  {i} ∈ F holds for all 1 ≤ i < j ≤ n and all F ∈ F with F ∩ {i, j} = {j}. The basic idea in the proof of Lemma 10.4 is the following: suppose that F is a shifted t-intersecting family with the maximum size but its head size does not satisfy the conclusion of the lemma; then we can modify F to get F˜ which is still t-intersecting and |F˜ | > |F|. We start by verifying some basic properties of the families Fi− and Fi+ . Lemma 11.1. Let F ⊂ 2[n] be a shifted family of subsets with head size h = h(F). (i) If F ∈ F − then h + 1 ∈ F . (ii) If F ∈ F − and j ∈ F ∩ [h] then Fj,h+1 ∈ F.

  (iii) Let F ∈ Fi− . Then H  F t ∈ Fi− for all H ∈ [h] i . h − t  h  − t − − + (iv) If Fi = ∅ then |Fi | = i |(Fi ) | and |Fi | = i−1 |(Fi ) |. (v) If G ∈ Fi+ then Gh = [h]. Proof. (i) If F ∈ F − then Fi,h+1 ∈ F for some 1 ≤ i ≤ h, and |F ∩ {i, h + 1}| = 1. If F ∩ {i, h + 1} = {h + 1}, then Fi,h+1 = (F \ {h + 1}) ∪ {i} is obtained from F by shifting, and since F is shifted it follows that Fi,h+1 ∈ F, a contradiction. This proves that F ∩ {i, h + 1} = {i}, and h + 1 ∈ F . (ii) If F ∈ F − then Fi,h+1 ∈ F for some 1 ≤ i ≤ h, and F ∩{i, h+ 1} = {i} by (i). If j = i then we are done. So let j = i and suppose, to the contrary, that F  := Fj,h+1 ∈ F. Since i, j ≤ h and Fi,j = F we  ∈ F. On the other hand, we have that F  = (F \{j})∪{h+1} have Fi,j

11.2. Structure of intersecting families

63

 and Fi,j = (F  \ {i}) ∪ {j} = (F \ {i}) ∪ {h + 1} = Fi,h+1 ∈ F, a contradiction.   be given. Since F is exchange (iii) Let F ∈ Fi− and H ∈ [h] i stable on [h], we have F  := H  F t ∈ F. Choose j  ∈ H = (F  )h and j ∈ F h arbitrarily. To show F  ∈ Fi− it suffices to show Fj ,h+1 ∈ F. By the exchange stability, (F  )h \ {j  } is obtained from F h \ {j} by applying exchange operations, and Fj ,h+1 is also obtained from Fj,h+1 . So Fj,h+1 ∈ F would imply Fj ,h+1 ∈ F. Indeed, by (ii) we have Fj,h+1 ∈ F.

(iv) If Fi− = ∅ then, by (iii), we can write   − t Fi− = {H  T : H ∈ [h] i , T ∈ (Fi ) }.   This gives us that |Fi− | = hi |(Fi− )t |. Then using (ii) we get Fi+ = {Fj,h+1 : j ∈ [h], F ∈ Fi− , Fj,h+1 ∈ F}  [h]  , T ∈ (Fi− )t }, = {H   {h + 1}  T : H  ∈ i−1  h  − t |(Fi ) |. and |Fi+ | = i−1 (v) By definition of Fi+ there is an F ∈ Fi− such that Fj,h+1 = G with j ∈ [h]. By (i) it follows that F ∩{j, h+1} = j. Thus j ∈ Gh . 

11.2. Structure of intersecting families Now we are going to consider t-intersecting families. From now on we assume the following. Assumption 1. Let n, k, and t be integers with k ≥ t ≥ 2 and be a shifted t-intersecting family with n > 2k − t. Let A ⊂ [n] k |A| = M (n, k, t) and h(A) = h. The goal of this section is to prove Claims 11.6 and 11.9, from which Lemma 10.4 follows easily. To this end we need some preliminaries. + Claim 11.2. We have that A− i = Ai = ∅ for 1 ≤ i < t.

Proof. Suppose, to the contrary, that A− i = ∅ for some i < t. Choose h A ∈ A− and a ∈ A . By (ii) of Lemma 11.1, Aa,h+1 ∈ A. Thus, i by the maximality, there exists B ∈ A such that |Aa,h+1 ∩ B|
|A|, which contradicts the maximality assumption. Suppose the opposite: |A˜i,j | ≤ |A| and − + − |A˜j,i | ≤ |A|, that is, |A+ i | ≤ |Aj | and |Aj | ≤ |Ai |. By Lemma 11.1, these inequalities hold if and only if 

 − t    h  − t h − t t |(Aj ) | ≤ hi |(A− i ) | and i−1 |(Ai ) | ≤ j |(Aj ) |.

h j−1

 h  h  hh Then we get i−1 j−1 ≤ i j . This simplifies to (h + 1)(i + j + 1 − h) ≤ 0, that is, (h + 1)(t + 1) ≤ 0, which is clearly false. Case i = j. In this case we have i = j = h+t 2 and h + t is even. Since h + t and h − t have the same parity (even), it follows from (11.2) that h ≤ t+2r−2. By Claim 11.9, C˜i is t-intersecting. We show that |C˜i | > |A|, or equivalently |Ci | > |Ci |. By Claim 11.7 it can be rewritten as

68

11. Pushing-pulling method



   h+1   t t |Ti | > hi (|(A− |Ti | > hi |(A− i ) | − |Ti |), that is, i ) |. Thus, i using the last inequality in Claim 11.7, it suffices to show that     h h+1 k−i > . i i n−h−1 h i−1

Using i =

h+t 2

n
|A| if n < (k − t + 1)(2 + t−1 r ), and this last condition is exactly (11.1). This contradicts the maximality assumption of |A|.  For more about the methods due to Ahlswede and Khachatrian, we recommend an excellent survey by Bey and Engel [11].

Chapter 12

Uniform measure versus product measure

Many problems in extremal set theory ask the maximum size of families satisfying a given constraint. Now suppose that each subset has a weight. Then, instead of counting the number of subsets in a family, one can consider the sum of weights, or a measure of families. In this chapter we introduce two important measures which are closely related to each other.

12.1. Uniform measure and product measure Let w : 2[n] → [0, 1] be given. That is, w(F ) is a weight assigned to a subset F ⊂ [n]. For a family of subsets F ⊂ 2[n] , we define its measure (total weight) by μ(F) =



w(F ).

F ∈F

We say that μ is a probability measure if μ(∅) = 0 and μ(2[n] ) = 1. For a given probability measure μ we are interested in determining the maximum measure Mn (μ) of intersecting families, that is, Mn (μ) = max{μ(F) : F ⊂ 2[n] is intersecting}. 69

70

12. Uniform measure versus product measure

There are two important measures in this setting. One is a uniform measure μk , where k is an integer with 0 ≤ k ≤ n. This is defined by a weight wk : 2[n] → [0, 1], where  −1 n if |F | = k, k wk (F ) = 0 if |F | = k,   for F ∈ 2[n] . Clearly we have μk (2[n] ) = μk ( [n] k ) = 1. The other is a product measure μp , where p is a fixed real number with 0 < p < 1. This is defined by a weight wp : 2[n] → [0, 1], where wp (F ) = p|F | (1 − p)n−|F | .

(12.1)

To understand the meaning of the weight, let us consider the following random process. For each i = 1, 2, . . . , n we decide whether we pick i with probability p or not pick i with probability 1 − p. We make each decision independently. After n rounds we eventually obtain F with probability precisely wp (F ). In view of this μp (2[n] ) = 1 is evident. We can also show this from the definition. Indeed, we have   wp (F ) = p|F | (1 − p)n−|F | μp (2[n] ) = =

F ∈2[n] n  

F ∈2[n]

p

|F |

(1 − p)

k=0 F ∈([n])

n−|F |

n    n k = p (1 − p)n−k k k=0

k

= (p + (1 − p))n = 1. Next we compute the measures of the intersecting family F = {F ∈ 2[n] : 1 ∈ F }. n−1   For the uniform measure we note that |F ∩ [n] k | = k−1 . Then     n−1 n k μk (F) = / = . k−1 k n For the product measure we note that F = {{1}∪G : G ∈ 2Y }, where Y = [2, n]. By definition we have  p|{1}∪G| (1 − p)n−|{1}∪G| μp (F) = G∈2Y

=p



G∈2Y

p|G| (1 − p)|Y |−|G| = p.

12.3. Some extensions

71

This suggests a correspondence between μk and μp by relating k/n with p.

12.2. Two versions of the EKR With language from the previous section we can restate the Erd˝ os– Ko–Rado Theorem (EKR) in the following form: Theorem 12.1 (EKR reprise). If

k n

≤ 12 , then Mn (μk ) =

k n.

The above statement has a measure counterpart, which was first obtained by Ahlswede and Katona in [4]: Theorem 12.2 (Product measure EKR). If p ≤ 12 , then Mn (μp ) = p. We include a proof due to Filmus [38]. Proof. Let F be an intersecting family. Let C be a circle of circumference 1. By an arc we mean a directed arc on C, say in the clockwise direction, of a fixed length p. Choose an arc I at random, and pick n independent random points x1 , . . . , xn uniformly distributed on C. Let XI = {i ∈ [n] : xi ∈ I}. Then Pr(XI ∈ F) = μp (F). Now fix n points x1 , . . . , xn on C, and fix an arc I0 such that XI0 ∈ F. Then any arc I = I0 for which XI ∈ F intersects I0 . Since p ≤ 12 , exactly one of the endpoints of I is in I0 . For a point x in I0 \{x1 , . . . , xn } let Sx (resp. Ex ) be the arc starting (resp. ending) at x. Then only at most one of XSx and XEx is in F. Thus there is an injection from the set of arcs I with XI ∈ F to I0 . So, conditioned on the positions of xi , Pr(XI ∈ F | x1 , . . . , xn ) ≤ p. This is true for every configuration of n points, and we conclude that Pr(XI ∈ F) ≤ p.  These two theorems seem to be “equivalent” in some sense. Problem 12.3. Find a general theorem concerning Mn (μ) that includes both of the above two statements as special cases.

12.3. Some extensions In the rest of this chapter we present some results and problems which extend the above two theorems in various ways.

72

12. Uniform measure versus product measure

Recall that a family F ⊂ 2[n] is t-intersecting if |F ∩ F  | ≥ t for all F, F  ∈ F. For example, F(n, t, i) := {F ⊂ [n] : |F ∩ [t + 2i]| ≥ t + i} is a t-intersecting family. The Ahlswede–Khachatrian Theorem has its measure version as well: Theorem 12.4 (Product measure Ahlswede–Khachatrian Theorem). Let n ≥ t ≥ 1 and p ∈ (0, 1/2]. Then Mn (μp ) = max μp (F(n, t, i)). i

The above result was first obtained by Ahlswede and Khachatrian [6] for rational p; see also Bey and Engel [11]. Then, Dinur and Safra [26] and Tokushige [108] deduced Theorem 12.4 from the Ahlswede– Khachatrian Theorem on k-uniform t-intersecting families. See also [39]. Exercise 12.5. Show that limn→∞ Mn (μp ) = 1 for p > 1/2. (Hint: Consider the product measure of F = {F ⊂ [n] : |F | > n+t 2 }.) Simple computation shows that if p ≤

1 t+1

then

max μp (F(n, t, i)) = μp (F(n, t, 0)) = pt . i

So Theorem 12.4 in this case reads as follows. Corollary 12.6. If n ≥ t ≥ 1 and p
t + 1 we can choose  > 0 so that p+ > t + 1. Fix 1 this  and let I = ((p − )n, (p + )n) ∩ N. Then n > p+ k > (t + 1)k holds for all k ∈ I.

Let F ⊂ 2[n] be a t-intersecting family with the maximum measure. For a lower bound of the measure we clearly have M (μp ) = μp (F) ≥ μp (F(n, t, 0)) = pt . For an upper bound we have   

 n [n]

k n−k

+ (12.2) μp (F) ≤ pk (1−p)n−k .

F ∩ k p (1−p) k k∈I

k∈I

12.3. Some extensions

73

n−t   Let k ∈ I. Then, by Theorem 10.2, we have |F ∩ [n] k | ≤ k−t . We n−t n also have that k−t / k ≤ (k/n)t < (p + )t . So the first term in the RHS of (12.2) is at most  n  n − t k n−k t < (p + ) p (1 − p) pk (1 − p)n−k ≤ (p + )t . k−t k k∈I

k∈I

On the other hand, by the de Moivre–Laplace limit theorem, the second term in the RHS of (12.2) goes to 0 as n → ∞. Thus we have that limn→∞ Mn (μp ) < (p + )t . Since  > 0 can be chosen arbitrarily small and since Mn (μp ) ≥ pt we can conclude limn→∞ Mn (μp ) = pt . Now we show that Mn+1 (μp ) ≥ Mn (μp ). To see this we just notice that if F ⊂ 2[n] is t-intersecting, then F  := F ∪ {F  {n + 1} : F ∈ F} ⊂ 2[n+1] is also t-intersecting, and these two families have the same measure. In fact μp (F  ) = μp (F)(1 − p) + μp (F)p = μp (F). Consequently we  have mn (μp ) = pt for all n ≥ t as needed. Friedgut [68] found another proof of the above result, which is a counterpart of Wilson’s spectral proof [115] of the Erd˝os–Ko–Rado Theorem for the case n ≥ (t + 1)(k − t + 1). We will use one of his ideas in Chapter 30. Recall the definition of μp . One can rewrite the RHS of (12.1) as  i∈F p j∈[n]\F (1 − p). Here we assigned the same probability p to each vertex (an element in [n]). Fishburn et al. extended this definition in such a way that each vertex can have a different probability as follows. For a vector p = (p(1) , p(2) , . . . , p(n) ) ∈ (0, 1)n and a family F ⊂ 2[n] , let us define    μp (F) := p(i) (1 − p(j) ). 

F ∈F i∈F

j∈[n]\F

If p(1) = · · · = p(n) = p, then μp coincides with the product measure μp . Theorem 12.7 (Fishburn et al. [40]). Let μp be the extended product measure defined above, and assume that p(1) = max{p(l) : l ∈ [n]} and that p(l) ≤ 1/2 for l ≥ 2. If F ⊂ 2[n] is intersecting, then

74

12. Uniform measure versus product measure

μp (F) ≤ p(1) . Moreover, if p(1) > p(l) for l ≥ 2, then equality holds if and only if F = {F ⊂ [n] : 1 ∈ F }. We will deduce Theorem 12.7 from a stronger result concerning cross intersecting families in Chapter 30. Let [m]n denote the set of integer sequences (chosen from [m]) of length n, that is, [m]n = {(a1 , . . . , an ) : ai ∈ [m]}. We say that a family of integer sequences A ⊂ [m]n is t-intersecting if #{i : ai = ai } ≥ t uredi used a down for all (a1 , . . . , an ), (a1 , . . . , an ) ∈ A. Frankl and F¨ shift introduced by Kleitman [83] to show the following. Theorem 12.8 ([55]). The maximum size of t-intersecting families (of integer sequences) in [m]n is given by the maximum of mn μp (F), where p = 1/m and F runs over all t-intersecting families (of subsets) in 2[n] . Combining the above result with Theorem 12.4, one can obtain the maximum size of t-intersecting families in [m]n . In part−1 , then the maximum size is ticular, if n ≥ t + 2i, where i =  m−2 n m μ1/m (F(n, t, i)); see [6, 11, 64]. Let r ≥ 2 and t ≥ 1 be integers. We say that r families of subsets F1 , . . . , Fr ⊂ 2[n] are r-cross t-intersecting if |F1 ∩ F2 ∩ · · · ∩ Fr | ≥ t for all Fi ∈ Fi , 1 ≤ i ≤ r. If, moreover, F1 = F2 = · · · = Fr =: F, then we say that F is r-wise t-intersecting. For simplicity we omit r and/or t in the case of r = 2 or t = 1. So “cross t-intersecting” means “2-cross t-intersecting,” and “r-wise intersecting” means “rwise 1-intersecting.” We introduce the family F(n, r, t, i) defined by F(n, r, t, i) := {F ⊂ 2[n] : |F ∩ [t + ri]| ≥ t + (r − 1)i} and its k-uniform subfamily F (k) (n, r, t, i) defined by   F (k) (n, r, t, i) := F(n, r, t, i) ∩ [n] k . Exercise 12.9. Show that F(n, r, t, i) is r-wise t-intersecting.

12.3. Some extensions

75

These families F(n, r, t, i) and F (k) (n, r, t, i) play an important role when we construct a large intersecting structure in general. Here is one of the most fundamental problems on intersecting families: Problem 12.10. Let k1 , . . . , kr be positive integers. Determine r  |Fi |, max i=1

where the  maximum is taken over all r-cross t-intersecting families [n] Fi ⊂ ki , 1 ≤ i ≤ r. The corresponding measure counterpart is as follows. Problem 12.11. Let p1 , . . . , pr ∈ (0, 1) be given. Determine max

r 

μpi (Fi ),

i=1

where the maximum is taken over all r-cross t-intersecting families F1 , . . . , Fr ⊂ 2[n] . For the case t = 1 we conjecture as follows. Conjecture 12.12. Let (r − 1)/r ≥ max{k1 /n, . . . , kr /n}, and let   Fi ⊂ [n] ki for 1 ≤ i ≤ r. If F1 , . . . , Fr are r-cross intersecting, then r 

μk (Fi ) ≤

i=1

r 

(ki /n).

i=1

Conjecture 12.13. Let (r − 1)/r ≥ max{p1 , . . . , pr }, and let Fi ⊂ 2[n] for 1 ≤ i ≤ r. If F1 , . . . , Fr are r-cross intersecting, then r r   μpi (Fi ) ≤ pi . i=1

i=1

We have some partial answers to these conjectures and will discuss some of them in later chapters.

Chapter 13

Kleitman’s correlation inequality

Let p be a fixed real number with 0 < p < 1. For a family F ⊂ 2X let μX p (F) denote the product measure of F, that is,  μX p|F | (1 − p)|X|−|F | . p (F) = F ∈F

We often write μp instead of μX p when X is fixed. n If p = 1/2 and X = [n], then μX p (F) is just |F|/2 . This can be viewed as the probability of F in the following sense. Consider the uniform distribution on 2X where each subset of X has the same probability 1/2n . In this probability space the probability that a randomly chosen set is in F is exactly |F|/2n .

In a probability space two events with probabilities p1 and p2 are positively (resp. negatively) correlated if the probability that they occur simultaneously is at least (resp. at most) p1 p2 . Recall that we say that F ⊂ 2X is an upset (resp. a downset) if F ∈ F and F ⊂ F  (resp. F  ⊂ F ) imply F  ∈ F. The following lemma is an extension of a result from [84]. Lemma 13.1 (Kleitman’s correlation inequality). Let F, G ⊂ 2[n] . If F is an upset and G is a downset, then μp (F ∩ G) ≤ μp (F)μp (G). 77

78

13. Kleitman’s correlation inequality

To prove the lemma we list some easy facts which immediately follow from the definition of the product measure. We note that if F ⊂ 2X is an upset (resp. downset) then F(x) ⊃ F(x) (resp. F(x) ⊂ F(x)) for x ∈ X, where F(x) and F(x) are defined by (0.1) and (0.2). [n−1]

Claim 13.2. Let F ⊂ 2[n] , fn = μp

[n−1]

(F(n)), and fn¯ = μp

(F(¯ n)).

[n]

(i) μp (F) = pfn + (1 − p)fn¯ . (ii) If F is an upset then fn ≥ fn¯ . (iii) If F is a downset then fn ≤ fn¯ . Proof of Lemma 13.1. We prove the statement by induction on n. It is easy to check for the case n = 1. Let n ≥ 2. Since F is an upset, F(n) and F(¯ n) are both upsets. Similarly, G(n) and G(¯ n) are both downsets. We define fn and fn¯ as in the previous claim, and define gn and gn¯ for G similarly. Let q = 1 − p. It follows from (i) of the claim that (13.1)

[n−1] μ[n] ((F ∩ G)(n)) + qμ[n−1] ((F ∩ G)(¯ n)). p (F ∩ G) = pμp p

Since (F ∩ G)(n) = F(n) ∩ G(n) and (F ∩ G)(¯ n) = F(¯ n) ∩ G(¯ n), it follows from the induction hypothesis that the RHS of (13.1) is ≤ pfn gn + qfn¯ gn¯ . On the other hand, we have (fn − fn¯ )(gn − gn¯ ) ≤ 0 by (ii) and (iii) of the claim. Thus it follows that [n] μ[n] ¯ )(pgn + qgn ¯) p (F)μp (G) = (pfn + qfn

≥ (pfn + qfn¯ )(pgn + qgn¯ ) + pq(fn − fn¯ )(gn − gn¯ ) = pfn gn + qfn¯ gn¯ , [n−1]

and, by the induction hypothesis, the RHS is ≥ pμp [n−1] [n] qμp (F(n) ∩ G(n)) = μp (F ∩ G).

(F(n)∩G(n))+ 

Exercise 13.3. Show that if F, G ⊂ 2[n] are both upsets, then μp (F ∩ G) ≥ μp (F)μp (G). 

(Hint: Note that G = 2[n] \ G is a downset, and apply Lemma 13.1 to F and G  .)

13. Kleitman’s correlation inequality

79

Kleitman invented the correlation inequality in order to prove the following result, conjectured by Erd˝os. Theorem 13.4. Let 0 < p ≤ 1/2 be a real number and q = 1 − p, and let n ≥ r be positive integers. Suppose that F1 , . . . , Fr ⊂ 2[n] are intersecting families. Then μp (F1 ∪ · · · ∪ Fr ) ≤ 1 − q r . The above inequality is sharp. To see this let Fi = {F ⊂ [n] : i ∈ F }. Then this is an intersecting family as well as an upset. Since 2[n] \ (F1 ∪ · · · ∪ Fr ) = 2[r+1,n] , it follows that μp (F1 ∪ · · · ∪ Fr ) = 1 − q r . Proof of Theorem 13.4. If we replace F with the upset generated by F, that is, {G ⊂ [n] : F ⊂ G for some F ∈ F}, then the product measure only increases while the intersection property is preserved. So we may assume that all Fi are upsets. We prove the statement by induction on r. The case r = 1 follows from Theorem 12.2. Let r ≥ 2. Suppose that we know the statement is true for r − 1 and let us prove it for r. Let G = F1 ∪ · · · ∪ Fr−1 . Applying Lemma 13.1 to G and Fr , we obtain μp (G ∩ Fr ) ≥ μp (G)μp (Fr ). Thus 1 − μp (G ∪ Fr ) = 1 − (μp (G) + μp (Fr ) − μp (G ∩ Fr )) ≥ 1 − μp (G) − μp (Fr ) + μp (G)μp (Fr ) = (1 − μp (G))(1 − μp (Fr )). Then, using the induction hypothesis μp (G) ≤ 1 − q r−1 and μp (Fr ) ≤ 1 − q, we finally get (1 − μp (G))(1 − μp (Fr )) ≥ q r−1 q = q r .  A family F ⊂ 2X is called intersecting-union if F is intersecting and also union, that is, F ∩ F  = ∅ and F ∪ F  = X for all F, F  ∈ F. Exercise 13.5. Let F ⊂ 2[n] be an intersecting-union family.

80

13. Kleitman’s correlation inequality (1) Let p = 1/2. Show that μp (F) ≤ 1/4. (Hint: Define an intersecting upset F ∗ and a union downset F∗ by F ∗ = {G ⊂ [n] : F ⊂ G for some F ∈ F}, F∗ = {G ⊂ [n] : G ⊂ F for some F ∈ F}, and apply Lemma 13.1 to these families.) (2) Let p = 1/2. Show that lim max μp (F) = min{p, 1 − p},

n→∞

where the maximum is taken over all intersecting-union families F ⊂ 2[n] . (Hint: Let p < 1/2. For the lower bound consider the family {F ⊂ [n] : 1 ∈ F, |F | < n/2}. For the upper bound let Fk be the k-uniform subfamily of F, where k ∈ (p − , p + ) and p +  < 1/2. Then  EKR. Note that the difference be|Fk | ≤ n−1 k−1 by the  tween μp (F) and k∈(p−,p+) μp (Fk ) can be made arbitrarily small by choosing n large enough.) We say that a family F ⊂ 2X is r-wise intersecting-union if F1 ∩ · · ·∩Fr = ∅ and F1 ∪· · ·∪Fr = X for all F1 , . . . , Fr ∈ F. In this case it is known that μp (F) ≤ p(1−p) provided r ≥ 2 and 1/r ≤ p ≤ 1−1/r; see [101]. Let us restate Theorem 4.5 in its original, intersecting form. Define G(n, t) by  if n + t is even, {G ⊂ [n] : |G| ≥ n+t 2 } G(n, t) = (n−1)+t {G ⊂ [n] : |G ∩ [n − 1]| ≥ } if n + t is odd. 2 Exercise 13.6. Show that the family G(n, t) is isomorphic to the complement family of K(n, n − t) defined in (4.2). Exercise 13.7. Verify that G(n, t) is a t-intersecting family. Theorem 13.8. If F ⊂ 2[n] is t-intersecting, then |F| ≤ |G(n, t)|. Exercise 13.9. Deduce the above theorem from Theorem 4.5. Considering G(n, t) as a family in 2[n+1] , it is both t-intersecting and union. Katona conjectured that it is maximal for this property.

13. Kleitman’s correlation inequality

81

The proof of this conjecture became the first published paper of the first author. Let us present here a simpler proof using Lemma 13.1. Theorem 13.10. Let n and t be positive integers with n ≥ t. If F ⊂ 2[n+1] is t-intersecting and 1-union, then (13.2)

|F| ≤ |G(n, t)|.

Proof. In view of Proposition 2.9 we may assume that F is shifted. Let H = F|[n] be the trace of F on [n], that is, H = {F ∩ [n] : F ∈ F}. We claim that H is t-intersecting. To see this, suppose the contrary. Then there are H1 , H2 ∈ H such that |H1 ∩ H2 | < t. Let Fi = Hi  {n + 1}. Since F is t-intersecting, we have F1 , F2 ∈ F and |F1 ∩ F2 | = t. Since F is union, we can find x ∈ F1 ∪ F2 and, in particular, x = n+1. By the shiftedness, F1 = (F1 \{n+1}){x} ∈ F. Then |F1 ∩ F2 | < t, a contradiction. Therefore we may assume that if n + 1 ∈ F ∈ F then F \ {n + 1} ∈ F, because we are interested in the maximum size of F. In particular we may assume that |F| = 2|F|[n] |. We define F ∗ , F∗ ⊂ 2[n+1] as in Exercise 13.5. Then, since H is t-intersecting, F ∗ |[n] is also t-intersecting. Thus it follows from Theorem 13.8 that |F ∗ |[n] | ≤ |G(n, t)| and |F ∗ | ≤ 2|G(n, t)|. On the other hand, F∗ has the union property, implying |F∗ | ≤ 2n . Applying Lemma 13.1 to F ∗ and F∗ with p = 1/2, we get |F ∗ ∩ F∗ | |F ∗ | |F∗ | |G(n, t)| ≤ ≤ . 2n+1 2n+1 2n+1 2n+1 This, together with F ⊂ F ∗ ∩ F∗ , yields |F| ≤ |G(n, t)|.



Examining the proof, it is easy to show that for shifted families and t ≥ 2 the family G(n, t) is the only one to achieve equality in (13.2). In [42] this is proved in general, that is, for non-shifted families as well. This can be done also by showing that Si,j (F) = G(n, t) implies F ∼ = G(n, t) as long as F is both t-intersecting and union. Exercise 13.11. Let c(n, t) denote the maximum of |F||G| where F, G ⊂ 2[n] are cross t-intersecting. Use the above proof to show that if A, B ⊂ 2[n+1] are cross union and cross t-intersecting, then |A||B| ≤ c(n, t).

82

13. Kleitman’s correlation inequality It is known that c(n, t) = |G(n, t)|2

if n + t is even and c(n, t) = max{|G(n, t)|2 , |G(n, t − 1)||G(n, t + 1)|} if n + t is odd; see [91].

Chapter 14

r-Cross union families

Let [n]n, k, and r be positive integers. We say that families F1 , . . . , Fr ⊂ are r-cross intersecting if F1 ∩ · · · ∩ Fr = ∅ for all Fi ∈ Fi , k 1 ≤ i ≤ r. In this chapter we prove the following extension of the Erd˝os–Ko–Rado Theorem.   Theorem 14.1 ([62]). Let (r − 1)n ≥ rk and let F1 , . . . , Fr ⊂ [n] k r  r . be r-cross intersecting families. Then i=1 |Fi | ≤ n−1 k−1   G ∪· · ·∪ Recall that families G1 , . . . , Gr ⊂ [n] l are r-cross union if  1 Gr = [n] for all Gi ∈ Gi , 1 ≤ i ≤ r. Note that F1 , . . . , Fr ⊂ [n] k are rcross intersecting if and only if the complement families F1c , . . . , Frc ⊂ [n] are r-cross union, where n = k + l. So considering r-cross interl secting families is equivalent to considering the corresponding r-union families. The former is more popular, but sometimes the latter is easier to handle in notation. So we will consider the r-union version of Theorem 14.1 in a stronger form. To state the result let us introduce [n] an important invariant, which is a key to xthe proof. For G ⊂ l choose a unique real x ≥ l so that |G| = l , and let Gl := x.   be r-cross Theorem 14.2. Let n ≤ rl and let G1 , . . . , Gr ⊂ [n] l union families. Then we have the following. r (i) i=1 Gi l ≤ r(n − 1). r  r . (ii) i=1 |Gi | ≤ n−1 l 83

84

14. r-Cross union families

Theorem 14.1 follows from (ii) of Theorem 14.2 by considering the complement families. Note that the condition n ≤ rl  in the theorem for n > rl.) cannot be weakened. (Consider, say, r copies of [n] l The proof of Theorem 14.2 is simple and is taken from [62]. In fact a referee of the paper said “The study of cross-intersecting families has a long history . . . It is therefore somewhat shocking that a statement as simple as Theorem 14.1 is being proven for the first time only in 2010. One’s surprise only increases after reading the proof: it uses little more than the ‘circle method’ of Katona, plus the AM-GM inequality and a simple induction on n.” This is a nice summary of the proof except that the induction is actually on rl − n, which is a tricky point of the proof. We also use the Lov´asz version of the Kruskal–Katona Theorem, that is, Theorem 2.14 with Remark 2.17. For convenience we restate it here. Theorem 14.3. Let 0 ≤ p 0. Suppose that (i) is true for the case rl − n  = s, and now we consider the case rl − n = s + 1. be r-cross union families, and let xi = Gi l for Let G1 , . . . , Gr ⊂ [n] l 1 ≤ i ≤ r.   by Here is the tricky part. Define Hi = Gi  Di ⊂ [n+1] l Di = {D  {n + 1} : D ∈ σl−1 (Gi )}.

 xi  , and Then, by Theorem 14.3, we have |Di | = |σl−1 (Gi )| ≥ l−1       xi xi xi + 1 |Hi | = |Gi | + |Di | ≥ , + = l l l−1 that is, zi := Hi l ≥ xi + 1.   are r-cross union families and rl − Since H1 , . . . , Hr ⊂ [n+1] l (n + 1) = s, we can apply the induction hypothesis to these families. Therefore we obtain r  (14.4) zi ≤ r((n + 1) − 1) = rn.

(14.3)

i=1

Finally it follows from (14.3) and (14.4) that This completes the proof of the theorem.

r i=1

xi ≤ r(n − 1). 

Chapter 15

Random walk method

Suppose that we are given a family F of subsets which satisfies some conditions, say, t-intersecting, and we want to bound the size of F. To this end we relate subsets in F with walks in the plane. These walks should satisfy some corresponding conditions, say, hitting a certain line. Then, by counting the number of such walks, we get some information about |F|. This is the core idea of the random walk method. The idea looks simple, but combined with the shifting operation it becomes a strong, far-reaching tool for studying intersecting families. In this chapter we present some simple applications of the method. Notation: For integers a and b we use [a, b]2 to denote the set {a + 2i ∈ Z : 0 ≤ i ≤  b−a 2 }.

15.1. Walks and some basic facts For a subset F ⊂ [n] we define its walk in R2 . This is an n-step walk starting from the origin, and the ith step is up if i ∈ F or right if i ∈ F. More precisely, suppose that after (i−1) steps we are at (x, y), then at the ith step we go from (x, y) to (x, y + 1) if i ∈ F , or we go from (x, y) to (x + 1, y) if i ∈ F . For example, if F = {2, 3} ⊂ [4], then its walk consists of four line segments (each segment has length one) connecting (0, 0), (1, 0), (1, 1), (1, 2), and (2, 2) in this order. We 87

88

15. Random walk method

identify a subset with its walk. For example, we say that F = {2, 3} hits the line y = x + 1 only at (1, 2). Exercise 15.1. Let x0 , y0 , and t be nonnegative integers with t ≤ y0 ≤ x0 + t. Show that the number x0 +yof0 walks from (0, 0) to (x0 , y0 ) which hit the line y = x + t is y0 −t . (Hint: Use the reflection principle. Find a bijection between the set of those walks and the set of walks from (−t, t) to (x0 , y0 ).) Exercise 15.2. Let F ⊂ [n]. Suppose that F hits the line y = x + t. Show that |F ∩ [i]| ≥ i+t 2 for some i. (Hint: Suppose that F hits the line at (x0 , y0 ) and put i = x0 + y0 .) We introduce some definitions and notation. For A ⊂ [n], let (A)i be the ith smallest element of A. For A, B ⊂ [n], we say that A shifts to B, denoted by A  B, if |A| ≤ |B| and (A)i ≥ (B)i for each i ≤ |A|. In other words, the walk of A is in the lower right area with respect to the walk of B, that is, in the area bounded by the walk of B and the lines y = 0 and x + y = n. For example, {2, 4, 6}  {1, 4, 5, 7}. We say that G ∈ F is the shift-end in F if F  G for all F ∈ F. Exercise 15.3. Let F be a shifted family in 2[n] . Show that F ∈ F and F  F  imply F  ∈ F provided |F | = |F  | or F is an upset. Claim 15.4. Let F ⊂ 2[n] be a shifted t-intersecting family and let F ∈ F. Then F hits the line y = x + t. Proof. Suppose to the contrary that F ∈ F does not hit the line. Let |F | = k. Define F  ⊂ [n] with |F  | = k by F  = [t − 1]  {t + 1, t + 3, . . .} = [t − 1]  [t + 1, 2k − t + 1]2 . Then F  is the shift-end in the k-uniform family whose walks do not hit the line. Since F is shifted, we have F  F  and F  ∈ F. Let F  = [t − 1]  {t, t + 2, . . .} = [t − 1]  [t, 2k − t]2 . Then F   F  and F  ∈ F. But |F  ∩F  | = t−1, a contradiction.



Walks in a shifted t-intersecting family hit the line y = x + t, and we can bound the size by counting such walks as follows.

15.1. Walks and some basic facts Theorem  n 15.5. Let n ≥ 2k − t. If F ⊂ . |F| ≤ k−t

89 [n] k

is t-intersecting, then

Proof. We may assume that F is shifted. Then by Claim 15.4 all F ∈ F hit the line y = x + t. Apply Exercise 15.1 with x0 = n− k and n .  y0 = k, to get that the number of such walks is at most k−t  n  is not sharp, but it is not too bad and we get it The bound k−t easily. This already shows the potential of the random walk method, and actually we can go further. If n ≥ (t + 1)(k − t + 1) then n−t  the exact upper bound for the size of t-intersecting families is k−t , which we call the Full Erd˝os–Ko–Rado Theorem (Full EKR). The first serious application of the random walk method, due to Frankl [45], was to obtain this sharp bound for the case t ≥ 15. In the next section we refine the original proof to get the corresponding result in the measure version. Our proof works for t ≥ 6, and one can get the corresponding Full EKR in a similar way; c.f. [58]. We recall some basic facts to deal with the measure version. Let p be a real number with 0 < p < 1, and let q = 1 − p. The product measure of a family F ⊂ 2[n] is defined by  μp (F) = p|F | q n−|F | . F ∈F

Exercise 15.6. Let p < 1/2. Consider an infinite random walk in the plane starting at the origin, where the ith step is up with probability p and right with probability q. For a positive integer t let P (t) denote the probability that the walk hits the line y = x+t. Show that P (t) = (p/q)t . (Hint: After the first step we are at (0, 1) with probability p, or at (1, 0) with probability q. This gives P (t) = pP (t−1)+qP (t+1). Assume that P (t) is given in the form xt and solve the equation.) Theorem 15.5 has the following corresponding measure version. Theorem 15.7. Let p < 1/2. If F ⊂ 2[n] is t-intersecting, then μp (F) < (p/q)t . Proof. We may assume that F is shifted. Then the walks in F hit the line y = x + t. Thus μp (F) is at most the product measure of the set of n-step walks hitting the line. The latter is precisely the

90

15. Random walk method

probability that the random walk defined in Exercise 15.6 hits the  line in the first n steps, and it is less than (p/q)t . 1 This bound (p/q)t is not sharp. In fact, if p ≤ t+1 then pt is the exact upper bound for μp (F), and this is a special case of Theorem 12.4. There are several ways to prove the result; for example, Friedgut [68] gave a spectral proof; see also Chapter 30. In the next section we present a proof using the random walk method. For the proof we gather some useful facts about intersecting families.

Lemma 15.8. Let F ⊂ 2[n] be a shifted t-intersecting family. For every F, G ∈ F there is some i such that |F ∩ [i]| + |G ∩ [i]| ≥ i + t. Proof. Suppose the contrary, and choose a pair of counterexamples F, G ∈ F so that |F ∩ G| is minimal. Let j be the tth element of F ∩ G. Then |F ∩ [j]| + |G ∩ [j]| < j + t = j + |F ∩ G ∩ [j]|, and |(F ∪ G) ∩ [j]| < j. Thus we can find i ∈ [j] \ (F ∪ G). By the shiftedness, G = (G \ {j})  {i} ∈ F, and F and G are also counterexamples because |G ∩ [j]| = |G ∩ [j]|. But |F ∩ G | < |F ∩ G|, contradicting the minimality of the choice.  Exercise 15.9. Let fn = max{μp (F) : F ⊂ 2[n] is t-intersecting}. Show that fn ≤ fn+1 . (Hint: If F ⊂ 2[n] is t-intersecting, then F  = F  {F  {n + 1} : F ∈ F} ⊂ 2[n+1] is t-intersecting as well.) Exercise 15.10. Let G, H ⊂ 2[n] . Show that if G is a shifted upset and H ∈ H \ G is the shift-end in H, then G ∩ H = ∅. In the next section we will bound the measure of a t-intersecting shifted upset G ⊂ 2[n] . For this we use the following simple fact. Let A, W, W  ⊂ [n], and suppose that W ∈ G and W  ∈ G. Then A ∈ G if |A ∩ W | < t or A  W  . If G satisfies some additional conditions, then by choosing W and W  appropriately we get a good estimation for the measure of G, as we will see. In Chapter 12 we mentioned the relation between the uniform   and the product measure measure μk of t-intersecting families in [n] k μp of those in 2[n] , where p = k/n. For walks hitting the line y = x+t we also have the correspondence. That is, the family of walks W ⊂

15.2. The measure of t-intersecting families

91

[n]

 n  n hitting the line has the uniform measure μk (W) = k−t / k , while the corresponding family of walks W  ⊂ 2[n] has the product  t measure  nμp (W ) αt+I ≥ μp (G \ F1t ). Proof. To show the first inequality, let H be the set of walks H satisfying the following conditions: (i) H hits all of (1, t − 1), (1, t + 1), and (I + 2, t + 1); (ii) after t + I + 3 steps it does not hit the line L : y = x + t − I. We note that H ⊂ F1t \ G, which follows from (iii) H ⊂ F1t and (iv) H ∩ G = ∅. Indeed (iii) follows from (i), and (iv) follows from the fact that WI+1 ∈ H \ G is the shift-end in H; see Exercise 15.10. Now we estimate μp (H). For (i) there are t ways from (0, 0) to (1, t − 1), and part (i) (the first I + t + 3 steps) consists of t + 1 up steps and I + 2 right steps. So (i) contributes tpt+1 q I+2 to μp (H). The probability that the random walk starting from (I + 2, t + 1) hits the line L is at most α, and (ii) contributes at least 1 − α to μp (H). Thus we get μp (F1t \ G) ≥ μp (H) ≥ tpt+1 q I+2 (1 − α). To show the last inequality, let E = [t]  [t + 3, t + I + 3]  [t + I + 5, n]2 . Since WI ∈ G and WI ∩E = [t−1], we have E ∈ G. If G ∈ G\F1t then G hits neither (0, t + 1) nor (1, t). Thus G must hit the line y = x + t + I. (Otherwise G  E and E ∈ G, a contradiction.) This means that all walks G \ F1t hit y = x + t + I, and μp (G \ F1t ) ≤ αt+I . Finally we show the middle inequality, that is, tpq t+2 (1 − α)(q 2 /p)I > 1.

94

15. Random walk method

Since q 2 /p > 1 for t ≥ 2, it suffices to check the inequality at I = 1,  that is, tq t+1 (1 − α) > 1, and this is true for t ≥ 6. By the claim we have μp (F1t \G) > μp (G\F1t ), that is, μp (G) < μp (F1t ). Next let s = 0. We can argue similarly to the previous case, and the present case is even easier. This time let Wi = [t]  [t + i + 2, n]2 . If Wimax = [t] ∈ G then clearly G ⊂ F0t . Otherwise we can define I := max{i : Wi ∈ G}. Then one can show that μp (F0t \ G) ≥ pt q I+1 (1 − α) > αt+I ≥ μp (G \ F0t ) for t ≥ 5. This yields μp (G) < μp (F0t ). Consequently we have shown that if G is shifted and t ≥ 8, then one of the following holds: (a) μp (G) < pt , (b) G ⊂ F0t , or (c) G ⊂ F1t . Moreover, in view of Exercise 15.11, even in the cases (b) and (c) we have μp (G) ≤ pt , and equality holds only if G = F0t or p = 1/(t + 1) and G = F1t . Finally we use Exercise 10.10, which tells us that if we start from G and obtain Fst by shifting, then G is isomorphic to Fst . This completes the proof.  Exercise 15.15. Improve (15.1) so that μp (G) < pt for t ≥ 5 and s ≥ 2 in the following way. (i) Let s ≥ 3. Prove that μp (G) < pt for t ≥ 5 by showing that  t+6   t + 6 j t+6−j t ˜ + αt+1 < pt . μp (F3 ) + μp (W) ≤ p q j j=t+3 (ii) Let s = 2. First show that μp (G) ≤ μp (F2t \ G) + μp (G) ≤ μp (F2t \ W) + μp (W) (t + 4)(t + 1) t+2 2 p q (1 − α + ) + αt+1 , 2 where  → 0 as n → ∞. Next show that the RHS is < pt for t ≥ 4 (assuming  > 0 is small enough). Finally use Exercise 15.9 to conclude that μp (G) < pt for all n ≥ t ≥ 4. ≤

The random walk method is applicable to cross t-intersecting families, r-wise t-intersecting families, and so on, see [58, 87] and references therein.

Chapter 16

L-systems

In terms of L-systems one can describe many important objects in combinatorics, including t-intersecting families, Steiner systems, projective planes over finite fields, and so on. In this chapter we introduce L-systems and some invariants which characterize the largest size of them. We also present some fundamental general results. We will discuss their proofs and some other interesting special cases of L-systems in later chapters.

16.1. Exponent of (n, k, L)-systems Let n and k be positive integers with n ≥ k, and let L ⊂ [0, k − 1] :=   {0, 1, . . . , k − 1}. We say that a family of k-element subsets F ⊂ [n] k is an (n, k, L)-system if |F ∩ F  | ∈ L holds for all distinct F, F  ∈ F. It is also called an L-system or a (k, L)-system for short. Let m(n, k, L) denote the maximum size of (n, k, L)-systems. In general it is very difficult to determine m(n, k, L) exactly. For example, to determine m(157, 13, {0, 1}) we need to settle the existence problem of a projective plane of order 12. So instead of the exact maximum size of L-systems we mainly deal with the exponent α(k, L) defined below. If there exist positive real numbers 95

96

16. L-systems

α, c, and c depending only on k and L such that cnα < m(n, k, L) < c nα , then we define α(k, L) = α, and we say that the (k, L)-systems have exponent α. It is conjectured that exponents exist for every k and L, but this is also a difficult open problem. From now on we will only consider pairs (k, L) such that the corresponding exponents exist (and if the conjecture is true, then this is not a restriction). Frankl showed in [49] that for every rational number q ≥ 1 there are infinitely many choices of k and L such that α(k, L) = q. We will discuss the main idea of the proof of this result in Chapter 17. On the other hand, no irrational exponent is known. Now let us take a brief look at one of the most interesting (n, k, L)systems. Let t < k and let L = [0, t − 1]. If F is one of the (n, k, L)  there is at most one F ∈ F such systems, then for every T ∈ [n] t that T ⊂ F . (In fact if T is contained in two distinct F, F  ∈ F, then |F ∩ F  | ≥ |T | ≥ t and |F ∩ F  | ∈ L, a contradiction.) Counting t-element subsets contained  F, in other words counting  in edges of the t-shadow of F, we get nt ≥ |F| kt and     n k (16.1) m(n, k, [0, t − 1]) ≤ / . t t   We say that F ⊂ [n] is a Steiner system S(t, k, n) if for every T ⊂ k [n] there is a unique F ∈ F such that T ⊂ F . Thus S(t, k, n) is an t     (n, k, [0, t − 1])-system of size exactly nt / kt . So if S(t, k, n) exists, then equality holds in (16.1). This is one of the rare cases where we know the exact value of m(n, k, L). On the other hand, if n is large enough for fixed t and k, then the bound in (16.1) is always almost tight. Indeed, R¨ old proved in [99] that     n k m(n, k, [0, t − 1]) = (1 − o(1)) / . t t We will return to this deep result and its application in Chapter 20. Next we present a general upper bound for m(n, k, L) due to Deza, Erd˝ os, and Frankl from [22].

16.2. Rank of (k, L)-systems

97

Theorem 16.1. For n > n0 (k, L), m(n, k, L) ≤

 n−l . k−l

l∈L

This upper bound is tight in some cases. For example,     if L = [0, t − 1] then the RHS of the above inequality is nt / kt . So if moreover an S(t, k, n) exists, then the upper bound coincides with m(n, k, L) itself. By Theorem 16.1 we have α(k, L) ≤ |L|, that is, the exponent cannot exceed the size of L. Actually, in most cases, exponents can be much smaller. We can improve the upper bound for exponents of L-systems by introducing a combinatorial invariant called the rank, which we will discuss in the next section. We will prove Theorem 16.1 in Chapter 18.

16.2. Rank of (k, L)-systems In this section we define the rank of an L-system and show that it gives an upper bound for the exponent. Let k ∈ Z>0 and L ⊂ {0, 1, . . . , k − 1} be given. A non-empty family I ⊂ 2[k] is called a closed L-system if I ∩ I  ∈ I and |I ∩ I  | ∈ L for all (not necessarily distinct) I, I  ∈ I. The rank of I is defined by rank(I) = min{r ∈ Z>0 : σr (I) =

[k] r },

where σr denotes the r-shadow. By definition 1 ≤ rank(I) ≤ k, and if r = rank(I) then every (r − 1)-element subset is contained in some I ∈ I, but there exists an r-element subset R such that no I ∈ I contains R. The rank of a (k, L)-system is defined by rank(k, L) = max{rank(I) : I ⊂ 2[k] is a closed L-system}. We say that I ⊂ 2[k] is an intersection structure of a (k, L)-system if I is a closed L-system. If moreover I satisfies rank(I) = rank(k, L), then it is called a proper intersection structure. Some (k, L)-systems

98

16. L-systems

have several non-isomorphic proper intersection structures1 . A generator set I ∗ of I is the collection of all maximal elements of I, that is, I ∗ = {I ∈ I :  ∃I  ∈ I such that I ⊂ I  , I = I  }. We can retrieve I from I ∗ by taking all possible intersections. To relate a (k, L)-system and its intersection structure we need   and an edge F ∈ F, some more preparation. For a family F ⊂ [n] k define the restriction of F on F by F|F = {F ∩ F  : F  ∈ F \ {F }} ⊂ 2F . Moreover, if F is k-partite with k-partition [n] = X1  · · ·  Xk , that is, if |F ∩ Xi | = 1 for all F ∈ F and 1 ≤ i ≤ k, then we define the ˜ → [k], where projection π : X (16.2)

˜ := {G ⊂ [n] : |G ∩ Xi | ≤ 1 for all i}, X

by π(G) := {i : |G ∩ Xi | = 1} and write π(F|F ) for {π(G) : G ∈ F|F }. F¨ uredi [69] (see also [71]) proved the following fundamental result, which was conjectured by Frankl. It roughly says that every (n, k, L)-system F contains a large “homogeneous” k-partite subfamily F ∗ in the sense that every projection of an edge of F ∗ determines the same closed L-system. Theorem 16.2 (F¨ uredi’s structure theorem). For given k ≥ 2 and L ⊂ {0, 1, . . . , k − 1}, there exists a positive constant c = c(k, L) such that every (n, k, L)-system F contains a k-partite subfamily F ∗ ⊂ F with k-partition [n] = X1  · · ·  Xk satisfying (i)–(iii). (i) |F ∗ | > c|F|. (ii) If F1 , F2 ∈ F ∗ , then π(F ∗ |F1 ) = π(F ∗ |F2 ). IS(F ∗ ) for this common family in 2[k] .

We write

(iii) IS(F ∗ ) is a closed L-system. 1 For example, the (9, {0, 1, 3, 4})-system has exactly three non-isomorphic proper intersection structures. It is known that α(0, {0, 1, 3, 4}) = 3, and this exponent comes from only one of the proper intersection structures (the Stiener system S(2, 3, 9)); see [60] for more details.

16.2. Rank of (k, L)-systems

99

In the above situation, we say that IS(F ∗ ) is the intersection structure 2 of F ∗ . We also say that F ∗ is a canonical (n, k, L)-system or a canonical family. It is an immediate consequence of Theorem 16.2 that the rank is an upper bound for the exponent, as stated in the next theorem. Theorem 16.3. Let k and L ⊂ {0, 1, . . . , k − 1} be given. If the exponent α = α(k, L) exists, then α(k, L) ≤ rank(k, L). Proof. Let F be an (n, k, L)-system with |F| = Θ(nα ). Choose a canonical family F ∗ with k-partition X1 · · ·Xk from Theorem 16.2. Let I = IS(F ∗ ) ⊂ 2[k] be the intersection structure of F ∗ and let   r = rank(I). By the definition of rank there is an R ∈ [k] r such that R ∈ σr (I). Then R ⊂ I for all I ∈ I, which is a key ingredient of ˜ where X ˜ is defined by (16.2), with the proof. Thus for every G ∈ X, π(G) = R (so |G∩Xi | = 1 for i ∈ R) there is at most one F ∈ F ∗ such that G ⊂ F . Otherwise there would be distinct F, F  ∈ F ∗ such that G ⊂ F ∩ F  , and then R = π(G) ⊂ π(F ∩ F  ) ∈ I, a contradiction. Thus the size |F ∗ | is at most the number of choices for G projected to R, and  |Vi | = O(nr ). |F ∗ | ≤ i∈R

Then (i) of Theorem 16.2 yields |F| = O(nr ), and α(k, L) ≤ r.



Remark 16.4. By the proof of Theorem 16.3 we see that if F ∗ is a canonical (n, k, L)-system with intersection structure of rank r, then |F ∗ | = O(nr ). On the other hand, F¨ uredi conjectures3 the following in [71]: Conjecture 16.5. α(k, L) > rank(k, L) − 1. If true, then this conjecture would give a strong lower bound. We list some supporting results here. The conjecture is true if rank(k, L) = 2 (cf. [71]), as we will see shortly. It is also true if k ≤ 12 for all L; see [60, 109]. If there exists a Steiner system 2 This is not necessarily one of the proper intersection structures of the (k, L)system. We only have that rank(IS(F ∗ )) ≤ rank(k, L). 3 He writes in [71] that this is “perhaps too optimistic.”

100

16. L-systems

S(t, b, k), then we have rank(k, L) = t + 1 for L = [0, t − 1] ∪ {b}. R¨odl and Tengan [100] found a construction which shows that α(k, L) > t in this situation, and we will present the construction in Chapter 20. On the other hand, nobody has succeeded in finding a general lower bound (of any kind) for α(k, L) in terms of rank(k, L) yet. Theorem 16.6. If rank(k, L) = 2 then α(k, L) ≥ 1 +

1 . k−1

Proof. Let I ⊂ 2[k] be a closed L-system with rank(I) = 2. Using I we will construct an (n, k, L)-system F of size Ω(nk/(k−1) ). Choose I1 , . . . , Ig ∈ I with g ≤ k such that I1 ∪ · · · ∪ Ig = [k]. For given g−1 ≤ n < k(m + 1)g−1 . Let n choose an integer m such that km  V |V | = n and define a family F ⊂ k of size mg by F = {F (t1 , . . . , tg ) : (t1 , . . . , tg ) ∈ [m]g }, where F (t1 , . . . , tg ) = {(j, xj1 , xj2 , . . . , xjg ) : 1 ≤ j ≤ k},  0 if j ∈ Ii , j xi = ti otherwise. This family F is k-partite with partition V = V1  · · ·  Vk where Vj = {(j, xj1 , xj2 , . . . , xjg ) : xji = 0 if j ∈ Ii , xji ∈ [m] if j ∈ Ii }. A key observation is that F sits on (at most) kmg−1 vertices instead  of kmg vertices. The reason is as follows. Since Ii = [k], for each j ∈ [k], there is some i ∈ [g] such that j ∈ Ii . Then xji = 0 and each vertex in Vj can be written as (j, xj1 , . . . , xji−1 , 0, xji+1 , . . . , xjg ). g−1 So we only need at ≤ n vertices in total, and we have V most km verified that F ⊂ k with |V | ≤ n. Now using (n/k)1/(g−1) < m + 1 and g ≤ k we get  g |F| = mg > (n/k)1/(g−1) − 1 = Ω(ng/(g−1) ) ≥ Ω(nk/(k−1) ).

16.2. Rank of (k, L)-systems

101

Finally we verify that F is an L-system. Choose two arbitrary distinct edges F, F  ∈ F with F = F (t1 , . . . , tg ) = {(j, xj1 , . . . , xjg ) : 1 ≤ j ≤ k}, F  = F (t1 , . . . , tg ) = {(j, x 1 , . . . , x g ) : 1 ≤ j ≤ k}. j

j

We need to show that |F ∩ F  | ∈ L. Let Λ = {i : ti = ti }. Since F = F  , the index set Λ is non-empty. We note that the following j four statements are equivalent: (i) F and F  meet on Vj , (ii) xji = x i  for all i, (iii) i ∈ Λ implies j ∈ Ii , and (iv) j ∈ i∈Λ Ii . Therefore we have    π F ∩ F  = i∈Λ Ii , where π is the projection. Since I is a closed L-system, it follows that   i∈Λ Ii ∈ I and | i∈Λ Ii | ∈ L. Thus we get  |F ∩ F  | = |π(F ∩ F  )| = | i∈Λ Ii | ∈ L, which completes the proof.



Chapter 17

Exponent of a (10, {0, 1, 3, 6})-system

Frankl proved that every rational q ≥ 1 occurs as the exponent of some (k, L)-system. In this chapter we consider the case q = 5/2 and prove that α(10, {0, 1, 3, 6}) = 5/2. This concrete example will illustrate some main ideas used in the proof for the general case. Where do the numbers 0, 1, 3, 6 and 10 come from? They come  from the sequence 2l , l = 1, 2, . . . , 5, and this is a key to the construction showing that α(10, L) ≥ 5/2, where L = {0, 1, 3, 6}. Let p be a positive integer. We construct a 10-uniform family F on the  [p] vertex set V = 2 by  

  [p] A . :A∈ F= 5 2     For any distinct F, F  ∈ F, where F = A2 and F  = A2 with   |A∩A |  A, A ∈ [p] |,which is one of 0, 1, 3, and 5 , we have |F ∩ F | = | 2  6. So F is a (10, L)-system on n := |V | = p2 vertices of size   √ p 2 5/2 |F| = ∼ n = Θ(n5/2 ). 5 30 This gives the desired lower bound for the exponent. 103

104

17. Exponent of a (10, {0, 1, 3, 6})-system

For later purposes it is useful to find a 10-partite L-system. By deleting some edges from F we can obtain a 10-partite subfamily F  10! of size at least 10 10 |F| (see Lemma 19.1 in Chapter 19). One can, however, get a larger 10-partite (10, L)-system G with intersection structure   [5] (17.1) I = { A2 : A ⊂ [5], A = [5]} ⊂ 2( 2 )   as follows. Note that we consider I on the vertex set [5] 2 instead of on [10]. So the vertex set V of F is partitioned into 10 classes as (17.2)

V = V{1,2}  V{1,3}  · · ·  V{4,5} .

Let P = P1  · · ·  P5 with |Pi | = n, and let V{i,j} = Pi × Pj (so N := |V | = 10n2 ). Then define   G := { {x1 ,x2 ,x23 ,x4 ,x5 } : xi ∈ Pi (i = 1, 2, . . . , 5)}, which is a 10-partite (10, L)-system of size |G| = n5 = (N/10)5/2 = Θ(N 5/2 ). Now let us verify that the intersection structureof G coincides by π(x) = with I. To this end we define a projection π : V → [5] 2 {i, j} if x ∈ V{i,j} . For G ∈ G we extend the definition and let   π(G) = {π(x) : x ∈ G}. For example, if G = {x1 ,x2 ,x23 ,x4 ,x5 } and   G = {x1 ,y2 ,x23 ,x4 ,y5 } are two members of G with x2 = y2 and x5 = y5 ,     . In general, if then G ∩ G = {x1 ,x23 ,x4 } and π(G ∩ G ) = {1,3,4} 2    G, G ∈ G satisfy G ∩ G = B2 for some B ⊂ i∈A Pi with |B| = |A|,   then π(G ∩ G ) = A 2 . This means that I is an intersection structure of G. The above construction is the largest one among canonical families. Indeed we have the following. Theorem 17.1. Let H be a 10-partite canonical (10, {0, 1, 3, 6})system on the vertex set (17.2) with |V{i,j} | = n2 for all i, j; then |H| ≤ n5 . Proof. It follows from computer search that I, defined by (17.1), is the unique closed (10, L)-system of rank 3 for L = {0, 1, 3, 6}. So we may assume that I is the intersection structure for H. Otherwise H would have an intersection structure of rank at most 2, and it follows

17. Exponent of a (10, {0, 1, 3, 6})-system

105

from Remark 16.4 that |H| = O(N 2 ) = O(n4 ). We also assume that the vertex partition (17.2) is consistent with I; in other words, we assume the following. Claim 17.2. For every pair of distinct H, H  ∈ H there is an A ⊂ [5]  A such that π(H ∩ H  ) = 2 . For A ⊂ [5] let H|A denote the restriction of H to A, that is,  H|A = {H ∩ VJ : H ∈ H}. J∈(A 2) For example, by H|[4] we mean that we look at H only on the vertex set V{1,2}  V{1,3}  V{1,4}  V{2,3}  V{2,4}  V{3,4} . Claim 17.3. |H|[4] | ≤ n4 . Proof. If K, K  ∈ H|[4] meet on both V{1,2} and V{3,4} , then |K ∩   K  | ≥ 2 and so |K ∩ K  | = 3 or 6. But there is no A ∈ [5] 3 containing both {1, 2} and {3, 4}, so it follows from Claim 17.2 that |K ∩K  | = 6, that is, K = K  . Thus we have |H|[4] | ≤ |V{1,2} ||V{3,4} | = n4 .  Let q = |H|[3] | and H|[3] = {L1 , L2 , . . . , Lq }. For each 1 ≤ i ≤ q let H|[4] (Li ) = {K ∈ H|[4] : Li ⊂ K} and ai = |H|[4] (Li )|. Since each K ∈ H|[4] uniquely determines Li ∈  H|[3] such that Li ⊂ K, we get a partition H|[4] = qi=1 H|[4] (Li ), and q |H|[4] | = i=1 ai . Thus, by Claim 17.3, we have a1 + a2 + · · · + aq ≤ n4 . By the symmetry of I, the same holds not only for H|[4] but also for any H|C with |C| = 4, say C = {1, 2, 3, 5}. So let bi be the number of subsets in H|{1,2,3,5} that contain Li . Then it follows that b1 + b2 + · · · + bq ≤ n4 . Thus we have (17.3)

q q    2 ai bi ≤ (ai + bi ) ≤ 2n4 . i=1

i=1

17. Exponent of a (10, {0, 1, 3, 6})-system

106

Let H(Li ) be the family of subsets in H containing Li . Claim 17.4. |H(Li )| ≤ min{ai bi , n2 }. Proof. In fact, ai counts the number of H ∈ H containing Li on V{1,4}  V{2,4}  V{3,4} , while bi counts the number on V{1,5}  V{2,5}  V{3,5} . Thus ai bi is an upper bound for |H(Li )|. On the other hand, for every v ∈ V{4,5} there is at most one H ∈ H such that Li ∪ {v} ⊂  H. Thus |H(Li )| ≤ |V{4,5} | = n2 .  Since |H| ≤ qi=1 |H(Li )| we have |H| ≤

(17.4)

q 

min{ai bi , n2 }.

i=1

Now we deduce |H| ≤ n from (17.3) and (17.4). Let ci = (17.3) we have 5

q 

(17.5)

√ ai bi . By

ci ≤ n 4 .

i=1

For each 1 ≤ i ≤ q, if ci > n then we replace ci with n; in other words we assume that ci ≤ n. After the replacement (17.5) is still valid, and  (17.4) reads |H| ≤ qi=1 c2i . Consequently we have |H| ≤

q  i=1

c2i ≤

q  i=1

nci = n

q 

ci ≤ n 5 .

i=1

This completes the proof of Theorem 17.1.



We have seen that α(10, {0, 1, 3, 6}) = 5/2. Actually, this construction and the proof for the upper bound come from a more general result due to Frankl, which we explain briefly here. As for the construction, for any integers s ≥ d ≥ 1 one can construct an (n, k, L)-system of size Ω(ns/d ) as follows. Exercise 17.5. Let s, d, a0 , a1 , . . . , ad be nonnegative integers with s ≥ d ≥ 1 and ad ≥ 1. Define an integer-valued polynomial p(x) of    degree d by p(x) = di=0 ai xi . Let b ≥ s and let Z be a finite set   consisting of ai disjoint copies of [b] i for i = 0, 1, . . . , d. So |Z| = p(b).   For A ⊂ [b] let φ(A) ⊂ Z be a subset consisting of ai copies of Ai for i = 0, 1, . . . , d, Note that |φ(A)| = p(|A|).

17. Exponent of a (10, {0, 1, 3, 6})-system

107

(i) Show that if A, A ⊂ [b] then φ(A ∩ A ) = φ(A) ∩ φ(A ).   (ii) Show that {φ(A) : A ∈ [b] s } is a (p(b), p(s), L)-system of  b size s , where L = {p(0), p(1), . . . , p(s − 1)}. (iii) Show that m(n, p(s), L) = Ω(ns/d ) by setting b = Θ(n1/d ). The proof for the upper bound is more difficult and we refer the reader to [49]. Using a certain monotonicity lemma one can show that under some additional conditions on a1 and p(s − 1), the above construction is best possible up to a constant factor.

Chapter 18

The Deza–Erd˝ os–Frankl Theorem

In this chapter we prove the following result, which clearly implies the Deza–Erd˝ os–Frankl Theorem (Theorem 16.1). Theorem 18.1. Let n ≥ 2k k3 , and let F be an (n, k, L)-system, where L = {l1 , l2 , . . . , ls } with 0 ≤ l1 < l2 < · · · < ls < k. Then we have the following.  n−l . (i) |F| ≤ k−l l∈L    (ii) If |F| ≥ 2k k2 ns−1 then | F| ≥ l1 , where F := F ∈F F . (iii) If s ≥ 2 and |F| ≥ 2k k2 ns−1 , then (l2 − l1 )|(l3 − l2 )| · · · |(ls − ls−1 )|(k − ls ), where a|b means that a divides b. The original results obtained by Deza, Erd˝ os, and Frankl are slightly stronger than the above version; that is, they proved (ii) and (iii) under a weaker assumption, |F| ≥ 2s−1 k2 ns−1 . Definition. For a family G ⊂ 2[n] and a positive integer t ≥ 3, we say that {G1 , G2 , . . . , Gt } ⊂ G form a t-sunflower with center A in G if Gi ∩ Gj = A all 1 ≤ i < j ≤ t. 109

110

18. The Deza–Erd˝ os–Frankl Theorem

Proof of Theorem 18.1. Let us apply induction on |L|. If L = ∅, then we have |F| = 1 and (i) holds. Let |L| = s ≥ 1. We distinguish two cases: 0 ∈ L and 0 ∈ L. Case I. 0 ∈ L. Let x ∈ [n]. Then F(x) := {F \ {x} : x ∈ F ∈ F} is an (n − 1, k − 1, L )-system, where L = {l − 1 : l ∈ L \ {0}}. By the induction hypothesis we have  n−l  (n − 1) − l . = |F(x)| ≤  (k − 1) − l k−l l ∈L l∈L\{0}  Using x∈[n] |F(x)| = k|F|, we get  n−l 1  n  n−l = , |F| = |F(x)| ≤ k k k−l k−l x∈[n]

l∈L\{0}

l∈L

which proves (i) in this case. (If you are only interested in the proof of (i), then you can go to Case II.) Since l1 = 0 in this case, (ii) holds automatically. Now we show (iii). Let s ≥ 2 and suppose that |F| ≥ 2k k2 ns−1 . Let d = 2k−1 k2 ns−2 and F 0 = F. We will define a sequence of families F 0 , F 1 , F 2 , . . . as follows. Suppose that F i is defined. If there is an x ∈ [n] such that 0 < degF i (x) ≤ d, then let F i+1 := F i \ {F ∈ F i : x ∈ F }. Otherwise let F  = F i and we stop. After at most n steps we get F  . Then degF  (x) = 0 or > d for each x ∈ [n]. At each step we deleted at most d edges. Therefore we deleted at most nd = 2k−1 k2 ns−1 edges in total. Thus we have (18.1)

|F  | ≥ |F| − nd ≥ 2k−1 k2 ns−1 .

Let X  = {x ∈ [n] : degF  (x) > d}. For each x ∈ X  let F  (x) := {F \ {x} : x ∈ F ∈ F  }, which is an (n − 1, k − 1, {l2 − 1, . . . , ls − 1})-system with |F  (x)| > d. Since n − 1 ≥ 2k k3 − 1 ≥ 2k−1 (k − 1)3 and (18.2)

|F  (x)| = degF  (x) > d > 2k−1 (k − 1)2 (n − 1)s−2 ,

18. The Deza–Erd˝ os–Frankl Theorem

111

we can apply the induction hypothesis for (ii) to F  (x). Then it   follows that | F  (x)| ≥ l2 − 1. So there exists Dx ⊂ F  (x) with |Dx | = l2 − 1. For each x ∈ X  let Dx = Dx ∪ {x}. Claim 18.2. For every y ∈ Dx we have Dx = Dy . Proof. Let Fx ∈ F  (x). Then, by the definition of Dx , Dx ⊂ Fx . Let y ∈ Dx and Gy := (Fx ∪ {x}) \ {y}. Since Gy ∪ {y} = Fx ∪ {x} ∈ F we have Gy ∈ F  (y) and Dy ⊂ Gy . Thus it follows that Dx ∪ Dy ⊂ Fx ∪ Gy = Fx ∪ {x}, and (18.3)

Fx ⊃ Ex := Dx ∪ (Dy \ {x}).

Since (18.3) is true for all Fx ∈ F  (x), F  (x) is |Ex |-intersecting. Notice that |Ex | ≥ |Dx | = l2 − 1. If |Ex | ≥ l2 , then F  (x) is l2 intersecting, and l2 −1 does not appear as the size of pairwise intersections in F  (x). Thus F  (x) is an (n−1, k−1, {l3 −1, l4 −1, . . . , ls −1})system. Then, by the induction hypothesis, we can conclude that s s−2 i < d, which is a contradiction. Con|F  (x)| ≤ i=3 n−l k−li ≤ n sequently |Ex | = l2 − 1 must hold. Thus we have Dx ⊃ Dy \ {x}. Since |Dx | = |Dy |, x ∈ Dx , and y ∈ Dy , this is possible only when  Dy = (Dx \ {y}) ∪ {x}, that is, Dx = Dy . By the claim, for x, y ∈ X  we have either Dx = Dy (if y ∈ Dx ) or Dx ∩ Dy = ∅ (if y ∈ Dx ). Since for every x ∈ X  there exists Dx that contains x, we get a partition X  = Dx 1  Dx 2  · · · . So each F ∈ F  is a disjoint union of some Dx . Indeed, if x ∈ F ∈ F  then Dx ⊂ F . This shows that l2 |k, and if s = 2 then this concludes the proof of (iii) (assuming l1 = 0). Now let s ≥ 3 (and so k ≥ 3). Claim 18.3. For every 3 ≤ i ≤ s, there are F, G ∈ F  such that |F ∩ G| = li . Proof. Suppose, to the contrary, that li does not appear as the size of pairwise intersections in F  . Then F  is an (n, k, L \ {li })-system.  s−1 i , By the induction hypothesis, we have |F  | ≤ l∈L\{li } n−l k−li ≤ n which contradicts (18.1). 

112

18. The Deza–Erd˝ os–Frankl Theorem

If F, G ∈ F  , then each of F and G is a disjoint union of some and F ∩ G is also a disjoint union of some Dx , where |Dx | = l2 . So the above claim implies that l2 |li for 3 ≤ i ≤ s. In particular, we have

Dx ,

(18.4)

l2 = (l2 − l1 )|(l3 − l2 ).

On the other hand, F  (x) is an (n−1, k −1, {l2 −1, . . . , ls −1})-system satisfying (18.2). Applying the induction hypothesis, we get ((l3 − 1) − (l2 − 1))|((l4 − 1) − (l3 − 1))| · · · |((k − 1) − (ls − 1)), that is, (l3 − l2 )|(l4 − l3 )| · · · |(k − ls ). This, together with (18.4), gives the desired property, which completes the proof of (iii) in the case 0 ∈ L. Case II. 0 ∈ L.

  Fix H ∈ F and let A = H l1 . Then |H ∩ F | ≥ l1 for every F ∈ F, and so F contains some A ∈ A. Thus we have  |F(A)|, (18.5) |F| ≤ A∈A

where F(A) := {F \ A : A ⊂ F ∈ F}. Subcase II-a. There is an A ∈ A such that F(A) contains k + 1 pairwise disjoint sets. In other words, there is a (k + 1)-sunflower {F1 , . . . , Fk+1 } ⊂ F with center A. Claim 18.4 (Deza). For each F ∈ F there is an i with 1 ≤ i ≤ k + 1 such that F ∩ Fi = F ∩ A. Proof. Since F1 \ A, F2 \ A, . . . , Fk+1 \ A are k + 1 disjoint sets, no k-element set F can intersect all of them. Thus if F ∈ F, then there is some i such that F ∩ (Fi \ A) = ∅, that is, F ∩ Fi = F ∩ A.  Let F ∈ F be arbitrarily fixed, and choose Fi ∈ F by the above claim. Since F is an L-system, it follows that l1 ≤ |F ∩ Fi | = |F ∩ A| ≤ |A| = l1 and A ⊂ F . So (ii) holds in this subcase, and F(A) is an

18. The Deza–Erd˝ os–Frankl Theorem

113

(n − l1 , k − l1 , L )-system, where L = {0, l2 − l1 , . . . , ls − l1 }. Now we have 0 ∈ L and n ≥ 2k k3 > 2k−1 (k − 1)3 + k > 2k−l1 (k − l1 )3 + l1 , that is, n − l1 ≥ 2k−l1 (k − l1 )3 . Then, by applying the result in Case I to F(A), we obtain |F| = |F(A)| ≤

s  (n − l1 ) − (li − l1 ) i=1

(k − l1 ) − (li − l1 )

=

s  n − li i=1

k − li

,

which proves (i). If, moreover, |F| ≥ 2k k2 ns−1 , then |F(A)| ≥ 2k k2 ns−1 > 2k−l1 (k − l1 )2 (n − l1 )s−1 and, by Case I, ((l2 − l1 ) − 0)|((l3 − l1 ) − (l2 − l1 ))| · · · |((k − l1 ) − (ls − l1 )), that is, (l2 − l1 )|(l3 − l2 )| · · · |(k − ls ), which proves (iii) in this subcase. Subcase II-b. F(A) contains at most k disjoint sets for each A ∈ A. Let {G1 , . . . , Gr } ⊂ F(A) be a collection of maximal disjoint sets, where r ≤ k, and let Y = G1  · · ·  Gr . Then |Y | = r(k − l1 ) < k2 . By the maximality we have G ∩ Y = ∅ for all G ∈ F(A). For y ∈ Y let F(A ∪ {y}) = {F \ (A ∪ {y}) : A ∪ {y} ⊂ F ∈ F}; then  |F(A ∪ {y})|. (18.6) |F(A)| ≤ y∈Y

Since F(A ∪ {y}) is an (n − l1 − 1, k − l1 − 1, L )-system, where L = {l2 −l1 −1, l3 −l1 −1, . . . , ls −l1 −1} and n−l1 −1 ≥ 2k−l1 −1 (k−l1 −1)3 , it follows from the induction hypothesis that |F(A ∪ {y})| ≤

s s  (n − l1 − 1) − (li − l1 − 1)  n − li = . (k − l1 − 1) − (li − l1 − 1) i=2 k − li i=2

This, together with (18.6) and |Y | < k2 , yields |F(A)| ≤

s s   n − li n − li < k2 . k − li k − li i=2 i=2

y∈Y

114

18. The Deza–Erd˝ os–Frankl Theorem

Using this with (18.5) and |A| = (18.7)

|F| < 2k k2

k l1

s  n − li i=2

= 2k k2

< 2k , we get

k − li

s k − l1  n − li . n − l1 i=1 k − li

 k−l1 Since n ≥ 2k k3 we have 2k k2 n−l < 2k k2 nk ≤ 1, and |F| < l∈L n−l k−l . 1 That is, we obtain (i) without equality. By (18.7) we have |F| < 2k k2 ns−1 , and there is nothing to show for (ii) and (iii) in this subcase. This completes the proof of Theorem 18.1. 

Chapter 19

F¨ uredi’s structure theorem

In this chapter we prove F¨ uredi’s structure theorem (Theorem 16.2). For the proof we need some basic properties of hypergraphs. If G is a graph with m edges, then we can find a bipartite subgraph H with at least m/2 edges. To see this, consider a random 2-coloring of the vertices. We say that an edge {u, v} is properly colored if the colors of u and v are different. Then the expected number of properly colored edges is m/2. This means that there is a bipartite subgraph of G with at least m/2 edges. Similarly one can show the following. Lemma 19.1 (Erd˝ os–Kleitman [34]). Let G be a k-uniform hypergraph. Then there is a k-partite subhypergraph H of G such that |H| ≥ (k!/kk )|G|. Exercise 19.2. Prove Lemma 19.1. Let G = (V (G), E(G)) be a graph. A subset C ⊂ V (G) is called a cover of G if every edge in E(G) has non-empty intersection with C. If G contains no t pairwise disjoint edges, then we can find a cover C ⊂ V (G) with |C| ≤ 2(t − 1). In fact, the union of any maximal pairwise disjoint edges has this property. Similarly one can show that if G is a k-uniform hypergraph containing no t pairwise disjoint edges, 115

116

19. F¨ uredi’s structure theorem

then there is a cover C ⊂ V (G) with |C| ≤ (t − 1)k. Using this, we have the following. Lemma 19.3. Let G be a k-uniform hypergraph containing no t pairwise disjoint edges. Then G can be partitioned into (t − 1)k parts such that



G = G1  G2  · · ·  G(t−1)k   Gi =  ∅ or Gi = ∅, where Gi := {G : G ∈ Gi } = ∅.

Proof. Let {G1 , . . . , Gl } be a collection of maximal pairwise disjoint  edges in G. Then l < t. Let C = Gi = {v1 , v2 , . . . , vlk } be the corresponding cover. Let G1 = {G ∈ G : v1 ∈ G}. Having defined G1 , . . . , Gi−1 , let Gi := {G ∈ G \ (G1  · · ·  Gi−1 ) : vi ∈ G}. In this way  we get a partition G = G1  · · ·  Glk such that Gi = ∅ for 1 ≤ i ≤ lk. Finally we just let Gi = ∅ for lk < i ≤ (t − 1)k.  We also need a simple but useful lemma about sunflowers. Lemma 19.4. Suppose that {F1 , . . . , Fk+1 } and {G1 , . . . , Gk+1 } are both (k + 1)-sunflowers in a k-uniform family with centers A and B, respectively. Then there exist i and j such that Fi ∩ Gj = A ∩ B. Proof. Since Fi \A (i = 1, . . . , k+1) are disjoint, at least k+1−|B| ≥ 1 of them are disjoint from B, say, (Fi \ A) ∩ B = ∅. Similarly, Gj \ B (j = 1, . . . , k + 1) are disjoint, and at least k + 1 − |Fi \ A| ≥ 1 of them are disjoint from Fi \ A, say, (Gj \ B) ∩ (Fi \ A) = ∅. This means that Fi and Gj are disjoint in their petals; in other words, they can meet  only in centers, that is, Fi ∩ Gj = A ∩ B. [n] Definition. For a family G ⊂ k let G¯ denote the collection of all subsets A ⊂ [n] such that A is obtained as an intersection of some members of G. In other words, G¯ is the minimal intersection-closed family containing G. We call G¯ the intersection closure of G.   be a (k, L)-system. Now we prove Theorem 16.2. Let F ⊂ [n] k Using Lemma 19.1 we may assume that F is k-partite. We are going to find a subfamily F ∗ ⊂ F which satisfies (i)–(iii) of Theorem 16.2 and also the following property: (iv) For every I ∈ IS(F ∗ ) and A ∈ F¯ ∗ with π(A) = I, A is a center of a (k + 1)-sunflower in F ∗ .

19. F¨ uredi’s structure theorem

117

We say that F ∗ is k-homogeneous if (ii) and (iv) are satisfied. We understand that a family consisting of one edge is k-homogeneous for technical reasons. Claim 19.5. If F ∗ ⊂ F is k-homogeneous, then IS(F ∗ ) is intersection closed; that is, (ii) and (iv) imply (iii). Proof. Let I1 , I2 ∈ IS(F ∗ ). We need to show that I1 ∩ I2 ∈ IS(F ∗ ). Let A1 , A2 ∈ F¯ ∗ be such that π(Ai ) = Ii ∈ IS(F ∗ ). Then, by (iv), they are both centers of a (k + 1)-sunflower in F ∗ . By Lemma 19.4 there are F, G ∈ F ∗ such that A1 ⊂ F , A2 ⊂ G, and F ∩G = A1 ∩A2 . ∗ ). On the other hand, we have Then π(A1 ∩ A2 ) = π(F ∩ G) ∈ π(F|F ∗ I1 ∩ I2 = π(A1 ∩ A2 ), and π(F|F ) = IS(F ∗ ) by (ii). Thus we get I1 ∩ I2 ∈ IS(F ∗ ).  Claim 19.6. If F ∗ ⊂ F is not k-homogeneous, then there is some A ⊂ F ∈ F ∗ such that A is not a center of a (k + 1)-sunflower in F ∗ . Proof. This clearly holds if F ∗ does not satisfy (iv). Suppose that ∗ ) = F ∗ does not satisfy (ii). Then there are F, G ∈ F ∗ such that π(F|F ∗ π(F|G ). Without loss of generality we may assume that there is an ∗ ∗ A ⊂ F such that π(A) ∈ π(F|G ) \ π(F|F ). Then A is not a center of a (k + 1)-sunflower in F ∗ .    Recall that F ⊂ [n] k is a k-partite (k, L)-system. We need to find a k-homogeneous subfamily F ∗ ⊂ F with |F ∗ | > c|F|. For G ⊂ F define C(G) ⊂ J (G) ⊂ 2[k] by C(G) := {C ⊂ [k] : C = π(A) for some A which is a center of a (k + 1)-sunflower in G}, J (G) := {J ⊂ [k] : J ∈ π(G|G ) for some G ∈ G}. Let m := |J (F)| ≤ 2k and J (F) = {J1 , . . . , Jm }. Note that G  ⊂ G implies J (G  ) ⊂ J (G). The outline of the proof is as follows. We will partition F into khomogeneous subfamilies, where the number of parts is independent of n (and depends only on k and L), and so one of the subfamilies will be our F ∗ . This partition will be obtained by a recursive procedure. Starting from F we delete subsets violating the homogeneous

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property one by one until we get a homogeneous family. Then the deleted subsets are partitioned so that each subfamily, say, F  , satisfies |J (F)| > |J (F  )|. We apply the same procedure to all such F  . Since |J (F)| ≤ 2k , this process will terminate after at most 2k steps, and in the end we will get a partition of F consisting of khomogeneous families only, and the number of families is independent of n. Claim 19.7. There is a partition of F into 1 + m parts, F = Fhom  FJ1  FJ2  · · ·  FJm , such that Fhom is k-homogeneous and Jj ∈ C(FJj ) for 1 ≤ j ≤ m. 0 Proof. Let Fhom := F and FJ0 := ∅ for all J ∈ J (F). Suppose that i i i Fhom and FJ are defined for some i ≥ 0. If Fhom is k-homogeneous, i and FJ := FJi for J ∈ J (F). then we stop and define Fhom := Fhom i such that A Otherwise, by Claim 19.6, there is some A ⊂ F ∈ Fhom i is not a center of a (k + 1)-sunflower in Fhom . Then we throw away i into one of the garbage cans labeled by the reason (in F from Fhom this case it is π(A)). Formally, we define i+1 i Fhom := Fhom \ {F },  FJi ∪ {F } if J = π(A), FJi+1 := FJi if J = π(A).

The key observation is that (19.1)

i+1 A is not a center of a (k + 1)-sunflower of Fπ(A) .

i0 +1 contains a superset of A for the first To see this, suppose that Fπ(A) time for some i0 ≤ i. This happens because A was not a center of a i0 . Then A cannot be a center of a (k + 1)-sunflower sunflower in Fhom j in Fπ(A) for any j ≥ i0 even after we add all supersets of A. This i+1 shows (19.1), or equivalently that π(A) ∈ C(Fπ(A) ). Since this is true

whenever we update a garbage can, it follows that J ∈ C(FJi+1 ) for every J ∈ J (F). We continue this procedure for i = 0, 1, 2, . . . , and then it stops after at most |F| rounds. 

19. F¨ uredi’s structure theorem

119

Claim 19.8. Suppose that J ∈ C(FJ ). Then there is a partition (1)

(k2 )

(2)

FJ = FJ  FJ  · · ·  FJ

(19.2)

(l)

such that J ∈ J (FJ ) for 1 ≤ l ≤ k2 . Proof. Let A be such that π(A) = J. Then FJ (A) := {F \ A : A ⊂ F ∈ FJ } contains no k + 1 pairwise disjoint sets. Otherwise A is a center of a (k + 1)-sunflower in FJ , contradicting J ∈ C(FJ ). So it follows from Lemma 19.3 that there exist L ≤ k2 and a partition (1)

(L)

FJ (A) = FJ (A)  · · ·  FJ (A) (l)

such that each FJ (A) is intersecting for 1 ≤ l ≤ L. Then (l)

(l)

FJ [A] := {A  G : G ∈ FJ (A)} is (|A| + 1)-intersecting. So if J  ∈ J (FJ [A]) then |J  | ≥ |A| + 1 = (l) |J| + 1. Thus we have that J ∈ J (FJ [A]). Finally, for 1 ≤ l ≤ L let  (l) (l) FJ := FJ [A], (l)

where the union is taken over all A such that π(A) = J, and for (l) L < l ≤ k2 just let FJ := ∅. Then we have the partition (19.2) with (l)  J ∈ J (FJ ). We recursively apply the above two claims. (Step 1) We get a partition F = Fhom 

 m  k2

 (l) FJj

,

j=1 l=1 (l)

(l)

where Jj ∈ J (FJj ) for 1 ≤ j ≤ m and 1 ≤ l ≤ k2 . We call FJj a garbage family. Then we have one k-homogeneous family and g1 := mk2 garbage families. (l)

(Step 2) We repeat this procedure for each FJj to subdivide the (1)

partition. For example, if G := FJ1 = ∅, then J1 ∈ J (F) \ J (G) and

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19. F¨ uredi’s structure theorem

we partition G into G = Ghom 

 m  k2

(l) GJj

 .

j=2 l=1

Note that |J (G)| ≤ |J (F)| − 1 = m − 1 and there is no garbage can labeled by J1 . At the end of this step we get mk2 new k-homogeneous (l) families corresponding to FJi for 1 ≤ j ≤ m and 1 ≤ l ≤ k2 , and we 2 get g2 := g1 × (m − 1)k garbage families. In general we do the following. (Step i) Apply the same procedure to gi−1 garbage families, and subdivide the partition. This creates gi−1 new k-homogeneous families and gi := gi−1 × (m − i + 1)k2 garbage families. Each garbage family G appearing in the partition satisfies |J (G)| ≤ m − i + 1. In the end, after m steps of this procedure, all the garbage families vanish and we are left with only k-homogeneous families. (At the last step, newly created k-homogeneous families have size one.) The total number K of the k-homogeneous families is K = 1 + g1 + g2 + · · · + gm−1 , which is independent of n. Therefore one of the k-homogeneous families F  has size at least |F|/K, and F  satisfies the condition (i). Since F  is k-homogeneous, it satisfies the conditions (ii) and (iv), and hence (iii) as well by Claim 19.5. That is, F  is the family satisfying all the conditions. This completes the proof of Theorem 16.2. 

Chapter 20

R¨ odl’s packing theorem

In this chapter we present a sufficient condition for a hypergraph to have an almost perfect matching1 . Then we show two applications: one is R¨ odl’s packing theorem and the other is a construction of a large L-system whose intersection structure is a Steiner system.

20.1. R¨ odl’s packing theorem There are only a few known exact or almost exact values of m(n, k, L). The case L = [0, t − 1] = {0, 1, . . . , t − 1} is one of them. Suppose that F is an (n, k, [0, t − 1])-system. Then any t-element subset of [n] is contained in at most one of the members of F, so     n  k (20.1) m(n, k, [0, t − 1]) ≤ . t t Erd˝os and Hanani conjectured that the ratio of the two sides tends to 1 as n → ∞ for fixed t and k. About 20 years later R¨odl proved this conjecture using the probabilistic method. His proof technique was further extended by Frankl and R¨odl [61] to obtain the following result stating that “nearly regular” hypergraphs possess almost perfect matchings; see also [96]. 1 A matching is a set of pairwise disjoint edges in a hypergraph. A matching is said to be perfect if it covers all the vertices.

121

122

20. R¨ odl’s packing theorem

Theorem 20.1. Let h be a fixed positive integer. Let {Xn }n=1,2,... be a sequence of finite sets with h ≤ |X1 | <  2 | < · · · , and let  |X {Hn }n=1,2,... be a sequence of hypergraphs in Xhn satisfying the following conditions. (i) Hn is Dn -regular; that is, there is Dn such that #{H ∈ Hn : x ∈ H} = Dn for all x ∈ Xn . Moreover, Dn is strictly increasing, and in particular Dn → ∞ as n → ∞.   (ii) For all {x, y} ∈ X2n , #{H ∈ Hn : {x, y} ⊂ H} = o(Dn ) as n → ∞. Then there exist pairwise disjoint H1 , . . . , Hm ∈ Hn with m ∼ |Xn |/h as n → ∞. See [3] for a proof of even more general cases. We use Theorem 20.1 to give lower bounds for m(n, k, L) when L = [0, t − 1] in this section, and when L = {0, 1, . . . , a − 1, b} in the next section. Theorem 20.2 (R¨ odl’s packing theorem [99]). For fixed t and k, lim

n→∞

m(n, k, [0, t − 1]) = 1. nk t

t

Proof. Because of the trivial inequality (20.1), it suffices to show that for every  > 0 there is n0 such that ifn> n0 then there is an (n, k, [0, t − 1])-system F with |F| > (1 − ) nt / kt .     Let Xn = [n] and h = kt . Define an h-uniform hypergraph on t   Xn    the vertex set Xn by Hn := { Ft : F ∈ [n] h . Then Hn is k } ⊂ n−t Dn -regular, where Dn = k−t . Moreover, for a pair {x, y} ⊂ Xn , we have |x ∪ y| ≥ t + 1 and   n−t−1 k−t = Dn = o(Dn ). #{H ∈ Hn : {x, y} ⊂ H} ≤ k−t−1 n−t Thus, by Theorem   we have a matching H1 , . . . , Hm ∈ Hn with   20.1, m ∼ |Xn |/h = nt / kt . For 1 ≤ i ≤ m we can write Hi = Fti for   some Fi ∈ [n] k . Then |Fi ∩ Fj | < t for i = j, and F = {F1 , . . . , Fm } is one of the desired (n, k, [0, t − 1])-systems. 

20.2. R¨ odl–Tengan construction

123

One can obtain a vector space version of Theorem 20.2 in a similar way. To make it precise we need some definitions. Let F be a finite   field and let Vn denote an n-dimensional vector space over F. Let Vrn     denote the set of r-dimensional subspaces of Vn , and let nr = | Vrn |. Exercise 20.3. Let r and s be given. Show that for every  > 0 all n > n0 one can find a family H ⊂ Vrn there is an n0 such that for    satisfying |H| > (1−) ns / rs and dim(H ∩H  ) < s for all H, H  ∈ H.

20.2. R¨ odl–Tengan construction of an L-system on Steiner systems R¨ odl and Tengan found a construction supporting Conjecture 16.5. In particular, they verified the conjecture when the corresponding closed (k, L)-systems are obtained from Steiner systems. Recall that     is a Steiner system S(a, b, k) if for every A ∈ [k] there is F ⊂ [k] b a k  b  exactly one F ∈ F such that A ⊂ F . So |F| = a / a . Let L = {0, 1, . . . , a − 1, b}, and let I = I(a, b, k) be the closed (k, L)-system obtained from one of the Steiner systems S(a, b, k), that is,       [k] [k] [k] (20.2) I=   ···  S(a, b, k). 0 1 a−1 Exercise 20.4. Show that if S(a, b, k) exists, then rank(k, L) = a+1. Theorem 20.5 ([100]). Suppose that a Steiner system S(a, b, k) exists. Then there is an  > 0 and a sequence {Fn } of k-partite (kn, k, L)-systems with |Fn | = Ω(na+ ), where L = {0, 1, . . . , a−1, b}. In the proof we will use the following notation. Let G ⊂ 2V be a k-partite family with vertex partition V = V1  · · ·  Vk . For G ∈ G let π(G) = {i : G ∩ Vi = ∅} denote the projection. We write V (G) for the vertex set V and Vi (G) for the ith partite set Vi . The l-shadow of G is denoted by σl (G). For a map h : X → Y and a subset Z ⊂ X, let h(Z) = {h(z) : z ∈ Z}. Proof of Theorem 20.5. Let I = {I1 , I2 , . . . , Ir } be defined by (20.2). First we construct a simple k-partite k-uniform family S = {S0 , S1 , . . . , Sr }

124

20. R¨ odl’s packing theorem

by S0 := {(i, 0) : 1 ≤ i ≤ k} and, for 1 ≤ e ≤ r, Se := {(i, 0) : i ∈ Ie }  {(j, e) : j ∈ [k] \ Ie }. For example, if Ie = [a] then Se = {(1, 0), (2, 0), . . . , (a, 0), (a + 1, e), (a + 2, e), . . . , (k, e)}.  Let V (S) = S∈S S ⊂ {(i, e) : 1 ≤ i ≤ k, 0 ≤ e ≤ r}, and for 1 ≤ i ≤ k let Vi (S) be the collection of vertices in V (S) of type (i, ∗). Clearly |Vi (S)| ≤ 1 + r, and by the symmetry of S it follows that this number is independent of i. So let v = |Vi (S)|. a−1 k−1 k−b Exercise 20.6. Show that v = 1 + j=0 + k |S(a, b, k)|. j By construction any two members of S intersect only in S0 , and these intersections induce I. Thus S is a (k, L)-system. Now we have the following key observation.   Claim 20.7. For each J ∈ [k] a , |{T ∈ σa (S) : π(T ) = J}| = r. We remark that the number counted above is r instead of r + 1. This is because for fixed J, there is a unique Ie ∈ S(a, b, k) with J ⊂ Ie , and so there is T ⊂ S0 ∩ Se such that π(T ) = J. This fact is essential for the proof of the theorem. Using Claim 20.7 we can apply Theorem 20.1 to get the following “seed” family F. Once we get a seed family, then we can blow it up by taking a direct product. In the end the family will have the required exponent a + . Claim 20.8. There are N > 0 and a k-partite k-uniform family F with |Vi (F)| = N such that |F| > N a and π(F|F ) ⊂ I for all F ∈ F, where F|F = {F ∩ F  : F  ∈ F \ {F }}. Proof. Let N be chosen sufficiently large as we will specify later. Let V˜ = V˜1  · · ·  V˜k with |V˜i | = N , and let s = |V (S)| = vk. We will embed all possible copies of S into V˜ , and then, using Theorem 20.1, we will choose some of them to construct a (kN, k, L)-system F of size > N a such that π(F|F ) ⊂ I for all F ∈ F. For this we first construct an auxiliary family HN on the vertex set   ˜ V XN = A ∈ : |A ∩ V˜i | ≤ 1 for 1 ≤ i ≤ k a

20.2. R¨ odl–Tengan construction

125

as follows. To each injection h : V (S) → V˜ with h(Vi (S)) ⊂ V˜i for 1 ≤ i ≤ k we assign the member H ∈ HN defined by H = {h(T ) : T ∈ σa (S)}.   By Claim 20.7 we have |H| = |σa (S)| = r ka for H ∈ HN . So HN     is an r ka -uniform family on |XN | = ka N a vertices. The size of H is the number of injections h : V (S) → V˜ . Noting that V (S) = kN i=1 Vi (S) and |Vi (S)| = k for each i (see Exercise 20.6), it follows that |HN | equals the k-fold product of the number of injections from [v] to [N ], that is,  k |HN | = N (N − 1) · · · (N − v − 1) . Thus |HN | = Θ(N vk ) = Θ(N s ) as N → ∞.   Fix J ∈ [k] a . Let us count the number of pairs (A, H) with π(A) = J and A ∈ H ∈ HN in two ways. On the one hand, there are N a choices for A, and for each A there are degHN (A) choices for H because degHN (A) = degHN (A ) if π(A) = π(A ) by construction. On the other hand, there are |HN | choices for H, and for each H there are exactly r choices for A by Claim 20.7. Thus we have N a degHN (A) = |HN | r. This shows that HN is DN -regular, where DN = |HN |r/N a = Θ(N s−a ). If A, A ∈ XN are distinct, then |A ∪ A | ≥ a + 1 and the number of injections h with A ∪ A ⊂ h(V (S)) is O(N s−a−1 ), so |{H ∈ HN : {A, A } ⊂ H}| = O(N s−a−1 ) = o(DN ). Thus, by Theorem 20.1, there are disjoint H1 , . . . , Hm ∈ H, where m ∼ |XN |/|Hi | = N a /r as N → ∞. Let h1 , . . . , hm be injections corresponding to the disjoint edges H1 , . . . , Hm in HN . Define a k-partite k-uniform family F on V˜ by F = {hi (S) : 1 ≤ i ≤ m, S ∈ S}. We verify that π(F|F ) ⊂ I for F ∈ F. Let S, S  ∈ S, and let F = hi (S) and F  = hj (S  ). We need to show that π(F ∩ F  ) ∈ I. If i = j, then Hi ∩ Hj = ∅, and hi (T ) = hj (T  ) for T, T  ∈ σa (S). So |F ∩ F  | = |hi (S) ∩ hj (S  )| < a and π(F ∩ F  ) ∈ I by (20.2). If

126

20. R¨ odl’s packing theorem

i = j, then F ∩ F  = hi (S) ∩ hi (S  ) = hi (S ∩ S  ) by the fact that hi is injective. Then, since S is an L-system with intersection structure I, it follows that π(F ∩ F  ) = π(S ∩ S  ) ∈ I. Finally we have |F| = m|S| ∼ (N a /r)(r + 1), and |F| > N a for N sufficiently large. Note that N depends only on a, b, and k. In particular N can be bounded by a function of k. This completes the proof of Claim 20.8.  The next exercise explains the direct product construction. We will apply this operation to the family F obtained above. Then F × · · · × F will be the desired family for Theorem 20.5. Exercise 20.9. Let I be a closed (k, L)-system, and let F and G be k-partite k-uniform families satisfying π(F|F ), π(G|G ) ⊂ I for all F ∈ F and G ∈ G. Define H = F × G with Vi (H) = Vi (F) × Vi (G) (1 ≤ i ≤ k) by {(f1 , g1 ), . . . , (fk , gk )} ∈ H if and only if (f1 , . . . , fk ) ∈ F and (g1 , . . . , gk ) ∈ G. Show that H is a k-partite k-uniform family satisfying π(H|H ) ⊂ I for all H ∈ H. Finally we are ready to finish the proof of Theorem 20.5. Let a, b, and k be given. Choose a constant N from Claim 20.8, and let F be the corresponding seed family with |V (F)| = kN and |F| > N a . Choose  = (a, N ) > 0 so that N a + 1 > N a+ . Now let n be given. Choose j so that N j ≤ n < N j+1 . We are going to construct a (kn, k, L)-system Fn with |Fn | = Θ(na+ ). Define Gj by taking the j-fold product of F: Gj = F × · · · × F. Then Gj is a k-partite k-uniform family on kN j vertices of size |Gj | = |F|j ≥ (N a + 1)j > N (a+)j . Moreover, by Claim 20.8 and Exercise 20.9, it follows that π(Gj |G ) ⊂ I for all G ∈ Gj . Since |Vi (G)| = N j ≤ n for 1 ≤ i ≤ k, by renaming the vertices, we can define Fn with Vi (Fn ) = [n] and the same members of Gj . Then Fn is a (kn, k, L)-system, and |Fn | = |Gj | > N (a+)j = (N j+1 /N )a+ > (n/N )a+ . This completes the proof of Theorem 20.5.



Chapter 21

Upper bounds using multilinear polynomials

Linear algebra methods are strong tools for combinatorics. In this chapter we present a method for bounding the size of an L-system by using polynomials. The basic idea is simple and is as follows. Let H be an L-system. To each H ∈ H we associate a polynomial fH . Suppose that the polynomials in P = {fH : H ∈ H} are linearly independent. Then the size of H cannot exceed the dimension of the vector space of polynomials V containing P ; that is, |H| ≤ dim V . Based on this idea we prove the Ray-Chaudhuri–Wilson Theorem and some related results.

21.1. Ray-Chaudhuri–Wilson Theorem Theorem 21.1 (Ray-Chaudhuri–Wilson Theorem [98]). Let F be an   (n, k, L)-system with |L| = s. Then |F| ≤ ns . The following proof, which uses the space of multilinear polynomials, is due to Alon, Babai, and Suzuki [1]. Proof. Let Ω = {0, 1}n be the n-dimensional cube. We can identify 2[n] and Ω by the correspondence between I ⊂ [n] and its characteristic vector x = (x1 , . . . , xn ) ∈ Ω where xi = 1 if i ∈ I and xi = 0 if i ∈ I. In this case, for f ∈ RΩ , we write f (I) to mean f (x). 127

128

21. Upper bounds using multilinear polynomials

 For I ⊂ [n] let xI = i∈I xi ∈ RΩ ; in particular, x∅ = 1. Then, for J ⊂ [n], it follows that  1 if I ⊂ J, (21.1) xI (J) = 0 otherwise. Lemma 21.2. Let f ∈ RΩ be such that f (I) = 0 for all I ⊂ [n] with |I| ≤ r. Then the set {xI f : |I| ≤ r} ⊂ RΩ is linearly independent. Proof. Suppose that there is a non-trivial linear combination  λI xI f = 0. |I|≤r

Let I0 be an inclusion-minimal subset such that λI0 = 0. Substi tute I0 into the above linear combination: |I|≤r λI xI (I0 )f (I0 ) = 0. If I  I0 , then λI = 0 by the minimality. If I ⊂ I0 , then xI (I0 ) = 0 by (21.1). Consequently we get λI0 xI0 (I0 )f (I0 ) = 0.  Since xI0 (I0 )f (I0 ) = 0 we have λI0 = 0, a contradiction. Let F = {F1 , . . . , Fm } be an (n, k, L)-system. For x, y ∈ Rn let  x · y = ni=1 xi yi denote the standard inner product. For 1 ≤ i ≤ m let vi ∈ Rn be the characteristic vector of Fi , and define fi ∈ RΩ by  fi (x) = (vi · x − l). l∈L

Since vi · vj = |Fi ∩ Fj |, (21.2)

fi (vj )

 = 0 if i = j, = 0 otherwise.

This showsthat  the fi are linearly independent. We claim that we can  n add s−1 i=0 i more functions while keeping the linear independency.  Claim 21.3. Let g = nj=1 xj − k ∈ RΩ . The set of functions (21.3)

{fi : 1 ≤ i ≤ m} ∪ {xI g : I ⊂ [n], |I| ≤ s − 1}

in RΩ is linearly independent. Proof. Consider a linear combination m   (21.4) λi fi + μI xI g = 0. i=1

|I|≤s−1

21.1. Ray-Chaudhuri–Wilson Theorem

129

n Notice that the function g = j=1 xj − k vanishes on k-element sets. So, for each j, substituting vj into (21.4) gives λj = 0 by (21.2). Thus  only the second term in (21.4) remains, that is, μI xI g = 0. Since s ≤ k and g vanishes only on a k-element set, it follows that g(I) = 0 for all I with |I| ≤ s − 1. Thus, by Lemma 21.2, the functions xI g (|I| ≤ s − 1) are linearly independent, and therefore μI = 0 for all I.  n  independent polynomials In (21.3) there are m+ s−1 i=0 i linearly    of degree at most s. On the other hand, si=0 ni is the dimension of the space of polynomials in of degree at most  n variables ns.  Thus s n  n ≤ , that is, m ≤ it follows that m + s−1 i=0 i i=0 i s . This completes the proof of Theorem 21.1.  Frankl and Wilson proved the modular version of Theorem 21.1. We write a ∈ L (mod p) if a ≡ l (mod p) for some l ∈ L. Theorem 21.4 (Frankl–Wilson [67]). Let n > k ≥ s be positive integers, and let p be a prime. Let L ⊂ [0, p − 1] = {0, 1, . . . , p − 1}   satisfies the following be a set of s integers. Suppose that F ⊂ [n] k conditions: (i) k ∈ L (mod p). (ii) If F, F  ∈ F with F = F  , then |F ∩ F  | ∈ L (mod p).   Then |F| ≤ ns . The condition that p is a prime cannot be dropped in general. We will prove Theorem 21.4 in a slightly stronger form in Chapter 23; see Theorem 23.3.   Exercise 21.5. Let G := {G ∈ [m] Then we have 11 : {1, 2, 3} ⊂ G}. G   |G ∩ G | ∈ [3, 10] for distinct G, G ∈ G. Let F := { 2 : G ∈ G} be   11 a k-uniform family on n := m 2 vertices, where k = 2 = 55 ≡ 1 (mod 6). Verify that   |F ∩ F  | ∈ { 2i : 3 ≤ i ≤ 10} ≡ {0, 3, 4} (mod 6)   for distinct F, F  ∈ F, and that |F| = |G| = m−3 = Θ(n4 ). Show 8 that Theorem 21.4 fails if p = 6 and s = 3.

130

21. Upper bounds using multilinear polynomials

Exercise 21.6. Prove Theorem 21.4 for the case s = 1 without assuming that p is a prime. (Hint: Go through the proof of Theorem 21.1 using Z/pZ in place of R, where p is not necessarily a prime.) Exercise 21.7. Suppose that F satisfies the assumptions in Theorem 21.4 and, moreover, that (iii) k ∈ [0, s − 1] (mod p). Then the proof of Theorem 21.1 still works almost verbatim. Examine of R) and show that |F| ≤ the  proof in this setting (using Fp in place n n Ω . (Hint: It follows from (iii) that g = j=1 xj − k ∈ Fp satisfies s g(I) = 0 for |I| ≤ s − 1. So we can apply Lemma 21.2.) By taking p > n in the above exercise we can deduce Theorem 21.1.

21.2. Deza–Frankl–Singhi Theorem Let p be a prime and let L ⊂ Fp . We say that a family of subsets H = {C1 , . . . , Cm } ⊂ 2[n] is (p, L)-intersecting if it satisfies the following two conditions: • |Ci ∩ Cj | ∈ L (mod p) for 1 ≤ i < j ≤ m. • |Ci | ∈ L (mod p) for 1 ≤ i ≤ m. Theorem 21.8 (Deza–Frankl–Singhi Theorem [24]). Let s = |L|. If a family H ⊂ 2[n] is (p, L)-intersecting, then       n n n |H| ≤ + + ···+ . 0 1 s For the proof we need some preparation. As usual we identify 2[n] and {0, 1}n by the correspondence between a subset and its characteristic vector. A polynomial is said to be multilinear if it has degree at most 1 in each variable. Let f be a polynomial in n variables of degree s over Fp . Then there exists a unique multilinear polynomial f¯ of degree at most s such that f (x) = f¯(x)

21.2. Deza–Frankl–Singhi Theorem

131

for all x ∈ {0, 1}n . In fact, since x2 = x for x ∈ {0, 1} we get f¯ just by replacing monomials in f with the corresponding products of distinct variables. For example, if f (x, y) = x4 + 2x3 y 2 + 3xy 5 , then we have f¯(x, y) = x + 5xy. The set of multilinear polynomials in variables x1 , . . . , xn of degree s constitutes a vector space V over Fp : 

V = span xi : |I| ≤ s, I ⊂ [n] . i∈I

We have dim V =

s    n . i i=0

For example, if n = 3 and s = 2, then V = span{1, x1 , x2 , x3 , x1 x2 , x1 x3 , x2 x3 },       3 3 3 dim V = + + = 7. 0 1 2 Proof of Theorem 21.8. Let X = {0, 1}n . To each subset Ci (1 ≤ i ≤ m) we assign a polynomial fi : X → Fp by  fi (x) := (x · vi − l), l∈L

where vi ∈ X denotes the characteristic vector of Ci . Then, by the (p, L)-intersecting property, we have  = 0 if i = j, fi (vj ) = 0 if i = j. Thus f1 , . . . , fm ∈ FX p are independent over Fp . Here is the plan for the proof. If we can find a vector space V ⊂ FX p satisfying • span{f1 , . . . , fm } ⊂ V , and    • dim V = si=0 ni ,

132

21. Upper bounds using multilinear polynomials

then it follows that

s    n m = dim span{f1 , . . . , fm } ≤ dim V = , i i=0

which will complete the proof. By definition the polynomial fi can be written as fi (x) := (x · vi − l1 )(x · vi − l2 ) · · · (x · vi − ls ), where L = {l1 , . . . , ls }. Now let x = (x1 , . . . , xn ). Then fi is a polynomial in n variables x1 , . . . , xn of degree s. This is a map from X = {0, 1}n to Fp . Then we get the corresponding multilinear polynomial f¯i of degree s. So f¯i is an element of a vector space

 xi : |I| ≤ s, I ⊂ [n] , V := span i∈I

and the dimension is

s    n . dim V = i i=0

This is exactly what we needed.



21.3. Snevily’s Theorem We have seen some simple applications of multilinear polynomials. In this last section we present a more delicate usage of the method due to Snevily to show the following strong result, which includes some earlier results obtained by Fisher, Bose, de Bruijin and Erd˝ os, Majumdar, Frankl and Wilson, Ramanan, and others. Theorem 21.9 (Snevily’s Theorem [102]). Let L be a set of s positive |F ∩ F  | ∈ L for all distinct F, F  ∈ F, integers. If F ⊂ 2[n] satisfies  s n−1 then |F| ≤ i=0 i . The upper bound for |F| is sharp. We give two examples. The first example has L = [s] and F = {F ⊂ [n] : 1 ∈ A, |A| ≤ s + 1}. The second one consists of L = {1} and a Steiner system S(2, 3, 7), that is, F ⊂ 2[7] is defined by F = {{1, 2, 3}, {3, 4, 5}, {1, 5, 6}, {1, 4, 7}, {3, 6, 7}, {2, 5, 7}, {2, 4, 6}}.

21.3. Snevily’s Theorem

133

The idea of the proof is as follows. First we assign a polynomial to each member of the family F and show that these polynomials are linearly independent. The dimension of the space of the polynomials n−1  s n more is i=0 i . Then we show that we can still add s−1 i=0 i polynomials to the same space without violating the linear indepen   n−1 s n−1  = i=0 i . dence. Thus we get |F| ≤ si=0 ni − s−1 i=0 i Proof. To deal with multilinear polynomials we need some more notation. Let X ∗ = {x1 , x2 , . . . , xn } denote the set of n variables, and  ∗ let Xk denote the set of multilinear monomials of degree k, that is,  ∗     X [n] xi : I ∈ , = k k X ∗ 

i∈I

where we understand 0 = {1}. Let σ(X ∗ , k) denote the basic sym metric function of degree k in X ∗ , that is, f ∈(X ∗ ) f . For example, k

if X ∗ = {x1 , x2 , x3 } then σ(X ∗ , 2) = x1 x2 + x2 x3 + x1 x3 . Finally, for a subset F ⊂ [n] let F ∗ = {xi : i ∈ F } and let v(F ) = (v1 , . . . , vn ) be the characteristic vector of F , that is, vi = 1 if i ∈ F and vi = 0 if i ∈ F . Notice that if we substitute v(F ) into σ(X ∗ , k) then we get   the value |Fk | . Also, if we substitute v(F  ), where F  ⊂ [n], into    | σ(F ∗ , k) then we get the value |F ∩F . k Let L be a set of s positive integers. We define a polynomial in y of degree s by  (y − l). g(y) = l∈L

We can uniquely determine s + 1 real numbers c0 , c1 , . . . , cs by g(y) =  s y . Then for x = (x1 , . . . , xn ) we define c i=0 i i g ∗ (x) =

s 

ci σ(X ∗ , i),

i=0

and for F ⊂ [n] let gF∗ (x)

=

s 

ci σ(F ∗ , i),

i=0 ∗

where we understand σ(F , i) = 0 if |F | < i. By definition we get the following.

134

21. Upper bounds using multilinear polynomials

Claim 21.10. For F, F  ⊂ [n] we have that gF∗ (v(F  )) = g ∗ (v(F ∩ F  )) = g(|F ∩ F  |). Now suppose that F ⊂ 2[n] satisfies |F ∩ F  | ∈ L for all distinct F, F  ∈ F. Since g(|F ∩ F  |) = 0 we get gF∗ (v(F  )) = 0. Claim 21.11. Polynomials in the set {gF∗ : F ∈ F} are linearly independent. Proof. Consider a linear combination  αF gF∗ = 0, (21.5) F ∈F

where αF ∈ R. By substituting v(F ) we obtain αF gF∗ (v(F )) = 0. So if gF∗ (v(F )) = 0, or equivalently if |F | ∈ L, then we get αF = 0. In particular, αF = 0 for all F ∈ F with |F | > max L. But we need to show that all αF are zero. We prove this by induction on |L|. The initial step is L = {l}. Let F ∈ F. If |F | = l then αF = 0. If |F | = l, then F is the only l-element subset in F. Thus (21.5) reads αF gF∗ = 0. Since gF∗ is not a zero polynomial, we have αF = 0. We proceed to the induction step. Let L = {l1 , . . . , ls } with l1 < · · · < ls , and let L = L \ {ls }. To apply the induction hypothesis let F  = F \ {F ∈ F : |F | > ls }. Then |F ∩ F  | ∈ L for all distinct F, F  ∈ F  . Since αF = 0 for all F ∈ F with |F | > ls , (21.5) can be  rewritten as F ∈F  αF gF∗ = 0. By the hypothesis applied to F  with  L we obtain that αF = 0 for all F ∈ F  . Let L = {l1 , . . . , ls } with 1 ≤ l1 < · · · < ls and F = {F1 , . . . , Fm }. If |Fi | = l1 for some i, then Fi is the only l1 -element subset in F and we may assume that n ∈ Fi by renaming the vertices if necessary. Let r ≥ 0 be the number of edges in F not containing n. We may assume that n ∈ Fi for 1 ≤ i ≤ r and n ∈ Fi for r + 1 ≤ i ≤ m. We may also assume that l1 < |Fr+1 | ≤ |Fr+2 | ≤ · · · ≤ |Fm |. We are going to assign a polynomial pi to Fi for 1 ≤ i ≤ m. To this end let  fFi (x) = (v(Fi ) · x − lj ), lj j, where we used |Fi | > l1 for the first case i = j. This means that pr+1 , . . . , pm are linearly independent. Now consider a linear combination (21.6)

m 

αi pi = 0.

i=1

Suppose that there is a nonzero scalar. Then there are at least two of them: one is αi0 with i0 ≤ r and the other is αj0 with j0 > r. So we may assume that αj0 = 0 and αj = 0 for r + 1 ≤ j < j0 . Substituting v(Fj0 ) into (21.6), we get αj0 pj0 (v(Fj0 )) = 0 and so  αj0 = 0, a contradiction. s   At this point we already have that |F| = m ≤ dim V = i=0 ni .  s−1  Next we will introduce N = i=0 n−1 more polynomials. Let i       [n − 1] [n − 1] [n − 1] {B1 , B2 , . . . , BN } =  ···  . 0 1 s−1 We may assume that |B1 | ≤ |B2 | ≤ · · · ≤ |BN |. Then for 1 ≤ i ≤ N we define  xj , qi (x) = (xn − 1) j∈Bi

136

21. Upper bounds using multilinear polynomials

where we understand q1 (x) = xn − 1. For 1 ≤ j ≤ i ≤ N it follows that  = 0 if i = j, qi (v(Bj )) = 0 if i > j. This means that q1 , . . . , qN are linearly independent. Claim 21.13. The m + N polynomials p1 , . . . , pm , q1 , . . . , qN are linearly independent. Proof. Consider a linear combination m N   αi pi + βj qj = 0. (21.7) i=1

j=1

First suppose that αi = 0 for all r + 1 ≤ i ≤ m. In this case let j0 be such that βj0 = 0 and βj = 0 for 1 ≤ j < j0 , and let t = |Bj0 |. We substitute y = (y1 , . . . , yn ) into (21.7), where yi = y if i ∈ Bj0  {n} and yi = 0 otherwise. By the definition of gF∗ i it follows that pi (y) (1 ≤ i ≤ r) is a polynomial in y of degree at most t. On the other hand, qj0 (y) = (y − 1)y t , and this is the only term containing y t+1 ; that is, the LHS of (21.7) reads βj0 y t+1 + O(y t ). This must be a zero polynomial, so βj0 = 0, and this is a contradiction. Next suppose that αi0 = 0 for some r + 1 ≤ i0 ≤ m. Since v(Fi0 ) = (∗, . . . , ∗, 1) we get qj (v(Fi0 )) = 0 for all j. Then, by substituting v(Fi0 ) into (21.7), we have αi0 pi0 (v(Fi0 )) = 0 and hence αi0 = 0, a contradiction.  We have found m + N linearly independent polynomials in V . Consequently we obtain    s   s−1  s    n n−1 n−1 |F| = m ≤ dim V − N = − = , i i i i=0 i=0 i=0 as promised.



Chapter 22

Application to discrete geometry

In some problems in discrete geometry one can express the geometric constraints in terms of intersections in hypergraphs. Then the problems may be solved by applying the corresponding results on L-systems. In this chapter we present some such examples.

22.1. Sylvester’s problem and Fisher’s inequality The following result is called Fisher’s inequality, which can be obtained from Theorem 21.9 by setting s = 1. Theorem 22.1 ([41]). Let C1 , C2 , . . . , Cm be distinct subsets of [n], and let λ ≥ 1. If |Ci ∩ Cj | = λ for all i = j, then m ≤ n. We present an application of Fisher’s inequality to a collinearity problem in discrete geometry. We say that points are collinear if they are on the same line. Two points in the plane determine a line. So m points in the plane determine m 2 lines if those points are in general position, that is, no three points are collinear. Of course, m collinear points determine only one line. Now suppose that we have m non-collinear points in the plane. Then at least how many lines are determined by those points? This problem was posed by Sylvester, 137

138

22. Application to discrete geometry

solved by Gallai, and popularized by Erd˝ os through his article in the American Mathematical Monthly [31]. Theorem 22.2. Any set of m non-collinear points in the plane determine at least m lines. The idea of the following proof goes back to de Bruijn and Erd˝os [16]. Proof. Let p1 , . . . , pm be the given m points. Suppose that they determine n lines, and let L = {l1 , l2 , . . . , ln } be the set of the lines. Let Ci ⊂ L be the subset of lines passing through pi , and let H = {C1 , . . . , Cm } ⊂ 2L . Then, for all i = j, Ci ∩Cj contains just one line, that is, the line connecting two points pi and pj . Thus |Ci ∩ Cj | = 1, and by the Fisher’s inequality we get |H| = m ≤ |L| = n.  Chv´ atal posed a generalization of this problem in a metric space. We say that three points x, y, and z in a metric space with metric d are collinear1 if they satisfy d(x, z) = d(x, y) + d(y, z). For example, a graph can be viewed as a metric space by the usual shortest-path metric. Then we can define a line in the graph in the sense above. Such lines behave rather differently from the usual lines in Euclidean space. Nevertheless, Chen and Chv´ atal conjectured the following. Conjecture 22.3 (Chen–Chv´ atal [18]). In any metric space, m noncollinear points determine at least m lines.

22.2. Chromatic number of the unit-distance graph A family of subsets H ⊂ 2[n] is called t-avoiding if |H ∩ H  | = t for all H, H  ∈ H. Applying Theorem 21.8 to the case n = 4p − 1, s = p − 1, we get the following. 1

This is called collinearity by betweeness.

L = {0, . . . , p − 2},

22.2. Chromatic number of the unit-distance graph

139

  Corollary 22.4 ([67]). Let p be a prime. If a family H ⊂ [4p−1] 2p−1 is (p − 1)-avoiding, then       4p − 1 4p − 1 4p − 1 |H| ≤ + + ··· + . 0 1 p−1   4p−1     Exercise 22.5. Show that 4p−1 + 1 + · · · + 4p−1 < 2 4p−1 0 p−1 p−1 . n s n (Hint: Use induction on s to deduce that i=0 i < 2 s for n ≥ 3s.) Now we present a geometric application of the above result. The distance-d graph in n-dimensional Euclidean space has vertex set Rn and two points are adjacent if their Euclidean distance is d. If d = 1 then it is called the unit-distance graph and denoted by Gn . In symbols: • V (Gn ) = Rn , • E(Gn ) = {{x, y} : x − y = 1},  where x = x21 + · · · + x2n for x = (x1 , . . . , xn ) ∈ Rn . Note that distance-d graphs in a fixed dimension are isomorphic to each other for all d > 0. It is difficult to determine the chromatic number of Gn , and even for the planar case (n = 2) we only know that 5 ≤ χ(G2 ) ≤ 7. For the general case Larman and Rogers [86] proved that χ(Gn ) < (3 + o(1))n . On the other hand, Frankl and Wilson [67] obtained the following lower bound using Corollary 22.4. Theorem 22.6. Let p be a prime, and let n = 4p − 1. Then we have χ(Gn ) > 1.1n . √ Proof. Let k = 2p − 1 and d = 2p. As usual we identify 2[n] and Ω = {0, 1}n . We are going to find a large distance-d structure in the n-dimensional cube Ωn . To this end we look at the points having  . If exactly k 1’s in the coordinate, that is, we look at subsets in [n] k   satisfy two subsets X, Y ∈ [n] k |X ∩ Y | = p − 1,

140

22. Application to discrete geometry

then the corresponding characteristic vectors satisfy x − y = d, where we used x − y2 = |XY | = 2(k − |X ∩ Y |) = 2p.   Now suppose that we can color [n] so that any two points with k distance d get different colors. Then a family of subsets having the same color is (p − 1)-avoiding, and, by Corollary  n  22.4 with Exer. So in order to cise 22.5, the size of the family is less than 2 p−1 color the distance-d to color G2 , we need at   n  graph,p or equivalently n /(2 p−1 ) > 1.1 colors2 .  least 2p−1 With a little bit more effort one can show that χ(Gn ) > 1.2n for sufficiently large n ∈ N; see [67].

2 n

1.1 .



 n  n  2p−1 /(2 p−1 )

=

1 (3p)···(2p+1) 2 (2p−1)···p

=

3p (3p−1)···(2p+1) 2p (2p−1)···(p+1)

> (3/2)p = (3/2)

n+1 4

>

Chapter 23

Upper bounds using inclusion matrices

For two hypergraphs F and G we can define an |F| × |G| matrix M whose (F, G)-entry is determined by a given function m(F, G). Obviously rank M ≤ min{|F|, |G|}. If the rows of M are linearly independent, then we get rank M = |F| ≤ |G|. In this chapter we present some applications of this inequality.

23.1. Bounds for s-independent families [n]   For 0 ≤ i ≤ k ≤ n, F ⊂ [n] k , and G ⊂ i , define the inclusion matrix M (F, G) as follows. This is an |F| × |G| matrix whose (F, G)entry m(F, G), where F ∈ F and G ∈ G, is defined by  1 if F ⊃ G, m(F, G) = 0 if F ⊃ G. For F ⊂ (23.1)

[n] k

and 0 ≤ j ≤ i ≤ k, simple counting yields   [n] [n] k−j  [n] M (F, [n] i )M ( i , j ) = i−j M (F, j ).

In fact, the (F, J)-entry of (23.1), where F ∈ F and J ∈ #{I ∈

[n] i

[n] j , counts

: J ⊂ I ⊂ F }. 141

142

23. Upper bounds using inclusion matrices

For a matrix M , the column space is a vector space spanned by its column vectors. Let colsp M denote the column space of M . Then (23.1) also shows the following.   Lemma 23.1. Let 0 ≤ j ≤ i ≤ k and F ⊂ [n] k . Then [n]   colsp M (F, j ) ⊂ colsp M (F, [n] i ). [n]   We say that F ⊂ [n] k is s-independent if the rows of M (F, s ) are linearly independent,that  is, the inclusion matrix has full rown rank. In this case, |F| ≤ s immediately follows. Lemma 23.2. Let p be a prime   degree s. Let F ⊂ [n] k . Suppose  ≡ 0  (23.2) f (|F ∩ F |) ≡0

and let f ∈ Q[x] be a polynomial of that F satisfies (mod p)

if F = F  ,

(mod p)

if F = F 

for F, F  ∈ F. Then F is s-independent. Proof. Since f is of degree s, it can be uniquely represented as   s  x f (x) = αi , i i=0

  where αi ∈ Q. Let Mi = M (F, [n] i ) be the inclusion matrix. Then  T the (F, F )-entry of Mi Mi counts the number of i-element subsets    | . Let contained in both F and F  , that is, |F ∩F i (23.3)

A=

s 

αi Mi MiT .

i=0 

Then the (F, F )-entry of A is f (|F ∩F  |), and A (mod p) is a diagonal matrix with no zero diagonal entries. So A is non-singular because det A ≡ 0 (mod p), and rank A = |F|. By Lemma 23.1 it follows that colsp Mi ⊂ colsp Ms for 0 ≤ i ≤ s. This, together with (23.3), yields colsp A ⊂ colsp Ms , and hence rank A ≤ rank Ms . Thus Ms has full row-rank, that is, F is s-independent.  Theorem 23.3 (Frankl–Wilson [67]). Let p be a prime and let L ⊂   [0, p − 1] = {0, 1, . . . , p − 1} with |L| = s. Suppose that F ⊂ [n] k

23.1. Bounds for s-independent families satisfies 

|F ∩ F |

 ∈ L

(mod p)

143

if F = F  ,

if F = F    for F, F  ∈ F. Then F is s-independent and |F| ≤ ns . ∈L

(mod p)

Proof. Define a polynomial f ∈ Q[x] by  f (x) = (x − l). l∈L

Then f satisfies (23.2) in Lemma 23.2. Thus F is s-independent and   |F| ≤ ns . Theorem 23.4 (Frankl–Wilson [67]). Let p be a prime and let q = pe   satisfies be a prime power. Suppose that F ⊂ [n] k |F ∩ F  | ≡ k (mod q)  n  . for distinct F, F  ∈ F. Then F ≤ q−1 Proof. Choose rational numbers αi (0 ≤ i ≤ q − 1) so that     q−1  x x−k−1 αi = , i q−1 i=0 and define a matrix A=

q−1 

αi Mi MiT ,

i=0

 [n]  ) is the inclusion matrix. Clearly we have where Mi = M (F, q−1   n . rank A ≤ rank Mi ≤ q−1 We show that A (mod p) is a diagonal matrix without zero diagonal entries. This would give rank A = |F|,   completing the proof. Recall that the (F, F  )-entry of Mi MiT is xi , where x = |F ∩ F  |.   =: a. Thus the (F, F  )-entry of A is given by x−k−1 q−1   −1 = (−1)(−2)···(−q+1) If F = F  then a = q−1 = (−1)q−1 . So p (q−1)!  divides no diagonal entries of A. Next let F = F and we show that a (mod p) ≡ 0. Since F = F  implies x ≡ k (mod q), q does not divide

144

23. Upper bounds using inclusion matrices

 x−k x−k−1 (x−k)a  = q = q x − k. On the other hand, x−k shows that q q−1 q divides (x − k)a. Thus p must divide a, as needed.  Exercise 23.5. Let n ≥ k ≥ 2l + 1, and let k − l be a prime power.   Suppose that F ⊂ [n] k satisfies  |F ∩ F | = l for all distinct F, F ∈ n F. Show that |F| ≤ k−l−1 . (Hint: Let q = k − l and apply Theorem 23.4.) We mention that Theorem 23.3 has been extended to non-uniform families as follows by Babai and Frankl [9]. Theorem 23.6. Let p be a prime and let L ⊂ [0, p − 1] with |L| = s. Let k be an integer with k ∈ L (mod p) and s + k ≤ n. Suppose that F ⊂ 2[n] satisfies  ≡ k (mod p) if F = F  ,  |F ∩ F | ∈ L (mod p) if F = F    for F, F  ∈ F. Then |F| ≤ ns . We present a quick application of Theorem 23.3 due to Gottlieb [75], which is also known as Kantor’s lemma. Theorem 23.7. Let 0 ≤ i ≤ j ≤ n. Then it follows that   [n] n n rank M ( [n] i , j ) = min{ i , j }.   Proof. First suppose that i + j ≥ n. Let F = [n] j . Then, for F, F  ∈ F, |F ∩ F  | ≥ 2j − n ≥ j − i. Thus F is an (n, j, L)-system, where L = {j − i, . . . , j − 1}. Choose a prime p > n arbitrarily. By Theorem 23.3 it follows that F is s-independent, where s = |L| = i. n   [n] Thus rank M ( [n] i , j ) = i . Next suppose that i + j < n, and let   [n] i = n − i and j  = n − j. Note that M ( [n] i , j ) is the transpose [n] [n]   of M ( j  , i ) and i + j > n. So we apply the first case and the     corresponding rank is jn = nj . 

23.2. Shadows in a t-intersecting family The exact lower bound for the size of u-shadows in a k-uniform family is given by the Kruskal–Katona Theorem (Theorem 6.2). But if the family is t-intersecting, then the bound should be much larger.

23.2. Shadows in a t-intersecting family

145

Katona found a sharp inequality in this case. We will present a proof of the result as an application of inclusion matrices. Theorem 23.8 (Katona’s intersecting shadow theorem [81]). Let   1 ≤ t ≤ k ≤ n and let F ⊂ [n] be t-intersecting. Then, for k − t ≤ k u ≤ k, we have     2k − t 2k − t . / (23.4) |σu (F)|/|F| ≥ k u To prove the theorem we use the following lemma due to Frankl and F¨ uredi [57].   Lemma 23.9. Let 0 ≤ s ≤ u ≤ k and let F ⊂ [n] k be s-independent. Then     k+s k+s (23.5) |σu (F)|/|F| ≥ / . u k Proof. For each x ∈ [n] let Wx = [n] \ {x} and  Wx  : {x} ∪ G ∈ F} = {F \ {x} : x ∈ F ∈ F}. Fx = {G ∈ k−1  Wx  Claim 23.10. Fx ⊂ k−1 is s-independent, that is,   rank M (Fx , Wsx ) = |Fx |. We postpone the proof of Claim 23.10, and we first prove the lemma by induction on k assuming Claim 23.10. Inequality (23.5) trivially holds for the following three cases: s = 0, u = s, and u = k. So let 1 ≤ s < u < k and assume that (23.5) is true for k − 1. By  Wx , Claim 23.10 we can apply the induction hypothesis to Fx ⊂ k−1 and we get (k−1)+s (23.6)

u−1 |σu−1 (Fx )| ≥ |Fx | (k−1)+s . (k−1)

By counting #{(x, F ) ∈ [n] × F : x ∈ F } in two ways, that is, by counting the number of edges in the corresponding bipartite graph from each side, we have  (23.7) |Fx | = k |F|. x∈[n]

146

23. Upper bounds using inclusion matrices

Similarly, by counting #{(x, G) ∈ [n] × σu (F) : x ∈ G}, we have  |σu−1 (Fx )| = u |σu (F)|. (23.8) x∈[n]

Using (23.8), (23.6), and (23.7), we get

k−1+s 1  1  u−1 |σu−1 (Fx )| ≥ |Fx | k−1+s |σu (F)| =  u u k−1 x∈[n] x∈[n] k−1+s k+s k u−1 u = |F| k−1+s = |F| k+s . u k−1 k

This shows that (23.5) is true for k as well and completes the induction.   into two Now we prove Claim 23.10. Fix x ∈ [n]. Divide [n] Wx   Wsx  parts C  D, where C = s and D = {{x}  T : T ∈ s−1 }. Let   F x = {F ∈ F : x ∈ F } ⊂ [n] k .   We divide the columns of M (F x , [n] s ) into two blocks:     x x (23.9) M (F x , [n] s ) = M (F , C) | M (F , D) .  Wx  ). Then, by Lemma 23.1, By definition M (F x , D) = M (F x , s−1  Wx    x x Wx colsp M (F , s−1 ) ⊂ colsp M (F , s ), and   (23.10) colsp M (F x , D) ⊂ colsp M (F x , Wsx ) = colsp M (F x , C).   x By (23.9) and (23.10) we have colsp M (F x , [n] s ) ⊂ colsp M (F , C) and   x (23.11) rank M (F x , [n] s ) ≤ rank M (F , C). The opposite inequality is trivial, so we have equality in (23.11). On x the other hand, since  F is xs-independent, F is also s-independent x [n] and rank M (F , s ) = |F |. Thus equality in (23.11) gives us that rank M (F x , C) = |F x |. Finally, noting that |F x | = |Fx | and M (F x , C) = M (Fx , C), we have rank M (Fx , C) = |Fx |, as needed. This completes the proof of Claim 23.10 and Lemma 23.5. 

23.2. Shadows in a t-intersecting family

147

Proof of Theorem 23.8. Choose a prime p > k and define a polynomial f ∈ Q[x] by f (x) = (x − t)(x − t − 1) · · · (x − k + 1). Let s = k − t and L = {t, t + 1, . . . , k − 1}, and apply Lemma 23.2. Then F is s-independent. Now the inequality (23.4) follows from Lemma 23.9.  One can apply the same proof technique to obtain a vector space version of Theorem 23.8; see [65].

Chapter 24

Some algebraic constructions for L-systems

Designs, codes, and finite geometries are useful sources for constructing large L-systems. In this chapter we present some L-systems constructed by blowing up these structures.

24.1. Algebraic constructions In this section we construct some L-systems related to Steiner systems and codes. Here let us briefly recall some basic definitions concerning codes. Let V = Fnq be an n-dimensional vector space over the qelement field. We say that a subset C ⊂ V is an [n, k, d]q -code if C is a k-dimensional subspace of V and every x ∈ V has at least d nonzero coordinates. An (n − k) × n matrix A is called a parity check matrix of C if C = {x ∈ V : Ax = 0}. As an illustrative example, let us construct a (7, {0, 1, 3})-system. Let A be the following 3 × 7 matrix over F2 : ⎡

(24.1)

⎤ 1 0 0 1 1 0 1 A = ⎣0 1 0 1 0 1 1⎦ . 0 0 1 0 1 1 1 149

150

24. Some algebraic constructions for L-systems

This is a parity check matrix of a [7, 4, 3]2 -code. This matrix has the following properties. (i) Any two columns are linearly independent over F2 . (ii) For any two columns cp , cq of A, the subspace spanned by cp , cq contains precisely three columns cp , cq , and cr of A, where cr = cp + cq . Note that (ii) implies that any four columns of A span the entire 3dimensional space F32 . The triples {p, q, r} in (ii) (the indices of three columns corresponding to 2-dimensional subspaces) form the Steiner triple system S(2, 3, 7), that is, {{1, 2, 4}, {1, 3, 5}, {1, 6, 7}, {2, 3, 6}, {2, 5, 7}, {3, 4, 7}, {4, 5, 6}}.   We construct a {0, 1, 3}-system F ⊂ X7 using the matrix A. This is a 7-partite hypergraph on the vertex set X = V1  · · ·  V7 , where each partite set Vi is a distinct copy of Fd2 . So we may understand F ⊂ (Fd2 )7 . Using A we define a linear map F : (Fd2 )3 → (Fd2 )7 as follows. For a, b, c ∈ Fd2 , with an abuse of notation, let F (a, b, c) ∈ (Fd2 )7 be the ordered 7-tuple defined by F (a, b, c) = (a, b, c)A = (a, b, c, a + b, a + c, b + c, a + b + c). Then define (24.2)

F = {F (a, b, c) : a, b, c ∈ Fd2 }.

A key observation is as follows. Suppose that two members F and F  in F meet on at least two vertices, say they meet on V1  V2 . Then they necessarily meet on V4 as well. Thus |F ∩ F  | ≥ 3. Moreover, if they meet on at least four vertices, then they necessarily coincide. Exercise 24.1. Let n = |X| = 7 · 2d . Show that the family F defined by (24.2) is an (n, 7, {0, 1, 3})-system with |F| = (n/7)3 . Let c ∈ Fsq and y ∈ (Fdq )s . We write y = (y1 , . . . , ys ), where s yj ∈ Fdq for 1 ≤ j ≤ s. Define c  y = j=1 cj yj ∈ Fdq . Exercise 24.2. Let ci ∈ F32 denote the ith column vector of A defined by (24.1). Show that if F (a, b, c) and F (a , b , c ) meet on Vi ∼ = Fd2 then ci  y = 0, where y = (a − a , b − b , c − c ) ∈ (Fd2 )3 .

24.1. Algebraic constructions

151

Now we generalize the above construction. Let t < b < k. We say that a (t + 1) × k matrix over Fq is a (t, b, k)q -matrix if it satisfies the following two conditions concerning columns of the matrix: (P1) Any t columns are linearly independent over Fq . (P2) For any t columns, the t-dimensional subspace spanned by these columns contains precisely b columns. For a (t, b, k)q -matrix, the number of (t+1)×b minor matrices of rank    t is kt / bt . Each such minor matrix gives a b-element subset (block) as indices of corresponding columns. The set of these blocks forms a Steiner system S(t, b, k). In this case we say that a (t, b, k)q -matrix supports S(t, b, k). Theorem 24.3. If there exists a (t, b, k)q -matrix, then there exists an (n, k, L)-system of size (n/k)t+1 where L = {0, 1, . . . , t − 1, b}. Proof. Let A = (ai,j ) be a (t, b, k)q -matrix. We construct a (k, L)system F which is k-partite on the vertex set X = V1 · · ·Vk , where each Vi is a distinct copy of Fdq . To each (t + 1)-tuple (x1 , . . . , xt+1 ), where xi ∈ Fdq , we assign a k-tuple F (x1 , . . . , xt+1 ) ∈ V1 × · · · × Vk by F (x1 , . . . , xt+1 ) = (x1 , . . . , xt+1 )A =

 t+1

ai1 xi , . . . ,

i=1

t+1 

 aik xi .

i=1

We understand this k-tuple as an edge of F; that is, we define F = {F (x1 , . . . , xt+1 ) : (x1 , . . . , xt+1 ) ∈ (Fdq )t+1 }. It is clear from the construction that F is k-partite and k-uniform. We need to verify that F is an L-system. Let F = F (x1 , . . . , xt+1 ) and F  = F (x1 , . . . , xt+1 ) be distinct edges of F. Let l = |F ∩ F  |    be such that F and F  meet on i∈I Vi . Let and let I ∈ [k] l y = (x1 − x1 , . . . , xt+1 − xt+1 ) and let ci denote the ith column of A. Then F − F  = yA and the following four statements are equivalent: (i) i ∈ I; (ii) F and F  meet on Vi ; (iii) the ith entry of F − F  is 0; (iv) ci  y = 0.

152

24. Some algebraic constructions for L-systems

We want to show that l ∈ L. If l n0 (k), then   n/p mp [n + r, k + r, {r}] = . k/p Exercise 25.4. Deduce Theorem 25.3 for the case r = 0 and p|n from Theorem 16.1. (Hint: In this case we can write n = pn and k = pk , and we have mp [n, k, {0}] = m(n, k, {0, p, 2p, . . . , (k − 1)p})       k −1 k −1  n n/p n − pi n −i ≤ = = = . k/p k − pi k − i k i=0 i=0 Recall that m(n, k, L) is defined in Chapter 16.) Proof of Theorem 25.3. Let n = pa + q and k = pb (0 ≤ q < p). First we show the lower bound by construction. For the case r = 0 let [n − q] = A1  A2  · · ·  Aa be a partition, where |Ai | = p (1 ≤ i ≤ a), and define    . Ai : I ∈ [a] Fn,k = b i∈I

This is an [n, k, {0}]p -system of size case r > 0 define

a b

=

n/p k/p

= Θ(nb ). For the

Gn+r,k+r = {F ∪ [n + 1, n + r] : F ∈ Fn,k }.

158

25. Oddtown and eventown problems

This is an [n + r, k + r, {r}]p -system of the same size as Fn,k . Consequently we have shown that   n/p = Θ(nb ) mp [n + r, k + r, {r}] ≥ |Fn,k | = k/p for all 0 ≤ r < p. Next we show the upper bound. We claim that it suffices to prove the bound for the case r = 0: (25.3)

|Fn,k | ≥ mp [n, k, {0}].

To see this let r > 0 and suppose that F is an [n+r, k+r, {r}]p -system with |F| = mp [n + r, k + r, {r}]. Then the size of (distinct) pairwise intersections is one of r, p + r, . . . , (b − 1)p + r. By Theorem 21.1 b we have |F| ≤ n+r b . Since |F| ≥ |Gn+r,k+r | we have |F| = Θ(n ). Then, by (ii) of Theorem 18.1, there is an r-element set R which is contained in all F ∈ F. In this case F  = {F \ R : F ∈ F} is an [n, k, {0}]p -system, and we have (25.4)

mp [n, k, {0}] ≥ |F  | = |F| = mp [n + r, k + r, {r}] ≥ |Gn+r,k+r | = |Fn,k |.

If (25.3) holds, then we have equalities everywhere in (25.4), and mp [n, k, {0}] = mp [n + r, k + r, {r}] = |Fn,k |. This would complete the proof of the theorem. Thus all we need is to prove (25.3). Now let F be one of the largest [n, k, {0}]p -systems, and we prove (25.3) by showing that |F| ≤ |Fn,k |. We say that A ⊂ [n] with |A| ≥ p is an atom of F if A ⊂ F or A ∩ F = ∅ for any F ∈ F and moreover A is inclusion maximal with this property. Notice that atoms are  be the set of atoms pairwise disjoint. Let A = {A1 , . . . , At } ⊂ [n] p of size p, and let X1 ⊂ [n] be the union of all atoms in A. Thus X1 = A1  · · ·  At is a partition. Then t ≤ a because n = pa + q and q < p. Let X0 = [n] \ X1 , FX1 = {F ∈ F : F ⊂ X1 }, and FX0 = F \ FX1 . It follows from the definition of X1 that for every F ∈ F, p divides |F ∩ X1 |, and hence p divides |F ∩ X0 | as well.

25.2. Uniform eventown problems

159

   If FX0 = ∅, then |F| ≤ bt ≤ ab = |Fn,k |, and we are done. We will show that FX0 = ∅ indeed holds. For each x ∈ X0 let F(x) = {F \ {x} : x ∈ F ∈ F}. This is an (n − 1, k − 1, L )-system, where L = {p − 1, 2p − 2, . . . , (b − 1)p − 1}. So |F(x)| = O(nb−1 ) follows from Theorem 18.1, but we can improve the bound as follows. Claim 25.5. |F(x)| = O(nb−2 ) for x ∈ X0 . Proof. Let x ∈ X0 . First suppose that there is an atom A ⊂ X0 with x ∈ A. Let c = |A|. Since all atoms of size p are in X1 , we have c > p. Let F(A) = {F \ A : A ⊂ F ∈ F}. Then |F(x)| = |F(A)| and F(A) is an (n − c, k − c, {2p − c, 3p − c, . . . , (b − 1)p − c})-system. Thus, by Theorem 18.1, the size is O(nb−2 ). Next suppose that there is no atom containing x. Let Ix =  G∈F (x) G. If |Ix | < p − 1 = min L , then, by (ii) of Theorem 18.1,





we have |F(x)| = O(n|L |−1 ) = O(nb−2 ). Let |Ix | ≥ p − 1, and choose Y ⊂ Ix with |Y | = p−1. Since Y {x} is a p-element subset in X0 , it is not an atom. So there is an F1 ∈ F such that 0 < |F1 ∩(Y {x})| < p. By definition Y ⊂ G for all G ∈ F(x), and x ∈ F1 implies Y ⊂ F1 . In this case |F1 ∩ (Y  {x})| = p, a contradiction. Thus we have (25.5)

x ∈ F1 and F1 ∩ Y = ∅.

Fix Y and F1 . For z ∈ F1 \ Y define G(z) = {G \ (Y  {z}) : Y  {z} ⊂ G ∈ F(x)}. Then G(z) is an (n − p − 1, k − p − 1, {p − 1, 2p − 1, . . . , (b − 2)p − 1})system, and |G(z)| = O(nb−2 ) follows from Theorem 18.1. Now suppose that for every G ∈ F(x), (F1 \ Y ) ∩ G = ∅.

(25.6)

Then for each G there is a z ∈ F1 \ Y such that G \ (Y  {z}) ∈ G(z). Thus we get  |G(z)| = |F1 \ Y | O(nb−2 ) = O(nb−2 ), |F(x)| ≤ z∈F1 \Y

160

25. Oddtown and eventown problems

completing the proof provided (25.6) is true. To prove (25.6) let G ∈ F(x). Since both F1 and F2 := G{x} are in F, p divides |F1 ∩F2 |. Moreover, |F1 ∩F2 | = |F1 ∩G| ≥ |F1 ∩Y | > 0 by (25.5), so |F1 ∩ G| ≥ p. Since Y ⊂ Ix ⊂ G and |Y | = p − 1 we  have (F1 \ Y ) ∩ G = ∅. By the claim we have  |FX0 | ≤ |F(x)| = O(|X0 |nb−2 ). x∈X0

 This, together with |FX1 | ≤ bt , implies that (25.7)    Θ(nb ) = ab = |Fn,k | ≤ |F| = |FX0 | + |FX1 | ≤ O(|X0 |nb−2 ) + bt . Note that |X0 | = n − |X1 | = n − pt = O(n) and |X0 |nb−2 = O(nb−1 ). If |X0 | = Ω(n), say n − pt = n for some  > 0, then and

a b

t = (1 − )n/p  (n − q)/p = a, t  b , which contradicts (25.7).

If |X0 | = o(n), then it follows from (25.7) that  a b−1 ) + bt . (25.8) b ≤ o(n Compare the coefficients of nb−1 in both sides. Since t ≤ a = (n − q)/p, (25.8) is possible only when t = a. Consequently |X1 | = pa and |X0 | = n − pa = q < p. Thus |F ∩ X0 | = 0 for every F ∈ F because  p divides |F ∩ X0 |. This means that FX0 = ∅.

Chapter 26

Tensor product method

Suppose that we want to bound the size of a family F of subsets. If we find a map from F to a vector space V such that all members of F are sent to linearly independent vectors, then we get |F| ≤ dim V . We have seen such examples in earlier chapters. In this chapter we repeat the same procedure, but this time V is not only a vector space but also a tensor space which has richer structure and may reflect some delicate combinatorial structures in F. In the first section we introduce some basic concepts and properties concerning tensor spaces. The remaining sections focus on applications to problems for two families with intersection constraints. We mention that [9] and [92] also deal with similar topics and are highly recommended. We also mention that Kalai used tensor products to define algebraic shifting. We do not cover this topic here, but it is an important tool for studying simplicial complexes; see, for example, [80]. In this chapter, by a vector space we mean a vector space over R for simplicity. See [9] for replacing the base field R with a finite field of reasonably large characteristic.

161

162

26. Tensor product method

26.1. Some basic facts about tensor spaces In this section we gather some basic properties of tensor products and construct two structures called the symmetric algebra and the exterior algebra, which are useful for our combinatorial purposes (and useful in many other branches of mathematics, of course). We will need only a small part of the mechanism, and we may introduce the facts with proofs. In fact, the last miniature in the book [92] is written in such a minimalistic way, and it is one nice way to prepare the ingredients we need. On the other hand, the essence of considering tensors is perhaps in an abstract conceptual framework rather than in a concrete setting. So in this section we will show a moderate “whole picture” of the mechanism without proofs. It is more than we will need for our applications, but it will be helpful to understand the motivations for using tensors. Most of the omitted proofs are routine and one can find them in any standard textbook dealing with multilinear algebra, such as [113]. 26.1.1. Multilinear map and tensor product. Let V1 , V2 , and W be vector spaces. We say that a map Φ : V1 × V2 → W is bilinear if it is linear with respect to each variable, that is, Φ(αv1 + βv1 , v2 ) = αΦ(v1 , v2 ) + βΦ(v1 , v2 ), Φ(v1 , αv2 + βv2 ) = αΦ(v1 , v2 ) + βΦ(v1 , v2 ) hold for all v1 , v1 ∈ V1 , v2 , v2 ∈ V2 , and α, β ∈ R. First we define a tensor product of V1 and V2 , which is based on the following fact. Theorem 26.1. Let V1 and V2 be vector spaces. Then there exist a vector space W and a bilinear map Φ : V1 × V2 → W which satisfy the following two conditions. (i) W is spanned by Φ(V1 × V2 ) = {Φ(v1 , v2 ) : v1 ∈ V1 , v2 ∈ V2 }. (ii) For every vector space U and every bilinear map F : V1 × V2 → U

26.1. Some basic facts about tensor spaces

163

there exists a unique linear map f : W → U such that F = f ◦ Φ. We write V1 ⊗ V2 for W and call it a tensor product of V1 and V2 , and we call an element of W a tensor. Note that Φ(V1 ×V2 ) = V1 ⊗V2 in general. We also write v1 ⊗ v2 for Φ(v1 , v2 ), where v1 ∈ V1 and v2 ∈ V2 . Then the condition (i) means that every tensor w ∈ W can be represented in the form  w= v1 ⊗ v2 . v1 ∈V1 , v2 ∈V2

We remark that this representation is not necessarily unique in general. Let e1 , . . . , em (resp. f1 , . . . , fn ) be a basis for V1 (resp. V2 ). Then ei ⊗ fj

(1 ≤ i ≤ m, 1 ≤ j ≤ n)

is a basis for V1 ⊗ V2 . In particular, dim V1 · dim V2 = dim V1 ⊗ V2 . Using the basis, every w ∈ W can be uniquely written as  αi,j ei ⊗ fj , w= i,j

where αi,j ∈ R and i and j run over 1 ≤ i ≤ m and 1 ≤ j ≤ n. The condition (ii) is called universality of a tensor product. We defined a tensor product of two vector spaces, and we can easily extend the concept to any finite number of vector spaces. So let V1 , . . . , Vr and W be vector spaces. We say that a map Φ : V1 × · · · × Vr → W is r-linear if it is linear with respect to each variable. Then we have the following. Theorem 26.2. Let V1 × · · · × Vr be vector spaces. Then there exist a vector space W and an r-linear map Φ : V1 × · · · × Vr → W which satisfy the following two conditions. (i) W is spanned by Φ(V1 × · · · × Vr ).

164

26. Tensor product method (ii) For every vector space U and every r-linear map F : V1 × · · · × Vr → U there exists a unique linear map f : W → U such that F = f ◦ Φ.

We write V1 ⊗ · · · ⊗ Vr for W and v1 ⊗ · · · ⊗ vr for Φ(v1 , . . . , vr ), where vi ∈ Vi (1 ≤ i ≤ r). 26.1.2. Symmetric tensors and alternating tensors. Now we consider the case where all Vi are the same, that is, V1 = · · · = Vr = V . In this case let T r (V ) = V ⊗ · · · ⊗ V

(r times),

and we sometimes write T r for T r (V ) for simplicity. We will define two important subspaces S r (V ) (symmetric tensors) and Ar (V ) (alternating tensors) of T r (V ). Let σ ∈ Sr , where Sr denotes the symmetric group of order r, and apply Theorem 26.2 with U = T r and Φ defined by Φ(v1 , . . . , vr ) = vσ(1) ⊗ · · · ⊗ vσ(r) . Then we get a unique isomorphism fσ : T r → T r such that (26.1)

fσ (v1 ⊗ · · · ⊗ vr ) = vσ(1) ⊗ · · · ⊗ vσ(r) .

We say that a tensor t ∈ T r is symmetric if fσ (t) = t for all σ ∈ Sr , and alternating if fσ (t) = sgn(σ)t for all σ ∈ Sr , where sgn(σ) is 1 if σ is an even permutation and −1 if σ is an odd permutation. Finally we define S r (V ) = {t ∈ T r (V ) : t is symmetric}, Ar (V ) = {t ∈ T r (V ) : t is alternating}. Then it is easy to verify that both S r (V ) and Ar (V ) are subspaces of T r (V ). The dimensions of these subspaces are given as follows.

26.1. Some basic facts about tensor spaces

165

Theorem 26.3. Let n = dim V . Then we have   r+n−1 r , dim S (V ) = r   n dim Ar (V ) = . r In particular, Ar (V ) = {0} for r > n. To explain the reason let us find a basis for each subspace. We introduce two operators S r (symmetrizer) and Ar (alternatizer) defined by 1  S r (t) = fσ (t), r! σ∈Sr

1  Ar (t) = sgn(σ)fσ (t), r! σ∈Sr

where fσ satisfies (26.1). Note that each of them is a linear map on T r . Then one can show that S r ◦ S r = S r and Ar ◦ Ar = Ar , from which it follows that S r (V ) = {S r (t) : t ∈ T r (V )}, Ar (V ) = {Ar (t) : t ∈ T r (V )}. In other words, S r (resp. Ar ) is a projection from T r (V ) to S r (V ) (resp. Ar (V )). Now it is not too difficult to show the following. Claim 26.4. Let n = dim V and let e1 , . . . , en be a basis for V . Then a basis for S r (V ) is given by S r (ei1 ⊗ · · · ⊗ eir ),

1 ≤ i1 ≤ i2 ≤ · · · ≤ in ≤ r,

and a basis for Ar (V ) is given by Ar (ei1 ⊗ · · · ⊗ eir ),

1 ≤ i1 < i2 < · · · < in ≤ r.

The dimension formulas immediately follow from the above claim. Let V and U be vector spaces. We say that an r-linear map F : V × ···× V → U is symmetric if F (v1 , . . . , vr ) = F (vσ(1) , . . . , vσ(r) )

166

26. Tensor product method

for all σ ∈ Sr , and alternating if F (v1 , . . . , vr ) = sgn(σ) F (vσ(1) , . . . , vσ(r) ) for all σ ∈ Sr . Then we have the following universality. Theorem 26.5. Let V and U be vector spaces and let F : V × · · · × V → U be an r-linear map. If F is symmetric, then there exists a unique linear map f : S r (V ) → U such that F = f ◦ S r ◦ Φ, where Φ is defined in Theorem 26.2. If F is alternating, then there exists a unique linear map g : Ar (V ) → U such that F = g ◦ Ar ◦ Φ. Exercise 26.6. Let F be an alternating r-linear map. (i) Show that F (. . . , vi , . . . , vj , . . .) = −F (. . . , vj , . . . , vi , . . .). (ii) Let span{u1 , . . . , ur } = span{v1 , . . . , vr }. Show that there is a real number λ = 0 such that F (u1 , . . . , ur ) = λF (v1 , . . . , vr ). 26.1.3. Symmetric algebra and exterior algebra. Let S 0 (V ) = R and let S(V ) = S 0 (V ) ⊕ S 1 (V ) ⊕ S 2 (V ) ⊕ · · · . The RHS is a set of formal sums v0 + v1 + · · · , where vi ∈ S i (V ) for i = 0, 1, . . . , and we assume that the number of i such that vi = 0 is    finite. As usual, for v, v  ∈ S(V ), where v = vi and v  = v , we   i define v + v  = (vi + vi ), and for α ∈ R we define αv = (αvi ). Then S(V ) is a (possibly infinite-dimensional) vector space. Let us define a product in S(V ). For s ∈ S i (V ) and s ∈ S j (V ) we define the product by s · s = S i+j (s ⊗ s ) ∈ S i+j (V ). We extend the definition, and for s, s ∈ S(V ), where s =  s = j sj , we define  s · s = si · sj . i,j



i si

and

26.2. Using symmetric products

167

The important feature of the product is that it is associative, that is, (s · s ) · s = s · (s · s ) for all s, s , s ∈ S(V ). We call S(V ) the symmetric algebra over V . Similarly we define the exterior algebra A(V ) as follows. Let A(V ) = A0 (V ) ⊕ A1 (V ) ⊕ A2 (V ) ⊕ · · · , where A0 (V ) = R. Then A(V ) is a vector space. Since Ar (V ) = {0} $n r for r > n := dim V , we may think of A(V ) as r=0 A (V ). For i  j a ∈ A (V ) and a ∈ A (V ) we define another product a ∧ a by a ∧ a = Ai+j (a ⊗ a ) ∈ Ai+j (V ).      For a general case with a = i ai and a = j aj , let a ∧ a =   i,j ai ∧ aj . This product is called the wedge product, and it is also associative. The next result explains how the wedge product reflects linear independence. Theorem 26.7. Let n = dim V , and let v1 , . . . , vn ∈ A1 (V ) = V . Then v1 ∧· · ·∧vn = 0 if and only if v1 , . . . , vn are linearly independent.

26.2. Using symmetric products As an application of symmetric products we prove the following result. Theorem 26.8. Let A1 , . . . , Am be r-element subsets and B1 , . . . , Bm be s-element subsets such that (i) Ai ∩ Bi = ∅ for 1 ≤ i ≤ m, (ii) Ai ∩ Bj = ∅ for 1 ≤ i < j ≤ m. Then m≤

  r+s . s

This result was first obtained by Bollob´as [13] under stronger conditions (i) and (ii ) Ai ∩ Bj = ∅ for i = j. Then Lov´ asz [88, 89] proved Theorem 26.8 using alternating products. We will discuss the proof in the next section. Here we give an alternative proof using symmetric products due to Frankl [46].

168

26. Tensor product method

  Proof. Let X = ( i Ai ) ∪ ( j Bj ). We may choose X ⊂ Rs+1 in general position, that is, any s + 1 points in X are linearly independent. One way to ensure this is to take X on the moment curve; for example, let X = {x1 , . . . , xn } where xi = (1, i, i2 , . . . , is ) ∈ Rs+1 . Exercise 26.9. Show that the n points in the X defined above are linearly independent. (Hint: Use Vandermonde’s determinant.) Let V be an (s + 1)-dimensional vector space consisting of all linear maps from Rs+1 to R. This is the dual space of Rs+1 , and we may identify v ∗ ∈ V and v ∈ Rs+1 such that v ∗ (x) = (v, x) for all x ∈ Rs+1 , where (v, x) denotes the standard inner product. Let U be a vector space consisting of all linear maps from (Rs+1 )r = Rs+1 × · · · × Rs+1 to R. We define F : V r → U by F (v1∗ , . . . , vr∗ ) =

1  ∗ ∗ vσ(1) · · · vσ(r) . r! σ∈Sr

This function acts on (x1 , . . . , xr ) ∈ (Rs+1 )r by 1  F (v1∗ , . . . , vr∗ )(x1 , . . . , xr ) = (vσ(1) , x1 ) · · · (vσ(r) , xr ). r! σ∈Sr

Note that F is symmetric by definition. For A = {a1 , . . . , ar } ∈ let FA = F (a∗1 , . . . , a∗r ) ∈ U .

X  r

Claim 26.10. FA1 , . . . , FAm are linearly independent. Proof. For each i with 1 ≤ i ≤ m let ui ∈ Rs+1 be a normal vector of the hyperplane (in Rs+1 ) containing the origin and s points in Bi . Then, by the condition of the theorem and our assumption that X is in general position, it follows that  = 0 if i = j, FAi (uj , . . . , uj ) = 0 if i < j. Consider a linear combination m  (26.2) ci FAi = 0. i=1

26.3. Using alternating products

169

Suppose that there is a nonzero scalar and let i0 be the index such that c1 = · · · = ci0 −1 = 0 but ci0 = 0. We substitute (ui0 , . . . , ui0 ) into (26.2) and get ci0 FAi0 (ui0 , . . . , ui0 ) = 0, which implies that  FAi0 (ui0 , . . . , ui0 ) = 0, a contradiction. By the claim we have m ≤ dim F (V r ). Since F is symmetric, it follows from Theorem 26.5 and Theorem 26.3 that dim F (V r ) = dim((f ◦ S r ◦ Φ)(V r )) = dim((f ◦ S r )(T r (V ))) = dim f (S r (V )) ≤ dim S r (V )     r + dim V − 1 r+s = = . r r   Thus we get m ≤ dim F (Vr ) ≤ r+s r , as required.



26.3. Using alternating products In this section we use alternating products to prove Theorem 26.8 and its easy consequence of a vector space version. These proofs are due to Lov´ asz [88, 89].   Proof of Theorem 26.8. Let X = ( i Ai ) ∪ ( j Bj ) and let V = Rr+s . We may choose X ⊂ V in general position, that is, any r+ s points in X are linearly independent. For F = {f1 , . . . , fk } ∈ Vk let vF = f1 ∧ · · · ∧ fk ∈ Ak (V ). Then, by the conditions (i), (ii) of Theorem 26.8 with Theorem 26.7, we have  = 0 if i = j, (26.3) vAi ∧ vBj = 0 if i < j. This shows that vA1 , . . . , vAm ∈ Ar (V ) are linearly independent. by Theorem 26.3.  Thus we get m ≤ dim Ar (V ) = r+s r We give a quick application of the above proof to families of vector spaces. Corollary 26.11. Let V be an (r + s)-dimensional vector space. Let A1 , . . . , Am be r-dimensional subspaces of V , and let B1 , . . . , Bm be

170

26. Tensor product method

s-dimensional subspaces of V . Suppose that (i) dim(Ai ∩ Bi ) = 0 for 1 ≤ i ≤ m, (ii) dim(Ai ∩ Bj ) = 0 for 1 ≤ i < j ≤ m.   r+s m≤ . s   Proof. For a subspace F ∈ Vk choose a basis f1 , . . . , fk , and let Then

vF = f1 ∧ · · · ∧ fk ∈ Ak (V ). Then we have (26.3) because vF is independent of the choice of basis except the scalar (see Exercise 26.6). So we can argue as in the previous proof.  The above proof works only when dim V = r + s, but we can drop this condition by general position arguments as we will see in the next section.

26.4. Extension by general position arguments F¨ uredi [70] extended Theorem 26.8 to the following threshold version. Theorem 26.12. Let r, s, t be integers with 0 ≤ t ≤ min{r, s}. Let A1 , . . . , Am be r-element subsets, and let B1 , . . . , Bm be s-element subsets. Suppose that (i) |Ai ∩ Bi | ≤ t for 1 ≤ i ≤ m, (ii) |Ai ∩ Bj | > t for 1 ≤ i < j ≤ m. Then

  r + s − 2t m≤ . s−t

We will deduce the above result from the following vector space version. Theorem 26.13. Let r, s, t be integers with 0 ≤ t ≤ min{r, s}. Let A1 , . . . , Am be r-dimensional subspaces of a vector space V , and let B1 , . . . , Bm be s-dimensional subspaces of V . Suppose that (i) dim(Ai ∩ Bi ) ≤ t for 1 ≤ i ≤ m,

26.4. Extension by general position arguments

171

(ii) dim(Ai ∩ Bj ) > t for 1 ≤ i < j ≤ m. Then m≤

  r + s − 2t . s−t

Proof of Theorem 26.12 via Theorem 26.13. Suppose that Ai and Bi (1 ≤ i ≤ m) satisfy the conditions (i) and (ii) of Theorem 26.12. Let   X = ( Ai ) ∪ ( Bj ). i

j

We may assume that X = [n]. We translate the situation in the setup of Theorem 26.13. So let V = Rn , and let v1 , . . . , vn be the basis. Define a map φ : X → V by φ(i) = vi . Let Ai be an r-dimensional subspace of V spanned by {φ(a) : a ∈ Ai }, and let Bi be an sdimensional subspace spanned by {φ(b) : b ∈ Bi }. Since the subsets Ai and Bi satisfy (i) and (ii) of Theorem 26.12, the subspaces Ai and Bi satisfy (i)  and (ii) of Theorem 26.13. Thus, by Theorem 26.13, we .  have m ≤ r+s−2t s−t F¨ uredi used so-called general position arguments to prove Theorem 26.13. The key is the following lemma. Lemma 26.14. Let n ≥ k and t = n − k. Let V be an n-dimensional vector space, and let C1 , . . . , Cm be subspaces of V with n > dim Ci for all i. Then there is a k-dimensional subspace V  such that dim(Ci ∩ V  ) = max{dim Ci − t, 0} for all 1 ≤ i ≤ m. The subspace V  guaranteed by the above lemma is said to be in general position with respect to C1 , . . . , Cm . Note that V  is chosen so that the dimension Ci ∩ V  is as small as possible for all i. Proof. Let C be a subspace of V , and let u1 , . . . , ur be a basis of C. Let x1 , . . . , xk be vectors in V which we will use as variables. We treat these r + k vectors as column vectors in Rn and let A be an n × (r + k) matrix consisting of u1 , . . . , ur , x1 , . . . , xk . Note that n ≤ r + k if r ≥ t and n > r + k if r < t. We show that there is a nonzero polynomial f in kn variables such that f (x1 , . . . , xk ) = 0 implies

172

26. Tensor product method

rank A = min{n, r+k}, that is, A has full rank. Here, by f (x1 , . . . , xk ) we understand f (x11 , . . . , x1n , . . . , xk1 , . . . , xkn ), where xi = (xi1 , . . . , xin ). If n ≤ r+k, then let f (x1 , . . . , xk ) be the determinant of the minor n × n matrix of A consisting of the first n columns. If n > r + k then there is a non-singular r × r minor matrix B in the first r columns, and let f be the determinant of any (r + k) × (r + k) minor matrix of A containing B. Note that in both cases f is not the zero polynomial and we can choose vectors v1 , . . . , vk so that f (v1 , . . . , vk ) = 0. Then det A = 0 and A has full rank. For 1 ≤ i ≤ k let fi be the polynomial corresponding to Ci defined as above, and let h(x1 , . . . , xk ) be the polynomial defined by the determinant of an n×n matrix consisting of n vectors e1 , . . . , et , x1 , . . . , xk , where ei denotes the ith standard basis vector of Rn . Note that if h = 0 then x1 , . . . , xk are linearly independent. Finally let g = f1 · · · fm h. Since g is not the zero polynomial, we can find vectors v1 , . . . , vk such that g(v1 , . . . , vk ) = 0. Let V  be the subspace spanned by these vectors. Then dim V  = k because h = 0. For 1 ≤ i ≤ m let Ai be the n × (dim Ci + k) matrix consisting of the basis of Ci (as column vectors) and v1 , . . . , vk . Then Ai has full rank because fi = 0, and we have dim(Ci + V  ) = rank Ai = min{n, dim Ci + k}. Thus it follows that dim(Ci ∩ V  ) = dim Ci + dim V  − dim(Ci + V  ) = max{dim Ci − t, 0}, as needed.



Proof of Theorem 26.13. We use Lemma 26.14 to modify the situation so that we can apply Corollary 26.11. Let n = dim V . By Lemma 26.14 we can find an n − t dimensional subspace V  in general position with respect to all Ai , Bj , and Ai ∩ Bj (1 ≤ i ≤ j ≤ m). Let n = n − t, r  = r − t, s = s − t, Ai = Ai ∩ V  , and Bi = Bi ∩ V  . Then dim Ai = r  , dim Bi = s , and dim(Ai ∩ Bj ) = max{dim(Ai ∩ Bj ) − t, 0}. In view of (i) and (ii) we get (i ) dim(Ai ∩ Bi ) = 0 for 1 ≤ i ≤ m,

26.4. Extension by general position arguments

173

(ii ) dim(Ai ∩ Bj ) > 0 for 1 ≤ i < j ≤ m. Once again, by Lemma 26.14 we can find an (n −(r  +s ))-dimensional subspace V  < V  in general position with respect to all Ai , Bj , and Ai ∩Bj . Then all Ai ∩V  , Bj ∩V  , and Ai ∩Bj ∩V  have dimension 0. Let W be the orthogonal subspace to V  in V  , and let Ai = Ai ∩ W   and Bj = Bj ∩ W . (Or one can take any linear map φ : V  → Rr +s with ker φ = V  and define Ai = φ(Ai ) and Bj = φ(Bj ).) Then we have dim W = r  + s , dim Ai = r  , dim Bj = s , and (i ) dim(Ai ∩ Bi ) = 0 for 1 ≤ i ≤ m, (ii ) dim(Ai ∩ Bj ) > 0 for 1 ≤ i < j ≤ m. Now we can apply Corollary 26.11 to Ai and Bj in W , and we get      r + s − 2t r + s = , m≤ s−t s which completes the proof of the theorem.



Chapter 27

The ratio bound

The independence number is one of the most important graph invariants. The ratio bound gives an upper bound for this invariant by using the eigenvalues of an adjacency matrix. In this chapter we explain the ratio bound and how to use it to prove the Erd˝ os–Ko–Rado Theorem.

27.1. Bounding the independence number by a positive semidefinite matrix For a graph G = (V (G), E(G)), we call a subset U ⊂ V (G) independent if no two vertices in U are adjacent. The maximum size of the independent subsets is called the independence number of G. We denote it by α(G), that is, α(G) = max{|U | : U ⊂ V (G) is an independent set}. We say that G is an n-vertex graph if |V (G)| = n. In this case we assume that V (G) = [n] unless stated otherwise. Suppose that A is an n×n real symmetric matrix. We say that A is positive semidefinite (PSD) if uT Au ≥ 0 for all u ∈ Rn . Exercise 27.1. Show that a real symmetric matrix A is PSD if and only if every eigenvalue of A is nonnegative. 175

176

27. The ratio bound

We can bound the independence number of a given graph by finding a PSD matrix as in the following lemma, and this is a key idea for the ratio bound. Lemma 27.2. Let G be an n-vertex graph, and let A = (ai,j ) be its adjacency matrix. Suppose that 1 A+I − J c is PSD for some c > 0. Then α(G) ≤ c. Proof. Let u ∈ {0, 1}n be the characteristic vector of a maximum independent set U . If {i, j} ∈ E(G) then ai,j = 0, and if {i, j} ∈ E(G) then ui uj = 0 because U is independent. Thus we have  (27.1) uT Au = ai,j ui uj = 0. i,j

We also have uT Iu = α, and uT Ju = α2 . Using the PSD property we have   1 T u A + I − J u ≥ 0, c and rearranging this gives α ≤ c.



27.2. Pam and the ratio bound In the proof of Lemma 27.2 we used ai,j = 0 if {i, j} ∈ E(G) in (27.1), but we did not use the information on ai,j for {i, j} ∈ E(G) at all. This motivates us to define a pseudo-adjacency matrix (pam). Definition. An n×n real symmetric matrix A is a pseudo-adjacency matrix for an n-vertex graph G if (i) ai,j = 0 whenever {i, j} ∈ E(G), (ii) A has a positive constant row sum, that is, A1 = λ1 1 for some λ1 > 0.

27.2. Pam and the ratio bound

177

Let A be a pam with A1 = λ1 1. Since A is a real symmetric matrix, it can be diagonalized and all the eigenvalues are real numbers. Moreover, we may choose the eigenvectors to be pairwise perpendicular with respect to the standard inner product. In particular, if v is a eigenvector with v = 1, then we may assume that Jv = 0. Let λ2 be the minimum eigenvalue. Recall that the sum of the eigenvalues is equal to the trace of A, which is 0. Since λ1 > 0 we have λ2 < 0. Claim 27.3. Let A be a pam for an n-vertex graph with A1 = λ1 1. Let λ2 be the minimum eigenvalue. Then the matrix 1 1 A+I − J M= −λ2 c is PSD, where −λ2 c= n. λ1 − λ2 Proof. We show that M and A have the same eigenvectors and that all eigenvalues of M are nonnegative. On the one hand,   n λ1 M1 = − + 1 − 1 = 01, λ2 c that is, 0 is an eigenvalue of M with eigenvector 1. On the other hand, if Av = λv with v = 1, then Jv = 0 and λ ≥ λ2 . Thus   λ M v = − + 1 v, λ2 that is, − λλ2 + 1 ≥ 0 is an eigenvalue of M corresponding to the eigenvector v.  We note that Lemma 27.2 still holds if we replace the adjacency 1 A, where A is matrix with any pam. Apply the lemma to a pam −λ 2 taken from Claim 27.3. Then we obtain the following result, which is due to Hoffman [77] and Delsarte [21]. Theorem 27.4 (The ratio bound). Let G be an n-vertex graph, let A be a pam for G with A1 = λ1 1, and let λ2 be the minimum eigenvalue. Then −λ2 α(G) ≤ n. λ1 − λ2

178

27. The ratio bound

Exercise 27.5. Let G be an n-vertex λ1 -regular graph, and let A be the (usual) adjacency matrix with minimum eigenvalue λ2 . Let u be the characteristic vector of a maximum independent set U ⊂ V (G) with |U | = α. Let B = A − λ2 I be a PSD matrix, and let w = u− α n 1. Deduce the ratio bound by expanding and rearranging w T Bw ≥ 0. This approach is due to Godsil and Newman; see [95] for more extensions.

27.3. Application Let us define a Kneser graph for t-intersecting families G(n, k, t) =   (V, E) by V = [n] and {u, v} ∈ E if and only if |u ∩ v| < t. Recall k that U ⊂ V is a t-intersecting family if |u ∩ v| ≥ t for all u, v ∈ U . Exercise 27.6. Show that in G(n, k, t), U ⊂ V is a t-intersecting family if and only if U is an independent set. Let us consider the case t = 1. Let n ≥ 2k and let A be the adjacency matrix of the Kneser graph G(n, k, 1). Exercise 27.7. Show that the set of eigenvalues of A is given by  

i n−k−i : i = 0, 1, . . . , k (−1) k−i n  n  with corresponding multiplicities i − i−1 . (See, e.g., [73] §6.5.) N -vertex λ1 -regular graph with N = n Note that  n−kG(n, k, 1) is an n−k−1 , λ , and λ , where A1 = λ1 1 and λ2 is the = = − 1 2 k k k−1 minimum eigenvalue. Then it follows from Theorem 27.4 that   n−1 −λ2 α(G(n, k, 1)) ≤ N= . k−1 λ1 − λ2 So this gives us an alternative proof of the Erd˝os–Ko–Rado Theorem for the 1-intersecting case. It is much more difficult to deal with the t-intersecting case for t ≥ 2, but Wilson succeeded in finding a suitable pam for G(n, k, t) in general, as we will briefly explain below. (For more details, see the original paper [115] or Chapter 8 in [73].) Let n > (t + 1)(k − t + 1). Then the Erd˝ os–Ko–Rado Theorem   n−t states that k−t is the maximal size of t-intersecting families in [n] k .

27.3. Application

179

To show this, it suffices to find a pam A for the N -vertex λ1 -regular   −1 Kneser graph G = G(n, k, t) with λ1 = nk n−t − 1 and λ2 = −1, k−t n where N = k , A1 = λ1 1, and λ2 is the minimum eigenvalue. Then the ratio bound yields   n−t −λ2 N= (27.2) α(G) ≤ , λ1 − λ2 k−t as needed. The pam A is defined as follows:   −1 t−1  k−1−i n−k−t+i (−1)t−1−i Bk−i , A := k−t k−t i=0   [n]     whose where Bk−i is an nk × nk matrix indexed by [n] k × k  (F, F )-entry is given by  

   [n] |F \ F | # J∈ : J ∩ F = ∅ and J ⊂ F  = . k−i k−i ¯ T ¯ Equivalently n n we can write Bj = (Wjk ) Wjk , where Wjk (resp. Wjk ) is an j × k matrix whose (J, K)-entry is 1 if J ⊂ K (resp. J ∩K = ∅) and 0 otherwise. Let Uj be the row space of Wjk . Then we have U0 ⊂ U1 ⊂ · · · ⊂ Uk , and Uk is the entire vector space E spanned by all characteristic vectors of the k-subsets of [n]. This follows from   W Wij Wjk = k−i ik for i ≤ j ≤ k. Let V0 = U0 , and for j ≥ 1 let Vj j−i be the orthogonal complement of Uj−1 in Uj , so that Uj = Vj ⊕ Uj−1 . Then we have an orthogonal decomposition E = V0 ⊕V1 ⊕· · ·⊕Vk and the corresponding eigenvalues θ0 , θ1 , . . . , θk of the matrix A such that Av = θj v for all v ∈ Vj . Moreover, some (not so easy) computation shows the following. Lemma 27.8 ([115]). Let n > (t + 1)(k − t + 1). Then we have   −1 θ0 = nk n−t − 1, θ1 = θ2 = · · · = θt = −1, 0 < θt+1 < θ0 , and k−t 1 > |θt+2 | > |θt+3 | > · · · > |θk |.   −1 This means that λ1 = θ0 = nk n−t − 1 and λ2 = θ1 = −1, from k−t which the required ratio bound (27.2) follows.

Chapter 28

Measures of cross independent sets

In this chapter we will extend the ratio bound in two ways. First we consider the measure of an independent set instead of an independence number, and second we consider cross independent sets in a bipartite graph, which is a natural extension of an independent set in a graph. These extensions enable us to apply the method to a much broader scope of problems.

28.1. Bounding measures using singular values Let V be a finite set. Recall that μ : 2V → [0, 1] is a probability measure on V if μ satisfies μ(∅) = 0, μ(V ) = 1, and μ(U ) =



μ({u})

u∈U

for U ⊂ V . With an abuse of notation we write μ(v) for v ∈ V to mean μ({v}). This measure μ induces an inner product on the set of functions from V to R, denoted by RV , as follows. For f, g ∈ RV , define  f (v)g(v)μ(v). !f, g"μ = v∈V

181

182

28. Measures of cross independent sets

Let G be a bipartite graph with a vertex partition V (G) = V1 V2 . We say that a |V1 | × |V2 | real matrix A = (au,v ) is a bipartite pseudoadjacency matrix (bpam) if it is indexed by V1 and V2 , with a constant row sum (but column sums can differ), and au,v = 0 whenever u ∈ V1 and v ∈ V2 are not adjacent. For example, one can always obtain a bpam by setting au,v = 1/ degG (u) for every adjacent pair {u, v}. We say that U1 ⊂ V1 and U2 ⊂ V2 are cross independent if there is no edge between U1 and U2 in G. Exercise 28.1 (Singular decomposition). Let A be an m × n real matrix, let k = min{m, n}, and let λ1 , λ2 , . . . , λk (and m − k 0’s if m > k) be the eigenvalues of AAT . Show that A can be decomposed into A = P DQT , where P is an m × m orthogonal matrix (with respect to the standard inner product), D is an m × n matrix with diagonal entries √ √ λ1 , . . . , λk (and 0 elsewhere), and Q is an n × n orthogonal matrix. (See, e.g., Theorem 2.6.3 in [78].) Let G be a bipartite graph with a vertex partition V1  V2 , and let A be a bpam of G. For i = 1, 2, fix a measure μi on Vi , which induces the inner product !·, ·"μi on RVi . Let m = |V1 |, n = |V2 |, and k = min{m, n}. We say that nonnegative real numbers σ1 , . . . , σk are the singular values of (μ1 , μ2 , A) if there exist an orthonormal1 basis u1 , . . . , um of RV1 and an orthonormal basis v1 , . . . , vn of RV2 such that (S1) u1 = 1m (the all-ones vector in Rm ) and v1 = 1n , (S2) σ2 = max{σ2 σ3 , . . . , σk } ≥ 0,  σi ui for 1 ≤ i ≤ k, (S3) Avi = 0 for k < i ≤ n. Now we present an extension of the ratio bound, which is found in [29], where the authors attribute one of the origins of the result to [8]. See also [28, 110, 111]. 1

Here, by orthonormal we mean ui , uj μ1 = δij and vi , vj μ2 = δij .

28.2. Proof of the theorem

183

Theorem 28.2. Let G be a bipartite graph with a vertex partition V1  V2 . Let A be a bpam, and let μi be a measure on Vi for i = 1, 2. Suppose that U1 ⊂ V1 and U2 ⊂ V2 are cross independent. Then  σ2 (28.1) μ1 (U1 )μ2 (U2 ) ≤ , σ1 + σ2 where σ1 and σ2 are the two singular values of (μ1 , μ2 , A) satisfying (S1)–(S3).

28.2. Proof of the theorem For the proof we need the following technical lemma. Lemma 28.3. Let k be a positive integer. Let x1 , . . . , xk , y1 , . . . , yk , and λ1 , . . . , λk be real numbers satisfying k k (L1) x1 ≥ i=1 x2i and y1 ≥ i=1 yi2 , (L2) |λ2 | = max{|λ2 |, |λ3 |, . . . , |λk |} ≥ 0, k (L3) i=1 λi xi yi = 0. Then |λ2 | . λ1 + |λ2 |

√ x1 y1 ≤

Proof. We estimate the LHS of (L3) in terms of λ1 , |λ2 | and x1 y1 . We start with (28.2)

k 

λi xi yi = λ1 x1 y1 +

i=1

k 

λi xi yi ≥ λ1 x1 y1 − |λ2 |

i=2

i=2

where we used (L2) in the last inequality. Then

(28.3)

k  i=2

(28.4)

 |xi ||yi | ≤

k  i=2

k 

 12  x2i

k  i=2

 12 yi2

% % ≤ x1 − x21 y1 − y12  √ √ = x1 y1 1 − x1 1 − y1 ,

|xi ||yi |,

184

28. Measures of cross independent sets

where we used the Cauchy–Schwarz inequality for (28.3) and (L1) for (28.4). Using the AM-GM inequality twice, we get √

(28.5)

1 − x1

 (1 − x1 ) + (1 − x2 ) 1 − y1 ≤ 2 x1 + y1 √ =1− ≤ 1 − x1 y1 . 2

Consequently we have k 

√ √ λi xi yi ≥ λ1 x1 y1 − |λ2 | x1 y1 (1 − x1 y1 ).

i=1

This, together with (L3), gives us the desired inequality.



Proof of Theorem 28.2. Let x = (x(u) : u ∈ V1 ) ∈ {0, 1}|V1 | be the characteristic vector of U1 , that is, x(u) = 1 if u ∈ U and x(u) = 0 m if u ∈ U . We expand x using the basis: x = i=1 xi ui , where these basis vectors satisfy (S1), (S2), and (S3). Then we have ' &m m m    xi ui , xi ui = x2i μ1 (U1 ) = !x, x"μ1 = i=1

and μ1 (U1 ) = !x, 1m "μ1 =

i=1

&m 

μ1

i=1

' xi ui , u1

i=1

= x1 . μ1

Thus we have (28.6)

μ1 (U1 ) =

m 

x2i = x1 .

i=1

In the same way, the characteristic vector y = isfies (28.7)

μ2 (U2 ) =

n 

n

j=1 yj vj

of U2 sat-

yi2 = y1 .

j=1

So (L1) in Lemma 28.3 holds with equality. Set λi := σi for 1 ≤ i ≤ k. Then (S2) implies (L2). Next we verify (L3). To this end we compute !x, Ay"μ1 in two ways. On the one hand, by the definition of the inner product, we

28.2. Proof of the theorem have !x, Ay"μ1 =



185

x(u)(Ay)(u)μ1 (u)

u∈V1

=



 x(u)

u∈V1

=



 au,v y(v) μ1 (u)

v∈V2

 

au,v x(u)y(v)μ1 (u).

u∈V1 v∈V2

If u and v are not adjacent, then au,v = 0 by the definition of A. If u and v are adjacent, then x(u)y(v) = 0 by the fact that U1 and U2 are cross independent. This means that au,v x(u)y(v) = 0 for all u and v. Thus we get !x, Ay"μ1 = 0.

(28.8)

On the other hand, it follows that ⎛ ⎞ n n k    Ay = A ⎝ yj vj ⎠ = yj (Avj ) = σj yj uj , j=1

j=1

j=1

where we used (S3) in the last equality, and ' &m k   (28.9) !x, Ay"μ1 = xi ui , σj yj uj i=1

j=1

= μ1

k 

σi xi yi .

i=1

By (28.8) and (28.9) we have k 

σi xi yi = 0,

i=1

that is, (L3) holds. Now we can apply Lemma 28.3 and get σ2 √ x1 y1 ≤ . σ1 + σ2  Moreover, the LHS is μ1 (U1 )μ2 (U2 ) by (28.6) and (28.7), and we are done.  Exercise 28.4. Suppose that equality holds in (28.1). Show that x (resp. y) is contained in a subspace spanned by u1 and u2 (resp. v1 and v2 ), where we are using the notation in the proof of Theorem 28.2.

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28. Measures of cross independent sets

(Hint: For the equality we need equalities in (28.5) and (28.3), which imply x1 = y1 and |xi | = |yi | for 2 ≤ i ≤ k. Then we also need equality in (28.2), which yields σi = σ2 or xi = yi = 0 for all 3 ≤ i ≤ k; cf. [79, 110].)

28.3. Application As the first application of Theorem 28.2 let us consider the bipartite version of the ratio bound. Let G be a bipartite graph with a vertex partition V1  V2 , where |V1 | = m and |V2 | = n. Suppose that G is (d1 , d2 )-biregular , that is, all vertices in V1 have the same degree d1 and all vertices in V2 have the same degree d2 (possibly d1 = d2 ). By counting the number of edges we get d1 m = d2 n. Let A = (au,v ) ∈ {0, 1}m×n be the usual bipartite adjacency matrix, that is, au,v = 1 if u ∈ V1 and v ∈ V2 are adjacent and au,v = 0 otherwise. Then A1n = d1 1m , AT 1m = d2 1n , and thus AAT 1m = A(d2 1n ) = (d1 d2 )1m , which means that d1 d2 is one of the eigenvalues of AAT . Now let ˜D ˜ V˜ T A=U ˜ i (resp. v˜i ) be be a singular decomposition from Exercise 28.1. Let u ˜ (resp. V˜ ). Then the decomposition above the ith column vector of U √ ˜ i for 1 ≤ i ≤ k and A˜ vi = 0 for k < i ≤ n, means that A˜ vi = λ i u where k = min{m, n} and λ1 , . . . , λk are the eigenvalues of AAT . Moreover, we may assume that λ1 = d1 d2 ,

1 ˜ 1 = √ 1m , u m

1 v˜1 = √ 1n , n

and λ2 = max{λ2 , λ3 , . . . , λk }. √ √ ˜ i for 1 ≤ i ≤ m and vj = n˜ vj for 1 ≤ j ≤ n. Next let ui = mu Then u1 , . . . , um (resp. v1 , . . . , vn ) is an orthogonal basis of RV1 (resp.

28.3. Application

187

RV2 ) with respect to the uniform measure μ1 (u) = (resp. μ2 (v) = n1 for all v ∈ V2 ). Finally, set , n λi σi = m

1 m

for all u ∈ V1

for 1 ≤ i ≤ k. Then all the conditions (S1), (S2), and (S3) are satisfied, and σ1 , . . . , σk are the singular values of (μ1 , μ2 , A). Therefore by Theorem 28.2 we have √ |U1 ||U2 | λ2 σ2 √ . = μ1 (U1 )μ2 (U2 ) ≤ = √ |V1 ||V2 | σ1 + σ2 λ1 + λ2 In summary we obtain the following. Corollary 28.5. Let G be a (d1 , d2 )-biregular bipartite graph with a vertex partition V1  V2 , and let A be the bipartite adjacency matrix. If U1 ⊂ V1 and U2 ⊂ V2 are cross independent, then √ |U1 ||U2 | λ2 √ , ≤√ |V1 ||V2 | λ1 + λ2 where λ1 = d1 d2 and λ2 is the maximum eigenvalue of AAT except λ1 . Exercise 28.6. Replace the condition (L2) in Lemma 28.3 with (L2 ) λ2 = min{λ2 , λ3 , . . . , λk } ≤ 0. Show that (L1), (L2 ) and (L3) imply −λ2 . λ1 − λ2 Use this inequality to deduce the ratio bound. √ x1 y1 ≤

Finally we mention two more applications of Theorem 28.2. Let 0 < p < 1 be fixed. Recall that the product measure μp on 2[n] is defined by  μp (U ) = p|u| (1 − p)n−|u| u∈U

for U ⊂ 2 . Let G be a bipartite graph with a vertex partition V1  V2 , where V1 and V2 are distinct copies of 2[n] , and u ∈ V1 and v ∈ V2 are adjacent if and only if |u ∩ v| < t. Then two families U1 , U2 ⊂ 2[n] are cross t-intersecting if and only if U1 ⊂ V1 and [n]

188

28. Measures of cross independent sets

U2 ⊂ V2 are cross independent in G. One can find a bpam A of G such that the singular values for (μp , μp , A) satisfy σ1 = p−t − 1 and σ2 = 1 provided p < 1 − √t12 . Thus Theorem 28.2 implies the following. Theorem 28.7 ([111]). Let n ≥ t ≥ 1 be integers, p ∈ (0, 1 − 1 [n] √ . If U1 and U2 are cross t-intersecting, then t ), and U1 , U2 ⊂ 2 2 μp (U1 )μp (U2 ) ≤ p2t with equality holding if and only if U1 = U2 =   {F ⊂ [n] : T ⊂ F } for some T ∈ [n] t .   Let V be an n-dimensional vector space over Fq . Recall that Vk V  n denotes the set of k-dimensional subspaces of V , and k = | k |.   Two families of subspaces F1 , F2 ⊂ Vk are called cross t-intersecting if dim(F1 ∩ F2 ) ≥ t for all F1 ∈ F1 and F2 ∈ F2 . In this case we consider a bipartite graph G with a vertex partition V1  V2 , where V1 and V2 are distinct copies of Vk , and F1 ∈ V1 and F2 ∈ V2 are adjacent if and only if dim(F1 ∩ F2 ) < t. Then one can find a bpam with optimal singular values, which yields the following. Theorem 28.8 ([110]). Let n ≥ k ≥ t ≥ 1 be integers,  V be an V let n-dimensional vector space over Fq , and let F1 , F2 ⊂ k . Suppose that F1 and F2 are cross t-intersecting; then we have the following.  2 (i) If n > 2k then |F1 ||F2 | ≤ n−t equality holding if k−t  with  and only if F1 = F2 = {F ∈ Vk : T < F } for some   T ∈ Vt , where T < F means that T is a subspace of F . 2  (ii) If 2k − t < n < 2k then |F1 ||F2 | ≤ 2k−t with equality k  V  W  . holding if and only if F1 = F2 = k for some W ∈ 2k−t n−t2 2k−t2 (iii) If n = 2k then |F1 ||F2 | ≤ k−t = k with equality   holding if and only if either F1 = F2 = {F ∈ Vk : T < F }    V    for some T ∈ Vt or F1 = F2 = W k for some W ∈ 2k−t . These results use the spectral information of the corresponding bipartite graphs from [66, 68].

Chapter 29

Application of semidefinite programming

Semidefinite programming (SDP) is a strong tool for some combinatorial problems1 (and of course for many other theoretical and practical problems). As an illustrative example we use SDP to bound the measures of cross independent sets in a bipartite graph, which yields Erd˝os–Ko–Rado-type inequalities for cross intersecting families. The material in this chapter is mainly taken from [105].

29.1. Weak duality in SDP Since SDP is an extension of linear programming (LP), we briefly recall LP and one of its basic properties called weak duality. The LP problem in primal form is (P): minimize cT x, subject to Ax = b, x ≥ 0, 1 For example the Lov´ asz theta function is efficiently computable by solving a corresponding SDP problem; see, e.g., the excellent exposition in [72] (Chapter 3). See also M28 and M29 in [92] for a related topic on Shannon capacity.

189

190

29. Application of semidefinite programming

where A ∈ Rm×n , b ∈ Rm , and c ∈ Rn are given, and x ∈ Rn is the variable. By x ≥ 0 we mean that x is nonnegative, that is, every entry of x is nonnegative. We say that x is feasible in (P) if x satisfies the constraints Ax = b and x ≥ 0. The corresponding dual form is (D): maximize bT y, subject to y T A ≤ cT , where y ∈ Rn is the variable. These two problems are closely related; in particular, we have the following easy but useful fact. Proposition 29.1 (Weak duality for LP). If x is feasible in (P) and y is feasible in (D), then cT x ≥ bT y. Proof. Indeed, cT x ≥ (y T A)x = y T (Ax) = y T b = bT y.



We can use this weak duality in the following way. Suppose that x is feasible in (P) and y is feasible in (D). Suppose, moreover, that x and y happen to satisfy cT x = bT y. Then, by the weak duality, we see that both x and y are optimal. In applications it often happens that one easily finds a feasible solution x for (P) which is a candidate for an optimal solution. To prove that x is in fact optimal, it suffices to find a feasible solution y for (D) satisfying cT x = bT y. Although finding such y is usually more difficult, it is indeed possible in many interesting cases. Now we consider SDP. One of the big differences between LP and SDP is the space of variables. In LP variables are taken from real numbers, whereas in SDP variables are taken from the set of real symmetric matrices of a fixed order, say n, denoted by SRn×n . For two matrices A = (ai,j ) and B = (bi,j ) in SRn×n , the inner product is denoted by A • B. More precisely, let  ai,j bi,j = trace(AT B). A • B := i,j

We write A # 0 (resp. A $ 0) to mean that A is symmetric and positive semidefinite (resp. positive definite). The SDP problem in primal form is

29.2. SDP for measures of cross independent sets

191

(P): minimize C • X, subject to Ai • X = bi , i = 1, 2, . . . , m, X # 0, where Ai ∈ SRn×n , b ∈ Rm , and C ∈ SRn×n are given, and X ∈ SRn×n is the variable. The corresponding dual form is (D): maximize bT y, m subject to i=1 yi Ai + S = C,

S # 0,

where y ∈ Rm and S ∈ SRn×n are the variables. Just as in LP we have the following weak duality in SDP. Proposition 29.2 (Weak duality for SDP). If X is feasible in (P) and (y, S) is feasible in (D), then C • X ≥ bT y. Exercise 29.3. Prove Proposition 29.2. See, for example, [72, 107] for more about semidefinite programming in general.

29.2. An SDP problem for measures of cross independent sets In this section, we consider an SDP problem corresponding to the problem of bounding the measures of cross independent sets in a bipartite graph. Let Ω1 and Ω2 be finite sets, and let Ω = Ω1  Ω2 . Let G be a bipartite graph with a vertex partition V (G) = Ω = Ω1  Ω2 . For two vertices x, y ∈ Ω, we write x ∼ y if and only if x and y are adjacent. For i = 1, 2, let μi be a probability measure on Ωi . Let RΩ×Ω be the set of real matrices with rows and columns indexed by Ω, and let RΩ be the set of real column vectors with coordinates indexed by Ω. The sets RΩi ×Ωj and RΩi are similarly defined. Let Ji,j ∈ RΩi ×Ωj be the all-ones matrix. For x ∈ Ωi and y ∈ Ωj , let Ex,y ∈ RΩi ×Ωj be the matrix with a 1 in the (x, y)-entry and 0 elsewhere. Let Δi ∈ RΩi ×Ωi be the diagonal matrix whose (x, x)-entry is μi ({x}) for x ∈ Ωi . Let SRΩ×Ω be the set of symmetric matrices in RΩ×Ω .

192

29. Application of semidefinite programming

Now we consider the following SDP problem in primal form: . / 1 0 Δ1 J1,2 Δ2 2 (P): maximize • X, 1 Δ J Δ1 0 / . / . 2 2 2,1 0 0 Δ1 0 •X = • X = 1, subject to 0 0 0 Δ2 . / 0 Ex,y • X = 0 for x ∈ Ω1 , y ∈ Ω2 , x ∼ y, Ey,x 0 X # 0, X ≥ 0, where X ∈ SRΩ×Ω is the variable. This problem is set up so that a feasible solution comes from cross independent sets U1 ⊂ Ω1 and U2 ⊂ Ω2 in G. Let xi ∈ RΩi be the characteristic vector of Ui , and let ⎤⎡ ⎤T ⎡ √ 1 x1 √ 1 x1 μ (U ) μ (U ) 1 1 1 1 ⎦⎣ ⎦ ∈ SRΩ×Ω . XU1 ,U2 = ⎣ √ 1 x2 √ 1 x2 μ2 (U2 )

μ2 (U2 )

Exercise 29.4. Verify that XU1 ,U2 is a feasible solution to (P) with objective value . / 1  0 2 Δ1 J1,2 Δ2 • X μ1 (U1 )μ2 (U2 ). U1 ,U2 = 1 0 2 Δ2 J2,1 Δ1 The corresponding problem in dual form is (29.1) (D): minimize α + β, . / αΔ1 − 12 Δ1 J1,2 Δ2 subject to S := − 12 Δ2 J2,1 Δ1 βΔ2 . /  0 Ex,y + γx,y − Z # 0, Ey,x 0 x∼y Z ≥ 0, where α, β, γx,y ∈ R and Z ∈ SRΩ×Ω are the variables, and the sum is over x ∈ Ω1 and y ∈ Ω2 with x ∼ y. Exercise 29.5. Verify that Δi • (xi xT i ) = μi (Ui ) for i = 1, 2, and T that Ex,y • (x1 x2 ) = 0 for x ∼ y. The problems (P) and (D) above are slightly different from the standard form in Proposition 29.2, but we can directly verify the weak

29.2. SDP for measures of cross independent sets

193

duality as follows. Suppose that X is feasible in (P) and (α, β, γx,y , Z) is feasible in (D). Then it follows that . / 1 Δ1 J1,2 Δ2 0 2 α+β− 1 •X 0 2 Δ2 J2,1 Δ1 / . − 12 Δ1 J1,2 Δ2 αΔ1 •X = − 12 Δ2 J2,1 Δ1 βΔ2   . /  0 Ex,y ≥ − γx,y +Z •X Ey,x 0 x∼y = Z • X ≥ 0. Thus α + β gives an upper bound for marize what we have proved.

 μ1 (U1 )μ2 (U2 ). Let us sum-

Theorem 29.6 ([105]). Let G be a bipartite graph with a vertex partition V (G) = Ω1  Ω2 , and let μi be a probability measure on Ωi for i = 1, 2. Suppose that U1 ⊂ Ω1 and U2 ⊂ Ω2 are cross independent in G. If (α, β, γx,y , Z) is feasible in (D), then μ1 (U1 )μ2 (U2 ) ≤ (α + β)2 . Suda and Tanaka proved the vector space version of the Erd˝os– Ko–Rado Theorem for cross intersecting families as stated below. Theorem 29.7 ([104]). Let V be an n-dimensional vector space over   Fq . For i = 1, 2, let n ≥ 2ki and Ui ⊂ kVi . If dim(x ∩ y) ≥ 1 for all x ∈ U1 and y ∈ U2 , then . /. / n−1 n−1 |U1 ||U2 | ≤ . k1 − 1 k2 − 1 To prove Theorem 29.7 one can use Theorem 29.6 by setting  V Ωi = ki and μi (U ) = |U |/|Ωi |. Suda and Tanaka succeeded in  n−1   finding a feasible solution in (D) such that (α + β)2 = kn−1 k2 −1 . 1 −1 We will give another application of Theorem 29.6 in the next section.

Chapter 30

A cross intersection problem with measures

This chapter is a continuation of the previous chapter. We use Theorem 29.6 to solve a cross intersection problem with measures. We actually construct a feasible optimal solution to the dual problem. The idea is simple, but the computation is somewhat involved. We mention that this SDP approach is the only known way to prove the result so far.

30.1. The theorem of FFFLO and its extension to two families Let us define a probability measure μ on Ω = 2[n] with respect to a probability vector p = (p(1) , p(2) , . . . , p(n) ) ∈ (0, 1)n as follows: For U ⊂ Ω let   p(l) (1 − p(k) ). μ(U ) = x∈U l∈x

k∈[n]\x

This measure was introduced by Fishburn, Frankl, Freed, Lagarias, and Odlyzko in [40], where they considered the maximum measure for intersecting families and obtained Theorem 12.7. We note that  if p(1) = · · · = p(n) then μ(U ) = x∈U p|x| (1 − p)n−|x| , and this is the product measure. In this chapter we present an SDP approach to 195

196

30. A cross intersection problem with measures

proving the following result, which is an extension of Theorem 12.7 to cross intersecting families. Theorem 30.1 ([105]). For i = 1, 2, let μi be a probability measure (1) (n) on Ωi = 2[n] with respect to a probability vector pi = (pi , . . . , pi ), and let Ui ⊂ Ωi . Suppose that U1 and U2 are cross intersecting. Then we have the following. (1)

(l)

(i) If pi = max{pi : 1 ≤ l ≤ n} < 1/2 for i = 1, 2, then (1) (1) μ1 (U1 )μ2 (U2 ) ≤ p1 p2 . (1) (1)

(l) (l)

(l)

(ii) If p1 p2 = max{p1 p2 : 1 ≤ l ≤ n} and pi ≤ 1/3 for (1) (1) all i = 1, 2 and 1 ≤ l ≤ n, then μ1 (U1 )μ2 (U2 ) ≤ p1 p2 . Borg considered a problem concerning cross intersecting integer sequences in [14] and obtained similar results to (i) of Theorem 30.1 using a shifting technique. On the other hand, we note that in (ii), (1) (l) pi is not necessarily the maximum of pi . This is a less convenient situation for shifting, but the SDP approach works well even in such settings.

30.2. Proof of the theorem The proof goes as follows. We translate our problem into the setting of Theorem 29.6 and then we construct a feasible optimal solution to the dual problem (D). For the construction we follow Friedgut [68] and first consider the case n = 1 carefully, and then we extend it to the general case using a tensor product construction. (1)

30.2.1. Setup. To simplify notation we write pi instead of pi . Let ( ) ( ) qi = 1 − pi for i = 1, 2 and  ∈ [n]. We also write qi instead of (1) qi . By symmetry we may assume p1 ≥ p2 . Let G be the bipartite Kneser graph. That is, V (G) = Ω1  Ω2 with Ωi = 2[n] (i = 1, 2), and x ∈ Ω1 and y ∈ Ω2 are adjacent if and only if x ∩ y = ∅. We apply Theorem 29.6 to this graph by finding a √ feasible solution to (D) with α = β = p1 p2 /2, which would yield (30.1)

μ1 (U1 )μ2 (U2 ) ≤ p1 p2 .

30.2. Proof of the theorem

197

30.2.2. Proof of (i). The case n = 1. We prove (30.1) by constructing a feasible solution, which looks more complicated than we actually need to deal with this trivial case. The point is that the construction can be easily extended to the general case n ≥ 2 later.  Let ci = pi /qi , and define 0 1 . / p p 1 − qji qji 1 0 Ai,j = , , Di,j = 0 −ci cj 1 0 . . / / 1 ci qi 0 Vi = = , Δ , i 1 − c1i 0 pi where the rows and columns are indexed in the order ∅, {1}. A direct computation shows that (30.2)

Ai,j Vj = Vi Di,j ,

ViT Δi Vi = I,

(Δi Ai,j )T = Δj Aj,i .

Using the first two equalities, we get ViT (Δi Ai,j )Vj = Di,j .

(30.3)

To construct a feasible solution to (D) in (29.1), let us find a PSD matrix S and a symmetric matrix Z with Z ≥ 0. To this end, let θ= and let (30.4) α = β = θ,

 x∼y

√ p1 p2 /2

. γx,y Ex,y = ηΔ1 A1,2 ,

Z=

1 Δ1 A1,1 0

0 2 Δ2 A2,2

/

where η, 1 , 2 ∈ R. By definition Z is symmetric. Also Z ≥ 0 is equivalent to i (1 − 2pi ) ≥ 0 for i = 1, 2. Since pi < 1/2, we need i ≥ 0 for i = 1, 2. By (29.1) we define a symmetric matrix . / θΔ1 − 1 Δ1 A1,1 ηΔ1 A1,2 − 12 Δ1 J1,2 Δ2 (30.5) S= . ηΔ2 A2,1 − 12 Δ2 J2,1 Δ1 θΔ2 − 2 Δ2 A2,2 We need to choose 1 ≥ 0, 2 ≥ 0, and η ∈ R so that S # 0.

198

30. A cross intersection problem with measures

By (30.2) and (30.3), we have ⎡ θ − 1 / . / . T ⎢ 0 0 V V1 0 S 1 =⎢ ⎣η − 1 0 V2T 0 V2 2 0 Thus, S # 0 if and only if . / θ − 1 η − 12 # 0, η − 12 θ − 2

.

0 θ + 1 c21 0 −ηc1 c2

θ + 1 c21 −ηc1 c2

η − 12 0 θ − 2 0

⎤ 0 −ηc1 c2 ⎥ ⎥. 0 ⎦ θ + 2 c22

/ −ηc1 c2 # 0, θ + 2 c22

if and only if (30.6)

0 ≤ 1 , 2 ≤ θ,

 2 1 (θ − 1 ) (θ − 2 ) ≥ η − , 2     θ + 1 c21 θ + 2 c22 ≥ η 2 c21 c22 .

By the above three conditions we obtain a one-parameter family of solutions p2 (p1 − p2 )p2 p2 q2 (30.7) 1 = 2 + , η=√ 2 + , 0 ≤ 2 ≤ θ. √ p1 2 p1 p2 p1 p2 2 √ This gives feasible solutions to (D) with objective value p1 p2 and proves the bound (30.1) for the case n = 1. The case n ≥ 2. The construction in this case is really similar to that in the previous case. We replace the 2 × 2 matrices used in the previous case with new n × n matrices by taking tensor products as explained below. % (l) (l) (l) Let ci = pi /qi < 1, and let ⎤ ⎡ (l) (l) pj pj 1 − (l) (l) (l) qi qi ⎦ , Ai,j = ⎣ 1 0 where the rows and columns are indexed in the order ∅, {}. Then we redefine [n]

(n)

(n−1)

Ai,j = Ai,j := Ai,j ⊗ Ai,j

(1)

⊗ · · · ⊗ Ai,j .

We identify 2{1} × · · · × 2{n} with 2[n] , and thus we view the rows (resp. columns) of Ai,j as indexed by Ωi (resp. Ωj ). Similarly we

30.2. Proof of the theorem [n]

199

[n]

[n]

redefine Di,j = Di,j , Vi = Vi , and Δi = Δi . Note that Δi is the diagonal matrix whose (x, x)-entry is μi ({x}) for x ∈ Ωi . These new matrices satisfy (30.2) and (30.3). We choose the variables by (30.4), where 1 , 2 , and η are given by (30.7). Note that for x ∈ Ωi and y ∈ Ωj , the (x, y)-entry of Ai,j is 0 if x ∩ y = ∅ and positive otherwise. Thus Z is a symmetric matrix with Z ≥ 0. Define the symmetric matrix S by (30.5). Then we have / . / . / . T 0 V1 0 θI1 − 1 D1,1 ηD1,2 − 12 E∅,∅ V1 (30.8) S = , ηD2,1 − 12 E∅,∅ θI2 − 2 D2,2 0 V2T 0 V2 where Ii ∈ RΩi ×Ωi denotes the identity matrix. The RHS of (30.8) is $ isomorphic to the direct sum z∈2[n] S (z) , where 0 1 . / (z) (z) θ − 1 c1,1 ηc1,2 θ − 1 η − 12 (∅) (z) S = (z = ∅), , S = (z) (z) η − 12 θ − 2 ηc2,1 θ − 2 c2,2  ( ) ( )   (z) (z) with ci,j = ∈z − ci cj . Note that |ci,j | < 1. We need to choose 1 ≥ 0, 2 ≥ 0, and η ∈ R so that S # 0, that is, for all z ∈ 2[n] , S (z) # 0.

(30.9)

We have already verified (30.9) for z = ∅ and {1} in the previous (z) case. So let z = ∅, {1}. By (30.6) with |ci,j | < 1 the diagonal entries of S (z) are nonnegative. Thus, S (z) # 0 if and only if #" #  " (z) (z) (z) 2 (30.10) θ − 1 c1,1 θ − 2 c2,2 ≥ ηc1,2 . Let us verify that (30.10) holds if 2 = 0. In this case, by (30.7), 1 = (p1 − p2 )p2 /(4θ) and η = q2 /2, and then (30.10) is rewritten as #

(z) " (z) (30.11) p1 p2 ≥ c1,1 q22 c2,2 + (−1)|z| (p1 − p2 )p2 . We also note that (z) |ci,i |

=



pi

l∈z

1 − pi



(l) (l)

≤

pi 1 − pi

|z|

because p/(1 − p) is increasing in p ∈ (0, 1/2). Thus, if |z| is odd, (z) then |ci,i | ≤ pi /qi and the RHS of (30.11) is   ≤ (p1 /q1 ) q22 (p2 /q2 ) − (p1 − p2 )p2 = p1 p2 .

200

30. A cross intersection problem with measures (z)

If |z| is even, then |ci,i | ≤ (pi /qi )2 and the RHS of (30.11) is   ≤ (p1 /q1 )2 q22 (p2 /q2 )2 + (p1 − p2 )p2 = p1 p2 (p1 /q1 )2 ≤ p1 p2 . Thus in both cases we have (30.11) and S # 0. This means that √ a feasible solution to (D) with objective value p1 p2 is obtained by setting 2 = 0. This completes the proof of (i). 30.2.3. Proof of (ii). If n = 1 then the result follows from (i). Let n ≥ 2. We proceed as in the proof of (i), but this time we choose 2 = θ. Then 1 = θ and η = 1/2 by (30.7). We need to show (30.10) for z = ∅, {1}. In this case it reads  (z)  (z)  1 − c2,2 1 − c1,1 ≥ 1. (30.12) p1 p2 (z) (z) |c1,1 | |c2,2 | We distinguish the cases according to the parity of |z|. (z)

If |z| is odd, then ci,i < 0, and (z)

1 − ci,i (z)

|ci,i |

=1+

1 (z)

|ci,i |

=1+

 q (l)

i . (l)

l∈z

pi

The RHS is minimized when |z| = 1, say z = {l}. Then the LHS of (30.12) is (l)

(l)

(l)

(l)

(l) (l)

≥ p1 p2 (1 + q1 /p1 )(1 + q2 /p2 ) = p1 p2 /(p1 p2 ) ≥ 1, as required. (z)

If |z| ≥ 2 is even, then ci,i > 0, and (z)

1 − ci,i (z)

|ci,i |

= −1 +

1 (z)

|ci,i |

= −1 +

 q (l)

i . (l)

l∈z

pi

The RHS is minimized when |z| = 2, say z = {k, l}, and the RHS is (k) (l)

≥ −1 +

q1 q1

(k) (l) p1 p1 (k)

(k)

=

(l)

1 − p1 − p1 (k) (l) p1 p1

(l)

(k)

≥

p1

(k) (l) p1 p1

=

1 (l)

,

p1

(k)

where we used 1 − p1 − p1 ≥ 1/3 ≥ p1 . Thus the LHS of (30.12) is (l) (l) (l) (l) ≥ p1 p2 (1/p1 )(1/p2 ) = p1 p2 /(p1 p2 ) ≥ 1. This completes the proof of (ii).



Chapter 31

Capsets and sunflowers

In this chapter we present a recent development of a polynomial method (the slice rank method). We give two applications. One is that a set containing no 3-term arithmetic progressions in Fn3 has size less than (2.8)n for n sufficiently large, and the other is that a family of subsets of [n] containing no sunflowers has size less than (1.9)n for n sufficiently large.

31.1. Things happened in May 2016 Finding large subsets of an abelian group G with no 3-term arithmetic progressions is one of the central problems in additive number theory. In May of 2016 Croot, Lev, and Pach [20] proved a remarkable result stating that the size of such subsets for the case G = (N/4N)n is at most cn for some c < 4. This was a starting point of successive breakthrough results all obtained in the same month. Soon after they put the preprint on arXiv, Ellenberg and Gijswijt independently found how to use the idea of [20] for the case G = Fn3 and obtained the following result. Theorem 31.1 (Ellenberg–Gijswijt [27]). Let X ⊂ Fn3 . Suppose that for x, y, z ∈ X, (31.1)

x + y + z = 0 if and only if x = y = z.

Then |X| < cn for some c < 3. 201

202

31. Capsets and sunflowers

Exercise 31.2. We say that three elements x, x + a, x + 2a ∈ Fn3 form a 3-term arithmetic progression if a = 0. Show that X ⊂ Fn3 contains no 3-term arithmetic progressions if and only if (31.1) holds. A subset X ⊂ Fn3 satisfying (31.1) is called a capset. Bounding the size of capsets is important in itself, and it is also related to the following two problems. One is a problem of determining the exponent ω of matrix multiplication; see [12]. The other is a conjecture concerning the size of families without sunflowers. Recall that three distinct subsets A, B, C ⊂ [n] form a sunflower if A ∩ B = B ∩ C = C ∩ A. The following result was conjectured by Erd˝os and Szemer´edi [37]. Theorem 31.3. Let F ⊂ 2[n] . If F contains no sunflower, then there exist c < 2 and n0 such that |F| < cn for all n > n0 . Alon, Shpilka, and Umans [2] had proved that if Theorem 31.1 were true (and we now know that it is indeed true), then it would imply Theorem 31.3. Tao reformulated the proof of Theorem 31.1 in his blog [106] by introducing a slice rank. Then using this slice rank method Naslund and Sawin [94] gave a direct proof of Theorem 31.3 with a better constant c than the constant obtained from [2] with [27]. In the following sections we explain the slice rank and its basic property, and then we apply the property to prove Theorems 31.1 and 31.3. Our presentation is partially based on a set of lecture notes by Zeeuw [17], which is also recommended for other topics including the density Hales–Jewett Theorem.

31.2. Slice rank A diagonal matrix without zero diagonal entries is of full rank. We will extend this fact to a “hypermatrix” in some sense. For this purpose let us introduce the slice rank of a function. Let X be a finite set, and let F be a field. Let f : X k → F be a function in k variables. We say that f is sliced if f can be written as a product of two functions a and b, where a is in one variable and b is in the other k − 1 variables, for example, f (x1 , x2 , . . . , xk ) = a(x3 ) b(x1 , x2 , x4 , . . . , xk ).

31.2. Slice rank

203

We can always write f as a sum of at most |X| sliced functions; indeed,  az (x1 )bz (x2 , . . . , xk ), f (x1 , . . . , xk ) = z∈X

where az (x) = δz,x and bz (x2 , . . . , xk ) = f (z, x2 , . . . , xk ). Now the slice rank of f , denoted by sr(f ), is the least number of sliced functions such that f is represented as the sum of them, that is, r   gi , where g1 , . . . , gr are sliced functions . sr(f ) = min r : f = i=1

As we just noted, sr(f ) ≤ |X|. Lemma 31.4 (Tao [106]). Let X be a finite set, F be a field, and k ≥ 2 be an integer. Let f : X k → F be a function such that f (x1 , . . . , xk ) = 0 if and only if x1 = · · · = xk . Then sr(f ) = |X|. Proof. We prove the statement by induction on k. Let k = 2. It suffices to show that sr(f ) ≥ |X|. Suppose, to the contrary, that r := sr(f ) < |X|. Then we can write f (x, y) =

r 

ai (x)bi (y)

i=1

for some nonzero functions ai and bi . Let F (resp. Fi ) be an |X| × |X| matrix whose (x, y)-entry is f (x, y) (resp. ai (x)bi (y)). Then F =

r 

Fi .

i=1

Since the usual matrix rank of Fi is one, it follows that rank F ≤

r 

rank(Fi ) = r.

i=1

On the other hand, the assumption that f (x, y) = 0 if and only if x = y implies that F is a diagonal matrix without zero diagonal entries. Thus rank F = |X| > r, a contradiction. Next we move on to the induction step, but to make things notationally easier we consider the case k = 3. Let f : X 3 → F and

204

31. Capsets and sunflowers

suppose that r := sr(f ) < |X|. Then we can write (31.2)    ai (x)bi (y, z) + ai (y)bi (x, z) + ai (z)bi (x, y) f (x, y, z) = i∈I

i∈J

i∈K

for some nonzero functions ai and bi , where I  J  K = [r]. Without loss of generality we may assume that I = ∅. Let V be a vector space consisting of all functions v : X → F  such that x∈X v(x)ai (x) = 0 for all i ∈ I. Thus V is defined by |I| linear equations in |X| variables, and (31.3)

dim V ≥ |X| − |I| > r − |I|.

Choose v ∈ V so that the support S = {x ∈ X : v(x) = 0} is maximal. We claim that |S| ≥ dim V.

(31.4)

In fact, if |S| < dim V then we can find a nonzero w ∈ V such that w(x) = 0 for all x ∈ S. But then v + w has a strictly larger support than S, and this contradicts the maximality of the choice of S.  Define a function g : S 2 → F by g(y, z) = x∈X v(x)f (x, y, z). We estimate the slice rank of g in two ways, which will give rise to a contradiction, showing that the earlier assumption sr(f ) < |X| is false. Compute g from the RHS of (31.2). For the first term we have    v(x)ai (x) bi (y, z) = 0. i∈I

x∈X

The second and third terms we can rewrite as   ai (y)ci (z) + ai (z)ci (y), i∈J

where ci (z) =

i∈K



v(x)bi (x, z) for i ∈ J,

x∈X

ci (y) =



v(x)bi (x, y) for i ∈ K.

x∈X

Thus we get g(y, z) =

 i∈J

ai (y)ci (z) +

 i∈K

ai (z)ci (y),

31.2. Slice rank

205

that is, g is represented by |J| + |K| sliced functions, and (31.5)

sr(g) ≤ |J| + |K| = r − |I|.

Finally we claim that g(y, z) = 0 if and only if y = z. In fact,  if y = z then f (x, y, z) = 0 and g(y, z) = x v(x)f (x, y, z) = 0. If  y = z, then g(y, y) = x v(x)f (x, y, y) = v(y)f (y, y, y) = 0 for y ∈ S, as needed. Thus by the induction hypothesis with (31.3) and (31.4) it follows that sr(g) = |S| ≥ dim V > r − |I|, 

which contradicts (31.5).

The next lemma gives us an upper bound for the slice rank of a function. We will use the lemma to solve the capset problem and the sunflower problem. Lemma 31.5. Let F be a field, let Y ⊂ F be a finite set, and let X ⊂ Y n , where n ∈ Z>0 . Define a function f : X 3 → F by (31.6)

f (x, y, z) =

n  

 (xi + yi + zi )s − t ,

i=1

where t ∈ F and s ∈ Z>0 . Let g(x) = x− 3 (1 + x + · · · + xs ) be a realvalued function defined on the interval (0, 1), and let α be a unique root of g  (x) = 0. Then s

sr(f ) < 3(g(α))n . In particular the following hold. (i) If s = 2, t = 1, and F = F3 , then sr(f ) < 3(2.76)n . (ii) If s = 1, t = 2, and F = R, then sr(f ) < 3(1.89)n . Proof. By expanding the RHS of (31.6) we can write f as a sum of monomials in the form xi11 · · · xinn y1j1 · · · ynjn z1k1 · · · znkn ,   with I + J + K ≤ sn, where I = nl=1 il , J = nl=1 jl , n and K = l=1 kl . For simplicity we write xI to mean xi11 · · · xinn and define y J and z K similarly. Since one of the I, J, and K

206

31. Capsets and sunflowers

is at most sn/3, we can divide f into three parts f = fx + fy + fz as follows. First we collect all monomials in f with I ≤ sn/3 and let    I J K x cIJK y z . fx = I≤sn/3

J,K

Next we collect all monomials in f − fx with J ≤ sn/3 and let    yJ cIJK xI z K . fy = J≤sn/3

J,K

Finally let fz = f − fx − fy , which is represented as    fz = zK cIJK xI y J . K≤sn/3

I,J

We note that fx is a sum of #I sliced functions with I ≤ sn/3. Moreover, #I ≤ N , where N is the number of integer solutions (i1 , . . . , in ) satisfying sn i1 + · · · + in ≤ 3 with 0 ≤ il ≤ s for all 1 ≤ l ≤ n. Suppose that (i1 , . . . , in ) is one of the solutions, and let au = |{l : il = u, 1 ≤ l ≤ n}| for 0 ≤ u ≤ s. Then (P) a0 + · · · + as = n, sn . 3 On the other hand, for a fixed feasible (a0 , . . . , as ) the number of the corresponding solutions (i1 , . . . , in ) is      n n − a0 n − (a0 + · · · + as−1 ) n! . ··· = a0 as a1 a0 !a1 ! · · · as ! (Q) a1 + 2a2 + · · · + sas ≤

Thus we have



n! , a0 !a1 ! · · · as ! where the sum is taken under the conditions of (P) and (Q). Clearly N is also an upper bound for the number of sliced functions in fy as well as in fz . Consequently f is a sum of at most 3N sliced functions, and sr(f ) ≤ 3N. N=

31.2. Slice rank

207

Let x ∈ (0, 1). By the multinomial expansion we have x−

sn 3

(1 + x + · · · + xs )n =

 (P)

>

n! sn xa1 +2a2 +···+sas − 3 a0 ! · · · as !

 (P), (Q)

≥

 (P), (Q)

n! sn xa1 +2a2 +···+sas − 3 a0 ! · · · as ! n! = N. a0 ! · · · as !

Since the LHS is (g(x))n , it follows that (g(x))n > N for all x ∈ (0, 1). By standard calculus we see that g  (x) = 0 has a unique root α in (0, 1), and g(x) takes its minimum at x = α. Therefore we get sr(f ) ≤ 3N < 3(g(α))n . (1 + x + x2 ) is For (i) let s = 2 and √t = 1. Then g(x) = x− 3 √ 1 1 3 minimized at x = α := 8 ( 33 − 1), and g(α) = 8 (33 33 + 207) 3 < 2.76. 2

For (ii) let s = 1 and t = 2. Then g(x) = x− 3 (1 + x) is minimized  at x = α := 1/2, and g(α) = 3/22/3 < 1.89. 1

The slice rank method used in this chapter is summarized as follows. Suppose that we want to bound the size of a set X and we know that X k satisfies some condition. On the one hand, using the condition, we construct a function f : X k → F with the property that f (x1 , . . . , xk ) = 0 if and only if x1 = · · · = xk . By Lemma 31.4 it follows that sr(f ) = |X|. On the other hand, we expand the function f as a sum of sliced functions and count the number N of them using Lemma 31.5. By definition of the slice rank we get sr(f ) ≤ N . Combining these two things, we obtain an upper bound on |X|, that is, |X| ≤ N . In the final section we apply this method to the capset problem and the sunflower problem.

208

31. Capsets and sunflowers

31.3. Proof of the theorems Proof of Theorem 31.1. Let X ⊂ Fn3 be a capset, that is, X satisfies (31.1). Define a function f : X 3 → F3 by f (x, y, z) =

n    (xi + yi + zi )2 − 1 . i=1

Then it follows from (31.1) that f (x, y, z) = 0 if and only if x = y = z. In fact, if x = y = z then f (x, y, z) = f (x, x, x) = (−1)n = 0. If x, y, and z are not all equal, then x + y + z = 0, so xi + yi + zi = 0 for some i, and in this case (xi + yi + zi )2 = 1, yielding f (x, y, z) = 0. Thus by Lemma 31.4 it follows that (31.7)

sr(f ) = |X|.

On the other hand, by (i) of Lemma 31.5, we have (31.8)

sr(f ) < 3(2.76)n .

By (31.7) and (31.8) we get |X| = sr(f ) < 3(2.76)n . Then there exists N such that 3(2.76)n < 2.8n for all n ≥ N . Since Fn3 itself is not a capset, we trivially have that |X| ≤ 3n − 1, and we can choose 2.8 < c < 3 so that 3N − 1 < cN . Consequently we have |X| < cn for all n.  Proof of Theorem 31.3. Recall that a characteristic vector x ∈ {0, 1}n of a subset F ⊂ [n] is defined by xi = 1 if i ∈ F and xi = 0 if i ∈ F . Let F ⊂ 2[n] be sunflower-free and let X ⊂ {0, 1}n be the set   of characteristic vectors of F. For 0 ≤ k ≤ n let Fk = F ∩ [n] and let k Xk ⊂ X be the corresponding subset, so if x ∈ Xk then i xi = k. The sunflower-free condition is translated into the following condition1 for Xk : For any three vectors x, y, z ∈ Xk not all the same and w := x + y + z, there is some i such that wi = 2. Define a function f : Xk3 → R by f (x, y, z) =

n 

(xi + yi + zi − 2).

i=1 1 We remark that this condition does not hold in X in general. The reason is that a sunflower-free family can have two distinct subsets A  B with corresponding characteristic vectors x and y, and in this case w := x + x + y satisfies wi = 2 for all i.

31.3. Proof of the theorems

209

Then it follows that f (x, y, z) = 0 if and only if x = y = z. Thus by Lemma 31.4 we have sr(f ) = |Xk |.

(31.9)

On the other hand, by (ii) of Lemma 31.5, we have sr(f ) < 3(1.89)n .

(31.10)

By (31.9) and (31.10) we get |F| = |X| =

n 

|Xk | = (n + 1) sr(f ) < (n + 1) · 3(1.89)n .

k=0

Thus there exist c < 2 and n0 such that |F| < cn for all n > n0 .



Chapter 32

Challenging open problems

In the last chapter we present some challenging open problems. For F ⊂ 2[n] and i ∈ [n], let / F ∈ F}. F(i) = {F \ {i} : i ∈ F ∈ F}, F(i) = {F : i ∈

32.1. Chv´ atal conjecture Conjecture 32.1. Let F ⊂ 2[n] be a downset and F0 ⊂ F an intersecting subfamily. Then |F0 | ≤ max |F(i)|. i∈[n]

The inequality asserts that no intersecting subfamily can surpass the trivial subfamilies. In some sense the conjecture asserts that the Erd˝os–Ko–Rado Theorem is true for every downset. Chv´ atal [19] proved this conjecture for the special case where F is shifted.

32.2. Frankl’s union-closed conjecture Conjecture 32.2. Let F ⊂ 2[n] be a union-closed family, that is, F, F  ∈ F imply F ∪ F  ∈ F. Suppose further that F = {∅}. Then there exists an element i such that |F(i)| ≥ |F|/2. 211

212

32. Challenging open problems

Frankl stated this conjecture in 1979 in a letter to Ron Graham, who popularized it. The simplicity of the statement drew a lot of attention. Richard Stanley calls it in his book Enumerative combinatorics “the diabolic conjecture of Frankl” because it eluded various approaches. Tim Gowers has a Polymath project running on this conjecture. We refer the reader to it for current information.

32.3. Maximal families without s + 1 pairwise disjoint sets Let s ≥ 1 be an integer and define G ⊂ 2[n] by G = {G ⊂ [n] : G ∩ [s] = ∅}. For n ≥ s, G has the property that though it contains no s+1 pairwise disjoint members, G  {H} does for all H ∈ 2[n] \ G. Let n ≥ s ≥ 1 and let F ⊂ 2[n] have the above property; then Erd˝os and Kleitman [35] conjectured that |F| ≥ 2n (1 − 2−s ) = |G|. Let us note that this is true if s = 1. For s ≥ 2, the best known bound so far is |F| ≥ 2n (1 − 1/s) due to Buci´c et al., see [15].

32.4. The Erd˝ os matching conjecture We have devoted Chapter 9 to this conjecture, but let us state it here once more. Conjecture 32.3. Let n, k, and s be positive integers, with n ≥ k(s+   has no s + 1 pairwise disjoint members. 1). Suppose that F ⊂ [n] k Then      

k(s + 1) − 1 n n−s |F| ≤ max , − . k k k Erd˝os made this conjecture more than half a century ago, but a complete solution is still not in sight.

32.6. Diversity of intersecting hypergraphs

213

32.5. Kleitman matching problem Let s ≥ 3 be an integer, let n > s, and let F ⊂ 2[n] be a family without s pairwise disjoint members. Let k(n, s) denote max |F| subject to the above conditions. Kleitman determined k(n, s) for n ≡ 0 or −1 (mod s). In the case n ≡ −2 (mod s), k(n, s) was determined by Quinn if s = 3 and by Frankl and Kupavskii for all s. However, to determine k(n, s) in general seems to be very difficult. Let us present two conjectures. Recall that k(n + l, s) ≥ 2l k(n, s). Indeed, if F ⊂ 2[n] has no s pairwise disjoint members, then neither does F  = {F ⊂ [n + l] : F ∩ [n] ∈ F}, and obviously |F  | = 2l |F|. Note that by (7.1) and Theorem 7.2, k(n, s) = 2k(n − 1, s) if s divides n. Conjecture 32.4. Let s ≥ 4. If n ≡ 1 (mod s), then k(n, s) = k(n − 2, s). Let a1 ≥ a2 ≥ · · · ≥ an ≥ 0 be rational numbers such that a1 + · · · + an = 1. Define the threshold family F = Fs (a1 , . . . , an ) by  ai > 1/s}. F = {F ⊂ [n] : i∈F

It should be clear that there are no s pairwise disjoint subsets in F. Conjecture 32.5. k(n, s) = max{|F| : F is a threshold family}. 1 In Conjecture 32.4 one can set ai = n−2 for 1 ≤ i ≤ n − 2 and set an−1 = an = 0. We should mention that even knowing that Conjecture 32.5 is true, it is not easy to determine k(n, s).

32.6. Diversity of intersecting hypergraphs   Let n and k be positive integers with n > 2k, and let F ⊂ [n] k be an intersecting family. Set Δ(F) = maxi |F(i)| (the maximum degree) and define the diversity ρ(F) of the family by ρ(F) = |F| − Δ(F).  Obviously, ρ(F) ≥ 1 if and only if F ∈F F = ∅. Without loss of generality we may assume that ρ(F) = |F(¯1)|. Since F(1) and F(¯1) are cross intersecting, by the Lov´asz version of the Kruskal–Katona

214

32. Challenging open problems

 x  and n − 1 > x ≥ n − 1 − k imply |F(1)| ≤ Theorem |F(¯ 1)| = n−1−k n−1  x  − , and therefore k−1 k−1 (32.1)

      n−1 x x |F| ≤ − + . k−1 k−1 n−1−k

This was used by Frankl and by Kupavskii and Zakharov to obtain upper bounds on |F|. Let us note that for x = n − 3 and x = n − 4 the value of the RHS of (32.1) is the same. Conjecture 32.6 (Frankl). Suppose that n > 3(k − 1) and F ⊂ is intersecting. Then  (32.2)

ρ(F) ≤

[n] k

 n−3 . k−2

  : |F ∩ [5]| ≥ 3} shows that (32.2) The example F = {F ∈ [n] k fails for n ≤ 3(k − 1). If F is shifted, then F(¯1) is 2-intersecting and (32.2) follows from the Full EKR due to Frankl and Wilson. Very recently Frankl proved the conjecture for n ≥ 6k2 . We have less evidence but let us still make the following more general conjecture. Conjecture 32.7 (Frankl). Let n > (t + 2)(k − t). If F ⊂   t-intersecting, then ρ(F) ≤ n−t−2 k−t−1 .

[n] k

is

Again, if F is shifted then F(¯ 1) is (t + 1)-intersecting on [2, n].   , and the conThus by the Full EKR we have |F(¯1)| ≤ (n−1)−(t+1) k−(t+1) jecture is true. The bound on n cannot be lowered, as shown by the example  

[n] F ∈ : |F ∩ [t + 4]| ≥ t + 2 . k   : |F ∩ [t + 2]| ≥ t + 1} we have Note also that for F = {F ∈ [n] k n−t−2 ρ(F) = k−t−1 , which shows that the inequality in the conjecture is, if true, best possible.

32.7. Missing intersections

215

32.7. Missing intersections   satisfies |F ∩ F  | = l for Conjecture 32.8. Suppose that F ⊂ [n] k some l, 2l < k. Then   n (32.3) |F| ≤ . k−l−1 Note that this is true in the case where k − l is a prime power; see Exercise 23.5. Let us also mention that, if true, for k = 2l + 1 equality holds in (32.3) for infinitely many values of n for fixed k; see [47]. The construction depends on the existence of Steiner systems, that is, on the recent breakthrough by Keevash [82].

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Index

L-system, 95 k-partite, 98 k-uniform, 1 m(n, k, L), 95 r-cross t-intersecting, 74 r-cross union, 28 r-wise intersecting union, 80 r-wise union, 29 s-independent family, 142 t-avoiding, 138 t-intersecting, 9, 53 alternating tensor, 164 antichain, 1 bilinear map, 162 bipartite pseudo-adjacency matrix (bpam), 182 biregular bipartite graph, 186 capset, 202 cascade representation, 33 center, 109 chain, 1 characteristic vector, 127 closed L-system, 97 colex ordering, 31 column space, 142 complement family, viii, 6 cover, 115 cross t-intersecting, 187

cross intersecting families, 74, 83 cross union families, 28, 83 degree, viii Deza–Erd˝ os–Frankl Theorem, 109 Deza–Frankl–Singhi Theorem, 130 diameter, 7 downset, 7, 77 dual form, 190 Erd˝ os t-antichain theorem, 27 Erd˝ os matching conjecture, 47 Erd˝ os–Ko–Rado Theorem, 19 eventown rules, 155 exchange operation, 56 exponent of a family, 95 exterior algebra, 167 F¨ uredi’s structure theorem, 98, 115 feasible, 190 Fisher’s inequality, 137 Frankl–Wilson, 129, 142 Full Erd˝ os–Ko–Rado Theorem, 89 Gottlieb’s theorem, 144 hereditary family, 7 Hilton–Milner Theorem, 43 homogeneous family, 117 immediate shadow, 2, 32 inclusion matrix, 141

223

224 independence number, 175 independent set in a graph, 175 intersecting family, 9 intersecting-union family, 79 intersection closure, 116 intersection structure (IS), 97 isomorphic families, viii Katona union theorem, 20 Katona’s intersecting shadow theorem, 145 Kleitman diameter theorem, 22 Kleitman’s correlation inequality, 77 Kneser graph, 59, 178 Kruskal–Katona Theorem, 10, 32 Kruskal–Katona Theorem (Lov´ asz version), 214 Kruskal–Katona Theorem (Lov´ asz version), 12, 84 linear programming (LP), 189 matching, 121 matching number, 47 multilinear map, 163 multilinear polynomial, 130 oddtown rules, 156 positive semidefinite (PSD), 175 primal form, 189 probability measure, 69, 181 product measure, 70, 77, 89 pseudo-adjacency matrix (pam), 176 pushing-pulling method, 61 R¨ odl’s packing theorem, 121, 122 R¨ odl–Tengan construction, 123 rank of an L-system, 97 ratio bound, 177 Ray-Chaudhuri–Wilson Theorem, 127 regular, viii, 122 semidefinite programming (SDP), 189 shadow, 9 shift-end, 88

Index shifting operation, 8 shifts to, 88 singular decomposition, 182 slice rank, 203 sliced function, 202 Snevily’s Theorem, 132 Sperner’s Theorem, 2, 26 squash operation, 6 Steiner system, 96, 123, 132, 150 sunflower, 109, 202 symmetric algebra, 167 symmetric tensor, 164 Takagi function, 36 tensor product, 163 trace, 7, 13 uniform family, 1 uniform measure, 70 union family, 9 unit-distance graph, 139 upset, 38, 77 walk, 87 weak duality for LP, 190 weak duality for SDP, 191 Yamamoto’s inequality, 25

One of the great appeals of Extremal Set Theory as a subject is that the statements are easily accessible without a lot of mathematical background, yet the proofs and ideas have applications in a wide range of fields including combinatorics, number theory, and probability theory. Written by two of the leading researchers in the subject, this book is aimed at mathematically mature undergraduates, and highlights the elegance and power of this field of study. The first half of the book provides classic results with some new proofs including a complete proof of the Ahlswede– Khachatrian theorem as well as some recent progress on the Erd˝os matching conjecture. The second half presents some combinatorial structural results and linear algebra methods including the Deza–Erd˝os–Frankl theorem, application of Rödl’s packing theorem, application of semidefinite programming, and very recent progress (obtained in 2016) on the Erd˝os–Szemerédi sunflower conjecture and capset problem. The book concludes with a collection of challenging open problems.

For additional information and updates on this book, visit www.ams.org/bookpages/stml-86

STML/86