Extremal Graph Theory

Lectured by A. Thomason

Lent Term 2013

1 The Erd˝os-Stone Theorem 1 2 Stability 4 3 Supersaturation 6 4 Szemer´edi’s Regularity Lemma 9 5 A couple of applications 13 6 Hypergraphs 16 7 The size of a hereditary property 19 8 Containers 22 9 The Local Lemma 25 10 Tail Estimation 26 11 Martingales Inequalities 29 12 The Chromatic Number of a Random Graph 31 13 The Semi-Random Method 33

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Last updated: Sat 7th May, 2016 Course description

Extremal graph theory is an umbrella title for the study of graph properties and their dependence on the values of graph parameters. This course builds on the material introduced in the Part II Graph Theory course, in particular Tur´an’s theorem and the Erd˝os-Stone theorem, as well as developing the use of randomness in combinatorial proofs. Further techniques and extensions to hypergraphs will be discussed. It is intended to cover some reasonably large subset of the following.

The Erd˝os-Stone theorem and stability. Supersaturation. Szemer´edi’s Regularity Lemma, with applications. The number of complete subgraphs.

Hypergraphs. Erd˝os’s r-partite theorem. Instability. The Fano plane. Razborov’s ﬂag algebras. Hereditary properties and their sizes.

Probabilistic tools: the Local Lemma and concentration inequalities. The chromatic number of a random graph. The semi-random method, large independent sets and the Erd˝os-Hanani problem. Dependent random choice.

Pre-requisite Mathematics

A knowledge of the basic concepts, techniques and results of graph theory, such as that aﬀorded by the Part II Graph Theory course.

Literature

No book covers the course but the following can be helpful.

B. Bollob´as, Modern graph theory, Graduate Texts in Mathematics 184, Springer-Verlag, New York (1998), xiv+394 pp.

N. Alon and J. Spencer, The Probabilistic Method, Wiley, 3rd ed. (2008) Lecture 1 1. The Erd˝os-Stone Theorem

Recall:

Tur´an’s theorem. If G = n, e(G) > tr(n) and G Kr+1, then G = Tr(n), the r-partite Tur´an graph of order| | n. 6⊃

Note. T (n) is the complete r-partite graph with class sizes n/r and n/r , and t (n) is r ⌈ ⌉ ⌊ ⌋ r e(Tr(n)).

This answers the extremal problem for Kr+1, and there is a unique extremal graph. The structure of Tr(n) invites many proofs by induction.

In general, we are interested in ex(n, F ) for some ﬁxed graph F , where

ex(n, F ) = max e(G): G = n, F G | | 6⊂ 1 n So Tur´an’s theorem states that ex(n,K )= t (n) 1 . r+1 r ≈ − r 2 > 1 n Comment. It is true that tr(n) 1 r 2 . This is equivalent to the average degree being > 1 − > 1 1 r (n 1). But in fact the minimum degree is 1 r (n 1), as can be − − − − 1 seen by looking at a vertex in the largest class and noting it misses at most r (n 1) vertices as neighbours. Equality holds only if there is precisely one lar gest class, in which− case the average degree is greater than the minimum anyway, so in fact we always have t (n) > 1 1 n . r − r 2 For general F there might be several extremal graphs and the extremal function might be hard to evaluate exactly.

Denote by Kr(t) the complete r-partite graph with t vertices per class.

So Kr = Kr(1) and Kr(t)= Tr(rt).

Lemma 1.1. Let r > 0 be an integer and ε> 0. Then there exist d = d′(r, ε) and n1 = n1(r, ε) such that, if G = n > n and, if r > 1, | | 1 1 δ(G) > 1 + ε n , − r then G K (t), where t = d log n . ⊃ r+1 ⌊ ⌋ Proof. If r =0 or ε > 1/r then the assertion is trivial. We proceed by induction on r.

By the induction hypothesis, we may assume that G has a subgraph K = Kr(T ), where εr 1 T = 2t/εr . (This requires only that d′(r, ε) < 3 d′ r 1, r(r 1) .) ⌈ ⌉ − − Now, each vertex of K sends at least 1 1 + ε)n K edges to G K. Let U be the set − r − | | − of vertices of G K having at least 1 1 + ε K neighbours in K. − − r 2 | | Writing e(G K,K) for the number of edges between G K and K, we have − − 1 1 ε K 1 + ε n K 6 e(G K,K) 6 U K + n U 1 + K | | − r − | | − | || | − | | − r 2 | |

1 ε 1 ε εrT 1 + K (r 1)T = 1 + rT (r 1)T = > t − r 2 | |− − − r 2 − − 2 T r vertices in each class of K, and so is joined to some Kr(t) in K. But there are only t many K (t) in K, and, recalling that n 6 ( en )k, we have r k k r tr rd log n T eT 3e εrn U 6 6 6 6 | | t t εr 3t t

if d′(r, ε) is small and n1(r, ε) is large.

Hence there exists W U, with W > t, joined to the same K (t) in K. ⊂ | | r Hence K (t) G. 2 r+1 ⊂

Lemma 1.2. Let c,ε > 0. Then there exists n2 = n2(c,ε) with the following property.

Suppose that G = n > n and e(G) > (c + ε) n . Then G has a subgraph H such that | | 2 2 δ(H) > c H and H > ε1/2n. | | | | 1/2 Proof. If not, there is a sequence G = Gn Gn 1 Gn 2 Gs, where s = ε n , ⊃ − ⊃ − ⊃···⊃ ⌊ ⌋ such that Gj = j and the only vertex in Gj not in Gj 1 has degree less than cj. Then | | − n n n n +1 s +1 εn2 s e(G ) > (c + ε) cj = (c + ε) c > > s 2 − 2 − 2 − 2 2 2 j=s+1 X

provided n2 is large enough. Contradiction. 2

Lecture 2 Theorem 1.3 (Erd˝os-Stone, 1946). Let r > 0 be an integer and ε > 0. Then there exist d = d(r, ε) and n0 = n0(r, ε) such that, if G = n > n0 and if for r > 1 we have e(G) > 1 1 + ε n , then G K (t) where| |t = d log n . − r 2 ⊃ r+1 ⌊ ⌋ Proof. Provided n > n 1 1 + ε , ε , we may apply Lemma 1.2 to G to obtain a subgraph 0 2 − r 2 2 H with δ(H) > 1 1 + ε H and H > ε1/2n. −r 2 | | | | > 1/2 ε Provided n0 ε− n1(r, 2 ), we may apply Lemma 1.1 to H to obtain Kr+1(t) with > ε 1/2 t d1(r, 2 ) log ε n .

1 1 ε 2 Provided n0 > ε , if we take d(r, ε)= 2 d1(r, 2 ), we are done.

As observed by Erd˝os and Simonovits in the mid-1960s, we can determine ex(n, F ) asymptoti- cally for every F .

2 Theorem 1.4. Let F be a ﬁxed graph with chromatic number r = χ(F ). Then

ex(n, F ) 1 lim =1 n n − r 1 →∞ 2 − Proof. Since χ(Tr 1(n)) = r 1, we have F Tr 1(n), hence − − 6⊂ − 1 n ex(n, F ) > e(Tr 1(n)) > 1 . − − r 1 2 − > 1 n On the other hand, given ε > 0, if G = n and e(G) 1 r 1 + ε 2 , then G K ( F ) F if n is large enough, by the| | Erd˝os-Stone theorem.− − ⊃ r | | ⊃ ex(n, F ) 1 Thus for every ε> 0, we have lim sup 6 1 + ε . 2 n − r 1 2 − Here’s a pretty consequence of Erd˝os-Stone.

Define the upper density of an infinite graph to be the supremum of densities of large finite subgraphs: F ud(G) = lim sup x : F G, F > n,e(F ) > x | | n ∃ ⊂ | | 2 →∞ Corollary 1.5. ud(G) 0, 1 , 2 , 3 ,... 1 . ∈{ 2 3 4 }∪{ } Can we strengthen Erd˝os-Stone to obtain larger t?

Theorem 1.6 Given r N, there exists εr > 0 such that, if ε<εr, there exists n(r, ε) such that for all n > n∈(r, ε) there is a graph G of order n with e(G) > 1 1 + ε n and − r 2 G K (t) where t = 3 log n . 6⊃ r+1 log(1/ε) Proof. Let W be a largest vertex class of Tr(n) with W = w = n/r . Form G by adding ε n edges within W so that G[W ] K (t) and hence| | G K ⌈ (t).⌉ 2 6⊃ 2 6⊃ r+1 To see that this addition is possible, choose edges inside W independently with probability 2 2 6 p =3εr . Take εr = (3r )− . So p< 1.

Let X be the number of edges chosen and Y the number of K2(t) formed by them. Then

w 1 w w t 2 E(X Y )= EX EY = p − pt − − 2 − 2 t t Now, t 1 t+1 t 1 2t 2 t2 1 2 t+1 − 2 5 (t+1) 2 5/2 − 1 w − p − = w p < w ε 6 < w n− < . 2 h i h i h i p w p w Hence E(X Y ) > . So there is a choice with X Y > . − 2 2 − 2 2 p w n Remove an edge from each K (t) to leave at least X Y > >ε edges with no 2 − 2 2 2 K2(t). 2 ε This shows dependence on n is correct. Our argument gives d(r, ε) > . 2r(r 1)! −

3 log n Bollob´as, Erd˝os, Simonovits (1976) proved that t>c . r log(1/ε)

log n Chvatal, Szemer´edi (1978) proved t> . log(1/ε)

Lecture 3 2. Stability

An extremal problem is stable if all nearly-optimal examples have the same structure.

> 1 n Theorem 2.1. Let t, r 2 be ﬁxed and suppose that G Kr+1(t). If e(G)= 1 r + o(1) 2 then 6⊃ − (a) there exists T (n) on V (G) with E(G) E(T (n)) = o(n2) r | △ r | (b) G contains an r-partite subgraph of size 1 1 + o(1) n − r 2 (c) G contains an r-partite subgraph of minimum degree 1 1 + o(1) n − r Global statement. Throughout, unless otherwise stated, G = n. | | Proof. Note that (a) (b) and (c) = (b). Note too that we may jettison o(n) vertices should we wish to.⇐⇒ In particular, we may⇒ assume that δ(G) > 1 1 +o(1) n,so(b) = (c): − r ⇒ for otherwise, there exist εn vertices of degree at most 1 1 + η)n for some small ε< 1 η, r 4 − 1 ε2 (1 ε)n and their removal leaves a graph of order (1 ε)n and size > 1 + − , so by r r 2 Erd˝os-Stone we would have a K (t). − − r+1 th Now, G K = Kr(s) where log n > s = s(n) . Let Ci be the i class of K, and ⊃ → ∞ s r let X be the set of vertices joined to at least t in every class of K. THere are t many r K (t) in J, and since K (t) G, we have X 6 t s = o(n). Jettison X. r r+1 ⊂ | | t s Let Y be the set of vertices joined to fewer than (r 1)s 2t + t vertices of K. Since > 1 > − − δ(G) 1 r + o(1) n, we have e(K, G K) s(r 1+ o(1))n. Because X has been jettisoned,− we have − − s (n Y ) (r 1)s + t + Y (r 1)s + t > s r 1+ o(1) n − | | − | | − − 2t − So Y = o(n). Jettison Y . | |

The remaining vertices are partitioned by Vi, 1 6 i 6 r, where Vi is the set of vertices joined to fewer than t of C . Since δ(G) > 1 1 + o(1) n, it is enough to show that i − r e G[V ] = o(n2) for each i, for then V , . . ., V are the parts of the r-partite graph we seek. i 1 r

2 Suppose instead that e G[Vi] > εn , say. Then G[V1] K2(t) by Erd˝os-Stone with r = 1. s ⊃ Each vertex of Ks(t) is joined to at least s 2t + 1 vertices in each class Ci. Hence Ci s − contains a set of s 2t 2t 1 >t vertices joined to all of K2(t). But then G Kr+1(t), a contradiction. − − ⊃ 2 Corollary 2.2. Let χ(F )= r + 1 and let G be extremal for F – i.e., F G, e(G)= ex(n, F ). Then δ(G)= 1 1 + o(1) n. 6⊂ − r 6 1 Proof. If not, by Theorem 1.4 we have δ(G) 1 r ε n. By Theorem 2.1(c) with t = F , G − 1− | | has F vertices x1, . . ., x F joined to m = 1 + o(1) n common neighbours y1, . . ., ym. | | | | − r Form G∗ from G v (where v is a vertex of lowest degree) by adding a new vertex u joined −

4 to y1, . . ., ym. Then e(G∗) > e(G), so G∗ F . This copy of F in G∗ contains u but not x , say. But then (F u) x contains F ⊃in G. Contradiction. 2 i − ∪ i Stability can sometimes be used as a bootstrapping device to obtain exact results. For example, ex(n, C5)= t2(n) for n > 6. In fact, ex(n, C2k+1)= t2(n) if n is large. This is a special case of the following theorem.

Theorem 2.3 (Simonovits). Let F be (r+1)-edge-critical, i.e. χ(F )= r+1 but χ(F e)= r − for all edges e E(F ). Then ex(n, F )= tr(n) for large n, and Tr(n) is the unique extremal graph. ∈

Proof. Let G be an extremal graph of order n. Select, by Theorem 2.1(c) using t = F , an r-partite subgraph H of minimum degree 1 1 + o(1) n. Necessarily, each part of| |H is − r of order 1 + o(1) n. Assign the o(n) vertices of G H to the parts of H with the fewest r neighbours. − Lecture 4 Suppose a vertex x is joined to εn vertices in its own class. Then it has εn neighbours in each part of H. These r sets of εn vertices span 1 1 + o(1) rεn edges because − r 2 δ(H)= 1 1 + o(1) n. − r By Erd˝os-Stone the neighbours of x span Kr( F ) which contains F v for any v V (F ). Hence G F , a contradiction. | | − ∈ ⊃ So each vertex has only o(n) neighbours in its own part. By Corollary 2.2, we know 1 δ(G)= 1 r + o(1) n. So each vertex of G is joined to all but o(n) vertices in each other class. Suppose− xy is an edge inside some class. Pick a set Z of F vertices in this class, including x, y . | | { } Then all but o(n) vertices of the other classes are common neighbours of Z. Hence these common neighbours span Kr 1( F ), either by Erd˝os-Stone or directly. But Z together − | | with Kr 1( F ) contains F , a contradiction. − | |

Hence G is r-partite, and since Tr(n) is the unique r-partite graph of maximum size, we have G = Tr(n). 2

Here is another example.

Theorem 2.4. Let r, s be ﬁxed. For large n, the unique extremal graph for sKr+1 (i.e., s disjoint copies of Kr+1) is Ks 1 + Tr(n s + 1) (where ‘+’ means ‘join all to all’). − − Proof. The proof is similar to that of Theorem 2.3. Proceed by induction on s. The case s =1 is Tur´an’s theorem.

As before, choose H and assign G H to the classes of H. Once again, if some vertex − x has εn neighbours in its own class, then the neighbours of x span some Kr(s(r + 1)). But G x cannot contain (s 1)Kr+1, hence e(G x) 6 e Ks 2 + Tr(n s + 1) , so − − − − − e(G) 6 e Ks 1 + Tr(n s + 1) . − − So equality holds in both cases, so G x = Ks 2 +Tr(n s+1) by the induction hypothesis, − − − so G = Ks 1 + Tr(s n + 1). − − Hence we may suppose that each vertex is joined to o(n) in its own class. Suppose some class contains s independent edges. Let Z be the 2s endvertices of these edges. As in the previous proof, the common neighbours of Z span Kr 1(s(r 1)), in which case G sKr+1. − − ⊃

5 th Thus no class contains s independent edges, so the j class contains a set Aj of 2(s 1) th − vertices such that every edge in the j class meets Aj . But each vertex in Aj has o(n) neighbours in the jth class.

Hence e(G) 6 2r(s 1)o(n)+ tr(n)

Supersaturation is the study of how many copies of F must exist in G if e(G) > ex(n, F ). The basic theorem holds in a general context.

Recall that an ℓ-uniform hypergraph is a pair G = (V, E) where

E V (ℓ) = Y V : Y = ℓ . ⊂ ⊂ | | For a class of ℓ-uniform hypergraphs, deﬁne F ex(n, ) = max e(G): G = n, G is ℓ-uniform, G contains no F F | | ∈F and ex(n, ) π( ) = lim F n n F →∞ ℓ Exercise. This limit exists.

Theorem 3.1.Let H be an ℓ-uniform hypergraph. Then for all ε> 0 there exists δ = δ(H,ε) > 0 such that every ℓ-uniform hypergraph G with G = n and e(G) > π(H)+ ε n contains | | ℓ δn H copies of H. | | Lecture 5 Proof. For each m-set M V (m), let G[M] be the hypergraph induced by G on M. If there n ∈ > ε m are η m sets M with e(G[M]) π(H)+ 2 ℓ then

n e G[M] η n m + (1 η) n π(H)+ ε m (π(H)+ ε) 6 e(G)= M 6 m ℓ − m 2 ℓ ℓ n ℓ n ℓ P m− ℓ m− ℓ − − So, assuming n > m > ℓ, we have

ε π(H)+ ε 6 η + (1 η) π(H)+ ε or η > 2 > 0 . − 2 1 π(H) ε − − 2 Pick m so that ex(m,H) < π(H)+ ε m . Then H G[M] for each of η n subsets M. 2 ℓ ⊂ m 1 1 n n H − n m − So G contains > η − | | distinct copies of H, i.e., > η copies. m m H H H − | | | || | n 1 n H Pick n > m so that > | | . 0 H 2 H ! | | | | 1 1 1 m − If n > n then pick δ 6 small enough that δn H works for n 6 n . 2 0 2 H ! H | | 0 | | | |

6 Back to graphs.

Let kp(G) denote the number of copies of Kp in G. Ramsey’s theorem says kp(G)+ kp(G) > 0, if G is large (where G is the complement of G).

Theorem 3.2 (Lorden, 1962). Let G have degree sequence d1, . . ., dn. Then

n n d k (G)+ k (G)= + (n 2)e(G)+ i . 3 3 3 − 2 i=1 X

di n 1 di Proof. The number of paths of length 2 in G and G is 2 + − 2− .

A complete or empty vertex triple contains 3 suchP paths, andP every other triple contains exactly 1. So n d n 1 d +2 k (G)+ k (G) = i + − − i 3 3 3 2 2 X d X n = 2 i 2(n 2)e(G)+3 . 2 − − 3 X 2

Corollary 3.3 (Goodman, 1959). k (G)+ k (G) > 1 n(n 1)(n 5). 3 3 24 − − n 2m/n Proof. Let m = e(G). Then k (G)+ k (G) > (n 2)m + n . 2 3 3 3 − − 2

These results show that k3(G)+ k3(G) depends only on the degree sequence, and the minimum density of monochromatic triangles is at least 1/4, as attained by a random colouring. No such result holds for K4 – it is known that the density can be < 1/33. m Corollary 3.4. k (G) > (4m n2) where n = G , m = e(G). 3 3n − | |

n d Proof. Theorem 3.2 implies that k (G)+k (G)= (n 2)m+ i , where m = e(G) 3 3 3 − − 2 and (di) is the degree sequence of G. X

But 3k (G) 6 di . So using m = n m, we have 3 2 2 − P n 2 2m/n m k (G) > (n 2)m + n = (4m n2) . 3 3 − − 3 2 3n − 2

This bound is tight only for regular graphs containing no triples with just one edge. This means that G is a union of complete graphs, so G is complete multipartite. Such graphs are rare: they are T (n) where r n. r |

Given F , let iF (G) be the number of induced subgraphs of G isomorphic to F .

So, e.g., i (G)= k (G). Kp p

Theorem 3.5. Let f(G) = F αF iF (G), the sum being over a ﬁnite collection of F , each of which is complete multipartite, with αF R, and αF > 0 unless F is complete. P ∈

7 Then, amongst graphs G of given order, f(G) is maximized on a complete multipartite

graph. Moreover, if αK3 > 0, there are no other maxima.

Proof. We may suppose that αK3 > 0, the case αK3 = 0 following by a limiting argument.

Choose a graph G of order n that maximizes f(G). Suppose G is not complete multipartite. Then G contains two non-adjacent vertices x, y whose neighbourhoods X, Y diﬀer.

The number iF (α) contains contributions from four kinds of F : those containing, respec- tively, x but not y, y but not x, both x and y, neither x nor y. The ﬁrst contribution depends only on X, and the second only on Y . Moreover the third depends only on X Y and V (X Y ) where V = V (G), since F is complete multipartite. ∩ − ∪

This fourfold partition of iF (G) means we can write f(G)= g(X)+ g(Y )+ h(X Y, V (X Y )) + C, ∩ − ∪ where C is independent of X and Y .

Lecture 6 Note that h(A, B) 6 h(A′,B′) if A A′, B B′, because F makes no contribution to h if ⊂ ⊂ F is complete, and otherwise αF > 0. Moreover, if also B = B′, then h(A, B) 0 and i (G) contributes exactly α B to h(6A, B). K3 K3 K3 | | We may suppose that g(X) > g(Y ) and, if g(X) = g(Y ), that X 6 Y and so X = X Y . Therefore 6 | | | | 6 ∪ g(X)+ h(X, V X) >g(Y )+ h(X Y, V (X Y )) − ∩ − ∪

Form H from G by removing all edges y to Y and inserting all edges y to X. Then f(H)=2g(X)+ h(X, V X)+ C>f(G). − Contradiction. 2

Note that this transformation does not increase the clique size or the chromatic number.

6 6 6 6 n Let 1 p r. For 0 x p , let ψ(x) be the maximal convex function deﬁned by ψ(0) = 0, ψ k (T (n)) = k (T (n)). (Where q = r 1, r, r +1,...) p q r q − Theorem 3.6 (Bollob´as, 1976). Let G be a graph of order n. Then kr(G) > ψ(kp(G))

Proof. Let f(G) = kp(G) ckr(G) for some real c > 0. Since ψ is convex, it suﬃces to prove that f is maximized on− a Tur´an graph. So let G be a graph on which f is maximized.

✓q✓ ✓ ★★ ✓ Suppose there exists G below ψ. ✓ ★ Consider the straight line of gradient ★ ✓q ✛ ★q c > 0. The intercept on the axis is ✦✦ ✓ ✦✦ ✓ kp(G) ckr(G). ✦q ✓ − kp(Tr−1) kp(Tr ) kp(Tr+1)

By Theorem 3.5, we know that f is maximized on a complete-partite graph, with q classes of sizes 0 r, else Tr 1(n) maximizes f. ··· − Now, f(G) = a a A ca a B + C, where A, B, C are non-zero rationals depending on 1 q − 1 q a2,a3, . . ., aq 1 and a1 + aq. −

8 We may assume without loss of generality that c is irrational, so A cB =0. If A cB < 0, then replace a by 0 and a by a + a to increase f. If A cB >−0 and6 a 6 a − 2, then 1 q 1 q − 1 q − replace a1 by a1 + 1 and aq by aq 1 to increase f. Thus a1 > aq 1 and so G is a Tur´an graph. − − 2

The exact value of min k3(G): G = n, k2(G)= m is unknown. It is conjectured to be given by r-partite graphs where r is minimal.| |

q✓✓ k3(T3(n)) I think. I foolishly tried to draw ★★ ★ this during the lecture, and so it ★ might not be right...... q ...... ★ k3(T4(n)) ...... ✦...... ✦...... ✦...... ✛ ...... Cor. 3.4 .. ✦...... q.✦......

k2(T2(n)) k2(T3(n)) =n2/4 =n2/3

The continuous envelope here for n2/4 6 m 6 n2/3 was posed as a lower bound by Fisher 6 n (1989). The whole range m 2 was proved (again in the limit) by Razborov, who introduced the method of ﬂag algebras (2002), a lowbrow view of this being a massive generalisation of Cauchy-Schwarz, using CSP to ﬁnd optimal quadratic forms.

Nikiforov (2008) did likewise for n = 4. Reiher (2012) did likewise for all r, for p = 2.

Open problem. On inducibility: what is max i (G) ? { length 3 paths }

Lecture 7 4. Szemer´edi’s Regularity Lemma

A graph having the (large scale) property that its (induced) subgraphs all have roughly the same density, the graph itself can be regarded as “pseudo-random” in a sense that can be made precise. Consider the following bipartite version.

Let U, W be disjoint subsets of the vertex set of some graph. The number of edges between U and W is denoted by e(U, W ).

e(U, W ) The density d(U, W ) is . U W | || | Deﬁnition 4.1. Let 0 <ε< 1. The pair (U, W ) is said to be ε-uniform (or ε-regular) if for U ′ U, W ′ W , ⊂ ⊂ d(U ′, W ′) d(U, W ) <ε whenever U ′ >ε U , W ′ >ε W − | | | | | | | |

(Clearly a lower bound on U ′ is necessary – e.g., U ′ = 1 is hopeless.) | | | | An ε-uniform pair is roughly regular.

Lemma 4.2. Let (U, W ) be ε-uniform and d(U, W )= d. Then

u U : Γ(u) W > (d ε) W > (1 ε) U ∈ | ∩ | − | | − | | and u U : Γ(u) W < (d + ε) W > (1 ε) U ∈ | ∩ | | | − | |

9 (Recall that Γ(u) is the neighbourhood of u.)

Proof. Let X = u U : Γ(u) W 6 (d ε) W . Then e(X, W ) 6 (d ε) X W , so d(X, W ) 6 d{ ε.∈ By the| definition∩ | of density,− |X |}6 ε U , which proves the− first| inequality.|| | − | | | | The proof of the other is similar. Alternatively, observe that in the complementary graph, (U, W ) is ε-uniform with density 1 d, and apply the first inequality. 2 − Repeated applications give lots of information about graph structure. For example, given a graph H with vertices v1, . . ., vk and subsets V1, . . ., Vk of V (G) such that (Vi, Vj ) is ε-uniform, we may find a copy of H in G. More is true.

Lemma 4.3. Let H be a graph with ∆(H) = d, and suppose that H has an r-colouring in which no colour is used more than s times. Let G be a graph containing disjoint vertex subsets V , . . ., V with V = u such that for all i, j, (V , V ) is ε-uniform and d(V , V ) > λ. 1 r | i| i j i j Suppose that (d + 1)ε 6 λd and s 6 εu . Then G contains a copy of H. ⌈ ⌉ Proof. Let c : V (H) 1, . . ., r be an r-colouring of H in which no colour is used more than s →{ } times. Let V (H)= v1, . . ., vk . We will select vertices x1, . . ., xk in G so that xixj E(G) if v v E(H). { } ∈ i j ∈ We claim that for 0 6 ℓ 6 k, vertices x , . . ., x can be chosen so that x V and, 1 ℓ i ∈ c(vi) for ℓ < j 6 k there is a set xℓ V of candidates for x at stage ℓ, meaning that j ⊂ c(vj ) j x y E(G) for every y Xℓ and every x N(j, ℓ)= x :1 6 i 6 ℓ and v v E(H) . i j ∈ j ∈ j i ∈ { i i j ∈ }

ℓ N(j,ℓ) Moreover, X > (λ ε)| | V . | j | − | c(vj )| 0 The claim clearly holds for ℓ = 0 – just take Xj = Vc(vj ). Proceed by induction on ℓ. In general, for each t T = j>ℓ + 1 : v v E(H) , let ∈ { ℓ+1 j ∈ } Y = y Xℓ : Γ(y) Xℓ 6 (λ ε) Xℓ . t ∈ ℓ+1 | ∩ t | − | t | Let m = N(ℓ +1,ℓ), and note that m + T 6 d. | | Since d(Y ,Xℓ) 6 λ ε 6 d(V , V ) ε and t t − c(vℓ+1) c(vt) − ℓ d 1 d 1 X > (λ ε) − V > (λ − (d 1)ε) V > ε V | t | − | c(vt) − − | c(vt)| | c(vt)| it follows that Y 6 ε V . | t| | c(vℓ+1)| Therefore,

Xℓ Y > (λ ε)m V (d m) V ℓ+1 − t − | c(vℓ+1) − − | c(vℓ+1)| t T [∈ > λm mε (d m)ε V − − − | c(vℓ+1)| > εu > s ⌈ ⌉ ℓ At most s 1 vertices of Xℓ+1 t T Yt have been used for x1, . . ., xℓ, so we may select − − ∈ x from this set. Take Xℓ+1 Xℓ Γ(x ) for t T , and Xℓ+1 = Xℓ for t / T . 2 ℓ+1 t − St ∩ ℓ+1 ∈ t t ∈ k 1 Lecture 8 Corollary 4.4. Let H be a graph with vertex set v1, . . ., vk. Let0 <λ,η < 1 satisfy kη 6 λ − . Let G be a graph with vertex set V V , where the V are disjoint sets of size y > 1. 1 ∪···∪ k i

Suppose that each pair (Vi, Vj ) is η-uniform, that d(Vi, Vj ) > λ if vivj E(H), and that d(V , V ) 6 1 λ if v v / E(H). ∈ i j − i j ∈

10 Then there exist vertices x1, . . ., xk (xi Vi) such that the map vi xi gives an isomorphism between H and G[ x , . . ., x ]. ∈ 7→ { 1 k} Proof. Note that, by replacing the set of V V edges by the complementary set if v v / E(H), i − j i j ∈ we may assume that H is complete and d(Vi, Vj ) > λ for all i, j.

The result is then immediate from Lemma 4.3 on taking ε = η, r = k, d = k 1, and s = 1. − 2

It is a remarkable fact that every graph can be partitioned into a bounded number of pieces, almost all pairs of which are ε-uniform. This result is due to Szemer´edi and is used in his proof that sets of integers with positive density contain arbitrarily long arithmetic progressions.

An equipartition of V (G) into k parts is a partition V , . . ., V such that n/k 6 V 6 n/k 1 k ⌊ ⌋ | i| ⌈ ⌉ for all 1 6 i 6 k, where n = V (G) . The partition is ε-uniform if (Vi, Vj ) is ε-uniform for all k 6| | 6 but at most ε 2 pairs, for 1 i < j k.

Theorem 4.5 (Szemer´edi’s Regularity Lemma). Let 0 <ε< 1 and let ℓ be a natural number. Then there exists L = L(ℓ,ε) such that every graph has an ε-uniform equipartition into m parts, for some ℓ 6 m 6 L.

Before proving the lemma, we establish two useful facts.

Lemma 4.6. Let U ′ U and W ′ W satisfy U ′ > (1 δ) U and W ′ > (1 δ) W . ⊂ ⊂ | | − | | | | − | |

Then d(U ′, W ′) d(U, W ) 6 2δ. −

Proof. Let d = d(U, W ) and d′ = d(U ′, W ′). Then

e(U, W ) e(U ′, W ′) U ′ W ′ 2 d = d(U, W )= > = d(U ′, W ′)| || | > d′(1 δ) U W U W U W − | || | | || | | || | 2 So d′ d 6 d′ 1 (1 δ) 6 2δd′ 6 2δ. − − − By considering the complementary graph, (1 d′) (1 d) 6 2δ, i.e., d d′ 6 2δ. 2 − − − − n m 1 1 Lemma 4.7. Let x1, . . ., xn be real numbers such that X = n xi. Let x = m xi. Then i=1 i=1 P P n 1 m m x2 > X2 + (x X)2 > X2 + (x X)2 n i n m − n − i=1 X − Proof. We have

n m n 1 1 1 x2 = x2 + x2 n i n i n i i=1 i=1 i=m+1 X X X m n m nX mx 2 > x2 + − − n n n m m − = X2 + (x X)2 n m − − by two applications of Cauchy-Schwarz. 2

11 Proof of Theorem 4.5. Deﬁne the index ind(P ) of an equipartition P with k parts V1, . . ., Vk to be 1 d2(V , V ) . k2 i j i k16k and 4kε5 > 100, and P is not ε-uniform, then there exists an equipartition Q into k4k parts with ind(Q) > ind(P )+ ε5/8.

This will be enough to prove the theorem. For, choose t > ℓ with 4tε5 > 100. Deﬁne f(j) 5 N f(0) = t and f(j +1) = f(j)4 . Let N = f( 4ε− ). Let L = N16 . Then if n 6 L, take an equipartition into n single vertices. Otherwise,⌈ ⌉ begin with an equipartition P into t parts.

So long as the current partition into k parts is not ε-uniform, replace it with one into k4k 6 1 5 parts and larger index. Since ind(P ) 2 , replacement can occur at most 4ε− times, so an ε-uniform partition is found with 6 L parts. ⌈ ⌉

For each (V , V ) that is not ε-uniform, select witness sets X V and X V such i j ij ⊂ i ji ⊂ j that Xij > ε Xi and Xji > ε Xj , and d(Xij ,Xji) d(Vi, Vj ) > ε. For each i, the sets X partition| | |V into| at| most| 2k| 1 atoms| . − ij i −

Let m = n/k4k and let n = k4km + ak + b where 0 6 a < 4k and 0 6 b 6 k. Then n/k =4⌊km + a⌋, so the parts of P have size 4km + a or 4km + a + 1, with B parts have the⌊ larger.⌋

Lecture 9 Partition each part of P into 4k sets of size m or m + 1. Any such partition is a partition of G into k4k sets of sizes m or m + 1, i.e. an equipartition.

Choose such a partition Q whose parts are as much as possible inside atoms: that is, every atom is a union of part of Q together with at most m extra vertices. All that remains is to check that ind(Q) > ind(P )+ ε5/8.

k q Let the setes of Q within Vi be Vi(s), for 1 6 s 6 q =4 . That is, Vi = s=1 Vi(s). Note m that e Vi(s), Vj (t) = e(Vi, Vj ) and Vi > q Vi(s) for each s, and so 16s,t6q | | | | m+1 S P 2 1 m d V (s), V (t) > d(V , V ) . q2 i j m +1 i j 16s,t6q X Since n > k16k, we have

m 2 2 2 ε5 > 1 > 1 > 1 . m +1 − m − 4k − 50 Therefore

2 1 C.S. 1 ε5 d2 V (s), V (t) > d V (s), V (t) > d2(V , V ) q2 i j q2 i j i j − 25 16s,t6q 16s,t6q ! X X

The main point is that we can improve this bound if (Vi, Vj ) is not uniform. Let Xij∗ be the largest subset of Xij that is a union of parts of Q. Because we choose parts of Q as much as possible within atoms, and because V 6 nl 4km + a > 4km, we have | i| ⌊ ⌋ k 1 k 1 2 − m 1 ε X∗ > X 2 − m > X 1 > X 1 > X 1 ij ij − ij − ε V ij − ε2k ij − 10 i | |

12 so by Lemma 4.6, we have ε d(X∗ ,X∗ ) d(X ,X ) 6 . ij ji − ij ji 5

We may assume that Xij∗ = Vi(s) for some number ri. 16s6ri S By a similar argument to that above,

1 ε5 d Vi(s), Vj (t) d(Xij∗ ,Xji∗ ) 6 rirj − 50 16s6ri 16t6rj X X

Recalling that d(X ,X ) d(V , V ) > ε, we see that | i j − i j | 1 3ε d Vi(s), Vj (t) d(Vi, Vj ) > rirj − 4 16s6ri 16t6rj X X

Applying Lemma 4.7 in the above, use of Cauchy-Schwarz with n = q2, m = r r gives i j 1 ε5 r r 9ε2 ε5 ε4 d2 V (s), V (t) > d2(V , V ) + i j > d2(V , V ) + q2 i j i j − 25 q2 16 i j − 25 3 16s,t6q X because

ri 1 m +1 1 1 Xij∗ 1 ε Xij 4ε > 1 ri > 1 | | > 1 1 | | > . q − m m q − m Vi − m − 10 Vi 5 | | | | Therefore, 1 ind(Q) = d2 V (s), V (t) k2q2 i j 16i,j6k 16s,t6q X X 1 ε5 1 k ε4 > d2(V , V ) + ε k2 i j − 25 k2 2 3 6 6 1 Xi,j k ε5 > ind(P )+ 8 completing the proof. 2

22 22 5 The proof gives something like L =2 , with ε− many 2s in the tower.

1/16 Gowers (1997) showed a tower of height ε− is necessary. Indeed, deﬁne an equipartition k V , . . ., V to be (ε, δ, η)-uniform if for all but η of the pairs (V , V ), it holds that d(V ′, V ′) 1 k 2 i j i j − d(V , V ) 6 ε whenever V > δ V . i j i′ i | | | |

Gowers proved that for small δ there is a graph such that every (1 δ1/16, δ, 1 20δ1/16)-uniform 2 − − 22 1/16 partition has at least 2 (with δ− many 2s) parts.

Lecture 10 5. A couple of applications

The simplest non-trivial case of Ramsey’s theorem asserts that there is a smallest integer R(k) such that if the edges of KR(k) are coloured red or blue then we get a monochromatic Kk. It is known that 2k/2 6 R(k) 6 4k. The existence of R(k) implies that for any graph G there is a minimal integer r(G) such that any red/blue colouring of of Kr(G) contains a monochromatic G. Clearly r(G) 6 R( G ). | |

13 Theorem 5.1. Given an integer d, there exists a number c(d) such that r(G) 6 c(d) G if ∆(G) 6 d. | |

Remark. The same is conjectured to be true for every graph with e(H) 6 d H for all H G. | | ⊂ It is known for a wider class than in the theorem, including planar graphs.

Proof. Let t = R(d + 1). Choose ε 6 min 1 , 1 . t 2d(d+1) } Let ℓ > t2 and let L = L(ℓ,ε) be the number given by Szemer´edi’s Regularity Lemma (SRL). Finally, let c = L/ε.

Let G be a gaph a maximum degree 6 d and let the edges of K , where n > c G , be n | | coloured red/blue. Apply Szemer´edi’s Lemma to R, the red subgraph of Kn, with ℓ, ε as above.

Let H be the graph with vertex set V1, . . ., Vm , where V1, . . ., Vm is the partition of { } 2 R given by SRL. Let ViVj E(H) if (Vi, Vj ) is ε-uniform. Then H = m > t and m ∈ | | e(H) 6 ε . So H Kt. [Or else, by Tur´an’s theorem, there exist integers m1, . . ., mt 1 2 − ⊃ mi m/(t 1) m 1 2 with m = m and e(H) > > (t 1) − >ε where ε 6 and m > t .] i 2 − 2 2 t P P Thus we may assume that every pair (Vi, Vj ), 1 6 i < j 6 t is ε-uniform. Colour the edges > 1 of Kt green if d(Vi, Vj ) 2 , and yellow otherwise.

Since t = R(d + 1), there exists a monochromatic Kd+1, and we may assume that it is > 1 6 6 1 spanned by V1, . . ., Vd+1. Therefore, d(Vi, Vj ) 2 for 1 i < j d +1, or d(Vi, Vj ) < 2 for 1 6 i < j 6 d + 1.

We will show that in the ﬁrst case we have R G, and that in the other case we obtain ⊃ similarly a blue G in Kn.

Take a vertex colouring of G with d + 1 colours in which no colour is used more than G times. Lemma 4.3 applied with H(there) = G(here), G(there) = (subgraph of R | | > > > 1 spanned by V1, . . ., Vd+1), u = Vi n/L c G /L G /ε, s = G , r = d + 1, λ = 2 , (d + 1)ε 6 1/2d. | | | | | | | |

This completes the proof. 2

Theorem 5.1 was proved by Chvatál, Rödl, Szemerédi, Trotter in 1983. It was extended to a larger class (using a modification of Lemma 4.3) by Chen and Schelp in 1993. Graham, Rödl, 2 Rucinski showed in 1999 that c(d) 6 2Cd log d (obviously without SRL).

Observe that the Erd˝os-Stone theorem follows from SRL and Tur´an’s theorem. Indeed, if e(G) > 1 n 1 r + ε 2 , form the “reduced” graph H whose edges correspond to pairs of positive density. Then− e(H) > t ( H ) so H K , and so G K by Lemma 4.3. r | | ⊃ r+1 ⊃ r+1 We make this precise, but, more interestingly, we recover stability too. We use an argument of Erd˝os.

Theorem 5.2 (Erd˝os, 1970). Let G be a Kr+1-free graph. Then there is a (complete) r- partite graph H with V (H)= V (G) and d (v) > d (v) for all v V (G). H G ∈ Proof. By induction on r. The case r = 1 is trivial.

14 In general, let x be a vertex of maximum degree. Then G′ = G[Γ(x)] is Kr-free, so by induction there is an (r 1)-partite graph H′ with V (H′) = V (G′) = Γ(x), and − d ′ (v) > d ′ (v) for all v Γ(x). H G ∈ Form H by joining every vertex of V (G) Γ(x) to every vertex in Γ(x). Then H is r- partite and if v / Γ(x) then d (x) = Γ(\x) = d (x) > d (v), while if v Γ(x) then ∈ H | | G G ∈ d (v)= d ′ (v)+ V (G) Γ(x) > d ′ (v)+ V (G) Γ(x) > d (v). 2 H H | \ | G | \ | G

Lecture 11 Theorem 5.3 (F¨uredi, 2010). Let G be a Kr+1-free graph. Then there is a complete r-partite graph H with V (H)= V (G) and E(G) E(H) 6 1 E(H) E(G) . | \ | 2 | \ | In particular, if e(G) > t ( G ) k then E(H) E(G) 6 3k. r | | − | △ | Proof. Construct H as in the previous proof. The proof is via induction, the case r = 1 being trivial.

6 1 Now E(G′) E(H′) 2 E(H′) E(G′) . Let there be e edges from G inside V (G) Γ(x), and f| edges\ missing| from| G between\ Γ(| x) and V (G) Γ(x). Then E(G) E(\H) = \ | \ | E(G′) E(H′) + e, and E(H) E(G) = E(H′) E(G′) + f, so it suﬃces to show that | 6 1 \ | | \ | | \ | e 2 f.

For each v V (G) Γ(x), let e(v) be the number of edges of G inside V (G) Γ(x) meeting v, and let ∈f(v) be\ the number of edges missing from G between Γ(x) and\ V (G) Γ(x) 1 1 \ meeting v. Then e = 2 e(v) and f = 2 f(v). v v P P But for each v, d (x) > d (v)= d (x) f(v)+ e(v), so e(v) 6 f(v), and we are done. G G G −

Finally, if e(G) > tr( G ) k, then tr( G ) > e(H)= e(G) E(G) E(H) + E(H) E(G) > 1 | | − | | 6 −| \ | | 6 \ | tr(G) k + 2 E(H) E(G) , so E(H) E(G) 2k, hence E(G) E(H) k, and so E(H)− E(G)| 6 3k.\ | | \ | | \ | 2 | △ |

Deﬁnition 5.4. Let G be a graph with an ε-uniform partition P into m parts V1, . . ., Vm. Let λ R. The reduced graph G(P, λ) is the graph of order M with vertex set V1, . . ., Vm and∈ edge set V V : (V , V ) is ε-uniform and d(V , V ) 6 λ . { } { i j i j i j } > G Lemma 5.5. Let G, P be as in this deﬁnition. If e(G) c | 2 | then 1 m e(G(P, λ)) 6 c ε λ , − − − m 1 2 − provided G > 2m2. | | Proof. Let e(G(P, λ)) = d m and n = G . Then 2 | | m n 2 m n 2 m n 2 n +1 e(G) 6 d +1 + ε +1 + λ +1 + m m 2 m 2 m 2 m 2 1 m n 2 6 d + ε + λ + +1 m 1 2 m − 1 n 6 d + ε + λ + m 1 2 − 2

Theorem 5.6. Let r, t N and let ε> 0. Then there exists n0(r,t,ε) such that, if G = n > n0 ∈ n | | and G K (t), then G has a subgraph G′ with e(G′) > e(G) ε and G′ K . 6⊃ r+1 − 2 6⊃ r+1 15 rt 2 Proof. Let λ = ε/2. Pick η so that (rt + 1)η 6 λ and η 6 ε/4. Let n0 =2L(ℓ,n) , where ℓ is 1 ε 6 picked so that ℓ 1 < 4 . Then t ε n/L (if not, make n0 bigger). − ⌊ ⌋ Take an η-uniform partition P of G into ℓ 6 m 6 L parts, as given by Theorem 4.5. If K G(P, λ), then by Lemma 4.3, G K (t). Hence K G(P, λ). r+1 ⊂ ⊃ r+1 r+1 6⊂

Let G′ be the subgraph of G consisting of all edges between pairs Vi and Vj where ViVj E(G(P, λ)). Then G′ Kr+1, for such a subgraph would have vertices in distinct∈ parts of P . 6⊃

The edges of G not in G′ are those in non-uniform pairs, pairs of density < λ, and inside parts, so

2 2 n m n m n m +1 e(G) e(G′) 6 η +1 + λ +1 + m − 2 m 2 m 2 1 n n 6 η + λ + < ε m 1 2 2 − 2

Theorem 5.8. Let r, t N and let ε> 0. Then there exists n0(r,t,ε) such that, if G = n > n0, G K (t) and e∈(G) > 1 1 + ε n , then there is a complete r-partite| graph| T with 6⊃ r+1 − r 2 V (T )= V (G) and E(T ) E(G) 6 11ε n . | △ | 2 Remarks. 1. This is essentially Theorem 2.1. 6 n 6 1 n 2. Hence e(G) e(T )+11ε 2 1 r + 12ε 2 , which is the Erd˝os-Stone theorem (without the log n). − > 1 n Proof. By Lemma 5.6, choose G′ G with e(G′) 1 r 2ε 2 and G′ Kr+1. Then n ⊂ − − ⊃ e(G′) > t (n) 3ε . r − 2 n Apply Theorem 5.3 to G′ to obtain T with E(G′) E(T ) 6 9ε . | △ | 2 Then E(G) E(T ) 6 11ε n . 2 | △ | 2 Lecture 12 6. Hypergraphs

Let F be an ℓ-uniform hypergraph. Deﬁne

ex(n, F ) = max e(G): G is ℓ-uniform, G = n, G F { | | 6⊃ } and ex(n, F ) π(F ) = lim n n →∞ ℓ ℓ r ℓ Denote by Kr the complete ℓ-uniform hypergraph of order r with ℓ edges. The value of π(Kr) is unknown for r>ℓ> 3. Tur´an conjectured that π(K3)=5/9. Erd˝os oﬀered $5000 for a 4 solution in his memory.

1 √ 1 √ Best bounds are 6 ( 1+ 21) = 0.5971 (Giraud, 1989), 12 (3 + 17) = 0.5935 (Chung & Lu, 1999), 0.561555 (Razborov,− using ﬂag algebras, 2008).

16 3 n n/2 n/2 Tur´an further conjectured ex(n,K5 ) = 3 ⌊ 3 ⌋ ⌈ 3 ⌉ , but this is false for odd n > 9 (Sidorenko, 1990s). − − The situation appears unstable.

π(F ) is known only in one or two simple cases; more recently, in a couple of less trivial examples, e.g. the Fano plane. But stability holds here. (F¨uredi, Simonovits, Pikhurko, Sudakov, Keevadi.)

ℓ 1 Clearly π(Kr) 6 1 r . The best known general bound is as follows. − ℓ

l Lemma 6.1. Let G be an ℓ-uniform hypergraph with nr copies of Kr (for r > ℓ), where n nℓ = e(G) and nℓ 1 = ℓ 1 , and n = G . Then provided nr 1 > 0, − − | | − n r2 n (ℓ 1)(n r)+ r r+1 > r − − nr (r ℓ + 1)(r + 1) × nr 1 − (r ℓ + 1)(r + 1) − − −

Proof. Let A1, . . ., An −1 be an enumeration of the Kr 1’s, and B1,...,Bn enumerate the Kr’s. r − r Let ai be the number of Kr’s containing Ai and bj be the number of Kr+1’s containing Bj . Then ai = rnr and bj = (r + 1)nr+1. P P Let N be the number of pairs (S,T ) where S is the vertex set of a Kr, T is an r-set not K , and S T = r 1. Clearly r | ∩ | − 2 2 r nr N = ai(n r +1 ai) = (n r + 1)rnr ai 6 (n r + 1)rnr − − − − − − nr 1 X X − On the other hand, for each Bj we can ﬁnd n r bj vertices x such that B x doesn’t span K . So there is an (ℓ 1)-set Y B −such− that Y x is not an edge.∪{ } r+1 − ⊂ j ∪{ } Then each z B Y gives a pair (S,T ) = (B ,B z x ). Thus ∈ j \ j i \{ }∪{ } N > (n r b )(r ℓ +1) = (r ℓ + 1) (n r)n (r + 1)n − − j − − − r − r+1 j X Comparing bounds on N gives the result. 2

n 1 n r +1 Corollary 6.2 (de Caen, 1983). ex(n,Kℓ) 6 1 − . r l r 1 n ℓ +1 − ℓ−1 × ! − − 1 ℓ 6 In particular, π(Kr) 1 r 1 . − −ℓ

ℓ n 6 Proof. Let G be maximal with no Kr. Then nℓ 1 = ℓ 1 and ns > 0 for ℓ s < r. − − We show by induction that

n n ℓ +1 s 1 n n s +1 1 s > − − ℓ 1+ − n s ℓ 1 n n ℓ +1 s 1 s 1 ℓ − × ℓ−1 ! − − − − holds for ℓ 6 s < r, which proves the Corollary since nr = 0.

n The case s = ℓ holds automatically since ns 1 = s 1 . − −

17 Writing qs = ns/ns 1 the desired inequality is − n ℓ +1 s 1 n n s +1 q > − − ℓ 1 + − s s ℓ 1 n − s − ℓ ! which follows from (s 1)2 (ℓ 1)(n s +1)+ s 1 qs > − qs 1 − − − (s ℓ)2 − − (s ℓ)s − − (from Lemma 6.1) and induction. 2

The next theorem (Erd˝os, 1964) shows that ℓ-partite, ℓ-uniform hypergraphs have extremal function o(nℓ), i.e. π = 0.

An ℓ-uniform hypergraph H is ℓ-partite if V (H)= V1 ... Vℓ disjoint Vi, and each edge has ∪ ∪ ℓ one vertex in each class. The complete ℓ-partite ℓ-uniform K (t1,...,tℓ) has Vi = ti and all possible edges. | |

ℓ ! Lecture 13 Theorem 6.3. Let G be an ℓ-uniform hypergraph of order n and size pn /ℓ . Let t1 6 t2 6 t1...tℓ−1 2 1 ... 6 tℓ be positive integers and suppose that p >T n− , where T = ti.

1 t1...tℓ T ℓ P Then G contains at least 2T ! p n copies of K (t1, . . ., tℓ).

Remarks. 1. Note that this quantity is similar to the expected number if the edges were chosen randomly.

ℓ ℓ 1/t1...t 2. This implies that ex(n,K (t1, . . ., tℓ)) 6 cT n − ℓ .

Proof. Let χ : V ℓ 0, 1 be the characteristic (indicator) function of edges. →{ } l Then χ(x1, . . ., xl)= pn . x1,...,xℓ P 1 χ i1 iℓ Let f(t1, . . ., tℓ)= T (x1 xℓ ). n t ··· t i ··· i ··· x1,...,x 1 x1,...,x ℓ x 1 x ℓ 1 X 1 ℓ X ℓ Y1 Yℓ

T ℓ Note that n f(t1, . . ., tℓ) is the number of labelled homomorphic copies of K (t1, . . ., tℓ) in G (copies where vertices in the same class might coincide). I.e., f() is the probability that ℓ a randomly chosen collection of T vertices spans a homomorphic copy of K (t1,...,tℓ). Now 1 χ i1 iℓ−1 iℓ f(t1,...,tℓ) = T t (x1 , . . ., xℓ 1 , xℓ ) n − ℓ ··· ··· − 1 t1 tℓ−1 i1 iℓ−1 1 tℓ x x ,...,x x − ,...,x x x ,...,x ℓ 1 X 1 ℓ 1 X ℓ−1 Y1 xYℓ−1 ℓ X ℓ Y

tℓ 1 1 χ i1 iℓ−1 = T t (x1 , . . ., xℓ 1 , xℓ) n − ℓ ··· ··· n − 1 t1 tℓ−1 i1 iℓ−1 x ! x ,...,x x − ,...,x x ℓ 1 X 1 ℓ 1 X ℓ−1 Y1 xYℓ−1 X

tℓ

use Jensen’s 1 1 i1 iℓ−1 > χ(x , . . ., x , xℓ)  T tℓ 1 ℓ 1  inequality n − ··· ··· n − 1 t1 tℓ−1 i1 iℓ−1 xℓ x1,...,x1 x −1,...,x x x  X ℓ X ℓ−1 Y1 Yℓ−1 X   tℓ  = f(t1, . . ., tℓ 1, 1) − tℓ−1tℓ t1t2...tℓ t1...tℓ > f(t1, . . ., tℓ 2, 1, 1) > > f(1, . . ., 1) = p − ···

18 T i j T 1 The contribution to n f(t1, . . ., tℓ) from terms where, say, x1 = x1 is n − f(t1 1,t2, .., tℓ), so the contribution from terms where all variables are distinct, i.e. the number− of labelled ℓ copies of K (t1,...,tℓ) is at least

ℓ t1...t T ti t1...(t 1)...t T 1 t1...t T T t1...t −1 1 p ℓ n p i− ℓ n − > p ℓ n 1 p− ℓ n− − 2 − 2 i=1 X 1 > pt1...tℓ nT 2 2

7. The size of a hereditary property

A(n ℓ-uniform hyper)graph property is a class of (ℓ-uniform hyper)graphs closed under isomorphism. P

It is non-trivial if it is not the class of all graphs.

It is hereditary if it is closed under taking induced subgraphs: if G and H is an induced subgraph of G, then H . Thus is hereditary if and only if it is∈P closed under the removal of vertices. ∈P P

is monotone if it is closed under taking any subgraph – i.e., it is closed under the removal ofP edges and vertices. Thus monotone = hereditary. ⇒ Examples. “3-colourable” is monotone • “planar” is monotone • “K -free” is monotone • 4 “no induced C ” is hereditary • 4

Observe that a hereditary property can be written = Forb( ), where is a class of graphs and Forb( ) is the class of graphs containingP no memberP of asF an inducedF subgraph. To see this, we couldF take to be = graphs not in . More usefully,F we have = Forb( ), where F F1 P P F0 0 = minimal elements of 1 with respect to inclusion, i.e. 0 = F/ : every proper induced subgraphF of F is in . F F { ∈P P} For monotone properties, it is more natural to write = Forb ( ) where “induced” is removed from the above. P ∗ F

Examples. “2-colourable” = Forb (odd cycles) • ∗ “no induced C ” = Forb(C ) • 4 4

Observe that, for any , Forb( ) is hereditary, and Forb ( ) is monotone. F F ∗ F How “large” is ? P Let be the graphs in with labelled vertex set [n]. Pn P

19 1 n t3(n) 1 3 +o(1) (2) Example: let = Forb(K4) = Forb (K4). Then n > 2 =2 − , because we can P ∗ |P | take a Tur´an graph T3(n) and all its subgraphs.

Lecture 14 We extend Deﬁnition 5.4 slightly.

Deﬁnition 7.1. Let G be a graph with an ε-uniform partition P into m = P parts V1, . . ., Vm. Let λ, µ R. The reduced graph G(P,λ,µ) is the graph of order| m| with vertex set V , . . ., V∈ and edge set V V : (V , V ) is ε-uniform, λ 6 d(V , V ) 6 1 µ . { 1 m} { i j i j i j − } Hence G(P, λ)= G(P, λ, 0).

> 2 1 Lemma 7.2. Let V1, . . ., Vm be a partition of [n], where n 2m . Let 0 <ε,λ< 4 . Let H be a graph with vertex set V , . . ., V , where e(H)= d m . { 1 m} 2 Then the number of graphs having vertex set [n] and having an ε-uniform partition P into parts V1, . . ., Vm such that G(P, λ, λ)= H is at most

d+ε+8λ log 1 + 1 n 2( λ m−1 )(2)

Proof. Consider a graph G of this kind. The number of possible edges of [n](2) between parts corresponding to edges of G(P, λ, λ), or to non-uniform pairs, or within parts, is (as in the 1 n proof of Lemma 5.5), at most d + ε + m 1 2 . − d+ε+ 1 n Hence there are at most 2( m−1 )(2) ways to choose edges of G from these. In how many ways can we choose edges of G between pairs where we must have d(Vi, Vj ) 6 λ or d(V , V ) > 1 λ? i j −

n 2 Let E = m +1 . The number of choices between some ﬁxed pair (Vi, Vj ) is at most

λE 2 2λE E E E e 8λ log 1 E 2 6 2(λE + 1) 6 6 6 2( λ ) i λE λE λ i=0 X n 6 en k (using k k ).

m 6 n Since E 2 2 , the total number of choices for G is at most

d+ε+ 1 n 8λ log 1 n 2( m−1 )(2) 2( λ )(2) × 2

Theorem 7.3 (Erd˝os, Frankl, R¨odl, 1936). Let = Forb ( ) and let r + 1 = min χ(F ): F . P ∗ F { ∈ F} 1 1 +o(1)) n Then =2( − r (2). |Pn| (1 1 +o(1))(n) Proof. Since every subgraph of T (n) is in , we have > 2 − r 2 . r P |Pn|

To prove the converse inequality, let ε > 0. We will show that, if n is large, then n 6 1 1 +ε) n |P | 2( − r (2).

Pick F with χ(F )= r +1. Let t = F . Thus if G then G Kr+1(t). Pick λ so ∈F1 ε ε | | ∈Prt 6⊃ 1 ε that 8λ log λ < 5 . Pick 0 <η< 5 so that (rt + 1)η < λ . Pick ℓ so that ℓ 1 < 5 and − t (ℓ) < 1 1 + ε ℓ . r − r 5 2 20 By Theorem 4.5, each G has an η-uniform partition P into ℓ 6 m 6 L parts, where ∈Pn L = L(ℓ, η). Since Kr+1(t) G, then Kr+1 G(P, λ, λ) if n is large, by Lemma 4.3. So e(G(P, λ, λ)) 6 t (m) 6 1 6⊂1 + ε m . 6⊂ r − r 5 2

L n ( 2 ) There are at most L choices for V1, . . ., Vm and at most 2 choices for H such that L G(P, λ, λ)= H, and at most 2(2 ) ways to specify which are the non-uniform pairs, and at L most 2(2 ) ways to specify whether a non-edge has density < λ or > 1 λ. − By Lemma 7.2, the number of G is at most ∈Pn 3(L) n 1 1 + ε + ε + ε + ε ) n 1 1 +ε) n 2 2 L 2( − r 5 5 5 5 (2) < 2( − r (2) 2

How about Forb( )? Can we ﬁnd some simple hereditary property to play the role of r- colourable graphsF in the previous proof?

Deﬁnition 7.4. A graph is (r, s)-colourable if its vertex set can be partitioned into r classes, s of which span complete graphs, and the others of which contains no edges.

Denote by (r, s) the class of (r, s)-colourable graphs. C Note. (r, 0) is the class of r-colourable graphs. (r, r) is their complements. And (r, s) is hereditary.C C C

Lecture 15 Deﬁnition. For a non-trivial property , let r( ) = max r : s (r, s) . P P { ∃ C ⊂ P} Note that r( ) is well-deﬁned. For there exists some graph H/ and H (r, s) for all r > H andP all s. Thus r( ) < H . On the other hand, if contains∈ P arbitrarily∈ C large graphs and| is hereditary,| then by Ramsey’sP | | theorem either (1, 0) P or (1, 1) . Thus r( ) > 1. C ⊂P C ⊂P P

Lemma 7.6. Let t > 3 and let R = R(t) be the Ramsey number of t. Let Kn, n > R, be 1 n edge-coloured red, blue and grey, with fewer than Rt 2 grey edges. Then there exists a red or blue K . t Proof. Suppose we recolour the grey edges red. Every set of R vertices contains a red or blue 1 1 n n R − n R − Kt, so we have at least R R− t = t t of these. The number of these containing − 1 1 n n 2 1 t n n R − 2 an originally grey edge is at most Rt 2 t −2 = Rt 2 t < t t . − N 6 6 1 N Lemma 7.7. Let r, t and 0 λ 4 . Then there exists n0 and η0 > 0 such that the following holds. ∈ ∈

Let G be a graph with vertex set r+1 V , there V being disjoint and V > n . Suppose, i=1 i i | i| 0 for 1 6 i < j 6 r + 1, then λ 6 d(Vi, Vj ) 6 1 λ, and (Vi, Vj ) is η0-uniform. Then S − there exists 0 6 s 6 r + 1 such that G contains every graph in (r +1,s)t as an induced subgraph. C

1 1 λ t Proof. Let R = R(t). Pick ε< Rt , t+1 2 . Let L = L(R,ε). Take an ε-uniform partition 6 6 Q of V1 into R m L parts.n o

Colour the edges of the reduced graph G[V1](Q, 0, 0) red or blue according to whether the 6 1 1 density of the pair is 2 or < 2 . Then, by Lemma 7.6, there is a monochromatic Kt. So we may pick subsets V1p,1 6 p 6 t, in V1, so that (V1p, V1q) is ε-uniform, 1 6 p, q 6 t, and

21 6 1 6 6 1 6 6 either d(V1p, V1q) 2 for 1 p, q t or d(V1p, V1q) > 2 for 1 p, q t.

We call V1 a “red” class in the ﬁrst case, and a “blue” class in the second case. Repeat this on the classes V2, . . ., Vr+1. Let there be s blue classes, wlog V1, . . ., Vs.

V ε λ Note that V > | i| for all i,p. Let η = min , . | ip| L +1 0 L +1 2

Then for all U Vip, W Vjq with U > ε Vip > η0 Vi , W > ε Vjq > η0 Vj , we have d(U, W ) d(V ⊂, V ) < η ⊂. In particular,| | d(|V ,| V ) | d|(V| , V| ) <| η .| Since| η| < ε , we | − i j | 0 | ip jq − i j | 0 0 2 have that (Vip, Vjq ) is ε-uniform.

λ 6 6 6 6 λ 6 6 Moreover, 2 λ η0 d(Vip, Vjq ) 1 λ + η0 1 2 for all 1 i < j r + 1, 1 6 p, q 6 t. − − −

Let H (r +1,s) . Then, provided that n is large enough, Corollary 4.4 shows that ∈ C t 0 we can ﬁnd H between some t of the classes Vip. Indeed, H has a partition into s classes of sizes t , . . ., t spanning complete graphs and r s classes of sizes t ,...,t spanning 1 s − s+1 r empty graphs, ti = t. Apply Corollary 4.4 to Vip, 1 6 i 6 r +1, 1 6 p 6 ti. All pairs > λ 6 λ are ε-uniform, with densities 2 where an edge is needed, and densities 1 2 where a non-edge is needed.P − 2

The following theorem was proved by Pr˝omel & Steger (1991) for = Forb(F ), by Bollob´as & Thomason and by Alekseev (1992) in general. The proof here is dueP to Montgomery (2011).

(1 1 +o(1))(n) Theorem 7.8. Let be a hereditary property. Then =2 − r 2 , where r = r( ). P |Pn| P Proof. By definition of r( ) there exists s such that (r, s) . Let H be any subgroup of P C ⊂ P Tr(n). Fill in s of the classes of H so that they are complete. This graph is in (r, s), so 1 n C tr(n) (1 +o(1))( ) in . There are 2 choices of H, so > 2 − r 2 . P |Pn| 1 1 +ε) n To prove the converse inequality, let ε> 0. We show 6 2( − r (2) for all large n. |Pn| (1 1 +ε)(n) Suppose not. Then there are infinitely many n with > 2 − r 2 . Pick t N. By |Pn| ∈ applying the argument of Theorem 7.8, where η is picked with η < η0, we see that there is an infinite number of n for which some G n has Kr+1 G(P, λ, λ). But then, by Lemma 7.7, (r +1,s) . ∈ P ⊂ C t ⊂Pn ⊂P Hence for each t N, ther exists s with (r +1,s) . But there are only finitely many ∈ C t ⊂P possible s, so for some s, (r +1,s)t for infinitely many t. But is hereditary, so for this s, (r +1,s) . ThisC contradicts⊂P the definition of r( ). P 2 C ⊂P P

Lecture 16 8. Containers

Let H be an ℓ-uniform hypergraph. Let r = e(H). The r-uniform hypergraph G(N,H) has (ℓ) vertex set [N] where B = v1, . . ., vr is an edge whenever B, regarded as an ℓ-graph on vertex set [N], is isomorphic to{ H. }

Notice that H-free ℓ-uniform hypergraphs on vertex set [N] correspond exactly to independent sets in G(N,H) (i.e., sets I V (G) containing no edge of G). Hence Theorem 7.3 amounts to counting independent sets in⊂G(N,H).

22 How many independent sets can there be in a graph? If it’s d-regular? Can be more than 2n/2 n (beautiful entropy argument of Kahn, 2001, shows that the maximum is found in 2d Kd,d).

How many maximal independent sets? Still many (add 1 factor to each class in previous example gives > 2n/4).

For many applications we would need fewer. Sometimes it is enough to have instead a small set of containers.

Deﬁnition 8.1. A set of containers for a hypergraph G is a collection of subsets C V (G) such that, for every independentC set I, there exists C with I C. ⊂ ∈C ⊂ We would like a small collection of containers so each C is “not too big”. C ∈C

We do this for graphs. It is not too optimistic to require C not too big – e.g., Kd,n d needs a container with C > n d. | | − | | − Deﬁnition 8.2. Let G be a graph with average degree d and of order n. The degree measure of a subset S V (G) is ⊂ 1 µ(S)= d(v) . nd v S X∈ Theorem 8.3. Let G be a graph of order n and average degree d. Let ζ > 0. Then there is a function C : [n] [n] such that, for every independent set I [n], there exists T I with P →P ⊂ ⊂

(a) I C(T ) ⊂ (b) µ(T ) 6 2/ζd (c) T 6 2n/ζ2d | | (d) µ(C(T )) 6 1 +2ζ + µ(T ) for all T [n] 2 ⊂P n n log d Remark. This gives a collection of containers with 6 2 6 e d and each container |C| 2n/ζ d is not big.

Proof. Order the vertices v1, . . ., vn by decreasing degree. Let m = max j : d(vj ) > ζd . Given I, run the following algorithm to ﬁnd T . Begin with sets T = ,Γ={ . } ∅ ∅ For j =1 to m : do if v I j ∈ if i > j : v v E(G) but v / Γ > ζd(v ), add v to T i j ∈ i ∈ j j for i = j +1 to n : do if Γ(v ) T > d(vi) , add v to Γ i ∩ d i

Suppose instead we are given some set T . Run the next algorithm, beginning with sets D = [n], Γ = . ∅ For j =1 to m : do if i > j : v v E(G) but v / Γ > ζd(v ), remove v from D i j ∈ i ∈ j j for i = j +1 to n : do if Γ(v ) T v , . . ., v > d(v i) , add v to Γ i ∩ ∩{ 1 j } d i

Observe that the set Γ constructed at each stage is the same. Observe that if T came from a run of the ﬁrst algorithm, then I D T where D is produced from the second ⊂ ∪

23 algorithm. Also, in this case, T I, and each vertex of Γ has a neighbour in T , so I Γ= . ⊂ ∩ ∅ Deﬁne C(T ) = (D T ) Γ via the second algorithm. The ﬁrst algorithm shows (a) is true. ∪ \

d(vi) Note that when a vertex vi enters Γ, it has at most d + 1 neighbours in T , since it had d(vi) fewer than d + 1 neighbours before. So, counting edges between T and Γ, d(v ) ζndµ(T )= ζ d(v ) 6 1+ i = n + n =2n j d vj T v [n] X∈ Xi∈ so (b) follows.

Since v T means j 6 m, so d(v ) > ζd and j ∈ j T ζd 6 d(v)= ndµ(T ) | | v T Xj ∈ and (c) follows.

Y = v v E(G): j

6 ζd(vj )+ ζd 6 ζnd + ζnd j=1 j=m+1 X X Therefore µ(D Γ) 6 1 +2ζ, so µ(C) 6 µ(D Γ)+ µ(T ) giving (d). 2 \ 2 \ This can be made to work for hypergraphs. Applied to G(N,H) it gives the following, where

e(H′) 1 m(H)= max − H′ H,e(H)>1 v(H ) ℓ ⊂ ′ −

Theorem 8.4. Let ε > 0. Then there exists c > 0 and n0 such that, for N >n0, there is a collection of ℓ-uniform hypergraphs on vertex set [N] such that (a) every H-free ℓ-uniform hypergraph is a subgraph of some C ∈C (b) for every C , e(C) 6 π(H)+ ε N ∈C ℓ ℓ 1/m(H) (c) log 6 cN − log N |C|

(π(H)+ε) N Corollary 8.5. The number of H-free ℓ-graphs is 2 ( ℓ ).

24 9. The Local Lemma

P n Suppose we have n events A1, . . ., An and we want j=1 Aj > 0, i.e. it is possible that no A occurs. j T 

Suppose P(Aj )= p, 1 6 j 6 n, 0

If the A are independent, then P( A )=(1 p)n > 0. i i − T Otherwise, if p< 1/n, then P( Ai) 0. S T What if p > 1/n but the Ai are “somewhat independent”? For each Ai, choose Ji [n] such that A is independent of the system A : j / J . Note there is no unique minimal⊂ choice for i { j ∈ i} Ji.

Theorem 9.1 (Local Lemma: Erd˝os, Lov´asz, 1975). Let A1, . . ., An and J1, . . ., Jn be as P above. Suppose there exist 0 <γi < 1 such that (Ai) 6 γi j J (1 γj ). ∈ i −

n n Q Then P A > (1 γ ) > 0. i − j i=1 j=1 T Q 1 n Corollary 9.2. Suppose J 6 ∆ for all i and that P(A ) 6 . Then P A > 0. | i| i e(∆ + 1) i i=1 T P n 1 Proof of 9.1. By induction on n, we may assume that i=1− Ai > 0, so we may condition P n 1 on this event. We shall prove (by induction) that ATn i=1− Ai 6 γn. T P This implies, for any S [n 1] and i / S, that Ai j S Aj 6 γi. ⊂ − ∈ ∈ T n n n The theorem follows because P A = P A A > (1 γ ). Now i i j − i i=1 i=1 j

T Q T Q P P An j

By deﬁnition of J , the top is P(A ). Relabelling so that J = 1, 2, . . ., d , the bottom is n n n { } d n 1 d n 1 d − − P A A = P A A > (1 γ )= (1 γ )  j j   i j  − i − j j=1 j=d+1 i=1 j=i+1 i=1 j J \ \ Y \ Y Y∈ n     2

Eek. Fiddly. I hope that at least some of that was transcribed correctly. . . Let me know if not.

25 Lecture 18 Theorem 9.3. Every r-regular r-uniform hypergraph is 2-colourable if r > 9 (i.e., the vertices can be coloured red/blue so that no edge is monochromatic).

6 6 nr Proof. Let the hypergraph have order n. Take a random 2-colouring. Let Ai, 1 i r = n, be the event that edge i is monochromatic. This is independent of Aj : edge j is disjoint from edge i . { } So we can take J = A : edge j meets edge i and J 6 r(r 1), since 1 6 1 i { j } | i| − 2i e(r(r 1)+1) if r > 9. − 2

We can use the Local Lemma to improve lower bounds for Ramsey. A simple argument gives t 3/2 R(3,t)=Ω log t . t 2 Theorem 9.4. R(3,t) > 1+ o(1) . log t t 2 1 ε Proof. Fix ε> 0. Let n = (1 ε) . Colour the edges of K red with probability p = − − log t n √n and blue otherwise. For α [n](3), let A be “G[α] is red”, and for β [n](t), let B be “G[β] is blue”. ∈ α ∈ β

We can take J = A ′ ,B ′ : α α′ > 2, α β′ > 2 , and J likewise. So α { α β | ∩ | | ∩ | } β n t t n J 6 3(n 3)+ and J 6 + (n t)+ | α| − t | β| 3 2 − t By the Local Lemma, it suﬃces to ﬁnd γ and δ such that

3 3(n 3) (n) p 6 γ(1 γ) − (1 δ) t − − (t) (t)+(t)(n t) (n) (1 p) 2 6 δ(1 γ) 3 2 − (1 δ) t − − −

s + t 2 t +1 Erd˝os-Szekeres showed R(3,t) 6 − = . s 1 2 − t2 Ajtai-Koml´os-Szemer´edi (1980) and Shearer (1986) showed R(3,t)= O . log t t2 Kim (1995) showed R(3,t)=Θ . log t 10. Tail Estimation

P 6 Var X Let X be a random variable with mean µ. Chebychev’s bound ( X µ > t) t2 gives a simple bound that X is far from µ, but it is weak. | − |

1 X n/2 n 2ε2n E.g., if X Bin(n, ), then − N(0, 1), so P X > εn 6 e− . ∼ 2 √n/4 ∼ | − 2 | n 1 x2 P X > x n/4 e− . | − 2 | ∼ √2π p n 1 Chebychev gives only P X > εn < 2 . | − 2 | 4ε n 26 Theorem 10.1 (Chernoﬀ, 1952). Let I1, . . ., In be independent indicators with P(Ii = 1)= 1 > pi. Let p = n pi. Let S = Ii. Then, for h 0,

P P nh2/2a nh2/2b P S > (p + h)n 6 e− and P S 6 (p h)n 6 e− − where

a = max α(1 α): p 6 α 6 p + h and b = max β(1 β): p h 6 β 6 p { − } { − − } Corollary 10.2. Let X Bin(n,p). Then ∼ 2 P X 6 (1 ε)pn 6 eε pn/2 if p 6 1 − 2 2 P X > (1 + ε)pn 6 eε pn/4 if ε 6 1

Proof of 10.2. Let h = εp. By Theorem 10.1, we obtain the ﬁrst inequality. Likewise the second. 2

Proof of 10.1. Let z R, z > 1. Then ∈ P S > γn = P zS > zγn E(zS) 6 by Markov’s inequality zγn n γn = z− (1 p )+ zp − i i i=1 γn Y n 6 z− (1 p)+ zp by AM-GM − γ (1 γ) n = (1 p)z− + pz − − Hn = e−

γ(1 p) Put z = − . Note γ > p, so z > 1. Write γ = p + h. p(1 γ) − H = (p + h) log 1+ h + (q h) log 1 h , where q =1 p. p − − q − h2 By Taylor’s theorem, H = , for some α with p 6 α 6 p + h. 2α(1 α) − The second inequality comes from the complementary P S 6 (q + h)n . 2

Lecture 19 Theorem 10.3. Let X be hypergeometrically distributed with parameters (n,N,pN). Then n X = i=1 Ii for some independent indicators Ii with some probabilities pi, such that 1 n pi = p. n i=1P InP particular, the bounds in Theorem 10.1 and Corollary 10.2 hold for X.

Proof. P X = k is the probability that, if n balls are chosen without replacement from an urn with{N balls,} pN of which are red, then k balls are red. Write R = pN.

1 R N R N − So P X = k = − , whence { } k n k n − 

27 1 X R N R N − k E = − ℓ k n k n ℓ k − X 1 R N − R ℓ N R = − − ℓ n k ℓ n k k − − 1 X R N − N ℓ = − ℓ n n ℓ 1− R n N − = ℓ ℓ ℓ

Therefore X E(zX ) = E (z 1)ℓ ℓ − ℓ X 1 n R N − = (z 1)ℓ ℓ ℓ ℓ − Xℓ 

1 R N − Noting that 6 pℓ, we see that ℓ ℓ n E(zX ) 6 pℓ(z 1)ℓ = [q + pz]n, for z > 1, ℓ − Xℓ from which we could jump in to the proof of Theorem 10.1. To get Theorem 10.3 in all its fullness,

R! n E(zX ) = (N ℓ)(N ℓ 1) (R ℓ + 1)(z 1)ℓ N! ℓ − − − ··· − − ℓ X N R R! 1 d − n N ℓ 1 = x − , where x = N! xR dxN R ℓ z 1 − ℓ − N R X R! 1 d − N 1 n = R N R x [1 + x− ] N! x dx − which has real roots. That is, E(zX ), which is a polynomial of degree n, has real roots.

E X n R E X Hence (z )= i=1(αi + βiz) with αi,βi . Since (1 ) = 1, αi + βi = 0 for all i, we X n ∈ X 6 may write E(z )= c (qi + piz) with qi + pi = 1. Again, E(1 ) = 1 means c = 1. Q i=1 Q X X Moreover, qi,pi > 0 or else there exists z > 0 with E(z ) = 0, contradicting E(z ) > 0 for all z > 0.

E X n Therefore (z )= i=1(qi+piz) with qi+pi = 1 and qi,pi > 0. But then X = I1+...+In, where there Ii are independent indicators, P(Ii =1)= pi. Q

Also, pi = EX = pn. 2 P

28 11. Martingales Inequalities

We would like bounds akin to Theorem 10.1 for variables not the sum of indicators. There are nowadays several inequalities available (Janson, Hoeffding, Talagrand) that give different information under different conditions. We look at Hoeffding.

A filter on a space Ω is a sequence , , . . ., of successively finer partitions of Ω. F0 F1 Fn A sequence of random variables X ,X ,...,X (with X defined on (Ω, ), meaning X constant 0 1 n i Fi i on parts of i) is a martingale if E(Xi+1 i) = Xi. Typically X is some random variable and X = E(FX ). | F i |Fi

The classical example is that of a gambler in a fair (zero expectation) game. If i is the game and X = current winnings, then E(X )= X . F i i+1 |Fi i (2) n Let f(G) be a graph-theoretic function. Label the edges of [n] as 1, 2,...,N = 2 . Insert the edges at random with probability p one by one to obtain a random graph G (n,p). ∈ G Let X = E f(G) the ﬁrst i edges are determined . i | E N Then X0 = (f(G)), XN = f(G) and (Xi)i=0 is a martingale. This is called the edge exposure martingale. (This is mildly ambiguous – depends on the edge labelling.)

We could instead, at stage i, add all edges between vertices 1,...,i 1 and vertex i. Xi = E f(G) all edges inside vertices 1,...,i are determined {. This is− the} vertex exposure martingale| and is a subsequence of{ (some) edge} martingale.

Lecture 20 Given a martingale, its diﬀerence sequence is Yi = Xi Xi 1. Thus E(Yi i 1) = 0. − − |F −

We aim to show that X is concentrated near its mean if the diﬀerences Xi Xi 1 are bounded (e.g., when X = χ(G)). | − − |

Lemma 11.1. Let Y be a random variable with r 6 Y 6 1 r, where r R+ and EY = 0. Then, for s > 0, we have − − ∈

sY sr s(1 r) s2/8 E(e ) 6 (1 r)e− + r − 6 e − sy sy sr s(1 r) Proof. Since e is a convex function of y, we have e 6 (1 r y)e− + (y + r)e − . Take expectations. − −

sr s For the second inequality, apply Taylor’s theorem to f(s) = log e− (1 r + re ) . 2 − Theorem 11.2 (Hoeﬀding, 1963). Let Y1, . . ., Yn be a martingale diﬀerence sequence (Xi = E(X )) with r 6 Y 6 1 r , 1 6 i 6 n. Let r = 1 n r , with 0 6 r 6 1. |Fi − i i − i n i=1 i P E P n nh2/2a P Then X > X + hn = i=1 Yi > hn 6 e−

Pn nh2/2b and P X 6 EX hn = P Y 6 hn 6 e− − i=1 i − where a = max x(1 x): r P6 x 6 r + h and b = max x(1 x): r h 6 x 6 r . { − } { − − }

For the equalities, recall: 0 = trivial, X0 = EX; n = complete partition, Xn = X. So X EX = X X = FY . F − n − 0 i P

29 Proof. We proceed in a similar way to Theorem 10.1. Thus, for s > 0, n n s P Yi shn P Yi > hn = P e i=1 > e i=1 ! X s Pn Y shn 6 E e i=1 i e−

n shn s P Yi = e− E e i=1 n 1 |F − n−1 shn s P Yi sYn = e− E e i=1 E e n 1 using E(A)= E(E(A B)) |F − | − shn s Pn 1 Y sr s 6 e− E e i=1 i e− n (1 r + r e ) − n n n shn sr s 6 e− e− i (1 r + r e ) by induction − i i i=1 shn Ysrn s n 6 e− e− (1 r + re ) by AM-GM − n s(h+r) s(1 h r) = (1 r)e− + re − − − nh2/2a 6 e− as before

The second inequality comes from considering the martingale ( Xi), with diﬀerence sequence ( Y ) and using the ﬁrst inequality. − 2 − i nh2/2a We could estimate e− as in Corollary 10.2, but the following is more useful.

Theorem 11.3. Let X = E(X ) be a martingale whose diﬀerence sequence satisﬁes a 6 i | Fi i Yi 6 ai + ci, where ai is a function on (Ω, i 1) and ci R, 1 6 i 6 n. F − ∈ Then, for t> 0, 2t2/ Pn c2 P X > EX + t 6 e− i=1 i 2t2/ Pn c2 PX 6 EX t 6 e− i=1 i − Yn Proof. Applying Lemma 11.1 to Y = E n 1 we obtain cn |F − 2 2 sYn cnsY s c /8 E e n 1 = E e 6 e n |F − Thus, following the previous proof,

2 ts s Pn c2 P (X > EX + t)= P Yi > t 6 e− e 8 i=1 i

n 2 X 2 Now take s =4t/ i=1 ci . P We can think of the ci as the maximum change in Xi at stage i.

t2/2 P c2 A weaker form of Theorem 11.3, known as Azuma’s inequality, gives the bound e− i , based on the assumption that Yi 6 ci. But in most applications, Yi is not symmetric about zero and so 11.3 is stronger. | |

A useful corollary follows.

Corollary 11.4. Let Z1,...,Zn be independent random variables with Zi taking values in space n R Ai, 1 6 i 6 n. Suppose that f : i=1 Ai satisﬁes f(z) f(z′) 6 ci whenever z,z′ th → | − | diﬀer only in the i coordinate. Let Z = f(Z1,...,Zn). Q 2t2/ P c Then P ( Z EZ > r) 6 2e− i . | − |

30 Proof. Define a filter on Ω = n A where partitions Ω into i A parts according to i=1 i Fi j=1 | j | the first i coordinates. Then Xi = E(Z i) is a martingale where X0 = EZ, Xn = Z. The conditions on f implyQ that | F Q

min Xi 6 Xi 1 = E(Xi) 6 max Xi 6 min Xi + ci z − z z

where z runs over Ai. Hence (Yi) satisfy Theorem 11.3 with ai = minz Xi Xi 1. 2 − −

Lecture 21 12. The Chromatic Number of a Random Graph

As usual, (n,p) is the space of random graphs on vertex set [n], with edges chosen independently at randomG with probability p (we take p constant).

We say that an event holds with high probability (whp) if P(A) 1 as n . (Strictly speaking, we have a sequence of events A (n,p).) → → ∞ n ⊂ G Martingale inequalities show that χ(G) is highly concentrated around its mean. But what is the mean?

Lemma 12.1. Let 0 1+ o(1) whp, where q =1 p. 2 log1/q n − Proof. Let ε> 0. Let d = (2 + ε) log)1/qn +1. Let X be the number of independent sets of size d in G. Then n n (d) (d 1)/2 d P χ(G) 6 6 P(X > 1) 6 EX = (1 p) 2 6 nq − 0 as n d d − → → ∞ This is true for all ε> 0. 2

It is not hard to show that the greedy algorithm uses 1+ o(1) n/ log1/q n colours. Chebychev’s bound shows that there exist independent sets of size 2 log1/q n whp, but not with very high probability.

Theorem 12.2 (Bollob´as, 1988). Let 0

1 1 ε 1 ε/2 Let 0 <ε< 20 , and let d = (2 ε) log1/q n . We shall ﬁnd m with n − 0, the theorem follows.

31 Let Pm = P H (n,p) has no independent d-set . Then the probability of A failing is at most ∈ G 1 n m m log n m log m P 6 n P = e P 6 e 1+ε P m m m m m m1+δ So it suﬃces to show that Pm 0.

m d Let X be the number of independent d-sets in H. Then EX = q(2). d 

2/d d 1 ε em q If m

d m qq/2 If m>n1 ε/2 then EX > > ndε/8 > n2. − d √ d If n is large. Clearly EX increases as m increases, and moving from m 1 to m the value m − increases by a factor m d < 2 for large n. It follows that there is some value of m for − which 2m7/4 6 EX 6 4m7/4.

Deﬁne the random variables Y = max number of independent d-sets with 6 1 vertex in common. Z = number of independent d-sets meeting every other in 6 1 vertex.

m Clearly Z 6 Y 6 X. Consider H as an element of the product space 0, 1 ( 2 ). The value of Y changes by at most 1 if we change one edge. So we can apply{ Corollary} 11.4 to Y m 2t2/(m) with c =1, 1 6 i 6 , so P(Y 6 EY t) 6 e− 2 . i 2 − If we can show that EY > m7/4 then we are done, because then

7/4 2m7/2/(m) 4m3/2 P = P(X = 0) 6 P(Y = 0) 6 P(Y 6 EY m ) 6 e− 2 6 e− m −

Lecture 22 Now EY > EZ, so it’s enough to show that EZ > m7/4. For 2 6 r 6 d 1, let − Wr = number of independent d-sets meeting another in r vertices

d 1 − Then X 6 Z + W2 + W3 + ... + Wd 1. So EZ > EX EWr. Then − − r=2 P m (d) d m d (d) (r) EW 6 q 2 − q 2 − 2 = EXf r d r d r r − d m d (d) (r) where f = − q 2 − 2 . r r d r − d 1 − 6 1 E > 7/4 So it suﬃces to show that fr 2 , since X 2m . r=2 P First,

2 1 2 1 d m m − d (r) d m − (r) f 6 q(2)− 2 = EX q− 2 r r d r r r 2 r d √q 9r/10 6 EX 6 EXm− mqr/2 

32 d/12 7/4 for r 6 d/12. Since EX 6 m , we have fr 6 1/4. r=2 P Secondly, writing s = d r, we have −

+1 s s d m d sd (s ) d s+1 1/2 13d/24 f = − q − 2 6 dmq − 2 6 dmq− q r s s for s 6 11d/12, i.e. for r > d/12.

1 13d 13 1 1 Now ε< , so > (1 ε) log n > 1+ log n > 1+ log m. 20 24 12 − 40 40 d 1 − 1/50 So fr 6 dm− 6 1/4. 2 r=d/12 P 13. The Semi-Random Method

In 1980, Ajtai-Komlós-Szemerédi proved that a triangle-free graph of order n and average degree n > n d has an independent set of size d log n (as opposed to d by Turán).

1 They achieved this by choosing a small random subset of size 2 X , such that the conditions in G X Γ(X) could be described, not much worse than in G. Repeat.| | This is better than − − (a) removing randomly one-by-one (b) removing a large set once

The method was used by R¨odl (then Frankl-R¨odl) to prove the Erd˝os-Hanani conjecture.

We use Chebychev’s inequality:

E (X EX)2 Var X P E > P E 2 > 2 6 − X X t = (X X) t 2 = 2 | − | − t t Note that E (X EX)2 = EX2 (EX)2. − − 2 Note that if X = Iα, then EX = E IαIβ = P(Iα and Iβ). α,β α,β P P P Call an r-uniform hypergraph δ-semiregular (of degree d) if

(1 δ)d 6 d(v) 6 (1 + δ)d for all but δ G vertices − | | d(v) 6 erd for all v d(u, v) 6 δd for all u, v with u = v 6 where d(u, v) is the codegree, the number of edges containing u, v . { }

Lemma 13.1. For all r > 2 and 0 <ε< 1, there exists δ0 such that, if δ <δ0 and G is 2 r-uniform δ-semiregular, then G has a subhypergraph G′ of order at most (1 ε + ε ) G 1/4 2 − | | that is δ -semiregular, and V (G) V (G′) is covered by at most (ε + ε ) G /r edges. − | |

33 Lecture 23 Proof. Let G be δ-semiregular of degree d. Call the vertices not satisfying (1 δ)d 6 d(v) 6 (1 + δ)d bad. Let B be the set of bad vertices. Choose a set X of edges independently− at random with probability ε/d. Let G′ be the subgraph on vertex set V (G) V (X) B. − − 1/4 2 We show that, with positive probability, G′ is δ -semiregular and X + B 6 (ε+ε )n/r, | | | | where n = G . This will prove the lemma, because our edge cover for V (G) V (G′)... (something| I missed,| sorry - anyone help?) −

Note δd > 1, so we can make d, and hence also n, as large as we like, by making δ0 small.

(a) e(G) 6 1 (1 δ)n(1 + δ)d + δnerd 6 1+ ε nd is δ is small. r − 2 r 0 By Corollary 10.2, 2 2 ε n 3ε n ε2 nε 1 P X > EX + 6 P X > ε + 6 e− 16 4r < | | 4 r | | 4 r 3 if δ is small. Hence P( ) > 2/3. ( B 6 δn 6 ε2n/4r.) 0 A | | (b) Let v be a good (i.e, not bad) vertex, and I be the event v V (G′). v ∈ So G′ = Iv. Now | | v P 1 ε (1+δ)d ε d(v) < 1 6 P(v G′) = 1 3 − d ∈ − d (1 δ)d ε − ε(1 δ) 2 2 6 1 6 e− − 6 1 ε + ε − d − 3 Var G′ 2 2 So by Chebychev, P( ) 6 | | , since E G′ 6 1 ε + ε n. B (ε2n/12)2 | | − 3 Let u, v V (G) B, u = v. Then ∈ − 6 ε d(v)+d(v) d(u,v) ε δd P(I and I ) = 1 − 6 1 − P(I )P(I ) u v − d − d u v So 2 2 ε 2 E( G′ ) = P(I and I ) = + 6 E( G′ )+ 1 E( G′ ) | | u v | | − d | | u,v u=v u=v X X X6 ε δd since 1 − 6 eεδ 6 1+2δε. So − d 2 2 2 2 Var ( G′ ) = E G′ (E G′ ) 6 E G′ +2δεE G′ 6 n +2δεn | | | | − | | | | | | n +2δεn2 1 Hence P(B) 6 < if δ is small. (ε2n/12)2 3 0 (c) See handout.

Theorem 13.2 (Frankl-R¨odl, 1985). Given r,ε > 0, there exists δ > 0 such that if G is d-regular r-uniform of maximum codegree < δd, then G can be covered by (1 + ε) G /r edges. | |

34 Remark. Hence there is a set of (1 rε) G /r independent edges, i.e. pairwise disjoint. For let W be the set of vertices in more− than| | one cover edge. Then G + W 6 (1 + ε) G , so W 6 ε G . Remove edges meeting W . | | | | | | | | | | | | 1+ η Proof. Choose η > 0 and k N so that + r(1 η + η2) < 1+ ε. ∈ 1 η − − 4k Then δ<δ0(r, η) will work. For Lemma 13.1 can be applied k times with r and η. The edges selected amount to at most

2 2 2 2 2 k 1 1+ η G (η + η ) 1+(1 η + η )+(1 η + η ) + + (1 η + η ) − G 6 | | − − ··· − | | 1 η r − and there remain 6 (1 η + η2)k G vertices, which can be covered by an edge apiece. 2 − | | Deﬁne:

m(n,k,ℓ) = maximum number of k-sets in [n] such that each ℓ-set is in at most one of them.

M(n,k,ℓ) = minimum number of k-sets in [n] such that each ℓ-set is in at least one of them.

Erd˝os-Hanani (1963) conjectured

n m(n,k,ℓ) ℓ M(n,k,ℓ) ∼ k ∼ ℓ