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External Graph Theory 2017-2018 Course Notes

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Extremal Graph Theory

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Part Extremal Graph Theory Based on lectures A. G. Thomason Notes taken Dexter Chua Michaelmas 2017 These notes are not endorsed the lecturers, and I have modified them (often significantly) after lectures. They are nowhere near accurate representations of what was actually lectured, and in particular, all errors are almost surely mine. theorem, giving the maximum size of a graph that contains no complete subgraph, is an example of an extremal graph theorem. Extremal graph theory is an umbrella title for the study of how graph and hypergraph properties depend on the values of parameters. This course builds on the material introduced in the Part II Graph Theory course, which includes theorem and also the theorem. The first few lectures will cover the theorem and stability. Then we shall treat Regularity Lemma, with some applications, such as to hereditary properties. Subsequent material, depending on available time, might include: hypergraph extensions, the flag algebra method of Razborov, graph containers and applications. A knowledge of the basic concepts, techniques and results of graph theory, as afforded the Part II Graph Theory course, will be assumed. This includes theorem, theorem, theorem and so on, together with applications of elementary probability. 1 Contents Extremal Graph Theory Contents 1 The theorem 3 2 Stability 9 3 Supersaturation 12 4 regularity lemma 17 5 Subcontraction, subdivision and linking 26 6 Extremal hypergraphs 37 Index 38 2 1 The theorem Extremal Graph Theory Lemma. Let c, 0. there exists n1 n1 (c, such that if n n1 n and e(G) (c 2 , then G has a subgraph H where and Proof. The idea is that we can keep removing the vertex of smallest degree and then we must eventually get the want. Suppose this gives us a suitable graph even after got vertices left. That means we can find a sequence G Gn Gs , where s nc, j and the vertex in Gj has degree cj in Gj . We can then calculate n X n j e(Gs ) (c 2 n (c 2 2 2 2 2 as n gets large (since c and s are fixed numbers). In particular, this is 2s . But Gs only has s vertices, so this is impossible. Using this, we can reduce the theorem to a version that talks about the minimum degree instead. Lemma. Let r 1 be an integer and 0. Then there exists a d1 d1 (r, and n2 n2 (r, such that if n n2 and 1 1 n, r then G (t), where t bd1 log nc. We first see how this implies the theorem: Proof of theorem. Provided n0 is large, say n0 n1 1 1r , , p we can apply the first lemma to G to obtain a subgraph H G where n, and 1 1r p We can then apply our final lemma as long as n is big enough, and obtain p (t) H G, with t d1 (r, log 2n . We can now prove the lemma. Proof of lemma. We proceed induction on r. If r 0 or 1r , the theorem is trivial. Otherwise, the induction hypothesis, we may assume G Kr (T ) for 2t T . Call it K Kr (T ). This is always possible, as long as 1 d1 (r, d1 r 1, . 2 r(r 1) 4 1 The theorem Extremal Graph Theory 1 1 1r , The exact form is not important. The crucial part is that which is how we chose the to put into d1 (r 1, Let U be the set of vertices in G K having at least 1 1r neighbours in K. Calculating 1 1 1 rT (r 1)T T (r 1)T t, r 2 r 2 2 we see that every element in U is joined to at least t vertices in each class, and hence is joined to some Kr (t) K. So we have to show that is large. If so, then a pigeonhole principle argument, it follows that we can always find enough vertices in U so that adding them to K gives a (t). Claim. There is some c 0 (depending on r and such that for n sufficiently large, we have cn. The argument to establish these kinds of bounds is standard, and will be used repeatedly. We write e(K, G K) for the number of edges between K and G K. the minimum degree condition, each vertex of K sends at least 1 1r n edges to G K. Then we have two inequalities 1 1 n e(K, G K) r 1 (n 1 r 2 Rearranging, this tells us 2 1 r 2 . Now note that log n, so for large n, the second term on the left is negligible, and it follows that U n. We now want to calculate the number of Kr in K. To do so, we use the simple inequality n en k . k k Then we have (t) log n 3e 3e T eT nrd t t Now if we pick d sufficiently small, then this tells us (t) grows sublinearly with n. Since grows linearly, and t grows logarithmically, it follows that for n large enough, we have t (t). Then the pigeonhole principle, there must be a set W U of size t joined to the same Kr (t), giving a (t). 5 1 The theorem Extremal Graph Theory as desired. To see that such an addition is possible, we choose edges inside W independently with probability p, to be determined later. Let X be the number of edges chosen, and Y be the number of K2 (t) created. If n2 , then this means there is an actual choice of edges with X Y n2 . We then an edge from each K2 (t) to leave a K2 graph with at least X Y n2 edges. So we want to actually compute Y Seeing that asymptotically, m is much larger than t, we have Y m 1 m m t t2 p 2 2 t t 2 1 1 pm2 m2t pt 2 2 2 1 pm2 (1 pt ) 2 1 pm2 (1 (m2 ). 2 We pick p and (3r2 . Then p , and m2 m2 (m2 1 . 2 Hence, we find that n 1 pm2 . 2 4 summarize what we have got so far. Let t(n, be the largest value of t such that a graph of order n with 1 1r n2 edges is guaranteed to contain (t). Then we know t(n, r, grows with n like log n. If we are keen, we can ask how t depends on the other terms r and We just saw that t(n, r, 3 log n . log So we see that the dependence on is at most logarithmic, and in 1976, os and Simonovits showed that t(n, r, c log n r log for some c. Thus, t(n, r, also grows (inverse) logarithmically with In our original bound, there is no dependence on r, which is curious. While and bound does, in 1978, and showed that 1 log n 500 log if n is large. So we know there actually is no dependence on r. 7 1 The theorem Extremal Graph Theory We can also ask about the containment of less regular graphs than Kr (t). In 1994, as and Kohayakawa adapted the proof of the theorem to show that there is a constant c such that for any 0 1, if e(G) 1 1r n2 and n is large, then we can find a complete (r subgraph with class sizes k j 3 log n log n , , r 1 times, 2 2 . r log log r We might wonder if we can make similar statements for hypergraphs. It turns out all the analogous questions are open. Somehow graphs are much easier. 8 2 Stability Extremal Graph Theory Let Vi be the set of vertices in G K joined to fewer than t of Ci . It is now enough to show that o(n2 ). We can then throw away the edges in each Vi to obtain an graph. Suppose on the contrary that , say, for some j. Then Stone with r 1 says we have K2 (t). Each vertex of K2 (t) has at s least s 2t 1 in each other Ci , for i j. So we see that K2 (t) has at least s s 2t 2t 1 t common neighbours in each Ci , giving (t) G. It now remains to show that (i) and follows from (ii). The details are left for the example sheet, but we sketch out the rough argument. (ii) (i) is straightforward, since an graph with that many edges is already pretty close to being a an graph. To deduce since we can assume 1 1r o(1) n, we note that if 1 did not hold, then we have vertices of degree 1 r n, and their removal leaves a graph of order (1 with at least 1 n 1 1 (1 1 o(1) 1 n 1 , r 2 r r 2 which would contain (t). So we are done. Corollary. Let ) r 1, and let G be for F , i. G F and e(G) F ). Then 1 1r o(1) n. Proof. our asymptotic bound on ex(n, F ), it cannot be that is greater than 1 1r o(1) n. So there must be a vertex v G with d(v) 1 1r n. We now apply version of the stability theorem to obtain an subgraph H of G. We can certainly pick vertices in the same part of H, which are joined to m 1 1r o(1) common neighbours. Form from G v adding a new vertex u joined to these m vertices. Then ) e(G), and so the maximality of G entails F . Pick a copy of F in . This must involve u, since G, and hence G v does not contain a copy of F . Thus, there must be some x amongst the vertices mentioned. But then we can replace u with x and we are done. Sometimes, stability and bootstrapping can give exact results. Example. ex(n, C5 ) t2 (n). In fact, ex(n, ) t2 (n) if n is large enough. Theorem (Simonovits). Let F be (r i. ) r 1 but ) e) r for every edge e of F . Then for large n, ex(n, F ) tr (n). and the only extremal graph is Tr (n). So we get a theorem like theorem for large n. Proof. Let G be an extremal graph for F and let H be an subgraph with 1 1 o(1) n. r Note that H necessarily have r parts of size 1r o(1) n each. Assign each of the o(n) vertices in V (G) V (H) to a class where it has fewest neighbours. 10 2 Stability Extremal Graph Theory Suppose some vertex v has neighbours in its own class. Then it has in each class. Pick neighbours in class of H. These neighbours span a graph with at least 1 1r o(1) edges. So (or arguing 2 directly), they span F w for some vertex w (contained in Kr Hence G F , contradiction. Thus, each vertex of G has only o(n) vertices in its own class. So it is joined to all but o(n) vertices in every other class. Suppose some class of G contains an edge xy. Pick a set Z of vertices in that class with Z. Now Z has 1r o(1) n common neighbours in each class, so ( or directly) these common neighbours span a But together with Z, we have a Kr but with an edge added inside a class. But our condition that F is for any e, this subgraph contains an F . This is a contradiction. So we conclude that no class of G contains an edge. So G itself is but the graph with most edges is Tr (n), which does not contain F . So G Tr (n). 11 3 Supersaturation Extremal Graph Theory Then H for at least n m choices of M . Hence G contains at least n n m copies of H, and we are done. (Our results hold when n is large enough, but we can choose small enough so that 1 when n is small) Let kP (G) be the number of copies of Kp in G. theorem tells us 0 if is large (where G is the complement of G). In the kp (G) kp (G) simplest case where p 3, it turns out we can count these monochromatic triangles exactly. Theorem (Lorden, 1961). Let G have degree sequence d1 , . . . , dn . Then n X di n (n 2)e(G) k3 (G) k3 (G) . 3 2 is precisely Proof. The number of paths of length 2 in G and G n n X X di n 1 di di n 2(n 2)e(G) 3 , 2 2 2 3 since to find a path of length 2, we pick the middle vertex and then pick the two edges. A complete or empty K3 contains 3 such Other sets of three vertices contain exactly 1 such path. Hence n number of paths of length 2. 2(k3 (G) k3 (G)) 3 Corollary (Goodman, 1959). We have k3 (G) k3 (G) 1 n(n 1)(n 5). 24 In particular, the Ramsey number of a triangle is at most 6. Proof. Let m e(G). Then k3 (G) k3 (G) n (n 2)m n . 3 2 Then minimize over m. This shows the minimum density of a monochromatic K3 in a colouring of Kn (for n large) is 41 . But if we colour edges monochromatic, then 1 4 is the probability of having a triangle being monochromatic. So the density is achieved a Recall also that the best bounds on the Ramsey numbers we have are obtained random colourings. So we might think the best way of colouring if we want to minimize the number of monochromatic cliques is to do so randomly. However, this is not true. While we do not know what the minimum density of monochromatic K4 in a colouring of Kn , it is already known to be 1 1 33 (while 32 is what we expect from a random colouring). It is also known 1 flag algebras to be 35 . So we are not very far. 13 3 Supersaturation Extremal Graph Theory Corollary. For m e(G) and n we have k3 (G) m (4m n2 ). 3n Proof. theorem, we know X n (n 2)e(G) k3 (G) k3 (G) , 3 2 But where is the degree sequence in G. 3k3 (G) X , 2 since the sum is counting the number of paths of length 2. So we find that n 2 , k3 (G) (n 2)m n 3 2 3 and m n2 m. Observe that equality is almost never attained. It is attained only for regular graphs with no subgraphs looking like So is an equivalence relation, so the graph is complete and regular. Thus it is Tr (n) with r n. Theorem. Let G be a graph. For any graph F , let iF (G) be the number of induced copies of F in G, i. the number of subsets M V (G) such that F . So, for example, iKp (G) kp (G). Define X f (G) iF (G), F with the sum being over a finite collection of graphs F , each being complete multipartite, with R and 0 if F is not complete. Then amongst graphs of given order, f (G) is maximized on a complete graph. Moreover, if 3 0, then there are no other maxima. Proof. We may suppose 3 0, because the case of 3 0 follows from a limit argument. Choose a graph G maximizing f and suppose G is not complete multipartite. Then there exist vertices x, y whose neighbourhoods X, Y differ. There are four contributions to iF (G), coming from (i) F that contain both x and (ii) F that contain y but not F that contain x but not (iv) F that contain neither x nor y. 14 3 Supersaturation Extremal Graph Theory If the theorem hold, then we can pick a G such that (kp (G), kr (G)) lies below Draw a straight line through the point with slope parallel to This has slope 1c 0 for some c. The intercept on the is then kp (G) ckr (G), which would be greater than f (any graph) convexity, a contradiction. Now the previous theorem immediately tells us f is maximized on some complete graph. Suppose this has q class, say of sizes a1 a2 aq . It is easy to verify q r 1. In fact, we may assume q r, else the maximum is on a an graph (n). Then we can write f (G) a1 aq A caq aq B C, where A, B, C are rationals depending only on a2 , . . . , and a1 aq (A and B count the number of ways to pick a Kp and Kr respectively in a way that involves terms in both the first and last classes). We wlog assume c is irrational. Hence aq aq A ca1 aq a1 aq (A cB) 0. If A cB 0, replace a1 and aq 0 and a1 aq . This would then increase f , which is impossible. If A cB 0 and a1 aq 2, then we can replace a1 , aq a1 1, aq 1 to increase f . Hence a1 aq 1. So G Tq (n). 16 4 4 regularity lemma Extremal Graph Theory regularity lemma regularity lemma tells us given a very large graph, we can always equipartition it into pieces that are in some sense. The lemma is arguably but it also has many interesting consequences. To state the lemma, we need to know what we mean Definition (Density). Let U, W be disjoint subsets of the vertex set of some graph. The number of edges between U and W is denoted e(U, W ), and the density is e(U, W ) d(U, W ) . Definition pair). Let 0 1. We say a pair (U, W ) is if 0 , W 0 ) d(U, W whenever U 0 U , W 0 W , and 0 0 Note that it is necessary to impose some conditions on how small U 0 and W 0 can be. For example, if 0 0 1, then d(U 0 , W 0 ) is either 0 or 1. So we cannot have a sensible definition if we want to require the inequality to hold for arbitrary U 0 , W 0 . But we might be worried that it is unnatural to use the same for two different purposes. This is not something one should worry about. The regularity lemma is a fairly robust result, and everything goes through if we use different for the two different purposes. However, it is annoying to have to have many different floating around. Before we state and prove regularity lemma, first try to understand why uniformity is good. The following is an elementary observation. Lemma. Let (U, W ) be an pair with d(U, W ) d. Then U : W (d (1 U : W (d (1 where is the set of neighbours of u. Proof. Let X U : W (d Then e(X, W ) (d So d(X, W ) d d(U, W ) So it fails the uniformity condition. Since W is definitely not small, we must have The other case is similar, or observe that the complementary bipartite graph between U and W has density 1 d and is What is good about pairs is that if we have enough of them, then we can construct essentially any subgraph we like. Later, regularity lemma says any graph large enough has equipartitions, and together, they can give us some pretty neat results. 17 4 regularity lemma Extremal Graph Theory Indeed, write m 1, Then m In particular, since the 0 case is trivial, we may assume m 1. So we can easily bound Thus, it must be the case that Therefore, we can bound Yt s 1. So we are done. Corollary. Let H be a graph with vertex set , . . . , vr Let 0 1 satisfy . Let G be a graph with disjoint vertex subsets V1 , . . . , Vr , each of size u 1. Suppose each pair (Vi , Vj ) is uniform, and d(Vi , Vj ) if vi vj E(H), and d(Vi , Vj ) 1 if vi vj E(H). Then there exists xi Vi so that the map vi xi is an isomorphism H , . . . , xr Proof. replacing the Vi edges the complementary set whenever vi vj E(H), we may assume d(Vi , Vj ) for all i, j, and H is a complete graph. We then apply the previous lemma with r 1 and s 1. showed that every graph that is sufficiently large can be partitioned into finitely many classes, with most pairs being The idea is simple whenever we see something that is not uniform, we partition it further into subsets that are more uniform. The of the proof is to come up with a measure of how far we are from being uniform. Definition (Equipartition). An equipartition of V (G) into k parts is a partition into sets V1 , . . . , Vk , where b nk c Vi d nk e, where n We say that the partition is if (Vi , Vj ) is for all but k2 pairs. Theorem regularity lemma). Let 0 1 and let be some natural number. Then there exists some L such that every graph has an equipartition into m parts for some m L, depending on the graph. This lemma was proved in order to prove his theorem on arithmetic progressions in dense subsets of integers. When we want to apply this, we usually want at least many parts. For example, having 1 part is usually not very helpful. The upper bound on m is helpful for us to ensure the parts are large enough, picking graphs with sufficiently many vertices. We first need a couple of trivial lemmas. 19 4 regularity lemma Extremal Graph Theory Lemma. Let U 0 U and W 0 W , where 0 (1 and 0 (1 Then 0 , W 0 ) d(U, W Proof. Let d d(U, W ) and d0 d(U 0 , W 0 ). Then e(U, W ) e(U 0 , W 0 ) 0 0 d0 d0 (1 . Thus, d0 d d0 (1 (1 ) The other inequality follows from considering the complementary graph, which tells us (1 d0 ) (1 d) Lemma. Let x1 , . . . , xn be real numbers with n 1X xi , n 1 X xi . m and let m Then n m m 1X 2 xi X 2 (x X)2 X 2 (x X)2 . n n If we ignore the second term on the right, then this is just Proof. We have n m n 1X 2 1X 2 1 X 2 xi xi x n n n i m 2 n m nX mx x n n m 2 (x X)2 two applications of We can now prove regularity lemma. Proof. Define the index ind(P) of an equipartition P into k parts Vi to be ind(P ) 1 X 2 d (Vi , Vj ). k 2 We show that if P is not then there is a refinement equipartition Q 5 into k4k parts, with ind(Q) ind(P) . 20

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External Graph Theory 2017-2018 Course Notes

Module: Extremal Graph Theory

3 Documents

Students shared 3 documents in this course

University: University of Cambridge

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Part III — Extremal Graph Theory

Based on lectures by A. G. Thomason

Notes taken by Dexter Chua

Michaelmas 2017

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Tur´an’s theorem, giving the maximum size of a graph that contains no complete

-vertex subgraph, is an example of an extremal graph theorem. Extremal graph theory

is an umbrella title for the study of how graph and hypergraph properties depend on

the values of parameters. This course builds on the material introduced in the Part

II Graph Theory course, which includes Tur´an’s theorem and also the Erd¨os–Stone

theorem.

The first few lectures will cover the Erd¨os–Stone theorem and stability. Then we

shall treat Szemer´edi’s Regularity Lemma, with some applications, such as to hered-

itary properties. Subsequent material, depending on available time, might include:

hypergraph extensions, the flag algebra method of Razborov, graph containers and

applications.

Pre-requisites

A knowledge of the basic concepts, techniques and results of graph theory, as afforded

by the Part II Graph Theory course, will be assumed. This includes Tur´an’s theorem,

Ramsey’s theorem, Hall’s theorem and so on, together with applications of elementary

probability.

External Graph Theory 2017-2018 Course Notes

Extremal Graph Theory

University of Cambridge

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External Graph Theory 2017-2018 Course Notes

Module: Extremal Graph Theory

University: University of Cambridge

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