# Rational exponents in extremal graph theory

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1 Rational exponents in extremal graph theory Boris Bukh David Conlon Abstract Given a family of graphs H, the extremal number ex(n, H) is the largest m for which there exists a graph with n vertices and m edges containing no graph from the family H as a subgraph. We show that for every rational number r between 1 and 2, there is a family of graphs H r such that ex(n, H r ) = Θ(n r ). This solves a longstanding problem in the area of extremal graph theory. 1 Introduction Given a family of graphs H, another graph G is said to be H-free if it contains no graph from the family H as a subgraph. The extremal number ex(n, H) is then defined to be the largest number of edges in an H-free graph on n vertices. If H consists of a single graph H, the classical Erdős Stone Simonovits theorem [8, 9] gives a satisfactory first estimate for this function, showing that ex(n, H) = where χ(h) is the chromatic number of H. ( 1 1 χ(h) 1 + o(1) ) ( n 2 When H is bipartite, the estimate above shows that ex(n, H) = o(n 2 ). This bound is easily improved to show that for every bipartite graph H there is some positive δ such that ex(n, H) = O(n 2 δ ). However, there are very few bipartite graphs for which we have matching upper and lower bounds. The most closely studied case is when H = K s,t, the complete bipartite graph with parts of order s and t. In this case, a famous result of Kővári, Sós and Turán [14] shows that ex(n, K s,t ) = O s,t (n 2 1/s ) whenever s t. This bound was shown to be tight for s = 2 by Erdős, Rényi and Sós [7] and for s = 3 by Brown [3]. For higher values of s, it is only known that the bound is tight when t is sufficiently large in terms of s. This was first shown by Kollár, Rónyai and Szabó [13], though their construction was improved slightly by Alon, Rónyai and Szabó [1], who showed that there are graphs with n vertices and Ω s (n 2 1/s ) edges containing no copy of K s,t with t = (s 1)! + 1. Alternative proofs showing that ex(n, K s,t ) = Ω s (n 2 1/s ) when t is significantly larger than s were later found by Blagojević, Bukh and Karasev [2] and by Bukh [4]. In both cases, the basic idea behind the construction is to take a random polynomial f : F s q F s q F q and then to consider the graph G between two copies of F s q whose edges are all those pairs (x, y) such that f(x, y) = 0. A further application of this random algebraic technique was recently given by Conlon [5], who showed that for every natural number k 2 there exists a natural number l such that, for every n, there is a graph Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA 15213, USA. Research supported in part by NSF grant DMS Mathematical Institute, Oxford OX2 6GG, United Kingdom. Research supported by a Royal Society University Research Fellowship. ), 1

2 Figure 1: Graphs in T 2 when (T, R) is a path of length 3 with rooted endpoints. on n vertices with Ω k (n 1+1/k ) vertices for which there are at most l paths of length k between any two vertices. By a result of Faudree and Simonovits [10], this is sharp up to the implied constant. We refer the interested reader to [5] for further background and details. In this paper, we give yet another application of the random algebraic method, proving that for every rational number between 1 and 2, there is a family of graphs H r for which ex(n, H r ) = Θ(n r ). This solves a longstanding open problem in extremal graph theory that has been reiterated by a number of authors, including Frankl [11] and Füredi and Simonovits [12]. Theorem 1.1 For every rational number r between 1 and 2, there exists a family of graphs H r such that ex(n, H r ) = Θ(n r ). In order to define the relevant families H r, we need some preliminary definitions. Definition 1.1 A rooted tree (T, R) consists of a tree T together with an independent set R V (T ), which we refer to as the roots. When the set of roots is understood, we will simply write T. Each of our families H r will be of the following form. Definition 1.2 Given a rooted tree (T, R), we define the pth power T p R of (T, R) to be the family of graphs consisting of all possible unions of p distinct labelled copies of T, each of which agree on the set of roots R. Again, we will usually omit R, denoting the family by T p and referring to it as the pth power of T. We note that T p is a family because we allow the unrooted vertices V (T ) \ R to meet in every possible way. For example, if T is a path of length 3 whose endpoints are rooted, the family T 2 contains a cycle of length 6 and the various degenerate configurations shown in Figure 1. The following parameter will be critical in studying the extremal number of the family T p. Definition 1.3 Given a rooted tree (T, R), we define the density ρ T of (T, R) to be The upper bound in Theorem 1.1 will follow from an application of the next lemma. e(t ) v(t ) R. 2

3 Figure 2: An unbalanced rooted tree T and two elements of T 2. Lemma 1.1 For any rooted tree (T, R) with at least one root, the family T p satisfies ex(n, T p ) = O p (n 2 1/ρ T ). It would be wonderful if there were also a matching lower bound for ex(n, T p ). However, this is in general too much to expect. If, for example, (T, R) is the star K 1,3 with two rooted leaves, T 2 will contain the graph shown in Figure 2 where the two central vertices agree. However, this graph is a tree, so it is easy to show that ex(n, T 2 ) = O(n), whereas, since ρ T = 3/2, Lemma 1.1 only gives ex(n, T 2 ) = O(n 4/3 ). Luckily, we may avoid these difficulties by restricting attention to so-called balanced trees. Definition 1.4 Given a subset S of the unrooted vertices V (T ) \ R in a rooted tree (T, R), we define the density ρ S of S to be e(s)/ S, where e(s) is the number of edges in T with at least one endpoint in S. Note that when S = V (T ) \ R, this agrees with the definition above. We say that the rooted tree (T, R) is balanced if, for every subset S of the unrooted vertices V (T ) \ R, the density of S is at least the density of T, that is, ρ S ρ T. In particular, if R 2, then this condition guarantees that every leaf in the tree is a root. With the caveat that our rooted trees must be balanced, we may now prove a lower bound matching Lemma 1.1 by using the random algebraic method. Lemma 1.2 For any balanced rooted tree (T, R), there exists a positive integer p such that the family T p satisfies ex(n, T p ) = Ω(n 2 1/ρ T ). Therefore, given a rational number r between 1 and 2, it only remains to identify a balanced rooted tree (T, R) for which 2 1/ρ T is equal to r. Definition 1.5 Suppose that a and b are natural numbers satisfying a 1 b < 2a 1 and put i = b a. We define a rooted tree T a,b by taking a path with a vertices, which are labelled in order as 1, 2,..., a, and then adding an additional rooted leaf to each of the i + 1 vertices 1, 1 + a i, a i,..., 1 + (i 1) a i For b 2a 1, we define T a,b recursively to be the tree obtained by attaching a rooted leaf to each unrooted vertex of T a,b a., a. 3

4 Figure 3: The rooted trees T 4,9 and T 4,10. Note that the tree T a,b has a unrooted vertices and b edges, so that ρ T = b/a. Now, given a rational number r with 1 < r < 2, let a/b = 2 r and let T p a,b be the pth power of T a,b. To prove Theorem 1.1, it will suffice to prove that T a,b is balanced, since we may then apply Lemmas 1.1 and 1.2 to T p a,b, for p sufficiently large, to conclude that ex(n, T p a,b ) = Θ(n2 a/b ) = Θ(n r ). Therefore, the following lemma completes the proof of Theorem 1.1. Lemma 1.3 The tree T a,b is balanced. All of the proofs will be given in the next section: we will prove the easy Lemma 1.1 in Section 2.1; Lemma 1.3 and another useful fact about balanced trees will be proved in Section 2.2; and Lemma 1.2 will be proved in Section 2.3. We conclude, in Section 3, with some brief remarks. 2 Proofs 2.1 The upper bound We will use the following folklore lemma. Lemma 2.1 A graph G with average degree d has a subgraph G of minimum degree at least d/2. With this mild preliminary, we are ready to prove Lemma 1.1, that ex(n, T p ) = O p (n 2 1/ρ T ) for any rooted tree (T, R). Proof of Lemma 1.1: Suppose that G is a graph on n vertices with cn 2 α edges, where α = 1/ρ T and c 2 min( T, p). We wish to show that G contains an element of T p. Since the average degree of G is 2cn 1 α, Lemma 2.1 implies that G has a subgraph G with minimum degree at least cn 1 α. Suppose that this subgraph has s n vertices. By embedding greedily one vertex at a time, the minimum degree condition allows us to conclude that G contains at least s cn 1 α (cn 1 α 1) (cn 1 α T + 2) (c/2) T 1 ( T 1)(1 α) sn labelled copies of the (unrooted) tree T. Since there are at most ( s R ) s R choices for the root vertices R, there must be some choice of root vertices R 0 which corresponds to the root set in at least (c/2) T 1 sn ( T 1)(1 α) s R (c/2) T 1 n ( T 1)(1 α) T 1 n R 1 = (c/2) distinct labelled copies of T, where we used that s n and α = 1/ρ T = ( T R )/( T 1). Since (c/2) T 1 p, this gives the required element of T p. 4

5 2.2 Balanced trees We will begin by proving Lemma 1.3, that T a,b is balanced. Proof of Lemma 1.3: Suppose that S is a proper subset of the unrooted vertices of T a,b. We wish to show that e(s), the number of edges in T with at least one endpoint in S, is at least ρ T S, where ρ T = b/a. We may make two simplifying assumptions. First, we may assume that a 1 b < 2a 1. Indeed, if b 2a 1, then the bound for T a,b follows from the bound for T a,b a, which we may assume by induction. Second, we may assume that the vertices in S form a subpath of the base path of length a. Indeed, given the result in this case, we may write any S as the disjoint union of subpaths S 1, S 2,..., S p with no edges between them, so that e(s) = e(s 1 S 2 S p ) = e(s 1 ) + e(s 2 ) + + e(s p ) ρ T ( S 1 + S S p ) = ρ T S. Suppose, therefore, that S = {l, l + 1,..., r} is a proper subpath of the base path {1, 2,..., a} and b a = i. As the desired claim is trivially true if i = 1, we will assume that i 0. In particular, it follows from this assumption that vertex 1 of the base path is adjacent to a rooted vertex. Let R be the number of rooted vertices adjacent to S. For 0 j i 1, the jth rooted vertex is adjacent to S precisely when l 1 + j ( a i ) < r + 1, which is equivalent to (l 1) i a j < r i a. Therefore, if a is not contained in S, it follows that R S i b a a = S a. Furthermore, if l = 1, then R = S b a a. Finally, if r = a and i > 0, then, using a 1 + j a (i j) a i i, it follows that S is adjacent to the jth root whenever i S /a > i j, and so R S b a a. Case 1: i = 0. Since S is a proper subpath, it is adjacent to at least S = (b/a) S edges. Case 2: R S b a a. Then the total number of edges adjacent to S is at least R + S (b/a) S. Case 3: i > 0 and R < S b a. Then S is adjacent to S + 1 edges in the base path, for a total of S b a a a + S + 1 (b/a) S adjacent edges. Before moving on to the proof of Lemma 1.2, it will be useful to note that if T is balanced then every graph in T p is at least as dense as T. Lemma 2.2 If (T, R) is a balanced rooted tree, then every graph H in T s satisfies e(h) ρ T ( H R ). Proof: We will prove the result by induction on s. It is clearly true when s = 1, so we will assume that it holds for any H T s and prove it when H T s+1. Suppose, therefore, that H is the union of s + 1 labelled copies of T, say T 1, T 2,..., T s+1, each of which agree on the set of roots R. If we let H be the union of the first s copies of T, the induction 5

6 hypothesis tells us that e(h ) ρ T ( H R ). Let S be the set of vertices in T s+1 which are not contained in H. Then, since T is balanced, we know that e(s), the number of edges in T s+1 (and, therefore, in H) with at least one endpoint in S, is at least ρ T S. It follows that e(h) e(h ) + e(s) ρ T ( H R ) + ρ T S = ρ T ( H R ), as required. 2.3 The lower bound The proof of the lower bound will follow [4] and [5] quite closely. We begin by describing the basic setup and stating a number of lemmas which we will require in the proof. We will omit the proofs of these lemmas, referring the reader instead to [4] and [5]. Let q be a prime power and let F q be the finite field of order q. We will consider polynomials in t variables over F q, writing any such polynomial as f(x), where X = (X 1,..., X t ). We let P d be the set of polynomials in X of degree at most d, that is, the set of linear combinations over F q of monomials of the form X a 1 1 Xat t with t i=1 a i d. By a random polynomial, we just mean a polynomial chosen uniformly from the set P d. One may produce such a random polynomial by choosing the coefficients of the monomials above to be random elements of F q. The first result we will need says that once q and d are sufficiently large, the probability that a randomly chosen polynomial from P d contains each of m distinct points is exactly 1/q m. Lemma 2.3 Suppose that q > ( m 2 ) and d m 1. Then, if f is a random polynomial from Pd and x 1,..., x m are m distinct points in F t q, P[f(x i ) = 0 for all i = 1,..., m] = 1/q m. We also need to note some basic facts about affine varieties over finite fields. If we write F q for the algebraic closure of F q, a variety over F q is a set of the form W = {x F t q : f 1 (x) = = f s (x) = 0} for some collection of polynomials f 1,..., f s : F t q F q. We say that W is defined over F q if the coefficients of these polynomials are in F q and write W (F q ) = W F t q. We say that W has complexity at most M if s, t and the degrees of the f i are all bounded by M. Finally, we say that a variety is absolutely irreducible if it is irreducible over F q, reserving the term irreducible for irreducibility over F q of varieties defined over F q. The next result we will need is the Lang Weil bound [15] relating the dimension of a variety W to the number of points in W (F q ). It will not be necessary to give a formal definition for the dimension of a variety, though some intuition may be gained by noting that if f 1,..., f s : F t q F q are generic polynomials then the dimension of the variety they define is t s. Lemma 2.4 Suppose that W is a variety over F q of complexity at most M. Then W (F q ) = O M (q dim W ). Moreover, if W is defined over F q and absolutely irreducible, then W (F q ) = q dim W (1 + O M (q 1/2 )). 6

7 We will also need the following standard result from algebraic geometry, which says that if W is an absolutely irreducible variety and D is a variety intersecting W, then either W is contained in D or its intersection with D has smaller dimension. Lemma 2.5 Suppose that W is an absolutely irreducible variety over F q and dim W 1. Then, for any variety D, either W D or W D is a variety of dimension less than dim W. The final ingredient we require says that if W is a variety which is defined over F q, then there is a bounded collection of absolutely irreducible varieties Y 1,..., Y t, each of which is defined over F q, such that t i=1 Y i(f q ) = W (F q ). Lemma 2.6 Suppose that W is a variety over F q of complexity at most M which is defined over F q. Then there are O M (1) absolutely irreducible varieties Y 1,..., Y t, each of which is defined over F q and has complexity O M (1), such that t i=1 Y i(f q ) = W (F q ). We can combine the preceding three lemmas into a single result as follows: Lemma 2.7 Suppose W and D are varieties over F q of complexity at most M which are defined over F q. Then one of the following holds, for all q that are sufficiently large in terms of M: W (F q ) \ D(F q ) q/2, or W (F q ) \ D(F q ) c, where c = c M depends only on M. Proof: By Lemma 2.6, there is a decomposition W (F q ) = t i=1 Y i(f q ) for some bounded-complexity absolutely irreducible varieties Y i defined over F q. If dim Y i 1, Lemma 2.5 tells us that either Y i D or the dimension of Y i D is smaller than the dimension of Y i. If Y i D, then the component does not contribute any point to W (F q ) \ D(F q ) and may be discarded. If instead the dimension of Y i is smaller than the dimension of Y i, the Lang Weil bound, Lemma 2.4, tells us that for q sufficiently large W (F q ) \ D(F q ) Y i (F q ) Y i (F q ) D q dim Y i O(q dim Y i 1 2 ) O(q dim Y i 1 ) q/2. On the other hand, if dim Y i = 0 for every Y i which is not contained in D, Lemma 2.4 tells us that W (F q ) Y i (F q ) = O(1), where the sum is taken over all i for which dim Y i = 0. We are now ready to prove Lemma 1.2, that for any balanced rooted tree (T, R) there exists a positive integer p such that ex(n, T p ) = Ω(n 2 1/ρ T ). Proof of Lemma 1.2: Let (T, R) be a balanced rooted tree with a unrooted vertices and b edges, where R = {u 1,..., u r } and V (T ) \ R = {v 1,..., v a }. Let s = 2br, d = sb, N = q b and suppose that q is sufficiently large. Let f 1,..., f a : F b q F b q F q be independent random polynomials in P d. We will consider the bipartite graph G between two copies U and V of F b q, each of order N = q b, where (u, v) is an edge of G if and only if f 1 (u, v) = = f a (u, v) = 0. Since f 1,..., f a were chosen independently, Lemma 2.3 with m = 1 tells us that the probability a given edge (u, v) is in G is q a. Therefore, the expected number of edges in G is q a N 2 = N 2 a/b. 7

8 Suppose now that w 1, w 2,..., w r are fixed vertices in G and let C be the collection of copies of T in G such that w i corresponds to u i for all 1 i r. We will be interested in estimating the s-th moment of C. To begin, we note that C s counts the number of ordered collections of s (possibly overlapping or identical) copies of T in G such that w i corresponds to u i for all 1 i r. Since the total number of edges m in a given collection of s rooted copies of T is at most sb and q is sufficiently large, Lemma 2.3 tells us that the probability this particular collection of copies of T is in G is q am, where we again use the fact that f 1,..., f a are chosen independently. Suppose that H is an element of T s def = T 1 T 2 T s. Within the complete bipartite graph from U to V, let N s (H) be the number of ordered collections of s copies of T, each rooted at w 1,..., w r in the same way, whose union is a copy of H. Then E[ C s ] = N s (H)q ae(h), H T s while N s (H) = O s (N H R e(h) ). Since (T, R) is balanced, Lemma 2.2 shows that H R ρ T = b a for every H T s. It follows that E[ C s ] = N s (H)q ae(h) = ( O s N H R ) q ae(h) H T s = O s H T s q b( H R ) q ae(h) = O s (1). H T s By Markov s inequality, we may conclude that P[ C c] = P[ C s c s ] E[ C s ] c s = O s(1) c s. Our aim now is to show that C is either quite small or very large. To begin, note that the set C is a subset of X(F q ), where X is the algebraic variety defined as the set of (x 1,..., x a ) F ba q satisfying the equations f i (w k, x l ) = 0 for all k and l such that (u k, v l ) T and f i (x k, x l ) = 0 for all k and l such that (v k, v l ) T for all i = 1, 2,..., a. For each k l such that v k and v l are on the same side of the natural bipartition of T, we let D kl = X {(x 1,..., x a ) : x k = x l }. We put D def = D kl. k,l That is, the D kl capture those elements of X which are degenerate and so not elements of C. As a union of varieties is a variety, the set D is a variety that captures the degenerate elements of X. Furthermore, the complexity of D is bounded since the number and complexity of D kl is bounded. 8

9 By Lemma 2.7, we see that that there exists a constant c T, depending only on T, such that either C c T or C q/2. Therefore, by the consequence of Markov s inequality noted earlier, P[ C > c T ] = P[ C q/2] = O s(1) (q/2) s. We call a sequence of vertices (w 1, w 2,..., w r ) bad if there are more than c T copies of T in G such that w i corresponds to u i for all 1 i r. If we let B be the random variable counting the number of bad sequences, we have, since s = 2br and q is sufficiently large, E[B] 2N r Os(1) (q/2) s = O s(q br s ) = o(1). We now remove a vertex from each bad sequence to form a new graph G. Since each vertex has degree at most N, the total number of edges removed is at most BN. Hence, the expected number of edges in G is N 2 a/b E[B]N = Ω(N 2 a/b ). Therefore, there is a graph with at most 2N vertices and Ω(N 2 a/b ) edges such that no sequence of r vertices has more than c T labelled copies of T rooted on these vertices. Finally, we note that this result was only shown to hold when q is a prime power and N = q b. However, an application of Bertrand s postulate shows that the same conclusion holds for all N. 3 Concluding remarks We have shown that for any rational number r between 1 and 2, there exists a family of graphs H r such that ex(n, H r ) = Θ(n r ). However, Erdős and Simonovits (see, for example, [6]) asked whether there exists a single graph H r such that ex(n, H r ) = Θ(n r ). Our methods give some hope of a positive solution to this question, but the difficulties now lie with determining accurate upper bounds for the extremal number of certain graphs. To be more precise, given a rooted tree (T, R), we define T p to be the graph consisting of the union of p distinct labelled copies of T, each of which agree on the set of roots R but are otherwise disjoint. Lemma 1.2 clearly shows that ex(n, T p ) = Ω(n 2 1/ρ T ) when T is a balanced rooted tree. We believe that a corresponding upper bound should also hold. Conjecture 3.1 For any balanced rooted tree (T, R), the graph T p satisfies ex(n, T p ) = O p (n 2 1/ρ T ). The condition that (T, R) be balanced is necessary here, as may be seen by considering the graph in Figure 2, namely, a star K 1,3 with two rooted leaves. Then T 2 contains a cycle of length 4, so the extremal number is Ω(n 3/2 ), whereas the conjecture would suggest that it is O(n 4/3 ). In order to solve the Erdős Simonovits conjecture, it would be sufficient to solve the conjecture for the collection of rooted trees T a,b with a < b and (a, b) = 1. However, even this seems surprisingly difficult and the only known cases are when a = 1, in which case T is a star with rooted leaves and T p is a complete bipartite graph, or b a = 1, when T is a path with rooted endpoints and T p is a theta graph. Acknowledgements. We would like to thank Jacques Verstraete for interesting discussions relating to the topic of this paper. 9

10 References [1] N. Alon, L. Rónyai and T. Szabó, Norm-graphs: variations and applications, J. Combin. Theory Ser. B 76 (1999), [2] P. V. M. Blagojević, B. Bukh and R. Karasev, Turán numbers for K s,t -free graphs: topological obstructions and algebraic constructions, Israel J. Math. 197 (2013), [3] W. G. Brown, On graphs that do not contain a Thomsen graph, Canad. Math. Bull. 9 (1966), [4] B. Bukh, Random algebraic construction of extremal graphs, arxiv: [math.co]. [5] D. Conlon, Graphs with few paths of prescribed length between any two vertices, arxiv: [math.co]. [6] P. Erdős, On the combinatorial problems which I would most like to see solved, Combinatorica 1 (1981), [7] P. Erdős, A. Rényi and V. T. Sós, On a problem of graph theory, Studia Sci. Math. Hungar. 1 (1966), [8] P. Erdős and M. Simonovits, A limit theorem in graph theory, Studia Sci. Math. Hungar. 1 (1966), [9] P. Erdős and A. H. Stone, On the structure of linear graphs, Bull. Amer. Math. Soc. 52 (1946), [10] R. J. Faudree and M. Simonovits, On a class of degenerate extremal graph problems, Combinatorica 3 (1983), [11] P. Frankl, All rationals occur as exponents, J. Combin. Theory Ser. A 42 (1986), [12] Z. Füredi and M. Simonovits, The history of degenerate (bipartite) extremal graph problems, in Erdős Centennial, , Bolyai Soc. Math. Stud., 25, Springer, Berlin, [13] J. Kollár, L. Rónyai and T. Szabó, Norm-graphs and bipartite Turán numbers, Combinatorica 16 (1996), [14] T. Kővári, V. T. Sós and P. Turán, On a problem of K. Zarankiewicz, Colloq. Math. 3 (1954), [15] S. Lang and A. Weil, Number of points of varieties in finite fields, Amer. J. Math. 76 (1954),

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