The Turán Problem for Hypergraphs of Fixed Size

We obtain a general bound on the Turán density of a hypergraph in terms of the number of edges that it contains. If F is an r-uniform hypergraph with f edges we show that π(F) < f−2 f−1 − (1 + o(1))(2r!2/rf3−2/r)−1, for fixed r ≥ 3 and f → ∞. Given an r-uniform hypergraph F , the Turán number of F is the maximum number of edges in an r-uniform hypergraph on n vertices that does not contain a copy of F . We denote this number by ex(n,F). It is not hard to show that the limit π(F) = limn→∞ ex(n,F)/ ( n r ) exists. It is usually called the Turán density of F . There are very few hypergraphs with r > 2 for which the Turán density is known, and even fewer for the exact Turán number. We refer the reader to [10, 11, 12, 13, 14, 15, 16] for recent results on these problems. A general upper bound on Turán densities was obtained by de Caen [3], who showed π(K (r) s ) ≤ 1 − ( s−1 r−1 )−1 , where K (r) s denotes the complete r-uniform hypergraph on s vertices. A construction showing π(K (r) s ) ≥ 1− ( r−1 s−1 )r−1 was given by Sidorenko [17] (see also [18]); better bounds are known for large r. We refer the reader to Sidorenko [18] for a full discussion of this problem. For a general hypergraph F Sidorenko [19] (see also [20]) obtained a bound for the Turán density in terms of the number of edges, showing that if F has f edges then π(F) ≤ f−2 f−1 . In this note we improve this as follows. Theorem 1 Suppose F is an r-uniform hypergraph with f edges. (i) If r = 3 and f ≥ 4 then π(F) ≤ 1 2 ( √ f 2 − 2f − 3 − f + 3). (ii) For a fixed r ≥ 3 and f → ∞ we have π(F) < f−2 f−1 − (1 + o(1))(2r!f 3−2/r)−1. We start by describing our main tool, which is Sidorenko’s analytic approach. See [20] for a survey of this method. Consider an r-uniform hypergraph H on n vertices. It is convenient to regard the vertex set V as a finite measure space, in which each vertex v has μ({v}) = 1/n, so that μ(V ) = 1. We write h : V r → {0, 1} for the symmetric function the electronic journal of combinatorics 12 (2005), #N11 1 h(x1, · · · , xr) which takes the value 1 if {x1, · · · , xr} is an edge of H and 0 otherwise. Then ∫ h dμ = r!e(H)n−r = d + O(1/n), where d = ( n r )−1 e(H) is the density of H. Now consider a fixed forbidden r-uniform hypergraph F with f edges on the vertex set {1, · · · , m}. We associate to vertex i the variable xi, and to an edge e = {i1, · · · , ir} the function he(x) = h(xi1 , · · · , xir), where x denotes the vector (x1, · · · , xm). The configuration product of F with respect to h is the function hF (x) = ∏ e∈F he(x). Then ∫ hFdμ m = n−mhom(F ,H) = n−mmon(F ,H)+O(n−1) = n−maut(F)sub(F ,H)+O(n−1), where hom(F ,H) is the number of homomorphisms (edge-preserving maps) from F to H, mon(F ,H) is the number of these that are monomorphisms (injective homomorphisms), aut(F) is the number of automorphisms of F and sub(F ,H) is the number of F -subgraphs of H. Also, Erdős-Simonovits supersaturation [6] implies that for any δ > 0 there is > 0 and an integer n0 so that for any r-uniform hypergraph H on n ≥ n0 vertices with ( n r )−1 e(H) > π(F) + δ we have n−msub(F ,H) > . It follows that π(F) = inf >0 lim inf |V |→∞ max h:V r→{0,1}, R hF dμm< ∫ h dμ. (1) We say that F is a forest if we can order its edges as e1, · · · , ef so that for every 2 ≤ i ≤ f there is some 1 ≤ j ≤ i − 1 so that ei ∩ ( ∪i−1 t=1et ) ⊂ ej . Sidorenko [20] showed that if F is a forest with f edges then ∫ hF dμ m ≥ (∫ h dμ )f . (2) Now we need a lemma on when a hypergraph contains a forest of given size. Lemma 2 (i) An r-uniform hypergraph with at least r!(t − 1) edges contains a forest with t edges. (ii) Let F be a 3-uniform hypergraph. Then either (a) F contains a forest with 3 edges, or (b) π(F) = 0, or (c) F ⊂ K 4 , or (d) F = F5 = {abc, abd, cde}. Proof. (i) This is immediate from the result of Erdős and Rado [5] that such a hypergraph contains a sunflower with t petals, i.e. edges e1, · · · , et for which all the pairwise intersections ei ∩ ej are equal. A sunflower is in particular a forest. (ii) Consider a 3-uniform hypergraph F that does not contain a forest with 3 edges. We can assume that F is not 3-partite (Erdős [4] showed that this implies π(F) = 0) so F has at least 3 edges. Clearly F cannot have two disjoint edges, as then adding any other edge gives a forest. Suppose there is a pair of edges that share two points, say e1 = abc and e2 = abd. Any other edge must contain c and d, or together with e1 and e2 we have a forest. Consider another edge e3 = cde. If there are no other edges then either F = F5 or F ⊂ K 4 (if e the electronic journal of combinatorics 12 (2005), #N11 2 equals a or b). If there is another edge e4 = cdf then the same argument shows that e1 and e2 both contain e and f , i.e. F = K 4 and there can be no more edges. The other possibility is that every pair of edges intersect in exactly one point. Then there are at most 2 edges containing any point, or we would have a forest with 3 edges. Consider three edges, which must have the form e1 = abc, e2 = cde, e3 = efa. There can be at most one more edge e4 = bdf . But this forms a 3-partite hypergraph (with parts ad, be, cf), a case we have already excluded. This proves the lemma. Proof of Theorem. Let F be an r-uniform hypergraph with f edges that contains a forest T with t edges. Label the edges e1, · · · , ef , where e1, · · · , et are the edges of T . Suppose that H is an r-uniform hypergraph on a vertex set V of size n. Define the measure μ and the function h : V r → {0, 1} as before. Observe the inequality hF(x) ≥ hT (x) + f ∑ i=t+1 he1(x)(hei(x) − 1). This holds, as the second term is non-positive (since he(x) ∈ {0, 1}), so it could only fail for some x if hF(x) = 0 and hT (x) = 1. But then we have he1(x) = · · · = het(x) = 1 and hei(x) = 0 for some i > t, and the term he1(x)(hei(x) − 1) = −1 cancels hT (x), so the inequality holds for all x. Integrating gives ∫ hF (x) dμ m ≥ ∫ hT (x) dμ m + f ∑ i=t+1 ∫ he1(x)hei(x)− he1(x) dμ ≥ p + (f − t)(p − p), where we write p = ∫ h dμ and apply the inequality (2) for the forests T and {e1, ei}, t + 1 ≤ i ≤ f . By equation (1) we deduce that the Turán density π = π(F) satisfies π + (f − t)(π − π) ≤ 0. Writing g(x) = xt−1 + (f − t)(x − 1) we either have π = 0 or g(π) ≤ 0. Now g(0) = −(f − t) ≤ 0, g(1) = 1 and dg dx = (t − 1)xt−2 + f − t ≥ 0 for 0 < x < 1 so g has exactly one root α in [0, 1], and π ≤ α. First we consider the case r = 3. If f ≥ 5 then by the lemma we can take t = 3. Solving the quadratic g(x) = x +(f −3)(x−1) = 0 gives π ≤ α = 1 2 ( √ f 2 − 2f − 3−f +3). This also holds when f = 4, as then by the lemma we may suppose that F = K 4 . Chung and Lu [2] showed that π(K (3) 4 ) ≤ 3+ √ 17 12 which is less than 1 2 ( √ 5 − 1). Now consider the case when r ≥ 3 is fixed and f → ∞. By the lemma we can take t = (f/r!). Write α = 1− . Since g(α) = 0 we have (f−t) = (1− )t−1 < 1, so < 1/(f−t). From the Taylor expansion of (1− )t−1 we have (f − t) > 1− (t−1) + ( t−1 2 ) 2 − ( t−1 3 ) . Also ( t−1 3 ) 3 < 1 6 ( t−1 f−t )3 < 1 6 (t/f) (since f > t) so ( t−1 2 ) 2 − (f − 1) + 1 − 1 6 (t/f) < 0. the electronic journal of combinatorics 12 (2005), #N11 3 Writing ∆ = (f − 1) − 4 ( t−1 2 ) (1− 1 6 (t/f)) for the discriminant of this quadratic we have > f − 1 − ∆ (t − 1)(t − 2) = 2(1 − 1 6 (t/f)) f − 1 + ∆1/2