Two questions of ErdH{o}s on hypergraphs above the Tur{a}n threshold}

Recall that the Turán number t(n, s, k) is the maximum number of edges in a k-uniform hypergraph on n vertices which does not have the complete k-uniform hypergraph on s vertices as a subgraph. Rademacher proved, but did not publish, see [Erd62] that any graph with t(n, 3, 2) + 1 edges contains at least n/2 triangles. Erdős later [Erd62] published a proof of this result and extended it by proving that for 0 ≤ q ≤ c 1 n/2, for some constant c 1 , any graph with t(n, 3, 2) + q edges contains at least qn/2 triangles. This result was later extended to graphs with density higher than t(n, 3, 2)/ n 2 , recently culminating in [Raz08] where the minimal number of triangles was determined for all densities.

Erdős and Dirac also observed that any graph with t(n, s, 2) + 1 edges has K - s , the complete graph with one edge removed, as a subgraph. This provides a strengthening of the result for triangles from [Erd62] in the sense that it shows that the graph contains two copies of K 3 which share two vertices.

In [Erd94] Erdős asked if these results could be generalized to 3-uniform hypergraphs as well Problem 1.1. Does every 3-uniform hypergraph on t(n, s, 3) + 1 edges contain two K 3 s ? Problem 1.2. Does every 3-uniform hypergraph on t(n, s, 3) + 1 edges contain K 3- s+1 ? In [CG98] these questions, with an affirmative answer to both, were formulated as conjectures,

The aim of this note is to show that the answer to both questions is no. For the first question we believe that the answer is yes for sufficiently large n, but as we shall see the second question fails even in an asymptotic sense.

We first construct a family of hypergraphs which give a negative answer to the two questions for certain small values of n.

We now have Theorem 2.2. Let H k be define as in the example.

1. The vertex set of any K k 2k-2 in H k cannot have one of the two non-edges as a subset. Let A and B be subsets of V (H k ) of size 2k -2. Since any vertex subset of size 2k -2 misses one vertex of H k at least one of A and B must have one of the two non-edges as a subset.

2. Since every subset of V (H k j) of size 2k -2 can have at most 2k-2 k -1 edges, we find by averaging that

The number of edges in

Corollary 2.3. H 3 gives a negative answer to Problem 1.1 and since there are two non-edges in H k it does not contain a K 3- 5 either, thereby giving a negative answer to Problem 1.2 as well.

Note that H k also provides a negative answer to a generalisation of both problems to k-graphs for k ≥ 3. However we believe that Question 1 should have a positive answer for sufficiently large values of n Conjecture 2.4. There is an n 0 (s) such that every 3-uniform hypergraph on n ≥ n 0 (s) vertices and t(n, s, 3) + 1 edges contains two K 3 s For problem 1.2 the failure is not a small n phenomenon, as our next theorem will imply. Recall that the Turán density π

, where t(n, G) is the maximum number of edges in a k-uniform hypergraph on n vertices which does not have G as a subgraph. We note that a positive answer to Problem 1.2 requires that π(K 3 s ) = π(K 3- s+1 ). For K 3- 5 we have the following lower bound, Theorem 2.5. π(K 3- 5 ) ≥ 46 81 Proof. Let H 3 be the 3-uniform hypergraph defined in our earlier example and let H 3 (n), for n divisible by 9, be the blow-up of H 3 with n/9 copies of vertex 1 in H 3 and 2n/9 copies of each of the other vertices in H 3 . Three vertices v i , u j , w k form an edge in H 3 (n) if {v, u, w} is an edge in H 3 . It is easy to see that H 3 (n) does not contain a K 3- 5 and a simple calculation shows that the number of edges in

Next, let D be the directed graph on the same vertex set as H 3 shown in Figure 1. We now define a 3-uniform hypergraph G 0 (n) on the same Figure 1: The auxiliary graph D vertex set as H 3 (n) where {v i , v j , u k } is an edge if, i < j and there is an edge from v to u in D. An edge in D not incident with vertex 1 gives rise to ( 8 243 +o(1)) n 3 edges. An edge leading to vertex 1 gives rise to ( 4 243 +o(1)) n 3 edges, and an edge leading from vertex 1 gives rise to ( 2 243 + o(1)) n 3 edges. Finally we let G(n) be the graph on the same vertex set as H 3 (n) with all edges from H 3 (n) and G 0 (n). The number of edges in G(n) is

In order to prove the theorem we now need to show that G(n) does not contain K 3- 5 . Let A be a set of 5 vertices in G(n). If A corresponds to five distinct vertices in H 3 then A is not a K 3- 5 . If A contains four or five vertices corresponding to the same vertex in H 3 then there are at least 4 non-edges in A, and A is not a K

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