Unavoidable order-size pairs in hypergraphs -- positive forcing density

Erd\H{o}s, F\"uredi, Rothschild and S\'os initiated a study of classes of graphs that forbid every induced subgraph on a given number $m$ of vertices and number $f$ of edges. Extending their notation to $r$-graphs, we write $(n,e) \to_r (m,f)$ if every $r$-graph $G$ on $n$ vertices with $e$ edges has an induced subgraph on $m$ vertices and $f$ edges. The \emph{forcing density} of a pair $(m,f)$ is $$ \sigma_r(m,f) =\left. \limsup\limits_{n \to \infty} \frac{|\{e : (n,e) \to_r (m,f)\}|}{\binom{n}{r}} \right. .$$ In the graph setting it is known that there are infinitely many pairs $(m, f)$ with positive forcing density. Weber asked if there is a pair of positive forcing density for $r\geq 3$ apart from the trivial ones $(m, 0)$ and $(m, \binom{m}{r})$. Answering her question, we show that $(6,10)$ is such a pair for $r=3$ and conjecture that it is the unique such pair. Further, we find necessary conditions for a pair to have positive forcing density, supporting this conjecture.


Introduction
The Turán function ex(n, H) is the maximum number of edges in an H-free n-vertex r-graph.The Turán density of H, denoted by π(H), is defined as follows Determining the Turán function for graphs and hypergraphs is a central topic in extremal graph theory with many challenging open problems, trying to identify what graph density forces the Axenovich and Weber [1] asked whether there are pairs (m, f ) for which not only σ r (m, f ) = 0, but a stronger statement holds.A pair (m, f ) is absolutely r-avoidable if there is n 0 such that for each n > n 0 and for every e ∈ {0, . . ., n r }, (n, e) → r (m, f ).In [1] it was shown that for r = 2 there are infinitely many absolutely avoidable pairs.Moreover, there is an infinite family of absolutely avoidable pairs of the form (m, m 2 /2) and for every sufficiently large m, there exists an f such that (m, f ) is absolutely avoidable.In [15] this result was extended to higher uniformities to show that for every r ≥ 3, there exists m 0 such that for every m ≥ m 0 either (m, ⌊ m r /2⌋) or (m, ⌊ m r /2⌋ − m − 1) is absolutely avoidable.
While there are many pairs (m, f ) for which σ r (m, f ) = 0, not a single (non-trivial) pair with positive forcing density was known for r-graphs when r ≥ 3. We denote by K r t the r-graph on t vertices where every r-set is an edge.Note that σ r (r, 1) = σ r (r, 0) = 1 and for f = 0, σ r corresponds to the Turán density, i.e., σ r (m, 0) = σ r (m, m r ) = π(K r m ), where the best currently known general bounds on the Turán density are , due to Sidorenko [12] and de Caen [2].Weber [15] asked whether for m > r ≥ 3, there is any f with 0 < f < n r such that σ r (m, f ) > 0 and suggested the pair (6, 10) as a candidate.We answer this question in the affirmative and prove σ 3 (6, 10) > 0.
Given families of r-graphs F, G, we denote by ex(n, ind F, G) the maximum number of edges in an n-vertex r-graph not containing any F ∈ F as an induced copy and also not any G ∈ G as a copy.Further, denote by π( ind F, G) the limit π( ind F, G) = lim sup n→∞ ex(n, ind F, G) n r .
We mostly consider 3-graphs in this paper.When clear from context, we shall write abc for the set {a, b, c} corresponding to an edge in a 3-graph.Denote by [n] = {1, 2, . . ., n} the set of the first n integers.The 3-graph on vertex set [4] with edgeset {123, 124, 124} is denoted by K 3− 4 .Let F 10 6 be the family of 6-vertex 3-graphs containing exactly 10 edges.
We do not know whether other pairs (m, f ) with m > 3, 0 < f < m 3 exist, such that σ 3 (m, f ) > 0. It seems plausible that for r = 3 there are indeed no other pairs with positive forcing density.

Conjecture 1.2. Let m and f be positive integers
The following result provides evidence for this conjecture to be true.
Thus, in particular if there are no other non-trivial solutions except for m = 6, x 1 = 5, x 2 = 5, x 3 = 3, to the above Diophantine equation, then Conjecture 1.2 is true.A computer search for suitable solutions of (1) did not give a result for m ≤ 10 6 .
This paper is organized as follows: In Section 2 we prove Theorem 1.1.In Section 3 we prove Theorem 1.3.Finally, in Section 4 we make concluding remarks and state open problems.

Proof of Theorem 1.1
We say a 3-graph G induces (6,10) if G contains an induced copy of some F ∈ F 10  6 .If G does not contain any F ∈ F 10  6 as an induced copy, we say G is (6, 10)-free, i.e., a 3-graph is (6, 10)-free if no 6-vertex set induces exactly 10 edges.

Proof idea
Before proving Theorem 1.1, we give a short sketch of the proof.We shall show that for every ǫ > 0 there is n 0 such that for every n > n 0 if G is an n-vertex 3-graph satisfying e(G) then G induces (6,10).Then we first use a standard Ramsey type argument to partition most of the vertices of G into many large homogeneous sets.First, we rule out the case that there is a large clique and a large independent set that are disjoint.Thus, most of the vertex set of G or its complement G c can be partitioned into large independent sets.Due to the symmetry of the problem, if we find a (6, 10)-set in G c , we also find a (6, 10)-set in G. Thus, without loss of generality, we can assume that most of the vertices of G can be partitioned into many large independent sets.Using a classical supersaturation result and the density assumption on G, we find many copies of K 3− 4 in G and thus, in particular, four large independent sets spanning many transversal copies of K 3− 4 .Using a final cleaning argument, we find a (6, 10)-set in this substructure.
On the other hand, we fix an arbitrary 3-graph G on ex(n, ind F 10 6 , {K 3− 4 }) edges that is (6, 10)free and K 3− 4 -free.Then every set of 6 vertices spans at most 9 edges, so there is a graph on n vertices and e edges, for any e ≤ ex(n, ind F 10 6 , {K 3− 4 }), that is (6, 10)-free.By taking complements, there also is a graph on n vertices and e edges for every e ≥ n 3 − ex(n, ind F 10 6 , {K 3− 4 }), that is (6, 10)-free.

Definitions, notations, and construction
An independent set in an r-graph is a vertex subset containing no edges.A clique in an r-graph is a vertex subset in which every r-set is an edge.A homogeneous set in an r-graph is a clique or an independent set.
Let G be a 3-graph and let X, Y, Z ⊆ V (G), not necessarily disjoint from each other.Then, let If the 3-graph G is clear from the context, we might omit the index and simply write E(X, Y, Z).Given a set S ⊆ V (G), the induced subhypergraph G[S] is the r-graph whose vertex set is S and whose edge set consists of all of the edges in E(G) that have all endpoints in S.
Let H be an r-graph and t ∈ N. The t-blow-up of H, denoted by H(t), is the r-graph with its vertex set partitioned in |V (H)| sets V 1 , V 2 , . . ., V |V (H)| , each of size t and edge set {{a 1 , . . ., a r } : a j ∈ V i j , j = 1, . . ., r, {i 1 , . . ., i r } ∈ E(H)}.Informally, H(t) is obtained from H by replacing each vertex i with an independent set V i and each hyperedge e of H with a complete r-partite hypergraph with parts corresponding to the vertices of e.
We say that a 3-graph G is a weak t-blowup of H, which we also call weak Note that we do not impose any condition on 3-tuples of vertices with exactly two vertices in some part V i .
Denote by r r (t, t) the Ramsey number of K r t versus K r t , i.e., the minimum number of vertices m such that every 2-coloring of the edges of K r m contains a monochromatic K r t .Erdős, Hajnal and Rado [5] showed that there exists constants c > 0 such that r 3 (t, t) < 2 2 ct .
Next, we shall provide a construction of a (6, 10)-free graph that we shall use to provide an upper bound in Theorem 1.1.

Construction of the 3-graph H it n
Let H be the 3-graph with vertex set [6] and edges 123, 124, 345, 346, 561, 562, 135, 146, and 236.Note that adding the edge 245 to H results in a 5-regular 3-graph on 6 vertices, which is K 3− 4 -free and the basis for the construction for the lower bound on π(K 3− 4 ) by Frankl and Füredi [8].
We define the following iterated unbalanced blow-up of this graph.Denote by H n the 3-graph on n vertices where the vertex set is partitioned into six sets A 1 , A 2 , A 3 , A 4 , A 5 , A 6 , where The 3-graph H n consists of all triples xyz, where x ∈ A i , y ∈ A j and z ∈ A k and ijk ∈ E(H).Now, let H it n be the 3-graph constructed from H n by iteratively adding a copy of ) edges such that every 6 vertices in H it n induce at most 9 edges.In particular, H it n is (6, 10)-free.
We present the proof of this lemma in the appendix.

Lemmas
The following lemma shows that every sufficiently large 3-graph can be partitioned into many large homogeneous sets.
Lemma 2.2.Let t > 0. Then there exists n 0 = n 0 (t) such that for every Since n ≥ r 3 (t, t), there exists a homogeneous set of size t in G.Call it D 0 and define We iteratively repeat this process.Define The following Lemma analyses the structure "between" two large vertex sets.This is partly motivated by a result by Fox and Sudakov [7] for 2-graphs.Lemma 2.3.Let t ≥ 0. Then there exists n 0 such that for all n ≥ n 0 the following holds.Let G be a 3-graph with vertex set , b, else.Note that by the standard bound on the diagonal Ramsey number r 2 (s, s) ≤ 4 s , each 2-colored 2-clique on k vertices contains a monochromatic clique of size log 4 (k). Let ) is either empty or complete.Thus, there exists a subset , such that the set E(A ′′ , B m , B m ) is either empty or complete.Now we repeat this process with vertices in is either empty or complete.The sets A ′ , B ′ satisfy the conditions of the lemma, completing the proof.
The next lemma shows that in a (6, 10)-free 3-graph there cannot be a large independent set and a large clique that are disjoint.

Lemma 2.4.
There exists t 0 > 0 such that for all t ≥ t 0 the following holds.Let G be a 2t-vertex 3-graph with vertex set is an independent set.Then G induces (6,10).
Proof.By Lemma 2.3, for sufficiently large t, we can we find subsets ) and E(A ′ , B ′ , B ′ ) are either empty or complete.
If E(A ′ , B ′ , B ′ ) is complete, then any vertex from A ′ together with the 5 vertices from B ′ induces (6,10).If E(A ′ , A ′ , B ′ ) is empty, then any vertex from B ′ together with the five vertices from A ′ induces (6,10).Hence, we may assume that E(A ′ , B ′ , B ′ ) is empty and E(A ′ , A ′ , B ′ ) is complete.But then three arbitrary vertices from A ′ together with three arbitrary vertices from B ′ induce (6, 10).Lemma 2.2 together with Lemma 2.4 immediately implies the following lemma.Lemma 2.5.There exists t 0 such for all t ≥ t 0 the following holds.There exists n 0 = n 0 (t) such that for all n ≥ n 0 , if G is a (6, 10)-free n-vertex 3-graph, then either G or G c contains at least n/t − √ n pairwise disjoint independent sets of size t.
Lemma 2.6.Let t ′ > 0. Then there exists t 0 > 0 such that for all t ≥ t 0 the following holds.Let G be a (6, 10)-free 2t-vertex 3-graph with vertex set Proof.We apply Lemma 2.3 for t ′ .Then there exists t 0 such that for t ≥ t 0 , we find A ′ ⊆ A, B ′ ⊆ B, such that the two sets E(A ′ , A ′ , B ′ ) and E(A ′ , B ′ , B ′ ) are either empty or complete.Assume the set E(A ′ , A ′ , B ′ ) is complete.Then we find induced (6, 10) by taking any 5 vertices from A ′ and 1 vertex from B. By symmetry the same holds for the set is the empty graph.
. By iteratively applying Lemma 2.6 to all of the tuples (V i , V j ), 1 ≤ i < j ≤ 4, we obtain an induced copy is empty for all i j, the sets E(X i , X j , X k ) are complete for {i, j, k} ∈ [4]  3 except for E(X 2 , X 3 , X 4 ), which is empty.Let (6,10).Now assume there is a weak By iteratively applying Lemma 2.6 to all of the tuples (V i , V j ), 1 ≤ i < j ≤ 4, we obtain an induced copy is empty for all i j and the sets E(X i , X j , X k ) are complete for all {i, j, k} ∈ [4]  3 .Let is a 6-vertex 3-graph spanning exactly 10 edges.Lemma 2.8.Let t > 0 be an integer and δ > 0. Then there exists m 0 = m 0 (t, δ) such that for all m ≥ m 0 the following holds.Let G be a 3-graph on 4m vertices such that the vertex set of G can be partitioned into four independent sets V 1 , V 2 , V 3 , V 4 of size m each and the number of copies of K 3− 4 with one endpoint from each of the V ′ i s is at least δm 4 .Then G contains an induced copy of a weak K 3 4 (t) or a weak K 3− 4 (t).Proof.Define the auxiliary 4-graph H on 4m vertices where a 4-set spans an edge iff the corresponding four vertices in G form a copy of K 3− 4 .We 5-color the edges of H in the following way: and it is colored with color 5 if {v 1 , v 2 , v 3 , v 4 } induces a K 3  4 in G.By pigeonhole principle, there exists (δ/5)m 4 edges of the same color.Erdős [3] proved that π(K 4 4 (t)) = 0 and thus, there exists a monochromatic copy of K 4 4 (t) in H. Denote by T the vertex set of this monochromatic copy.The 3-graph G[T ] is a weak K 3 4 (t) or weak K 3− 4 (t).
We will use a supersaturation result discovered by Erdős and Simonovits [6].The proof presented below follows a proof given by Keevash (Lemma 2.1.in [10]).Lemma 2.9.For ε > 0 and families F, G of r-graphs, there exists constants δ > 0 and n 0 > 0 so that if G is an r-graph on n > n 0 vertices with e(G) > (π Proof.Let G be an r-graph on sufficiently many vertices n with e(G) > (π Otherwise, we would have but we also have a contradiction.By the choice of k, each of these k-sets K contains an induced copy of some H ∈ F or a copy of some H ∈ G.By the pigeonhole principle, there exists H 1 ∈ F such that at least ε 2(|F|+|G|) n k of these k-sets K contain an induced copy of H 1 , or there exists H 2 ∈ G such that at least ε 2(|F|+|G|) n k of these k-sets K contain a copy of H 2 .Thus, in the first case, the number of induced copies of H 1 is at least Similarly, in the second case, the number of copies of H 2 is at least

Proof of Theorem 1.1.
Proof of Theorem 1.1.Let ε > 0. Fix an integer t whose existence is guaranteed by Lemma 2.7, such that every weak K 3 4 (t) and also every weak By Lemma 2.5 either G or G c contains at least n ′ := n/m 1 − √ n pairwise disjoint independent sets, each of size m 1 .Since the density assumption is symmetric, and since G induces (6,10) if and only if G c induces (6, 10), we can assume, without loss of generality, that G contains at least By Lemma 2.9, G contains at least 2δ n 4 (not necessarily induced) copies of K 3− 4 .We call a 4-set transversal in G if each of the four vertices is in a different V i .A copy of K 3−  4 in G is called transversal if the vertex set of the copy is transversal in G.The number of 4-sets which are not transversal in G is at most set, called base set, and we have the additional edges i∈S E i , where ) is just a clique on k vertices and n − k isolated vertices, and G [2] (n, k) is the complete graph on n vertices with a clique of size n − k removed.For an illustration of G({2}, n, k) see Figure 1.
We call a 3-graph G m-sparse if every subset of m vertices in G induces at most m edges.We say that a 3-graph G is canonical plus with parameters (S, n, k), or simply canonical plus if G is a 3-graph obtained as a union of G(S, n, k) and an m-sparse graph whose vertex set is the base independent set of G(S, n, k).A 3-graph G is canonical minus with parameters (S, n, k), or simply canonical minus, if G is the complement of a canonical plus graph with parameters ([2]−S, n, n−k).Note that a canonical minus graph with parameters (S, n, k) is obtained from the graph G(S, n, k) by removing edges of a copy of an m-sparse graph from the clique A. We see that (letting y x = 0 for y < x), that ).Note that any induced subgraph of a canonical plus 3-graph with parameters (S, n, k) is a canonical plus 3-graph with parameters (S, n ′ , k ′ ), for some n ′ and k ′ .A similar statement holds for canonical minus graphs.Thus, these two classes of graphs are hereditary.We see that if an m-vertex 3-graph is canonical plus with parameters (S, m, x), then the number of edges in such a graph is in the interval [f (S, m, x), f (S, m, x) + m].Similarly, the number of edges in a canonical minus graph with parameters (S, m, x) is in the interval [f (S, m, x) − m, f (S, m, x)].Thus, if f is the number of edges of a graph that could be represented as both a canonical plus and a canonical minus graph with first parameter S and m vertices, then f ∈ F (S, m), where

Proof idea
We are using the following general principle: Here, we use two classes C 1 and C 2 of 3-graphs that are canonical plus and canonical minus with the same first parameter S. Specifically, the main idea of the proof of Theorem 1.3 is that for any sufficiently large n, any S ⊆ [2], and any e in the interval [c n 3 , (1 − c) n 3 ] for 0 < c < 1/2, there is a canonical plus 3-graph G + c,S and a canonical minus 3-graph G − c,S with first parameter S, on n vertices and e edges.If, for a pair (m, f ), f F (S, m) for some S ⊆ [2], then the pair (m, f ) is not representable as a canonical plus or canonical minus graph with first parameter S. Then in particular, G + c,S and G − c,S are (m, f )-free and (n, e) → (m, f ).Letting c be arbitrarily small, we conclude that σ 3 (m, f ) = 0 for such a pair (m, f ).Finally, we derive number theoretic conditions for a pair (m, f ) not being representable by a canonical plus or a canonical minus graph.

Lemmas
In the following lemmas, n, m, f, e are non-negative integers with m > 3, 0 < f < m 3 .In [15] it was shown that for any m ≤ 15 and for any 0 < f < m 3 such that (m, f ) (6, 10), σ 3 (m, f ) = 0. Thus, we can assume that m ≥ 16.The following folklore result can be obtained by a standard probabilistic argument.Lemma 3.2.Let m > 0. Then for any sufficiently large n there exists an n-vertex 3-graph with Ω(n 2+ 1 m+1 ) edges which is m-sparse.
For a proof of Lemma 3.2 see e.g.[15].The next lemma is a generalization of a similar statement proven in [4] for graphs.Proof.Let n be a given sufficiently large integer.Let k be a non-negative integer such that either f (S, n, k) ≤ e ≤ f (S, n, k + 1) or f (S, n, k) ≤ e ≤ f (S, n, k − 1) holds.Without loss of generality assume that f (S, n, k) ≤ e ≤ f (S, n, k + 1).Let . The existence of G ′ is guaranteed by Lemma 3.2.Define G ′′ to be the 3-graph obtained as a union of G(S, n, k) and a copy of G ′ on the vertex set that is the base independent set of G(S, n, k).Then Here, the second inequality holds since f (S, n, k + 1) − f (S, n, k) = O(n 2 ).Finally, let G 1 (n, e) be a subgraph of G ′′ with e edges, obtained from G ′′ by removing some edges of G ′ .
For the second part of the lemma, take G 2 (n, e) to be the complement of G 1 (n, n 3 − e) with first parameter [2] − S, guaranteed by the first part of the lemma.
In particular, we have where in the last step we used that m−1 2 > 2m.Thus, σ 3 (m, f ) = 0. Proof.Consider m and f as given in the statement of the lemma and let S = {2}.By Lemma 3.4, it is sufficient to prove that f F (S, m) and in particular, it is sufficient to show that In particular, we have

Lemma 3 . 3 .
Let S ⊆[2] and c be a constant, 0 < c < 1/2.For n ∈ N sufficiently large and any e where c < e < (1 − c)n  3 , there exist 3-graphs G 1 (n, e) and G 2 (n, e) on n vertices and e edges that are canonical plus and canonical minus respectively, with first parameter S.

Lemma 3 . 4 .Lemma 3 . 5 .
Let S ⊆[2].If f F (S, m), then σ 3 (m, f ) = 0.Proof.Assume we have integers m, f as above, some S ⊆[2] and f F (S, m).Let c be a constant, 0 < c < 1/10, n ≥ n 0 , and e be any integer satisfyingc n 3 ≤ e ≤ (1 − c) n 3 .Define graphs G 1 = G 1 (n, e) and G 2 = G 2 (n,e) whose existence is guaranteed by Lemma 3.3.Any induced subgraph of G 1 on m vertices is canonical plus with parameters (S, m, x) for some x and thus, its number of edges is in m−1 x=0 [f (S, m, x), f (S, m, x) + m].Any induced subgraph of G 2 on m vertices is canonical minus with parameters (S, m, x) for some x and thus, its number of edges is in m x=1 [f (S, m, x) − m, f (S, m, x)].Since f F (S, m), we get that G 1 and G 2 are (m, f )-free.Letting c go to zero, we see that σ 3 (m, f ) = 0.In the following lemmas we shall use the set S = ∅, S = {1}, or S = {2}, to claim that for many pairs (m, f ), σ 3 (m, f ) = 0. Let m ≥ 7 and 0
− c) n r , there is a graph G i ∈ C i on n vertices and e edges for all i = 1, . . ., k.If for any sufficiently large n and some Proposition 3.1.Let C 1 , . . ., C k be hereditary classes of r-graphs such that for any c, 0 < c < 1/2, any sufficiently large n, and any e with c n r ≤ e ≤ (1