Asymptotics of the Hypergraph Bipartite Turán Problem

For positive integers s, t, r, let Ks,t(r)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$K_{s,t}^{(r)}$$\end{document} denote the r-uniform hypergraph whose vertex set is the union of pairwise disjoint sets X,Y1,⋯,Yt\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$X,Y_1,\dots ,Y_t$$\end{document}, where |X|=s\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$|X| = s$$\end{document} and |Y1|=⋯=|Yt|=r-1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$|Y_1| = \dots = |Y_t| = r-1$$\end{document}, and whose edge set is {{x}∪Yi:x∈X,1≤i≤t}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\{\{x\} \cup Y_i: x \in X, 1\le i\le t\}$$\end{document}. The study of the Turán function of Ks,t(r)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$K_{s,t}^{(r)}$$\end{document} received considerable interest in recent years. Our main results are as follows. First, we show that 1exn,Ks,t(r)=Os,rt1s-1nr-1s-1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \textrm{ex}\left( n,K_{s,t}^{(r)}\right) = O_{s,r}\left( t^{\frac{1}{s-1}}n^{r - \frac{1}{s-1}}\right) \end{aligned}$$\end{document}for all s,t≥2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$s,t\ge 2$$\end{document} and r≥3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r\ge 3$$\end{document}, improving the power of n in the previously best bound and resolving a question of Mubayi and Verstraëte about the dependence of ex(n,K2,t(3))\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\textrm{ex}(n,K_{2,t}^{(3)})$$\end{document} on t. Second, we show that (1) is tight when r is even and t≫s\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$t \gg s$$\end{document}. This disproves a conjecture of Xu, Zhang and Ge. Third, we show that (1) is not tight for r=3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r = 3$$\end{document}, namely that ex(n,Ks,t(3))=Os,t(n3-1s-1-εs)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\textrm{ex}(n,K_{s,t}^{(3)}) = O_{s,t}(n^{3 - \frac{1}{s-1} - \varepsilon _s})$$\end{document} (for all s≥3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$s\ge 3$$\end{document}). This indicates that the behaviour of ex(n,Ks,t(r))\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\textrm{ex}(n,K_{s,t}^{(r)})$$\end{document} might depend on the parity of r. Lastly, we prove a conjecture of Ergemlidze, Jiang and Methuku on the hypergraph analogue of the bipartite Turán problem for graphs with bounded degrees on one side. Our tools include a novel twist on the dependent random choice method as well as a variant of the celebrated norm graphs constructed by Kollár, Rónyai and Szabó.


Introduction
Let H be an r-uniform hypergraph.The Turán function ex(n, H) of H is the largest number of edges in an r-uniform hypergraph on n vertices with no copy of H.The study of the function ex(n, H) for various hypergraphs H is one of the central problems of extremal combinatorics.In the graph case r = 2, the Turán function is fairly well understood unless H bipartite.On the other hand, for r ≥ 3, our understanding of the Turán function is much worse and there is only a small number of tight results.For example, determining the answer for the 3-uniform clique on 4 vertices is still open.Given the difficulty of the problem even for hypergraph cliques, various hypergraphs originating from graphs have been considered for which better bounds can be obtained.Mubayi [11] studied a hypergraph extension of the graph clique and his result was refined by Pikhurko [13].A different hypergraph extension of the triangle was introduced by Frankl [6] who determined the asymptotics of its Turán number and an exact answer was given by Keevash and Sudakov [9].Sidorenko [15] asymptotically determined the Turán number of a hypergraph extension of trees.We refer the interested reader to an extensive survey of Keevash [8] on Turán problems for non-r-partite r-uniform hypergraphs.
It is well-known that for r-partite H, one has ex(n, H) = O(n r−ε ) for some ε = ε(H) > 0 and the main goal here is to determine or estimate the best possible ε(H).One of the very old such Turán-type questions for hypergraphs is a problem of Erdős [3], asking for the maximum number f r (n) of edges in an r-uniform hypergraph on n vertices which does not have four distinct edges A, B, C, D satisfying A ∪ B = C ∪ D and A ∩ B = C ∩ D = ∅.Note that in this problem forbidden hypergraphs originate quite naturally from a four-cycle.Erdős in particular asked whether f r (n) = O(n r−1 ).This was answered affirmatively by Füredi [7], who showed that f r (n) ≤ 3.5 n r−1 .Mubayi and Verstraëte [12] extended Erdős's question by considering the following family of r-uniform hypergraphs which generalize complete bipartite graphs: for positive integers r, s, t, let K s,t is just the s × t complete bipartite graph.Observe that the edges of 2,2 ) (this is in fact an equality for r = 3).Mubayi and Verstraëte [12] proved that ex(n, , improving the constant in Füredi's result.Pikhurko and Verstraëte [14] improved the coefficient of r−1 is a lower bound can be seen by considering the star, i.e. the hypergraph consisting of all edges containing a fixed vertex.
Mubayi and Verstraëte [12] initiated the study of ex(n, K s,t ) for general s, t, and proved that ex(n, K s,t ) ≤ C s,t n 3−1/s as well as that ex(n, K s,t ) ≥ c t n 3−2/s for t > (s − 1)!.For small values of s, they obtained more accurate estimates.Namely, for s = 3, they improved their bound to ex(n, K 3,t ) ≤ C t n 13/5 , while for s = 2, they showed that ex(n, K 2,t ) on t.Ergemlidze, Jiang and Methuku [4] improved the upper bound to ex(n, K 2,t ) ≤ (15t log t + 40t)n 2 , leaving a log t gap from the lower bound of Ω(tn 2 ).For r > 3, little is known.Ergemlidze, Jiang and Methuku found a construction showing ex(n, K (4) 2,t ) = Ω(tn 3 ).Xu, Zhang and Ge [16,17] proved a tight bound on ex(n, K (r) s,t ) when s is much larger than t, using a standard application of the random algebraic method of Bukh [2].
Our first result, Theorem 1.1, achieves two goals.First, it resolves the problem of Mubayi and Verstraëte by proving that ex(n, K 2,t ) = Θ(tn 2 ).And second, it improves the upper bound of [12] on ex(n, K s,t ) for every s and t by reducing the exponent of n from 3 − 1 s to 3 − 1 s−1 .We also obtain an analogous result for every r ≥ 3. The proof of this bound relies on a new weighted variant of the dependent random choice method (see [5] for a description of the technique and a brief history).
Theorem 1.1.For any s, t ≥ 2 and r ≥ 3 there is a constant C s depending only on s such that ex(n, K 2,t ) ≤ Ctn 2 for some absolute constant C.
Our next result shows that, somewhat surprisingly, the bound in Theorem 1.1 is tight in terms of both n and t if the uniformity r is even and t ≫ s.Our construction uses as building blocks a variation of the norm graphs, introduced by Kollár, Rónyai and Szabó [10], which might be of independent interest.Theorem 1.2.For any positive integers s ≥ 2 and k, there is a positive constant c = c(k, s) such that for every integer t > (s − 1)!, if n is sufficiently large, then By combining Theorems 1.1 and 1.2, we see that ex(n, K s,t ) = Θ r,s (t if r ≥ 4 is even and t > (s−1)!. (Here and elsewhere in the paper, Θ r,s means that the implied constants can depend on r and s.)In the special case s = 2, this gives ex(n, K (r) 2,t ) = Θ r (tn r−1 ) for even r ≥ 4.This partially answers a question of Ergemlidze, Jiang and Methuku [4], who asked to determine the dependence of ex(n, K (r) 2,t ) on t.Also, Theorem 1.2 disproves a conjecture of Xu, Zhang and Ge [17,Conjecture 5.1] which stated that ex(n, K (r) s,t ) = Θ r,s,t (n r−2/s ) for all 2 ≤ s ≤ t.
It is natural to ask whether the bound in Theorem 1.1 is tight for odd uniformities as well.Our next theorem shows that this is not the case for r = 3.This indicates that, perhaps surprisingly, the parity of r may play a role.
Theorem 1.3.For any s ≥ 3, there exists some ε > 0 such that for any t, ex(n, K Theorems 1.2 and 1.3 together show that ex(n, K s,t )/n r−1 has a different order of magnitude for even r ≥ 4 and for r = 3.Indeed, for even r the function ex(n, K (r) s,t )/n r−1 is asymptotically Θ r,s,t (n 1− 1 s−1 ), assuming s ≪ t, while for r = 3 this function is smaller by at least a factor of n ε .This runs contrary to a claim made in [4, Proposition 1] that ex(n, K (r) s,t ).One can check that the proof suggested in [4] is incorrect.Moreover, as we now see, the statement itself is disproved by Theorems 1.2 and 1.3.
It would be very interesting to determine if Theorem 1.3 can be extended to all odd uniformities r.If so, then this would be a rare example of an extremal problem where the answer depends on the parity of the uniformity.(See [9] for another hypergraph Turán problem where the extremal construction depends heavily on number theoretic properties of the parameters.)The first open case is r = 5: is it true that ex(n, K We end with some results for a more general family of hypergraphs.Let G be a bipartite graph with an ordered bipartition (X, Y ), Y = {y 1 , . . ., y m }.Following [4], we define G (r) X,Y to be the r-uniform hypergraph whose vertex set consists of disjoint sets X, Y 1 , . . ., Y m , Note that Theorem 1.2 shows that this bound can be attained whenever r is even.
Finally, we consider the hypergraph 2t for this hypergraph.In an unpublished work, Jiang and Liu showed that Ω r (tn r−1 ) ≤ ex(n, C (r) 2t ) ≤ O r (t 5 n r−1 ).The lower bound is obtained by taking all edges which contain one of t − 1 vertices.This hypergraph has cover number t − 1, so it cannot contain C (r) 2t , which has cover number t. Ergemlidze, Jiang and Methuku [4] improved the upper bound to ex(n, C (r) 2t ) ≤ O r (t 2 (log t)n r−1 ).We determine the correct dependence on t.
Theorem 1.5.For every t ≥ 2 and r ≥ 3, we have that ex(n, C (r) 2t ) = Θ r (tn r−1 ).The rest of the paper is organized as follows.In Section 2, we prove a general result which implies Theorems 1.1, 1.4 and 1.5.In Section 3, we prove Theorem 1.2.In Section 4, we prove Theorem 1.3.In Section 5, we give some concluding remarks.

Upper bounds
In what follows, for an r-uniform hypergraph G and a set We omit the subscript when the hypergraph is clear.Given a vertex v ∈ V (G), the link hypergraph of v (with respect to G) is the (r − 1)-uniform hypergraph containing all (r − 1)sets which together with v form an edge in G.
Definition 2.1.In an r-uniform hypergraph G, we call a set S ⊂ V (G) t-rich if there are sets T 1 , T 2 , . . ., T t ⊂ V (G) of size r − 1 such that S, T 1 , . . ., T t are pairwise disjoint and {u} ∪ T i ∈ E(G) for every u ∈ S and i ∈ [t].
Note that if G has any t-rich set of size s, then it contains K (r) s,t as a subgraph.
Theorem 2.2.Let α > 1 be a real number and let r ≥ 3, s ≥ 2, t and n be positive integers.Then there is a constant C which depends only on s such that the following holds.If G is an n-vertex r-uniform hypergraph with at least Cα Observe that the conclusion of this theorem implies that G contains K (r) s,t as a subgraph, so Theorem 1.1 follows immediately (by taking α = 2, for example).Moreover, as we will see shortly, Theorem 2.2 also implies Theorems 1.4 and 1.5 fairly easily.
The proof of Theorem 2.2 uses a novel variant of the dependent random choice method.The rough idea is to choose random vertices v 2 , v 3 , . . ., v r ∈ V (G) and take A to be the set of . However, we add two major twists to this.Firstly, we only put into Secondly, the vertices v 2 , . . ., v r are not chosen uniformly at random, but with probability proportional to 1/d(v 2 , . . ., v r ).
Proof of Theorem 2.2.Choose C such that C ≥ 4s and C s−1 ((r−1)!) s 2 s r!s s −r 2 ≥ 1.Since s ≥ 2, C can be chosen to be independent from r.Let G be an n-vertex r-uniform hypergraph with e(G) ≥ Cα By successively deleting all edges containing a set of size r −1 which lies in less than D edges, we obtain a subhypergraph G ′ (on the same vertex set) with e(G ′ ) ≥ e(G) For the rest of the proof, we let d(S) .
By the definition of G ′ , if d(v 2 , . . ., v r ) > 0, then d(v 2 , . . ., v r ) ≥ D, so we have p ≤ 1.Let us define a random set A ⊂ V (G ′ ) as follows.With probability 1 − p, we let A = ∅.With probability p, we choose a random (r − 1)-tuple (v 2 , . . ., v r ) of distinct vertices in G ′ in a way that the probability that v i = v i for every i is . ., v r ) > 0 and 0 otherwise.Set A = A v 2 ,...,vr .
Proof of Claim 2. Using Hölder's inequality for three functions with parameters p 1 = s, p 2 = s, p 3 = s/(s − 2), we get where each sum is over all (r−1)-tuples of distinct vertices (v 2 , . . ., v r ) with where the second inequality used Claim 1.
Claim 3. Let u 1 , u 2 , . . ., u s , v 2 , . . ., v r−1 be distinct vertices in G ′ .Then the probability that u 1 , u 2 , . . ., u s ∈ A and . Clearly, there are at most d(u 1 , v 2 , . . ., v r−1 ) choices for v r satisfying these two properties, and for each such choice, the probability that . Hence, summing over all possibilities for v r proves the claim.Claim 4. Let u 1 , u 2 , . . ., u s and v 2 be distinct vertices in G ′ .Then the probability that u 1 , . . ., u s ∈ A and v 2 = v 2 is at most D/n 2 .
Proof of Claim 4. This follows from Claim 3 and the union bound over all choices for v 3 , . . ., v r−1 .
Claim 5. Suppose that u 1 , u 2 , . . ., u s are distinct vertices in G ′ such that {u 1 , . . ., u s } is not t-rich in G ′ .Then the probability that u j ∈ A for every j ∈ [s] is at most (r−1) 2 (t−1)D n 2 .Proof of Claim 5. Since {u 1 , . . ., u s } is not t-rich, the common intersection of the link hypergraphs of u 1 , . . ., u s does not contain a matching of size t.Hence, there is a set T ⊂ V (G ′ ) of size at most (r − 1)(t − 1) with the property that whenever u j v 2 . . .v r ∈ E(G ′ ) for every j ∈ [s], we have v i ∈ T for some 2 ≤ i ≤ r.Therefore, if u j ∈ A for every j ∈ [s], then v i ∈ T for some 2 ≤ i ≤ r.By Claim 4, the probability that u j ∈ A for every j ∈ [s] and v 2 ∈ T is at most |T |D/n 2 .By symmetry, for any fixed 2 ≤ i ≤ r, the probability that u j ∈ A for every j ∈ [s] and v i ∈ T is also at most |T |D/n 2 .The claim follows.
Let b be the number of sets of size s in A which are not t-rich.It follows from Claim 5 that D s , so using that x s ≥ (x/s) s for x ≥ s, we have Hence, It follows that there is an outcome for which s−1 and the proportion of t-rich sets of size s in A is at least 1 − α −1 , as desired.
It is now not hard to deduce Theorem 1.4 and Theorem 1.5.
Proof of Theorem 1.4.We may assume that there is a vertex in Y of degree exactly s in G (else, we can replace s by a smaller number).Let t = |V (G Let G be an n-vertex r-uniform hypergraph with at least C ′ n r− 1 s−1 edges.By Theorem 2.2, there is a set A ⊂ V (G) of size at least α Proof of Theorem 1.5.The lower bound was justified in the paragraph before the statement of the theorem, so it is enough to prove the upper bound.Let C be the constant provided by Theorem 2.2 with s = 2, and let G be an n-vertex r-uniform hypergraph with at least 100Crtn r−1 edges.By Theorem 2.2 applied with α = 100, s = 2 and rt in place of t, there is a set A ⊂ V (G) of size at least 100rt such that the proportion of rt-rich sets of size 2 in A is at least 99/100.
We claim that there is a set A ′ ⊂ A of size at least 4t such that for each u ∈ A ′ , the number of v ∈ A ′ for which {u, v} is rt-rich is at least 3|A ′ |/4.Indeed, let A 0 = A and, recursively for every i: • if there is some u ∈ A i such that the number of vertices v ∈ A i for which {u, v} is rt-rich is less than 3|A i |/4, then choose such a vertex and let A i+1 = A i \ {u}, • else terminate the process and let A ′ = A i .
Clearly, we obtain a set A ′ such that for each u ∈ A ′ , the number of v ∈ A ′ for which {u, v} is rt-rich is at least 3|A pairs in A which are not rt-rich.This is a contradiction, so indeed |A ′ | ≥ 4t.

Lower bounds
In this section we prove Theorem 1.2.The key ingredient is the following lemma.Lemma 3.1.Let A and B be two disjoint sets of size n.Assume that there exist pairwise edge-disjoint bipartite graphs G 1 , G 2 , . . ., G m with parts A and B such that for any distinct vertices x 1 , x 2 , . . ., x s ∈ A∪B, there are fewer than t vertices y ∈ A∪B for which there exists i ∈ [m] (that may depend on y) with x 1 y, x 2 y, . . ., Proof.Let X 1 , X 2 , . . ., X 2k be pairwise disjoint sets of size n.For every 1 ≤ p ≤ m, we define a 2k-partite 2k-uniform hypergraph G(p) with parts X 1 , X 2 , . . ., X 2k as follows.For Hence, there exists some p for which e(G(p)) ≥ e k /m.
We now want to show that for m ≈ (n/t) 1 s−1 , one can almost completely cover the edge set of K n,n with pairwise edge-disjoint graphs G 1 , . . ., G m satisfying the property described in Lemma 3.1.The bound in Lemma 3.1 will then give ex(2kn, K The following lemma provides a suitable collection of subgraphs under some mild divisibility conditions.Lemma 3.2.Let s ≥ 2 and h be positive integers and let p be a prime congruent to 1 modulo h.Let m = (p − 1)/h.Then there are pairwise edge-disjoint bipartite graphs G 1 , . . ., G m with the same parts A and B such that and for any distinct vertices x 1 , . . ., x s ∈ A ∪ B there are at most h s−1 (s − 1)! vertices y ∈ A ∪ B for which there exists i ∈ [m] with x 1 y, x 2 y, . . ., x s y ∈ E(G i ).
In the proof, we make use of the following result of Kollár, Rónyai and Szabó which was used to obtain their celebrated lower bound for the Turán number of complete bipartite graphs; see also [1] for a refinement.Lemma 3.3 ([10,Theorem 3.3]).Let K be a field and let a i,j , b i ∈ K for 1 ≤ i, j ≤ t such that a i,j 1 = a i,j 2 for any i ∈ [t] and j 1 = j 2 .Then the system of equations and note that N (xy) = N (x)N (y) for any x, y ∈ F p s−1 .For x ∈ A, y ∈ B and i ∈ [m], let xy be an edge in G i if and only if N (x + y) ∈ S i .Since S 1 , . . ., S m partition F × p and N (z) = 0 if and only if z = 0, it follows that G 1 , G 2 , . . ., G m are pairwise edge-disjoint and We are left to show that for any distinct vertices x 1 , . . ., x s ∈ A ∪ B there are at most h s−1 (s − 1)! vertices y ∈ A ∪ B for which there exists i ∈ [m] with x 1 y, x 2 y, . . ., x s y ∈ E(G i ).We may assume without loss of generality that x j ∈ A for each j ∈ [s].Suppose that for some y ∈ B there is i ∈ [m] with x 1 y, x 2 y, . . ., x s y ∈ E(G i ).This means that N (x j + y) ∈ S i holds for each j ∈ [s].Then N ( Claim.Let x 1 , . . ., x s be distinct elements of F p s−1 and let λ 1 , . . ., λ s−1 ∈ H. Then there are at most (s − 1)! elements y ∈ F p s−1 such that N ( Proof of Claim.Note that N ( xs+y , a j = 1 x j −xs and b j = λ j /N (x j − x s ), the problem is reduced to counting the number of solutions to the system of equations in the variable z.Since N (z + a j ) = (z + a j )(z p + a p j ) . . .(z p s−2 + a p s−2 j ), we can apply ) to see that (2) has at most (s − 1)! solutions for z, completing the proof of the claim.Remark 3.4.One can prove a variant of Lemma 3.2 using the random algebraic method of Bukh (see [2] for a detailed example of how this method is applied).More precisely, one can take a uniformly random polynomial f : F 2s−2 p → F p of a given (large) degree and set where S i are defined as in the proof of Lemma 3.2.The proof above uses essentially the same construction for the explicit choice f (x, y) = N (x + y) (with the minor difference that the parts there are identified with F p s−1 rather than F s−1 p ).
We are now in a position to prove Theorem 1.2.
4 An improved upper bound for r = 3 In this section we prove Theorem 1.3.The following definition will be crucial in the proof.
Here and in the rest of this section, we will ignore floor and ceiling signs whenever doing so does not make a substantial difference.Definition 4.1.Let s ≥ 3 be an integer and let G be a 3-uniform 3-partite hypergraph with parts X, Y and Z of size n each.We call a vertex z ∈ Z s-nice in G if there exist partitions X = X 1 ∪ • • • ∪ X s−1 such that if xyz ∈ E(G) for some x ∈ X i , then y ∈ Y i .We define s-nice vertices in X and Y analogously.
Observe that if some vertex z is s-nice in G, then it is also s-nice in any subhypergraph of G.
The proof of Theorem 1.3 will consist of two main steps.First, we prove the following structural result which states that (under a mild condition on the maximum degree) if a K (3) s,t -free hypergraph has close to n 3− 1 s−1 edges, then it contains a subgraph with a similar number of edges in which all vertices in two of the parts are nice.Lemma 4.2.Let s ≥ 3, let t be a positive integer, let ε > 0 and let n be sufficiently large.Let G be a K The second step is showing that (again under some mild conditions on the degrees) such a structured hypergraph must contain K Lemma 4.3.Let s ≥ 3, let t be a positive integer, let 0 < ε < 1 4s+6 and let n be sufficiently large.Let G be a 3-uniform 3-partite hypergraph with parts X, Y and Z of size n each.Assume that any pair of vertices in G is contained in either 0 or in at least n 1− 1 s−1 −ε hyperedges, but every pair of vertices is in at most n 1− 1 s−1 +ε hyperedges.Assume that xyz ∈ E(G) for some x ∈ X, y ∈ Y and z ∈ Z and that every vertex in Z ∪ {x} is s-nice in G. Then G contains a copy of K We will give the proof of Lemmas 4.2 and 4.3 in Subsections 4.1 and 4.2, respectively.Now let us see how these lemmas imply Theorem 1.3.The last ingredient is a lemma that shows that we can assume that no pair of vertices belongs to many hyperedges.Lemma 4.4.Let s ≥ 3, let t be a positive integer, let 0 < ε < 1/2, let n be sufficiently large and let G be a K denote the r-uniform hypergraph whose vertex set consists of disjoint sets X, Y 1 , . . ., Y t , where |X| = s and|Y 1 | = • • • = |Y t | = r − 1, and whose edge set is {{x} ∪ Y i : x ∈ X, 1 ≤ i ≤ t}.Note that K (r)s,t is r-partite and for r = 2, K

2
for infinitely many n.They further asked to determine the correct dependence of ex(n, K t with X being the part of size s and Y being the part of size t, then G (r)X,Y = K (r) s,t .Ergemlidze, Jiang and Methuku[4] asked whether it is true that if all vertices in Y have degree at most 2 in G, then ex(n, G (r) X,Y ) = O(n r−1 ) where the implied constant depends only on G and r.Here we resolve this conjecture in greater generality.Theorem 1.4.Let s ≥ 2, r ≥ 3 and let G be a bipartite graph with an ordered bipartition (X, Y ) such that every vertex in Y has degree at most s.Then ex(n, G (r) X,Y ) ≤ Cn r− 1 s−1 where C only depends on G and r.

1 s−1 t 1 s− 1 .
)| and let α = t s .Let C be the constant provided by Theorem 2.2 and let C ′ = Cα Note that C ′ is a constant that depends only on G and r.
Moreover, the proportion of t-rich sets in A is greater than 1 − |X| s −1 , so A has a subset A ′ of size |X| in which all s-sets are t-rich.This implies that G contains G (r) X,Y as a subgraph.Indeed, using that t = |V (G (r) X,Y )|, we can construct a copy of G (r) X,Y by embedding X arbitrarily into A ′ and then embedding the sets Y 1 , . . ., Y m from the definition of G (r) X,Y greedily one by one.
x j +y xs+y ) = λ j for each j ∈ [s − 1].Since there are h s−1 ways to choose the possible values of N ( x j +y xs+y ) for j ∈ [s − 1] from H, the claim implies the lemma.

( 3 )
s,t -free 3-uniform 3-partite hypergraph with parts X, Y and Z of size n each.Assume that e(G) ≥ n 3− 1 s−1 −ε and that every pair of vertices belongs to at most n 1− 1 s−1 +ε hyperedges.Then G has a subhypergraph H (on the same vertex set) such that e(H) ≥ n 3− 1 s−1 −225s 4 ε and every vertex in X ∪ Y , Y ∪ Z or Z ∪ X is s-nice in H.

( 3 )
s,t -free 3-uniform hypergraph with 3n vertices and at least n 3− 1 s−1 −ε hyperedges.Then G has a 3-partite subgraph H with parts of size n such that e(H) ≥ n 3− 1 s−1 −2ε and any pair of vertices belongs to at most n 1− 1 s−1 +2ε hyperedges in H.
t ) . . .(z t − a t,t ) = b t has at most t! solutions (z 1 , z 2 , . . ., z t ) ∈ K t .Proof of Lemma 3.2.Let A and B be disjoint copies of the field F p s−1 .Let H be a subgroup of F × p of order h, where F × p denotes the multiplicative group of F p .Let S 1 , S 2 , . . ., S m be the cosets of H in F × p .Recall that the norm map N :