A linear hypergraph extension of Tur´an’s Theorem

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Introduction
The r-uniform hypergraph (or r-graph) H = (V (H), E(H)) consists of a set V (H) of vertices and a set E(H) of edges, where each edge is an r-element subset of V (H). A graph is a 2-uniform hypergraph, and the 3-uniform hypergraphs are called triple systems. Given a hypergraph H and a family of hypergraphs F, we say H is F-f ree if H does not contain any member of F as a subgraph. The Turán number, denoted by ex r (n, F), is the maximum number of edges of an F-free r-graph on n vertices. The F-free r-graphs with n vertices and ex r (n, F) edges is called extremal hypergraphs of F.
To obtain the r-graph T r (n, l), l ! r, partition n vertices into l almost equal parts (that is, of sizes ! n l " or # n l $ ) and take all those edges which intersect every part in at most one vertex. Let t r (n, l) be the number of edges of the T r (n, l). Denote by K l the complete graph with l vertices. The classical Turán's Theorem is Theorem 1. ( [25]) Fix l ! 2. Then ex 2 (n, K l+1 ) = t 2 (n, l).
Moreover, equality is achieved only by the Turán graph T 2 (n, l).
A dozen years ago, Mubayi [19] and Pikhurko [21] gave a hypergraph extension of Turán's theorem. Given a graph F and positive integer r ! 3, the r-expansion of F is the r-graph F + obtained from F by enlarging each edge of F with a vertex set of size r −2 disjoint from V (F ) such that distinct edges of F are enlarged by disjoint sets. Mubayi [19] proved that if l ! r is fixed, then ex r % n, K + l+1 & = t r (n, l) + o (n r ) and Pikhurko [21] improved this to an exact result for n sufficiently large.
Moreover, equality is achieved only by the r-graph T r (n, l).
There is a vast amount of literature on the Turán problem in graphs and hypergraphs. We refer the reader to the surveys of recent results [10,17,20].
In this paper, we focus on the Turán problem in linear hypergraphs. An r-graph is linear if every two edges have at most one common vertex. Similar to the Turán number, given a family of r-graphs F, the linear Turán number is the maximum number of edges of an F-free linear r-graph on n vertices. We denote it by ex lin r (n, F) and simply write ex lin r (n, F ) instead of ex lin r (n, {F }) when F = {F }. Interestingly, the linear Turán problem is closely related to the function f r (n, v, e), where f r (n, v, e) is the maximum number of edges in an n-vertex r-uniform hypergraph not carrying e edges on v vertices, where r ! 3. The study of f r (n, v, e) was initiated by Brown, Erdős, and Sós [1] in 1970's. In one of the classical results in extremal combinatorics, Ruzsa and Szemerédi [22] showed that n 2−o(1) " f 3 (n, 6, 3) = o(n 2 ). This result was extended by Erdős, Frankl and Rödl [5] to n 2−o(1) " f r (n, 3r − 3, 3) = o(n 2 ).
Define T D r (n, l) be the linear r-graph with n vertices partitioning into l almost equal parts, and edges intersecting every part in at most one vertex such that each pair of vertices from distinct parts are contained in one edge exactly. One should note that the linear r-graph T D r (n, l) is equivalent to the so-called group divisible design in the design theory. Given a triple (r, n, l), determining the existence of T D r (n, l) is still a very open problem in design theory. It can be found in [2] that the T D r (n, r) exists for sufficiently large n and r divides n. For l|n, Hanani [16] proved that T D 3 (n, l) exists if and only if: ≡ 0 (mod 6). We refer the reader to [2,26] for more knowledge about the group divisible design.
Usually, the linear r-graph T D r (n, l) is not unique when it exists. In addition, it is easy to see that |T D r (n, l)| = t 2 (n, l)/ % r 2 & . The following theorem is a linear hypergraph extension of the Turán's theorem.
Moreover, if T D r (n, l) exists, then equality holds and the extremal hypergraph is T D r (n, l).
Gao, Chang and Hou [13] gave a spectral version of Theorem 3 when l = r. They prove that if an n-vertex linear r-graph H is K + r+1 -free, then the spectral radius of the adjacency tensor of H is no more than n/r. Clearly, Theorem 3 implies that ex lin for l ! r ! 3 and sufficiently large n. However, given an arbitrary graph F , the inequality ex lin r (n, F + ) " ex 2 (n, F )/ does not generally hold. We give an example in the following theorem.
Theorem 4. Fix l ! 3. Let n be sufficiently large and 6l|n. Then Given a graph F , let χ(F ) be the chromatic number of F . Although ex lin r (n, F + ) " does not generally hold, we prove that it holds asymptotically when r = 3 and χ(F ) ! 4.
Note that Theorem 5 holds for graphs with chromatic number at least 4. However, for graphs with small chromatic number, we have Theorem 6. Let r ! 3 and F be a graph with χ(F ) " r. Then Note that ex lin r (n, C + 3 ) = o(n 2 ) can be deduced from Theorem 6. Since f r (n, 3r−3, 3) = ex lin r (n, C + 3 ) for sufficiently large n, Theorem 6 can be regarded as an extension of Ruzsa, Szemerédi's result [22] and Erdős, Frankl, Rödl's result [5]. Let W k be the wheel graph obtained from the joint of a vertex and a cycle of length k. Mubayi and Verstraëte [20] raise a problem that whether ex 3 (n, W + 2k ) = O(n 2 ) or not? It is still an open problem even for W 4 . Since χ(W 2k ) = 3, by Theorem 6, we can immediately obtain the following corollary.
The rest of this paper is organized as follows. In next Section, we give some preliminaries. We present the proof of Theorem 3 in the Section 3. Theorem 4 is proven in Section 4. Finally, we prove Theorem 5 and Theorem 6 in the Section 5.

Preliminaries
Before presenting our main results, we need to introduce some basic notations and known results.
For a set V , let % V r & be the set of r-element subsets of V . Let H be an r-graph. For any vertex set S ⊂ V (H), the subgraph of H induced by S is the r-graph with vertex set S and edge set {e We use ∆(H) and δ(H) to denote the maximal degree and minimal degree of a vertex of H, respectively. The shadow graph of H is the graph with vertex set V (H) and edge set ( Given an r-graph F and a positive integer t, the t-blow-up F (t) is an r-graph obtained by replacing each vertex of F by t copies of itself and each edge by corresponding complete r-partite r-graph of these copies. For convenience, the pair {a, b} and the triple {a, b, c} are sometimes referred to as ab and abc, respectively.
Next we will introduce some known results which will be used to prove our results. An r-graph H is l-partite if V (H) can be partitioned into l parts such that every edge in H intersects each part in at most one vertex. An l-partite r-graph is complete if we take all those edges which intersect every part in at most one vertex.
. . , t l ) be the complete (l + 1)-partite graph with class sizes 1 " t 1 " t 2 " . . . " t l . Fix l and t, let if t is even and ⌊ n l ⌋ is even, if t is even and ⌊ n l ⌋ is odd.
Moreover, if G is a K l+1 (1, t 1 , . . . , t l )-free graph with n vertices and t 2 (n, l) Lemma 11. ( [5]) Let F be a graph with f vertices. For every ε > 0, there exists δ = δ(f, ε) > 0 such that if one has to delete at least εn 2 edges from a graph G to make it F -free, then G has at least δn f copies of F .

Proof of Theorem 3
Let H be a linear r-graph. If the vertex pair ab ⊂ V (H) is contained in some edges of H, then ab is exactly contained in one edge of H. We use h ab to denote the edge of H containing ab.
Lemma 12. Fix integers r ! 3, s ! 2 and t ! r 2 s 3 . Let H be a linear r-graph and F be a graph with s vertices. If the shadow graph ∂(H) contains a copy of F (t), then H contains a copy of F + .
Proof. Let H be a linear r-graph and F (t) ⊂ ∂(H). Our goal is to find a copy of F + in H. Let X be the set of vertices in F + which correspond to the vertices in F and Y = V (F + )\X. We will use the following steps to find the vertex sets X and Y in H.
We take a copy of where V i is the t copies of the vertex v i , 1 " i " s. Let X 1 = ∅ and Y 1 = ∅. At the first step, choose x 1 ∈ V 1 and x 2 ∈ V 2 and let ∃y ∈ X 2 ∪ Y 2 and y ∕ = x 2 such that x 2 yv is contained in an edge of H}. Since H is a linear r-graph, we have |B 3 ). Then Choose an x 3 ∈ V ′′ 3 and let X 3 = X 2 ∪ {x 3 }. If both x 1 x 3 and x 2 x 3 are not an edge of F (t), then we let Y 3 = Y 2 . If x 1 x 3 is an edge and x 2 x 3 is not an edge of F (t), then we let It is easy to see that |X 3 | = 3 and |Y 3 | " 3(r − 2).
At the k-th step, where 3 " k " s − 1, we will use the following method to add one vertex of V k+1 into X k to obtain X k+1 and add at most (r − 2)k vertices into Y k to obtain Y k+1 . Hence, |X k | = k and |Y k | " (r − 2) : ∃y ∈ X k ∪ Y k and y ∕ = x such that xyv is contained in an edge of H}.
Since H is a linear r-graph, we have |B the electronic journal of combinatorics 29(4) (2022), #P4.41 The last inequality holds since r ! 3 and s ! k ! 3. Choose an x k+1 ∈ V ′′ k+1 and let X k+1 = X k ∪ {x k+1 }. For each x ∈ X k , if xx k+1 is an edge of F (t), then we add those vertices of h xx k+1 \{x, x k+1 } into Y k . We denote the obtained vertex set by Y k+1 .
Finally, at the (s − 1)-th step, we obtain vertex sets X s and Y s . One can easily check that F + ⊆ H[X s ∪ Y s ], as desired.
Combining Corollary 13 and Lemma 10, we can easily prove Theorem 3.
and H is not a T D r (n, l). It implies that ∂(H) ≇ T 2 (n, l). By Lemma 10, we can deduce that T 2 (n, l) is the unique extremal graph for K l+1 (1, 1, t).
= t 2 (n, l) and ∂(H) ≇ T 2 (n, l), we have that K l+1 (1, 1, t) ⊂ ∂(H). Again by Corollary 13, we can deduce that H contains a copy of K + l+1 , which is a contradiction. Thus, the equality holds if and only if H is a T D r (n, l). The proof is completed.

Proof of Theorem 4
Let K l+1 (1, 3, t) be the complete (l+1)-partite graph with first part one vertex, the second part three vertices and the other parts t vertices. Let H be a linear 3-graph. For the vertex pair ab ⊂ V (H), we still use h ab to denote the edge of H containing ab. Lemma 14. Fix integers l ! 3 and t ! 4(l + 1) 3 . Let H be a linear 3-graph. If the shadow graph ∂(H) contains a copy of K l+1 (1, 3, t), then H contains a copy of K + l+1 (1, 2).
Proof. Let H be a linear 3-graph and K l+1 (1, 3, t) ⊂ ∂(H). Our goal is to find a copy of K + l+1 (1, 2) in H. Let X be the set of vertices in K + l+1 (1, 2) which correspond to the vertices in K l+1 (1, 2), and Y = V (K + l+1 (1, 2))\X. We will use the following steps to find the vertex sets X and Y in H.

At the second step, let
At the k-th step, where 3 " k " l + 1, we will use the following method to add two vertices of V k into X k−1 to obtain X k and add 4k − 6 vertices into Y k−1 to obtain Y k . Hence, we have |X k−1 | = 2k − 3 and |Y k−1 | = 2(k − 2) 2 . For 1 " i " k − 1 and 1 " j " 2, Since H is a linear 3-graph, we have |B Since t ! 4(l + 1) 3 and k " l + 1, we have Then |X ′ k−1 | = 2k − 2 and |Y ′ k−1 | = 2(k − 2) 2 + 2k − 3. And then for x ∈ X k−1 , we let Since H is a linear 3-graph, we have |B Since t ! 4(l + 1) 3 and k " l + 1, we have Thus we have that V ′′ k is not empty. Choose an x k2 ∈ V ′′ k and let X k = X ′ k−1 ∪ {x k2 } and Finally, at the (l + 1)-th step, we obtain vertex sets X l+1 and Y l+1 . One can easily check that K + l+1 (1, 2) is a subhypergraph of H[X l+1 ∪ Y l+1 ], as desired. Combining Lemma 10 and Lemma 14, we can easily prove Theorem 4.
That is, ex lin 3 (n, F + ) " The proof is completed.
Proof of Theorem 6. Let F be a graph with χ(F ) " r and H be an F + -free linear r-graph with n vertices.
Suppose that H has cn 2 edges, where c is a positive constant number. Since H is linear, we have that ∂(H) contains at least cn 2 copies of edge-disjoint K r 's. Hence, ∂(H) has to delete at least cn 2 edges to make it K r -free. By Lemma 11, there is a constant δ > 0 such that ∂(H) has at least δn r copies of K r 's. Now we construct an auxiliary r-graph H ′ as follows: The vertex set of H ′ is V (H). If there is a K r ⊆ ∂(H), we let these r vertices of K r be an edge of H ′ . Thus the number of edges of H ′ is at least δn r . By Lemma 8, we deduce that H ′ contains a complete r-partite r-graph with each class of size r 2 (|V (F )|) 4 . It implies ∂(H) contains a complete r-partite subgraph with each class of size r 2 (|V (F )|) 4 . Since χ(F ) " r, we have that ∂(H) contains a copy of F (t), where t = r 2 (|V (F )|) 3 . By Lemma 12, H contains a copy of F + , a contradiction. Hence, we have |E(H)| = o(n 2 ). This completes the proof.