Tur´an problems in pseudorandom graphs

Given a graph F , we consider the problem of determining the densest possible pseudorandom graph that contains no copy of F . We provide an embedding procedure that improves a general result of Conlon, Fox, and Zhao which gives an upper bound on the density. In particular, our result implies that optimally pseudorandom graphs with density greater than n − 1 / 3 must contain a copy of the Peterson graph, while the previous best result gives the bound n − 1 / 4 . Moreover, we conjecture that the exponent 1 / 3 in our bound is tight. We also construct the densest known pseudorandom K 2 , 3 -free graphs that are also triangle-free. Finally, we obtain the densest known construction of clique-free pseudorandom graphs due to Bishnoi, Ihringer and Pepe in a novel way and give a diﬀerent proof that they have no large clique.


Introduction
Given a family F of graphs we say a graph G is F-free if it does not contain any member in F as a subgraph.A fundamental problem in extremal graph theory is to determine the maximum number ex(n, F) of edges in an F-free graph on n vertices.Here ex(n, F) is called the Turán number of F, and the limit π(F ) = lim n→∞ ex(n, F)/ n 2 , whose existence was proved by Katona, Nemetz, and Simonovits [20], is called the Turán density of F.
For a graph G we use V (G) to denote the vertex set of G, and use v(G) and e(G) to denote the number of vertices and edges in G, respectively.For a set S ⊂ V (G) we use e G (S) to denote the number of edges in the induced subgraph G[S].Given two vertex sets X, Y ⊂ V (G) we use e G (X, Y ) to denote the number of edges in G that have one vertex in X and one vertex in Y (here edges with both vertices in X ∩ Y are counted twice, hence e G (X, X) = 2e G (X)).We will omit the subscript G if it is clear from the context.Informally, we say that a graph is pseudorandom if its edge distribution behaves like a random graph.In this note we use the following notation, which was firstly introduced by Thomason in his fundamental papers [33,34], to quantify the randomness of a graph.
For two real numbers p ∈ [0, 1] and α ≥ 0, we say a graph G is (p, α)-jumbled if it satisfies for all X, Y ⊂ V (G).
A special family of (p, α)-jumbled graphs are the well-known (n, d, λ)-graphs.A graph G is an (n, d, λ)-graph if it is a d-regular graph on n vertices and the second largest eigenvalue in absolute value of its adjacency matrix is λ.The well-known Expander mixing lemma (e.g.see [24,Theorem 2.11]) implies that an (n, d, λ)-graph is (d/n, λ)-jumbled.Conversely, Bilu and Linial [8] proved that an n-vertex d-regular (p, α)-jumbled graph is an (n, d, λ)graph with λ = O(α log(d/α)).
Constructions of dense pseudorandom graphs that avoid a certain graph as a subgraph are extremely useful for many problems.In particular, the second author and Verstaëte [28] recently showed that for every fixed integer t ≥ 3, the existence of K t -free (n, d, λ)-graphs for the off-diagonal Ramsey numbers, and this matches the best known upper bound in exponent.More generally, [28] shows that the existence of dense F -free pseudorandom graphs implies a good lower bound for the Ramsey number R(F, n).This motivates us to consider the following pseudorandom version of the Turán problem.
Let F be a family of graphs and C > 0 be a real number.Let ex rand (n, C, F) be the maximum number of edges in an n-vertex (p, α)-jumbled F-free graph with α ≤ C √ np.Note that in the definition of ex rand (n, C, F) we do not have any restriction on p.
In many applications, it suffices to know the exponent of ex rand (n, C, F).So we let In other words, exp(F) is the supremum of β such that there exist a constant C and a sequence Using the Expander mixing lemma one can prove that for every integer t ≥ 3 we have exp(K t ) ≤ 1 − 1 2t−3 .Alon's construction [2] shows that this bound is tight for t = 3, that is, exp(K 3 ) = 2  3 .It is a major open problem to determine exp(K t ) in general.Alon and Krivelevich proved in [4] that exp(K t ) ≥ 1 − 1 t .Recently, Bishnoi, Ihringer, and Pepe [9] improved their bound and proved the following result.
Theorem 1.1 (Bishnoi-Ihringer-Pepe [9]).Suppose that t ≥ 4 is an integer.Then Mattheus and Pavese [27] give a different construction of K t -free pseudorandom graphs which also matches the bound in Theorem 1.1.In Section 4, we will present a construction that is isomorphic to the construction of Bishnoi, Ihringer, and Pepe [9], and give a new proof to Theorem 1.1.
For bipartite graphs, the pseudorandom version of the Turán problem does not appear to differ much from the ordinary Turán problem since many constructions for the lower bound are pesudorandom.For example, for complete bipartite graphs, the projective norm graphs (see [22,5]) are optimally pseudorandom (see [31]) and do not contain K s,t with t ≥ (s − 1)! + 1.Therefore, together with the well known Kövari-Sós-Turán Theorem [23], we know that exp(K s,t ) = 1 − 1 s for all positive integers s, t with t ≥ (s − 1)! + 1.For even cycles, constructions from generalized polygons [25] and an old result of Bondy and Simonovits [12] imply that exp(C 6 ) = 4  3 and exp(C 10 ) = 6  5 .The value of exp(C 2k ) for k = 2, 3, 5 are still unknown due to the lack of constructions.For non-bipartite graphs, exp(F ) is completely different from the ordinary Turán problem (indeed, exp(F ) < 2 while π(F ) > 0) and there are very graphs F for which exp(F ) is known.For example, for odd cycles, a construction due to Alon and Kahale [3] together with Proposition 4.12 in [24] implies that exp(C ℓ ) = 2 ℓ for all odd integers ℓ ≥ 3. The first general upper bound on exp(F ), is due to Kohayakawa Rödl, Schacht, Sissokho, and Skokan [21].They prove that exp(F ) ≤ 1 − 1 2ν(F )−1 for every triangle-free graph F .Here ν(F ) = 1 2 (d(F ) + D(F ) + 1), where D(F ) = min {2d(F ), ∆(F )} and d(F ) is the degeneracy of F .This was improved via the following result of Conlon, Fox, and Zhao in [14].
Theorem 1.2 (Conlon-Fox-Zhao [14]).For every graph F , we have exp(F ) ≤ 1 − 1 2d 2 (F )+1 , where d 2 (F ) is the minimum real number d such that there is an ordering of the vertices v 1 , . . ., v m of F so that In some cases the bound provided by Theorem 1.2 is sharp (e.g. it is conjectured to be sharp for cliques), but we speculate that for most graphs F it can be improved.Below we give an improvement that holds for many graphs.
Theorem 1.3.Let F be a fixed graph and d be such that the following holds.There exists an ordering v 1 , v 2 , . . ., v m of the vertices of F and a 1 ≤ k ≤ m such that: • for all edges v i v j ∈ F with i < j, i < k, we have N <i (v i ) + N <i (v j ) ≤ 2d, and Remark.Roughly speaking, Theorem 1.3 says that if there exists an induced forest on an interval of some ordering of V (F ), then it is possible to improve the bound of Theorem 1.2 by treating this forest as one edge.In fact, we will see later that the proof of Theorem 1.3 can be extended easily to get a more general result.
As an application of Theorem 1.3, we study exp(P), where P is the Petersen graph.The Petersen graph was considered by several researchers in related contexts.For example, Tait and Timmons [32] proved that the Erdős-Rényi orthogonal polarity graphs [17] (henceforth the Erdős-Rényi graph), which are optimally pseudorandom C 4 -free graphs, contain the Petersen graph as a subgraph.Conlon, Fox, Sudakov, and Zhao asked in [13] whether there is a counting lemma for the Petersen graph in an n-vertex C 4 -free graph with Ω(n 3/2 ) edges.We know very little about exp(P), for example, it is not known whether exp(P) ≥ 1 2 .The only lower bound we have is exp(P) ≥ exp(C 5 ) = 2 5 .In the other direction, the previous best upper bound is exp(P) ≤ 3 4 that follows from [14] (it is not too difficult to prove that d 2 (P) = 3  2 ).We conjecture that exp(P) = 2 3 .We think that the construction of K 3 -free pseudorandom graphs due to Kopparty does not contain the Petersen graph as a subgraph.If this is true, then it will prove the lower bound exp(P) ≥ 2 3 .For completeness, we include his construction here.
Let p = 3 be a prime, and let F q be a finite field with where q = p h for some integer h ≥ 1. Recall that the absolute trace function Tr : q , T = {x ∈ F q : Tr(x) ∈ {1, −1}}, and S ⊂ F 3 q be a subset defined as Using some simple linear algebra one can show that G is triangle-free, and using some results about finite fields and abelian groups one can prove that G is an (n, d, λ)-graph with n = q 3 , d = Θ( q 2 p ), and λ = Θ( q p ). Remark.Ferdinand Ihringer informed us that the construction above contains an induced copy of the Petersen graph when p = 2 and h = 3, and he thinks that, in general, Kopparty's construction contains many copies of the Petersen graph.Nevertheless, it still might be true that exp(P) = 2 3 .Our next result about K 2,3 was motivated by an old problem of Erdős [16], which asks if is true.A construction due to Parsons [29] for the lower bound comes from the Erdős-Rényi graph by removing half of its vertices.Since the Erdős-Rényi graph is optimally pseudorandom, Parsons' construction also implies that ex rand (n, for some absolute constant C. In [1], Allen, Keevash, Sudakov, and Verstraëte proved that the extremal constructions for ex(n, {K 3 , K 2,t }) cannot be bipartite for every t ≥ 3 by constructing a {K 3 , K 2,t }-free graph whose number of edges is greater than the maximum number of edges in a {K 3 , K 2,t }-free bipartite graph.However, their construction is (t − 1)-partite, and therefore it does not give a lower bound for ex rand (n, C, {K 3 , K 2,3 }).The previous best lower bound is that follows from Parsons' construction.We improve this and present a construction of the densest known {K 3 , K 2,3 }-free pseudorandom graphs.
Theorem 1.5.We have ex rand (n, 2, Remark.Thang Pham pointed out to us that the following construction also provides a lower bound for ex rand (n, 2, {K 3 , K 2,3 }).Let q be an odd prime power.The distance graph D on F 2 q is a graph whose vertex set is F 2 q , and two points (x 1 , x 2 ), (y 1 , y 2 ) are adjacent iff (x 1 − y 1 ) 2 + (x 2 − y 2 ) 2 = 1.The K 2,3 -freeness of D follows from the fact that any two cycles have at most two points in the intersection.The K 3 -freeness of D follows from results in [7].The pseudorandomness of D follows from results in [19].
In Section 2, we prove Theorems 1.3 and 1.4.In Section 3, we prove Theorem 1.5.In Section 4 we present a new proof of Theorem 1.1.Throughout the paper we will omit the use of floors and ceilings to make the presentation cleaner.
2 Upper bound for the Petersen graph 2.1 Proof of Theorem 1.4 We prove Theorem 1.4 in this section.In the next section, we will show that Theorem 1.4 follows immediately from the more general Theorem 1.3, but we think it is instructive to see an independent proof of Theorem 1.4 first.
Let us present first two standard lemmas.We start with the following direct consequence of the definition of a jumbled graph.
. By the definition of jumbleness, we get The next lemma is a simple cleaning procedure which is useful in problems concerning (p, α)-jumbled graphs.
Lemma 2.2.Let G be a (p, α)-jumbled graph on n vertices.Then, for all sets X, Y ⊆ V (G) such that |X||Y | ≥ 100(α/p) 2 the following holds.There exist subsets Proof.Consider the following process.Start with X 0 := X, Y 0 := Y and at step i ≥ 0, do the following.Take On the other hand, the definition of a (p, α)-jumbled graph implies that To prove Theorem 1.4, it suffices to show that for every C 1 > 0 there exist C 2 > 0 and n 0 > 0 such that if n > n 0 and G is an n-vertex graph that is (p, α)-jumbled with α ≤ C 1 √ pn and pn ≥ C 2 n 2/3 , then G contains the Peterson graph.This follows from the following theorem.Proof.Let G be a (p, α)-jumbled graph on n vertices such that α ≤ p 2 n/200, so that p 2 n 2 /5000 ≥ 4(α/p) 2 .We will find an embedding of the Petersen graph with vertices v 1 , . . ., v 10 which correspond to the labelling in the right drawing of Figure 1.First, let v 1 be a vertex of G of degree at least pn/2 (which is guaranteed by Lemma 2.1 with q = 2) and let X denote a set of pn/2 of its neighbours.Let also Y denote the rest of the vertices, that is, Y := V \ ({v 1 } ∪ X), which is of size at least n/2.By Lemma 2.2, there exist subsets X ′ ⊆ X, Y ′ ⊆ Y of size at least 9pn/20 and 9n/20 respectively, with .Let v 6 v 9 be an edge contained in these 2n/5 vertices, which is guaranteed by Lemma 2.1, so that both v 6 , v 9 have at least p 2 n/500 neighbours in Z 7,8 .Let Z 7 be a set of p 2 n/1000 such neighbours of v 9 and Z 8 be a set of p 2 n/1000 such neighbours of v 6 so that Z 7 ∩ Z 8 = ∅.Now, recall that since v 6 , v 9 ∈ Y ′ , both of them have at least p|X ′ |/10 ≥ 9p 2 n/200 ≥ p 2 n/25 neighbours in X ′ and so, let Z 2 ⊆ X ′ be a set of p 2 n/50 such neighbours of v 6 and Z 4 be a set of p 2 n/50 such neighbours of v 9 so that are all edges.To finish, we note that if there exists an edge in E[Z ′ 5 , Z ′ 10 ], then the Petersen graph can be embedded.Indeed let v 5 v 10 be such an edge with v 5 ∈ Z ′ 5 and v 10 ∈ Z ′ 10 .In particular, we have that v 5 v 7 , v 10 v 8 are edges.Further, by the definition of Z 5 , Z 10 , there exist v 2 ∈ Z 2 , v 4 ∈ Z 4 such that v 2 v 5 and v 4 v 10 are edges.Furthermore, by definition, we also have that v 2 v 6 , v 2 v 1 , v 4 v 9 , v 4 v 1 are edges and thus, one can check that all the edges in the Petersen graph are present.To conclude then, note that there exists an edge in
Let us prove the following embedding lemma for forests first.
Lemma 2.4.Suppose that T is a forest on [m] and G is an n-vertex (p, α)-jumbled graph.Let X 1 , . . ., X m ⊂ V (G) be nonempty pairwise disjoint subsets of V (G) that satisfy for all edges ij in T .Then there exists an embedding of f : Proof.We prove this lemma by induction on m.The base case m = 1 is clear since X 1 is nonempty.So we may assume that m ≥ 2. Without loss of generality, we may assume that the vertex m is a leaf of T and the vertex m − 1 is its neighbor in T .Let T Now apply the induction hypothesis to the sets X 1 , . . ., X m−2 , X ′ m−1 , we obtain an embedding f : By the definition of X ′ m−1 , there exists v ∈ X m such that {f (m − 1), v} ∈ G. Hence we can extend f to get an embedding of T to G by setting f (m) = v.This completes the proof of Lemma 2.4.Now we are ready to prove Theorem 1.3.
Proof of Theorem 1.3.Let G be a (p, α)-jumbled graph with α < p d+1 n C for an arbitrarily large constant C > m4 m .We will show that G contains a copy of F .This implies that exp(F ) and our result will imply the theorem.
Consider an ordering v 1 , v 2 , . . ., v m of the vertices of F and a 1 ≤ k ≤ m such that: We will first embed a copy of F 1 using (b).At the same time, we will also ensure by (c), that the candidate sets for the vertices v k , . . ., v m are still large enough so that the forest F 2 can be embedded in them, thus giving an embedding of F .

Take a partition
Then there exist vertices u j ∈ V j for all j ∈ [s] such that G[{u 1 , . . ., u s }] contains a copy of F [{v 1 , . . ., v s }] and The proof is by induction on s.For the base case s = 1, first observe that for all j ∈ I 1 .Hence we can apply Lemma 2.1 to V 1 and V j for all j ∈ I 1 with q = 2 to obtain a vertex Observe that for every j ∈ I s we have In the second last inequality we used (b), and in the last inequality we used the assumption that α ≤ p d+1 n/(m4 m ).So we may apply Lemma 2.1 to U s and U j for all j ∈ I s with q = 2 and obtain u s ∈ U s such that d(u s , U j ) ≥ p|U j |/2 for all j ∈ I s .Now by (2), for every i ∈ I s we have On the other hand, by (2), for every i ∈ [s + 1, m] \ I s , we have Finally, it is clear that G[{u 1 , u 2 , . . ., u s }] contains a copy of F [{v 1 , v 2 , . . ., v s }], so the proof of the claim is complete.
Applying Claim 2.5 with s = k − 1 we obtain . Now that the first portion of the graph has been embedded, it remains only to embed a forest on the given candidate sets X i := V i,k−1 .If we find an embedding f : contains a copy of F .Similar to (3), by (c) and Claim 2.5, for every {v i , v j } ∈ F 2 we have Applying Lemma 2.4 with T = F 2 and the sets X k , . . ., X m , we know that such an embedding f exists.This completes the proof of Theorem 1.3.
We remark that there are some graphs to which the precise statement of the above theorem cannot be applied in order to get a tight result -for example, odd cycles.However, the proof can be slightly adapted to deal with them.For odd cycles we take k = 2, so that F [v k , . . ., v m ] is a path; this F 2 can then be embedded in a different way than in the general theorem above, in particular, using also the expansion properties of (α, p)-jumbled graphs.
We now give a further generalization of Theorem 1.3 where instead of partitioning the graph into two parts which are dealt with separately, we partition the graph into several parts.
For every graph F on m vertices, let d2 (F ) denote the smallest number d for which there exists an ordering v 1 , v 2 , . . ., v m of V (F ) such that the following statements hold for some ℓ ∈ N and 1 = Theorem 2.6.For every graph F we have exp(F ) ≤ 1 − Remark.One can extend Theorem 2.6 to get a counting result for F in pseudorandom graphs that improves Theorem 1.14 in [14] (by replacing d 2 (F ) there with d2 (F ) here).This could result in some improvements for the corresponding Turán and Ramsey problems in pseudorandom graphs (see Theorems 1.4, 1.5, and 1.6 in [14]).
Let λ 1 ≥ • • • ≥ λ n be the eigenvalues of the adjacency matrix A G of G. Since G is regular, we have λ 1 = |S| = p − 1.
First we prove that G is K 3 -free.Suppose to the contrary that there exist three vertices u, v, w ∈ Z 2 p that form a copy of K 3 in G. Assume that v − u = (a, a 3 ), w − v = (b, b 3 ), and u − w = (c, c 3 ).Then a + b + c = 0, Since p = 3, we must have 0 ∈ {a, b, c}, a contradiction.
Next we prove that G is K 2,3 -free.It is equivalent to show that every pair of vertices {u, v} ⊂ Z 2 p has at most two common neighbors.Let a = u 1 − v 1 and b = u 2 − v 2 .A common neighbor of u and v implies that there exist x, y ∈ Z p \ {0} such that y − x = a, and These two equations imply that (x+a) 3 = x 3 +b, which simplifies to 3ax 2 +3a 2 x+a 3 −b = 0. Since (a, b) = (0, 0) and p = 3, this quadratic equation in x has at most two solutions in Z p \ {0}.Therefore, u and v have at most two common neighbors.
First, it is easy to see that the character χ : Z p → C × defined by χ(α) = ω α p for all α ∈ Z p has order p.On the other hand, since (a 1 , a 2 ) = (0, 0) and p = 3, the polynomial f (X) = a 1 X + a 2 X 3 is not of the form c (g(X)) p for any c ∈ Z p and for any polynomial g(X).Therefore, it follows from Theorem 3.3 that This implies that |λ i | ≤ 2 √ p + 1 for all i ∈ [n] \ {1}, and hence completes the proof of Theorem 3.2.

K t -free pseudorandom graphs
In this section we present a construction that is isomorphic to the construction of Bishnoi, Ihringer, and Pepe [9], and give a new proof to Theorem 1.1.

x
A proof of the following theorem can be found in [15].
Theorem 4.3.Suppose that q is an odd prime power and t ≥ 3 is an odd integer.Then the induced subgraph of Proof.Suppose to the contrary that there exists a set S of t distinct points x 1 , . . ., x t ∈ X ⊠ /∼ such that the induced subgraph of AK(t − 1, q) on S is complete.Then it follows from Fact 4.1 that there exists a nonzero element a ∈ F q such that x Therefore, by (4), we have In the last equality we used the fact that x i • x j = 0 for all i = j.Applying the quadratic character χ(•) to both sides of the equation above, we obtain For the case that t ∈ N is even we use a different argument.
Proposition 4.4.For every vertex v ∈ PG(t, q) the induced subgraph of AK(t, q) on N (v) is isomorphic to AK(t − 1, q).
Proof.Let e 1 , . . ., e t be the standard orthonormal basis of the t-dimensional space F t q .Fix a vector v ∈ F t+1 q \ X 0 and let e ′ 1 , . . ., e ′ t be an orthonormal basis of the t-dimensional space v ⊥ , where v ⊥ = w ∈ F t+1 q ∈ : w • v = 0 .Define the map φ : F t q → v ⊥ by sending t i=1 a i e i to t i=1 a i e ′ i .Clearly, the map φ is linear and induces a bijection between F t q /∼ and v ⊥ /∼.Moreover, φ sends absolute points to absolute points.Now suppose that u 1 = t i=1 a i e i and u 2 = t i=1 b i e i are two distinct points in F t q .Then ψ(u 1 ) • ψ(u 2 ) = This implies that the map φ preserves the orthogonality of two vectors, and hence, it sends edges (resp.non-edges) in AK(t, q) to an edge (resp.non-edge) in the induced subgraph of AK(t, q) on v ⊥ .Therefore, φ induces an isomorphism between AK(t − 1, q) and the induced subgraph of AK(t, q) on N (v).
Lemma 4.5.Suppose that α ∈ (0, 1) is a constant and V 1 ⊂ PG(t, q) is a subset of size α • |PG(t, q)| in the graph AK(t, q).Then there exists a vertex v ∈ V 1 such that Proof.Suppose to the contrary that there exists an absolute constant ǫ > 0 such that ≤ (1 − ǫ)α for all v ∈ V 1 and for all q.Choose q to be sufficiently large.Let n = |PG(t, q)| be the number of vertices in AK(t, q), and let d be the degree of AK(t, q).Then, it follows from our assumption that Since ǫα 2 d ≫ √ d, this contradicts the fact that AK(t, q) is ( d n , Θ( √ d))-jumbled.
Now we are ready to prove Theorem 1.1 for even t.Our construction will be an induced subgraph of AK(t, q) on a subset of the neighborhood of a vertex.
Proof of Theorem 1.1 for even t.Let t ∈ N be an even number.Let V denote the vertex set of AK(t, q).Let .By Proposition 4.4, the induced subgraph of AK(t, q) on the set N ( v ) is isomorphic to AK(t − 1, q), which is (p, α)-jumbled with p = Θ(m − 1 t−1 ) and α = Θ( √ mp), where m = |PG(t − 1, q)|.On the other hand, by Theorem 4.3, the induced subgraph of AK(t, q) on the set X ⊠ /∼ is K t+1 -free.Therefore, the induced subgraph of AK(t, q) on the set U is K t -free.This proves Theorem 1.1 for even t.

Acknowledgment
We would like to thank Anurag Bishnoi, Ferdinand Ihringer, and Thang Pham for their insightful comments.

Theorem 2 . 3 .
Let G be a (p, α)-jumbled graph on n vertices such that α ≤ p 2 n/200 and p > 10n −1/3 .If n is sufficiently large, then G contains the Petersen graph.Remark.Theorem 2.3 and some simple calculations show that for every C > 0 if n is sufficiently large and G is an n-vertex (p, α)-jumbled graph with α ≤ C √ pn and pn ≥ (200C + 10) 2/3 n 2/3 , then G contains the Petersen graph as a subgraph.

a
i b i = t i=1 a i e i • t i=1 b i e i = u 1 • u 2 .