A Clique-Free Pseudorandom Subgraph of the Pseudo Polarity Graph

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[AS16, Corollary 9.2.5]) that the edge density of an optimally pseudorandom Kk-free graph is O(n−1/(2k−3)). If optimally pseudorandom graphs with this edge density exist, it follows from the results of Mubayi and Verstra¨ete [MV19], and the multicolour generalization by He and

Wigderson [HW20], that the determination of the off-diagonal Ramsey number r(k1,...,km, l) for fixed k ,...,k 3 and l is essentially solved. Therefore the question of finding 1 m ≥ → ∞ optimally pseudorandom Kk-free graphs with highest possible edge density is one of the most interesting and challenging open problems in the theory of pseudorandom graphs. −1/(2k−3) Optimally pseudorandom Kk-free graphs with edge density d/m = Θ(n ) are only known to exist in the case when k = 3, by a construction of Alon [Alo94]. In [AK97], Alon and d −1/(k−2) Krivelevich obtained optimally pseudorandom Kk-free graphs with n = Ω n , by con- Fk−1 sidering a symmetric non–alternating bilinear form of q , q even. Recently, Bishnoi, Ihringer Fk and Pepe [BIP20], by using a symmetric bilinear form of q , q odd, provided a construction of d −1/(k−1) optimally pseudorandom Kk-free graphs with n = Ω n and this is currently the best known construction in terms of edge density for optimally pseudorandom Kk-free graphs, k 4. d −1/(k−≥1) Here we give a new family of optimally pseudorandom Kk-free graphs with n =Ω n , Fk arising from a symmetric non-alternating bilinear form of q , q even. Although the underlying construction is different, this family of graphs extends the result for k = 3 in [MPS19] to higher dimensions.

2 Polarity graphs

Before we get to the construction, we first introduce some terminology from finite geometry. Let PG(k 1,q) be the (k 1)–dimensional projective space over the finite field F . Points − − q and hyperplanes of PG(k 1,q) can be described as follows. Define an equivalence relation − Fk on the set of all nonzero vectors (x1,...,xk) in q , by saying that two vectors are equivalent if one is a multiple of the other by an element of F . The points of PG(k 1,q) as well as q − the hyperplanes can be represented by the equivalence classes of this equivalence relation. A k point (x1,...,xk) belongs to the hyperplane (y1,...,yk) if and only if i=1 xiyi = 0. Hence the hyperplane (a , . . . , a ) consists of the set of points satisfying the relation a X + +a X = 0. 1 k P 1 1 · · · k k A polarity of PG(k 1,q) is associated to a non–degenerate reflexive sesquilinear form β ⊥ − of Fk. In particular, given any representative vector x of a point in PG(k 1,q), the polarity q − sends x to the hyperplane x⊥ = y PG(k 1,q): β(x, y) = 0 , whereas for a projective { ∈ − } subspace S = x ,..., x , we define S⊥ = x⊥. In this way, is an involutory bijective map h 1 mi ∩i i ⊥ sending points to hyperplanes and vice versa, reversing incidence. A point x is called absolute with respect to if x x⊥. If β is alternating or Hermitian, then the polarity is called ⊥ ∈ ⊥

2 symplectic or unitary; if q is odd and β is symmetric, then is said orthogonal, whereas if q is ⊥ even and β is symmetric non–alternating, then is pseudo-symplectic. A map g of GL(Fk) is ⊥ q an isometry of β if β(xg, yg) = β(x, y), for all x, y Fk. Similarly two sesquilinear forms β, γ ∈ q of Fk are equivalent if there is a map g GL(Fk) and a F× such that γ(x, y) = aβ(xg, yg), q ∈ q ∈ q for all x, y Fk. To any linear map g GL(Fk) corresponds a projectivity of PG(k 1,q), i.e. a ∈ q ∈ q − bijection between projective subspaces of PG(k 1,q) of given dimension preserving incidences. − A projectivity of PG(k 1,q) associated with an isometry of β preserves the polarity . We − ⊥ Fk shall ﬁnd it helpful to represent linear maps of q with members of GL(k,q) acting on the left. The polarity graph (k, ) of PG(k 1,q) with respect to a polarity is the graph having Gq ⊥ − ⊥ as vertex set the points of PG(k 1,q) and two distinct points x, y are adjacent if and only − if x y⊥. Polarity graphs provide a good basis for clique-free pseudorandom graphs for three ∈ reasons.

Reason 1. The eigenvalues of (k, ) are d = (qk−1 1)/(q 1) and q(k−1)/2 = O(√d). Gq ⊥ − − ± Proof. It suﬃces to see that the square of its adjacency matrix A satisﬁes

qk−2 1 A2 = qk−2I + − J, q 1 − where I and J denote the identity matrix and the all-one matrix of appropriate size respectively.

Reason 2. Polarity graphs have edge density d/n = Θ(1/q)=Θ(n−1/(k−1)).

Proof. The number of vertices is (qk 1)/(q 1) and each vertex is adjacent to (qk−1 1)/(q 1) − − − − vertices, giving the result.

This implies that they are pseudorandom and quite dense, but we have to be careful as they are not simple graphs. This can be ﬁxed by taking a subgraph that avoids the vertices with loops, which correspond to absolute points. A nice property of pseudorandom graphs is that after taking a large subgraph, we again ﬁnd a pseudorandom graph of the same edge density.

Proposition 2.1. Consider a dH -regular induced subgraph H of a pseudorandom (n, d, λ)-graph G of positive density, i.e. V (H) := n = αn, α > 0. Then H is again pseudorandom and | | H equally dense, i.e. Θ(d/n)=Θ(dH /nH ).

Proof. The proof is another application of the expander mixing lemma which we mentioned before and interlacing of eigenvalues, for which [Hae95] is a standard reference.

Now upon taking the subgraph of non-absolute points, we even ﬁnd that it is clique-free.

Reason 3. The subgraph of non-absolute points of (k, ) is K -free. Gq ⊥ k+1 Proof. Suppose that x ,..., x are pairwise orthogonal points in PG(k 1,q), then we obtain 1 k+1 − Fk that the corresponding vectors in q are linearly independent. Therefore, we ﬁnd k + 1 linearly independent vectors in a k-dimensional vector space, which is a contradiction.

3 Note that if is symplectic, then the subgraph of non-absolute points of (k, ) is empty, ⊥ Gq ⊥ whereas if k and q are odd, and is orthogonal, then the subgraph of non-absolute points ⊥ of (k, ) is not regular. In all the remaining cases it can be checked that the subgraph of Gq ⊥ non-isotropic points of (k, ) is an optimally pseudorandom K -free graph with d/n = Gq ⊥ k+1 Θ(n−1/(k−1)). These three reasons, along with the construction of the subgraph of non-absolute points are also contained in [AK97]. The main insight of [BIP20] is that it is possible, when q is odd and orthogonal, to take a large subgraph on non-absolute points which reduces the clique number ⊥ even further. In this way, they ﬁnd a subgraph of the same density d/n = Θ(n−1/(k−1)), but which is K -free. We will employ the same idea for even q and pseudo-symplectic . k ⊥ 3 Construction

Let q = 2h and denote by U the point of PG(k 1,q) represented by the vector having one in i − the i-th position and zero elsewhere. Consider the pseudo-polarity associated with the bilinear ⊥ form β deﬁned as follows:

x y + x y + + x y + x y + x y if k is odd, 1 2 2 1 · · · k−2 k−1 k−1 k−2 k k x y + x y + + x y + x y + x y if k is even. 1 2 2 1 · · · k−1 k k k−1 k k The set of absolute points of are all the points of the hyperplane H : X = 0 and H⊥ equals ⊥ ∞ k ∞ Uk or Uk−1, according as k is odd or even, respectively. Let T = a Fq Tr(a) = 1 be the elements of absolute trace one, recalling Tr : x { ∈ | h−1 } ∈ F x + x2 + x4 + + x2 F . It is easily seen that T = q/2. Fix an element t T and q 7→ · · · ∈ 2 | | 0 ∈ deﬁne the hyperplane H : t X + tX = 0, where t T . If k 3, its image H⊥ lies on ℓ, t 0 k−1 k ∈ ≥ t where U U if k is odd, ℓ = k−2 k (Uk−1Uk if k is even.

When k is odd, we observe that the line ℓ is contained in the hyperplane Xk−1 = 0. When k is even, ℓ intersects H , t T , in one point. t ∈ Let (k,q) be the induced subgraph of (k, ) on the vertex set H (H ℓ). We H Gq ⊥ ∪t∈T t \ ∞ ∪ will prove in a few steps that this graph is indeed a Kk-free pseudorandom graph. Since the deﬁnition depends on the parity of k, our proofs will also be split into these two cases. Lemma 3.1. There is a group of projectivities of PG(k 1,q), k 3, which preserves and − ≥ ⊥ acts transitively on points of H (H ℓ), t T . t \ ∞ ∪ ∈ Proof. If k is odd, the linear maps given by

1 0 . . . 0 a1 0 0 1 . . . 0 a2 0  ......  ......   , a a . . . 1 a 0  2 1 k−2   0 0 . . . 0 1 0    0 0 . . . 0 0 1     4 where a , . . . , a F are isometries of β. Similarly, if k is even, the following linear maps are 1 k−2 ∈ q isometries of β 0 0 A . .  . . , 0 0 . . . 1 0   0 0 . . . 0 1   where A is an isometry of the alternating bilinear form x y + x y + + x y + x y . 1 2 2 1 · · · k−3 k−2 k−2 k−3 Note that induces a pseudo-symplectic polarity on H , say , whose absolute points are ⊥ t ⊥|Ht those of H H and H H = (ℓ H )⊥|Ht . The projectivities preserving are associated t ∩ ∞ t ∩ ∞ ∩ t ⊥|Ht with the isometries given above and the group of these projectivities is transitive on H (ℓ H ). t\ ∪ ∞ Hence, in both cases, the group of projectivities associated with these maps is transitive on points of H (ℓ H ), t T . t \ ∪ ∞ ∈ Proposition 3.2. If k 3, the graph (k,q) is an (n, d, λ)-graph where λ = O(√d) and ≥ H qk−1 qk−2 , if k is odd, 2 2 (n, d)=   qk−1 q qk−2  − , if k is even. 2 2  k−2  ′ ′ Proof. Since H H = q and H H ′ = H H = H ′ H , for t,t T , t = t , it | t \ ∞| t ∩ t t ∩ ∞ t ∩ ∞ ∈ 6 follows that H H = qk−1/2. Recall that when k is odd, the line ℓ is contained in |∪t∈T t \ ∞| H : X = 0 which has no impact on V ( (k,q)), as 0 / T . On the other hand for k even, one 0 k−1 H ∈ point is removed from every H ,t T . t ∈ Now consider a point x V ( (k,q)); hence there exists t T such that x H . By ∈ H 1 ∈ ∈ t1 Lemma 3.1, we may assume without loss of generality that x equals (0,..., 0,t1,t0), if k is odd, or (0,..., 0, 1,t ,t ), if k is even. In the former case x⊥ : t X + t X = 0 and x⊥ 1 0 1 k−2 0 k ∩ (H (H ℓ)), t T , consists of the qk−3 points ( ,..., ,t2,tt ,t t ). In the latter case x⊥ : t \ ∞ ∪ ∈ ∗ ∗ 0 1 0 1 X + t X + (t + t )X = 0 and x⊥ (H (H ℓ)), t T , consists of the qk−3 points k−3 0 k−1 0 1 k ∩ t \ ∞ ∪ ∈ ( ,..., ,t (t + t + t ), ,t,t ). Note that t = t + t , otherwise 1 = Tr(t)=Tr(t + t ) = ∗ ∗ 0 0 1 ∗ 0 6 0 1 0 1 Tr(t0)+Tr(t1) = 0, a contradiction. The fact that λ = O(√d) follows from Proposition 2.1.

Proposition 3.3. The graph (k,q) is K -free. H k Proof. By induction on k. The statement holds true for k = 2. Indeed, in this case no two points of H (H ℓ) are orthogonal with respect to β. To see this fact, let x = (t ,t ), x′ = ∪t∈T t \ ∞ ∪ 1 0 (t ,t ) V ( (2,q)) and observe that β(x, x′)= t (t + t + t ) = 0, since t ,t ,t T . 2 0 ∈ H 0 0 1 2 6 0 1 2 ∈ Assume that (k,q) is K -free. We show that the claim holds true for k+1. Let us consider a H k point x V ( (k + 1,q)). Hence there is t T such that x (H (H ℓ)). By Lemma 3.1, ∈ H 1 ∈ ∈ t1 \ ∞ ∪ we may assume that x is the point (0, 0,..., 0,t1,t0) or (0, 0,..., 1,t1,t0), according as k is even or odd respectively. Thus x⊥ H has equations ∩ t

5 t1Xk−1 + t0Xk+1 = 0 (3.1) (t0Xk + tXk+1 = 0

Xk−2 + t0Xk + (t0 + t1)Xk+1 = 0 (3.2) (t0Xk + tXk+1 = 0 according as k is even or odd respectively. The polarity induced by on x⊥ has bilinear form ⊥ t 2 x y + x y + + x y + x y + 1 x y if k is even, (3.3) 1 2 2 1 · · · k−1 k k k−1 t k−1 k−1 0 x y + x y + + x y + x y + t (x y + x y ) 1 2 2 1 · · · k−4 k−3 k−3 k−4 0 k k−1 k−1 k + (t0 + t1)(xk+1yk−1 + xk−1yk+1)+ xkyk+1 + xk+1yk + xk+1yk+1 if k is odd. (3.4)

By applying the invertible linear map

′ ′ Xi = Xi, 1 i k 3, Xi = Xi, 1 i k 2, ′ ≤ ≤ − ≤ ≤ − X = t0Xk−1 + Xk+1, X′ = t0 X ,  k−2  k−1 t1 k  ′ t0+t1 X = Xk + Xk+1, X′ = t1 X ,  k−1 t0  k t0 k−1  ′  ′ Xk = Xk+1, X = Xk+1, k+1 X′ = X ,   k+1 k−2     if k is even and ifk is odd,

Fk it is easily seen that the bilinear forms (3.3), (3.4) are equivalent to the bilinear forms of q given by

x′ y′ + x′ y′ + + x′ y′ + x′ y′ + x′ y′ if k is even, 1 2 2 1 · · · k−1 k k k−1 k k x′ y′ + x′ y′ + + x′ y′ + x′ y′ + x′ y′ if k is odd, 1 2 2 1 · · · k−1 k−2 k−2 k−1 k k and the hyperplanes x⊥ H (3.1), (3.2) of x⊥ are mapped to the hyperplanes of PG(k 1,q) ∩ t − given by

′ ′ t1Xk−1 + tXk =0 if k is even, ′ ′ t0Xk−1 + (t0 + t1 + t)Xk =0 if k is odd.

As t T and t t T = t + t + t t T respectively, the assertion follows by the 1 ∈ { | ∈ } { 0 1 | ∈ } induction hypothesis.

4 Conclusion

In this short note, we gave a new construction for optimally pseudorandom Kk-free graphs of edge density d/n = Θ(n−1/(k−1)) from polarity graphs, where the characteristic of the underlying

6 ﬁnite ﬁeld is even. It would be interesting to provide geometrical constructions leading to even denser pseudorandom graphs, ideally breaking the barrier of Θ(n−1/k). The idea of taking large subgraphs of polarity graphs applied here and in [BIP20] seems to reach its limits, at least for case of k = 3 on which the induction relies: a K3-free graph obtained from any polarity graph (3, ) has at most α( (3, )) = Θ(q3/2) vertices as shown by Mubayi and Williford [MW07]. Gq ⊥ Gq ⊥ Acknowledgments. This work was supported by the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA– INdAM).

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