On a correspondence between maximal cliques in Paley graphs of square order
aa r X i v : . [ m a t h . C O ] F e b On a correspondence between maximal cliques in Paleygraphs of square order
Sergey Goryainov a , Alexander Masley c , Leonid V. Shalaginov a,b a Chelyabinsk State University, Brat’ev Kashirinyh st. 129Chelyabinsk 454021, Russia b Krasovskii Institute of Mathematics and Mechanics, S. Kovalevskaja st. 16Yekaterinburg 620108, Russia c Technion - Israel Institute of Technology, CS Taub Building, Haifa 3200003, Israel
Abstract
Let q be an odd prime power. Denote by r ( q ) the value q modulo 4. In thispaper we establish a correspondence between two types of maximal cliques oforder q + r ( q )2 in the Paley graph of order q . Keywords:
Paley graph; maximal clique; M¨obius transformation
1. Introduction
Let q be an odd prime power, q ≡ Paley graph P ( q ) is thegraph whose vertices are the elements of the finite field F q with two verticesbeing adjacent whenever their difference is a square in the multiplicative group F ∗ q .In this paper we study maximal cliques in Paley graphs of square order P ( q ),where q is an odd prime power. The Paley graphs P ( q ) are known to be self-complementary strongly regular graphs with parameters ( q , q − , q − , q − ).It follows from the Delsarte bound that a maximum clique (as well as a maxi-mum independent set) in P ( q ) has size q . For any odd prime power q , the field F q can be naturally viewed as the affine plane A (2 , q ). From this point of view,all lines in A (2 , q ) can be divided into two classes: quadratic lines and non-quadratic lines. For a quadratic line (resp. non-quadratic line) the differencebetween any two distinct elements from this line is a square (resp. a non-square).Thus, the parallel classes of quadratic (resp. non-quadratic) lines give examplesof partitions of vertices of P ( q ) into maximum cliques (resp. maximum inde-pendent sets). In particular, the subfield F q in F q , being a quadratic line, is a Email addresses: [email protected] (Sergey Goryainov), [email protected] (Alexander Masley), (Leonid V. Shalaginov)
Preprint submitted to Arxiv February 9, 2021 lique of size q in P ( q ), which defines a parallel class consisting of quadraticlines in A (2 , q ). In [1], Blokhuis showed that a clique (resp. an independent set)of size q in P ( q ) is necessarily a quadratic (resp. non-quadratic) line. Problem 1.
What are maximal but not maximum cliques in Paley graphs ofsquare order?
Let r ( q ) denote the reminder after division of q by 4. In [2], for any oddprime power q , a maximal (but not maximum) clique in P ( q ) of size q + r ( q )2 wasconstructed.In [4], for any odd prime power q , a maximal clique in P ( q ) of the same size q + r ( q )2 was constructed. This clique was shown to have a remarkable connectionwith eigenfunctions of P ( q ) that have minimum cardinality of support q + 1. Problem 2.
Given a prime power q , what are eigenfunctions of P ( q ) that haveminimum cardinality of support? It was found with use of computer that the cliques from [2] and [4] are theonly maximal cliques of size q + r ( q )2 under the action of the automorphism groupwhenever 25 q
83 holds. On the other hand, there are extra cliques of size q + r ( q )2 for 9 q Problem 3.
Given a prime power q , are there maximal cliques in P ( q ) whosesize is greater than q + r ( q )2 and less than q ? Problem 4.
Given a prime power q , what are maximal cliques in P ( q ) of size q + r ( q )2 ? In this paper paper we make a contribution to the solution of Problem 4.
2. Preliminaries
In this section we list some useful notation and results. Let q be an oddprime power. A (2 , q )Denote by A (2 , q ) the point-line incidence structure, whose points are thevectors of 2-dimensional vector space V (2 , q ) over F q , and the lines are theadditive shifts of 1-dimensional subspaces of V (2 , q ). It is well-known that A (2 , q ) satisfies the axioms of a finite affine plane of order q . In particular, eachline contains q points and there exist q + 1 lines through a point. An oval in theaffine plane A (2 , q ) is a set of q + 1 points such that no three are on a line. Aline meeting an oval in one point (resp. in two points) is called tangent (resp. secant ). For any point of an oval there exists a unique tangent at this point and q secants. By Qvist’s theorem (see, for example, [3, p. 147]), given an oval in aprojective plane (an affine plane) of odd order and a point that does not belongto the oval, there are either 0 or 2 tangents to the oval through this point.2 .2. Finite fields of square order Let d be a non-square in F ∗ q . The elements of the finite field of order q canbe considered as F q = { x + yα | x, y ∈ F q } , where α is a root of the polynomial f ( t ) = t − d . Since F q is a 2-dimensional vector space over F q , we can assumethat the points of A (2 , q ) are the elements of F q and a line l is presented bythe elements { x + y α + c ( x + y α ) } , where x + y α ∈ F q , x + y α ∈ F ∗ q are fixed and c runs over F q . The element x + y α is called the slope of line l . A line l is called quadratic (resp. non-quadratic ), if its slope is a square(resp. non-square) in F ∗ q . Note that the slope is defined up to multiplicationby a constant c ∈ F ∗ q . Let β be a primitive element of the finite field F q . Sincethe elements of F ∗ q = h β q +1 i are squares in F ∗ q , the difference between any twopoints of quadratic (resp. non-quadratic) line is a square (resp. non-square) in F ∗ q . In this setting, two distinct vertices are adjacent in P ( q ) if and only if theline through these points is quadratic. Lemma 1.
For any point of A (2 , q ) , there exists ( q + 1) / quadratic and ( q +1) / non-quadratic lines through this point. Proof.
Without loss of generality, we can consider the lines through the pointwith coordinates (0 , q − neighbours in P ( q ). For every square γ in F ∗ q , the corresponding point in A (2 , q ) defines aquadratic line through (0 , q − F ∗ q are squares. Thus, there are q − · q − = q +12 quadraticlines through (0 , q +12 lines are non-quadratic. (cid:3) For any γ = x + yα ∈ F ∗ q define the norm mapping N by N ( γ ) := γ q +1 = γγ q = ( x + yα )( x − yα ) = x − y d . The norm mapping is a homomorphismfrom F ∗ q to F ∗ q with Im ( N ) = F ∗ q . Thus, the kernel Ker ( N ) is the subgroup oforder q + 1 in F ∗ q . Since the kernel Ker ( N ) is defined by the quadratic equation x − y d = 1, each line of A (2 , q ) has at most 2 points (elements) of Ker ( N ).Thus, the points of Ker ( N ) form an oval by definition.Now we make some remarks on squares in finite fields. Lemma 2. (1)
The element − is a square in F ∗ q if and only if q ≡ . (2) For any non-square d in F ∗ q the element − d is a square in F ∗ q if and only if q ≡ . Proof. (1) It follows from the fact that − δ q − , where δ is a primitiveelement in F q .(2) It follows from (1) and the fact that the product of two non-squareelements in F ∗ q is a square. (cid:3) The following lemma can be used to test whether an element γ = x + yα ∈ F ∗ q is a square. Lemma 3.
An element γ = x + yα ∈ F ∗ q is a square if and only if N ( γ ) is asquare in F ∗ q . roof. Let γ = ( γ ) for some γ ∈ F ∗ q . Since the norm mapping is ahomomorphism, we have N ( γ ) = N (( γ ) ) = ( N ( γ )) .Let us prove the converse. Since | Ker ( N ) | = q + 1 and | Im ( N ) | = q − q + 1 preimages for every element from F ∗ q . We have already shownthat the image of a square from F ∗ q is a square in F ∗ q . Thus, the q − squaresfrom F ∗ q are preimages of q − square elements from F ∗ q . It implies that the q − non-squares from F ∗ q are preimages of q − non-square elements from F ∗ q . (cid:3) Lemma 4.
The element α is a square in F ∗ q if and only if q ≡ . Proof.
It follows from Lemma 2, Lemma 3 and the fact that N ( α ) = − d . (cid:3) There are two special lines in A (2 , q ) for which we can decide in general ifthey are quadratic. Lemma 5.
The following statements hold. (1)
The line { c | c ∈ F q } is quadratic. (2) The line { cα | c ∈ F q } is quadratic if and only if q ≡ . Proof. (1) It follows from F ∗ q = h β q − i and the fact that q − N ( cα ) = − c d . (cid:3) The following lemma shows what is the automorphism group of P ( q ). Lemma 6 ([5, Theorem 9.1]).
The automorphism group of P ( q ) acts arc-transitively, and the equality Aut( P ( q )) = { v av γ + b | a ∈ S, b ∈ F q , γ ∈ Gal( F q ) } holds, where S is the set of square elements in F ∗ q . Lemma 7.
The automorphism group of P ( q ) has a subgroup that stabilises thequadratic line F q and acts faithfully on the set of points that do not belong to F q . Proof.
This subgroup is given by the affine transformations x ax + b , where a ∈ F ∗ q and b ∈ F q . (cid:3) ( F q , α ) -construction of maximal cliques In this section, we recall and reformulate the geometric construction of max-imal cliques of size q + r ( q )2 in P ( q ) found in [2]. This construction depends onthe choice of a line ℓ and a point P that is not on this line. Since any line isuniquely determined by a pair of distinct points, in view of Lemma 6, we maytake any line. In view of Lemma 7, given a line, we may take any point that isnot on the line. 4ake the quadratic line F q and the point α / ∈ F q . Let c , . . . , c q − bethe points where quadratic lines through α intersect F q (note that one of thequadratic lines through α is parallel with F q ). It means that the neighbours of α in F q are c , . . . , c q − . By Lemma 3, the vertex − α has the same neighbours in F q as α . Thus, { α, c , . . . , c q − } and {− α, c , . . . , c q − } are cliques of size q +12 .Consider the line { cα | c ∈ F q } through α and − α ; by Lemma 5(2), this lineis quadratic if and only if q ≡ q + r ( q )2 . Construction 1. If q ≡ , then { α, c , . . . , c q − } is a maximal clique in P ( q ) of size q +12 ; if q ≡ , then { α, − α, c , . . . , c q − } is a maximal clique in P ( q ) of size q +32 . Let us call Construction 1 as ( F q , α )- construction . It has the property thatin both cases q ≡ q ≡ r ( q ). ( α F q , -construction of maximal cliques Take the line α F q and the point 1 / ∈ α F q .Suppose the case q ≡ q ≡ c α, . . . , c q − α bethe points where non-quadratic (resp. quadratic) lines through 1 intersect α F q (note that one of the non-quadratic (resp. quadratic) lines through α is parallelwith α F q ). By Lemma 3, the vertex − α F q as 1.Thus, { , c α, . . . , c q − α } and {− , c α, . . . , c q − α } are independent sets (resp.cliques) of size q +12 . Consider the line F q through 1 and −
1; this line is quadratic.In this setting, we formulate the following construction of maximal independentsets (resp. cliques) of size q + r ( q )2 . Construction 2. If q ≡ , then { , c α, . . . , c q − α } is a maximal indepen-dent set in P ( q ) of size q +12 ; if q ≡ , then { , − , c α, . . . , c q − α } is amaximal clique in P ( q ) of size q +32 . Let us call Construction 2 as ( α F q , construction . It has the property thatin the case q ≡ q ≡ q + r ( q )2 . Q -construction of maximal cliques In this section we recall the construction of maximal cliques proposed in [4].Let β be a primitive element of the finite field F q . Put ω := β q − . Notethat ω is a square in F ∗ q . Then Q := h ω i is the subgroup of order q + 1 andhence Q = Ker ( N ). Put Q := h ω i and Q := ω h ω i . We have Q = Q ∪ Q .5ince the kernel Ker ( N ) is defined by the quadratic equation x − y d = 1, eachline of A (2 , q ) has at most 2 points (elements) of Ker ( N ). Thus, the points of Q form an oval by definition. Construction 3 ([4, Theorem 1]). If q ≡ , then Q and Q are maximalindependent sets of size q +12 in the graph P ( q ) ; if q ≡ , then Q ∪ { } and Q ∪ { } are maximal cliques of size q +32 in the graph P ( q ) . This construction has the property that in the case q ≡ q ≡ αQ -construction of maximal cliques The following construction is a corollary of Construction 3 and Lemma 4.
Construction 4. If q ≡ , then αQ and αQ are maximal cliques of size q +12 in the graph P ( q ) ; if q ≡ , then αQ ∪ { } and αQ ∪ { } are maximalcliques of size q +32 in the graph P ( q ) . Construction 4 has the property that in both cases q ≡ q ≡
3. Correspondence between maximal cliques
Consider the mapping ϕ : F q F q defined by the rule: ϕ ( γ ) := ( γ +1 γ − if γ = 1 , γ = 1 . Proposition 1.
For any γ = x + yα ∈ Q , γ = 1 , the equality ϕ ( γ ) = yx − α holds. Proof.
Since x + yα ∈ Q , we have x − y d = 1 and ϕ ( γ ) = x + yα + 1 x + yα − x + yα + 1)( x − − yα )( x − + y d = x − (1 + yα ) x − x + 1 − y d == x − − yα − y d − x = yx − α. (cid:3) It follows from Proposition 1 that ϕ stabilises 1 and maps the oval Q to theset { } ∪ { cα | c ∈ F q } . Further, we investigate what are the images of Q .Let γ = x + yα be an arbitrary element from Q . Then γ = x + y d | {z } x + 2 xy |{z} y α represents an arbitrary element from Q . In this setting, we formulate thefollowing proposition. 6 roposition 2. For any γ = x + yα ∈ Q , γ = 1 , the equality ϕ ( γ ) = xyd α holds. Proof.
In view of Proposition 1, we have ϕ ( γ ) = ϕ ( x + y α ) = y x − α = 2 xyx + y d − α = 2 xy y d α = xyd α. (cid:3) The following theorem establishes a correspondence between Construction 3and Construction 2.
Theorem 1. If q ≡ , then ϕ ( Q ) coincides with the independent set of size q +12 given by Construction 2; if q ≡ , then ϕ ( Q ∪ { } ) coincides with themaximal clique of size q +32 given by Construction 2. Proof.
Note that ϕ swaps the elements 0 and − γ = x + yα be an arbitrary element from Q . Then γ represents anarbitrary element from Q . By Proposition 2, we have ϕ ( γ ) = xyd α . Let uscheck when the element xyd α is adjacent to 1. By Lemma 3, we have N (1 − xyd α ) = 1 − x dy d = 1 − x y d = y d − x y d = − y d . Let us consider two cases. If q ≡ − F ∗ q and − y d is a non-square in F ∗ q , which means that 1 is not adjacent with xyd α in P ( q ).If q ≡ − F ∗ q and − y d is a square in F ∗ q , whichmeans that 1 is adjacent with xyd α in P ( q ). (cid:3) Consider the mapping ψ : F q F q defined by the rule: ψ ( γ ) := ( αγ + dγ − α if γ = α,α if γ = α. Now we are going to establish a correspondence between Construction 4 andConstruction 1.Let γ = x + yα ∈ Q be an arbitrary element. Then αγ = yd + xα representsan arbitrary element from αQ . Proposition 3.
For any γ = x + yα ∈ Q, γ = 1 , the equality ψ ( αγ ) = ydx − holds. Proof.
In view of Proposition 1, we have ψ ( αγ ) = α γ + dαγ − α = α γ + 1 γ − αϕ ( γ ) = ydx − . (cid:3) orollary 1. For any γ = x + yα ∈ Q, γ = 1 , the equality ψ ( αγ ) = αϕ ( γ ) holds. It follows from Proposition 3 that ψ stabilises α and maps the oval αQ tothe set { α } ∪ F q . Further, we investigate what are the images of αQ .Let γ = x + yα be an arbitrary element from Q . Then αγ representsan arbitrary element from αQ . In this setting, we formulate the followingproposition. Proposition 4.
For any γ = x + yα ∈ Q , γ = 1 , the equality ψ ( αγ ) = xy holds. Proof.
By Corollary 1 and Proposition 2, we have ψ ( αγ ) = αϕ ( γ ) = xy . (cid:3) The following theorem establishes a correspondence between Construction 4and Construction 1.
Theorem 2. If q ≡ , then ψ ( αQ ) coincides with the maximal clique ofsize q +12 given by Construction 1; if q ≡ , then ψ ( αQ ∪ { } ) coincides withthe maximal clique of size q +32 given by Construction 1. Proof.
Note that ψ swaps the elements 0 and − α .Let γ = x + yα be an arbitrary element from Q . Then αγ represents anarbitrary element from αQ . By Proposition 4, we have ψ ( αγ ) = xy . Let uscheck when the element xy is adjacent to α . N ( xy − α ) = x y − d = x − y dy = 1 y , which means that α is adjacent with xy in P ( q ). (cid:3)
4. Concluding remarks
A finite field of square order can be viewed as an analogue of the complexfield. In this paper, we establish a correspondence between maximal cliques inPaley graphs P ( q ), which is given by finite analogues of the M¨obius transfor-mations of the complex plane. Further, it would be interesting to develop thisanalogy. Acknowledgments
Sergey Goryainov and Leonid Shalaginov are supported by RFBR accordingto the research project 20-51-53023. 8 eferences [1] A. Blokhuis,
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