We give an explicit classification of the arcs in PG (2, q) (q even) with a large conical suset and excess 2, i.e., that consist of q/2 + 1 points of a conic and two points not on that conic. Apart from the initial setup,the methods used are similar to those for the case of odd q, published earlier

Coolsaet, Kris

Sticker, Heide

English

Ghent University Academic Bibliography

Arcs with large conical subsetsin Desarguesian planes of even orderKris Coolsaet Heide StickerDepartment of Applied Mathematics, Computer Science and Statistics,Ghent University,Krijgslaan 281–S9, B–9000 Gent, Belgium{Kris.Coolsaet,Heide.Sticker}@UGent.beSubmitted: Nov 20, 2012; Accepted: Jan 18, 2014; Published: Jan 24, 2014Mathematics Subject Classification: 51E21AbstractWe give an explicit classification of the arcs in PG(2, q) (q even) with a largeconical subset and excess 2, i.e., that consist of q/2 + 1 points of a conic and twopoints not on that conic. Apart from the initial setup, the methods used are similarto those for the case of odd q, published earlier (Electronic Journal of Combinatorics,17, #R112).1 IntroductionConsider the Desarguesian projective plane PG(2, q) over the finite field of order q. Anarc is a set of points of PG(2, q) with the property that no three points of this set arecollinear. An arc is called complete if it is not contained in a bigger arc.When q is odd it is well known that an arc can be of size at most q + 1 and in thatcase always coincides with the set of points of some conic. When q is even, the maximumsize of an arc is q + 2 and again examples are known. (For instance, one can always addthe nucleus to the conic.) For further information on the geometrical and combinatorialproperties of arcs we refer to [2, Chapters 9–10].Removing some points from a conic again yields an arc, but this arc is obviously notcomplete. However, after removing a sufficient number of points it may be possible toextend the set thus obtained to a nonconical arc by adding a point that does not belongto the original conic. This new arc might not be complete, but can be made complete byadding yet more points.We shall be interested in arcs for which the conical part is as large as possible andthat contain (at least) two points not on that conic. A simple result by B. Segre [4] andthe electronic journal of combinatorics 21(1) (2014), #P1.14 1L. Lombado-Radice [3] shows that the size of the conical part can be at most (q + 3)/2when q is odd, and (q + 2)/2 when q is even.In [1] we classified all arcs of this type with two points not on the conic, for odd qonly. Our method made use of an initial step that needs an element of the field GF(q)that is not a square. We could therefore not immediately extend this technique to thecase where q is even.In this paper we show that this problem can be overcome by starting with a differentsetup, described in Section 2. The end result for q even turns out to be much simplerthan that for q odd. For instance, instead of three different types of arcs for the odd case,there is only one case to consider when q is even (cf. Section 3). This is a consequence ofthe fact that there is no distinction between internal and external points of a conic in theeven case.2 Notation and preliminary definitions(Where possible we have chosen notations in such a way that they conform to the notationsfor similar concepts in [1].)Let S denote an arc of PG(2, q). We define a conical subset of S to be any subset R ofS of the form R = S∩C where C is a conic. (Most of the time the conic C and the conicalsubset R will be clear from context. We will therefore usually leave out the reference toC when talking about tangents and secants of C and lines external to C.) The points ofS that do not belong to R will be called supplementary points. The number e = |S \ R|will be called the excess of the arc. We shall always assume that the arc S has at leastone supplementary point, i.e., that S is not fully contained in a conic.When q is even, all tangents to C go through a common point N (not on C) which iscalled the nucleus of C. Consider a point S of the plane. If Q does not belong to C andQ 6= N , then there is exactly one line through Q which is tangent to C, and there are q/2lines that intersect C in two points. Each of these may intersect R in at most one point,and hence |R| 6 q/2 + 1. A conical subset that attains this bound will be called large.Arcs with large conical subsets are the subject of this paper.Henceforth let K denote the finite field of order q, with q even.Without loss of generality we may fix C to be the conic with equation XZ = Y 2.Mapping t ∈ K to the point with (homogeneous) coordinates (1 : t : t2) and ∞ to thepoint with coordinates (0 : 0 : 1) defines a one–to–one relation between K ∪ {∞} and C.The nucleus N of C has coordinates (0 : 1 : 0). The tangent through the point (1 : t : t2)has equation Z = t2X. The tangent through the point (0 : 0 : 1) has equation X = 0.The subgroup of PGL(3, q) that stabilizes C is isomorphic to PGL(2, q). The elementof PGL(2, q) associated to the matrix(a bc d)maps the point (1 : t) of the projectiveline PG(1, q) onto the point (1 :b+ dta+ ct). This yields a corresponding action on C thatcan be extended to an action on any point (x : y : z) of the whole plane PG(2, q), whichthe electronic journal of combinatorics 21(1) (2014), #P1.14 2can be written as follows:(a bc d)↔ (x y z) 7→ (x y z) a2 ab b22ac ad+ bc 2bdc2 cd d2 (1)where 2 is obviously zero in even characteristic.Let L denote the quadratic extension field of K and let N(·) and Tr(·) be the normand trace function for this extension. Consider φ ∈ L such that φ has multiplicative orderq + 1 in L. Then N(φ) = φq+1 = 1 and Tr(φ) = φ + φq = T , with T ∈ K \ {0}. Thecharacteristic equation of φ is X2 + TX + 1 = 0. In particular φ2 + Tφ + 1 = 0. Notethat 1, φ, φ2, . . . runs through all elements of L with norm equal to 1. Since q is even, theonly element of K with norm equal to 1 is 1 itself.For any integer i definefidef=1TTr(φi) =1T(φi + φiq) =1T(φi + φ−i).We have the following properties:f0 = 0, f1 = 1, f2 = T, fi+2 = Tfi+1 + fi, (2)fi+q+1 = fi, f−i = fi, fi+1 + fi−1 = Tfi, fi+j + fi−j = Tfifj. (3)(Note that the index i of fi can be treated as an element of Zq+1.)From (3) we may easily computefi+j = fi−1fj + fifj+1 = fifj−1 + fi+1fj, (4)and hencef2i = (fi−1 + fi+1)fi = Tf 2if2i−1 = f 2i + f2i−1 = (fi + fi−1)2,f2i+1 = f2i + f2i+1 = (fi + fi+1)2.(5)Consider the following element of PGL(2, q) :M ′idef=(fi−1 fifi fi+1). (6)It follows from (2–4) thatM ′0 =(1 00 1), M ′i+j = M′iM′j, M′i = M′i1 , detM′i = 1.The equivalent action of PGL(3, q) has the following form : f 2i−1 fi−1fi f 2i0 1 0f 2i fi+1fi f2i+1 .the electronic journal of combinatorics 21(1) (2014), #P1.14 3Note that (1 : 0 : 1) is mapped to (f2i−1 : f2i : f2i+1). Moreover, all of these points lie onthe line ` with equation x+ Ty + z = 0, which is an external line to the conic C.We shall number the points of ` as Q0, Q1, . . . , Qq such that Qi has coordinates (fi−1 :fi : fi+1). Note that all these points are different and hence ` = {Q0, Q1, . . . , Qq}. Theindex i of Qi will be called the orbital index of Qi. Orbital indices can be treated aselements of Zq+1.The tangent through Q0(1 : 0 : 1) is the line with equation X + Z = 0 (i.e., X = Z)which intersects C in P0(1 : 1 : 1). Define Pi to be the image of P0 under M′i . Then Pihas coordinates(f2i−1 : f2i + 1 : f2i+1) ≈ (1 : ti : t2i ),by (5), withtidef=fi + fi+1fi−1 + fi.(The ‘≈’-sign denotes equality up to a scalar factor.)We havetq/2 = 0, tq/2+1 =∞, t0 = 1, t−i = 1/ti.Again, the index i of Pi will be called its orbital index, and again it can be treated as anelement of Zq+1.Lemma 1. Let i, j, k ∈ Zq+1. Then• Pi, Pj, Qk are collinear if and only if k ≡ i+ j (mod q + 1).• PiQk is a tangent to C if and only if k ≡ 2i (mod q + 1).Proof. Indeed∣∣∣∣∣∣f2i−1 f2i + 1 f2i+1f2j−1 f2j + 1 f2j+1fi+j−1 fi+j fi+j+1∣∣∣∣∣∣ =∣∣∣∣∣∣f2i−1 f2i + 1 Tf2if2j−1 f2j + 1 Tf2jfi+j−1 fi+j Tfi+j∣∣∣∣∣∣ = T∣∣∣∣∣∣f2i−1 1 f2if2j−1 1 f2jfi+j−1 0 fi+j∣∣∣∣∣∣= T∣∣∣∣ f2i−1 f2ifi+j−1 fi+j∣∣∣∣+ T ∣∣∣∣ f2j−1 f2jfi+j−1 fi+j∣∣∣∣and this can be expanded tof3i+j−1 + fi−j+1 + f3i+j−1 + fi−j−1 + f3j+i−1 + fj−i+1 + f3j+i−1 + fj−i+1 = 0.This proves that Pi, Pj, Qk are collinear when k ≡ i+j (mod q+1). Because the line PiPjintersects the line ` in exactly one point, and the orbital index of that point is uniquelydetermined (mod q + 1), Pi, Pj, Qk cannot be collinear for other values of k.The tangent through Pi has equation Z = t2iX which is clearly satisfied by the pointQ2i. Again, the intersection of this tangent with ` is a single point, and hence no otherpoint Qk can satisfy this property.the electronic journal of combinatorics 21(1) (2014), #P1.14 4As in [1], apart from the ‘rotations’M ′i defined in (6) we also introduce the ‘symmetries’Mi, as follows:Midef=(fi fi+1fi−1 fi),or represented as elements of PGL(3, q) : f 2i fi+1fi f 2i+10 1 0f 2i−1 fi−1fi f2i .Note that M0 interchanges X- and Z-coordinates, that Mi = M0M′i and M′iM0 = M−i.Lemma 2. Let H denote the subgroup of PGL(3, q) that leaves both the conic C and itsexternal line ` invariant (and hence is a subgroup of PGL(2, q)). Then H consists of theelements Mi,M′i , i ∈ Zq+1 and is a dihedral group of order 2(q + 1).The action of H on C and ` is given byMi : Pj 7→ Pi−j, Qj 7→ Q2i−j,M ′i : Pj 7→ Pj+i, Qj 7→ Qj+2i. (7)Proof. We leave it to the reader to prove that the action of H on the points of C and `is indeed as given in (7).Consider the action of a general element of PGL(2, q), as presented in (1), on the line` (represented by a column vector) : a2 ab b20 ad+ bc 0c2 cd d2−1 1T1 = d2 bd b20 ad+ bc 0c2 ac a2 1T1 = b2 + Tbd+ d2T (ad+ bc)a2 + Tac+ c2 (8)It follows that ` is stabilized by this action if and only ifa2 + Tac+ c2 = b2 + Tbd+ d2 = ad+ bc. (9)Note thatf 2i + Tfi+1fi + f2i+1 = f2i + Tfi−1fi + f2i−1 = f2i + fi−1fi+1 = 1,by (2–5). Hence Mi and M′i satisfy this condition. It is easily seen that the set of allelements Mi,M′i with i ∈ Zq+1 forms a group isomorphic to the dihedral group of order2(q + 1). It is therefore sufficient to show that no other elements leave both C and `invariant, or equivalently, that the stabilizer of H of the point Q0 ∈ ` has size at most 2.The fact that Q0(1 : 0 : 1) is stabilized is expressed as follows:a2 + c2 = b2 + d2, ab+ cd = 0. (10)the electronic journal of combinatorics 21(1) (2014), #P1.14 5This yields a+ b+ c+ d = 0 and ab = cd, and therefore0 = a(a+ b+ c+ d) = a2 + ab+ ac+ ad = a2 + cd+ ac+ ad = (a+ c)(a+ d),0 = b(a+ b+ c+ d) = ab+ b2 + bc+ bd = cd+ b2 + bc+ bd = (b+ c)(b+ d),0 = c(a+ b+ c+ d) = ac+ bc+ c2 + cd = ac+ bc+ c2 + ab = (a+ c)(b+ c),0 = d(a+ b+ c+ d) = ad+ bd+ cd+ d2 = ad+ bd+ ab+ d2 = (a+ d)(b+ d),and hence a = c or a = d, b = c or b = d, a = c or b = c and a = d or b = d. Andtherefore a = c and b = d, or a = d and b = c.Together with (9), the first case leads toa2 + Ta2 + a2 = b2 + Tb2 + b2 = ab+ ba,and then Ta2 = Tb2 = 0, implying a = b = c = d = 0, which is not allowed. The secondcase impliesa2 + Tab+ b2 = b2 + Tab+ a2 = a2 + b2,and then ab = 0. When a = 0 this yields transformation M0, when b = 0 this yieldsM ′0.3 Classification up to projective equivalenceAfter these preliminaries we can continue with the classification up to projective equiva-lence of arcs with large conical subsets much in the same way as when q is odd. Becauseof the similarity with [1] we shall try to be brief.First q must be large enough to ensure that the conic C is completely determined bythe arc S. Let S be an arc with conical subset R and excess |S \R| = e. If C ′ is anotherconic giving rise to the conical subset R′ = S ∩C ′ with |S \R′| = e′. Then by [1, Lemma3], we know that e′ > |R| − 4. As an immediate consequence we obtain the followingresult.Lemma 3. Let PG(2, q) be of even order. If q > 16 then an arc of PG(2, q) with a largeconical subset of excess e < 5 can contain at most one such conical subset.Proof. If S has a large conical subset of excess e, then by the above e′ > q/2 − 3 andhence e′ > 5 when q > 16. Hence, any other conical subset will have an excess that is toolarge.In this section we shall consider an arc S with a large conical subset of excess 2. Firstnote that the nucleus of the corresponding conic cannot be one of the supplementarypoints: otherwise the line joining the supplementary points would be a tangent to C andthen 3 points of that tangent would belong to S.Hence we may assume that the line joining the supplementary points is external toC, and without loss of generality we may assume that this line is the line ` introduced inthe previous section. In fact, we may even assume that one of the supplementary pointsis Q0, and then the other is of the form Q2a for some a ∈ Zq+1.Consider the following graph Γ:the electronic journal of combinatorics 21(1) (2014), #P1.14 6• Vertices are the elements of Zq+1,• Two different vertices i, j are adjacent if and only if the line PiPj contains eitherQ0 or Q2a, i.e., if and only if i+ j ≡ 0 or 2a (mod q + 1). (Cf. Lemma 1.)Note that the degree of a vertex of Γ is 2 in most cases, except for the vertices 0 anda which are of degree 1, because they lie on a tangent through Q0 or Q2a. It followsthat Γ is the disjoint union of one path and a number (possibily zero) of cycles. We mayenumerate the vertices of the cycle or path that contains i as follows:. . . , i,−i, 2a+ i,−2a− i, 4a+ i,−4a− i, . . . (11)For a cycle, this sequence eventually starts to repeat. For the path this sequence stops at0 and a.Define n to be the order of 2a (mod q + 1) and let d = gcd(a, q + 1) = (q + 1)/n. (nand d will always be odd.) Note that each of the vertices in (11) is equal to ±i (mod d).Also note that 0 and a are both divisible by d. Hence, if i 6≡ 0 (mod d), then (11) denotesa cycle, and not a path. The cycle either has length 2n with i ≡ 2na+ i (mod q + 1), orlength 2n+ 1 with i ≡ −2na− i (mod q + 1). The latter case is impossible as the graphonly has two types of edges and two consecutive edges never are of the same type, i.e.the number of edges in the cycle must be even. Hence the cycle has length 2n and mustcontain all vertices that are equal to ±i (mod d). The path has endpoints 0 and a andcontains all points that are multiple of 2a. This implies that the path has size n.This proves the following lemma.Lemma 4. Γ is the disjoint union of (d − 1)/2 cycles of length 2n and one path of nvertices, where n is the order of 2a (mod q+ 1) and d = (q+ 1)/n, i.e. d = gcd(a, q+ 1).Define the set Zkdef= k+ 2aZq+1 = k+ dZq+1. We call Zk a half cycle of Γ. The cyclesof Γ are precisely the sets Zk ∪ Z−k, with k in the range 1, . . . , (d− 1)/2.Denote by N(R) the set of orbital indices of vertices of the large conical subset R.Since S is an arc, N(R) must be an independent set of Γ. To have an independent set ofsize q/2 + 1 we need to take the largest possible independent set for each component of Γ.When n is odd, as is the case here, the largest independent set in a path with n verticesis unique and has size (n+ 1)/2. The largest independent set in a cycle is one of its twohalf cycles and has length n. This proves the following result.Theorem 5. Let a ∈ {1, . . . , q}. Let d = gcd(a, q + 1), n = (q + 1)/d. Let S =R ∪ {Q0, Q2a}, with R ⊂ C and |R| = q/2 + 1. Then S is an arc of PG(2, q) if and onlyif N(R) can be written as the union of pairwise disjoint sets, of the formN(R) = Π ∪ Z±1 ∪ . . . ∪ Z±(d−1)/2,with independent choices of sign, andΠ = {0,−2a,−4a, · · · , a}.the electronic journal of combinatorics 21(1) (2014), #P1.14 7Every arc listed in Theorem 5 can be denoted by its signature (a; 1, . . . , (d−1)/2),where k = ±1 depending on the choice made for the half cycle Z±k.Theorem 6. Let q be even, a ∈ {1, . . . , q}, d = gcd(a, q + 1) and n = (q + 1)/d. LetHa denote the subgroup of PGL(3, q) that leaves the conic C invariant and fixes the pair{Q0, Q2a}. Then Ha = {M ′0,Ma} and acts as follows on the components of Γ and the arcwith signature (a; 1, . . . , (d−1)/2)Element of Ha Image of Zk Image of(a; 1, . . . ,  d−12)M ′0 (the identity) Zk (a; 1, . . . ,  d−12)Ma Z−k (a;−1, . . . ,− d−12)Proof. Note that Ha must fix the line joining Q0 and Q2a, i.e., `, and hence is a subgroupof H. From (7) it follows that the identity is the only transformation that fixes both Q0and Q2a and that Ma is the only element of H that interchanges Q0 and Q2a.By (7), Ma maps Zk onto Za−k. Half cycle indices are determined modulo d, and a isa multiple of d. Hence Za−k = Z−kCorollary 7. Let q > 16, q even. Let HS denote the subgroup of PG(3, q) that leavesinvariant the arc S with signature (a; 1, . . . , d−1). If d = 1, then HS = Ha. Otherwise,HS is trivial.Lemma 8. Let Lq(a) denote the number of projectively inequivalent arcs S with a signa-ture of the form (a; 1, . . . , (d−1)/2), with d = gcd(a, q + 1). ThenLq(a) ={1, when d = 1,2d−32 , when d > 1.(12)Proof. If d = 1, then there is clearly the one signature (a). Otherwise, when d > 1, thenumber of signatures is equal to 2(d−1)/2 and the value of |SHa | = |Ha|/|HS| is equal to 2for each such arc. Hence, Lq(a) =2(d−1)/22for d > 1.Theorem 9. Let q > 16, q even. The number Lq of projectively inequivalent arcs S inPG(2, q) of size |S| = q/2 + 3, with a conical subset R = S ∩ C of size |R| = q/2 + 1, isgiven byLq =12∑d′φ(q + 1d)Lq(d) (13)where the sum is restricted to all proper divisors d of q + 1, φ denotes Eulers totientfunction, and Lq(d) is as given in Lemma 8.Proof. Note that M0 interchanges Q2a and Q−2a and fixes Q0, and hence to enumerate allarcs up to isomorphism, we should consider only one of a and q + 1− a. In other words,we may take a in the range 1, . . . , q/2.The total number of inequivalent arcs is then given by∑q/2a=1 Lq(a). Note that Lq(a)does not directly depend on a, but only on d = gcd(a, q + 1). The number of integers a,1 6 a 6 q/2 such that d = gcd(a, q + 1) is equal to 12φ((q + 1)/d).the electronic journal of combinatorics 21(1) (2014), #P1.14 8The table below lists the values of Lq for small values of q.q 16 32 64 128 256Lq 8 31 100 1048639 128For larger values of q the value of Lq becomes very large, unless q+1 is a (Fermat) prime,in which case Lq = q/2.For 32 6 q 6 256 we have checked by computer that all of these arcs are complete.For q = 16 four of the eight arcs are complete and there exist three inequivalent arcs ofexcess 3, all of them complete.AcknowledgementsWe would like to thank the anonymous referees for helpful comments and suggestions andfor correcting an error in the original proof of Lemma 2.References[1] K. Coolsaet and H. Sticker. Arcs with Large Conical Subsets. Electron. J. Combin.17 (2010), #R112[2] J. W. P. Hirschfeld. Projective geometries over finite fields (second edition) (1988).Clarendon Press, Oxford.[3] L. Lombardo-Radice. Sul problema dei k-archi completi in S2,q. Boll. Un. Mat. Ital.(3)11 (1956), pp. 178–181.[4] B. Segre. Introduction to Galois geometries. Atti Accad. Naz. Lincei Mem. Cl. Sci.Fis. Mat. Natur. Sez. I (8)8 (1967), pp. 133–236.the electronic journal of combinatorics 21(1) (2014), #P1.14 9

Arcs with large conical subsets in Desarguesian planes of even order

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Arcs with large conical subsets in Desarguesian planes of even order

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