IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VOL.23 (1) 2010 The Maximum Complete (k,n)-Arcs in the Projective Plane PG(2,4) By Geometric Method S. J. Kadhum Departme nt of Mathematics, College of Education, Ibn Al-Haitham, Unive rsity of Baghdad Abstract A (k,n)-arc A in a finite projective plane PG(2,q) over Galois fi eld GF(q), q=p ⁿ for same p rime number p and some inte ger n≥2, is a set of k p oints, no n+1 of which are collin ear. A (k,n)-arc is complete if it is not contained in a(k+1,n)-arc. In this p ap er, the maximum complete (k,n)-arcs, n=2,3 in PG(2,4) can be constructed from the equation of the conic. 1. Introduction [1], found the complete (k,2)-arcs in the p rojective planes over Galois f ield GF(p ⁿ) for some p rime number p and some integer n. [2] found an algebraic method for construction of (k,4)-arcs in the projective plane PG(2,4). In this p ap er, the maximu m comp lete (k,n)-arcs in PG(2,4), n=2,3 are obtained fro m the equation of the conic by geometric method. A p rojective plane PG(2,q) over GF(q) consists of 1+q+q 2 p oints and 1+q+q 2 of lines, every line contains 1+q p oints end every p oint is on 1+q lines. Any p oint of t he p lane has t he form of triple (x0, x1, x2), wher e x0, x1, x2 are elements of GF(q) with the exception of a triple consisting of three zero elements, two triples rep resent the same p oint if there exist s λ in GF(q)\ {0}, s.t . ( y 0,y 1,y 2 ) = λ( x0,x1, x2 ) . Similar ly, any line of the p lane has the from of a trip le [ x0,x1, x2 ], x0, x1, x2 are in GF(q) with the excep tion of a triple consisting three zero elements. Two lines [x0,x1, x2] and [y 0,y 1,y 2] rep resent the same line if there exists λ in GP(q)\{0} s.t . [x0,x1, x2] = λ[y 0,y 1,y 2] . The point (x0,x1, x2) is incident with line [y 0,y 1,y 2] if x0y 0+x1y 1+x2y 2 = 0. IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VOL.23 (1) 2010 2. Basi c Defini tions and Theorems 2.1 De fini tion [3] A (k,n)-arc in a finite p rojective p lane as a set of k p oints no n+1 of which are collinear. 2.2 De fini tion [4] A (k,n)-arc is a set of k p oint, no t hree of them are collinear, we denote this by k-arc. 2.3 De fini tion [5] A (k,n)-arc is said to be complete if it is not contained in a (k+1,n)-arc. We denote by m(2,p) the maximum number of p oints in PG(2,p ) that a (k,n)-arc can have. 2.4 The orem [6] A (k,n)-arc in PG(2,p ) is comp lete if and only if C0 = 0. Proof: (  ) Let a (k,n)-arc K be a comp lete arc in PG(2,p ) sup p ose that C0 ≠ 0, t hen there is at lest one point say N has index zero and N  K. T hen K  {N} is an arc in PG(2,p ). Hence K  K  {N}, which imp lies (k,n)-arc K is incomp lete (contradiction). (  ) Sup p ose that C0=0 for t he (k,n)-arc K, then there are no points of index zero, then the (k,n)- arc K is complete. 2.5 De fini tion [4] A k-arc is called an oval when k = (2,p ). 2.6 De fini tion [7] Let ℓ be any line in PG(2,p ) if ℓ intersects a k-arc in i-p oints, | ℓ ∩ K | = i, then ℓ is called an i-secant of k, Then 2-secont of K is called a bisecant of K Then 1-secont of K is called a unisecant of K Then 0-secont of K is called an external of K. 2.7 De fini tion [3] Let N be a point in PG(2,p ) and N is not on a k-arc, then we say N is a p oint of index i if there are exactly i-bisecant through N . 2.8 De fini tion [5] The set Ci consists of all p oints of index i Ci = | Ci | = # the number of points in Ci and # the number of index i. 2.9 Theorem [5] Let M be a point of the k-arc of PG(2,k) and t(M ) be the number of the unisecants of K t hrough M , then t(M ) = p +2-k = t . IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VOL.23 (1) 2010 Proof: The number of lines in PG(2,p ) through any p oint is p +1 there are exactly p +1 lines through M the point M with any other point of t he k-arc determine a bisecant of K since there are (k-1) p oints of K other then M , then there are exactly (k-1) lines determined from M and the other p oint of K which are the bisecants of K through M since any line through M is either a bisecant or a unisecant, then the number of unisecants of K through M = p +1-(k-1) = p +2-k = t . 2.10 Notati on: Ti = the number of i-secant of a k-arc, T2 = the number of bisecant lines of a k-arc. T1 = the number of unsecant lines of a k-arc. T0 = the number of external lines of a k-arc. 2.11 De fini tion [4] Let K be a k-arc which is an oval an external point t o an oval, is a point of intersection two unisecants of K. 2.12 The orem [5] m (2, p) =      ev en pfor 2p odd pfor 1p 2.13 The orem [5] The number of external p oints of an oval k-arc in PG(2,p ) is 2 )1( pp . Proof: If K is an 0val, then m (2, p) =      ev en pfor 2p odd pfor 1p P is odd  k = p +1 t = p +2-k = p +2-(p +1) = 1 T1 = kt = (p +1)*1 = p +1 = t he number of unisecants of k in PG(2,p ) , Since each two unisecants intersect in an external p oint, then the number of external p oints =        2 1p = !2)!21( )!1(   p p = 1*2)!1( )!1()1(   p ppp = 2 )1( pp . 2.14 De fini tion [3] An internal point to an oval if it is not any unisecant of t he oval which is not on the oval. 2.15 The orem [5] The number of the internal p oints to on oval is 2 )1( pp . Proof: The number of the external p oints + the number of the internal p oints = p 2+p +1- k. # of external p oints = 2 )1( pp IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VOL.23 (1) 2010 # of the internal points = p 2 + p + 1- (p +1) - 2 )1( pp = 2 22222 22 ppppp  = 2 2 pp  = 2 )1( pp . 2.16 The orem [7] In PG(2,p ) with p  4, there is a unique conic through a 5-arc . 2.17 The orem [4,5] Every conic in PG(2,p ) is a(p +1)-arc The converse of theorem is also satisfied. 2.18 The orem [3] In PG(2,p ), with p odd, every oval has a conic. 2.19 De fini tion [8] A complete quadrangle is a set of four p oints A, B, C and D in which no three of them are collinear, the p oints A, B, C and D are called the vertices of the quadrangle, the lines joining any two vertices are called the sides which are AB, AC, BD, BC, AD, CD. T wo sides are said to be op p osite if they have no vertex in common. The point of intersection of any two op p osite sides is called a diagonal p oint t he diagonal p oints; D1 = AB  CD, D2 = AC  BD and D3 = AD  BC. 3. Maximum Comple te (k,n)-Arcs i n PG(2,4) 3.1 The Additions and Mul tiplications O perations of GF(4) [9] To find addition and multiplication tables in GF(4), we have the order p airs (x1,x2) such that x1,x2 in GF(2), as follows: 0  (0,0), 1  (1,0), 2  (0,1), 3  (1,1). Put these points in one IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VOL.23 (1) 2010 orbit, (1,0) at the first p oint and by the p rinciple of (1,0)A i , i=0,1,2,3 and A=       11 01 , (1,0)A=(0,1) and (1,0)A 2 = (1,1) , So (1,0)       11 01 )1,1( )1,0( . Now, in the left of the following table, m is the operation of multiplication and in the right n is the operation of addition, in multiplication side we write the numeration of p oint as last , and the addition side takes t he normal sequence. m(*) (+)n = f(m) 1 (1,0) 0 2 (0,1) 1 3 (1,1) 2 Mod 3 In addition table, we have the following relation: (x1, x2) + (y 1,y 2) = (z1,z2) where zi = ( y i+ xi ) mod(2) for i =1,2 In multiplication table we have the followin g relation: ( (1,0) A f(m 1 ) ) A f(m 2 )  m1*m2 = m3 =(1,0) A (f(m 1 )+f(m 2 )) (m od 3) = (x1, x2) For example : 2*3 = 1  ( (1,0)A 1 )A 2 = (1,0)A 3 = (1,0)A 0 = (1,0), where (1,0) is equal to 1 in multiplication tables. The additions and multiplications op erations of GF(4) are in table ( 1 ). 3.2 The Projective Plane PG(2,4) The projective plane PG(2,4) contains 21 p oints, 21 lines, 5 p oints on every line and 5 lines through every p oint. Let Pi and Li ,i=1,2,---,21 b e p oints and the lin es of PG(2,4) resp ectively, the points and lines of PG(2,4) are in table ( 2 ). 3.3 The Construction of k-arc in PG(2,4) Let A={1,2,6,11}, be the reference and unit p oints of PG(2,4) where 1=(1,0,0), 2=(0,1,0), 6=(0,0,1), 11=(1,1,1). A is (4,2)-arc since it contains four p oints no three of them are collinear. There are six lines from the joining of these points which are ℓ1 = [1,2]={1,2,3,4,5} ℓ2 = [1,6]={1,6,7,8,9} ℓ3 = [1,11]={1,10,11,12,13} ℓ4 = [2,6]={2,6,10,14,18} ℓ5 = [2,11]={2,7,11,15,19} ℓ6 = [6,11]={3,6,11,16,21} The diagonal p oints of A are the p oints {3,7,10} where ℓ1  ℓ6 = 3, ℓ2  ℓ5 = 7, ℓ4  ℓ3 = 10. The points of PG(2,4) are classified with resp ect to the lines through the reference and unit p oints as follows: IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VOL.23 (1) 2010 1. The number of points on these lines is 19. 2. There exists t wo p oints of index zero for A, which are the p oint 17 and the point 20 not on any of these lines, t hen the 4-arc A is incomp lete. 3.4 The Conic in PG(2,4) Through the Reference and Uni t Points . The general equation of the conic is: 2 2 2 1 1 2 2 3 3 4 1 2 5 1 3 6 2 3 a x a x a x a x x a x x a x x 0      …..……….…(1) By substitut ing the points of the arc A in (1), we get: 1 = (1,0,0)  a1 = 0, 2 = (0,1,0)  a2 =0, 6 = (0,0,1)  a3 = 0, 11 = (1,1,1)  a4 + a5 + a6 = 0. So (1) becomes: a4 x1 x2 + a5 x1 x3 + a6 x2 x3 = 0 ………… …………………………...(2) If a4 = 0, then a5 x1 x3 + a6 x2 x3 = 0, and hence (a5 x1 + a6 x2) x3= 0 Then the conic is degenerated, therefore for a4  0, similarly a5  0 and a6  0. Dividing equation (2) by a4, we get: x1 x2 +  x1 x3 +  x2 x3 = 0 ………………………………………..…(3) where 5 6 1 2 1 3 2 3 4 4 a a x x x x x x 0 a a    where 5 6 4 4 a a , a a    , then  = – (1 + ) since 1 +  +  = 0 (mod.4) so x1 x2 +  x1 x3 – (1 + ) x2 x3 = 0 ………………………………..…(4) where   0 and   1 for if  = 0 or  = 1, we get a degenerated conics, i.e.  = 2, 3. 3.5 The Equati on of the Conics of PG(2,4) and the Comple te arcs For any value of  there is a unique conic contains the reference and the unit p oints. 1. If  = 2, then the equation of the conic C1 is x1 x2 + 2 x1 x3 + x2 x3 = 0, the points of C1 are {1,2,6,11,12,21}, which is not a comp lete (k,3)-arc, since there exist the p oints {4,5,7,8,9,14,15,17,18,19,20} which are the points of index zero for C1. Now, we add to C1 three p oints of index zero which are {4,7,8}. Then 1C = {1,2,6,11,12,21,4,7,8} is a comp lete (9,3)-arc, since C0 = 0 and 1C is maximum arc. 2. If  = 3, then the equation of the conic C2 is x1 x2 + 3 x1 x3 + 2 x2 x3 = 0, the points of C2 are {1,2,6,11,17}, which is not a comp lete (k,2)-arc, since there exist one p oint {20} which is t he point of index zero for C2. Now, we add to C2 one p oint of index zero {20}. Then 2C = {1,2,6,11,17,20} is a comp lete (6,2)-arc, since C0 = 0. Conclusion: 1. Each of C1 and C2 is not comp lete (k,2)-arc. 2. We add the p oints of index zero for each of them for comp leteness. 3. The p oints of index zero of PG(2,4) with resp ect to 4-arc A are in the same line = {17,20}. IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VOL.23 (1) 2010 3.6 The Construction of Complete and Maximum (k,3)-arc in PG(2,4) We will try to get a comp lete (k,3)-arc by taking the comp lete (k,2)-arc, say 2 C and denoted by B, we notice that B={1,2,6,11,17,20,5} is incomplete (k,3)-arc, since there exists the points {7,9,10,12,14,16,19,21} which are the points of index zero for B. Now, we add two p oints of index zero which are {9,10}. Then B={1,2,6,11,17,20,5,9,10} is a comp lete (9,3)-arc, since C0 = 0 and B is a maximum arc. 3.7 The Construction of Complete (k,4)-arc We try to get a comp lete (k,4)-arc by taking the union of two maximum comp lete (k,3)-arcs, say 1 C and B denoted by D, we notice that D = {1,2,6,11,12,21,4,7,18,17,20,5,9,10} is incomp lete (k,4)-arc, since there exists t wo p oints {15,16} which are p oints of index zero for D. Now, we add the two p oints of index zero. Then D = {1,2,6,11,12,21,4,7,18,17,20,5,9,10,15,16} is a complete (16,4)-arc since C0 = 0, and D is a maximum arc. Conclusion: 1. There exists one comp lete and maximum (16,4)-arc. 2. The points of index zero of (k,4)-arc with resp ect are in the same line. Re frences : 1. Salih, R. A., (1999), "Complete Arcs in Projective Plane Over Galois Field", M .Sc. Thesis, University of Baghdad, Iraq. 2. Hassan, A. S., (2001), "Const ruction of (k,3) – arcs in Projective Plane Over Galois Field GF(q), q = p h when p = 2 and h = 2, 3 and 4, M .Sc. Thesis, University of Baghdad, Iraq. 3. Hirschfeld, T. W. and Sadeh, A. R., (1984), “The Projective Plane Over Field of Eleven Elements“ Giessan . 4. Rutter, J. W., (2000), "Geometry of Curves", Chapman and Hall / CRC. 5. Hirschfeld, T. W., (1979), “Projective Geometrices Over Finite Field", Oxford Press. 6. Thas, J. A., (1987), " Complete arcs and Algebraic Curves in PG(2,p ) ", J. of Algebra, 106(2): 451-464. 7. Hughes, D. R. and Pip er, F. C., (1973), " Projective Planes ", Sp ringer- velag , new York Inc . 8. Veblen, O. and Young, J. W., (1910), " Projective Geometry " Volment, GINN. 9. Albert A. A., (1968), "An Introduction to Finite Projective Plane", Holt Rinehart and Winst on, Inc. IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VOL.23 (1) 2010 Table ( 1 )The addition's and multiplications operations of GF(4) Table ( 2 ) The points and lines of PG(2,4) Li Pi i 18 14 10 6 2 0 0 1 1 9 8 7 6 1 0 1 0 2 21 16 11 6 3 0 1 1 3 20 15 13 6 5 0 1 2 4 19 17 12 6 4 0 1 3 5 5 4 3 2 1 1 0 0 6 19 15 11 7 2 1 0 1 7 21 17 13 9 2 1 0 2 8 20 16 12 8 2 1 0 3 9 13 12 11 10 1 1 1 0 10 20 17 10 7 3 1 1 1 11 19 16 10 9 5 1 1 2 12 21 15 10 8 4 1 1 3 13 21 20 19 18 1 1 2 0 14 18 16 13 7 4 1 2 1 15 18 15 12 9 3 1 2 2 16 18 17 11 8 5 1 2 3 17 17 16 15 14 1 1 3 0 18 21 14 12 7 5 1 3 1 19 20 14 11 9 4 1 3 2 20 19 14 13 8 3 1 3 3 21 * 1 2 3 1 1 2 3 2 2 3 1 3 3 1 2 + 0 1 2 3 0 0 1 2 3 1 1 0 3 2 2 2 3 0 1 3 3 2 1 0 \ 2010) 1( 23المجلد مجلة ابن الھیثم للعلوم الصرفة والتطبیقیة بطریقة هندسیة PG)(2,4األقواس العظمى الكاملة في المستوى االسقاطي سوسن جواد كاظم قسم الریاضیات ،ابن الهیثم ،كلیة التربیة ، جامعة بغداد الخالصه إن ، إذ q=pⁿ و PG(q)حول حقل كالوا PG(2,4)قاطي منتهي في مستوي إس - (k,n) األقواس p ولعدد صحیح أوليعددn ≥2 , هو مجموعة مكونة منk یوجد من النقاط الn+1 منها تقع على .مستقیم واحد ) .(k+1,n - لم یكن محتوى في القوس إذایكون كامل ) (k,n –القوس من PG(2,4)في المستوي n=2,3,4العظمى الكاملة و ) k,n( – األقواسیتم بناء سفي هذا البحث .معادلة المخروط