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 Department of Mathematics, College of Education, Ibn Al-Haitham, University of Baghdad Abstract A (k,n)-arc A in a finite projective plane PG(2,q) over Galois field GF(q), q=pⁿ for same prime number p and some integer n≥2, is a set of k points, no n+1 of which are collinear. A (k,n)-arc is complete if it is not contained in a(k+1,n)-arc. In this paper, 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 projective planes over Galois field GF(pⁿ) for some prime 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 paper, the maximum complete (k,n)-arcs in PG(2,4), n=2,3 are obtained from the equation of the conic by geometric method. A projective plane PG(2,q) over GF(q) consists of 1+q+q 2 points and 1+q+q2 of lines, every line contains 1+q points end every point is on 1+q lines. Any point of the plane has the form of triple (x0,x1,x2), where x0,x1,x2 are elements of GF(q) with the exception of a triple consisting of three zero elements, two triples represent the same point if there exists λ in GF(q)\ {0}, s.t . ( y0,y1,y2 ) = λ( x0,x1,x2 ) . Similarly, any line of the plane has the from of a triple [ x0,x1,x2 ], x0,x1,x2 are in GF(q) with the exception of a triple consisting three zero elements. Two lines [x0,x1,x2] and [y0,y1,y2] represent the same line if there exists λ in GP(q)\{0} s.t . [x0,x1,x2] = λ[y 0,y1,y2] . The point (x0,x1,x2) is incident with line [y0,y1,y2] if x0y0+x1y1+x2y2 = 0. IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VOL.23 (1) 2010 2. Basic Definitions and Theorems 2.1 Definition [3] A (k,n)-arc in a finite projective plane as a set of k points no n+1 of which are collinear. 2.2 Definition [4] A (k,n)-arc is a set of k point, no three of them are collinear, we denote this by k-arc. 2.3 Definition [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 points in PG(2,p) that a (k,n)-arc can have. 2.4 Theorem [6] A (k,n)-arc in PG(2,p) is complete if and only if C0 = 0. Proof: ( ) Let a (k,n)-arc K be a complete arc in PG(2,p) suppose that C0 ≠ 0, then there is at lest one point say N has index zero and N  K. Then K  {N} is an arc in PG(2,p). Hence K  K  {N}, which implies (k,n)-arc K is incomplete (contradiction). ( ) Suppose that C0=0 for the (k,n)-arc K, then there are no points of index zero, then the (k,n)- arc K is complete. 2.5 Definition [4] A k-arc is called an oval when k = (2,p). 2.6 Definition [7] Let ℓ be any line in PG(2,p) if ℓ intersects a k-arc in i-points, | 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 Definition [3] Let N be a point in PG(2,p) and N is not on a k-arc, then we say N is a point of index i if there are exactly i-bisecant through N . 2.8 Definition [5] The set Ci consists of all points 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 through 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 point is p+1 there are exactly p+1 lines through M the point M with any other point of the k-arc determine a bisecant of K since there are (k-1) points of K other then M, then there are exactly (k-1) lines determined from M and the other point 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 Notation: 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 Definition [4] Let K be a k-arc which is an oval an external point to an oval, is a point of intersection two unisecants of K. 2.12 Theorem [5] m (2, p) =      even pfor 2p odd pfor 1p 2.13 Theorem [5] The number of external points of an oval k-arc in PG(2,p) is 2 )1( pp . Proof: If K is an 0val, then m (2, p) =      even 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 = the number of unisecants of k in PG(2,p) , Since each two unisecants intersect in an external point, then the number of external points =        2 1p = !2)!21( )!1(   p p = 1*2)!1( )!1()1(   p ppp = 2 )1( pp . 2.14 Definition [3] An internal point to an oval if it is not any unisecant of the oval which is not on the oval. 2.15 Theorem [5] The number of the internal points to on oval is 2 )1( pp . Proof: The number of the external points + the number of the internal points = p2+p+1- k. # of external points = 2 )1( pp IBN AL- HAITHAM J. FO R PURE & APPL. SC I. VOL.23 (1) 2010 # of the internal points = p2 + p + 1- (p+1) - 2 )1( pp = 2 22222 22 ppppp  = 2 2 pp  = 2 )1( pp . 2.16 Theorem [7] In PG(2,p) with p  4, there is a unique conic through a 5-arc . 2.17 Theorem [4,5] Every conic in PG(2,p) is a(p+1)-arc The converse of theorem is also satisfied. 2.18 Theorem [3] In PG(2,p), with p odd, every oval has a conic. 2.19 Definition [8] A complete quadrangle is a set of four points A, B, C and D in which no three of them are collinear, the points 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. Two sides are said to be opposite if they have no vertex in common. The point of intersection of any two opposite sides is called a diagonal point the diagonal points; D1 = AB CD, D2 = AC  BD and D3 = AD  BC. 3. Maximum Complete (k,n)-Arcs in PG(2,4) 3.1 The Additions and Multiplications Operations of GF(4) [9] To find addition and multiplication tables in GF(4), we have the order pairs (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 point and by the principle of (1,0)Ai, i=0,1,2,3 and A=       11 01 , (1,0)A=(0,1) and (1,0)A2= (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 point as last, and the addition side takes the 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) + (y1,y2) = (z1,z2) where zi = ( y i+ xi ) mod(2) for i =1,2 In multiplication table we have the following relation: ( (1,0) A f(m 1 ) ) A f(m2 )  m1*m2 = m3 =(1,0) A(f(m 1 )+f(m 2 )) (m od 3) = (x1,x2) For example : 2*3 = 1  ( (1,0)A1)A2 = (1,0)A3 = (1,0)A0 = (1,0), where (1,0) is equal to 1 in multiplication tables. The additions and multiplications operations of GF(4) are in table ( 1 ). 3.2 The Projective Plane PG(2,4) The projective plane PG(2,4) contains 21 points, 21 lines, 5 points on every line and 5 lines through every point. Let Pi and Li ,i=1,2,---,21 be points and the lines of PG(2,4) respectively, 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 points 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 points 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 points of A are the points {3,7,10} where ℓ1  ℓ6 = 3, ℓ2  ℓ5 = 7, ℓ4  ℓ3 = 10. The points of PG(2,4) are classified with respect to the lines through the reference and unit points 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 two points of index zero for A, which are the point 17 and the point 20 not on any of these lines, then the 4-arc A is incomplete. 3.4 The Conic in PG(2,4) Through the Reference and Unit 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 3a x a x a x a x x a x x a x x 0      …..……….…(1) By substituting 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 Equation of the Conics of PG(2,4) and the Complete arcs For any value of  there is a unique conic contains the reference and the unit points. 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 complete (k,3)-arc, since there exist the points {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 points of index zero which are {4,7,8}. Then 1C = {1,2,6,11,12,21,4,7,8} is a complete (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 complete (k,2)-arc, since there exist one point {20} which is the point of index zero for C2. Now, we add to C2 one point of index zero {20}. Then 2C = {1,2,6,11,17,20} is a complete (6,2)-arc, since C0 = 0. Conclusion: 1. Each of C1 and C2 is not complete (k,2)-arc. 2. We add the points of index zero for each of them for completeness. 3. The points of index zero of PG(2,4) with respect 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 complete (k,3)-arc by taking the complete (k,2)-arc, say 2C 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 points of index zero which are {9,10}. Then B={1,2,6,11,17,20,5,9,10} is a complete (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 complete (k,4)-arc by taking the union of two maximum complete (k,3)-arcs, say 1C and B denoted by D, we notice that D = {1,2,6,11,12,21,4,7,18,17,20,5,9,10} is incomplete (k,4)-arc, since there exists two points {15,16} which are points of index zero for D. Now, we add the two points 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 complete and maximum (16,4)-arc. 2. The points of index zero of (k,4)-arc with respect are in the same line. Refrences : 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), "Construction 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 Piper, F. C., (1973), " Projective Planes ", Springer- 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 Winston, 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