مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 The Construction and Reverse Construction of the Complete Arcs in the Projective 3-Space Over Galois Field GF(2) A. SH. Al-Mukhtar Department of Mathematics-Ibn-Al-Haitham College of Education - University of Baghdad Received in : 11 May 2011 Accepted in :16 June 2011 Abstract The main purpose of this work is to find the complete arcs in the projective 3-space over Galois field GF(2), which is denoted by PG(3,2), by two methods and then we compare between the two methods. Keywords: arcs, secant, quadrable. Introduction, [1,2] A projective space PG(3,q) over Galois field GF(q), q = p m , for some prime number p and some integer m, is a 3 – dimensional projective space. Any point in PG(3,q) has the form of a quadrable (x1, x2, x3, x4), where x1, x2, x3, x4 are elements in GF(q) with the exception of the quadrable consisting of four zero elements. Two quadrables (x1, x2, x3, x4) and (y1, y2, y3, y4) represent the same point if there exists  in GF(q) \ {0} such that (x1, x2, x3, x4) =  (y1, y2, y3, y4), this is denoted by (x1, x2, x3, x4)  (y1, y2, y3, y4). Similarly, any plane in PG(3,q) has the form of a quadrable [x1, x2, x3, x4], where x1, x2, x3, x4 are elements in GF(q) with the exception of the quadrable consisting of four zero elements. Two quadrables [x1, x2, x3, x4] and [y1, y2, y3, y4] represent the same plane if there exists  in GF(q)\{0} such that [x1, x2, x3, x4] =  [y1, y2, y3, y4], this is denoted by [x1, x2, x3, x4]  [y1, y2, y3, y4].. Also a point P(x1, x2, x3, x4) is incident with the plane  [a1, a2, a3, a4] iff a1 x1 + a2 x2 + a3 x3 + a4 x4 = 0. Every line in PG(3,q) contains q + 1 points and every point is on exactly q + 1 lines. Any plane in PG(3,q) contains exactly q 2 + q + 1 points and q2 + q + 1 lines. Every point is on q2 + q + 1 planes and is on q2 + q + 1 lines. Moreover PG(3,q) contains exactly q 3 + q 2 + q + 1 points and also contains exactly q3 + q2 + q + 1 p lanes. Definition 1: [1,3] A (k,n) – arc A in PG(3,q) is a set of k points such that at most n points of which lie in any plane, n  3. n is called the degree of the (k,n) – arc. مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 Definition 2: [1,3] In PG(3,q), if A is any (k,n) – arc, then an (n-secant) of A is a plane  such that   A= n. Definition 3: [1,3] Let Ti be the total number of the i – secants of a (k,n) – arc A, then the type of A denoted by (Tn, Tn – 1, , T0). Definition 4: [1,3] Let (k1,n) – arc A is of type (Tn, , T0) and (k2,n) – arc B is of type (Sn, , S0), then A and B are projectively equivalent iff Ti = Si. Definition 5: [1,3] If a point N not on a (k,n)-arc A has index i iff there are exactly i(n –secants) of A through N, one can denote the number of points N of index i by Ci. Definition 6: If (k,n)-arc A is not contained in any (k + 1,n)-arc, then A is called a complete (k,n)-arc. Remark: From definition 5, it is concluded that the (k,n)-arc is complete iff C0 = 0. Thus the (k,n)-arc is complete iff every point of PG(3,q) lies on some n-secant of the (k,n)-arc. 1- The Construction of Complete (k,n)-Arcs in PG(3,2) 1.1 The Construction of Complete (k,3)-arcs in PG(3,2): PG(3,q) contains 15 points and 15 planes such that each point is on 7 planes and every plane contains 7 points (see table 1). The set A = {1, 2, 3, 4, 13} is taken which is the set of unit and reference points: 1(1,0,0,0), 2(0,1,0,0), 3(0,0,1,0), 4(0,0,0,1), 13(1,1,1,1). This set contains five points no four of them are on a plane since A intersects any plane in at most three points. Thus A is a (5,3)-arc. A is a complete (5,3) – arc since every point of PG(3,2) not in A is on a 3-secant; that is, there are no points of index zero for A. This is equivalent to C0 = 0. 1.2 The Construction of Complete (k,4) – arcs in PG(3,2) : The distinct (k,4) –arcs can be constructed by adding to A in each time one point from the remaining ten points of PG(3,2) as follows: A1=A{5}, A2=A{6}, A3=A{7}, A4=A{8}, A5=A{9}, A6=A{10}, A7=A{11}, A8=A{12}, A9=A{14}, A10=A{15}. By definition 4 of projectively equivalent (k,n) – arcs, there is only one (6,4) – arc since the arcs A1, , A10 are projectively equivalent. For T0=0, T1=2, T2=3, T3=6, T4=4. Thus we have B=A{5}={1,2,3,4,5,13} is a complete (6,4) – arc, since every point not in B is on a 4 – secant and B intersects any plane in at most 4 points, that is C0 = 0. مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 1.3 The Construction of Complete (k,5) – arcs in PG(3,2) : The arc B is a complete (6,4) – arc. The distinct (k,5) – arcs can be constructed by adding to B in each time one of the remaining nine points as follows: B1=B{6}, B2=B{7}, B3=B{8}, B4=B{9}, B5=B{10}, B6= B{11}, B7=B{12}, B8=B{14}, B9=B{15}. By definition 4, there are only two projectively distinct (7,5) – arcs since the arcs B1, B4, B5, B7, B8, B9 are projectively equivalent, for T0=0, T1=1, T2=2, T3=5, T4=6, T5=1 and the arcs : B2, B3, B6 are projectively equivalent, for : T0=0, T1=0, T2=4, T3=5, T4=4, T5=2. Thus we have two projectively distinct (7,5) – arcs C=B{6}={1,2,3,4,5,6,13}, D = B  {7} ={1,2,3,4,5,7,13}. We try to show the completeness of these arcs. Each of C and D is not complete since there exist some points of index zero. We take the union of C and D. Then E=CD={1,2,3,4,5,6,7,13}, E is incomplete (8,5) – arc since there exists one point of index zero for E, which is the point (15). We add the point (15) to E, we obtain a complete (9,5) – arc F, F=E{15}={1,,7,13,15}. Thus every point not in F is on a (5 – secant) and F intersects any plane in at most 5 points. 1.4 The Construction of Complete (k,6) – arcs in PG(3,2) : The arc F={1,,7,13,15} is a complete (9,5) – arc. The distinct (k,6) – arcs can be constructed by adding to F in each time one of the remaining six points, then: F1=F{8}, F2=F{9}, F3=F{10}, F4=F{11}, F5=F{12}, F6=F{14}. By the definition 4, there are only two projectively distinct arcs since the arcs F1, F2, F5, F6 are projectively equivalent, For T0=T1=T2=0, T3=2, T4=4, T5=6, T6=3 and the arcs F3 and F4 are projectively equivalent, for T0=T1= 2, T3=2, T4=4, T5=7, T6=2. Thus we have two projectively distinct (10,6) – arcs G1={1,2,3,4,5,6,7,8,13,15}, G2={1,2,3,4,5,6,7,11,13,15} each of them is incomplete since there exist some points of index zero. We take the union of G1 and G2. G= G1G2={1,2,3,4,5,6,7,8,11,13,15}. G is incomplete (11,6) – arc since there exists one point of index zero, which is the point (9), then H=G{9}={1,…,9,11,13,15}. H is a complete (12,6) – arc, since every point not in H is on a 6 – secant and H intersects any plane in at most 6 points. 1.5 The Construction of Complete (k,7) – arcs in PG(3,2) : The arc H = {1,,9,11,13,15} is a complete (12,6) – arc. Adding all the remaining points to H, The complete (15,7) – arc can be obtained which is the maximal arc since it contains all points of PG(3,2), (see figure (1)). 2- The Reverse Construction of Complete (k,n)-Arcs in PG(3,2): Complete (k,n) – arcs in PG(3,2) can be constructed by eliminating some points from the complete arcs of degree m, where m = n + 1, 3  n  6, through the following steps: 2.1 The complete (k,7) – arc in PG(3,2) : The projective space PG (3,2) contains 15 points and 15 planes, each plane contains exactly 7 points, then the maximal complete (k,7) – arc A exists when k = 15. This arc contains all the points of PG(3,2) since it intersects every plane in exactly 7 points and hence there arc no points of index zero for A. So A = {1, , 15} is the complete (15,7) – arc. 2.2 The Construction of Complete (k,6) – arc in PG(3,2) : A complete (k,6) – arc B is constructed from the complete (15,7) – arc A by eliminating some points from A such that: 1. B intersects any plane in at most 6 points. مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 2. every point not in B is on at least one 6 – secant of B. The points 1, 2, 5 are eliminated from A, we obtain a complete (12,6) – arc B, since there are no points of index zero for B. B = {3, 4, 6, , 15}. 2.3 The Construction of Complete (k,5) – arc in PG(3,2) : A complete (k,5) – arc in PG (3,2) can be constructed from the complete (12,6) – arc B by eliminating some points from B, which are: 3,6,9. Then a complete (9,5) – arc C is obtained, C = {4, 7, 8, 10, 11, 12, 13, 14, 15} since each point not in C is on at least one 5 – secant, hence there are no points of index zero for C and C intersects any plane of PG(3,2) in at most 5 points. 2.4 The Construction of Complete (k,4) – arc in PG(3,2) : A complete (k,4) – arc in PG(3,2) can be constructed from the complete (9,5) – arc C by eliminating three points from C, which are the points 4, 7, 10, then a complete (6,4) – arc D is obtained, D = {8, 11, 12, 13, 14, 15} since each point not in D is on at least one 4 – secant of D and hence there are no points of index zero and D intersects each plane in at most 4 points. 2.5 The Construction of Complete (k,3) – arc in PG(3,2) : A complete (k,3) – arc in PG(3,2) can be constructed from the complete (6,4) – arc D by eliminating one point from D, which is the point : 15. A complete (5,3) – arc E is obtained, E = {8, 11, 12, 13, 14} since each point not in E is on at least one 3 – secant, hence there are no points of index zero for E and E intersects each plane in at most 3 points. See figure (2). 3- Results and Conclusion From the previous results of the two methods, we found that there is no differences between them, the numbers of the points of the complete (k,n) – arcs in the two methods given in table (2). References 1. Al-Mukhtar, A.Sh. (2008) Complete Arcs and Surfaces in three Dimensional Projective Space Over Galois Field, Ph.D. Thesis, University of Technology, Iraq. 2. Hirschfeld, J. W. P. (1998) Projective Geometries Over Finite Fields, Second Edition, Oxford University Press. 3. Mohammed, S. K. and Al-Mukhtar, A. Sh. (2009) Engineering and Technology Journal, On Projective 3-Space, Vol.27(8): مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 Table (1):The Points Pi and Planes i of PG(3,2) i Pi i 1 (1,0,0,0) 2 3 4 6 7 10 12 2 (0,1,0,0) 1 3 4 7 9 14 15 3 (0,0,1,0) 1 2 4 5 8 10 15 4 (0,0,0,1) 1 2 3 5 6 9 11 5 (1,1,0,0) 3 4 5 7 8 11 13 6 (0,1,1,0) 1 4 6 11 12 13 15 7 (0,0,1,1) 1 2 5 7 12 13 14 8 (1,1,0,1) 3 5 10 11 12 14 15 9 (1,0,1,0) 2 4 9 10 11 13 14 10 (0,1,0,1) 1 3 8 9 10 12 13 11 (1,1,1,0) 4 5 6 8 9 12 14 12 (0,1,1,1) 1 6 7 8 10 11 14 13 (1,1,1,1) 5 6 7 9 10 13 15 14 (1,0,1,1) 2 7 8 9 11 12 15 15 (1,0,0,1) 2 3 6 8 13 14 15 Table (2):The Maximum (k,n)-arcs in Two Methods n maximum (k,n)– arcs in the first method maximum (k,n)– arcs in the second method 3 5 5 4 6 6 5 9 9 6 12 12 7 15 15 مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 Fig. (1):All complete (kn,n) – arcs in PG(3,2), 3 ≤ n ≤ 7 مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 Fig. (2):All complete (kn,n) – arcs in PG(3,2), 3 ≤ n ≤ 7, by reverse construction مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn Al-Haitham Journal for Pure and Applied Science No. 1 Vol. 25 Year 2012 البناء والبناء العكسي لألقواس الكاملة للفضاء الثالثي االسقاطي حول حقل GF(2)كالوا آمال شهاب المختار جامعة بغداد، ابن الهیثم -كلیة التربیة ،قسم الریاضیات 2011 حزیران 16: قبل البحث في 2011 آیار 11:استلم البحث في الخالصة ، GF(2) الهدف االساسي من هذا البحث هو ایجاد االقواس الكاملة في الفضاء الثالثي االسقاطي حول حقل كالوا .، بطریقتین ومن ثم نقارن بین الطریقتینPG(3.2)والذي یرمز له