مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn A l-Haitham Journal f or 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 Galo is Field GF(2) A. SH. Al-Muk htar Departme nt of Mathematics-Ibn-Al-Haitham College of Education - Unive rsity of Baghdad Received in : 11 May 2011 Accepte d in :16 June 2011 Abstract The main p urp ose of this work is t o find the comp lete arcs in the p rojective 3-sp ace over Galois field GF(2), which is denoted by PG(3,2), by two methods and then we comp are between the two methods. Keywords: arcs, secant, quadrable. Introduction, [1,2] A p rojective sp ace PG(3,q) over Galois field GF(q), q = p m , for some p rime number p and some integer m, is a 3 – dimensional p rojective sp ace. Any p oint 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 t he excep tion of the quadrable consisting of four zero elements. Two quadrables (x1, x2, x3, x4) and (y1, y 2, y 3, y 4) represent t he same point if there exists  in GF(q) \ {0} such that (x1, x2, x3, x4) =  (y 1, y 2, y 3, y 4), this is denoted by (x1, x2, x3, x4)  (y 1, y 2, y 3, y 4). Similarly , any p lane 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 excep tion of the quadrable consisting of four zero elements. Two quadrables [x1, x2, x3, x4] and [y1, y 2, y 3, y 4] represent t he same plane if there exists  in GF(q)\{0} such that [x1, x2, x3, x4] =  [y 1, y 2, y 3, y 4], this is denoted by [x1, x2, x3, x4]  [y 1, y 2, y 3, y 4].. Also a p oint P(x1, x2, x3, x4) is incident with the p lane  [a1, a2, a3, a4] iff a1 x1 + a2 x2 + a3 x3 + a4 x4 = 0. Every line in PG(3,q) contains q + 1 p oints and every p oint is on exactly q + 1 lines. Any p lane in PG(3,q) contains exactly q 2 + q + 1 p oints and q 2 + q + 1 lines. Every p oint is on q 2 + q + 1 p lanes and is on q 2 + q + 1 lines. M oreover PG(3,q) contains exactly q 3 + q 2 + q + 1 p oints and also contains exactly q 3 + q 2 + q + 1 p lanes. De fini tion 1: [1,3] A (k,n) – arc A in PG(3,q) is a set of k p oints such that at most n p oints of which lie in any p lane, 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 A l-Haitham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 De fini tion 2: [1,3] In PG(3,q), if A is any (k,n) – arc, then an (n-secant) of A is a p lane  such that   A= n. De fini tion 3: [1,3] Let Ti be the total number of the i – secants of a (k,n) – arc A, t hen the ty p e of A denoted by (Tn, Tn – 1, , T0). De fini tion 4: [1,3] Let (k1,n) – arc A is of ty p e (Tn, , T0) and (k2,n) – arc B is of ty p e (Sn, , S0), then A and B are projectively equivalent iff Ti = Si. De fini tion 5: [1,3] If a p oint 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. De fini tion 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 comp lete iff every p oint of PG(3,q) lies on some n-secant of the (k,n)-arc. 1- The Construction of Complete (k,n)-Arcs i n PG(3,2) 1.1 The Construction of Complete (k,3)-arcs in PG(3,2): PG(3,q) contains 15 p oints and 15 p lanes such that each p oint is on 7 p lanes and every p lane contains 7 p oints (see table 1). The set A = {1, 2, 3, 4, 13} is taken which is the set of unit and reference p oints: 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 p oints no four of them are on a p lane since A intersects any p lane in at most three p oints. Thus A is a (5,3)-arc. A is a comp lete (5,3) – arc since every p oint of PG(3,2) not in A is on a 3-secant; that is, there are no p oints of index zero for A. This is equivalent t o 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 p oint 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}, A 9=A{14}, A 10=A{15}. By definition 4 of p rojectively equivalent (k,n) – arcs, there is only one (6,4) – arc since the arcs A1, , A10 are p rojectively 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 comp lete (6,4) – arc, since every p oint not in B is on a 4 – secant and B intersects any p lane in at most 4 p oints, that is C0 = 0. مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn A l-Haitham Journal f or 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 comp lete (6,4) – arc. The distinct (k,5) – arcs can be constructed by adding to B in each time one of the remaining nine p oints 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 p rojectively distinct (7,5) – arcs since the arcs B1, B4, B5, B7, B8, B9 are p rojectively equivalent, for T0=0, T1=1, T2=2, T3=5, T4=6, T5=1 and the arcs : B2, B3, B6 are p rojectively equivalent, for : T0=0, T 1=0, T 2=4, T 3=5, T 4=4, T 5=2. Thus we have two p rojectively 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 comp leteness of these arcs. Each of C and D is not comp lete 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 p oint of index zero for E, which is t he point (15). We add the p oint (15) to E, we obtain a comp lete (9,5) – arc F, F=E{15}={1,,7,13,15}. T hus every p oint not in F is on a (5 – secant) and F intersects any p lane in at most 5 p oints. 1.4 The Construction of Complete (k,6) – arcs in PG(3,2) : The arc F={1,,7,13,15} is a comp lete (9,5) – arc. The distinct (k,6) – arcs can be constructed by adding to F in each time one of the remaining six p oints, then: F1=F{8}, F 2=F{9}, F 3=F{10}, F 4=F{11}, F 5=F{12}, F 6=F{14}. By the definition 4, there are only two p rojectively distinct arcs since the arcs F1, F2, F5, F6 are p rojectively equivalent, For T0=T 1=T 2=0, T3=2, T 4=4, T 5=6, T 6=3 and the arcs F3 and F4 are p rojectively equivalent, for T0=T 1= 2, T3=2, T4=4, T5=7, T6=2. Thus we have two p rojectively 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 p oints 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 p oint of index zero, which is the point (9), t hen H=G{9}={1,…,9,11,13,15}. H is a comp lete (12,6) – arc, since every p oint not in H is on a 6 – secant and H intersects any p lane in at most 6 p oints. 1.5 The Construction of Complete (k,7) – arcs in PG(3,2) : The arc H = {1,,9,11,13,15} is a comp lete (12,6) – arc. Adding all the remaining p oints to H, The comp lete (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 i n PG(3,2): Complete (k,n) – arcs in PG(3,2) can be const ructed by eliminating some points from the comp lete arcs of degree m, where m = n + 1, 3  n  6, through the following st eps: 2.1 The complete (k,7) – arc in PG(3,2) : The p rojective sp ace PG (3,2) contains 15 p oints and 15 p lanes, each p lane contains exactly 7 p oints, then the maximal comp lete (k,7) – arc A exists when k = 15. This arc contains all the p oints of PG(3,2) since it intersects every p lane in exactly 7 p oints and hence there arc no p oints 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 comp lete (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 p lane in at most 6 p oints. مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn A l-Haitham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 2. every p oint not in B is on at least one 6 – secant of B. The p oints 1, 2, 5 are eliminated from A, we obtain a complete (12,6) – arc B, since there are no p oints of index zero for B. B = {3, 4, 6, , 15}. 2.3 The Construction of Complete (k,5) – arc in PG(3,2) : A comp lete (k,5) – arc in PG (3,2) can be constructed from the comp lete (12,6) – arc B by eliminating some points from B, which are: 3,6,9. Then a comp lete (9,5) – arc C is obtained, C = {4, 7, 8, 10, 11, 12, 13, 14, 15} since each p oint not in C is on at least one 5 – secant, hence there are no p oints of index zero for C and C intersects any p lane of PG(3,2) in at most 5 p oints. 2.4 The Construction of Complete (k,4) – arc in PG(3,2) : A comp lete (k,4) – arc in PG(3,2) can be constructed from the comp lete (9,5) – arc C by eliminating three p oints from C, which are the p oints 4, 7, 10, then a comp lete (6,4) – arc D is obtained, D = {8, 11, 12, 13, 14, 15} since each p oint 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 p lane in at most 4 p oints. 2.5 The Construction of Complete (k,3) – arc in PG(3,2) : A comp lete (k,3) – arc in PG(3,2) can be constructed from the comp lete (6,4) – arc D by eliminating one point from D, which is the point : 15. A comp lete (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 p oints of index zero for E and E intersects each plane in at most 3 p oints. See figure (2). 3- Results and Conclusion From the p revious 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). Re ferences 1. Al-M ukhtar, A.Sh. (2008) Complete Arcs and Surfaces in three Dimensional Projective Sp ace 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. M ohammed, S. K. and Al-M ukhtar, A. Sh. (2009) Engineering and Technology Journal, On Projective 3-Sp ace, Vol.27(8): مجلة إبن الھیثم للعلوم الصرفة و التطبیقیة 2012 السنة 25 المجلد 1 العدد Ibn A l-Haitham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 Table (1):The Points Pi and Plane s 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 maximu m (k,n)– arcs in the first method maximu m (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 A l-Haitham Journal f or 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 A l-Haitham Journal f or 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 A l-Haitham Journal f or 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