Qualifier [19]

A (k,ƪ)-set in P G ( n , q ) is a set of k spaces π_ƪ. A k-set is a(k,0)-set, that is a set of k points.

Definition [19]

A (k, ƪ; r, s, n, q)-set is a (k,ƪ)-set in P G ( n , q ) at most r spaces π_ƪ of which lie in any πs.

It is of great interest for applications to find maximal and particularly maximum such sets as k varies but the other parameters remain fixed. Thus a complete (k, ƪ; r, s; n, q)-set is one not contained in any (k + 1, ƪ; r, s, n, q)-set.

The definition of (k, ƪ; r, s; n, q)-set is specialized as follows.

A (k; r, s; n, q)-set is a (k, 0; r, s; n, q)-set;

A (k, r; n, q)-set is a (k; r, r-1; n, q)-set;

A (k; r)-cap is a (k; r, ƪ; n, q)-set with n ³ 3;

A k-cap is a (k, 2)-cap;

A k-arc is a (k; n, n-1; n, q)-set;

A plane (k; r)-arc is a (k; r, 1; 2, q)-set;

A (k,ƪ)-span is a (k, l; 1, 2ƪ; n, q)-set with ƪ ³ 1.

Thus a plane k-arc (or just k-arc) is a set of k points in P G ( 2 , q ) , no three of which are collinear, a maximum plane k-arc is an oval.

A k-cap is a set of k points in P G ( n , q ) with n ³ 3 such that no three are collinear.

A maximum k-cap in P G ( 3 , q ) is an ovaloid, denoted by m(n,q).

A (k,ƪ)-span is a set of k spaces π_ƪ no two of which intersect, a maximum (k,ƪ)-span is a sprea

Definition

In P G ( 3 , q ) , if K is any k-set, then an n-secant of K is a line (a plane) ƪ such that | l ∩ K | = n . In particular, the following terminology are also used

0-secant is an external line (plane).

1-secant is a unisecant line (plane).

2-secant is a bisecant line (plane).

3-secant is a trisecant line (plane).

Qualifier [6] [7]

A polynomial F in a polynomial ring K [ x 1 , ⋯ , x n ] is called homogenous or a form of degree d if all of its terms have the same degree d.

Qualifier [6] [7]

A subset v of P G ( n , K ) is a variety (over K) if there exist forms F 1 , F 2 , ⋯ , F r in K [ x 1 , ⋯ , x n ] , such that:

v = { P ( A )   in   P G ( n , K ) | F 1 ( A ) = F 2 ( A ) = ⋯ = F r ( A ) = 0 } = V n , K ( F 1 , F 2 , ⋯ , F r )

The points P(A) are points of v. When K = GF(q), the notation V n , q ( F 1 , F 2 , ⋯ , F r ) is used.

A variety V n , K ( F ) in P G ( n , K ) is a primal. A primal in P G ( 2 , K ) is a plane algebraic curve, a primal in P G ( 3 , K ) is a surface. The dimension of a primal is n − 1 ; in particular, a curve in P G ( 2 , K ) and a surface in P G ( 3 , K ) have dimensions one and two respectively.

Definition

A point N not on a (k,n)-set A has index i if there are exactly i (n-secants) of K through N, one can denote the number of points N of indexi by C_i.

It is concluded that the (k,n)-set is complete iff C 0 = 0 . Thus the k-set is complete iff every point of P G ( 3 , q ) lies on some n-secant of the (k,n)-set.

Qualifier

A(k,n)-arc A in P G ( 3 , q ) is a set of k points such that at most n points of which lie in any plane, n ³ 3. n is called the degree of the (k,n)-arc.

Definition [19]

Let T_i be the total number of the i-secants of a (k,n)-arc A, then the type of A w.r.t. its planes denoted by ( T n , T n – 1 , ⋯ , T 0 ) .

Qualifier [19]

Let (k₁,n)-arc A is of type ( T n , T n – 1 , ⋯ , T 0 ) and (k₂,n)-arc B is of type ( S n , S n – 1 , ⋯ , S 0 ) , then A and B have the same type iff T i = S i , for all i, in this case, they are projectively equivalent.

Notion

Let t(P) represents the number of unisecants (planes) through a point P of a (k,n)-arc A and let T_i represent the numbers of i-secants (planes) for the arc A in P G ( 3 , q ) , then:

1) t = t ( P ) = q 2 + q + 2 − k − ( k − 1 ) ( k − 2 ) 2 − ⋯                               − ( k − 1 ) ( k − 2 ) ⋯ ( k − ( n − 1 ) ) ( n − 1 ) !

2) T 1 = k t

3) T 2 = k ( k − 1 ) 2

4) T 3 = k ( k − 1 ) ( k − 2 ) 3 !

5) T n = k ( k − 1 ) ⋯ ( k − n + 1 ) n !

6) T 0 = q 3 + q 2 + q + 1 − k t − k ( k − 1 ) 2 − k ( k − 1 ) ( k − 2 ) 3 ! − ⋯     − k ( k − 1 ) ( k − 2 ) ⋯ ( k − n + 1 ) n !

Proof

1) there exist k − 1 bisecants to A through P and there exist ( k − 1 2 ) trisecants to A through P, and so there exist ( k − 1 n − 1 ) n-secants to A through P, and since there exist exactly q 2 + q + 1 planes through P, then the number of the unisecants through P:

t ( P ) = q 2 + q + 1 − ( k − 1 ) − ( k − 1 2 ) − ⋯ − ( k − 1 n − 1 ) = q 2 + q + 2 − k − ( k − 1 ) ( k − 2 ) 2 − ⋯ − ( k − 1 ) ( k − 2 ) ⋯ ( k − n + 1 ) ( n − 1 ) ! = t

2) T₁ = the number of unisecants to A, since each point of A has t unisecants and the number of the points of A is k, then T 1 = k t .

3) T₂ = the number of bisecants to A, which is the number of planes passing through any two points of A. Hence T 2 = ( k 2 ) = k ( k − 1 ) 2 .

4) T₃ = the number of trisecants of A, which is the number of planes passing through any three points of A. Hence T 3 = ( k 3 ) = k ( k − 1 ) ( k − 2 ) 3 ! .

5) T_n = the number of n-secants planes to A, T n = ( k n ) = k ( k − 1 ) ⋯ ( k − n + 1 ) n ! .

6) q 3 + q 2 + q + 1 represents the number of all planes, then in a (k,n)-arc of P G ( 3 , q ) , q 3 + q 2 + q + 1 = T 0 + T 1 + T 2 + T 3 + ⋯ + T n

T 0 = q 3 + q 2 + q + 1 − T 1 − T 2 − T 3 − ⋯ − T n = q 3 + q 2 + q + 1 − k t − k ( k − 1 ) 2 − k ( k − 1 ) ( k − 2 ) 3 ! − ⋯     − k ( k − 1 ) ( k − 2 ) ⋯ ( k − n + 1 ) n !

Theorem

Let T_i represent the total number of the i-secants for a (k,n)-arc A in P G ( 3 , q ) , then the following equations are satisfied:

1) ∑ i = 0 n T i = q 3 + q 2 + q + 1

2) ∑ i = 1 n i ! T i = k t + k ( k − 1 ) + k ( k − 1 ) ( k − 2 ) + ⋯ + k ( k − 1 ) ( k − n )

3) ∑ i = 2 n i ( i – 1 ) T i = k ( k − 1 ) + k ( k − 1 ) ( k − 2 ) + 1 / 2 k ( k − 1 ) ( k − 2 ) ( k − 3 )     + ⋯ + 1 / ( ( n − 2 ) ! ) k ( k − 1 ) ( k − n )

Proof

1) ∑ i = 0 n T i represents the sum of numbers of all i-secants to A, which is the number of all planes in the space. Hence ∑ i = 0 n T i = q 3 + q 2 + q + 1 .

2) T 1 = k t , t = q 2 + q + 2 − k − ( k − 1 ) ( k − 2 ) 2 − ⋯ − ( k − 1 ) ⋯ ( k − n + 1 ) ( n − 1 ) ! , T 2 = k ( k − 1 ) 2 , T 3 = k ( k − 1 ) ( k − 2 ) 3 ! , T 4 = k ( k − 1 ) ( k − 2 ) ( k − 3 ) 4 ! , ⋯ , T n = k ( k − 1 ) ⋯ ( k − n + 1 ) n !

∑ i = 1 n i ! T i = T 1 + 2 ! T 2 + 3 ! T 3 + ⋯ + n ! T n = k t + k ( k − 1 ) + k ( k − 1 ) ( k − 2 )     + ⋯ + k ( k − 1 ) ( k − n + 1 )

3) ∑ i = 2 n i ( i − 1 ) T i = 2 T 2 + 6 T 3 + 12 T 4 + ⋯ + n ( n − 1 ) T n = k ( k − 1 ) + k ( k − 1 ) ( k − 2 ) + 1 2 k ( k − 1 ) ( k − 2 ) ( k − 3 ) + ⋯     + 1 ( n − 2 ) ! k ( k − 1 ) ( k − n + 1 )

Notion

Let R i = R i ( P ) represents the number of the i-secants (planes) through a point P of a (k,n)-arc A, in P G ( 3 , q ) then the following equations are satisfied:

1) ∑ i = 1 n R i = q 2 + q + 1

2) ∑ i = 2 n ( i – 1 ) ! R i = ( k − 1 ) + ( k − 1 ) ( k − 2 ) + ⋯ + ( k − 1 ) ( k − 2 ) ( k − n − 1 ) = ∑ i = 1 n − 1 ( k − 1 ) ( k − i )

Proof

1) ∑ i = 1 n R i = R 1 + R 2 + ⋯ + R n , ∑ i = 1 n R i represents the sum of numbers of all the i-secants through a point P of the arc A, which is the number of the planes through P. Thus,

∑ i = 1 n R i = q 2 + q + 1 .

2) ∑ i = 2 n ( i − 1 ) ! R i = R 2 + 2 ! R 3 + 3 ! R 4 + ⋯ + ( n − 1 ) ! R n

R 2 = k – 1 , R 3 = ( k − 1 2 ) , R 4 = ( k − 1 3 ) , ⋯ , R n = ( k − 1 n − 1 )

R 3 = ( k − 1 ) ! 2 ! ( k − 3 ) ! , R 4 = ( k − 1 ) ! 3 ! ( k − 4 ) ! , ⋯ , R n = ( k − 1 ) ! ( n − 1 ) ! ( k − n ) !

R 3 = ( k − 1 ) ( k − 2 ) 2 , R 4 = ( k − 1 ) ( k − 2 ) ( k − 3 ) 3 ! , ⋯ , R n = ( k − 1 ) ⋯ ( k − ( n − 1 ) ) ( n − 1 ) !

∑ i = 2 n ( i − 1 ) ! R i = k − 1 + 2 ! ( k − 1 ) ( k − 2 ) 2 ! + 3 ! ( k − 1 ) ( k − 2 ) ( k − 3 ) 3 ! + ⋯     + ( n − 1 ) ! ( k − 1 ) ( k − 2 ) ⋯ ( k − ( n − 1 ) ) ( n − 1 ) ! = ( k − 1 ) + ( k − 1 ) ( k − 2 ) + ( k − 1 ) ( k − 2 ) ( k − 3 ) + ⋯     + ( k − 1 ) ( k − 2 ) ( k − ( n − 1 ) ) = ∑ i = 1 n − 1 ( k − 1 ) ( k − i )

Theorem

Let S i = S i ( Q ) represent the numbers of the i-secants (planes) of a (k,n)-arc A through a point Q in P G ( 3 , q ) such that Q not in A, then the following equations are satisfied:

1) ∑ i = 0 n S i = q 2 + q + 1

2) ∑ i = 1 n i S i = k

Proof

1) ∑ i = 0 n S i represents the sum of the total numbers of all i-secants to A through a point Q not in A, which is equal to the number of all planes through the point Q. Thus ∑ i = 0 n S i = q 2 + q + 1 .

2) ∑ i = 1 n i S i = S 1 + 2 S 2 + 3 S 3 + ⋯ + n S n .

S 1 , S 2 , ⋯ , S n represent the numbers of the i-secants of the arc A through the point Q not in A. S₁ is the number of the unisecants to A, each one passes through one point of A. S₂ is the number of the bisecants to A, each one passes through two points of A. S₃ is the number of the trisecants to A, each one passes through three points of A. Also, S_n is the number of the n-secants to A, each one passes through n points of A. Since the number of points of the (k,n)-arc A is k, then ∑ i = 1 n i S i = k .

Notion

Let C_i be the number of points of index i in S = P G ( 3 , q ) which are not on a complete (k,n)-arc A, then the constants C_i of A satisfy the following equations:

i) ∑ α β C i = q 3 + q 2 + q + 1 − k

ii) ∑ α β i C i = k ( k − 1 ) ⋯ ( k − n + 1 ) n ! ( q 2 + q + 1 − n )

where α is the smallest i for which C i ≠ 0 , β be the largest i for which C i ≠ 0 .

Proof

The equations express in different ways the cardinality of the following sets

i) { Q | Q ∈ S \ A }

ii) { ( Q , π ) / Q ∈ π \ A , π   an   n -secant   of   A } for in (i) ∑ α β C i represents all points in the space which are not in A, then ∑ α β C i = q 3 + q 2 + q + 1 − k , and in (ii) ∑ α β i C i represents all points in the space not in A, which are on n-secants of A, that is, each n-secant contains q 2 + q + 1 − n points, and the number of the n-secants is ( k n ) , then

∑ α β i C i = ( k n ) ( q 2 + q + 1 − n ) = k ( k − 1 ) ⋯ ( k − n + 1 ) n ! ( q 2 + q + 1 − n )

Theorem

If P is a point of a (k,n)-arc A in P G ( 3 , q ) , which lies on an m-secant (plane) of A, then the planes through P contains at most ( n − 1 ) q ( q + 1 ) + m points of A.

Proof

If P in A lies on an m-secant (plane), then every other plane through P contains at most n − 1 points of A distinct from P. Hence the q 2 + q + 1 planes through P contain at most ( n − 1 ) ( q 2 + q ) + m points of A.

P G ( 3 , 7 ) contains 400 points and 400 planes such that each point is on 57 planes and every plane contains 57 points, any line contains 8 points and it is the intersection of 8 planes, all the points, planes and lines of P G ( 3 , 7 ) are given in Table 1 and Table 2.

In Table 3 any two non-intersecting lines can be taken in PG(3,7).

In Table 3 any elements of the set ƫ_i = {ξ, ν, μ, ⋯ , Ѡ} except the first element can be representing by union of below set and non-intersecting of them.

Finally, the line Ѡ = {57, 104, 145, 186, 227, 268, 316, 357} cannot intersect any line of the set (ƫ_i) and (Ѡ) is (50,ƪ)-span, which is the maximum (k,ƪ)-span of P G ( 3 , 7 ) can be obtained. Thus Ѡ is called a Spread of fifty lines of P G ( 3 , 7 ) which partitions P G ( 3 , 7 ) ; that every point of P G ( 3 , 7 ) lies in exactly one line of ƫ_i. and every line are disjoint. From the above results the number of the planes in the projective space

P G ( 3 , 7 ) are 400 planes and each plane contains 57 lines, therefore the total number of the lines in P G ( 3 , 7 ) are 22,800. We found that the number of the lines do not intersect with some of them are fifty lines, these lines contains the whole points of the projective space P G ( 3 , 7 ) , and called him a (50,ƪ)-span, i.e.

(50,ƪ)-span = {ƪ₁, ƪ₂, ⋯ , ƪ₅₀} = PG(3,7) = {1, 2, 3, ⋯ , 400}

Moreover, we found that a (50,ƪ)-span is a maximum complete (k,ƪ)-span in P G ( 3 , 7 ) .