Complete Arcs and Surfaces in Three Dimensional Projective Space PG(3,7)
Abstract: The purpose of this thesis is to construct surfaces and complete arcs in the projective 3-space over Galois fields GF(q), q = 7. A(k,n)-arc in is a set of k points; no n + 1 of them are coplanar. A(k,n)-arc is complete if it is not contained in a (k + 1, n)-arc. In this work the (k,¶)-span are constructed in and it is found that, it exists in when k = 50. Moreover, the maximum (k,¶)-span, is called a spread.

1. Introduction

The study of finite projective spaces was at one time no more than an adjunct to algebraic geometry over the real and complex numbers. But, more recently, finite spaces have been studied both for their application to practical topics such as coding theory and design experiments, and for their illumination of more abstract mathematical topics such as finite group theory and graph theory. Perhaps the fastest growing area of modern mathematics is combinatorics that is concerned with the study of arrangement of elements into sets. These elements are usually finite in number, and the arrangement is restricted by certain boundary conditions imposed by the particular problem under investigation. Much of the growth of combinatory has gone hand in hand with the development of the computer. A major reason for this rapid growth of combinatorics is its wealth of applications, to computer science, communications, transportations, genetics, experimental design, and so on.

A recurring theme of this work is the characterization of algebraic varieties in $PG\left(n,q\right)$ as finite sets of points with certain combinatorial properties. Among these finite sets, the concept of (k,n)-arcs is in the n-dimensional projective space over Galois field GF(q), $q={p}^{m}$ for some prime number p and some integer m.

Hirschfeld, J.W.P. (1979) studies the basic definition and theorems of projective geometrics over finite fields [1] , Kareem view (k,ƪ)-span in $PG\left(\text{3},p\right)$ over Galois field GF(p), p = 4 in 2013 [2] . Al-Mokhtar studies the complete arcs and surface in three-dimensional projective space over Galois field GF(P), p = 2, 3 [3] . In three-dimensional projective space, the control problem is how to construct and find the whole space spread which is (50,ƪ)-span in $PG\left(\text{3},7\right)$ and prove it in general when $P\ge 2$ is ${p}^{2}+1$ . Complete arcs have important connections with a number of other objects, see [4] - [19] and the references therein. Hirschfeld, J.W.P. (1998) studies the basic definition and theorems of projective geometrics over finite fields, in 2008 [20] . This paper includes three sections. First section considers the preliminaries of projective 3-space which contains some definition and theorems for the concept, whereas the second section consists of the subspace in $PG\left(\text{3},p\right)$ . Finally, the third section constructs maximum complete (k,ƪ). A span in $PG\left(\text{3},7\right)$ is spreading, and in general proves Geometric rule in (Conclusions) $P\ge 2$ . The total number of (k,ƪ)-span in $PG\left(3,q\right)$ is ${p}^{2}+1,p\ge 2$ .

2.1. Sets of Subspaces

Qualifier [19]

A (k,ƪ)-set in $PG\left(n,q\right)$ is a set of k spaces πƪ. 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 $PG\left(n,q\right)$ at most r spaces πƪ 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 $PG\left(\text{2},q\right)$, no three of which are collinear, a maximum plane k-arc is an oval.

A k-cap is a set of k points in $PG\left(n,q\right)$ with n ³ 3 such that no three are collinear.

A maximum k-cap in $PG\left(\text{3},q\right)$ is an ovaloid, denoted by m(n,q).

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

Definition

In $PG\left(\text{3},q\right)$, if K is any k-set, then an n-secant of K is a line (a plane) ƪ such that $|\mathcal{l}\cap K|=n$ . 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).

2.2. Algebraic Curves and Surfaces

Qualifier [6] [7]

A polynomial F in a polynomial ring $K\left[{x}_{1},\cdots ,{x}_{n}\right]$ 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 $PG\left(n,K\right)$ is a variety (over K) if there exist forms ${F}_{1},{F}_{2},\cdots ,{F}_{r}$ in $K\left[{x}_{1},\cdots ,{x}_{n}\right]$, such that:

$\begin{array}{c}v=\left\{P\left(A\right)\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}PG\left(n,K\right)|{F}_{1}\left(A\right)={F}_{2}\left(A\right)=\cdots ={F}_{r}\left(A\right)=0\right\}\\ ={V}_{n,K}\left({F}_{1},\text{}{F}_{2},\cdots ,\text{}{F}_{r}\right)\end{array}$

The points P(A) are points of v. When K = GF(q), the notation ${V}_{n,q}\left({F}_{1},\text{}{F}_{2},\cdots ,\text{}{F}_{r}\right)$ is used.

A variety ${V}_{n,K}\left(F\right)$ in $PG\left(n,K\right)$ is a primal. A primal in $PG\left(2,K\right)$ is a plane algebraic curve, a primal in $PG\left(3,K\right)$ is a surface. The dimension of a primal is $n-1$ ; in particular, a curve in $PG\left(2,K\right)$ and a surface in $PG\left(3,K\right)$ 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 Ci.

It is concluded that the (k,n)-set is complete iff ${C}_{0}=0$ . Thus the k-set is complete iff every point of $PG\left(3,q\right)$ lies on some n-secant of the (k,n)-set.

Qualifier

A(k,n)-arc A in $PG\left(3,q\right)$ 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 Ti 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 $\left({T}_{n},{T}_{n–1},\cdots ,{T}_{0}\right)$ .

Qualifier [19]

Let (k1,n)-arc A is of type $\left({T}_{n},{T}_{n–1},\cdots ,{T}_{0}\right)$ and (k2,n)-arc B is of type $\left({S}_{n},{S}_{n–1},\cdots ,{S}_{0}\right)$, 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 Ti represent the numbers of i-secants (planes) for the arc A in $PG\left(\text{3},q\right)$, then:

1) $\begin{array}{l}t=t\left(P\right)={q}^{2}+q+2-k-\frac{\left(k-1\right)\left(k-2\right)}{2}-\cdots \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{\left(k-1\right)\left(k-2\right)\cdots \left(k-\left(n-1\right)\right)}{\left(n-1\right)!}\end{array}$

2) ${T}_{1}=kt$

3) ${T}_{2}=\frac{k\left(k-1\right)}{2}$

4) ${T}_{3}=\frac{k\left(k-1\right)\left(k-2\right)}{3!}$

5) ${T}_{n}=\frac{k\left(k-1\right)\cdots \left(k-n+1\right)}{n!}$

6) $\begin{array}{c}{T}_{0}={q}^{3}+{q}^{2}+q+1-kt-\frac{k\left(k-1\right)}{2}-\frac{k\left(k-1\right)\left(k-2\right)}{3!}-\cdots \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{k\left(k-1\right)\left(k-2\right)\cdots \left(k-n+1\right)}{n!}\end{array}$

Proof

1) there exist k − 1 bisecants to A through P and there exist $\left(\begin{array}{c}k-1\\ 2\end{array}\right)$ trisecants to A through P, and so there exist $\left(\begin{array}{c}k-1\\ n-1\end{array}\right)$ 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:

$\begin{array}{c}t\left(P\right)={q}^{2}+q+1-\left(k-1\right)-\left(\begin{array}{c}k-1\\ 2\end{array}\right)-\cdots -\left(\begin{array}{c}k-1\\ n-1\end{array}\right)\\ ={q}^{2}+q+2-k-\frac{\left(k-1\right)\left(k-2\right)}{2}-\cdots -\frac{\left(k-1\right)\left(k-2\right)\cdots \left(k-n+1\right)}{\left(n-1\right)!}\\ =t\end{array}$

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 ${T}_{1}=kt$ .

3) T2 = the number of bisecants to A, which is the number of planes passing through any two points of A. Hence ${T}_{2}=\left(\begin{array}{c}k\\ 2\end{array}\right)=\frac{k\left(k-1\right)}{2}$ .

4) T3 = the number of trisecants of A, which is the number of planes passing through any three points of A. Hence ${T}_{3}=\left(\begin{array}{c}k\\ 3\end{array}\right)=\frac{k\left(k-1\right)\left(k-2\right)}{3!}$ .

5) Tn = the number of n-secants planes to A, ${T}_{n}=\left(\begin{array}{c}k\\ n\end{array}\right)=\frac{k\left(k-1\right)\cdots \left(k-n+1\right)}{n!}$ .

6) ${q}^{3}+{q}^{2}+q+1$ represents the number of all planes, then in a (k,n)-arc of $PG\left(3,q\right)$, ${q}^{3}+{q}^{2}+q+1={T}_{0}+{T}_{1}+{T}_{2}+{T}_{3}+\cdots +{T}_{n}$

$\begin{array}{c}{T}_{0}={q}^{3}+{q}^{2}+q+1-{T}_{1}-{T}_{2}-{T}_{3}-\cdots -{T}_{n}\\ ={q}^{3}+{q}^{2}+q+1-kt-\frac{k\left(k-1\right)}{2}-\frac{k\left(k-1\right)\left(k-2\right)}{3!}-\cdots \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{k\left(k-1\right)\left(k-2\right)\cdots \left(k-n+1\right)}{n!}\end{array}$

Theorem

Let Ti represent the total number of the i-secants for a (k,n)-arc A in $PG\left(3,q\right)$, then the following equations are satisfied:

1) $\underset{i=0}{\overset{n}{\sum }}{T}_{i}={q}^{3}+{q}^{2}+q+1$

2) $\underset{i=1}{\overset{n}{\sum }}i!{T}_{i}=kt+k\left(k-1\right)+k\left(k-1\right)\left(k-2\right)+\cdots +k\left(k-1\right)\left(k-n\right)$

3) $\begin{array}{c}\underset{i=2}{\overset{n}{\sum }}i\left(i–1\right){T}_{i}=k\left(k-1\right)+k\left(k-1\right)\left(k-2\right)+1/2k\left(k-1\right)\left(k-2\right)\left(k-3\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\cdots +1/\left(\left(n-2\right)!\right)k\left(k-1\right)\left(k-n\right)\end{array}$

Proof

1) $\underset{i=0}{\overset{n}{\sum }}{T}_{i}$ represents the sum of numbers of all i-secants to A, which is the number of all planes in the space. Hence $\underset{i=0}{\overset{n}{\sum }}{T}_{i}={q}^{3}+{q}^{2}+q+1$ .

2) ${T}_{1}=kt$, $t={q}^{2}+q+2-k-\frac{\left(k-1\right)\left(k-2\right)}{2}-\cdots -\frac{\left(k-1\right)\cdots \left(k-n+1\right)}{\left(n-1\right)!}$, ${T}_{2}=\frac{k\left(k-1\right)}{2}$, ${T}_{3}=\frac{k\left(k-1\right)\left(k-2\right)}{3!}$, ${T}_{4}=\frac{k\left(k-1\right)\left(k-2\right)\left(k-3\right)}{4!}$, $\cdots$, ${T}_{n}=\frac{k\left(k-1\right)\cdots \left(k-n+1\right)}{n!}$

$\begin{array}{c}\underset{i=1}{\overset{n}{\sum }}i!{T}_{i}={T}_{1}+2!{T}_{2}+3!{T}_{3}+\cdots +n!{T}_{n}\\ =kt+k\left(k-1\right)+k\left(k-1\right)\left(k-2\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\cdots +k\left(k-1\right)\left(k-n+1\right)\end{array}$

3) $\begin{array}{c}\underset{i=2}{\overset{n}{\sum }}i\left(i-1\right){T}_{i}=2{T}_{2}+6{T}_{3}+12{T}_{4}+\cdots +n\left(n-1\right){T}_{n}\\ =k\left(k-1\right)+k\left(k-1\right)\left(k-2\right)+\frac{1}{2}k\left(k-1\right)\left(k-2\right)\left(k-3\right)+\cdots \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{1}{\left(n-2\right)!}k\left(k-1\right)\left(k-n+1\right)\end{array}$

Notion

Let ${R}_{i}={R}_{i}\left(P\right)$ represents the number of the i-secants (planes) through a point P of a (k,n)-arc A, in $PG\left(3,q\right)$ then the following equations are satisfied:

1) $\underset{i=1}{\overset{n}{\sum }}{R}_{i}={q}^{2}+q+1$

2) $\begin{array}{l}\underset{i=2}{\overset{n}{\sum }}\left(i–1\right)!{R}_{i}\\ =\left(k-1\right)+\left(k-1\right)\left(k-2\right)+\cdots +\left(k-1\right)\left(k-2\right)\left(k-n-1\right)\\ =\underset{i=1}{\overset{n-1}{\sum }}\left(k-1\right)\left(k-i\right)\end{array}$

Proof

1) $\underset{i=1}{\overset{n}{\sum }}{R}_{i}={R}_{1}+{R}_{2}+\cdots +{R}_{n}$, $\underset{i=1}{\overset{n}{\sum }}{R}_{i}$ 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,

$\underset{i=1}{\overset{n}{\sum }}{R}_{i}={q}^{2}+q+1$ .

2) $\underset{i=2}{\overset{n}{\sum }}\left(i-1\right)!{R}_{i}={R}_{2}+2!{R}_{3}+3!{R}_{4}+\cdots +\left(n-1\right)!{R}_{n}$

${R}_{2}=k–1,{R}_{3}=\left(\begin{array}{c}k-1\\ 2\end{array}\right),{R}_{4}=\left(\begin{array}{c}k-1\\ 3\end{array}\right),\cdots ,{R}_{n}=\left(\begin{array}{c}k-1\\ n-1\end{array}\right)$

${R}_{3}=\frac{\left(k-1\right)!}{2!\left(k-3\right)!},{R}_{4}=\frac{\left(k-1\right)!}{3!\left(k-4\right)!},\cdots ,{R}_{n}=\frac{\left(k-1\right)!}{\left(n-1\right)!\left(k-n\right)!}$

${R}_{3}=\frac{\left(k-1\right)\left(k-2\right)}{2},{R}_{4}=\frac{\left(k-1\right)\left(k-2\right)\left(k-3\right)}{3!},\cdots ,{R}_{n}=\frac{\left(k-1\right)\cdots \left(k-\left(n-1\right)\right)}{\left(n-1\right)!}$

$\begin{array}{c}\underset{i=2}{\overset{n}{\sum }}\left(i-1\right)!{R}_{i}=k-1+\frac{2!\left(k-1\right)\left(k-2\right)}{2!}+\frac{3!\left(k-1\right)\left(k-2\right)\left(k-3\right)}{3!}+\cdots \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{\left(n-1\right)!\left(k-1\right)\left(k-2\right)\cdots \left(k-\left(n-1\right)\right)}{\left(n-1\right)!}\\ =\left(k-1\right)+\left(k-1\right)\left(k-2\right)+\left(k-1\right)\left(k-2\right)\left(k-3\right)+\cdots \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(k-1\right)\left(k-2\right)\left(k-\left(n-1\right)\right)\\ =\underset{i=1}{\overset{n-1}{\sum }}\left(k-1\right)\left(k-i\right)\end{array}$

Theorem

Let ${S}_{i}={S}_{i}\left(Q\right)$ represent the numbers of the i-secants (planes) of a (k,n)-arc A through a point Q in $PG\left(3,q\right)$ such that Q not in A, then the following equations are satisfied:

1) $\underset{i=0}{\overset{n}{\sum }}{S}_{i}={q}^{2}+q+1$

2) $\underset{i=1}{\overset{n}{\sum }}i{S}_{i}=k$

Proof

1) $\underset{i=0}{\overset{n}{\sum }}{S}_{i}$ 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 $\underset{i=0}{\overset{n}{\sum }}{S}_{i}={q}^{2}+q+1$ .

2) $\underset{i=1}{\overset{n}{\sum }}i{S}_{i}={S}_{1}+2{S}_{2}+3{S}_{3}+\cdots +n{S}_{n}$ .

${S}_{1},{S}_{2},\cdots ,{S}_{n}$ 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, Sn 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 $\underset{i=1}{\overset{n}{\sum }}i{S}_{i}=k$ .

Notion

Let Ci be the number of points of index i in $S=PG\left(3,q\right)$ which are not on a complete (k,n)-arc A, then the constants Ci of A satisfy the following equations:

i) $\underset{\alpha }{\overset{\beta }{\sum }}{C}_{i}={q}^{3}+{q}^{2}+q+1-k$

ii) $\underset{\alpha }{\overset{\beta }{\sum }}i{C}_{i}=\frac{k\left(k-1\right)\cdots \left(k-n+1\right)}{n!}\left({q}^{2}+q+1-n\right)$

where α is the smallest i for which ${C}_{i}\ne 0$, β be the largest i for which ${C}_{i}\ne 0$ .

Proof

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

i) $\left\{Q|Q\in S\A\right\}$

ii) $\left\{\left(Q,\pi \right)/Q\in \pi \A,\pi \text{\hspace{0.17em}}\text{an}\text{\hspace{0.17em}}n\text{-secant}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}A\right\}$ for in (i) $\underset{\alpha }{\overset{\beta }{\sum }}{C}_{i}$ represents all points in the space which are not in A, then $\underset{\alpha }{\overset{\beta }{\sum }}{C}_{i}={q}^{3}+{q}^{2}+q+1-k$, and in (ii) $\underset{\alpha }{\overset{\beta }{\sum }}i{C}_{i}$ 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 $\left(\begin{array}{c}k\\ n\end{array}\right)$, then

$\begin{array}{c}\underset{\alpha }{\overset{\beta }{\sum }}i{C}_{i}=\left(\begin{array}{c}k\\ n\end{array}\right)\left({q}^{2}+q+1-n\right)\\ =\frac{k\left(k-1\right)\cdots \left(k-n+1\right)}{n!}\left({q}^{2}+q+1-n\right)\end{array}$

Theorem

If P is a point of a (k,n)-arc A in $PG\left(3,q\right)$, which lies on an m-secant (plane) of A, then the planes through P contains at most $\left(n-1\right)q\left(q+1\right)+m$ 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 $\left(n-1\right)\left({q}^{2}+q\right)+m$ points of A.

3. The Fundamental Theorem of Projective Geometry

In this section, the characterization of algebraic varieties in $PG\left(3,q\right)$ is given as finite sets of points with certain combinatorial properties.

Definition: “Plane π” [1]

A plane π in $PG\left(3,p\right)$ is the set of all points $P\left({X}_{1},{X}_{2},{X}_{3},{X}_{4}\right)$ satisfying a linear equation ${U}_{1}{X}_{1}+{U}_{2}{X}_{2}+{U}_{3}{X}_{3}+{U}_{4}{X}_{4}=0$ . This plane is denoted by $\pi \left[{U}_{1},{U}_{2},{U}_{3},{U}_{4}\right]$ . space which consists of points, lines and planes with the incidence relation between them [1] .

Notion [1]

A projective 3-space $PG\left(3,k\right)$ over a field K is a 3-dimensional projective $PG\left(3,k\right)$ satisfying the following axioms:

1) Any two distinct points are contained in a unique line.

2) Any three distinct non-collinear points, also any line and point not on the line are contained in a unique plane.

3) Any two distinct coplanar lines intersect in a unique point.

4) Any line not on a given plane intersects the plane in a uniquepoint.

5) Any two distinct planes intersect in a unique line.

A projective space $PG\left(3,p\right)$ over Galois field GF(p), where $p={q}^{m}$ For some prime number q and some integer m, is a 3-dimensional projective space.

Any point in $PG\left(3,p\right)$ has the form of a quadruple $\left({X}_{1},{X}_{2},{X}_{3},{X}_{4}\right)$, where ${X}_{1},{X}_{2},{X}_{3},{X}_{4}$ are elements in GF(p) with the exception of the quadrable consisting of four zero elements. Two quadrable $\left({X}_{1},{X}_{2},{X}_{3},{X}_{4}\right)$ and $\left({y}_{1},{y}_{2},{y}_{3},{y}_{4}\right)$ represent the same point if there exists λ in GF(p)/{0} such that $\left({X}_{1},{X}_{2},{X}_{3},{X}_{4}\right)=\lambda \left({y}_{1},{y}_{2},{y}_{3},{y}_{4}\right)$ . Similarly, any plane in $PG\left(\text{3},p\right)$ has the form of a quadrable $\left[{X}_{1},{X}_{2},{X}_{3},{X}_{4}\right]$, where ${X}_{1},{X}_{2},{X}_{3},{X}_{4}$, are elements in GF(p) with the exception of the quadrable consisting off our zero elements.

Two quadrable $\left[{X}_{1},{X}_{2},{X}_{3},{X}_{4}\right]$ and $\left[{y}_{1},{y}_{2},{y}_{3},{y}_{4}\right]$ represent the same plane if there exists λ in GF(p)\{0} such that $\left[{X}_{1},{X}_{2},{X}_{3},{X}_{4}\right]=\lambda \left[{y}_{1},{y}_{2},{y}_{3},{y}_{4}\right]$ .

Finally, a point $P\left({X}_{1},{X}_{2},{X}_{3},{X}_{4}\right)$ is incident with the plane $\pi \left[{a}_{1},{a}_{2},{a}_{3},{a}_{4}\right]$ iff

${a}_{1}{X}_{1}+{a}_{2}{X}_{2}+{a}_{3}{X}_{3}+{a}_{4}{X}_{4}=0$ .

Theorem [1] [19]

The points of $PG\left(3,p\right)$ have a unique forms which are $\left(1,0,0,0\right),\left(x,1,0,0\right),\left(x,y,1,0\right),\left(x,y,z,1\right)$ for all $x,y,z$ in GF(p).

There exists one point of the form $\left(1,0,0,0\right)$ .

There exists p points of the form $\left(x,1,0,0\right)$ .

There exists ${p}^{2}$ points of the form $\left(x,y,1,0\right)$ .

There exists ${p}^{3}$ points of the form $\left(x,y,z,1\right)$ .

Theorem [19]

The planes of $PG\left(3,p\right)$ have a unique forms which are: $\left[1,0,0,0\right],\left[x,1,0,0\right],\left[x,y,1,0\right],\left[x,y,z,1\right]$ for all $x,y,z$ in GF(p).

There exists one plane of the form $\left[1,0,0,0\right]$ .

There exists p planes of the form $\left[x,1,0,0\right]$ .

There exists ${p}^{2}$ planes of the form $\left[x,y,1,0\right]$ .

There exists ${p}^{3}$ planes of the form $\left[x,y,z,1\right]$ .

Notion [1]

In $PG\left(3,p\right)$ satisfies the following:

1) Every line contains exactly $p+1$ points and every point is on exactly $p+1$ lines.

2) Every plane contains exactly $p²+p+1$ points (lines) and every point is on exactly ${p}^{2}+p+1$ planes.

3) There exist ${p}^{3}+{p}^{2}+p+1$ of points and there exists ${p}^{3}+{p}^{2}+p+1$ of planes.

4) Any two planes intersect in exactly $p+1$ points and any line is on exactly $P+1$ planes. So, any two points are on exactly $p+1$ planes.

Notion [1]

There exists $\left({p}^{2}+1\right)\left({p}^{2}+p+1\right)$ of lines in $PG\left(3,p\right)$ .

Proof

In $PG\left(3,p\right)$, there exist ${p}^{3}+{p}^{2}+p+1$ planes, and each plane contains exactly ${p}^{2}+p+1$ lines, then the numbers of lines is equal to $\left({p}^{3}+{p}^{2}+p+1\right)\left({p}^{2}+p+1\right)$, but each line is on $p+1$ planes, then there exist exactly $\frac{\left({p}^{3}+{p}^{2}+p+1\right)\left({p}^{2}+p+1\right)}{p+1}=\left({p}^{2}+1\right)\left({p}^{2}+p+1\right)$ lines in $PG\left(3,p\right)$ .

Qualifier [1] [19]

A (k,ƪ)-span, ƪ ≥ 1 is a set of k spaces πƪ no two of which intersect.

Definition [1]

A maximum (k,ƪ)-span is a set of k spaces πƪ which are every points of $PG\left(3,p\right)$ lies in exactly one line of the, and every two lines of πƪ are disjoint

Qualifier [1] [19]

Every maximum (k,ƪ)-span is a spread.

4. The Projective Space and the (k,ƪ)-Span in PG(3,7)

4.1. The Projective Space in PG(3,7)

$PG\left(3,7\right)$ 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 $PG\left(3,7\right)$ are given in Table 1 and Table 2.

Table 1. Points and plane of PG(3,7).

Table 2. Plane and lines of PG(3,7).

4.2. The (k,ƪ)-Span in PG(3,p)

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

In Table 3 any elements of the set ƫi = {ξ, ν, μ, $\cdots$, 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 (ƫi) and (Ѡ) is (50,ƪ)-span, which is the maximum (k,ƪ)-span of $PG\left(3,7\right)$ can be obtained. Thus Ѡ is called a Spread of fifty lines of $PG\left(3,7\right)$ which partitions $PG\left(3,7\right)$ ; that every point of $PG\left(3,7\right)$ lies in exactly one line of ƫi. and every line are disjoint. From the above results the number of the planes in the projective space

$PG\left(3,7\right)$ are 400 planes and each plane contains 57 lines, therefore the total number of the lines in $PG\left(3,7\right)$ 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 $PG\left(3,7\right)$, and called him a (50,ƪ)-span, i.e.

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

Moreover, we found that a (50,ƪ)-span is a maximum complete (k,ƪ)-span in $PG\left(3,7\right)$ .

5. Conclusions

1) Complete Arcs and Surfaces are constructed in $PG\left(3,q\right)$ over Galois field GF(q), $q=7$ by two methods and some theorems are proved on the (k,n)-arc of $PG\left(3,q\right)$ and on a 3-dimensional projective space.

2) We prove, the number of spread in projective space $PG\left(3,p\right)$ where p is prime, and $P\ge 2$ is ${p}^{2}+1$ .

Proof

In $PG\left(3,p\right)$, there are ${p}^{3}+{p}^{2}+p+1$ planes, but each line is on $p+1$ planes; then there is exactly $\frac{{p}^{3}+{p}^{2}+p+1}{p+1}={p}^{2}+1$ spread in $PG\left(3,p\right)$ .

Cite this paper: Abdulla, A. and Yahya, N. (2020) Complete Arcs and Surfaces in Three Dimensional Projective Space PG(3,7). Open Access Library Journal, 7, 1-15. doi: 10.4236/oalib.1106071.
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