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A new lower bound for the smallest complete (kn)-arc in \(\mathrm {PG}(2,q)\)

  • S. Alabdullah
  • J. W. P. Hirschfeld
Open Access
Article
Part of the following topical collections:
  1. Special Issue: Coding and Cryptography
  2. Special Issue: Coding and Cryptography

Abstract

In \(\mathrm {PG}(2,q)\), the projective plane over the field \(\mathbf{F}_{q}\) of q elements, a (kn)-arc is a set \(\mathcal {K}\) of k points with at most n points on any line of the plane. A fundamental question is to determine the values of k for which \(\mathcal {K}\) is complete, that is, not contained in a \((k+1,n)\)-arc. In particular, what are the smallest and largest values of k for a complete \(\mathcal {K}\), denoted by \(t_n(2,q)\) and \(m_n(2,q)\)? Here, a new lower bound for \(t_n(2,q)\) is established and compared to known values for small q.

Keywords

Finite projective plane Arc Lower bound 

Mathematics Subject Classification

51E20 

1 Introduction and background

A projective plane of order q consists of a set of \(q^2+q+1\) points and a set of \(q^2+q+1\) lines, where each line contains exactly \(q+1\) points and two distinct points lie on exactly one line. It follows from the definition that each point is contained in exactly \(q+1\) lines and two distinct lines have exactly one common point.

The main focus of this paper is to find a lower bound for k of a (kn)-arc in \(\mathrm {PG}(2,q)\). First, some basic constants and their properties are summarised. See [8, Chap. 12] or [7, Chap. 12].

Definition 1.1

A (kn)-arc in \(\mathrm {PG}(2,q)\) is a set \(\mathcal {K}\) of k points, no \(n+1\) of which are collinear, but with at least one set of n points collinear. When \(n =2\), a (k, 2)-arc is a k-arc.

Definition 1.2

A (kn)-arc is complete if it is not contained in a \((k,n+1)\)-arc.

Notation 1.3

The maximum value of k for a (kn)-arc to exist is denoted by \(m_n(2,q)\).

Definition 1.4

A line \(\ell \) is an i-secant of \({\mathcal {K}}\) if \(|\ell \cap {\mathcal {K}}| =i\).

Notation 1.5

For a (kn)-arc \(\mathcal {K}\) in \(\mathrm {PG}(2,q),\) let
$$\begin{aligned} \tau _i= & {} \hbox {the total number of { i}-secants of} \mathcal {K},\\ \rho _i=\rho _i(P)= & {} \hbox {the number of { i}-secants through a point { P} of} \mathcal {K},\\ \sigma _i=\sigma _i(Q)= & {} \hbox {the number of} { i}-\hbox {secants through a point} { Q} \hbox {of } \mathrm {PG}(2,q)\backslash \mathcal {K}. \end{aligned}$$

Lemma 1.6

For a (kn)-arc \(\mathcal {K},\) the following equations hold : 
$$\begin{aligned} \sum _{i=0}^{n} \tau _i= & {} q^2+q+1; \end{aligned}$$
(1.1)
$$\begin{aligned} \sum _{i=1}^{n} i\tau _i= & {} k(q+1); \end{aligned}$$
(1.2)
$$\begin{aligned} \sum _{i=2}^{n} {\textstyle \frac{1}{2}}i(i-1)\tau _i= & {} {\textstyle \frac{1}{2}}k(k-1). \end{aligned}$$
(1.3)

Proof

See [8, Chap. 12]. \(\square \)

The constants \(\rho _i,\,\sigma _i\) are useful in investigations of the properties of (kn)-arcs, but are not required here.

Theorem 1.7

$$\begin{aligned} m_2(2,q)= {\left\{ \begin{array}{ll} q+2, &{} {for}\, q\, {even};\\ q+1, &{} {for}\, q\, {odd}. \end{array}\right. } \end{aligned}$$

Proof

See [8, Chap. 8]. \(\square \)

Theorem 1.8

  1. (1)
    $$\begin{aligned} m_n(2,q) {\left\{ \begin{array}{ll} =(n-1)q+n, &{} {for}\, q\, {even}\, {and}\, n\mid \, q;\\ <(n-1)q+n, &{} {for}\, q \,{odd}. \end{array}\right. } \end{aligned}$$
     
  2. (2)

    A (kn)-arc \(\mathcal {K}\) is maximal if and only if every line in \(\mathrm {PG}(2,q)\) is either an n-secant or a 0-secant.

     

Proof

See [8, Chap. 12]. \(\square \)

Lemma 1.9

If \(\mathcal {K}\) is a complete (kn)-arc,  then \((q+1 - n)\tau _n \ge q^2+q+1-k,\) with equality if and only if \(\sigma _n=1\) for all Q in \(\mathrm {PG}(2,q)\backslash \mathcal {K}\).

Proof

See [8, Chap. 12]. \(\square \)

Definition 1.10

The type of a point P in \(\mathrm {PG}(2,q)\) for a (kn)-arc is the \((n+1)\)-tuple \((\rho _0,\rho _1, \ldots ,\rho _n)\).

2 New lower bound

A lower bound for the smallest complete (kn)-arcs \(\mathcal {K}\) is established below.

Theorem 2.1

In \(\mathrm {PG}(2,q),\) a complete (kn)-arc does not exist for \(k \le n^*,\) where
$$\begin{aligned} n^*= \frac{(q+1-n^2)+\sqrt{(q+1-n^2)^2+4(n^2-n)(q+1-n)(q^2+q+1)}}{2(q+1-n)}. \end{aligned}$$

Proof

Let \(\mathcal {K}\) be a complete (kn)-arc. The number of n-secants through a point P in \(\mathcal {K}\) is at most \((k-1)/(n-1)\). Then, counting the set \(\{(P,\ell ) \}\), where \(\ell \) is an n-secant and P is a point of \(\mathcal {K}\) incident with \(\ell \) gives that
$$\begin{aligned} \tau _n \le \frac{k(k-1)}{n(n-1)}. \end{aligned}$$
(2.1)
On the other hand, Lemma 1.9 implies that
$$\begin{aligned} \tau _n \ge \frac{q^2+q+1-k}{q+1-n} \end{aligned}$$
(2.2)
Now, from Eqs. (2.1) and (2.2),
$$\begin{aligned} \frac{k^2-k}{n^2-n} = \frac{q^2+q+1-k}{q+1-n}. \end{aligned}$$
Hence
$$\begin{aligned}&(q+1-n)k^2-(q+1-n)k = (n^2-n)(q^2+q+1)-(n^2-n)k, \nonumber \\&(q+1-n)k^2-(q+1-n-n^2+n)k -(n^2-n)(q^2+q+1)= 0, \nonumber \\&(q+1-n)k^2-(q+1-n^2)k -(n^2-n)(q^2+q+1)= 0. \end{aligned}$$
(2.3)
Now, Eq. (2.3) implies that \(k= n^* > 0.\)\(\square \)
This can be applied to k-arcs and (k, 3)-arcs, as in Tables  1 and 2, with the notation \(n^* = b_n(2,q)\) and \(n=2,3\).
Table 1

Bounds for complete k-arcs for \(4\le q\le 23\)

q

4

5

7

8

9

11

13

16

17

19

23

\(b_2(2,q)\)

5

5

5

5

6

6

6

7

7

7

8

\(t_2(2,q)\)

6

6

6

6

6

7

8

9

10

10

10

\(m_2(2,q)\)

6

6

8

10

10

12

14

18

20

20

25

Table 2

Bounds for complete (k, 3)-arcs for \(4\le q\le 16\)

q

4

5

7

8

9

11

13

16

\(b_3(2,q)\)

7

8

9

9

9

10

11

12

\(t_3(2,q)\)

7

9

9

11

12

13

15

15

\(m_3(2,q)\)

9

11

15

15

17

21

23

28

3 Comparison with known results

Table 3 gives the comparison, for (k, 3)-arcs, between [9, 14] and Theorem 2.1 for the values of q with \(4\le q \le 16\).
Table 3

Lower bounds for complete (k, 3)-arcs for \(4\le q\le 16\)

k

[14]

[9]

Theorem

Exact result

4

7

6

7

7

5

8

6

8

9

7

9

7

9

9

8

9

8

9

11

9

10

8

10

12

11

10

9

10

13

13

11

10

11

15

16

12

11

12

15

Notes

Acknowledgements

Salam Alabdullah obtained a Ph.D. studentship funded by the Ministry of Higher Education and Scientific Research of the Government of Iraq via the University of Basra.

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Authors and Affiliations

  1. 1.Department of MathematicsUniversity of SussexBrightonUK

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