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

Abstract

In $\mathrm {PG}(2,q)$, the projective plane over the field $\mathbf{F}_{q}$ of q elements, a (k, n)-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.

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

Notation 1.3

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

• (2) A (k, n)-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 (k, n)-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 (k, n)-arc is the $(n+1)$-tuple $(\rho _0,\rho _1, \ldots ,\rho _n)$.

2 New lower bound

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

Theorem 2.1

In $\mathrm {PG}(2,q),$ a complete (k, n)-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 (k, n)-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 $4\le q\le 23$

Table 2 Bounds for complete (k, 3)-arcs for $4\le q\le 16$

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 $4\le q\le 16$