9 Designs 9.3 Finite linear spaces 9.5 Blocking Set

9.4 Projective Planes

A projective plane are linear spaces ${\mathcal{L}}$ such that every line contains exactly $q+1$ points. Equivalently, ${\mathcal{L}}$ is a projective plane if it is a symmetric Steiner design with $k=q+1$ . Hence the cardinality of points $v=q^{2}+q+1$ .

If we were to define afresh, ${\mathcal{L}}$ is a projective plane of order $q$ , if ${\mathcal{L}}\subset\mathcal{P}$ with $|\mathcal{P}|=q^{2}+q+1$ such that every line (elements of ${\mathcal{L}}$ ) contains $q+1$ points (elements of $X$ ) and any two points lie on a unique line.

As seen before triangle is the projective plane of order $1$ and Fano plane is the projective plane of order $2$ .

Proposition 9.11.

1.

Any point lies on $q+1$ lines i.e., the replication number is $q$ .
2.

There are $q^{2}+q+1$ lines.
3.

Any two lines meet at a unique point.

Proof.

By definition, projective plane is a $2-(v,q+1,1)$ design for $v=q^{2}+q+1$ . Thus we can check that it is a symmetric Steiner design and so the properties follow. For example, $r=k$ and so first two items follows. The third item follows from the equality case of linear spaces. ∎

Construction via $\mathbb{F}_{q}$ .

Suppose $q\geq 2$ is a prime power. Let $V\subset\mathbb{F}_{q}^{3}$ such that not all the elements (say $x_{0},x_{1},x_{2}$ ) are $0$ . Identify vectors which are non-trivial multiples i.e.,

[x_{0},x_{1},x_{2}]=\{(cx_{0},cx_{1},cx_{2}):x\in\mathbb{F}_{q}\setminus\{0\}\}.

Thus each $[x_{0},x_{1},x_{2}]$ is a set of $q-1$ vectors and since $V$ consists of $q^{3}-1$ elements, there are $q^{2}+q+1$ equivalence classes in general. These set of equivalence classes are our points i.e., the set $\mathcal{P}$ . For each $(a_{0},a_{1},a_{2})\in V$ , define the line $L(a_{0},a_{1},a_{2})$ as $[x_{0},x_{1},x_{2}]\in\mathcal{P}$ such that

a_{0}x_{0}+a_{1}x_{1}+a_{2}x_{2}=0.

(9.1)

Since $L(a_{0},a_{1},a_{2})=L(ca_{0},ca_{1},ca_{2})$ for any $c\neq 0,c\in\mathbb{F}_{q}$ , there are at most $q^{2}+q+1$ such lines. But if we show each line contains $q+1$ points and any two points are in a unique line, this will prove there are $q^{2}+q+1$ lines.

Firstly, we can assume $a_{0}\neq 0$ WLOG. Thus, for any $x_{1},x_{2}$ (both not zero), we can uniquely determine $x_{0}$ satisfying (9.1). Thus there are $q^{2}-1$ solutions $(x_{0},x_{1},x_{2})\in V$ satisfying (9.1). Since $[x_{0},x_{1},x_{2}]$ consists of $q-1$ vectors and each of them satisfies (9.1) if one of them does, there are at most $q+1$ solutions in $\mathcal{P}$ to (9.1). Thus there are at most $q+1$ points in a line.

Let $[x_{0},x_{1},x_{2}]$ and $[y_{0},y_{1},y_{2}]$ be distinct points. If a line $L(a_{0},a_{1},a_{2})$ contains both the points and say $x_{0}\neq 0$ (WLOG), then

a_{0}=-a_{1}x_{1}/x_{0}-a_{2}x_{2}/x_{0}.

So,

a_{1}(y_{1}-y_{0}x_{1}/x_{0})+a_{2}(y_{2}-y_{0}x_{2}/x_{0})=0.

If $y_{1}-y_{0}x_{1}/x_{0}=y_{2}-y_{0}x_{2}/x_{0}=0$ , then $(y_{0},y_{1},y_{2})=c(x_{0},x_{1},x_{2})$ for $c=y_{0}/x_{0}$ and necessarily $c\neq 0$ . This contradicts $[x_{0},x_{1},x_{2}]$ and $[y_{0},y_{1},y_{2}]$ are distinct points. Thus WLOG $y_{1}-y_{0}x_{1}/x_{0}\neq 0$ . Then for any $a_{2}\in\mathbb{F}_{q}$ , the above equation determines uniquely $a_{1}$ and hence $a_{0}$ by the previous equation. Also observe that $a_{0},a_{1},a_{2}$ determined by different representatives of $[y_{0},y_{1},y_{2}]$ and $[x_{0},x_{1},x_{2}]$ differ only by a constant.

The above constructed projective plane is denoted as $PG(2,q)$ .

9.4 Projective Planes

Proposition 9.11.

Proof.

Construction via 𝔽q.

Construction via $\mathbb{F}_{q}$ .