8 Finite Geometry

A sub-geometry of $\mathcal{S}$ is a geometry ${\mathcal{S}}^{\prime }=({\mathcal{P}}^{\prime },{\mathcal{L}}^{\prime },{\mathrm{I}}^{\prime })$ with ${\mathcal{P}}^{\prime }\subseteq P$, ${\mathcal{L}}^{\prime }\subseteq \mathcal{L}$ and ${\mathrm{I}}^{\prime }=\mathrm{I}\cap ({\mathcal{P}}^{\prime }\times {\mathcal{L}}^{\prime })$.

Link to original

Definition

An incidence geometry is a pair $\mathcal{G}=(\Omega ,I)$ where $\Omega$ is a set and $I\subseteq \Omega \times \Omega$ is symmetric and transitive.

Definition

Let $\mathcal{G}=(\Omega ,I)$ be an incidence geometry. A (maximal) flag in $\mathcal{G}$ is a (maximal) chain with $\mathcal{G}$ seen as a poset.

Definition

A geometry $\mathcal{G}=(\Omega ,I)$ has rank $r$ if we can partition $\Omega$ into $r$ antichains ${\Omega }_1,\cdots ,{\Omega }_r$.

Definition

A projective plane is an incidence geometry of points and lines with the following properties:

• Every two points are incident with a unique common line.
• Every two lines are incident with a unique common point.
• There are four points, no three collinear.

Note that i) and ii) are dual, and iii) implies its dual.

Proposition

• Every point in a finite projective plane is incident with $(n+1)$ lines.
• Dually, every line is incident with $(n+1)$ points.

$n$ is called the order of the projective plane $\mathcal{G}=(\mathcal{P},\mathcal{L},I)$.

\begin{proof} There exist a point $P$ and a line $L$ such that $P/IL$. Note that the number of points incident with $L$ equals the number of lines incident with $P$. There also exists point $Q$ such that $Q/IL$, so the number of lines incident with $P$ equals the number of lines incident with $Q$.

The $P,Q,L$ are arbitrarily chosen, so we finish the proof. \end{proof}

Proposition

A finite projective plane of order $n$ has $n^2+n+1$ points and $n^2+n+1$ lines.

\begin{proof} Each point $x$ has valency $n+1$ and each line incident with $x$ has $n+1$ points including $x$. Then $|S|=(n+1)n+1=n^2+n+1$. \end{proof}

Prime Order Conjecture

For a prime $n$, $PG(2,n)$ is the only projective plane of order $n$.

Prime Power Conjecture

The order of a projective plane is necessarily a prime power.

Later we will investigate projective planes in higher dimension. There all are $PG(n,q)$, where points are $1$-spaces of $V(n+1,q)$, lines are $2$-spaces and so on.

Bruck-Ryser-Chowla

If there is a projective plane of order $n$ and $n\equiv 1(\mathrm{mod}4)$ of $n\equiv 2(\mathrm{mod}4)$, then $n$ is the sum of two squares.

\begin{proof} See A course in combinatorics - 1992 - van Lint, Wilson.pdf. \end{proof}

Remark. For example, a projective plane of order $10$ does not exist, proved by Clement Lam in 1989. Furthermore, the existence of projective plane of order $12$ is still open.

One More Construction

Definition

Let $n$ be a positive integer. A set $\mathcal{D}$ of positive integer is called difference set of order $n$ if

• $|\mathcal{D}|=n+1$
• Each number $m$ in $[n^2+n]$ has a unique representation $m\equiv d-d^{\prime }(\mathrm{mod}{n}^2+n+1)$ for $d,d^{\prime }\in \mathcal{D}$.

Examples.

• ${\mathcal{D}}_2=\{1,2,4\}$
• ${\mathcal{D}}_3=\{1,2,4,10\}$
• ${\mathcal{D}}_4=\{1,2,5,15,17\}$

Theorem

Let $\mathcal{D}$ be a different set of order $n⩾2$. Then the following is a projective plane of order $n$:

• the points are $0,1,\cdots ,n^2+n$;
• the lines are $\mathcal{D}+i$ with $i\in \{0,\cdots ,n^2+n\}$, where $\mathcal{D}+i=\{d+i:d\in \mathcal{D}\}$ module $n^2+n+1$.

Remark. The projective plane induced from ${\mathcal{D}}_2$ is Fano plane.

\begin{proof} i) Let $x,x^{\prime }$ be points. By definition of $\mathcal{D}$, there exist unique $d,d^{\prime }\in \mathcal{D}$ such that $x-x^{\prime }\equiv d-d^{\prime }(\mathrm{mod}{n}^2+n+1)$. Put $i=x-d$, then $i=x^{\prime }-d^{\prime }$ module $n^2+n+1$. Thus $\mathcal{D}+i$ is a line through $x$ and $x^{\prime }$. Conversely, if $x,x^{\prime }\in \mathcal{D}+i^{\ast }$, then $x=d_1+i^{\ast }$ and $x^{\prime }=d_2+i^{\ast }$. It deduces that $x-x^{\prime }=d_1-d_2(\mathrm{mod}{n}^2+n+1)$, and so $d_1=d$, $d_2=d^{\prime }$. Therefore, $i^{\ast }=i$ and the line through $x$ and $x^{\prime }$ is unique.

ii) Let $\mathcal{D}+i$ and $\mathcal{D}+j$ be distinct lines. Then $x\in (\mathcal{D}+i)\cap (\mathcal{D}+j)$ iff there exists $d,d^{\prime }\in \mathcal{D}$ such that $x\equiv d+i\equiv d^{\prime }+j(\mathrm{mod}{n}^2+n+1)$. Then $d,d^{\prime }$ are determined by $j-i$ and so $x$ is uniquely determined. \end{proof}

Ovals

Definition

An arc in a projective plane is a set of points such that no three points are collinear.

Lemma

Let $G$ be an arc of a projective plane of order $n$. Then $|G|⩽n+2$ and $|G|⩽n+1$ if $n$ is odd.

\begin{proof} Take $P\in G$, then there are $n+1$ lines through $P$ and each line has at most one point in $G$. Hence $|G|⩽1+n+1=n+2$.

If the equality holds, then no line intersects $G$ in $1$ point. Take point $Q\notin G$. Let $m$ denote number of secant lines through $Q$. Then $n+2=|G|=2m$. It is impossible for $n$ odd. \end{proof}

Definition

• An arc of size $n+1$ is called an oval.
• An arc of size $n+2$ is called hyperoval.
• Conic is anything isomorphic to

$$\begin{array}{rl}G& =\{⟨x,y,z⟩:x^2=yz\}\\ & =\{⟨(0,1,0)⟩\}\cup \{⟨(t,t^2,1)⟩:t\in \mathrm{GF}(q)\}.\end{array}$$

Remark that conics are ovals when $q$ is odd.

Proposition

The set ${\mathcal{O}}_i=\{⟨(0,0,1)⟩\}\cup \{⟨(1,t,t^{2^i})⟩:t\in \mathrm{GF}(2^h)\}$ is an oval in $PG(2,2^h)$ iff $\gcd (i,h)=1$.

\begin{proof} Every line incident with $⟨(0,0,1)⟩$ of the form $y=ax$ is incident with precisely one more point of ${\mathcal{O}}_i$, namely, $⟨(1,a,a^{2^i})⟩$. The line $x=0$ is a tangent. The other lines are of the form $z=ax+by$. The line $z=ax+by$ is incident with $⟨(1,t,t^{2^i})⟩$ iff $t^2=bt+a$. If $u^{2^i}=bu+a$ and $v^{2^i}=bv+a$, then $u^{2^i}-v^{2^i}=b(u-v)$, but also $u^{2^i}-v^{2^i}=(u-v{)}^{2^i}$. Thus, $b=(u-v{)}^{2^i-1}$.

As $\gcd (i,h)=1$, there exists $m,n$ such that $m(2^i-1)+n(2^h-1)=1$. Hence $b^m=(u-v{)}^{1-n(2^h-1)}=u-v$. So $u,v$ determine $b$ and we do not have further solutions. \end{proof}