Embeded Plane

Definition: Embedded plane

If $\sigma=(\mathscr{P,L,I})$ and $\sigma'=(\mathscr{P',L',I'})$ are two planes then we say $\sigma$ is embedded to $\sigma'$ if

1. $\mathscr{P} \in \mathscr{P'}$
2. For every line $L \in \mathscr{L}$ there exists a line $L' \in \mathscr{L'}$ such that $L =L' \cap \mathscr{P}$

Definition: Subplane

If $\sigma=(\mathscr{P,L,I})$ and $\sigma'=(\mathscr{P',L',I'})$ are two planes then we say $\sigma$ is a subplane of $\sigma'$ if

1. $\sigma$ is embedded to $\sigma'$.
2. For every line $L' \in \mathscr{L'}$ and if $L' \cap \mathscr{P}$ contains two points then there exists a line $L \in \mathscr{L}$ such that $L =L' \cap \mathscr{P}$

Definition: Principal subplane

If $\sigma=(\mathscr{P,L,I})$ and $\sigma'=(\mathscr{P',L',I'})$ are two planes then we say $\sigma$ is a principal subplane of $\sigma'$ if

1. $\sigma$ is a subplane of $\sigma'$.
2. $\mathscr{P}=\mathscr{P'}/L'$ for some line $L' \in \mathscr{L'}$

Note: Principal subplane

A subplane $\sigma$ can be a principal subplane of $\sigma$ if and only if the plane differs by exactly one line and all belonging points on it.

Example:

A complete four-point is a principal subplane of fano-configuration because the configurations are differ by exactly a line and all belonging points on this line.

Theorem: Any principal sub-plane of a projective plane is an affine plane.

Proof:

Let $\sigma'=(\mathscr{P',L',I'})$ be a projective plane. Now we construct:

• $\mathscr{P}=\mathscr{P'}/D$ for some $D \in \mathscr{L'}$
• $M=\{p\in L' ∶ p \notin D\}$ for all $L\in \mathscr{L'}/D$
• $L=\{L:L' \in \mathscr{L'}/D\}$

Then $\sigma=(\mathscr{P,L,I})$ is a principal sub-plane of $\sigma'$ obtained by removing $D \in \mathscr{L'}$. Now we show that $\sigma$ is an affine plane.

1. Two points in $\sigma$ determine a line:
   Let $p,q\in \sigma$. By construction, $p,q\in \sigma'$. By [P1], $p$ and $q$ determine a line $L' \in \sigma'$. By construction, $p$ and $q$ determine a line $L \in \sigma$.
2. If $L$ is a line and $p \notin L$ is a point, there is exactly one line on $p$ parallel to $L$:
   Let $L$ is a line and $p \notin L$ be a point. By construction $L' \in \sigma'$ is a line not containing the point $p$. Since $D$ is already a line in $\sigma'$, by [P2], we say $D \cap L'=q$. By [P1], $p$ and $q$ determine a line, say $M' \in \sigma'$. By construction, $M'$ corresponds to a line $M \in \sigma$ not containing the point $q$. Hence, $M$ is the required line on $p$ parallel to $L$.
3. There is a set of three non-collinear points:
   Since $\sigma'$ is a projective plane, there is a four-point, say $a_0 a_1 a_2 a_3$. Now three cases arise:
   1. Case 1: If $D$ contains no points from $a_0, a_1, a_2, a_3$, then $\sigma$ contains the same four-point.
   2. Case 2: If $D$ contains one of the points from $a_0, a_1, a_2, a_3$, then the remaining three points are not on $D$ and so are in $\sigma$. Hence $\sigma$ contains three non-collinear points.
   3. Case 3: If $D$ contains any two points from $a_0, a_1, a_2, a_3$ say $a_0, a_1$, then by construction $a_2 a_3$ are not on $D$ and so are in $\sigma$. Also, the point $a_0 a_2 \cap a_1 a_3$ is not on $D$ and so in $\sigma$. Thus $\sigma$ contains three non-collinear points $a_2 a_3, a_0 a_2 \cap a_1 a_3$.

Thus $\sigma$ is an affine plane. This completes the proof.

Theorem: Show that every affine plane is a principal sub-plane of some projective plane.

Proof:

Let $\sigma=(\mathscr{P,L,I})$ be an affine plane. By [A2], for any line $L \in \sigma$ and a point $p \notin L$, there is exactly one line on $p$ parallel to $L$. Let such a parallel line on $p$ meet $L$ at an ideal point, say $p_L$. Now we construct:

• $L'=L \cup p_L$ for each $L$
• $L_\infty=\{p_L:L \in \mathscr{L}\}$
• $\mathscr{P'}= \mathscr{P} \cup L_\infty$ and
• $\mathscr{L'}=\{L':L \in \mathscr{L}\} \cup L_\infty$

Then $\sigma'=(\mathscr{P',L',I'})$ is an extended plane of $\sigma=(\mathscr{P,L,I})$ so as $\sigma=(\mathscr{P,L,I})$ is a principal subplane of $\sigma'=(\mathscr{P',L',I'})$. Now we show that $\sigma'=(\mathscr{P',L',I'})$ is a projective plane.

1. Two points on $\sigma'$ determine a line:
   Let $p$ and $q$ be two points in $\sigma'$
   1. Case 1: If $p,q \in \sigma$, by [A1], they determine a line in $\sigma$, say $L \in \sigma$. By construction, $p$ and $q$ determine a line $L' \in \sigma'$.
   2. Case 2: If any one point, say $q \notin \sigma$, by construction $q$ is an ideal point. By construction, there is a line on $p$ and $q$ in $\sigma'$.
   3. Case 3: If both points $p,q \notin \sigma$, then both $p$ and $q$ are ideal points so there is a line $pq=L_\infty \in \sigma'$.
   Hence two points in $\sigma'$ determine a line.
2. Two lines on $\sigma'$ always meet:
   Let $L',M' \in \sigma'$
   1. Case 1: If both $L'$ and $M'$ are ordinary lines in $\sigma'$. By construction $L, M \in \sigma$ and they do meet either in a finite point [if L and M do intersect] or in an ideal point [if L || M].
   2. Case 2: If any one line, say $L'$ is an ideal line in $\sigma'$. By construction they [$L'$ and $M'$] do meet at an ideal point.
   Hence two lines on $\sigma'$ always meet.
3. There is a four-point on $\sigma'$:
   Since $\sigma$ contains a four-point, $\sigma'$ contains the same four-point.

Hence $\sigma'=(\mathscr{P',L',I'})$ is a projective plane. This completes the proof.

Theorem: Show that $\alpha_R$ is a principal subplane of $\pi_R$.

Proof:

Let us remove a line $⟨0,0,1⟩$ and all its belonging points $[x,y,0]$ from $\pi_R$. Then all points in $\pi_R$ can be written as $[x,y,1]$ and all lines in $\pi_R$ are of the form $⟨a,b,c⟩$ with the property that $a,b$ both are not zero.

Now we define:

• $f(x,y) \to [x,y,1]$
• $F(ax+by+c=0) \to ⟨a,b,c⟩$

Which are both one-one onto.

This completes the proof.

Order of plane

Let $\sigma$ be a finite affine plane. The **order of $\sigma$** is the number of points on any line in $\sigma$. It is denoted by $O(\sigma)$. Since any affine plane $\sigma$ has the form
$\sigma: (n^2_{n+1}, n^2+n_n)$ for some $n \ge 2$
The order of an affine plane with this form is
$O(\sigma) = n$

Let $\pi$ be a finite projective plane. The **order of $\pi$** is the order of any principal sub-plane of $\pi$. It is denoted by $O(\pi)$.

Prove that the order of a finite projective plane $\pi$ with the form $(n^2+n+1_{n+1})$ for $n \ge 2$ is $n$.

Proof

Let $\sigma$ be a finite affine plane with the form $(n^2_{n+1}, n^2+n_n)$ for $n \ge 2$. By definition,
$O(\sigma) = n$
Now, we extend $\sigma$ to a projective plane $\pi$ so that $\sigma$ is a principal sub-plane of $\pi$. Then $\pi$ is a finite projective plane, in which $\pi$ has $n^2+n+1$ lines, each on $n+1$ points. Since every projective plane has duality, it has the form
$(n^2+n+1_{n+1})$ for $n \ge 2$
Hence, a projective plane with the form $(n^2+n+1_{n+1})$ for $n \ge 2$ has order $n$.

Note: Order

If a projective plane has order $n$, then it has the form $(n^2+n+1_{n+1})$ for $n \ge 2$ and conversely.

Example

The order of the fano configuration is 2.

If $F$ is a finite field with $q$ elements, then $O(\pi_F) = q$.

Proof

Let $F$ be a finite field with $q$ elements. Then $\pi_F$ is a finite projective plane. The points in $\pi_F$ are denoted by triples $[x_1, x_2, x_3]$ for $x_1, x_2, x_3$ not all zero, and proportionality is preserved. The first element of the triples can be chosen in $q$ different ways, and similarly for the second and third elements.

Since $[0,0,0]$ cannot be a point in $\pi_F$, there is a total of
$ \frac{q^3 - 1}{q - 1} = q^2 + q + 1 $
number of non-proportional and not all zero triples in $\pi_F$. Thus,
$O(\pi_F) = q$
This completes the proof.

Let $\pi = (\mathscr{P}, \mathscr{L}, \mathscr{I})$ be a projective plane of order $n$. Then the number of points/lines in it is $n^2+n+1$.

Proof

Let $\pi = (\mathscr{P}, \mathscr{L}, \mathscr{I})$ be a projective plane of order $n$. If $p$ is any point of the projective plane, then each of the $n+1$ lines through $p$ contains $n$ additional distinct points. Thus, we have altogether
$(n+1)n+1$ points.
By duality, this proves that the number of lines is also the same.
This completes the proof.