Counting points in affine planes, projective planes, and the first idea of projective completion.

1. Counting points in affine planes

In this entire section, let

\[ \alpha =(\mathcal {P},\mathcal {L},\mathcal {R}) \]

be an affine plane.

Given a line \(\ell \in \mathcal {L}\), we recall the notation

\[ R(\ell ):=\{p\in \mathcal {P}: p\,I\,\ell \}. \]

This is the set of points incident with \(\ell \).

During last class and this week’s homework, we proved the following proposition.

Using the proposition above, we will prove the following theorem.

By axiom I3, there exists a point

\[ p\in \mathcal {P} \]
such that
\[ p\mathrel {\not \! I} \ell . \]

Choose a point \(q\in R(\ell )\). Let \(m\) be the unique line incident with \(p\) and \(q\). This line exists and is unique by axiom I1.

Since \(p\mathrel {\not \! I}\ell \), we have

\[ m\neq \ell . \]
Moreover, the point \(q\) is the unique intersection point of \(\ell \) and \(m\). Indeed, if there were another point \(q'\neq q\) incident with both \(\ell \) and \(m\), then \(\ell \) and \(m\) would be two distinct lines incident with both \(q\) and \(q'\), contradicting axiom I1.

Since \(\#R(\ell )=n\), write the points of \(\ell \) as

\[ R(\ell )=\{q,q_1,\dots ,q_{n-1}\}. \]

For each \(1\leq i\leq n-1\), define \(m_i\) to be the unique line parallel to \(m\) and passing through \(q_i\).

This is possible by the Euclidean parallel property, because

\[ q_i\mathrel {\not \! I} m. \]
Indeed, \(m\cap \ell =\{q\}\), and \(q_i\neq q\).

We now observe that

\[ R(m_i)\cap R(\ell )=\{q_i\}. \]
Indeed, \(q_i\in R(m_i)\cap R(\ell )\). Moreover, since \(m_i\parallel m\), we have \(m_i\neq \ell \), and \(m_i\) cannot meet \(\ell \) in any point other than \(q_i\).

Let \(i\neq j\). Suppose, for contradiction, that there exists a point \(x\in \mathcal {P}\) such that

\[ x\,I\,m_i \qquad \text {and}\qquad x\,I\,m_j. \]
Since \(m_i\parallel m\), we have \(x \mathrel {\not \! I} m\). Indeed, if \(x\,I\,m\), then \(m_i\) and \(m\) would meet at \(x\), contradicting \(m_i\parallel m\).

Therefore \(m_i\) and \(m_j\) are both lines through \(x\) parallel to \(m\). By the Euclidean parallel property, there is a unique line through \(x\) parallel to \(m\). Hence

\[ m_i=m_j. \]

But \(m_i\) and \(m_j\) meet \(\ell \) at the unique points \(q_i\) and \(q_j\), respectively. Hence

\[ q_i=q_j. \]
This contradicts the fact that \(q_i\) and \(q_j\) are distinct. Therefore the lines \(m_i\) and \(m_j\) do not meet.

Therefore the lines

\[ m,m_1,\dots ,m_{n-1} \]
are pairwise parallel.

By the proposition from last class, each of these lines has exactly \(n\) points. Therefore the union

\[ R(m)\cup R(m_1)\cup \cdots \cup R(m_{n-1}) \]
contains exactly \(n^2\) points.

We now prove that this union is all of \(\mathcal {P}\).

Suppose, for contradiction, that there exists a point \(x\in \mathcal {P}\) such that

\[ x\mathrel {\not \! I} m \]
and, for every \(1\leq i\leq n-1\),
\[ x\mathrel {\not \! I} m_i. \]

Let \(t\) be the unique line incident with \(q\) and \(x\). This line exists and is unique by axiom I1.

We claim first that

\[ t\neq m. \]
Indeed, if \(t=m\), then \(x\,I\,m\), contradicting the choice of \(x\).

We also claim that \(t\) is not parallel to \(m\). Indeed, both \(t\) and \(m\) are incident with \(q\), so they meet at \(q\).

Now fix \(1\leq i\leq n-1\). We claim that

\[ t\neq m_i \qquad \text {and}\qquad t \text { is not parallel to } m_i. \]

First, \(t\neq m_i\), because \(x\mathrel {\not \! I} m_i\), whereas \(x\,I\,t\).

Second, suppose for contradiction that

\[ t\parallel m_i. \]
Since \(m_i\parallel m\), both \(t\) and \(m\) would be lines through \(q\) parallel to \(m_i\). This contradicts the uniqueness part of the Euclidean parallel property, because \(t\neq m\).

Therefore \(t\) and \(m_i\) are distinct and not parallel. Hence they meet in a unique point. Call this point \(x_i\):

\[ x_i\,I\,t \qquad \text {and}\qquad x_i\,I\,m_i. \]

Thus the line \(t\) is incident with the points

\[ q,x_1,\dots ,x_{n-1},x. \]

These \(n+1\) points are all distinct. Indeed, the points \(x_i\) lie on the distinct parallel lines \(m_i\), the point \(q\) lies on \(m\), and \(x\) was chosen to lie on none of the lines

\[ m,m_1,\dots ,m_{n-1}. \]

Therefore \(t\) is incident with at least \(n+1\) points. But by the proposition from last class, every line is incident with exactly \(n\) points. This is a contradiction.

Hence no such point \(x\) exists, and therefore

\[ \mathcal {P} = R(m)\cup R(m_1)\cup \cdots \cup R(m_{n-1}). \]

Since the lines

\[ m,m_1,\dots ,m_{n-1} \]
are pairwise parallel and each contains exactly \(n\) points, this union contains exactly
\[ n\cdot n=n^2 \]
points.

Therefore

\[ \#\mathcal {P}=n^2. \]

2. Projective planes

Recall that the elliptic parallel property says that any two distinct lines meet. In other words, if

\[ \ell ,m\in \mathcal {L} \qquad \text {and}\qquad \ell \neq m, \]

then there exists a point \(p\in \mathcal {P}\) such that

\[ p\,I\,\ell \qquad \text {and}\qquad p\,I\,m. \]

Example. The Fano plane is a projective plane with \(7\) points and \(7\) lines.

3. Activity: A point avoiding two lines

Show that the following statement holds in any projective plane:

For every pair of distinct lines \(\ell ,m\in \mathcal {L}\), there exists a point \(P\in \mathcal {P}\) such that

\[ P\mathrel {\not \! I} \ell \qquad \text {and}\qquad P\mathrel {\not \! I} m. \]

In other words, show that in a projective plane, given two distinct lines, there exists a point incident with neither of these two lines.

Let \(\ell ,m\in \mathcal {L}\) be two distinct lines.

By the elliptic parallel property, there exists a point \(x\in \mathcal {P}\) such that

\[ x\,I\,\ell \qquad \text {and}\qquad x\,I\,m. \]

Since we are in a projective plane, every line is incident with at least three distinct points. Therefore, choose distinct points

\[ x,a,b\in R(\ell ) \]
and distinct points
\[ x,c,d\in R(m). \]

Observe that \(a,c,d\) are distinct. Indeed, if one of \(c\) or \(d\) were equal to \(a\), then the two distinct lines \(\ell \) and \(m\) would have two distinct points in common, namely \(x\) and \(a\). This would contradict axiom I1. Similarly, \(b,c,d\) are distinct.

By axiom I1, there exists a unique line \(n\) incident with \(a\) and \(c\).

We note that

\[ n\neq \ell . \]
Indeed, if \(n=\ell \), then \(c\,I\,\ell \). Since \(c\,I\,m\) and \(c\neq x\), the lines \(\ell \) and \(m\) would have two distinct points in common, contradicting axiom I1.

Similarly,

\[ n\neq m. \]
Indeed, if \(n=m\), then \(a\,I\,m\), again contradicting the fact that \(\ell \) and \(m\) are distinct and already meet at \(x\).

Since \(n\neq \ell \), the point \(b\) is not incident with \(n\). Otherwise \(n\) and \(\ell \) would both contain \(a\) and \(b\), so \(n=\ell \).

Similarly, since \(n\neq m\), the point \(d\) is not incident with \(n\).

Also,

\[ x\mathrel {\not \! I} n. \]
Indeed, if \(x\,I\,n\), then \(n\) and \(\ell \) would both contain \(x\) and \(a\), so \(n=\ell \), a contradiction.

The line \(n\) must be incident with at least three points. It is already incident with \(a\) and \(c\), so there exists a point

\[ P\in R(n) \]
such that
\[ P\neq a \qquad \text {and}\qquad P\neq c. \]

We claim that

\[ P\mathrel {\not \! I}\ell \qquad \text {and}\qquad P\mathrel {\not \! I} m. \]

Indeed, if \(P\,I\,\ell \), then the lines \(n\) and \(\ell \) would both contain \(a\) and \(P\). Since \(P\neq a\), axiom I1 would imply

\[ n=\ell , \]
a contradiction.

Similarly, if \(P\,I\,m\), then the lines \(n\) and \(m\) would both contain \(c\) and \(P\). Since \(P\neq c\), axiom I1 would imply

\[ n=m, \]
also a contradiction.

Thus

\[ P\mathrel {\not \! I}\ell \qquad \text {and}\qquad P\mathrel {\not \! I} m. \]

4. Activity: All projective lines have the same number of points

Let

\[ \pi =(\mathcal {P},\mathcal {L},\mathcal {R}) \]

be a projective plane.

Given a line \(\ell \in \mathcal {L}\), define

\[ R(\ell ):=\{P\in \mathcal {P}: P\,I\,\ell \}. \]

This is the set of points incident with \(\ell \).

Show that if \(\ell \) and \(m\) are distinct lines, then there is a bijective map
\[ f:R(\ell )\longrightarrow R(m). \]

In other words, show that all lines in a projective plane have the same number of points on them.

If \(\ell =m\), then the identity map

\[ \mathrm {id}:R(\ell )\longrightarrow R(m) \]
is a bijection.

So assume that

\[ \ell \neq m. \]

By the previous activity, there exists a point \(P\in \mathcal {P}\) such that

\[ P\mathrel {\not \! I}\ell \qquad \text {and}\qquad P\mathrel {\not \! I} m. \]

Let \(x\in R(\ell )\). By axiom I1, there exists a unique line \(\ell _{xP}\) incident with both \(x\) and \(P\):

\[ x\,I\,\ell _{xP} \qquad \text {and}\qquad P\,I\,\ell _{xP}. \]

Since \(P\mathrel {\not \! I} m\), the line \(\ell _{xP}\) is not equal to \(m\).

Thus, by the elliptic parallel property, \(\ell _{xP}\) and \(m\) meet in a unique point. Define

\[ f(x) := \text {the unique point in }R(\ell _{xP})\cap R(m). \]

Equivalently,

\[ f(x)=\ell _{xP}\cap m. \]

This defines a map

\[ f:R(\ell )\longrightarrow R(m). \]

You will prove in section that this map is bijective.

5. Projective completion

Goal

The goal is to construct a projective plane from an affine plane.

Let

\[ \alpha =(\mathcal {P},\mathcal {L},\mathcal {R}) \]

be an affine plane.

The idea is the following: think of train rails disappearing toward the horizon. Parallel lines do not meet inside the affine plane, but they appear to meet at a common point at infinity.

For example, every line parallel to a fixed line \(\ell \) should meet at one common point at infinity. Every line parallel to another line \(m\) should also meet at a common point at infinity, but a different one if

\[ m\not \parallel \ell . \]

Thus, in the projective completion, each parallel class of affine lines will give one new point at infinity.