Affine planes and the number of points on lines.

Affine Planes

In the previous lecture, we defined affine planes as incidence geometries satisfying the Euclidean parallel property.

Recall that, if \(P\in \mathcal {P}\) and \(\ell \in \mathcal {L}\), we write

\[ P\, I\, \ell \]

to mean that the point \(P\) is incident with the line \(\ell \).

1. Activity

(1)
Show that in an affine plane, the following property holds:

For every pair of distinct and non-parallel lines \(\ell \) and \(m\), there exists a unique point \(P\) incident with both lines.

In other words, if \(\ell \neq m\) and \(\ell \) is not parallel to \(m\), then there exists a unique point \(P\in \mathcal {P}\) such that

\[ P\,I\,\ell \qquad \text {and}\qquad P\,I\,m. \]
(2)
Let
\[ \alpha =(\mathcal {P},\mathcal {L},\mathcal {R}) \]
be an affine plane. Given a line \(\ell \in \mathcal {L}\), we denote
\[ R(\ell ):=\{P\in \mathcal {P}: P\,I\,\ell \}. \]

The goal is to show that, given any two lines \(\ell \) and \(m\), there exists a bijection

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

2. Intersections of non-parallel lines

We first prove the property from Activity 1.

Since \(\ell \) and \(m\) are not parallel, by definition there exists at least one point \(P\in \mathcal {P}\) such that

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

We now prove uniqueness. Suppose that \(P\) and \(Q\) are two points incident with both \(\ell \) and \(m\). If \(P\neq Q\), then by axiom I1, there is a unique line incident with both \(P\) and \(Q\).

But both \(\ell \) and \(m\) are incident with \(P\) and \(Q\). Therefore \(\ell =m\), contradicting the assumption that \(\ell \neq m\).

Hence \(P=Q\). Therefore the point of intersection is unique.

3. All lines have the same number of points

Let

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

be an affine plane. For a line \(\ell \), recall that

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

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

\[ \mathrm {id}:R(\ell )\longrightarrow R(m) \]
is a bijection. So assume from now on that
\[ \ell \neq m. \]

Step 1: Choose two special points.

We first show that there exist two distinct points \(a,b\in \mathcal {P}\) such that

\[ a\,I\,\ell ,\qquad a\not I m, \]
and
\[ b\not I \ell ,\qquad b\,I\,m. \]

By axiom I2, there exist two distinct points on \(\ell \). If both of these points were also on \(m\), then by axiom I1 the lines \(\ell \) and \(m\) would be equal, contradicting the assumption that \(\ell \neq m\). Therefore at least one of these two points lies on \(\ell \) but not on \(m\). Call this point \(a\).

Similarly, by axiom I2, there exist two distinct points on \(m\). If both of these points were also on \(\ell \), then by axiom I1 the lines \(m\) and \(\ell \) would be equal, again contradicting \(\ell \neq m\). Therefore at least one of these two points lies on \(m\) but not on \(\ell \). Call this point \(b\).

In particular, \(a\neq b\).

Step 2: Define an auxiliary line.

Let \(n\) be the unique line incident with both \(a\) and \(b\). This line exists and is unique by axiom I1.

We claim that

\[ n\neq \ell \qquad \text {and}\qquad n\neq m. \]

Indeed, if \(n=\ell \), then \(b\,I\,\ell \), contradicting the choice of \(b\). Similarly, if \(n=m\), then \(a\,I\,m\), contradicting the choice of \(a\).

Step 3: Intersections of \(n\) with \(\ell \) and \(m\).

Since \(n\) is incident with \(a\), and \(a\,I\,\ell \), the lines \(n\) and \(\ell \) meet at \(a\). Since \(n\neq \ell \), they meet in exactly one point by the first problem. Hence

\[ R(n)\cap R(\ell )=\{a\}. \]

Similarly,

\[ R(n)\cap R(m)=\{b\}. \]

Step 4: Construct the map.

Let \(x\in R(\ell )\).

If \(x=a\), we will define

\[ f(a)=b. \]

Now assume that

\[ x\in R(\ell ) \qquad \text {and}\qquad x\neq a. \]

Since \(x\in R(\ell )\), \(x\neq a\), and

\[ R(n)\cap R(\ell )=\{a\}, \]
we have
\[ x\not I n. \]
Therefore, by the Euclidean parallel property, there exists a unique line \(n_x\) such that
\[ x\,I\, n_x \qquad \text {and}\qquad n_x\parallel n. \]

We now show that \(m\) and \(n_x\) meet in a unique point. First, note that

\[ n_x\neq m. \]
Indeed, if \(n_x=m\), then \(m\parallel n\), because \(n_x\parallel n\). But this is impossible, since \(m\) and \(n\) meet at the point \(b\).

Next, we show that \(m\) is not parallel to \(n_x\). Suppose, for contradiction, that

\[ m\parallel n_x. \]
Since \(n_x\parallel n\), both \(m\) and \(n\) would be lines through the point \(b\) parallel to \(n_x\). Also, \(m\neq n\), by step 2. This contradicts the uniqueness part of the Euclidean parallel property.

Therefore \(m\) and \(n_x\) are distinct and not parallel. By the previous proposition, they meet in a unique point. Let \(y\in \mathcal {P}\) be this unique point:

\[ y\,I\,m \qquad \text {and}\qquad y\,I\,n_x. \]

We define

\[ f(x)=y. \]

Thus

\[ f:R(\ell )\longrightarrow R(m) \]
is given by
\[ f(x)= \begin{cases} b, & \text {if } x=a,\\ \text {the unique point in } R(m)\cap R(n_x), & \text {if } x\neq a. \end{cases} \]
Step 5: The constructed map \(f:R(\ell )\longrightarrow R(m)\) is bijective.

Next HW.