1. Plan for the end of the course

Today, our main goal is to finish the proof of Pappus’ theorem, one of the classical theorems of incidence geometry. This proof will use several ideas from the last few weeks:

• affine and projective planes;
• homogeneous coordinates;
• transformations;
• normalization.

The final exam will focus only on the material covered after the midterm: incidence geometry, affine and projective planes, finite and infinite models, counting arguments, homogeneous coordinates, projective completion, and transformations.

2. Reminder: affine Pappus

Let \(\mathbb F\) be \(\mathbb Q\), \(\mathbb R\), \(\mathbb C\), or more generally any field. We want to prove Pappus’ theorem in the affine plane

Recall the statement. Let \(\ell \) and \(m\) be two distinct lines in \(\mathbb A^2(\mathbb F)\). Let

be three distinct points, and let

be three distinct points. Assume that none of these six points lies on \(\ell \cap m\). Assume also that the three intersection points

exist in the affine plane. Then

are collinear.

A direct affine proof is lengthy because there are cases to distinguish. The lines \(\ell \) and \(m\) may be parallel, or they may intersect. The point of passing to the projective plane is to remove this distinction.

3. From the affine plane to the projective plane

Recall the map

This map has two important properties:

• it sends affine lines to projective lines;
• points \(P,Q,R\in \mathbb A^2(\mathbb F)\) are collinear if and only if \(\varphi (P),\varphi (Q),\varphi (R)\) are collinear in \(\mathbb P^2(\mathbb F)\).

Thus, starting with a Pappus configuration in the affine plane, we may apply \(\varphi \) to everything. The affine lines \(\ell \) and \(m\) become projective lines

and the points \(A,B,C,A',B',C'\) become points

on \(\widetilde \ell \) and \(\widetilde m\).

If we can prove that

are collinear in \(\mathbb P^2(\mathbb F)\), then \(P,Q,R\) are collinear in \(\mathbb A^2(\mathbb F)\).

Let us justify this last implication directly. Suppose

belong to a projective line

Then

Therefore the affine points

all lie on the affine line

Hence \(P,Q,R\) are collinear.

4. Projective Pappus

We now prove the projective version.

Theorem (Projective Pappus). Let \(\ell \) and \(m\) be distinct lines in \(\mathbb P^2(\mathbb F)\). Let

be three distinct points, and let

be three distinct points. Assume that none of these six points is equal to

Define

Then \(P,Q,R\) are collinear.

Notice that in the projective plane, there is no need to assume that \(P,Q,R\) exist. Indeed, any two distinct projective lines meet.

There is also no need to distinguish the cases \(\ell \parallel m\) and \(\ell \cap m\neq \emptyset \), because in a projective plane any two distinct lines meet.

5. Why transformations help

The proof of Projective Pappus has two main steps.

(a)

Use a projective transformation to send the two lines \(\ell \) and \(m\) to the simple coordinate lines

\[ \{X=0\} \qquad \text {and}\qquad \{Y=0\}. \]

(b)

Prove Pappus’ theorem in this normalized configuration.

This is a powerful use of transformations. Instead of proving the theorem for two completely arbitrary lines, we transform the whole configuration to a simpler one. Since transformations send lines to lines and preserve collinearity, proving the theorem after transformation proves it before transformation.

More precisely, suppose \(F\) is a projective transformation. If

are collinear, then \(P,Q,R\) are collinear, because applying \(F^{-1}\) sends the line through \(F(P),F(Q),F(R)\) to a line through \(P,Q,R\).

6. Projective transformations from matrices

Let \(A\in GL_3(\mathbb F)\) be an invertible \(3\times 3\) matrix. It defines a map

by

This is well-defined because if two nonzero vectors are proportional, then their images under \(A\) are still proportional.

Since \(A\) is invertible, \(F_A\) is bijective. Also, \(F_A\) sends projective lines to projective lines. Indeed, projective lines in \(\mathbb P^2(\mathbb F)\) correspond to planes in \(\mathbb F^3\) passing through the origin, and an invertible linear map sends planes through the origin to planes through the origin.

Thus \(F_A\) is a transformation of \(\mathbb P^2(\mathbb F)\).

Example. Let

Then

So

7. Why translations do not work directly

In the affine plane, translations are natural transformations. In the projective plane, the situation is different.

Recall that points of \(\mathbb P^2(\mathbb F)\) correspond to lines in \(\mathbb F^3\) passing through the origin. If we translate all vectors in \(\mathbb F^3\) by a fixed nonzero vector, a line through the origin need not be sent to a line through the origin.

For example, let

Consider the translation

The line

passes through the origin. Its image under \(T_v\) is

which does not pass through the origin. Therefore this translation does not define a transformation of \(\mathbb P^2(\mathbb F)\).

8. The normalization step

We now prove the normalization step. Let \(\ell \) and \(m\) be distinct lines in \(\mathbb P^2(\mathbb F)\). Write

and

Let

Since \(\ell \) and \(m\) are distinct lines, the vectors \(L\) and \(M\) are linearly independent.

Choose a vector

such that \(L,M,N\) form a basis of \(\mathbb F^3\). Define

Then \(A\) is invertible.

We claim that the projective transformation \(F_A\) sends \(\ell \) to the line \(\{X=0\}\) and \(m\) to the line \(\{Y=0\}\).

Indeed, if \([X:Y:Z]\in \ell \), then

The first coordinate of

is \(aX+bY+cZ\), so the first coordinate of \(F_A([X:Y:Z])\) is zero. Hence

Conversely, since \(A\) is invertible, every point on \(\{X=0\}\) has a preimage, and the same computation applied to \(A^{-1}\) shows that this preimage lies on \(\ell \). Therefore

Similarly,

Thus it is enough to prove Projective Pappus in the normalized case

9. Beginning the normalized coordinate proof

Assume from now on that

First, we find the intersection point

Let

Since \(O\in \ell \), we have \(X=0\). Since \(O\in m\), we have \(Y=0\). Since \([X:Y:Z]\) is a projective point, the vector \((X,Y,Z)\) is not zero, so \(Z\neq 0\). Therefore

Now let \(A'\in m\). Write

Since \(A'\in m=\{Y=0\}\), we have \(Y=0\). We claim that \(X\neq 0\). If \(X=0\), then since \((X,Y,Z)\neq (0,0,0)\), we would have \(Z\neq 0\), and hence

But by hypothesis \(A'\neq O\). Therefore \(X\neq 0\). Dividing by \(X\), we can write

for some \(a'\in \mathbb F\).

The same argument gives

for some \(b',c'\in \mathbb F\). Since \(A',B',C'\) are distinct, the scalars

are distinct.

Similarly, because \(A,B,C\in \ell =\{X=0\}\) and none of them is equal to \(O\), we can write

where \(a,b,c\in \mathbb F\) are distinct.

This is the normalized coordinate setup for the rest of the proof.