1. Finishing the integer line

Recall the model \(\mathcal M_{\mathbb Z}\). Its points are the integers,

and its lines are the subsets

Last time, we introduced the translations

and the reflection

Each \(T_\ell \) is a transformation, and \(s\) is also a transformation. Since transformations form a group under composition, every map of the form

is a transformation.

The goal is to prove that these are all the transformations of \(\mathcal M_{\mathbb Z}\).

2. Normalization

Let

Set

We want to simplify \(f\) by composing it with transformations that we already understand. Define

Then

Since \(\{0,1\}\) is a line and \(g\) is a transformation,

is a line. The only lines containing \(0\) are \(\{-1,0\}\) and \(\{0,1\}\). Hence

Choose \(S\in \{\operatorname {id},s\}\) as follows:

Define

Then

This is the normalization step: starting from an arbitrary transformation \(f\), we constructed a simpler transformation \(h\) fixing both \(0\) and \(1\).

3. The normalized transformation is the identity

We now prove that if \(h\in \operatorname {Aut}(\mathcal M_{\mathbb Z})\), \(h(0)=0\), and \(h(1)=1\), then

From last time, we already proved by strong induction that

for every \(k\geq 0\). It remains to prove the same statement for negative integers.

First, since \(\{-1,0\}\) is a line,

is a line. Thus \(h(-1)=1\) or \(h(-1)=-1\). But \(h(1)=1\), and \(h\) is injective. Since \(-1\neq 1\), we cannot have \(h(-1)=1\). Therefore

We now prove by strong induction that

for every \(k\geq 0\). We already know this for \(k=0\) and \(k=1\).

Assume that \(n\geq 1\), and suppose that

We prove that

Since \(\{-(n+1),-n\}\) is a line, its image is a line:

Thus \(h(-(n+1))\) is adjacent to \(-n\). Hence

But \(h(-(n-1))=-(n-1)\) by the induction hypothesis, and \(h\) is injective. Since \(-(n+1)\neq -(n-1)\), we cannot have

Therefore

By strong induction,

for every \(k\geq 0\). Combining this with the result for nonnegative integers, we get

for every \(k\in \mathbb Z\). Hence

4. Classification of the transformations

Recall that

Since \(h=\operatorname {id}\), we have

We now solve this equation for \(f\). Composing on the left by \(S^{-1}\), we get

Composing on the left by \(T_L\), we get

Now \(S\) is either \(\operatorname {id}\) or \(s\). In both cases,

Indeed, \(\operatorname {id}^{-1}=\operatorname {id}\), and

so \(s^{-1}=s\). Therefore

Thus the transformations of \(\mathcal M_{\mathbb Z}\) are exactly the maps

5. Pappus’ theorem in the affine plane

We now turn to a classical theorem of incidence geometry.

Let \(\mathbb F\) be \(\mathbb Q\), \(\mathbb R\), \(\mathbb C\), or more generally any field. We work in the affine plane

Theorem (Affine Pappus). 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 the following non-degeneracy conditions:

• none of the six points \(A,B,C,A',B',C'\) lies on \(\ell \cap m\);

• the three intersection points

  \[ P=AB'\cap A'B,\qquad Q=AC'\cap A'C,\qquad R=BC'\cap B'C \]
  exist in the affine plane.

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

The condition that \(P,Q,R\) exist is important in the affine plane. For example, it may happen that the lines \(AB'\) and \(A'B\) are parallel. In that case, the point

does not exist in the affine plane.

However, if \(AB'\) and \(A'B\) are parallel, then they meet at a point at infinity in the projective completion. This suggests that the theorem should become cleaner in the projective plane.

6. Why pass to the projective plane?

There are two main reasons to prove Pappus’ theorem projectively.

First, in the affine plane, there are different cases. The two lines \(\ell \) and \(m\) may be parallel or they may meet. A direct affine proof has to treat these cases separately.

Second, some of the intersection points \(P,Q,R\) may fail to exist in the affine plane because the corresponding lines are parallel. In the projective plane, distinct lines always meet.

Thus the projective plane removes both difficulties: there is no parallel case, and the missing affine intersections become points at infinity.

7. From affine to projective

Recall the standard embedding

This map lets us view the affine plane as the part of the projective plane where the last homogeneous coordinate is nonzero.

A key property is the following.

Collinearity is preserved. For points \(P,Q,R\in \mathbb A^2(\mathbb F)\),

if and only if

Indeed, an affine line has an equation

and its projective completion has equation

If \((x,y)\) lies on the affine line, then \([x:y:1]\) lies on the corresponding projective line. Conversely, a projective line containing points with \(Z=1\) gives an affine line after setting \(Z=1\).

Therefore, to prove affine Pappus, it is enough to prove a projective version of Pappus.

8. Pappus’ theorem in the projective plane

Theorem (Projective Pappus). Let \(\ell \) and \(m\) be two 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 the point

Define

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

In the projective plane, the points \(P,Q,R\) always exist, because any two distinct projective lines meet. Thus the theorem has a cleaner statement than in the affine plane.