Is Richter-Gebert's Theorem: "A projective transformation maps collinear points to collinear points." really a theorem?

Question

Perspectives on Projective Geometry: A Guided Tour Through Real and Complex Geometry, by Jürgen Richter-Gebert.

Definition 2.1. A projective plane is a triple $\left(\mathcal{P},\mathcal{L}, \mathcal{I}\right)$. The set $\mathcal{P}$ consists of the points, and the set $\mathcal{L}$ consists of the lines of the geometry. The inclusion $\mathcal{I}\subseteq \mathcal{P}\times\mathcal{L}$ is an incidence relation satisfying the following three axioms:

• (i) For any two distinct points, there is exactly one line incident with both of them.
• (ii) For any two distinct lines, there is exactly one point incident with both of them.
• (iii) There are four distinct points such that no line is incident with more than two of them.

The only requirement for a projective transformation is that it is the mapping produced by a non-singular $3\times3$ matrix, and:

We obtain the group structure by requiring that collections of transformations be closed under reasonable operations.

Theorem 3.2. A projective transformation maps collinear points to collinear points.

Proof. It suffices to prove the theorem for a generic triple of points. Let $\left[a\right],\left[b\right],\left[c\right]\in\mathcal{P}_{\mathbb{R}}$ be three collinear points represented by homogeneous coordinates $a,b,c.$ In this case there exists a line $\left[l\right]\in\mathcal{L}_{\mathbb{R}}$ with $\left\langle l,a\right\rangle =\left\langle l,b\right\rangle =\left\langle l,c\right\rangle .$ We assume that all homogeneous coordinates are represented by column vectors. We have to show that under these conditions the points represented by $a^{\prime}=M\cdot a,b^{\prime}=M\cdot b,c^{\prime}=M\cdot c$, are also collinear. For this simply consider the line $\left[l^{\prime}\right]$ represented by $l^{\prime}:=\left(M^{-1}\right)^{T}l.$ We have $$ \left\langle l^{\prime},a^{\prime}\right\rangle =\left(L^{\prime}\right)^{T}a^{\prime}=\left(\left(M^{-1}\right)^{T}l\right)^{T}Ma=l^{T}\left(\left(M^{-1}\right)^{T}\right)^{T}Ma=l^{T}a=\left\langle l,a\right\rangle =0. $$ A similar calculation applies also to the other two points. Thus the line represented by $l^{\prime}$ is simultaneously incident to all three points represented by $a^{\prime},b^{\prime}$ and, $c^{\prime}.$ Hence these points are collinear.

I write this using classical tensor notation, treating the lines as (covariant) linear forms, and the points as (contravariant) coordinates as follows:

\begin{align*} \mathfrak{L}= & \begin{bmatrix}L_{1} & L_{2} & L_{3}\end{bmatrix}\\ \mathfrak{P}_{\iota}= & \begin{bmatrix}P_{\iota}^{1}\\ P_{\iota}^{2}\\ P_{\iota}^{3} \end{bmatrix}\\ 0= & \mathfrak{L}\mathfrak{P}_{\iota}=\left\{ L_{i}P_{\iota}^{i}\right\} \\ \overline{\mathfrak{T}}= & \begin{bmatrix}T^{\bar{1}}{}_{1} & T^{\bar{1}}{}_{2} & T^{\bar{1}}{}_{3}\\ T^{\bar{2}}{}_{1} & T^{\bar{2}}{}_{2} & T^{\bar{2}}{}_{3}\\ T^{\bar{3}}{}_{1} & T^{\bar{3}}{}_{2} & T^{\bar{3}}{}_{3} \end{bmatrix}\backepsilon\text{det}\overline{\mathfrak{T}}\ne0\\ \overline{\mathfrak{P}_{\iota}}= & \overline{\mathfrak{T}}\mathfrak{P}_{\iota}=\left\{ T^{\bar{i}}{}_{i}P_{\iota}^{i}\right\} \\ \underline{\mathfrak{T}}= & \begin{bmatrix}T^{1}{}_{\bar{1}} & T^{1}{}_{\bar{2}} & T^{1}{}_{\bar{3}}\\ T^{2}{}_{\bar{1}} & T^{2}{}_{\bar{2}} & T^{2}{}_{\bar{3}}\\ T^{3}{}_{\bar{1}} & T^{3}{}_{\bar{2}} & T^{3}{}_{\bar{3}} \end{bmatrix}=\overline{\mathfrak{T}}^{-1}\\ \underline{\mathfrak{L}}= & \mathfrak{L}\underline{\mathfrak{T}}=\left\{ L_{i}T^{i}{}_{\bar{i}}\right\} \\ 0= & \underline{\mathfrak{L}}\overline{\mathfrak{P}_{\iota}}=\mathfrak{L}\underline{\mathfrak{T}}\overline{\mathfrak{T}}\mathfrak{P}_{\iota}\\ = & \left\{ L_{j}T^{j}{}_{\bar{i}}T^{\bar{i}}{}_{i}P_{\iota}^{i}\right\} \\ = & \left\{ L_{j}\delta^{j}{}_{i}P_{\iota}^{i}\right\} \end{align*}

This is exactly what I would expect. Nonetheless, it doesn't seem to necessarily follow from the definitions and axioms. It is conceivable that components of points could transform under the same law as those of lines. Line space and point space would continue to coincide, but incident relations would not be preserved. The axioms (i),(ii) and (iii) would still hold, just not between the images of the objects being incident prior to transformation. The requirement that "the same" points and lines must remain incident after transformation doesn't appear to follow from the premises. If this requirement falls under the "inclusion $\mathcal{I}\subseteq \mathcal{P}\times\mathcal{L}$" which is an "incidence relation," then it seems to be part of the definition.

How would my above proposed transformation lead to a contradiction?

Answer 1

I'm not sure I exactly understood where the core of your question lies. Correct me if I'm wrong. The way I see it, you are asking whether one can really infer that projective transformations preserve incidence.

Let me distinguish several aspects here.

1. The axioms you quoted up front are for a general projective plane. They say nothing about transformations.

2. Somewhere between that and the cited theorem there is a section showing that the projective plane over a field, as obtained from homogeneous coordinate vectors, does indeed form a projective plane that satisfies said axioms. (Theorem 3.1 of the book.)

3. Then there is the definition of a projective transformation as a non-singular matrix (or rather an equivalence class of scalar multiples like you have for homogeneous coordinates). This transformation acts on points by matrix times vector product, turning one homogeneous coordinate vector into another.

4. Now comes the theorem statement. It speaks about collinear points.

   Three points are collinear iff there is a single line incident with all three of them. While this definition refers to lines and incidence, it's primarily a property of three points.

   The theorem claims that this property is preserved by the operations of the projective transformation, i.e. collinear triples of points map to collinear triples of points.

5. Note that the theorem statement says nothing about transformation of lines. It's only in the proof of the theorem that a transformation of lines is used.

   Richter-Gebert uses inverse transpose matrix acting on a column vector, but as far as I can tell that's very much equivalent to your inverse tensor acting on a row vector. The difference there is merely notational.

   The proof shows that with this transformation acting on lines, incidence is preserved so that iff there was a common line for the three preimage points, then there is a common transformed line for the three image points.

The wording of the theorem statement says collinearity is preserved, not incidence. But as a corollary of the proof we find a way to also transform lines in a way that incidence is in fact preserved. In the book, the first sentence after the proof is this:

Implicitly, the calculations in the last proof describe how a projective transformation $M: \mathcal P_{\mathbb R}\to\mathcal P_{\mathbb R}$ represented by a $3\times 3$ matrix $M$ acts on the space of lines $\mathcal L_{\mathbb R}$.

In that sense, the transformation of lines is neither an independently defined operation nor a conclusion of the theorem, but instead is a way of extending the concept of a transformation from points to lines in a way that makes sense and fits the algebraic framework.

It is conceivable that components of points could transform under the same law as those of lines. Line space and point space would continue to coincide, but incident relations would not be preserved.

Of course you could come up with some random transformation which maps points to points and lines to lines without preserving incidence. Doing the same matrix times vector multiplication for both, without using the inverse (transpose) for one kind of objects, most likely would achieve this. The results would still satisfy all the properties of the projective plane, because those properties talk about the objects of the plane, not about how they should behave under transformations. So as long as you map points to points and lines to lines, the axioms will remain satisfied.

But would such a transformation which breaks incidence be of any use? I can't think of one. Of course you can have useful transformations that don't preserve collinearity. Circle inversion or Möbius transformations come to mind as good examples (although both work even better in a $\mathbb C\mathbb P^1$ setup, see chapter 17 of the book). But you would only treat these as transformations that map points to points, you would not extend them to a transformation that maps lines to lines in some way that breaks incidence.

And could such a transformation which breaks incidence be called projective? No. In a narrow sense, a projective transformation is only defined to act on points. In a wider sense, the effect of that transformation on lines will be defined / assumed / implied to be in such a way that incidence is preserved. Any other convention would go against established terminology.

The theorem proves that every projective transformation is a collineation, i.e. a transformation that preserves collinearity. Somewhere later in that book (in section 5.4 and 5.5 to be precise) you will find the fundamental theorem of projective geometry discussing the converse, whether every collineation is a projective transformation. Since the answer depends on whether or not the underlying field has any non-trivial automorphisms, you'll get different answers for real and complex numbers.

Why does a set of collinear points map to a set of collinear points? Because the line they share transforms "contragrediently" to that set. Why do we have contragredient transformations? So that point-line incidence is preserved.

Essentially what you write is correct, although my take would be different. Can we extend the transformation from points to lines in a way that makes sense, i.e. that preserves incidence? If collinearity is preserved then we can and should. If collinearity is broken, we can not find the images of lines in any reasonable way and therefore need to treat the transformation as acting on points or some objects other than lines. For projective transformations the former is the case, so we have a concept of transforming lines. For some other classes of transformations we might not have that.

In other words, the surprising / non-obvious subject of the theorem here is not so much “we picked the definitions (of point and line transformation) to match one another, then show that they match” but instead it's “we can pick the definitions to match, therefore we do pick them thus”. The possibility of such a matching definition is the core of the theorem, the details of that pick are a side benefit.

Answer 2

A projective transformation is a map from the points of $\mathcal{P}_{\mathbb{R}}$ to the points of $\mathcal{P}_{\mathbb{R}}$ which is described by multiplication by a non-singular matrix. In this definition, there is nothing about doing anything to lines.

The theorem that says "a projective transformation maps collinear points to collinear points" means, formally: if three points in $\mathcal{P}_{\mathbb{R}}$ are collinear, then their images are also collinear. Or, to put it even more concretely:

Given three points $[a], [b], [c] \in \mathcal{P}_{\mathbb{R}}$, a line $[l]\in\mathcal{L}_{\mathbb{R}}$, and a non-singular matrix $M$, there exists some line $[l']\in\mathcal{L}_{\mathbb{R}}$ that passes through all three points $[Ma], [Mb], [Mc]$.

This still says nothing about transformations doing anything to lines.

However, the most straightforward way to prove the existence of $[l']$ is to construct $[l']$. And this proof constructs it by letting $l' = (M^{-1})^{\mathsf T} l$. This is not an operation that was baked into the definition of a projective transformation; it is simply something that we can do, because $l$ is a vector and $(M^{-1})^{\mathsf T}$ is a matrix, and it happens to work.

There are other proofs. For example, you could first prove the lemma that $[a]$, $[b]$, and $[c]$ are collinear if and only if $$\det\begin{bmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{bmatrix} = 0.$$ Then you could check that if $[a]$, $[b]$, and $[c]$ are collinear, then so are $[Ma]$, $[Mb]$, and $[Mc]$, because this corresponds to left-multiplying the matrix in the determinant by $M$. Now we do not have to talk at all about lines.