Definition of collineation

Question

In a book I am reading on transformational Euclidean geometry, the author defines a collineation as a bijection of the plane which takes lines to lines -- that is, for a collineation $F$, if $L$ is a line, then $F[L]$ is a line. A line is understood as a set of points satisfying a linear equation.

The author remarks that for a collineation $F$, $F[L]$ is a line and $F(p)$ lies on $F[L]$ if and only if point $p$ lies on line $L$.

My question is: is it possible to establish the "only if" here given that we are not assuming an "only if" in the definition of collineation?

I believe this remark is equivalent to each of the following:

• If F is a collineation, then the inverse of F is also a collineation
• If F is a collineation, then F is (naturally induces) a bijection on the set of all lines

I am trying to determine if this definition is incorrect or if I am overlooking something trivial.

Answer 1

Interesting question. If you assume that F maps (the set of points of) any one line onto (the set of points of) the image line, then your remark does indeed solve the problem. But it is sometimes claimed that a collineation has only to be one-one and onto for points, and that it should preserve the relation of collinearity.

However, this condition on its own will not guarantee that the inverse also preserves collinearity. For a simple counterexample, consider the following partial planes (actually also primitive planes, ie every line is incident with at least two points) and collineation.

Plane 1 consists of four points, call them 1, 2, 3, 4, with lines comprising (1,2), (2,3) and (1,3), with point 4 lying on no lines (this is OK for a partial plane). Map this onto the primitive plane also consisting of 4 points where this time there is just one line, (1,2,3), and point 4 is on no lines, with the obvious map of point i mapped onto point i. This map is a collineation but its inverse is not (1,2,3 are collinear in plane 2 but not plane 1).

It is however the case that any map from a PROJECTIVE plane which 1-1 and onto another projective plane which preserves collinearity automatically does have an inverse which preserves collinearity. To see this, suppose A, B, C are three non-collinear points in plane 1: we show that fA, fB and fC are also non-collinear in plane 2, which is enough.

Assume for a contradiction that fA, fB and fC are collinear in plane 2. Take a general point P in plane 1. We will show that P is mapped under f to a point on the line fAfBfC, which will be the required contradiction since then the whole of plane 1 is mapped onto a single line in plane 2, whereas we assumed it mapped onto the whole of plane 2.

If P lies on AB or BC we are done, since then fP must map into fAfB or fBfC.

If P lies on neither AB nor BC, then the line AP intersects BC in D, say. A, P, D are collinear, therefore so are fA, fP, fD. Also D is collinear with BC, so fD is collinear with fBfC, so the line fAfD = the line fAfBfC, therefore fP lies on fAfBfC, which gives the result as P is a general point.

I have not seen this proof in any of the books, but it is so simple that I guess somebody must have published it somewhere. Anyhow, I (re)discovered it for myself.

Answer 2

This was just me being silly. At the location of the remark, I was not sure if we could safely assume certain things (like that there is a unique line through any two distinct points), but I see now the author obviously intends these to just be derived from the analytic definitions presented later.

Given a collineation $F$ and a line $L$, we can fix distinct points $p$ and $q$ on $L$, and since $F$ is a bijection, $p'=F^{-1}(p)$ and $q'=F^{-1}(q)$ are distinct. Let $L'$ be the line through $p'$ and $q'$. Then $F[L']$ is a line through $p$ and $q$, so it must be $L$. This shows that $F$ is surjective on lines. It is clear that $F$ is also injective on lines, so it is bijective, which is equivalent to the remark as noted.