If I have some function F that fixes two points in a line

Question

How do I show that this function is the identity function? We are currently studying collineations and I do not know how to know the other points within the line are not permuting(or moving around). Basically, how can I show they are fixed point-wise?

Answer 1

A function that fixes two points on a projective line is not neccessarily the identity. Example in homogeneous coordinates:

$$\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}2x\\y\end{pmatrix}$$ fixes the origin $(0,1)^T$ and the point at infinity $(1,0)^T$.

For an affine line, you essentially imply that the point at infinity remains fixed as well, so you have three fixed points. But you still have to restrict your class of functions. If you only accept projective transformations, i.e. things you can express as matrix multiplication, then you can use the 2d version of this argument to show that three points and their images uniquely define a projective transformation of the line. This gives you the neccessary uniqueness of the identity transformation.

If you want to treat this completely affine, then you should probably investigate affine transformations $x\mapsto ax+b$. If you have two points and their images, you can deduce $a$ and $b$, and if you know the points agree with their images, you get $a=1,b=0$.

Note that for some projective planes there exist collineations which are not projective transformations. Similarly, there exist projective lines for which a non-identity transformation can have three or more fixed points. Most practical example would be $\mathbb{CP}^1$ where conjugation $z\mapsto\bar z$ fixes real points but is not the identity.