If $\ell_1$ parallel to $\ell_2$, prove $g(\ell1)$ parallel $g(\ell_2)$.

Question

I am working on this problem, but can not figure it out. I know that Automorphisms map parallel lines into parallel lines but I can not prove it.

Let $A$ be an Affine plane and $g$ an Automorphism of A. Given $\ell_1,\ell_2$ elements of $A$, if $\ell_1 \parallel \ell_2$, Prove $g(\ell_1) \parallel g(\ell_2)$.

Answer 1

It depends on how you define automorphism. One possible definition would be a bijective map from points to points and from lines to lines which preserves incidence. In that case, you can show this by contradiction. Suppose that the images were not parallel. Then they would intersect in at least one point. As the automorphism is bijective, that point of intersection has to have a preimage. As the automorphism has to preserve incidence, that preimage has to coincide with both the preimage lines. Which contradicts the assumption that the preimage lines were parallel.

Answer 2

We are free, by choosing an adapted origin and adapted units (see remark below), to consider that

• $\ell_1$ has equation $y=0$ (i.e., has points of the form $(t,0).$)

• $\ell_2$ has equation $y=1$ (i.e., has points of the form $(t,1).$).

Remark: it suffices to take 2 points $A,B \in \ell_1$ and one point $C \in \ell_2$; $(A,\vec{AB},\vec{AC})$ is an adequate affine system of coordinates.

With respect to these coordinates, the affine transformation $g$ can be described thus

$$ (x,y)\mapsto(x',y') \ \ \iff \ \ \cases{x'=ax+cy+e\\y'=bx+dy+f}$$

for certain constants $a,b,c,d,e,f$.

Thus the image of $\ell_1$ is made of points:

$$\tag{1}\cases{x'=at+e\\y'=bt+f} \ \ \ \iff \ \ \ \binom{x}{y}=t \binom{a}{b}+\binom{e}{f}$$

and the image of $\ell_2$ is made of points:

$$\tag{2}\cases{x'=at+c+e\\y'=bt+d+f} \ \ \ \iff \ \ \ \binom{x}{y}=t \binom{a}{b}+\binom{c+e}{d+f}$$

(1) and (2) are parametric equations of 2 lines with the same directing vector. Thus they are parallel.