A little question on certain parallel-lines-preserving maps

Question

Let $\alpha:\mathbb{R}^n\to\mathbb{R}^n$, $n\geq 2$, be a $\mathbb{Q}$-linear bijection with the following properties:

1) $\alpha$ sends straight affine $\mathbb{R}$-lines to straight affine $\mathbb{R}$-lines

2) If $r$ and $s$ are parallel straight affine $\mathbb{R}$-lines, then so are $\alpha(r)$ and $\alpha(s)$.

Question: does $\alpha$ have to be $\mathbb{R}$-linear?

This (perhaps not exactly research level...) question came to me when I tried to answer the following (also elementary - Edit: as the comment by Misha points out, it was not elementary indeed! Just classical from nineteenth Century) question: does a set-theoretical bijection $\alpha:\mathbb{R}^n\to\mathbb{R}^n$ that sends lines to lines and preserves parallellism have to be $\mathbb{R}$-affine? When $n=1$ the answer is vacuously no, as there is only one line and every set-theoretical bijection would do. But when $n\geq 2$, assuming $\alpha(0)=0$, one can prove${}^*$ that $\alpha$ is actually $\mathbb{Q}$-linear, and then -if one assumes continuity- also that $\alpha$ is $\mathbb{R}$-linear. But what if we don't assume continuity?

${}^*$: First let's prove $\alpha$ is additive, that is $\alpha(u+v)=\alpha(u)+\alpha(v)$. If $u$ and $v$ are not multiples of each other, then the nondegenerate parallelogram corresponding to the points $0,u,v,u+v$ is sent to the nondegenerate parallelogram $0,\alpha(u),\alpha(v),\alpha(u+v)$, hence $\alpha(u+v)$ is the vector sum of $\alpha(u)$ and $\alpha(v)$. If $u,v$ are multiples of each other, write $u=u_1+u_2$ and $v=v_1+v_2$ generically enough that none of the following pairs $(w,w')$ of points is such that $w,w'$ are multiples of each other: $(u_1+v_1, u_2+v_2)$, $(u_1,v_1)$, $(u_2,v_2)$, $(u_1,u_2)$, $(v_1,v_2)$. Now,

$$\alpha(u+v)=\alpha(u_1+u_2+v_1+v_2)=\alpha(u_1+v_1)+\alpha(u_2+v_2)=$$

$$\alpha(u_1)+\alpha(v_1)+\alpha(u_2)+\alpha(v_2)=\alpha(u_1+u_2)+\alpha(v_1+v_2)=\alpha(u)+\alpha(v).$$

So $\alpha$ is additive. For every $u$, by additivity $\alpha(u)=n \alpha (\frac{1}{n} u)$, so for every $u$, $\alpha(\frac{1}{n} u)=\frac{1}{n} \alpha(u)$. The above implies $\mathbb{Q}$-linearity.

Answer 1

The fundamental theorem of projective geometry proven by von Staudt around 1850 (at least, this is when he published the proof) states:

Let $n\ge 2$, $F$ be a field and $f: FP^n\to FP^n$ is a collineation, i.e., a bijection preserving collinearity of points. Then $f$ is a composition of an automorphism of the field $F$ and of an element of $PGL(n+1, F)$.

In the setting of the question of Qfwfq, the bijection induces a collineation of ${\mathbb R}P^n$, so it is a real-linear map, as ${\mathbb R}$ has no automorphisms. (The point is that every automorphism of real numbers has to be continuous as one can encode inequalities into equations, as $x^2\ge 0$ for every real number $x$.)

What's even more interesting is how von Staudt proved his theorem: He noticed that (once coordinate lines in the plane are fixed), one can encode algebraic operations in $F$ into point-line projective configurations (convert algebra into geometry). Von Staudt's theorem and its proof had many generalizations and applications. For instance, Tits' rigidity theorem for spherical buildings is a generalization of the von Staudt's result (Margulis' super-rigidity theorem in turn, is based on Tits theorem in measurable setting). Von Staudt's proof also has lead to Mnev's Universality Theorem, to universality results for character varieties here, and here, Ravi Vakil's Murphy's law in algebraic geometry here and so on.