Let $M=\mathbb R^4$ with the symmetric bilinear form $\langle\cdot,\cdot\rangle$ defined by $\langle(x_0,x_1,x_2,x_3),(y_0,y_1,y_2,y_3)\rangle=x_0y_0-x_1y_1-x_2y_2-x_3y_3$. Let $T:\mathbb R^4\to\mathbb R^4$ be a map satisfying the following conditions.
- $T$ is a bijection
- $T(0)=0$
- $\langle x,y\rangle=0\implies\langle Tx,Ty\rangle=0$
- $T$ maps inertial worldlines to inertial worldlines.
In (4), an inertial worldline is a set of the form $(x_0,x_1,x_2,x_3)+\mathbb R(1,v_x,v_y,v_z),$ where $v_x^2+v_y^2+v_z^2<1$. (If the first coordinate is thought of as time, then these are straight lines through $M$ representing particles moving at a constant velocity less than the speed of light, in units where $c=1.)$
A Lorentz transformation is a linear map on $M$ preserving $\langle\cdot,\cdot\rangle$ between all points. Must $T$ be a Lorentz transformation? Please give a proof, if yes, or a counterexample, if no.
There are lots of questions asking about whether maps from $\mathbb R^n\to\mathbb R^n$ preserving straight must be linear. The answer seems to be "yes" if the map is injective and $n\geqslant2$, but the result seems to be difficult to prove, and related to something called the fundamental theorems of affine/projective geometry.
This question, alternatively, considers only lines whose "slope" is bounded by $1$, along with some other conditions motivated by special relativity.
