Alexandrov–Zeeman theorem

Remarks

No continuity or linearity of $f$ is assumed: the causal order alone forces $f$ to be affine. Proof idea: the order determines the light cones, since $y$ lies on the future light cone of $x$ exactly when $y \ne x$, $x \not\ll y$, and $x \ll z$ for every $z \gg y$. So $f$ maps light cones to light cones and null lines to null lines. One then shows that $f$ maps lines to lines, so it is affine. Its linear part preserves the light cone, so it is a positive multiple of a Lorentz transformation. The hypothesis $n \ge 3$ is necessary. For $n = 2$, use the null coordinates $u = x^0 - x^1$ and $v = x^0 + x^1$. Then $x \ll y$ iff $u(x) < u(y)$ and $v(x) < v(y)$. So every map $(u, v) \mapsto (\phi(u), \psi(v))$, with $\phi, \psi$ increasing bijections of $\R$, preserves the order, and most such maps are not affine. The map $x \mapsto a x$ is a homothety; it preserves the order but not the metric, so the order alone cannot fix the scale.

Going deeper

For a student

The result says that the automorphism group of Minkowski space as a partially ordered set is generated by the orthochronous Poincaré group and the dilations $x \mapsto a x$ with $a > 0$. It is a rigidity theorem, like the fundamental theorem of affine geometry. That theorem says that real bijections sending lines to lines are affine in dimension at least $2$. It motivates approaches to spacetime, such as causal set theory, in which the causal order is the basic structure.