Theorem

There is an interpolating, past- and future-distinguishing causal-space structure on the smooth manifold ${\mathbb{R}}^3$ whose Alexandrov intervals form a basis for its usual topology, but which admits a chronological automorphism that is not differentiable.

Proof. Put

This is the proper-orthant construction. Midpoints give interpolation, and intervals are the open boxes ${\prod }_i(a_i,b_i)$. Strict futures ${\prod }_i(x_i,\infty )$ and strict pasts determine all coordinates, so both distinguishability conditions hold.

The strictly increasing bijection

induces

The piecewise-linear stretch $h$ induces a chronological automorphism of $({\mathbb{R}}^3,\ll )$ that fails to be differentiable on the plane $x_1=0$ (Figure fig:stretch). Indeed, strict monotonicity preserves each coordinate inequality in both directions, while the two one-sided derivatives in the first coordinate are $1$ and $2$.

For the open Alexandrov interval $U=(-1,1{)}^3$, one has

Writing $S=\mathrm{cl}U=[-1,1{]}^3$ gives

Thus the map preserves the interval basis, but not smoothness.