Proving the linearity of the transformations between inertial frames without assuming differentiability

Question

All the references that I could find on the proof of the linearity of the transformations assume that the map that connects the coordinates between two inertial frames is at least once differentiable. I was curious if there was a way to get rid of the differentiability assumption.

For example, it can be shown that any map that preserves a non-degenerate bilinear form is bound to be linear. So the transformations, that must preserve the Minkowski product, must be linear. This reasoning, however, seems to be circular since all the proofs that I have seen on the invariance of the Minkowski product assume linearity! However this reasoning shows that one could remove the differentiability hypothesis by showing that the product is preserved, but without assuming linearity.

Thanks in advance for any help.

Answer 1

There is a paper that might be of interest to you.
I haven't gone through it myself... so I am just providing a reference.

"Causality Implies the Lorentz Group"
E.C. Zeeman
J.Math.Phys. Vol 5, April 1964, p.490-493
https://doi.org/10.1063/1.1704140

I've transcribed the abstract

Causality is represented by a partial ordering on Minkowski space, and the group of all automorphisms that preserve this partial ordering is shown to be generated by the inhomogeneous Lorentz group and dilatations.

and some key statements (bolded by me) from the opening paragraphs:

(Where $M$ is Minkowski spacetime, $Q$ is the quadratic form [square-interval], $<$ is a partial ordering on $M$ [the causal ordering], and $G$ is the inhomogeneous Lorentz group...)

Let $f: M \rightarrow M$ be a function that is a one-to-one mapping
(we make no assumptions that $f$ is linear or continuous).
We call $f$ a causal automorphism if both $f$ and $f^{-1}$ preserve the partial ordering; in other words, $x < y \Leftrightarrow fx < fy$, all $x,y \in M$. The causal automorphisms form a group, which we call the causality group.

Let $G$ be the group generated by (i) the orthochronous Lorentz group (linear maps of $M$ that leave $Q$ invariant, and preserve time orientation, but possibly reverse space orientation), (ii) translations of $M$, and (iii) dilatations of $M$ (multiplication by a scalar).

Theorem. The causality group = G.