Why are isometries of Minkowski space necessarily linear?

The following paper shows that if chronological order on $\mathbb R^n$ is defined by cone (i.e., $x\in \mathbb R^n$ chronologically precedes $y\in \mathbb R^n$ iff $y − x$ belongs to some fixed cone) then any bijection which preserve the chronological order has to be linear.

• Alexandrov A.D. Contribution to chronogeometry Canad. J. Math. - 1967.- V.19, N.6. - P.1119-1128.

This statement is much stronger than you need. After Alexandrov, it was reproved independently 5 times or so.

Let's fix notation and define the bilinear form $\eta: \mathbb{R}^4 \times \mathbb{R}^4 \to \mathbb{R}$ by:

$\eta((x,y,z,t),(x',y',z',t')) = xx'+yy'+zz'- tt'$

Given a map $T:\mathbb{R}^4 \to \mathbb{R}^4$ which fixes $0$ and preserves $\nu$ we want to show that $T$ is linear.

Let $e_1,e_2,e_3,e_4$ be the canonical basis of $\mathbb{R}^4$. The first observation is that for any four vectors $v_1,v_2,v_3,v_4$ such that $\eta(v_i,v_j) = \eta(e_i,e_j)$ for all $i,j$ the linear map sending each $v_i$ to $e_i$ is invertible and preserves $\eta$.

Hence by composing $T$ with a linear invertible $\eta$ preserving map we may assume that $Te_i = e_i$ for $i = 1,2,3,4$.

Now we have for any $v \in \mathbb{R}^4$ that $\eta(v,e_i) = \eta(Tv,Te_i) = \eta(Tv, e_i)$ this implies that $T$ is the identity (since we can get each coordinate of $Tv$).

The following general result is described in the answer I posted on math.stackexchange here: if $V_1$ and $V_2$ are finite-dimensional vector spaces of equal dimension over an arbitrary field and they are both equipped with nondegenerate bilinear forms $B_1$ and $B_2$, then a function $\sigma \colon V_1 \rightarrow V_2$ such that $B_1(v,w) = B_2(\sigma(v),\sigma(w))$ for all $v$ and $w$ in $V_1$ must be an isomorphism of vector spaces (linear, injective, and surjective). The proof does not use non-degeneracy of $B_2$ on $V_2$, so that property actually follows from non-degeneracy of $B_1$ on $V_1$.

In the particular case you ask about, where $V_1 = V_2$ and $B_1 = B_2$, the result says: for a finite-dimensional vector space $V$ over an arbitrary field and a nondegenerate bilinear form $B$ on $V$, a function $\sigma \colon V \rightarrow V$ for which $B(v,w) = B(\sigma(v),\sigma(w))$ for all $v$ and $w$ in $V$ must be linear, injective, and surjective.