TITLE: Why are isometries of Minkowski space necessarily linear?
QUESTION [13 upvotes]: The Mazur-Ulam theorem says that any surjective isometry of normed vector spaces is affine. This argument doesn't seem to apply to Minkowski space (of special relativity) since the metric is indefinite. How would one show that the Poincaré group consists of affine maps? This seems really standard but I can't seem to find it anywhere.

REPLY [5 votes]: 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.