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.



Why are isometries of Minkowski space necessarily linear?
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 $\eta$ 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 $Tv$).


