Why are isometries of Minkowski space necessarily linear? 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.
 A: 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.
A: I haven't worked out the details, so I might have this all wrong, but couldn't you proceed as follows:


*

*Prove that an isometry is differentiable.

*Prove that an isometry is an infinitesimal isometry. In other words, the pull back of the   metric tensor is equal to the metric tensor.

*Prove that the composition of any linear co-ordinate function with the infinitesimal isometry has vanishing Hessian (this requires both metrics be flat) and therefore is also a linear co-ordinate function.

*Conclude that the map is linear, since it maps linear co-ordinate functions to linear co-ordinate functions.
It seems to me that the proof that an isometry is linear should be very similar to the proof that any flat metric is locally isometric to the standard one.
A: 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.
A: 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 $Tv$).
