The MazurUlam 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.

3$\begingroup$ There are a lot of ways to prove these kinds of theorems. One strategy is to argue isometries carry geodesics to geodesics  i.e. geodesics in the differentialgeometric sense are rectifiable curves that locally minimize length. $\endgroup$– Ryan BudneyDec 30, 2011 at 0:31

$\begingroup$ Offtopic question: what are some applications of the MazurUlam theorem? $\endgroup$– Tom LaGattaDec 30, 2011 at 4:59

1$\begingroup$ I guess what is confusing me is that usually isometries are defined to be smooth maps, but a priori they could just be continuous. Apparently the MyersSteenrod theorem says that all isometries of Riemannian manifolds are smooth, however this theorem requires the isometry to be surjective. $\endgroup$– Boaz HabermanDec 30, 2011 at 18:10

3$\begingroup$ @Boaz what's your definiton of an isometry of a PseudoRiemannian manifold? There is no notion of a distance or of arclength of a continuous curve here. $\endgroup$– Vitali KapovitchDec 30, 2011 at 22:02

1$\begingroup$ @Boaz and @Vitali: An isometry of a semiRiemannian (or pseudoRiemannian) manifold $(M,g)$ is a diffeomorphism $\phi$ that preserves the metric tensor, i.e., the pullback of $g$ by $\phi$ coincides with $g$: $\phi^*(g)=g$. I have never seen any other definitions in the literature. Although there is a notion of distance in Lorentzian manifolds (see e.g. the book of O'Neill or the one of Beem, Ehrlich and Easley), there are very few results analogous to the Riemannian case. $\endgroup$– Renato G. BettiolJan 14, 2012 at 18:43
4 Answers
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 $Tv$).

1$\begingroup$ What if we only know that $T$ preserves lengths? How do you polarize? $\endgroup$ Dec 30, 2011 at 18:25

6$\begingroup$ @Boaz preserves lengths of what? please state your definition of an isometry. One reasonable definition for the Minkowski space is a map that preserves Minkowski norm, i.e. $f(x)f(y)_{3,1}=xy_{3,1}$ for all $x,y\in\mathbb R^{3,1}$. this polarizes trivially so Pablo's answer is complete with that definition. Another reasonable definition is if you assume that $f$ is smooth and preserves the metric infinitesimally. Then the result is also elementary but judging by your other comment this is not the definition you want. $\endgroup$ Dec 30, 2011 at 21:59

2$\begingroup$ yes, the polarization is trivial. we are assuming $f(0)=0$ and $f(x)f(y)^2=xy^2$ for all $x,y$. Then the formula $2\langle x, y\rangle=x^2+y^2xy^2$ immediately implies that $\langle f(x), f(y)\rangle=\langle x, y\rangle $. $\endgroup$ Jan 1, 2012 at 3:49

1$\begingroup$ What is $\nu$? I don't see it defined in either the question or the answer. $\endgroup$ Jun 22, 2020 at 7:07

1$\begingroup$ @BenMcKay Assume $T$ fixes the origin. Since the exponential mapping at the origin is the identity and $T$ preserves geodesics, you deduce $T = T'(0)$. In general two isometries which coincide at a point up to order one are the same, see the proposition on page 91 of Barrett O'neill's book "SemiRiemannian geometry". $\endgroup$ Jun 26, 2020 at 0:15
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.
This statement is much stronger than you need. After Alexandrov, it was reproved independently 5 times or so.
The following general result is described in the answer I posted on math.stackexchange here: if $V_1$ and $V_2$ are finitedimensional 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 nondegeneracy of $B_2$ on $V_2$, so that property actually follows from nondegeneracy 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 finitedimensional 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.
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 coordinate function with the infinitesimal isometry has vanishing Hessian (this requires both metrics be flat) and therefore is also a linear coordinate function.
Conclude that the map is linear, since it maps linear coordinate functions to linear coordinate 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.