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.
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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 differential-geometric sense are rectifiable curves that locally minimize length. $\endgroup$– Ryan BudneyCommented Dec 30, 2011 at 0:31
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$\begingroup$ Off-topic question: what are some applications of the Mazur-Ulam theorem? $\endgroup$– Tom LaGattaCommented Dec 30, 2011 at 4:59
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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 Myers-Steenrod theorem says that all isometries of Riemannian manifolds are smooth, however this theorem requires the isometry to be surjective. $\endgroup$– Boaz HabermanCommented Dec 30, 2011 at 18:10
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3$\begingroup$ @Boaz what's your definiton of an isometry of a Pseudo-Riemannian manifold? There is no notion of a distance or of arc-length of a continuous curve here. $\endgroup$– Vitali KapovitchCommented Dec 30, 2011 at 22:02
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1$\begingroup$ @Boaz and @Vitali: An isometry of a semi-Riemannian (or pseudo-Riemannian) manifold $(M,g)$ is a diffeomorphism $\phi$ that preserves the metric tensor, i.e., the pull-back 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. BettiolCommented Jan 14, 2012 at 18:43
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4 Answers
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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 $Tv$).
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1$\begingroup$ What if we only know that $T$ preserves lengths? How do you polarize? $\endgroup$ Commented Dec 30, 2011 at 18:25
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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. $||f(x)-f(y)||_{3,1}=||x-y||_{3,1}$ 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. $\endgroup$ Commented Dec 30, 2011 at 21:59
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2$\begingroup$ yes, the polarization is trivial. we are assuming $f(0)=0$ and $||f(x)-f(y)||^2=||x-y||^2$ for all $x,y$. Then the formula $2\langle x, y\rangle=||x|^2+||y||^2-||x-y||^2$ immediately implies that $\langle f(x), f(y)\rangle=\langle x, y\rangle $. $\endgroup$ Commented Jan 1, 2012 at 3:49
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1$\begingroup$ What is $\nu$? I don't see it defined in either the question or the answer. $\endgroup$ Commented Jun 22, 2020 at 7:07
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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 "Semi-Riemannian geometry". $\endgroup$ Commented Jun 26, 2020 at 0:15
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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.
answered Dec 30, 2011 at 5:59
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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.
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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.
answered Dec 30, 2011 at 21:44