TITLE: How to prove that all Minkowski spacetime isometries (transformations in Poincare group) are compositions of translation and Lorentz Transformations?
QUESTION [4 upvotes]: It is said in wikipedia that Minkowski spacetime isometries, i.e. the transformation that preserves
$$
(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2-(t_1-t_2)^2
$$
between points, can be represented as $\mathbb{R}^{1,3}\rtimes O(1,3)$, meaning that it consists of transformations with the form
$$
x \mapsto \Lambda x+b,   \Lambda \in O(1,3) (\text{Lorentz group}), b\in \mathbb{R}^{1,3}.
$$
How can we prove this?
It seems that the techniques that is available for proving the form of Euclidean group is unavailable here since the "distance" is not positive definite.

REPLY [3 votes]: $\newcommand\R{\mathbb{R}}$In fact, the proof that a Euclidean isometry is a rigid motion works here, too. Here is the proof for the Minkowski case. No differential geometry is needed.
Given points $p_1 = (x_1,y_1,z_1,t_1)$ and $p_2 = (x_2,y_2,z_2,t_2)$, let
$$
Q(p_1,p_2) = x_1x_2 + y_1y_2 + z_1z_2 - t_1t_2.
$$
You should view $Q$ as the Minkowski analogue of the dot product. The key facts used below are that it is bilinear and nondegenerate.
A map $F: \R^{1,3} \rightarrow \R^{1,3}$ is an isometry, if for any points $p_1, p_2 \in \R^{1,3}$,
$$
  Q(F(p_2)-F(p_1),F(p_2)-F(p_1)) = Q(p_2-p_1,p_2-p_1).
$$
Observe that we make no other assumptions about the map $F$, not even that it is continuous.
Let $(e_1,e_2,e_3,e_4)$ be the standard basis. It satisfies
\begin{align*}
  Q(e_i,e_j) &= \begin{cases} 0 &\text{ if }i\ne j\\
    1 &\text{ if }i=j < 4\\
    -1 &\text{ if }i = j = 4
  \end{cases}.
\end{align*}
In particular, if $p = (x,y,z,t) = xe_1 + ye_2 + ze_3 + te_4$, then
\begin{align*}
  x &= Q(p,e_1)\\
  y &= Q(p,e_2)\\
  z &= Q(p,e_3)\\
  t &= - Q(p,e_4).
\end{align*}
More generally, we call $(f_1,f_2,f_3,f_4)$ an orthonormal basis if
\begin{align*}
  Q(f_i,f_j) &= \begin{cases} 0 &\text{ if }i\ne j\\
    1 &\text{ if }i=j < 4\\
    -1 &\text{ if }i = j = 4
  \end{cases}.
\end{align*}
Moreover, if $p = af_1 + bf_2 + cf_3 + df_4$, then
\begin{align*}
  a &= Q(p,f_1)\\
  b &= Q(p,f_2)\\
  c &= Q(p,f_3)\\
  d &= - Q(p,f_4).
\end{align*}
Now let $F$ be an isometry. Since translation by a vector $v$ is an isometry, the map
$$ \widetilde{F}(p) = F(p) - F(0) $$
is an isometry. So we assume $F(0) = 0$. This implies that
\begin{align*}
  Q(F(p),F(p)) &= Q(F(p)-F(0), F(p)-F(0))\\
               &= Q(p-0,p-0)\\
               &= Q(p,p).
\end{align*}
This in turn implies that
\begin{align*}
  Q(F(p),F(q)) &= -\frac{1}{2}(Q(F(p)-F(q),F(p)-F(q)) - Q(F(p),F(p)) - Q(F(q),F(q))\\
               &= -\frac{1}{2}(Q(p-q,p-q) - Q(p,p) - Q(q,q))\\
               &= Q(p,q).
\end{align*}
Now let $f_k = F(e_k)$, for $k = 1, 2, 3, 4$. Then, for any $1 \le j,k \le 4$,
\begin{align*}
  Q(f_j,f_k) &= Q(F(e_j),F(e_k))\\
             &= Q(e_j,e_k).
\end{align*}
This implies that $(f_1,f_2,f_3,f_4)$ is an orthnormal basis. Therefore, if $p = xe_1+ye_2+ze_3 + te_4$ and
$$ F(p) = af_1+ bf_2 + cf_3 + df_4, $$
then
\begin{align*}
  a &= Q(f_1, F(p)) = Q(F(e_1),F(p)) = Q(e_1,p) = x\\
  b &= Q(f_2, F(p)) = Q(F(e_2),F(p)) = Q(e_2,p) = y\\
  c &= Q(f_3, F(p)) = Q(F(e_3),F(p)) = Q(e_3,p) = z\\
  d &= -Q(f_4, F(p)) = -Q(F(e_4),F(p)) = -Q(e_4,p) = t.
\end{align*}
This implies that
$$
F(xe_1+ye_2+ze_3+te_4) = xF(e_1)+yF(e_2) + zF(e_3) + tF(e_4).
$$
In other words, $F$ is a linear map.
Finally, if $F$ is linear, then $F(p) = Ap$ for a matrix $A$. Moreover, for any $p, q \in \R^{1,3}$,
$$
Q(Ap,Aq) = Q(p,q).
$$
By definition, this means $A \in O(1,3)$.
Putting this all together we have shown that if $F$ is a Minkowski isometry, then it is of the form
$$
F(p) = Ax + b,
$$
where $A \in O(1,3)$.
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