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Consider theorem 1.7 from chapter III of 'Elementary differential geometry' by O'Neill. It says that:

Theorem 1.7: If $\phi$ is an isometry of $E^3 $, then there exists a unique translation $T$ and a unique linear transformation $\psi$ such that $\phi = T\psi$.

Now, consider the relativistic interval defined on $R^n$ as usual, i.e. $I(x,y) = \sqrt{\sum (x_i - y_i)^2 - c^2(x_n - y_n)^2}$. We say that $\phi$ is a minkowskian isometry from $R^n$ to $R^n$ just in case for any x,y $\in R^n$, $I(x,y) = I(\phi(x), \phi(y))$.

The question is: does the theorem 1.7 transfer immediately to the minkowskian isometry? That is, can we decompose uniquely the Minkowskian isometry into a translation $T'$ in Minkowski space and a linear transformation $\psi'$ (which would be represented by a Lorentz matrix) ?

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    $\begingroup$ Remark that your I(x,y) is not well-defined if what is under the square root is negative. But if you use I(x,y)^2, which is well-defined, to define minkowskian isometry, then the answer is yes. $\endgroup$
    – user25309
    Commented Nov 2, 2018 at 12:14
  • $\begingroup$ Physicists normally define the interval without the square root. I have never seen it defined with the square root. Most people these days also choose units such that $c=1$. The group of linear transformations you're talking about is called the Lorentz group: en.wikipedia.org/wiki/Lorentz_group . $\endgroup$
    – user21349
    Commented Nov 3, 2018 at 15:53
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    $\begingroup$ Yes, the resulting group is known as the Poincaré group. As a consequence of the property that you describe, it is a semidirect product of the group of translations and Lorentz transformations (just the Euclidean group is the semidirect product of translations and rotations). $\endgroup$
    – Igor Khavkine
    Commented Nov 3, 2018 at 18:05

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