# If Derivative at each point $x \in \Bbb R^n$ is an orthogonal matrix then $f(x) = Ox +b$ [duplicate]

Let $f : \Bbb R^n \to \Bbb R^n$ be a $C^1$ map such that it's Derivative at each point $x \in \Bbb R^n$ is an orthogonal matrix i.e. $Df_x \in O(n,\Bbb R) \text{ , } \forall x \in \Bbb R^n$ . Then prove that $$f(x) = Ox +b \text{ , for some fixed } O \in O(n,\Bbb R) \text{ ,} \forall x \in \Bbb R^n$$

I have no idea about this problem, so couldn't make any attempt.

• While the question clearly is a duplicate of the linked one, I would still like to see a direct elementary answer if there is one (Alexandre's refers to theory, and David's makes the extra assumption that $f\in C^2$). It could of course be posted as an answer to the earlier question. Aug 12, 2018 at 19:21
• If $\gamma(t)$ is a curve in $\mathbb{R}^n$, then the image of $\gamma(t)$ under $f$ has the same length as $\gamma(t)$. This shows every line segment is mapped to a line segment of equal length. Aug 12, 2018 at 21:34
• @Marco: don't get the conclusion. Every line segment is mapped to a curve of equal length, but why a line segment? The endpoints are not fixed. Aug 12, 2018 at 23:06
• This only shows the map is distance non-increasing. Since the volumes are preserved, would this be sufficient to finish the problem? Aug 13, 2018 at 1:13

As pointed out in the comments, this is a corollary of Liouville's Theorem on rigidity of conformal maps (in dimension $\geq 3$). Liouville proved his theorem with fairly high regularity requirements, but a proof under only the $C^1$ hypothesis was given by Hartman in 1958:
Since then, the theorem has been proved under progressively lower regularity assumption ($1$-quasiregular). None of the arguments are easy.