Global diffeomorphisms of $\mathbb R^n$

Question

I would like to describe the smooth global diffeomorphisms $\kappa:\mathbb R^n\rightarrow\mathbb R^n$ such that for all $x\in \mathbb R^n$, $$ \kappa'(x)\in O(n), \quad \text{i.e.}\quad ^t\!\kappa'(x)\kappa'(x)=I_n. $$ In particular, I would be interested in the existence of non-linear $\kappa$.

Answer 1

Since $\|D\kappa(x) \|\le1$ and $\|D\kappa^{-1}(x) \|\le1$, both $\kappa$ and $\kappa^{-1}$ are $1$-Lipschitz, hence isometries.

By (a particular case of) the Mazur-Ulam theorem, any isometry on $\mathbb{R}^n$ is affine, so $\kappa(x)=v+Ux$ with $U\in O(n)$, $v\in\mathbb{R}^n$.

Answer 2

$\kappa$ must be an affine isometry. If $\gamma:[0,1]\to\mathbb{R}^n$ is a smooth curve and $L(\gamma)$ denotes its length, then $$ L(\kappa\circ\gamma)=\int_0^1|D(\kappa\circ\gamma)(t)|\, dt= \int_0^1|D\kappa(\gamma(t))\gamma'(t)|\, dt=\int_0^1|\gamma'(t)|\, dt=L(\gamma). $$ This is because $D\kappa\in O(n)$ and hence this linear map preserves lengths of vectors. The above calculation shows that $\kappa$ preserves lengths of curves from which it easily follows that it preserves distances and hence it is an isometry of $\mathbb{R}^n$. To see that $\kappa$ preserves distamces we argue as follows: If $\gamma$ is a parametrization of a segment connecting $x$ to $y$, then $\kappa\circ\gamma$ connects $\kappa(x)$ to $\kappa(y)$ and hence its length is at least $|\kappa(x)-\kappa(y)|$ $$ |\kappa(x)-\kappa(y)|\leq L(\kappa\circ\gamma)=L(\gamma)=|x-y|. $$ Applying the same argument to the inverse diffeomorphism $\kappa^{-1}$ we have $$ |x-y|=|\kappa^{-1}(\kappa(x))-\kappa^{-1}(\kappa(y))|\leq |\kappa(x)-\kappa(y)| $$ and the above two inequalities show that $|x-y|=|\kappa(x)-\kappa(y)|$.

Answer 3

Even if one relaxes the condition to $D\kappa(x)$ being proportional to an element of $O(n)$, there are no nonlinear examples. These are conformal maps but the global condition maps $\infty$ to $\infty$ so one is back to similary transformations.