Is there any diffeomorphism $x:\mathbb{R}^n\to\text{Im }x=B_1\left(0\right)\subset\mathbb{R}^n$ such that

  1. $x$ is an orthogonal chart, i.e., the coordinate vector fields $X_i=\partial x/\partial u_i$ satisfy $X_i\left(u\right)\cdot X_j\left(u\right)=0$ for all $1\leq i\neq j\leq n$ at every point $u=\left(u_1,\dots,u_n\right)\in\mathbb{R}^n$;
  2. the Euclidean norm of the differential $dx$ satisfies $$\left\|dx\left(u\right)\right\|^2\mathrel{\mathop:}=\sum_{i=1}^n\left\|X_i\left(u\right)\right\|^2=\rho^2\left(x\left(u\right)\right)$$ at every point $u\in\mathbb{R}^n$, where $\rho:B_1\left(0\right)\to\mathbb{R}$ is given by $\rho\left(x\right)=\frac{1-\left\|x\right\|^2}{2}$. In particular, $dx$ vanishes at infinity.

The above assumptions lead to the following second-order PDE $$\Delta x-\rho\left(x\right)x=\sum_{i=1}^n\frac{\partial}{\partial u_i}\ln\left(\left\|X_i\right\|^2\right)X_i,$$ which might be useful somehow.

$\hspace{4pt}$ I believe that such a diffeomorphism does not exist. Actually, I can't even think of such an $x$ satisfying just condition 1. By Liouville's theorem we know that $x$ cannot be conformal. Moreover, radial diffeomorphisms typically used to show that $\mathbb{R}^n$ and $B_1\left(0\right)$ are diffeomorphic do not satisfy condition 1 nor 2.


1 Answer 1


This problem is just the classical problem of finding global Tchebychev coordinates on hyperbolic $n$-space. By Hilbert's Theorem, this is impossible when $n=2$. The problem remains open in higher dimensions, despite years of work.

Here is how one can see the reformulation: Let $J$ be the Jacobian of $x$ with respect to $u$, i.e., $\mathrm{d}x = J\,\mathrm{d}u$. Then condition 1 is that $J^TJ$ be diagonal, say, $J^TJ = \mathrm{diag}(\lambda_1^2,\ldots,\lambda_n^2)$, and condition 2 is that $$ \lambda_1^2+\cdots+\lambda_n^2 = (1-\|x\|^2)^2/4. $$ Writing $\lambda_i = \tfrac12(1{-}\|x\|^2) g_i$ yields $$ g_1^2+\cdots+g_n^2=1\tag1 $$ and the above equations become $$ (g_1\,\mathrm{d}u_1)^2 + \cdots +(g_n\,\mathrm{d}u_n)^2 = \frac{4}{(1-\|x\|^2)^2}\bigl(\mathrm{d}x_1^2+\cdots+\mathrm{d}x_n^2\bigr) \tag2 $$ The right hand side of (2) is the standard form of the conformal hyperbolic metric on the unit ball, so the question is asking for coordinates $u_i$ and functions $g_i>0$ on hyperbolic $n$-space satisfying equations (1) and (2). This is the classic problem of Tchebychev coordinates.

For example, when $n=2$, one can write $(g_1,g_2)=(\cos f,\sin f)$ for some function $f$ taking values in $(0,\tfrac12\pi)$, and then one is asking when the metric $$ ds^2 = \cos^2f\,\mathrm{d}u_1^2 + \sin^2f\,\mathrm{d}u_2^2 $$ has Gauss curvature $-1$. It is well-known that this holds if and only if $f$ satisfies the Sine-Gordon equation $$ \frac{\partial^2f}{\partial u_1^2} - \frac{\partial^2f}{\partial u_2^2} = \cos f\,\sin f.\tag3 $$ Hilbert proved that there do not exist global coordinates $u_i$ on the hyperbolic plane and a function $f(u_1,u_2)$ taking values in $(0,\tfrac12\pi)$ satisfying (3).

As I wrote, it is still not known whether global solutions (with $g_i>0$) exist in higher dimensions, but, as Cartan showed, there do exist many local solutions. Essentially, they depend on $n(n{-}1)$ functions of one variable.

Addendum: Of course, if one only wants global orthogonal coordinates, that is not hard: If you first apply inversion with respect to a point on the boundary of the ball $B_1(0)$ in $x$-space, you'll get an injective diffeomorphism $v:B_1(0)\to\mathbb{R}^n$ whose image is a half-space. Say $v\bigl(B_1(0)\bigr)$ is defined by $v_n>0$ (after some translation and rotation in $v$-space). Then define $$ (u_1,\ldots,u_{n-1},u_n) = \bigl(v_1,\ldots,v_{n-1},\log(v_n)\bigr). $$ You'll then have $u:B_1(0)\to\mathbb{R}^n$ being a bijective orthogonal diffeomorphism, and its inverse will be the diffeomorphism $x:\mathbb{R}^n\to B_1(0)$ that you desire.

When $n=2$, there are more options: Choose a conformal diffeomorphism $z:B_1(0)\to R\subset\mathbb{C}$ where $R = I_1\times I_2$ is a rectangle that is a product of two intervals $I_i\subset\mathbb{R}$, not both equal to $\mathbb{R}$, and let $z = v_1 + i v_2$ and choose any bijective diffeomorphisms $h_i:I_i\to\mathbb{R}$. Then $u_1 = h_1(v_1)$ and $u_2 = h_2(v_2)$ will provide the desired functions. Now let $x:\mathbb{R}^2\to B_1(0)$ be the inverse of $u:B_1(0)\to\mathbb{R}^2$.

  • $\begingroup$ Thank you for your answer. What about dropping condition 2? Do you know any orthogonal coordinate system on the open unit ball with the whole $\mathbb{R}^n$ as domain? $\endgroup$ Nov 11, 2017 at 13:52
  • $\begingroup$ @GiannidelFiore: First, can we clarify terminology? I would say that you are searching for orthogonal coordinates $u = (u_i)$ on the open ball $B_1(0)$ in $x$-space such that the range (not the domain) of $u:B_1(0)\to\mathbb{R}^n$ is all of $\mathbb{R}^n$. Would you agree with this formulation? $\endgroup$ Nov 11, 2017 at 14:20
  • $\begingroup$ I'm actually looking for a diffeomorphism $x:\mathbb{R}^n\to B_1\left(0\right)$ such that the coordinate vector fields $X_i=\partial_ix$ are orthogonal everywhere. Just my original question but without condition 2. $\endgroup$ Nov 11, 2017 at 15:06

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