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In order to define what it means for a map $f \colon \Omega \subseteq \mathbb R^n \to \mathbb R^n$ to be conformal, it is sufficient to require that $f$ is everywhere differentiable. Does conformality automatically implies that $f$ is $\mathcal C^1$ (hence real-analytic, see below)?

By complex analysis, we know the answer is positive when $n=2$.

In higher dimensions, Liouville's theorem characterizes conformal maps as Möbius transformations, but it is stated for $f \in W^{1,n}$ in Wikipedia. Is it known whether it also holds when $f$ is assumed everywhere differentiable?

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  • $\begingroup$ Can one retinterprete the concept of conformality in terms of certain elliptic PDE? hence regularity would imply real analyticity? In dimension 2 it is CR equation, an elliptic PDE. $\endgroup$ Commented Jan 27, 2020 at 23:35
  • $\begingroup$ Can one retinterprete the concept of conformality in terms of certain elliptic PDE? hence regularity would imply real analyticity? In dimension 2 it is CR equation, an elliptic PDE. $\endgroup$ Commented Jan 27, 2020 at 23:35

1 Answer 1

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Let $n\geq 3.$ Let $\Omega$ be an open connected subset of $\mathbb R^n,$ and let $f:\Omega\to\mathbb R^n$ be a function having a pointwise derivative $Df(x)$ everywhere satisfying $(Df)^T(Df)=g(x)I$ with $g(x)>0.$ Then $f$ is continuously differentiable.

By the inverse function theorem, $f$ is a local homeomorphism. By shrinking $\Omega$ we can assume $f$ maps $\Omega$ homeomorphically to $f(\Omega),$ and that $\Omega$ and $f(\Omega)$ are bounded. Note $$\|Df\|^2=c_ng(x)=c_n\det(g(x)I)^{1/n}=c_n|\det Df(x)|^{2/n}$$ where $\|\cdot\|$ is Frobenius norm, and $c_n$ is a constant depending on $n.$ By the change of variables formula (proof sketch), $$\int_\Omega |\det Df(x)|\;dx=\mu(f(\Omega))<\infty.$$ So $f\in W^{1,n}(\Omega,\mathbb R^n)$ and you can use the $W^{1,n}$ result you mentioned.

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  • $\begingroup$ Beautiful. Just to be sure, the point is that the integral is also equal to the $L^n$ norm of $Df$ (up to a constant)? $\endgroup$
    – seub
    Commented Jan 27, 2020 at 1:49
  • $\begingroup$ @seub: yes, $\|Df\|^n$ is integrable, so $Df$ is in $L^n.$ $\endgroup$
    – Dap
    Commented Jan 27, 2020 at 14:59
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    $\begingroup$ @seub Note that this answer assumes that $df$ is invertible at every point. Some people allow the differential to degenerate in their definition of conformal maps. $\endgroup$ Commented Jan 27, 2020 at 15:03
  • $\begingroup$ @AsafShachar I am not aware of this convention. As far as I am concerned, a conformal map is an angle-preserving map. In other words it is a map whose derivative at any point is a linear similarity. It does not make sense to me to include maps having critical points. $\endgroup$
    – seub
    Commented Jan 28, 2020 at 20:54

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