# Regularity of conformal maps

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?

• 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. Commented Jan 27, 2020 at 23:35
• 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. Commented Jan 27, 2020 at 23:35

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.

• 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)?
– seub
Commented Jan 27, 2020 at 1:49
• @seub: yes, $\|Df\|^n$ is integrable, so $Df$ is in $L^n.$
– Dap
Commented Jan 27, 2020 at 14:59
• @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. Commented Jan 27, 2020 at 15:03
• @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.
– seub
Commented Jan 28, 2020 at 20:54