Proposition to prove. Let $\tau\colon \mathbb S^n\to \mathbb S^n$ be a conformal map, meaning that $\tau^\star g= \Lambda^2 g$ for a scalar field $\Lambda$. (Here $g$ denotes the standard metric tensor on $\mathbb S^n$). Then there exists $\theta\in\mathbb R^{n+1}, |\theta|<1$ such that $$\Lambda(\omega)=(1-\theta\cdot \omega)^{-1},\qquad \forall \omega\in \mathbb S^n.$$

I have found this statement in some sources; for example, in this paper of Chen-Frank-Weth, pag.7, "since the Jacobian determinant...", without proof.

Can you show me a proof, or point me in the right direction in the literature?

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    $\begingroup$ $S^n$ with one point removed is conformal to ${\mathbb R}^n$. So the following should be relevant: en.wikipedia.org/wiki/… $\endgroup$
    – Yuval
    Commented Apr 21, 2019 at 2:07
  • $\begingroup$ @YuDing: It surely is relevant, thanks. Actually, I think I found the answer to the present question in this book (Dubovin-Fomenko-Novikov volume 1, theorem 15.3, pag. 143). It is based on the computation of the conformal transformations of subsets of $\mathbb R^n$, that is, the theorem of Liouville you mention. $\endgroup$ Commented Apr 25, 2019 at 10:09

1 Answer 1


I can prove something close to the statement in the question for 2d case. Let me know if there is anything wrong!

We want to consider automorphisms of $CP^1\cong \mathbb{S}^2$. The canonical coordinate of $CP^1$ is related to the spherical coordinate through the stereographic projection: $$ z = \cot\left(\frac{\theta}{2}\right)e^{i\phi}, $$ It follows that (Stereographic projection is conformal --- from the line element) $$ g = d\theta^2 + \sin^2\theta d\phi^2 $$ $$ = \frac{4dzd\overline{z}}{(1+|z|^2)^2}, $$

The automorphisms of $CP^1$ are Möbius transformations, which can be generated by

$$ z \rightarrow az + b, \quad z \rightarrow \frac{1}{z}, $$ where $a,b\in \mathbb{C}$. But $$ \frac{4d\frac{1}{z}d\frac{1}{\overline{z}}}{(1 + |\frac{1}{z}|^2)^2} = \frac{4dzd\overline{z}}{(1+|z|^2)^2} $$ so the metric is invariant under the inverse operation, we only need to consider translation and scaling. Continuing with the transformation $z \rightarrow az + b$, $$ \frac{4|a|^2dzd\overline{z}}{(1 + |az+b|^2)^2} = \frac{|a|^2(1+|z|^2)^2}{(1 + |az + b|^2)^2} \cdot g, $$ So the conformal factor $$ \Lambda = \frac{|a|^2(1+|z|^2)^2}{(1 + |az + b|^2)^2}\\ = \frac{|a|^2}{\sin^2\left(\frac{\theta}{2}\right)(1 + |a\cot\left(\frac{\theta}{2}\right)e^{i\phi} + b|^2)^2}, $$

Denote $\omega = \begin{pmatrix}\sin\theta\cos\phi\\\sin\theta\sin\phi\\\cos\theta\end{pmatrix}$ the Cartesian coordinate of a point on the sphere, we can rewrite the expression above as $$ \Lambda(\omega) = \frac{1}{(A + \theta^T\omega)^2}, $$ where $$ A = \frac{1}{2}(1 + \frac{1+|b|^2}{|a|^2}), $$ $$ \theta_1 = \frac{1}{2}\left(\frac{b}{a} + \frac{\overline{b}}{\overline{a}}\right), \quad \theta_2 = \frac{i}{2}\left(\frac{\overline{b}}{\overline{a}}-\frac{b}{a} \right), \quad \theta_3 = \frac{1}{2}\left(1 - \frac{1 + |b|^2}{|a|^2}\right). $$

  • $\begingroup$ I have not checked the details and I cannot right now, but I am sure this is the way to go. The key step here is that the automorphisms of $CP^1$ are the Möbius transformations. Have you got an idea of how to prove this? Because I think that an appropriate generalization of this must be true in higher dimension as well. (Thank you very much for your interest in my question). $\endgroup$ Commented Mar 11 at 12:28
  • $\begingroup$ @GiuseppeNegro Oh, it's quite well-known in complex analysis. You can see section 8.5 in math.stanford.edu/~eliash/Public/117-2024/116-117text-2024.pdf. Basically you just need to "projectivize" the linear transformation in $\mathbb{C}^2$ $\endgroup$
    – Max Snow
    Commented Mar 14 at 2:28
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    $\begingroup$ Hey Max, sorry for coming here only now and then. You clearly solved the case of $\mathbb S^2$. Great. Now, Yuval mentioned here a Liouville theorem for conformal maps of $\mathbb R^d$ with $d>2$. Since $\mathbb R^d$ is conformal to $\mathbb S^d\setminus\{N\}$, where $N$ is a point, that should be enough to conclude the general $d$ case as well. $\endgroup$ Commented Mar 20 at 12:27

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