Is every conformal transformation a stereographic projection of an isometry of the inverse projection?

Question

Here is a really nice video that shows the following:

Suppose we have a subset of $\mathbb{C}\cup \{\infty\}$. Then we can project it onto the 2-sphere via the inverse stereographic projection. If we rotate or translate the sphere in the surrounding $\mathbb{R}^3$ and apply the stereographic projection again, then the result amounts to a Möbius transformation of the initial subset. Here is a screenshot to illustrate this:

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Furthermore, on this wiki-page on conformal maps, the 4th paragraph of the subsection "Complex analysis" says that "A map of the extended complex plane (which is conformally equivalent to a sphere) onto itself is conformal if and only if it is a Möbius transformation."

Now this seems quite interesting to me because it seems to imply that if I take a subset of $\mathbb{R}^2$ and apply a conformal transformation $\Phi$ to it, then I can always write it as \begin{equation} \Phi = P\circ \phi\circ P^{-1} \end{equation} where $P$ is the stereographic projection and $\phi$ is a rotation and/or translation of the sphere in 3-space.

My first question is: Is that right?

My second question concerns whether this generalises to higher dimensions. I saw the wiki page on Liouville's theorem which states that every conformal mapping in $\mathbb{R}^n$, $n\ge 3$ is a Möbius transformation. However, I am not sure if a Möbius transformation in higher dimensions still has the same intuitive meaning as in 2 dimensions.

Answer 1

Here is a counterexample. Let $Q$ be the open first quadrant in $\mathbb{C}$, let $H$ be the open upper half plane of $\mathbb{C}$, let $f : Q \to H$ be the conformal transformation defined by $f(z)=z^2$. I'll abuse notation and write the inverse transformation $f^{-1} : H \to Q$ as $f^{-1}(z) = \sqrt{z}$. Consider the following transformation: $$\Phi : Q \xrightarrow{z \mapsto z^2} H \xrightarrow{z \to z+1} H \xrightarrow{z \mapsto \sqrt{z}} Q $$ and so we may write $\Phi(z) = \sqrt{z^2+1}$. This map $\Phi$ does not extend to a Möbius transformation of the extended complex plane. So, if you conjugate by stereographic projection to get $\phi = P^{-1} \circ \Phi \circ P$, then $\phi$ does not extend to any of the particularly nice functions on the 2-sphere that you ask about.

In general, conformal transformations are much wilder beasts than Möbius transformations. You might be interested in the Schwartzian Derivative, a quantity which, given a conformal transformation between two open subsets of the complex plane, let's you measure whether that transformation is "infinitesmally Möbius". Calculating the Schwartzian derivative of $\sqrt{z^2+1}$ will show that it is not even infinitesmally Möbius, let alone the restriction of a Möbius transformation.

What you seem to have overlooked is what was mentioned in the comment of @Rahul: that theorem quoted from the wiki-page only applies to conformal bijections of $\mathbb{C} \cup \{\infty\}$.

Answer 2

(This answer was written before the large edit of the question.)

In defining a Moebius transformation $T\in{\cal M}$, $\ T(z)={az+b\over cz+d}$, you have three complex, or six real, degrees of freedom, whereas in defining a rotation of $S^2$ you have just three real degrees of freedom: You can choose the axis, and then the rotation angle. This argument already shows that there are "many more" Moebius transformations than there are rotations of $S^2$. (It is unclear to me what you mean by a translation of $S^2$.)

Where are the three real degrees of freedom lost? Number one, a Moebius transformation may have any two points $a$, $b\in\bar {\mathbb C}$ as fixed points (there are even $T\in{\cal M}$ with just one fixed point), whereas the two fixed points of a rotation stereographically correspond to complex $a$, $b$ satisfying $a\bar b=-1$. Number two, for a general $T\in{\cal M}$ with two fixed points one is attracting, and the other repelling, whereas for a rotation the corresponding eigenvalue satisfies $|\lambda|=1$.