= Solution
Choose an evenly covered neighbourhood $U\subset R$ and a component $V$ of $\pi^{-1}(U)$. If $z:U\to\mathbb C$ is a complex chart, use
$$
z\circ\pi|_V:V\longrightarrow\mathbb C
$$
as a chart on $S$. On overlaps, the transition maps are exactly transition maps between charts of $R$, hence are holomorphic. These charts give the unique <complex structure lifted through a covering map> for which $\pi$ is a local biholomorphism, and therefore analytic. The Hausdorff assumption is already given; connectedness and the second-countability of a Riemann surface ensure the resulting covering surface has the required manifold properties.
A surface is simply connected when it is path-connected and every loop is null-homotopic, equivalently when its fundamental group is trivial. The <uniformization theorem> says that every simply connected Riemann surface is biholomorphic to exactly one of
$$
\mathbb C_\infty,\qquad \mathbb C,\qquad \mathbb D.
$$
Their analytic automorphism groups are
$$
\begin{aligned}
\operatorname{Aut}(\mathbb C_\infty)
&=\left\{z\mapsto\frac{az+b}{cz+d}:ad-bc\ne0\right\}/\mathbb C^*,\\
\operatorname{Aut}(\mathbb C)
&=\{z\mapsto az+b:a\ne0\},\\
\operatorname{Aut}(\mathbb D)
&=\left\{z\mapsto e^{i\theta}
\frac{z-a}{1-\bar a z}:|a|<1\right\}.
\end{aligned}
$$
A <covering space action> of $G$ on $X$ is an action by homeomorphisms such that every $x\in X$ has a neighbourhood $U$ satisfying
$$
gU\cap U=\varnothing\qquad(g\ne1).
$$
In particular the action is free. Every nonidentity Möbius transformation of $\mathbb C_\infty$ has a fixed point: solving $z=(az+b)/(cz+d)$ gives a root on the sphere. Consequently a subgroup $H\leq\operatorname{Aut}(\mathbb C_\infty)$ acting as a covering space action must be trivial. Its quotient is the sphere itself and is Hausdorff.
The Hausdorff conclusion fails on other Riemann surfaces. On
$R=\mathbb C^*\cong\mathbb R^2\setminus\{0\}$, let $G=\langle T\rangle$ with
$$
T(x,y)=(2x,y/2).
$$
Every orbit is discrete in $R$ and has trivial stabilizer; small enough neighbourhoods have disjoint nontrivial translates, so this is a covering space action. Yet the two orbits through
$$
p=(1,0),\qquad q=(0,1)
$$
cannot be separated in the quotient. Indeed, for
$$
z_n=(2^{-n},1),\qquad
T^nz_n=(1,2^{-n}),
$$
we have $z_n\to q$ and $T^nz_n\to p$. Thus every pair of quotient neighbourhoods of the two distinct orbits intersects, proving that $R/G$ is not Hausdorff.
Simple connectedness does not repair this. Lift $T$ to the universal cover $\widetilde R\cong\mathbb C$ of $\mathbb C^*$. Choose the lift preserving the angular interval from $0$ to $\pi/2$. The lifted points of $z_n$ still converge to a lift of $q$, while their $n$th translates converge to a lift of $p$; those two lifts belong to different orbits. The lifted action is again a covering space action, and its quotient is non-Hausdorff although $\widetilde R$ is simply connected.
Back to article page
