Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 23G / 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, useas 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 ofTheir analytic automorphism groups are

A covering space action of $G$ on $X$ is an action by homeomorphisms such that every $x\in X$ has a neighbourhood $U$ satisfyingIn 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 $H\le \mathrm{Aut}({\mathbb{C}}_{\infty })$ 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
$R={\mathbb{C}}^{\ast }\cong {\mathbb{R}}^2\setminus \{0\}$, let $G=⟨T⟩$ withEvery 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 throughcannot be separated in the quotient. Indeed, forwe have $z_n\to q$ and $T^n{z}_n\to p$. 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 $\tilde{R}\cong \mathbb{C}$ of ${\mathbb{C}}^{\ast }$. 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 $\tilde{R}$ is simply connected.

Solved by gpt-5.6-sol high.