Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 24G (Riemann Surfaces)

A Riemann surface is a connected, Hausdorff, second-countable topological space with an atlas to open subsets of $\mathbb{C}$ whose transition maps are holomorphic.

Every open connected subset inherits the restricted charts and the other topological properties, so it is a Riemann surface. Removing finitely many points gives an open subset because points are closed. It remains connected: a Riemann surface is path connected, and a path meeting finitely many deleted points can be perturbed inside coordinate discs around those points. Countably many points may also be removed successfully; for example, $\mathbb{C}\setminus \mathbb{Z}$ is open and connected and hence is a Riemann surface.

The sphere $S^2$ is the Riemann sphere. Stereographic projection from the north and south poles gives two complex charts, and their overlap map is $z\mapsto 1/z$ up to the chosen conjugation convention, which is holomorphic after choosing compatible orientations.

Projection onto the first two coordinates is a homeomorphismbecause the last two coordinates are uniquely recovered as $(x/r,y/r)$. Identifying $(x,y)$ with $z=x+iy$ transports the standard Riemann-surface structure of ${\mathbb{C}}^{\ast }$ to $X$. Topologically this is a cylinder.

The equation factors asThus $Y$ is the union of the two complex lines $z=2i$ and $w=i$, meeting at $(2i,i)$. A sufficiently small neighborhood of the intersection, with that point removed, has two connected components, whereas a punctured disc is connected. The intersection therefore has no neighborhood homeomorphic to an open subset of $\mathbb{C}$. Hence $Y$ cannot be a Riemann surface.