Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 24G (Riemann Surfaces)

The identity theorem says that if two holomorphic functions on a connected Riemann surface agree on a set having an accumulation point, then they agree everywhere.

Apply local coordinates to their difference $F$. The power-series proof in the plane shows that near any point, either $F$ vanishes identically or its zeros are isolated: if the first nonzero Taylor coefficient has order $m$, then $F(z)=(z-z_0{)}^{m}G(z)$ with $G(z_0)\ne 0$. An accumulation point of the zero set therefore has a neighbourhood on which $F=0$. The set of points having such a neighbourhood is nonempty and open. It is also closed: near a limit point, a coordinate neighbourhood contains an open set on which $F=0$, and the planar identity theorem makes $F$ vanish throughout that coordinate neighbourhood. Connectedness now makes this set the whole surface.

For a nonempty connected open set $U\subset {\mathbb{R}}^2$, a function $h:U\to \mathbb{R}$ is harmonic when $h\in C^2(U)$ andOn a sufficiently small disc, the one-formis closed and hence equals $dk$ for some $k$. The Cauchy--Riemann equations then make $h+ik$ holomorphic. Holomorphic functions are smooth, so $h\in C^{\infty }(U)$.

A real-valued function $H$ on a Riemann surface is harmonic when, in every holomorphic chart $z$, its coordinate expression is harmonic. This definition is chart-independent. If $w=w(z)$ is a holomorphic transition map, the chain rule and Cauchy--Riemann equations giveThus vanishing of the Laplacian is preserved by every change of holomorphic coordinate.

No. For example, letBoth functions are continuous and nonzero, but their supports lie in disjoint closed unit discs. Hence $f(z)g(z)=0$ for every $z\in \mathbb{C}$.

Yes. The setsare open. The equation $fg=0$ makes them disjoint, while the assumption that $f$ and $g$ never vanish simultaneously gives $U\cup V=\mathbb{C}$. Since $\mathbb{C}$ is connected, one of $U,V$ must be empty. Therefore one of $f,g$ is identically zero.

No. Define the standard smooth bumpand set, for $x\in {\mathbb{R}}^2$,These are nonzero smooth functions supported in disjoint unit discs, so $fg=0$ identically.

Yes. Suppose $f$ is not identically zero. Thenis a nonempty open set, and $fg=0$ forces $g=0$ on $U$. Harmonic functions satisfy unique continuation: locally they are real parts of holomorphic functions and hence are real analytic, so a harmonic function vanishing on a nonempty open subset of a connected domain vanishes everywhere. Thus $g\equiv 0$. Interchanging $f$ and $g$ proves that one factor must be identically zero.