# Constructing one-forms on a Riemann surface using the uniformization theorem

I found this statement about proving the existence of one-forms on a Riemann surface in an answer on MathOverflow:

This deep fact is essentially the same as the uniformization theorem. The problem is how to construct at least one holomorphic or meromorphic form with prescribed singularity. All known proofs use some Analysis, and none of them is simple. Once you have Uniformization, it is easy to construct holomorphic forms.

Why is this easy? Uniformization tells us that any Riemann surface has a universal cover that is $\mathbb D$, $\mathbb C$, or $\mathbb P^1$. It is easy to transfer forms from the base space to the covering space by pulling back. But how do we use the covering map to construct forms with prescribed singularities on the base space?

## 1 Answer

Note that the quote does not talk about one-forms, just about forms. Concentrating on the hyperbolic case, if the surface is $\mathbb{D}/\Gamma$, where $\Gamma$ is a group of Möbius transformations acting freely and properly discontinuously on $\mathbb{D}$, then it is easy to show that $\sum_{\gamma \in \Gamma} |\gamma'(z)|^2$ converges locally uniformly in $\mathbb{D}$ (essentially because $\mathbb{D}$ has finite Euclidean area.) Then the function $$F(z) = \sum_{\gamma\in\Gamma} \frac{\gamma'(z)^2}{\gamma(z)}$$ is meromorphic in $\mathbb{D}$ and satisfies $F(\gamma(z)) \gamma'(z)^2 = F(z)$ for all $z \in \mathbb{D}$ and $\gamma \in \Gamma$. Furthermore, it has simple poles on the $\Gamma$-orbit of $0$. This shows that $F$ defines a meromorphic quadratic differential form on $\mathbb{D}/\Gamma$ with one singularity.

It is now easy to modify this construction to move the singularity to any other point, and a quotient of two such quadratic differentials with singularities at different points will give a non-constant meromorphic function.