## Some background on (compact) Belyi surfaces

$\newcommand{\Ch}{\hat{\mathbb{C}}}$
A compact Riemann surface $X$ is called a **Belyi surface** if there exists a branched covering map $f:X\to \Ch$ such that $f$ is branched over at most three points of $\Ch$. Here $\Ch$ denotes the Riemann sphere; we can and will take the three points to be $0$, $1$ and $\infty$.

(Recall that $f$ is a *branched covering map* if, for every $a\in\Ch$, there is a simply-connected neighborhood $U$ of $a$ such that $f$ maps every component of $f^{-1}(U)$ like $z\mapsto z^d$ for some $d\geq 1$. The function is *branched over $a$* if $d>1$ for every such $U$.)

Equivalently, $X$ is a Belyi surface if it can be created by glueing together finitely many equilateral triangles together (defining a complex structure at the vertices in the obvious manner).

Every Belyi function is uniquely determined, up to a conformal change of variable, by its *line complex*, also known as a *dessin d'enfant*. This is a finite graph that essentially tells us how to form the surface by glueing together triangles.

In particular, the set of Belyi surfaces is countable. (This also follows from Belyi's famous theorem, which states that Belyi surfaces are exactly those that are definable over a number field.) So, in a nontrivial moduli space of Riemann Surfaces, such as the space of complex tori, most surfaces are *not* Belyi.

## Belyi functions on non-compact surfaces

It seems natural to extend this notion to noncompact surfaces.

**Definition.** Let $X$ be a non-compact Riemann surface. An analytic function $f:X\to\Ch$ is called a **Belyi function** if $f$ is a branched covering whose only branched points lie over $0$, $1$ and $\infty$, and if $f$ has no removable singularities at the punctures of $X$.

**Question.** On which non-compact surfaces do Belyi functions exist?

This question seems extremely natural and came up in discussions among Bishop, Epstein, Eremenko and myself. Again, a Belyi function is uniquely determined by its line complex, which is now an infinite graph. Since the space of these graphs is totally disconnected, one might expect that not every non-compact surface supports a Belyi function. However, we discovered that this initial intuition is wrong:

**Theorem.** For every non-compact Riemann surface $X$, there is a Belyi function $f:X\to\Ch$. In particular, every non-compact Riemann surface can be built by glueing together equilateral triangles.

(**EDIT.** The paper is now - finally - available: arxiv:2103.16702.)

Note that, for non-compact surfaces, as pointed out by Misha in the comments, the existence of a Belyi function is formally stronger than being built from triangles, assuming that we allow vertices to be incident to infinitely many triangles. (Such vertices would not correspond to points in the resulting surface, as we have no way of defining a complex structure near these.) To get a Belyi function, we should assume that every vertex is incident to only finitely many triangles, so that each triangle is compactly contained in the resulting surface.

(We can also prove the existence of what one might call "Shabat functions", which have two critical values and omit $\infty$.)

## My question:

Have such Belyi functions, and particular the problem of their existence on arbitrary non-compact surfaces, previously appeared in the literature?

(I would also be interested to hear whether our results might be of interest outside of one-dimensional complex function theory and complex dynamics. After all, classical Belyi functions and dessins d'enfants are relevant to many areas of mathematics.)

**EDIT.** Since I first asked the problem, Bowers and Stephenson raised an equivalent question in their work on conformal tilings; see Appendix B of "Conformal tilings I: foundations, theory, and practice", Conform. Geom. Dyn. 21 (2017), 1–63.