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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.

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.)

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Not that it matters a lot, but when you say "branched cover", do you really mean that it is a topological covering (to its image) away from a certain discrete subset, or simply that it is a local homeomorphism away from branch points? This is unclear from your description. Note that, in general, gluing a surface from equilateral triangles will give you a developing map which is only a "branched cover" in the latter sense. In Belyi's case, both definitions agree because of properness. –  Misha Jan 10 '13 at 17:49
    
I am not sure that I understand your comment correctly. Any holomorphic map is a "homeomorphism away from branch points". By "branched cover", I mean precisely the definition given in the second paragraph, which is stronger than both of your definitions. (The exponential map is a covering to its image on the plane, but it is not a Belyi function in my sense.) –  Lasse Rempe-Gillen Jan 11 '13 at 9:23
    
Perhaps I should have clarified that, when we build the Riemann surface from infinitely many triangles, we only include those corner points that are contained in only finitely many triangles. Near these, we define a Riemann surface structure in the obvious way. (Near the others, it isn't at all clear how we would define a Riemann surface structure.) –  Lasse Rempe-Gillen Jan 11 '13 at 9:30
    
@Misha, I think I see what you mean - for noncompact surfaces, being built from triangles is indeed (formally) weaker than having a Belyi function on it. To have a Belyi function, every corner should be adjacent to only finitely many triangles. I will add the question to clarify this. (Our theorem establishes the stronger property for all non-compact surfaces, and hence the weaker property also holds.) –  Lasse Rempe-Gillen Jan 11 '13 at 14:11
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2 Answers

up vote 1 down vote accepted

As you may possibly already be aware, there is a parallel phenomenon in circle packing riemann surfaces.

Those compact riemann surfaces admitting full circle packings are a countable dense subset of the moduli space, where by full circle packing a riemann surface, I mean finding a circle packing $C$ such that the carrier of the nerve graph $T_C$ coincides with our surface. This is due to R. Brooks, "Circle packings and co-compact extensions of Kleinian groups", Invent. Math. 86, 1986, 461-469.

But G. Brock Williams has also proven that all noncompact riemann surfaces are packable: "Noncompact surfaces are packable", J. D'Analyse Math, Vol 90, 2003, 243-255.

The first basic 'intuition' as to why noncompact riemann surfaces are packable is of course that any obstructions to completing a (partial) circle packing to a full packing can be `hidden' into the cusp (similar to how any noncompact surface admits lots of nonvanishing vector fields).

I would be very interested in reading your proof that noncompact riemann surfaces admit Belyi functions. In particular, I would like to know if your functions have the same "flat euclidean" structure up within the cusps as Williams' paper. Specifically, the final corollary 6.1 in Williams states "every noncompact riemann surface of finite type supports a circle packing asymptotic to the euclidean ball-bearing packing in the cusps". I have to admit that I do not find Williams' paper to be entirely educating, and am very anxious to see other approaches.

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Sorry for not seeing your very interesting answer before now! There may well be some connection between the two, and I shall have to try and take a look. I am not sure whether I entirely understand the point about the flat euclidean structure. Of course the cusps (or ends) are somehow an important part of the proof. The basic idea is that we build a quasiregular function somewhere close up to a cusp. Straightening this out will, of course, give us a different surface. –  Lasse Rempe-Gillen Nov 15 '13 at 7:23
    
But the point is that - using methods developed by Chris Bishop - we can make sure that the surface is quite close to the original one - close enough to ensure that we can actually compensate for it in the next step, and in the limit obtain a function actually defined on the desired surface. –  Lasse Rempe-Gillen Nov 15 '13 at 7:55
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I would like to point out the following very nice reference (unfortunately I cannot plug into MathSciNet right now and only have a preprint version on my computer, but you should be able to track it down if you're interested)

• K. Pilgrim, «Polynomial vector fields, dessins d'enfants, and circle packings»

where he establishes the relationship between usual Belyi polynomials on the Riemann sphere, and real flows of complex polynomial of the Riemann sphere $\dot{z}(t)=P(z(t))$. The topological classification of these flows has been done by A. Douady and P. Sentenac (a very nice paper unfortunalety yet unplished), and it boils down to building a graph whose vertices are the singular points of the system, and whose edges link two singular points lying in the limit set of a same trajectory. The graph in question is actually a tree and its combinatorial data is related to the good-bracketing problem. The invariant of the flow is the class of such a graph up to obvious combinatorial equivalence. The result of Pilgrim states that there is a very simple correspondance between $P$ and the Belyi polynomial $B$ associated to the dessin d'enfants embodied in the combinatorial invariant. You obtain $P$ by simply droping the exponents of the irreducible factors of $B$, and vice-versa.

Pilgrims's paper is also related to two questions of Douady:

  1. Is it possible to compute explicitely the combinatorial invariant being given a flow?

  2. Conversely, being given a combinatorial data is it possible to produce an explicit example of a flow with the corresponding invariant in the same combinatorial class?

To the best of my knowledge both questions remain open for degrees bigger than 4, as they are equivalent to the very same problem regarding dessins d'enfants, which is a difficult question if I understood things properly.

That being said, I would conjecture that the context of the OP's question is related (say, on $\mathbb{C}$ ) to the topological classification of entire flows $\dot{z}\left(t\right)=f\left(z\left(t\right)\right)$ with $f\in\mathcal{O}(\mathbb{C})$. This classification should also give a topological classification of flows in a neighborhood of an essential singularity. Such a classification is not done (again, to the best of my knowledge). A limiting procedure allows yet to build a locally finite graph which should embody the topological data. Then one would have to adapt the construction of Pilgrim to bind these objects and the objects defined by the OP. This is of course of a very speculative nature. I'm also aware of the fact that it does not answer the question directly, but this was too long a comment (and I hope it is nonetheless relevant regarding the question).

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Thanks Loic - I wasn't aware of Kevin's work on vector fields.As you say, this isn't quite what I was asking for, but I will take a look at this paper. –  Lasse Rempe-Gillen Jan 14 '13 at 8:54
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