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What are the best references for finding explicit formulas for Belyi maps for rational dessin d'enfants?

I am most interested in a formula for the Belyi map that corresponds to a specific rational dessin d'enfant: it is clean, with four black vertices, each of valency 3, and with ramification indices above the infinity of 6,3,2,and 1. (I believe that there is only one such dessin d'enfant). Ideally, it would be great to have tables or software for other similar examples.

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Every elliptic surface on $\mathbb P^1$ with four semistable fibers and no other singular fibers has a $j$ invariant with ramification of order $3$ around $j=0$ and ramification of order $2$ around $j=1728$. By Riemann-Roch, one can check that this is the only ramification, so this means the $j$ invariant is a Belyi map coming from a clean dessin with all vertices of valence $3$, with four points in the fiber over $\infty$.

Conversely, from such a Belyi map one can construct an elliptic surface, uniquely defined up to quadratic twist. One can choose the quadratic twist so that all fibers with $j\neq \infty$ are smooth and all fibers with $j=\infty$ are smooth.

Beauville (http://math1.unice.fr/~beauvill/pubs/ellss.pdf) classified such surfaces by classifying their Belyi map. (The diagram on page 659 is a combinatorial way of specifying a Belyi map - though I think a less visually attractive way than Grothendick's.) He found explicit equations for the surfaces, from which you can extract the Belyi map by taking the j invariant.

In particular your map, as a function of $t$, is the $j$ invariant of the curve with equation $$(X+Y)(Y+Z)(Z+X)+tXYZ=0$$ because according to his table this is the unique one with ramification at $\infty$ of order $1,2,3,6$. Computing the $j$ invariant from this is routine, at least in theory.

He also related them to modular curves. For instance, your map can be described as the covering map from the modular curve $X_1(6)$ to $X(6)$. One can check directly from the group theory of $SL_2(\mathbb Z)$ that this modular map has the desired ramification properties, and use the uniqueness of the dessins to conclude it equals yours.

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  • $\begingroup$ The map that I was looking for is the following: f(x)=(x^2+4x-8)^2 (x^4+8x^3+8x-8)^2 / (-1728 x^3(x+8)(x-1)^2) $\endgroup$
    – Alexander Borisov
    Commented Feb 26, 2017 at 18:56
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You are right that there is only one such dessin. To see that, type the following in Maple 2019:

with(GroupTheory):
FindDessins([3,3,3,3],[2,2,2,2,2,2],[6,3,2,1]);

You can find more information on: https://www.math.fsu.edu/~hoeij/Heun

In the main file there you can find:

   S3: ModuliField:    Q.
       Obstruction:    none
       Branch type:    0:[3^4]  1:[2^6]  infty:[1,2,3,6]
       Decomposition:  S47(deg 3) 1/S34(deg 4) 
       Galois group:   A4 x S3.  GAP transitive group ID = 12T43.  Order = 72

The name "S3: ..." refers to rational functions in the file: https://www.math.fsu.edu/~hoeij/Heun/BelyiMaps which gives this rational function:

$S_3 := {\frac {4\, \left( {x}^{3}-6\,x-6 \right) ^{3}{x}^{3}}{27\, \left( x+1 \right) ^{3} \left( 2\,x+3 \right) ^{2} \left( x-3 \right) }}$

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  • $\begingroup$ Given $f(x) := \frac{4(x^3-6x-6)^3x^3}{27(x+1)^3(2x+3)^2(x-3)},\,$ then $1-f(\frac{-x-2}2) = \frac{(x^2+4x-8)^2(x^4+8x^3+8x-8)^2}{-1728x^3(x+8)(x-1)^2}$. $\endgroup$
    – Somos
    Commented Oct 4, 2022 at 16:41

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