[Math] Shimura-Taniyama-Weil VS Grothendieck’s dessins

ag.algebraic-geometryco.combinatoricsnt.number-theoryreference-request

When listening to the beautiful lectures by Gilles Schaeffer at
the SLC68, the following (perhaps crazy) question occurred to me:
did anyone attempt (succeed?) to combinatorially prove modularity of elliptic curves using dessins d'enfant?

Of course I am not talking about a combinatorial proof of the general result due to Wiles, Taylor, Breuil, Conrad and Diamond. If such a thing existed, everyone and their dog would have heard about it. I am interested in learning about combinatorial proofs, if any, even for very modest examples. As I do not know anything about the subject,
references to the relevant literature would be appreciated.

This question can be broken down into the following three:

1) Can one tell `by looking at a dessin' if the corresponding curve is defined over $\mathbb{Q}$? If this is too hard, can one construct an explicit collection of dessins which catches all elliptic curves defined over $\mathbb{Q}$?

2) Does one know explicit dessins for all modular curves?

3) Let $X\rightarrow\mathbb{P}^1$ and $Y\rightarrow\mathbb{P}^1$ be two coverings given by dessins. Is there some sufficient criteria for the existence of a cover $X\rightarrow Y$?

Crazy addendum to a crazy question:

Can one `count' $H_{X,Y}$ the number of covers in question 3)? Again, I am talking about examples.

Best Answer

  1. It shouldn't be too hard to find some necessary conditions and some sufficient conditions, e.g., complex conjugation acts on dessins by reflection, so a dessin defined over Q should certainly have a mirror symmetry.

  2. One can certainly give an explicit dessin for all the modular curves, since they all have a map to $X(1) \cong \mathbb P^1$ ramified over only $3$ points (the elliptic points and the cusp), the precondition for a dessin. One can compute it by looking at the group action of $\Gamma/\Gamma(N)$.

  3. The existence of a map which factors through the map to $\mathbb P^1$ is an obvious sufficient condition. It is fairly painless to check but seems very unlikely to be strong enough. I would expect that the problem is extremely hard in general.

Related Question