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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.

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There were remarks in an expository article on the modularity theorem I read recently that dessins really come into play when the curve is defined over $\bar{\mathbb{Q}}$, rather than $\mathbb{Q}$. For a start on 1), one would have to consider the collection of dessins which the $Gal(\bar{\mathbb{Q}}|\mathbb{Q})$-action fixed pointwise. But I have no idea if this is enough (and I strongly doubt it, with no evidence, though a counterexample would be nice) to pin down curves defined over the rationals. –  David Roberts May 11 '12 at 0:43
    
@David: thanks for your input. Do you remember which expository article? –  Abdelmalek Abdesselam May 11 '12 at 1:06
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Possibly the one on the BCDT proof in the Notices of the AMS from the late 90s. As far as dessins and elliptic curves go, there is a little bit in this thesis: circle.ubc.ca/bitstream/handle/2429/14690/…;, in section 3.2. –  David Roberts May 11 '12 at 1:30
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@David, it's enough for an elliptic curve since each dasein gives a unique $j$-invariant, so if the dasein is Galois-invariant, $j$ is in $\mathbb Q$ and the elliptic curve is defined over $\mathbb Q$. I also believe it is enough in general. –  Will Sawin May 11 '12 at 18:36
    
In case, you didn't know this: In general, the subgroups, which turn up in Beyli's theorem are not congruence subgroups, whereas the modular form associated to an elliptic curve lives in a congruence setting. So the modular form does not live on the curve. –  Marc Palm May 30 '12 at 20:28

2 Answers 2

  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.

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@Will: thanks for the answer. It sounds more encouraging than I expected. If I understand you correctly 1) and 2) are not a problem, the difficulty is all in 3), right? Can the latter be formulated in a purely combinatorial way (i.e. using the kind of things published in journals with "combinatorics" in the name), even though such combinatorial problem might be "extremely hard in general"? Also, my question was not so much about the general case (that would be way too optimistic), but about examples. Has a combinatorial construction of covers for some examples of 3) been done? –  Abdelmalek Abdesselam May 11 '12 at 12:44
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1 might be a problem, I don't really know. All I claimed is that there are necessary and sufficient conditions. One can, for instance, construct infinite families of daseins that are defined over Q. 2 is not a problem, yes. I'm not sure if 3 can even be expressed in any kind of reasonable-sounding combinatorial terms. There is certainly at least one example: $X_0(11)$ is an elliptic curve, so its dasein is the dasein of an elliptic curve. –  Will Sawin May 11 '12 at 18:25
    
I am not sure if I currently have a good example of an infinite family of daseins over $\mathbb Q$. I neglected to check something. –  Will Sawin May 11 '12 at 18:32

This does not answer your question. But it was a bit too long to put as a comment.

Firstly, it seems that the following old question is of some relevance.

Families of curves for which the Belyi degree can be easily bounded

In fact, dessins $X\to \mathbf{P}^1$ are also called Belyi maps/morphisms/functions on $X$. I wanted to know of curves for which one has explicit bounds on the Belyi degree, i.e., the minimal degree of a dessin $X\to \mathbf{P}^1$. Here are the examples

  1. Fermat curves
  2. Modular curves (congruence or non-congruence)
  3. Hurwitz spaces (see JSE's answer to the above question)
  4. Galois Belyi curves = Wolfart-curves = Galois three-point covers
  5. Elkies' curves (see his answer to the above question).

Let me elaborate on 2. If $\Gamma\subset \mathrm{SL}_2(\mathbf{Z})$ is a finite index subgroup, you can consider the quotient $Y_\Gamma = \Gamma\backslash \mathbf{H}$, where $\mathbf{H}$ is the complex upper half-plane and $\mathrm{SL}_2(\mathbf{Z})$ acts on $\mathbf{H}$ by Mobius transformations. The curve $Y_\Gamma$ naturally inherits the structure of a connected Riemann surface from $\mathbf{H}$. We compactify $Y_\Gamma$ by adding "cusps". The compactification of $Y_\Gamma$ is usually denoted by $X_\Gamma$. Note that there is a natural map $Y_\Gamma \to Y_{\mathrm{SL}_2(\mathbf{Z})} = Y(1)$ induced by the inclusion $\Gamma\subset \mathrm{SL}_2(\mathbf{Z})$. This morphism extends to the compactifications $X_\Gamma \to X(1)$ and induces a dessin $X_\Gamma \to \mathbf{P}^1(\mathbf{C})$ after you compose with the isomorphism given by the $j$-invariant $j:X(1)\to\mathbf{P}^1(\mathbf{C}$. (The branch points are the elliptic points $0$, $1728$ and the cusp $\infty$ of $X(1)$.)

Let me adress your third question. The above is about your second question. I don't have much to say about your first question, unfortunately. What do you mean by a dessin which "captures" all elliptic curves over $\mathbf{Q}$?

Firstly, assume that $X\to \mathbf{P}^1$ is a dessin of prime degree. It's clear that this morphism will not factor.

I get the feeling (but I might be wrong) that you are interested in modular parametrizations of elliptic curves in the following sense. You want to know whether the above explicit dessins on $X_0(n)$ can be shown to factor through some elliptic curve. If this is the case, the answer is likely to be no for $n$ big.

Now, you can bound the number of dessins on a curve $X$ of given degree $d$ by the number of dessins of degree $d$, i.e., the number of topological covers of $\mathbf{P}^1-\{0,1,\infty\}$.

But your $H_{X,Y}$ will be zero or infinite.

In fact, if it not zero then there exists a dessin $X\to \mathbf{P}^1$ which factors through a dessin $Y\to \mathbf{P}^1$. But Belyi proved that for any finite set $B\subset \mathbf{P}^1(\overline{\mathbf{Q}})$ there exists a dessin $R:\mathbf{P}^1_{\mathbf{Q}}\to\mathbf{P}^1_{\mathbf{Q}}$ (defined over $\mathbf{Q}$ even!) such that $R$ sends $B$ to the set $\{0,1,\infty\}$. So from a given factorization $X\to Y\to \mathbf{P}^1$ you can construct an infinite number of really different dessins (and associated factorizations).

The former paragraph is just applying the fact that given a dessin $f:X\to \mathbf{P}^1$ you can construct an infinite number of dessins $g :X\to \mathbf{P}^1$ by composing $f$ with an arbitrary dessin on $\mathbf{P}^1$. (Belyi actually gave an algorithm to compute a dessin $R$ on $\mathbf{P}^1$ associated to $B$ as above.)

So to make sense of your last "crazy" question, you might want to fix a dessin $X\to \mathbf{P}^1$ on $X$ and try to look at possible factorizations, where $Y\to \mathbf{P}^1$ is a dessin and $Y$ is not fixed. Thus, let $H_{\pi}$ be the number of pairs $(Y,f)$ up to isomorphism, where $f:Y\to \mathbf{P}^1$ is a dessin and there exists a factorization $g:X\to Y$ such that $\pi = fg$.

I don't think it is possible to give a precise formula for $H_\pi$ easily, but it is certainly possible to bound this number in terms of the degree of your dessin.

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@Ariyan: this reformulation is quite different from my original question. I don't understand why you want 1) a single dessin to capture all elliptic curves over Q, 2) the map X->Y to form a commutative triangle with the maps X->P1 and Y->P1? –  Abdelmalek Abdesselam May 14 '12 at 16:16
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@Abdelmalek. Sorry for the late reply! I didn't see your comment. Let me also apologize for "wanting" a single dessin to capture all elliptic curves over Q. This was a "typo". I really meant to ask: what do you mean by a family of dessins that capture all elliptic curves over Q? I still don't understand what this means because I'm too used to thinking about Belyi maps. What would "a family of dessins that capture all elliptic curves over Q" mean in the Belyi maps terminology? Is it a Belyi map $f_E:E\to \mathbf{P}^1_\mathbf{Q}$ for every elliptic curve $E$ over $\mathbf{Q}$ satisfying some... –  Ari May 29 '12 at 19:44
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I see that I slightly misunderstood your last question. You are just asking for a cover $X\to Y$. I was assuming you wanted the cover $X\to \mathbf{P}^1$ to factor through your cover $\mathbf{Y}\to\mathbf{P}^1$. I wrote the above answer with this in mind. (By the way, Will Sawin wrote his answer before me and he also seems to have understood you want your map to factor through $Y\to \mathbf{P}^1$.) Finally, I think I understand what you mean by $H_{X,Y}$. Is $H_{X,Y} $ the number of finite morphisms $X\to Y$? If yes, then $H_{X,Y}$ is finite if the genus of $Y$ is at least $2$. As you.... –  Ari May 29 '12 at 19:52
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....probably know this is called de Franchis' theorem: en.wikipedia.org/wiki/De_Franchis_theorem I think this is again a very (unnecessarily) long comment. My apologies. –  Ari May 29 '12 at 19:52
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@Abdelmalek. I see what you were getting at right now. It's a shame there isn't a "de Franchis" where the target is an elliptic curve. Anyway, as a final comment (which won't help you again unfortunately) I just came across this "effective" version of de Franchis' theorem by accident: projecteuclid.org/DPubS/Repository/1.0/… You can probably find much more stuff on "effective" versions of de Franchis theorem if you're interested a bit in the situation for genus >1 curves. Again, sorry to not be able to say anything else useful. –  Ari May 29 '12 at 21:33

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