Shimura-Taniyama-Weil VS Grothendieck's dessins - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T12:14:41Z http://mathoverflow.net/feeds/question/96621 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96621/shimura-taniyama-weil-vs-grothendiecks-dessins Shimura-Taniyama-Weil VS Grothendieck's dessins Abdelmalek Abdesselam 2012-05-10T22:37:00Z 2012-05-30T19:26:25Z <p>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?</p> <p>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.</p> <p>This question can be broken down into the following three:</p> <p>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 $\mathbb{Q}$?</p> <p>2) Does one know explicit dessins for all modular curves?</p> <p>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 $X\rightarrow Y$?</p> <hr> <p>Crazy addendum to a crazy question:</p> <p>Can one count' $H_{X,Y}$ the number of covers in question 3)? Again, I am talking about examples.</p> http://mathoverflow.net/questions/96621/shimura-taniyama-weil-vs-grothendiecks-dessins/96637#96637 Answer by Will Sawin for Shimura-Taniyama-Weil VS Grothendieck's dessins Will Sawin 2012-05-11T02:14:54Z 2012-05-30T19:26:25Z <ol> <li><p>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.</p></li> <li><p>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 $\Gamma/\Gamma(N)$.</p></li> <li><p>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.</p></li> </ol> http://mathoverflow.net/questions/96621/shimura-taniyama-weil-vs-grothendiecks-dessins/96763#96763 Answer by Ariyan Javanpeykar for Shimura-Taniyama-Weil VS Grothendieck's dessins Ariyan Javanpeykar 2012-05-12T11:02:18Z 2012-05-12T11:08:09Z <p>This does not answer your question. But it was a bit too long to put as a comment.</p> <p>Firstly, it seems that the following old question is of some relevance.</p> <p><a href="http://mathoverflow.net/questions/68213/families-of-curves-for-which-the-belyi-degree-can-be-easily-bounded" rel="nofollow">http://mathoverflow.net/questions/68213/families-of-curves-for-which-the-belyi-degree-can-be-easily-bounded</a></p> <p>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 </p> <ol> <li>Fermat curves</li> <li>Modular curves (congruence or non-congruence)</li> <li>Hurwitz spaces (see JSE's answer to the above question)</li> <li>Galois Belyi curves = Wolfart-curves = Galois three-point covers</li> <li>Elkies' curves (see his answer to the above question).</li> </ol> <p>Let me elaborate on 2. If <code>$\Gamma\subset \mathrm{SL}_2(\mathbf{Z})$</code> is a finite index subgroup, you can consider the quotient <code>$Y_\Gamma = \Gamma\backslash \mathbf{H}$</code>, where $\mathbf{H}$ is the complex upper half-plane and <code>$\mathrm{SL}_2(\mathbf{Z})$</code> acts on $\mathbf{H}$ by Mobius transformations. The curve <code>$Y_\Gamma$</code> 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 <code>$Y_\Gamma \to Y_{\mathrm{SL}_2(\mathbf{Z})} = Y(1)$</code> induced by the inclusion <code>$\Gamma\subset \mathrm{SL}_2(\mathbf{Z})$</code>. 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 <code>$j:X(1)\to\mathbf{P}^1(\mathbf{C}$</code>. (The branch points are the elliptic points $0$, $1728$ and the cusp $\infty$ of $X(1)$.)</p> <p>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 $\mathbf{Q}$? </p> <p>Firstly, assume that <code>$X\to \mathbf{P}^1$</code> is a dessin of prime degree. It's clear that this morphism will not factor.</p> <p>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 <code>$X_0(n)$</code> can be shown to factor through some elliptic curve. If this is the case, the answer is likely to be no for $n$ big.</p> <p>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 <code>$\mathbf{P}^1-\{0,1,\infty\}$</code>.</p> <p>But your $H_{X,Y}$ will be zero or infinite. </p> <p>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 <code>$\{0,1,\infty\}$</code>. 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).</p> <p>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.)</p> <p>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 $\pi = fg$.</p> <p>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.</p>