The best rejected proposal ever

By lieven

Dessins d'enfants

The Oscar in the category The Best Rejected Research Proposal in Mathematics (ever) goes to … Alexander Grothendieck
for his proposal Esquisse d’un Programme, Grothendieck\’s research program from 1983, written as part of his application for a position at the CNRS, the French equivalent of the NSF. An English translation is available.

Here is one of the problems discussed : Give TWO non-trivial elements of $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ the absolute Galois group of the algebraic closure of the rational numbers $\overline{\mathbb{Q}}$ , that is the group of all $\mathbb{Q}$ -automorphisms of $\overline{\mathbb{Q}}$ . One element most of us can give (complex-conjugation) but to find any other element turns out to be an extremely difficult task.

To get a handle on this problem, Grothendieck introduced his ‘Dessins d’enfants’ (Children’s drawings). Recall from last session the pictures of the left and right handed Monsieur Mathieu

The left hand side drawing was associated to a map $\mathbb{P}^1_{\mathbb{C}} \rightarrow \mathbb{P}^1_{\mathbb{C}}$ which was defined over the field $\mathbb{Q} \sqrt{-11}$ whereas the right side drawing was associated to the map given when one applies to all coefficients the unique non-trivial automorphism in the Galois group $Gal(\mathbb{Q}\sqrt{-11}/\mathbb{Q})$ (which is complex-conjugation). Hence, the Galois group $Gal(\mathbb{Q}\sqrt{-11}/\mathbb{Q})$ acts faithfully on the drawings associated to maps $\mathbb{P}^1_{\mathbb{Q}\sqrt{-11}} \rightarrow \mathbb{P}^1_{\mathbb{Q}\sqrt{-11}}$ which are ramified only over the points $\{ 0,1,\infty \}$ .

Grothendieck’s idea was to extend this to more general maps. Assume that a projective smooth curve (a Riemann surface) X is defined over the algebraic numbers $\overline{\mathbb{Q}}$ and assume that there is a map $X \rightarrow \mathbb{P}^1_{\mathbb{C}}$ ramified only over the points $\{ 0,1,\infty \}$ , then we can repeat the procedure of last time and draw a picture on X consisting of d edges (where d is the degree of the map, that is the number of points lying over another point of $\mathbb{P}^1_{\mathbb{C}}$ ) between white resp. black points (the points of X lying over 1 (resp. over 0)).

Call such a drawing a ‘dessin d\’enfant’ and look at the collection of ALL dessins d’enfants associated to ALL such maps where X runs over ALL curves defined over $\overline{\mathbb{Q}}$ . On this set, there is an action of the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ and if this action would be faithful, then this would give us insight into this group. However, at that time even the existence of a map $X \rightarrow \mathbb{P}^1$ ramified in the three points $\{ 0,1,\infty \}$ seemed troublesome to prove, as Grothendieck recalls in his proposal

In more erudite terms, could it be true that every projective non-singular algebraic curve defined over a number field occurs as a possible ‚ modular curve‚ parametrising elliptic curves equipped with a suitable rigidification? Such a supposition seemed so crazy that I was almost embarrassed to submit it to the competent people in the domain. Deligne when I consulted him found it crazy indeed, but didn’t have any counterexample up his sleeve. Less than a year later, at the International Congress in Helsinki, the Soviet mathematician Bielyi announced exactly that result, with a proof of disconcerting simplicity which fit into two little pages of a letter of Deligne ‚ never, without a doubt, was such a deep and disconcerting result proved in so few lines!
In the form in which Bielyi states it, his result essentially says that every algebraic curve defined over a number field can be obtained as a covering of the projective line ramified only over the points 0, 1 and infinity. This result seems to have remained more or less unobserved. Yet, it appears to me to have considerable importance. To me, its essential message is that there is a profound identity between the combinatorics of finite maps on the one hand, and the geometry of algebraic curves defined over number fields on the other. This deep result, together with the algebraic- geometric interpretation of maps, opens the door onto a new, unexplored world within reach of all, who pass by without seeing it.

Belyi’s proof is indeed relatively easy (full details can be found in the paper Dessins d’enfants on the Riemann sphere by Leila Schneps). Roughly it goes as follows : as both X and the map are defined over $\overline{\mathbb{Q}}$ the map is only ramified over (finitely many) $\overline{\mathbb{Q}}$ -points. Let S be the finite set of all Galois-conjugates of these points and consider the polynomial

$f_0(z_0) = \prod_{s \in S} (z_0 -s) \in \mathbb{Q}[z_0]$

Now, do a resultant trick. Consider the polynomial $f_1(z_1) = Res_{z_0}(\frac{d f_0}{d z_0},f_0(z_0)-z_1)$ then the roots of are exactly the finite critical values of f_0 , f_1 is again defined over $\mathbb{Q}$ and has lower degree (in z_1 ) than f_0 in z_1 . Continue this trick a finite number of times untill you have constructed a polynomial $f_n(z_n) \in \mathbb{Q}[z_n]$ of degree zero.

Composing the original map with the maps f_j in succession yields that all ramified points of this composition are $\mathbb{Q}$ -points! Now, we only have to limit the number of these ramified $\mathbb{Q}$ -points (let us call this set T) to three.

Take any three elements of T, then there always exist integers $m,n \in \mathbb{Z}$ such that the three points go under a linear fractional transformation (a Moebius-function associated to a matrix in $PGL_2(\mathbb{Q})$ ) to $\{ 0,\frac{m}{m+n},1 \}$ . Under the transformation $z \rightarrow \frac{(m+n)^{m+n}}{m^m n^n}z^m(1-z)^n$ the points 0 and 1 go to 0 and $\frac{m}{m+n}$ goes to 1 whence the ramified points of the composition are one less in number than T. Continuing in this way we can get the set of ramified $\mathbb{Q}$ -points of a composition at most having three elements and then a final Moebius transformation gets them to $\{ 0,1,\infty \}$ , done!

As a tribute for this clever argument, maps $X \rightarrow \mathbb{P}^1$ ramified only in 0,1 and $\infty$ are now called Belyi morphisms. Here is an example of a Belyi-morphism (and the corresponding dessin d’enfants) associated to one of the most famous higher genus curves around : the Klein quartic (if you haven’t done so yet, take your time to go through this marvelous pre-blog post by John Baez).

One can define the Klein quartic as the plane projective curve K with defining equation in $\mathbb{P}^2_{\\mathbb{C}}$ given by X^3Y+Y^3Z+Z^3X = 0 K has a large group of automorphism, namely the simple group of order 168 $G = PSL_2(\mathbb{F}_7) = SL_3(\mathbb{F}_2)$ It is a classical fact (see for example the excellent paper by Noam Elkies The Klein quartic in number theory) that the quotient map $K \rightarrow K/G = \mathbb{P}^1_{\mathbb{C}}$ is ramified only in the points 0,1728 and $\infty$ and the number of points of K lying over them are resp. 56, 84 and 24. Now, compose this map with the Moebius transormation taking $\{ 0,1728,\infty \} \rightarrow \{ 0,1,\infty \}$ then the resulting map is a Belyi-map for the Klein quartic. A topological construction of the Klein quartic is fitting 24 heptagons together so that three meet in each vertex, see below for the gluing data-picture in the hyperbolic plane : the different heptagons are given a number but they appear several times telling how they must fit together)

The resulting figure has exactly $\frac{7 \times 24}{2} = 84$ edges and the 84 points of K lying over 1 (the white points in the dessin) correspond to the midpoints of the edges. There are exactly $\frac{7 \times 24}{3}=56$ vertices corresponding to the 56 points lying over 0 (the black points in the dessin). Hence, the dessin d\’enfant associated to the Klein quartic is the figure traced out by the edges on K. Giving each of the 168 half-edges a different number one assigns to the white points a permutation of order two and to the three-valent black-points a permutation of order three, whence to the Belyi map of the Klein quartic corresponds a 168-dimensional permutation representation of $SL_2(\mathbb{Z})$ , which is not so surprising as the group of automorphisms is $PSL_2(\mathbb{F}_7)$ and the permutation representation is just the regular representation of this group.

Next time we will see how one can always associate to a curve defined over $\overline{\mathbb{Q}}$ a permutation representation (via the Belyi map and its dessin) of one of the congruence subgroups $\Gamma(2)$ or $\Gamma_0(2)$ or of $SL_2(\mathbb{Z})$ itself.

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dessin d'enfants, Elkies, Galois, geometry, Grothendieck, groups, hyperbolic, Klein, Mathieu, modular, permutation representation, Riemann

4 comments
Posted in geometry, groups, modular
Written on Wed, 07 March 2007 at 8:17 pm
Tags: dessin d'enfants, Elkies, Galois, geometry, Grothendieck, groups, hyperbolic, Klein, Mathieu, modular, permutation representation, Riemann
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4 Responses to “The best rejected proposal ever”

anabelian geometry at neverendingbooks Says:
December 27th, 2007 at 12:42 pm
[...] Last time we saw that a curve defined over gives rise to a permutation representation of or one of its subgroups (of index 2) or (of index 6). As the corresponding monodromy group is finite, this representation factors through a normal subgroup of finite index, so it makes sense to look at the profinite completion of , which is the inverse limit of finite groups where N ranges over all normalsubgroups of finite index. These profinte completions are horrible beasts even for easy groups such as . Its profinite completion is [...]
recycled : dessins at neverendingbooks Says:
December 27th, 2007 at 3:57 pm
[...] The best rejected proposal ever on Grothendieck’s programme and Belyi maps. [...]
noncommutative curves and their maniflds | neverendingbooks Says:
January 2nd, 2008 at 8:14 pm
[...] Categories: geometry Digg This [?] Table of contents for Dessins d’enfantsMonsieur MathieuThe best rejected proposal everThe cartographers’ groupsthe cartographers’ groups (2)the noncommutative manifold of a [...]
permutation representations of monodromy groups | neverendingbooks Says:
January 2nd, 2008 at 8:14 pm
[...] geometry and groups Digg This [?] Table of contents for Dessins d’enfantsMonsieur MathieuThe best rejected proposal everThe cartographers’ groupsthe cartographers’ groups (2)the noncommutative manifold of a [...]