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

src="http://www.neverendingbooks.org/DATA/twomathieus.jpg" />

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 $f_1(z_1) $ 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.