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 the absolute Galois group of the algebraic closure of the rational numbers , that is the group of all -automorphisms of . 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 which was defined over the field 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 (which is complex-conjugation). Hence, the Galois group acts faithfully on the drawings associated to maps which are ramified only over the points .

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 and assume that there is a map ramified only over the points , 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 ) 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 . On this set, there is an action of the absolute Galois group 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 ramified in the three points 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 the map is only ramified over (finitely many) -points. Let S be the finite set of all Galois-conjugates of these points and consider the polynomial

Now, do a resultant trick. Consider the polynomial then the roots of are exactly the finite critical values of , is again defined over and has lower degree (in ) than in . Continue this trick a finite number of times untill you have constructed a polynomial of degree zero.

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

Take any three elements of T, then there always exist integers such that the three points go under a linear fractional transformation (a Moebius-function associated to a matrix in ) to . Under the transformation the points 0 and 1 go to 0 and 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 -points of a composition at most having three elements and then a final Moebius transformation gets them to , done!

As a tribute for this clever argument, maps ramified only in 0,1 and 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 given by K has a large group of automorphism, namely the simple group of order 168 It is a classical fact (see for example the excellent paper by Noam Elkies The Klein quartic in number theory) that the quotient map is ramified only in the points 0,1728 and 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 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 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 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 , which is not so surprising as the group of automorphisms is 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 a permutation representation (via the Belyi map and its dessin) of one of the congruence subgroups or or of itself.

2005 was the year that the DaVinci code craze hit Belgium. (I started reading Dan Brown’s Digital Fortress and Angels and Demons a year before on the way back from a Warwick conference and when I read DVC a few months later it was an anti-climax…). Anyway, what better way to end 2005 than with a fitting chess problem, composed by Noam Elkies

The problem is to give an infinite sequence of numbers, the n-th term of the sequence being the number of ways White can force checkmate in exactly n moves. With the DVC-hint given, clearly only one series can be the solution… To prove it, note that White’s only non-checkmating moves are with the Bishop traveling along the path (g1,h2,g3,h4) and use symmetry to prove that the number of paths of length exactly k starting from h2 is the same as those starting from g3…

If that one was too easy for you, consider the same problem for the position

Here the solution are the 2-powers of those of the first problem. The proof essentially is that White has now two ways to deliver checkmate : Na6 and Nd7… For the solutions and more interesting chess-problems consult Noam Elkies’ excellent paper New directions in enumerative chess problems. Remains the problem which sequences can arise on an $N \times N$ board with an infinite supply of chess pieces!

Noam Elkies is one of those persons I seem to bump into (figuratively speaking) wherever my interests take me. At the moment I’m reading (long overdue, I know, I know) the excellent book Notes on Fermat’s Last Theorem by Alf Van der Poorten. On page 48, Elkies figures as an innocent bystander in the 1994 April fools joke e-perpetrated by Henri Darmon in the midst of all confusion about ‘the gap’ in Wiles’ proof.

There has been a really amazing development today on Fermat’s Last Theorem. Noam Elkies has announced a counterexample, so that FLT is not true after all! He spoke about this at the institute today. The solution to Fermat that he constructs involves an incredibly large prime exponent (larger than $10^{20}$), but it is constructive. The main idea seems to be a kind of Heegner-point construction, combined with a really ingenious descent for passing from the modular curves to the Fermat curve. The really difficult part of the argument seems to be to show that the field of definition of the solution (which, a priori, is some ring class field of an imaginary quadratic field) actually descends to $\mathbb{Q}$. I wasn’t able to get all the details, which were quite intricate…

Elkies is also an excellent composer of chess problems. The next two problems he composed as New Year’s greetings. The problem is : “How many shortest sequences exists (with only white playing) to reach the given position?”

Here’s Elkies’ solution :

There are 2004 sequences of the minimal length 12. Each consists of the sin- gle move g3, the 3-move sequence c4,Nc3,Rb1, and one of the three 8-move sequences Nf3,Ne5,f3,Kf2,Ke3,Kd3(d4),Kc4(c5),Kb5. The move g3 may be played at any point, and so contributes a factor of 12. If the King goes through c5 then the 3- and 8-move sequences are independent, and can be played in $\binom{11}{3}$ orders. If the King goes through c4 then the entire 8-move sequence must be played before the 3-move sequence begins, so there are only two possibilities, depending on the choice of Kd3 or Kd4. Hence the total count is $12(\binom{11}{3}+2)=2004$ as claimed.

A year later he composed the problem

of which Elkies’ solution is :

There are 2005 sequences of the minimal length 14. This uses the happy coincidence $\binom{14}{4}=1001$. Here White plays the 4-move sequence f4,Kf2,Kg3,Kh3 and one of the five sequences Nc3,Na4,c3,Qc2,Qe4,d3,Bd2(e3,f4,g5,h6),Rc1,Rc2,Bc1 of length 10. If the Bishop goes to d2 or e3, the sequences are independent, and can be played in $\binom{14}{4}$ orders. Otherwise the Bishop must return to c1 before White plays f4, so the entire 10-move sequence must be played before the 4-move sequence begins. Hence the total count is $2 \binom{14}{4}+3 = 2005$.

With just a few weeks remaining, anyone in for a 2006 puzzle?

Klein’s quartic $X$is the smooth plane projective curve defined by $x^3y+y^3z+z^3x=0$ and is one of the most remarkable mathematical objects around. For example, it is a Hurwitz curve meaning that the finite group of symmetries (when the genus is at least two this group can have at most $84(g-1)$ elements) is as large as possible, which in the case of the quartic is $168$ and the group itself is the unique simple group of that order, $G = PSL2(\mathbb{F}7)$ also known as Klein\’s group. John Baez has written a beautiful page on the Klein quartic and its symmetries. Another useful source of information is a paper by Noam Elkies The Klein quartic in number theory.
The quotient map $X \rightarrow X/G \simeq \mathbb{P}^1$ has three branch points of orders $2,3,7$ in the points on $\mathbb{P}^1$ with coordinates $1728,0,\infty$. These points correspond to the three non-free $G$-orbits consisting resp. of $84,56$ and $24$ points.
Now, remove from $X$ a couple of $G$-orbits to obtain an affine open subset $Y$ such that $G$ acts on its cordinate ring $\mathbb{C}[Y]$ and form the Klein stack (or hereditary order) $\mathbb{C}[Y] \bigstar G$, the skew group algebra. In case the open subset $Y$ contains all non-free orbits, the one quiver of this qurve has the following shape Here, the three components correspond to the three non-free orbits and the vertices correspond to the isoclasses of simple $\mathbb{C}[Y] \bigstar G$ of dimension smaller than $168$. There are two such of dimension $84$, three of dimension $56$ and seven of dimension $24$ which I gave the non-imaginative names \’twins\’, \’trinity\’ and \’the dwarfs\’. As we want to spice up later this Klein stack to a larger group, we need to know the structure of these exceptional simples as $G$-representations. Surely, someone must have written a paper on the general problem of finding the $G$-structure of simples of skew-group algebras $A \bigstar G$, so if you know a reference please let me know. I used an old paper by Idun Reiten and Christine Riedtmann to do this case (which is easier as the stabilizer subgroups are cyclic and hence the induced representations of their one-dimensionals correspond to the exceptional simples).

Noam D. Elkies is a Harvard mathematician whose main research interests have to do with lattices and elliptic curves. He is also a very talented composer of chess problems. The problem to teh left is a proof game in 14 moves. That is, find the UNIQUE legal chess game leading to the given situation after the 14th move by black. Elkies has also written a beautiful paper On Numbers and Endgames applying combinatorial game theory (a la Winning Ways!) to chess-endgames (mutual Zugzwang positions correspond to zero positions) and a follow-up article Higher Nimbers in pawn endgames on large chessboards. Together with Richard Stanley he wrote a paper for the Mathematical Intelligencer called The Mathematical Knight which is stuffed with chess problems! But perhaps most surprising is that he managed to run his own course on Chess and Mathematics!