neverendingbooks

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.

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?”

$\\begin{position}
\\White(Kb5,Qd1,Rb1,Rh1,Nc3,Ne5,Bc1,Bf1,a2,b2,c4,d2,e2,f3,g3,h2)
\\end{position}{\\font\\mathbb{C}hess=chess10 \\showboard
}xc $

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

$\\begin{position}
\\White(Kh3,Qe4,Rc2,Rh1,Na4,Ng1,Bc1,Bf1,a2,b2,c3,d3,e2,f4,g2,h2)
\\end{position}{\\font\\mathbb{C}hess=chess10 \\showboard
}xd $

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, $G = PSL_2(\mathbb{F}_7)$ also known as
Klein\’s group. John Baez has written a [beautiful page](http://math.ucr.edu/home/baez/klein.html) on the Klein quartic and
its symmetries. Another useful source of information is a paper by Noam
Elkies [The Klein quartic in number theory](www.msri.org/publications/books/Book35/files/elkies.pd).
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](www.matrix.ua.ac.be/master/coursenotes/onequiver.pdf) of this
qurve has the following shape $\xymatrix{\vtx{} \ar@/^/[dd] \\
\\ \vtx{} \ar@/^/[uu]} $ $\xymatrix{& \vtx{} \ar[ddl] & \\
& & \\ \vtx{} \ar[rr] & & \vtx{} \ar[uul]} $ $\xymatrix{& &
\vtx{} \ar[dll] & & \\ \vtx{} \ar[d] & & & & \vtx{} \ar[ull] \\ \vtx{}
\ar[dr] & & & & \vtx{} \ar[u] \\ & \vtx{} \ar[rr] & & \vtx{} \ar[ur]
&} $ 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!