Monsieur Mathieu

Even a
virtual course needs an opening line, so here it is : Take your favourite
$SL_2(\mathbb{Z}) $-representation
Here is mine : the
permutation presentation of the Mathieu group(s). Emile Leonard Mathieu is remembered especially for his discovery (in 1861
and 1873) of five sporadic simple groups named after him, the Mathieu
groups $M_{11},M_{12},M_{22},M_{23} $ and $M_{24} $. These were studied in
his thesis on transitive functions. He had a refreshingly direct style
of writing. I'm not sure what Cauchy would have thought (Cauchy died in
1857) about this 'acknowledgement' in his 1861-paper in which Mathieu
describes $M_{12} $ and claims the construction of $M_{24} $.

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Also the opening sentenses of his 1873 paper are nice,
something along the lines of "if no expert was able to fill in the
details of my claims made twelve years ago, I'd better do it
myself".

However, even after this paper opinions remained divided on
the issue whether or not he did really achieve his goal, and the matter
was settled decisively by Ernst Witt connecting the Mathieu groups to
Steiner systems (if I recall well from Mark Ronan's book Symmetry and the monster)

As Mathieu observed, the quickest
way to describe these groups would be to give generators, but as these
groups are generated by two permutations on 12 respectively 24
elements, we need to have a mnemotechnic approach to be able to
reconstruct them whenever needed.

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Here is a nice approach, due to
Gunther Malle in a Luminy talk in 1993 on "Dessins d'enfants" (more
about them later). Consider the drawing of "Monsieur
Mathieu" on the left. That
is, draw the left-handed bandit picture on 6 edges and vertices,
divide each edge into two and give numbers to both parts (the actual
numbering is up to you, but for definiteness let us choose the
one on the left). Then, $M_{12} $ is generated by the order two permutation
describing the labeling of both parts of the
edges

$s=(1,2)(3,4)(5,8)(7,6)(9,12)(11,10) $

together with
the order three permutation obtained from cycling counterclockwise
around a trivalent vertex and calling out the labels one encounters. For
example, the three cycle corresponding to the 'neck vertex' is
$~(1,2,3) $ and the total permutation
is

$t=(1,2,3)(4,5,6)(8,9,10) $

A quick verification using
GAP tells that these elements do indeed generate a simple group of order
95040.


Similarly, if you have to reconstruct the largest Mathieu group
from scratch, apply the same method to the the picture above or to "ET Mathieu"
drawing on the left. This picture I copied from Alexander Zvonkin's paper How to draw a group as
well as the computational details below.

This is all very nice and well but what do these drawings have
to do with Grothendieck's "dessins d'enfants"? Consider the map from
the projective line onto
itself

$\mathbb{P}^1_{\mathbb{C}}$ \rightarrow
\mathbb{P}^1_{\mathbb{C}} $

defined by the rational
map

$f(z) =
\frac{(z^3-z^2+az+b)^3(z^3+cz^2+dz+e)}{Kz} $

where N. Magot
calculated that

$a=\frac{107+7 \sqrt{-11}}{486},
b=-\frac{13}{567}a+\frac{5}{1701}, c=-\frac{17}{9},
d=\frac{23}{7}a+\frac{256}{567},
e=-\frac{1573}{567}a+\frac{605}{1701} $

and finally

$K =
-\frac{16192}{301327047}a+\frac{10880}{903981141} $

One verifies
that this map is 12 to 1 everywhere except over the points ${
0,1,\infty } $ (that is, there are precisely 12 points mapping under f
to a given point of $\mathbb{P}^1_{\mathbb{C}} - { 0,1,\infty } $.
From the expression of f(z) it is clear that over 0 there lie 6
points (3 of which with multiplicity three, the others of multiplicity
one). Over $\infty $ there are two points, one with multiplicity 11
and one with multiplicity one. The difficult part is to compute the
points lying over 1. The miraculous fact of the given values is
that

$f(z)-1 =
\frac{-B(z)^2}{Kz} $

where

$B(z)=z^6+\frac{1}{11}(10c-8)z
^5+(5a+9d-7c)z^4+(2b+4ac+8e-6d)z^3+(3ad+bc-5e)z^2+2aez-be) $

and
hence there are 6 points lying over 1 each with mutiplicity
two.


Right, now consider the complex projective line
$\mathbb{P}^1_{\mathbb{C}} $ as the Riemann sphere $S^2 $ and mark the six points lying over 1 by a white vertex and
the six points lying over 0 with a black vertex (in the source
sphere). Now, lift the real interval $[0,1] $ in the target sphere
$\mathbb{P}^1_{\mathbb{C}} = S^2 $ to its inverse image on the source
sphere. As there are exactly 12 points lying over each real
number $0 \lneq r \lneq 1 $, this inverse image will consist of 12
edges which are noncrossing and each end in one black and one white
vertex. The obtained graph will look like the \"Monsieur Mathieu\"
drawing above with the vertices corresponding to the black vertices and
the three points over 1 of multiplicity three corresponding to the
trivalent vertices, those of multiplicity one to the three end-vertices.
The white vertices correspond to mid-points of the six edges, so that we
do get a drawing with twelve edges, one corresponding to each
number. From the explicit description of f(z) it is clear that
this map is defined over $\mathbb{Q}\sqrt{-11} $ which is also the
smallest field containing all character-values of the Mathieu group
$M_{12} $. Further, the Galois group of the
extension $Gal(\mathbb{Q}\sqrt{-11}/\mathbb{Q}) =
\mathbb{Z}/2\mathbb{Z} $ and is generated by complex
conjugation. So, one might wonder what would happen if we
replaced in the definition of the rational map f(z) the value of a
by $a = \frac{107-\sqrt{-11}}{486} $. It turns out that
this modified map has the same properties as $f(z) $ so again one can
draw on the source sphere a picture consisting of twelve edges each
ending in a white and black vertex.


If we consider the white vertices
(which incidentally each lie on two edges as all points lying over 0
are of multiplicity two) as mid-points of longer edges connecting the
black vertices we obtain a drawing on the sphere which looks like
\"Monsieur Mathieu\" but this time as a right handed bandit, and
applying our mnemotechnic rule we obtain _another_ (non conjugated)
embedding of $M_{12} $ in the full symmetric group on 12 vertices.
What is the connection with $SL_2(\mathbb{Z}) $-representations?
Well, the permutation generators s and t of $M_{12} $ (or $M_{24} $
for that matter) have orders two and three, whence there is a projection
from the free group product $C_2 \star C_3 $ (here $C_n $ is just the
cyclic group of order n) onto $M_{12} $ (respectively $M_{24} $). Next
time we will say more about such free group products and show (among
other things) that $PSL_2(\mathbb{Z}) \simeq C_2 \star
C_3 $ whence the connection with $SL_2(\mathbb{Z}) $. In a
following lecture we will extend the Monsieur Mathieu example to
arbitrary dessins d\'enfants which will allow us to assign to curves
defined over $\overline{\mathbb{Q}} $ permutation representations of
$SL_2(\mathbb{Z}) $ and other _cartographic groups_ such as the
congruence subgroups $\Gamma_0(2) $ and
$\Gamma(2) $.