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Tag: Mathieu

recap and outlook

Published May 7, 2007 by lievenlb

After a lengthy spring-break, let us continue with our course on noncommutative geometry and $SL_2(\mathbb{Z}) $-representations. Last time, we have explained Grothendiecks mantra that all algebraic curves defined over number fields are contained in the profinite compactification
$\widehat{SL_2(\mathbb{Z})} = \underset{\leftarrow}{lim}~SL_2(\mathbb{Z})/N $ of the modular group $SL_2(\mathbb{Z}) $ and in the knowledge of a certain subgroup G of its group of outer automorphisms. In particular we have seen that many curves defined over the algebraic numbers $\overline{\mathbb{Q}} $ correspond to permutation representations of $SL_2(\mathbb{Z}) $. The profinite compactification $\widehat{SL_2}=\widehat{SL_2(\mathbb{Z})} $ is a continuous group, so it makes sense to consider its continuous n-dimensional representations $\mathbf{rep}_n^c~\widehat{SL_2} $ Such representations are known to have a finite image in $GL_n(\mathbb{C}) $ and therefore we get an embedding $\mathbf{rep}_n^c~\widehat{SL_2} \hookrightarrow \mathbf{rep}_n^{ss}~SL_n(\mathbb{C}) $ into all n-dimensional (semi-simple) representations of $SL_2(\mathbb{Z}) $. We consider such semi-simple points as classical objects as they are determined by – curves defined over $\overline{Q} $ – representations of (sporadic) finite groups – modlart data of fusion rings in RCTF – etc… To get a feel for the distinction between these continuous representations of the cofinite completion and all representations, consider the case of $\hat{\mathbb{Z}} = \underset{\leftarrow}{lim}~\mathbb{Z}/n \mathbb{Z} $. Its one-dimensional continuous representations are determined by roots of unity, whereas all one-dimensional (necessarily simple) representations of $\mathbb{Z}=C_{\infty} $ are determined by all elements of $\mathbb{C} $. Hence, the image of $\mathbf{rep}_1^c~\hat{\mathbb{Z}} \hookrightarrow \mathbf{rep}_1~C_{\infty} $ is contained in the unit circle

and though these points are very special there are enough of them (technically, they form a Zariski dense subset of all representations). Our aim will be twofold : (1) when viewing a classical object as a representation of $SL_2(\mathbb{Z}) $ we can define its modular content (which will be the noncommutative tangent space in this classical point to the noncommutative manifold of $SL_2(\mathbb{Z}) $). In this way we will associate noncommutative gadgets to our classical object (such as orders in central simple algebras, infinite dimensional Lie algebras, noncommutative potentials etc. etc.) which give us new tools to study these objects. (2) conversely, as we control the tangentspaces in these special points, they will allow us to determine other $SL_2(\mathbb{Z}) $-representations and as we vary over all classical objects, we hope to get ALL finite dimensional modular representations. I agree this may all sound rather vague, so let me give one example we will work out in full detail later on. Remember that one can reconstruct the sporadic simple Mathieu group $M_{24} $ from the dessin d’enfant

This
dessin determines a 24-dimensional permutation representation (of
$M_{24} $ as well of $SL_2(\mathbb{Z}) $) which
decomposes as the direct sum of the trivial representation and a simple
23-dimensional representation. We will see that the noncommutative
tangent space in a semi-simple representation of
$SL_2(\mathbb{Z}) $ is determined by a quiver (that is, an
oriented graph) on as many vertices as there are non-isomorphic simple
components. In this special case we get the quiver on two points
$\xymatrix{\vtx{} \ar@/^2ex/[rr] & & \vtx{} \ar@/^2ex/[ll]
\ar@{=>}@(ur,dr)^{96} } $ with just one arrow in each direction
between the vertices and 96 loops in the second vertex. To the
experienced tangent space-reader this picture (and in particular that
there is a unique cycle between the two vertices) tells the remarkable
fact that there is **a distinguished one-parameter family of
24-dimensional simple modular representations degenerating to the
permutation representation of the largest Mathieu-group**. Phrased
differently, there is a specific noncommutative modular Riemann surface
associated to $M_{24} $, which is a new object (at least as far
as I’m aware) associated to this most remarkable of sporadic groups.
Conversely, from the matrix-representation of the 24-dimensional
permutation representation of $M_{24} $ we obtain representants
of all of this one-parameter family of simple
$SL_2(\mathbb{Z}) $-representations to which we can then perform
noncommutative flow-tricks to get a Zariski dense set of all
24-dimensional simples lying in the same component. (Btw. there are
also such noncommutative Riemann surfaces associated to the other
sporadic Mathieu groups, though not to the other sporadics…) So this
is what we will be doing in the upcoming posts (10) : explain what a
noncommutative tangent space is and what it has to do with quivers (11)
what is the noncommutative manifold of $SL_2(\mathbb{Z}) $? and what is its connection with the Kontsevich-Soibelman coalgebra? (12)
is there a noncommutative compactification of $SL_2(\mathbb{Z}) $? (and other arithmetical groups) (13) : how does one calculate the noncommutative curves associated to the Mathieu groups? (14) : whatever comes next… (if anything).
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anabelian geometry

Published March 22, 2007 by lievenlb

Last time we saw
that a curve defined over $\overline{\mathbb{Q}} $ gives rise
to a permutation representation of $PSL_2(\mathbb{Z}) $ or one
of its subgroups $\Gamma_0(2) $ (of index 2) or
$\Gamma(2) $ (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 $SL_2(\mathbb{Z}) $, which is the inverse limit
of finite
groups $\underset{\leftarrow}{lim}~SL_2(\mathbb{Z})/N $
where N ranges over all normalsubgroups of finite index. These
profinte completions are horrible beasts even for easy groups such as
$\mathbb{Z} $. Its profinite completion
is

$\underset{\leftarrow}{lim}~\mathbb{Z}/n\mathbb{Z} =
\prod_p \hat{\mathbb{Z}}_p $

where the right hand side
product of p-adic integers ranges over all prime numbers! The
_absolute Galois group_
$G=Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ acts on all curves
defined over $\overline{\mathbb{Q}} $ and hence (via the Belyi
maps ans the corresponding monodromy permutation representation) there
is an action of $G $ on the profinite completions of the
carthographic groups.

This is what Grothendieck calls anabelian
algebraic geometry

Returning to the general
case, since finite maps can be interpreted as coverings over
$\overline{\mathbb{Q}} $ of an algebraic curve defined over
the prime field $~\mathbb{Q} $ itself, it follows that the
Galois group $G $ of $\overline{\mathbb{Q}} $ over
$~\mathbb{Q} $ acts on the category of these maps in a
natural way.
For instance, the operation of an automorphism
$~\gamma \in G $ on a spherical map given by the rational
function above is obtained by applying $~\gamma $ to the
coefficients of the polynomials P , Q. Here, then, is that
mysterious group $G $ intervening as a transforming agent on
topologico- combinatorial forms of the most elementary possible
nature, leading us to ask questions like: are such and such oriented
maps ‚conjugate or: exactly which are the conjugates of a given
oriented map? (Visibly, there is only a finite number of these).
I considered some concrete cases (for coverings of low degree) by
various methods, J. Malgoire considered some others ‚ I doubt that
there is a uniform method for solving the problem by computer. My
reflection quickly took a more conceptual path, attempting to
apprehend the nature of this action of G.
One sees immediately
that roughly speaking, this action is expressed by a certain
outer action of G on the profinite com- pactification of the
oriented cartographic group $C_+^2 = \Gamma_0(2) $ , and this
action in its turn is deduced by passage to the quotient of the
canonical outer action of G on the profinite fundamental group
$\hat{\pi}_{0,3} $ of
$(U_{0,3})_{\overline{\mathbb{Q}}} $ where
$U_{0,3} $ denotes the typical curve of genus 0 over the
prime field Q, with three points re- moved.
This is how my
attention was drawn to what I have since termed anabelian
algebraic geometry, whose starting point was exactly a study
(limited for the moment to characteristic zero) of the action of
absolute Galois groups (particularly the groups Gal(K/K),
where K is an extension of finite type of the prime field) on
(profinite) geometric fundamental groups of algebraic varieties
(defined over K), and more particularly (break- ing with a
well-established tradition) fundamental groups which are very far
from abelian groups (and which for this reason I call
anabelian).
Among these groups, and very close to
the group $\hat{\pi}_{0,3} $ , there is the profinite
compactification of the modular group $Sl_2(\mathbb{Z}) $,
whose quotient by its centre ±1 contains the former as congruence
subgroup mod 2, and can also be interpreted as an oriented
cartographic group, namely the one classifying triangulated
oriented maps (i.e. those whose faces are all triangles or
monogons).

and a bit further, on page
250

I would like to conclude this rapid outline
with a few words of commentary on the truly unimaginable richness
of a typical anabelian group such as $SL_2(\mathbb{Z}) $
doubtless the most remarkable discrete infinite group ever
encountered, which appears in a multiplicity of avatars (of which
certain have been briefly touched on in the present report), and which
from the point of view of Galois-Teichmuller theory can be
considered as the fundamental ‚building block‚ of the
Teichmuller tower
The element of the structure of
$Sl_2(\mathbb{Z}) $ which fascinates me above all is of course
the outer action of G on its profinite compactification. By
Bielyi’s theorem, taking the profinite compactifications of subgroups
of finite index of $Sl_2(\mathbb{Z}) $, and the induced
outer action (up to also passing to an open subgroup of G), we
essentially find the fundamental groups of all algebraic curves (not
necessarily compact) defined over number fields K, and the outer
action of $Gal(\overline{K}/K) $ on them at least it is
true that every such fundamental group appears as a quotient of one
of the first groups.
Taking the anabelian yoga
(which remains conjectural) into account, which says that an anabelian
algebraic curve over a number field K (finite extension of Q) is
known up to isomorphism when we know its mixed fundamental group (or
what comes to the same thing, the outer action of
$Gal(\overline{K}/K) $ on its profinite geometric
fundamental group), we can thus say that
all algebraic
curves defined over number fields are contained in the profinite
compactification $\widehat{SL_2(\mathbb{Z})} $ and in the
knowledge of a certain subgroup G of its group of outer
automorphisms!

To study the absolute
Galois group $Gal(\overline{\mathbb{\mathbb{Q}}}/\mathbb{Q}) $ one
investigates its action on dessins denfants. Each dessin will be part of
a finite family of dessins which form one orbit under the Galois action
and one needs to find invarians to see whether two dessins might belong
to the same orbit. Such invariants are called _Galois invariants_ and
quite a few of them are known.

Among these the easiest to compute
are

  • the valency list of a dessin : that is the valencies of all
    vertices of the same type in a dessin
  • the monodromy group of a dessin : the subgroup of the symmetric group $S_d $ where d is
    the number of edges in the dessin generated by the partitions $\tau_0 $
    and $\tau_1 $ For example, we have seen
    before     that the two
    Mathieu-dessins

form a Galois orbit. As graphs (remeber we have to devide each
of the edges into two and the midpoints of these halfedges form one type
of vertex, the other type are the black vertices in the graphs) these
are isomorphic, but NOT as dessins as we have to take the embedding of
them on the curve into account. However, for both dessins the valency
lists are (white) : (2,2,2,2,2,2) and (black) :
(3,3,3,1,1,1) and one verifies that both monodromy groups are
isomorphic to the Mathieu simple group $M_{12} $ though they are
not conjugated as subgroups of $S_{12} $.

Recently, new
Galois invariants were obtained from physics. In Children’s drawings
from Seiberg-Witten curves
the authors argue that there is a close connection between Grothendiecks
programme of classifying dessins into Galois orbits and the physics
problem of classifying phases of N=1 gauge theories…

Apart
from curves defined over $\overline{\mathbb{Q}} $ there are
other sources of semi-simple $SL_2(\mathbb{Z}) $
representations. We will just mention two of them and may return to them
in more detail later in the course.

Sporadic simple groups and
their representations There are 26 exceptional finite simple groups
and as all of them are generated by two elements, there are epimorphisms
$\Gamma(2) \rightarrow S $ and hence all their representations
are also semi-simple $\Gamma(2) $-representations. In fact,
looking at the list of ‘standard generators’ of the sporadic
simples

(here the conjugacy classes of the generators follow the
notation of the Atlas project) we see that all but
possibly one are epimorphic images of $\Gamma_0(2) = C_2 \ast
C_{\infty} $ and that at least 12 of then are epimorphic images
of $PSL_2(\mathbb{Z}) = C_2 \ast
C_3 $.

Rational conformal field theories Another
source of $SL_2(\mathbb{Z}) $ representations is given by the
modular data associated to rational conformal field theories.

These
representations also factor through a quotient by a finite index normal
subgroup and are therefore again semi-simple
$SL_2(\mathbb{Z}) $-representations. For a readable
introduction to all of this see chapter 6 \”Modular group
representations throughout the realm\” of the
book Moonshine beyond the monster the bridge connecting algebra, modular forms and physics by Terry
Gannon. In fact, the whole book
is a good read. It introduces a completely new type of scientific text,
that of a neverending survey paper…

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permutation representations of monodromy groups

Published March 20, 2007 by lievenlb

Today we will explain how curves defined over
$\overline{\mathbb{Q}} $ determine permutation representations
of the carthographic groups. We have seen that any smooth projective
curve $C $ (a Riemann surface) defined over the algebraic
closure $\overline{\mathbb{Q}} $ of the rationals, defines a
_Belyi map_ $\xymatrix{C \ar[rr]^{\pi} & & \mathbb{P}^1} $ which is only ramified over the three points
$\\{ 0,1,\infty \\} $. By this we mean that there are
exactly $d $ points of $C $ lying over any other point
of $\mathbb{P}^1 $ (we call $d $ the degree of
$\pi $) and that the number of points over $~0,1~ $ and
$~\infty $ is smaller than $~d $. To such a map we
associate a _dessin d\’enfant_, a drawing on $C $ linking the
pre-images of $~0 $ and $~1 $ with exactly $d $
edges (the preimages of the open unit-interval). Next, we look at
the preimages of $~0 $ and associate a permutation
$\tau_0 $ of $~d $ letters to it by cycling
counter-clockwise around these preimages and recording the edges we
meet. We repeat this procedure for the preimages of $~1 $ and
get another permutation $~\tau_1 $. That is, we obtain a
subgroup of the symmetric group $ \langle \tau_0,\tau_1
\rangle \subset S_d $ which is called the monodromy
group of the covering $\pi $.

For example, the
dessin on the right is
associated to a degree $8 $ map $\mathbb{P}^1 \rightarrow
\mathbb{P}^1 $ and if we let the black (resp. starred) vertices be
the preimages of $~0 $ (respectively of $~1 $), then the
corresponding partitions are $\tau_0 = (2,3)(1,4,5,6) $
and $\tau_1 = (1,2,3)(5,7,8) $ and the monodromy group
is the alternating group $A_8 $ (use
GAP ).

But wait! The map is also
ramified in $\infty $ so why don\’t we record also a
permutation $\tau_{\infty} $ and are able to compute it from
the dessin? (Note that all three partitions are needed if we want to
reconstruct $C $ from the $~d $ sheets as they encode in
which order the sheets fit together around the preimages). Well,
the monodromy group of a $\mathbb{P}^1 $ covering ramified only
in three points is an epimorphic image of the fundamental
group of the sphere
minus three points $\pi_1(\mathbb{P}^1 – { 0,1,\infty
}) $ That is, the group of all loops beginning and
ending in a basepoint upto homotopy (that is, two such loops are the
same if they can be transformed into each other in a continuous way
while avoiding the three points).

This group is generated by loops
$\sigma_i $ running from the basepoint to nearby the i-th
point, doing a counter-clockwise walk around it and going back to be
basepoint $Q_0 $ and the epimorphism to the monodromy group is given by sending

$\sigma_1 \mapsto \tau_0~\quad~\sigma_2 \mapsto
\tau_1~\quad~\sigma_3 \mapsto \tau_{\infty} $

Now,
these three generators are not independent. In fact, this fundamental
group is

$\pi_1(\mathbb{P}^1 – \\{ 0,1,\infty \\}) =
\langle \sigma_1,\sigma_2,\sigma_3~\mid~\sigma_1 \sigma_2
\sigma_3 = 1 \rangle $

To understand this, let us begin
with an easier case, that of the sphere minus one point. The fundamental group of the plane minus one point is
$~\mathbb{Z} $ as it encodes how many times we walk around the
point. However, on the sphere the situation is different as we can make
our walk around the point longer and longer until the whole walk is done
at the backside of the sphere and then we can just contract our walk to
the basepoint. So, there is just one type of walk on a sphere minus one
point (upto homotopy) whence this fundamental group is trivial. Next,
let us consider the sphere minus two points


Repeat the foregoing to the walk $\sigma_2 $, that
is, strech the upper part of the circular tour all over the backside of
the sphere and then we see that we can move it to fit with the walk
$\sigma_1$ BUT for the orientation of the walk! That is, if we do this
modified walk $\sigma_1 \sigma_2^{\’} $ we just made the
trivial walk. So, this fundamental group is $\langle
\sigma_1,\sigma_2~\mid~\sigma_1 \sigma_2 = 1 \rangle =
\mathbb{Z} $ This is also the proof of the above claim. For,
we can modify the third walk $\sigma_3 $ continuously so that
it becomes the walk $\sigma_1 \sigma_2 $ but
with the reversed orientation ! As $\sigma_3 =
(\sigma_1 \sigma_2)^{-1} $ this allows us to compute the
\’missing\’ permutation $\tau_{\infty} = (\tau_0
\tau_1)^{-1} $ In the example above, we obtain
$\tau_{\infty}= (1,2,6,5,8,7,4)(3) $ so it has two cycles
corresponding to the fact that the dessin has two regions (remember we
should draw ths on the sphere) : the head and the outer-region. Hence,
the pre-images of $\infty$ correspond to the different regions of the
dessin on the curve $C $. For another example,
consider the degree 168 map

$K \rightarrow \mathbb{P}^1 $

which is the modified orbit map for the action of
$PSL_2(\mathbb{F}_7) $ on the Klein quartic.
The corresponding dessin is the heptagonal construction of the Klein
quartic

Here, the pre-images of 1 correspond to the midpoints of the
84 edges of the polytope whereas the pre-images of 0 correspond to the
56 vertices. We can label the 168 half-edges by numbers such that
$\tau_0 $ and $\tau_1 $ are the standard generators b
resp. a of the 168-dimensional regular representation (see the atlas
page ).
Calculating with GAP the element $\tau_{\infty} = (\tau_0
\tau_1)^{-1} = (ba)^{-1} $ one finds that this permutation
consists of 24 cycles of length 7, so again, the pre-images of
$\infty $ lie one in each of the 24 heptagonal regions of the
Klein quartic. Now, we are in a position to relate curves defined
over $\overline{Q} $ via their Belyi-maps and corresponding
dessins to Grothendiecks carthographic groups $\Gamma(2) $,
$\Gamma_0(2) $ and $SL_2(\mathbb{Z}) $. The
dessin gives a permutation representation of the monodromy group and
because the fundamental group of the sphere minus three
points $\pi_1(\mathbb{P}^1 – \\{ 0,1,\infty \\}) =
\langle \sigma_1,\sigma_2,\sigma_3~\mid~\sigma_1 \sigma_2
\sigma_3 = 1 \rangle = \langle \sigma_1,\sigma_2
\rangle $ is the free group op two generators, we see that
any dessin determines a permutation representation of the congruence
subgroup $\Gamma(2) $ (see this
post where we proved that this
group is free). A clean dessin is one for which one type of
vertex has all its valancies (the number of edges in the dessin meeting
the vertex) equal to one or two. (for example, the pre-images of 1 in
the Klein quartic-dessin or the pre-images of 1 in the monsieur Mathieu
example ) The corresponding
permutation $\tau_1 $ then consists of 2-cycles and hence the
monodromy group gives a permutation representation of the free
product $C_{\infty} \ast C_2 =
\Gamma_0(2) $ Finally, a clean dessin is said to be a
quilt dessin if also the other type of vertex has all its valancies
equal to one or three (as in the Klein quartic or Mathieu examples).
Then, the corresponding permutation has order 3 and for these
quilt-dessins the monodromy group gives a permutation representation of
the free product $C_2 \ast C_3 =
PSL_2(\mathbb{Z}) $ Next time we will see how this lead
Grothendieck to his anabelian geometric approach to the absolute Galois
group.

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The best rejected proposal ever

Published March 7, 2007 by lievenlb


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 $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.

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Monsieur Mathieu

Published March 1, 2007 by lievenlb

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} $.

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.

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) $.

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symmetry and the monster

Published June 2, 2006 by lievenlb

Mark
Ronan has written a beautiful book intended for the general public
on Symmetry and the Monster. The
book’s main theme is the classification of the finite simple groups. It
starts off with the introduction of groups by Galois, gives the
classifivcation of the finite Lie groups, the Feit-Thompson theorem and
the construction of several of the sporadic groups (including the
Mathieu groups, the Fischer and Conway groups and clearly the
(Baby)Monster), explains the Leech lattice and the Monstrous Moonshine
conjectures and ends with Richard Borcherds proof of them using vertex
operator algebras. As in the case of Music of the
Primes it is (too) easy to be critical about notation. For example,
whereas groups are just called symmetry groups, I don’t see the point of
calling simple groups ‘atoms of symmetry’. But, unlike du Sautoy,
Mark Ronan stays close to mathematical notation, lattices are just
lattices, characer-tables are just that, j-function is what it is etc.
And even when he simplifies established teminology, for example
‘cyclic arithmetic’ for modular arithmetic, ‘cross-section’
for involution centralizer, ‘mini j-functions’ for Hauptmoduln
etc. there are footnotes (as well as a glossary) mentioning the genuine
terms. Group theory is a topic with several colourful people
including the three Johns John Leech, John
McKay and John Conway
and several of the historical accounts in the book are a good read. For
example, I’ve never known that the three Conway groups were essentially
discovered in just one afternoon and a few telephone exchanges between
Thompson and Conway. This year I’ve tried to explain some of
monstrous moonshine to an exceptionally good second year of
undergraduates but failed miserably. Whereas I somehow managed to give
the construction and proof of simplicity of Mathieu 24, elliptic and
modular functions were way too difficult for them. Perhaps I’ll give it
another (downkeyed) try using ‘Symmetry and the Monster’ as
reading material. Let’s hope Oxford University Press will soon release a
paperback (and cheaper) version.

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sexing up curves

Published June 10, 2005 by lievenlb

Here the
story of an idea to construct new examples of non-commutative compact
manifolds, the computational difficulties one runs into and, when they
are solved, the white noise one gets. But, perhaps, someone else can
spot a gem among all gibberish…
[Qurves](http://www.neverendingbooks.org/toolkit/pdffile.php?pdf=/TheLibrary/papers/qaq.pdf) (aka quasi-free algebras, aka formally smooth
algebras) are the \’affine\’ pieces of non-commutative manifolds. Basic
examples of qurves are : semi-simple algebras (e.g. group algebras of
finite groups), [path algebras of
quivers](http://www.lns.cornell.edu/spr/2001-06/msg0033251.html) and
coordinate rings of affine smooth curves. So, let us start with an
affine smooth curve $X$ and spice it up to get a very non-commutative
qurve. First, we bring in finite groups. Let $G$ be a finite group
acting on $X$, then we can form the skew-group algebra $A = \mathbfk[X]
\bigstar G$. These are examples of prime Noetherian qurves (aka
hereditary orders). A more pompous way to phrase this is that these are
precisely the [one-dimensional smooth Deligne-Mumford
stacks](http://www.math.lsa.umich.edu/~danielch/paper/stacks.pdf).
As the 21-st century will turn out to be the time we discovered the
importance of non-Noetherian algebras, let us make a jump into the
wilderness and consider the amalgamated free algebra product $A =
(\mathbf k[X] \bigstar G) \ast_{\mathbf k G} \mathbfk H$ where $G
\subset H$ is an interesting extension of finite groups. Then, $A$ is
again a qurve on which $H$ acts in a way compatible with the $G$-action
on $X$ and $A$ is hugely non-commutative… A very basic example :
let $\mathbb{Z}/2\mathbb{Z}$ act on the affine line $\mathbfk[x]$ by
sending $x \mapsto -x$ and consider a finite [simple
group](http://mathworld.wolfram.com/SimpleGroup.html) $M$. As every
simple group has an involution, we have an embedding
$\mathbb{Z}/2\mathbb{Z} \subset M$ and can construct the qurve
$A=(\mathbfk[x] \bigstar \mathbb{Z}/2\mathbb{Z}) \ast_{\mathbfk
\mathbb{Z}/2\mathbb{Z}} \mathbfk M$ on which the simple group $M$ acts
compatible with the involution on the affine line. To study the
corresponding non-commutative manifold, that is the Abelian category
$\mathbf{rep}~A$ of all finite dimensional representations of $A$ we have
to compute the [one quiver to rule them
all](http://www.matrix.ua.ac.be/master/coursenotes/onequiver.pdf) for
$A$. Because $A$ is a qurve, all its representation varieties
$\mathbf{rep}_n~A$ are smooth affine varieties, but they may have several
connected components. The direct sum of representations turns the set of
all these components into an Abelian semigroup and the vertices of the
\’one quiver\’ correspond to the generators of this semigroup whereas
the number of arrows between two such generators is given by the
dimension of $Ext^1_A(S_i,S_j)$ where $S_i,S_j$ are simple
$A$-representations lying in the respective components. All this
may seem hard to compute but it can be reduced to the study of another
quiver, the Zariski quiver associated to $A$ which is a bipartite quiver
with on the left the \’one quiver\’ for $\mathbfk[x] \bigstar
\mathbb{Z}/2\mathbb{Z}$ which is just $\xymatrix{\vtx{}
\ar@/^/[rr] & & \vtx{} \ar@/^/[ll]} $ (where the two vertices
correspond to the two simples of $\mathbb{Z}/2\mathbb{Z}$) and on the
right the \’one quiver\’ for $\mathbf k M$ (which just consists of as
many verticers as there are simple representations for $M$) and where
the number of arrows from a left- to a right-vertex is the number of
$\mathbb{Z}/2\mathbb{Z}$-morphisms between the respective simples. To
make matters even more concrete, let us consider the easiest example
when $M = A_5$ the alternating group on $5$ letters. The corresponding
Zariski quiver then turns out to be $\xymatrix{& & \vtx{1} \\\
\vtx{}\ar[urr] \ar@{=>}[rr] \ar@3[drr] \ar[ddrr] \ar[dddrr] \ar@/^/[dd]
& & \vtx{4} \\\ & & \vtx{5} \\\ \vtx{} \ar@{=>}[uurr] \ar@{=>}[urr]
\ar@{=>}[rr] \ar@{=>}[drr] \ar@/^/[uu] & & \vtx{3} \\\ & &
\vtx{3}} $ The Euler-form of this quiver can then be used to
calculate the dimensions of the EXt-spaces giving the number of arrows
in the \’one quiver\’ for $A$. To find the vertices, that is, the
generators of the component semigroup we have to find the minimal
integral solutions to the pair of equations saying that the number of
simple $\mathbb{Z}/2\mathbb{Z}$ components based on the left-vertices is
equal to that one the right-vertices. In this case it is easy to see
that there are as many generators as simple $M$ representations. For
$A_5$ they correspond to the dimension vectors (for the Zariski quiver
having the first two components on the left) $\begin{cases}
(1,2,0,0,0,0,1) \\ (1,2,0,0,0,1,0) \\ (3,2,0,0,1,0,0) \\
(2,2,0,1,0,0,0) \\ (1,0,1,0,0,0,0) \end{cases}$ We now have all
info to determine the \’one quiver\’ for $A$ and one would expect a nice
result. Instead one obtains a complete graph on all vertices with plenty
of arrows. More precisely one obtains as the one quiver for $A_5$
$\xymatrix{& & \vtx{} \ar@{=}[dll] \ar@{=}[dddl] \ar@{=}[dddr]
\ar@{=}[drr] & & \\\ \vtx{} \ar@(ul,dl)|{4} \ar@{=}[rrrr]|{6}
\ar@{=}[ddrrr]|{8} \ar@{=}[ddr]|{4} & & & & \vtx{} \ar@(ur,dr)|{8}
\ar@{=}[ddlll]|{6} \ar@{=}[ddl]|{10} \\\ & & & & & \\\ & \vtx{}
\ar@(dr,dl)|{4} \ar@{=}[rr]|{8} & & \vtx{} \ar@(dr,dl)|{11} & } $
with the number of arrows (in each direction) indicated. Not very
illuminating, I find. Still, as the one quiver is symmetric it follows
that all quotient varieties $\mathbf{iss}_n~A$ have a local Poisson
structure. Clearly, the above method can be generalized easily and all
examples I did compute so far have this \’nearly complete graph\’
feature. One might hope that if one would start with very special
curves and groups, one might obtain something more interesting. Another
time I\’ll tell what I got starting from Klein\’s quartic (on which the
simple group $PSL_2(\mathbb{F}_7)$ acts) when the situation was sexed-up
to the sporadic simple Mathieu group $M_{24}$ (of which
$PSL_2(\mathbb{F}_7)$ is a maximal subgroup).

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