The AMS describes the importance of the red book as follows. “This is one of the few books that attempt to convey in pictures some of the highly abstract notions that arise in the field of algebraic geometry. In his response upon receiving the prize, Mumford recalled that some of his drawings from The Red Book were included in a collection called Five Centuries of French Mathematics. This seemed fitting, he noted: “After all, it was the French who started impressionist painting and isn’t this just an impressionist scheme for rendering geometry?””

These days it is perfectly possible to get a good grasp on difficult concepts from algebraic geometry by reading blogs, watching YouTube or plugging in equations to sophisticated math-programs. In the early seventies though, if you wanted to know what Grothendieck’s scheme-revolution was all about you had no choice but to wade through the EGA’s and SGA’s and they were notorious for being extremely user-unfriendly regarding illustrations…

So the few depictions of schemes available, drawn by people sufficiently fluent in Grothendieck’s new geometric language had no less than treasure-map-cult-status and were studied in minute detail. Mumford’s red book was a gold mine for such treasure maps. Here’s my favorite one, scanned from the original mimeographed notes (it looks somewhat tidier in the Springer-version)

It is the first depiction of $\mathbf{spec}(\mathbb{Z}[x]) $, the affine scheme of the ring $\mathbb{Z}[x] $ of all integral polynomials. Mumford calls it the”arithmetic surface” as the picture resembles the one he made before of the affine scheme $\mathbf{spec}(\mathbb{C}[x,y]) $ corresponding to the two-dimensional complex affine space $\mathbb{A}^2_{\mathbb{C}} $. Mumford adds that the arithmetic surface is ‘the first example which has a real mixing of arithmetic and geometric properties’.

Let’s have a closer look at the treasure map. It introduces some new signs which must have looked exotic at the time, but have since become standard tools to depict algebraic schemes.

For starters, recall that the underlying topological space of $\mathbf{spec}(\mathbb{Z}[x]) $ is the set of all prime ideals of the integral polynomial ring $\mathbb{Z}[x] $, so the map tries to list them all as well as their inclusions/intersections.

The doodle in the right upper corner depicts the ‘generic point’ of the scheme. That is, the geometric object corresponding to the prime ideal $~(0) $ (note that $\mathbb{Z}[x] $ is an integral domain). Because the zero ideal is contained in any other prime ideal, the algebraic/geometric mantra (“inclusions reverse when shifting between algebra and geometry”) asserts that the gemetric object corresponding to $~(0) $ should contain all other geometric objects of the arithmetic plane, so it is just the whole plane! Clearly, it is rather senseless to depict this fact by coloring the whole plane black as then we wouldn’t be able to see the finer objects. Mumford’s solution to this is to draw a hairy ball, which in this case, is sufficiently thick to include fragments going in every possible direction. In general, one should read these doodles as saying that the geometric object represented by this doodle contains all other objects seen elsewhere in the picture if the hairy-ball-doodle includes stuff pointing in the direction of the smaller object. So, in the case of the object corresponding to $~(0) $, the doodle has pointers going everywhere, saying that the geometric object contains all other objects depicted.

Let’s move over to the doodles in the lower right-hand corner. They represent the geometric object corresponding to principal prime ideals of the form $~(p(x)) $, where $p(x) $ in an irreducible polynomial over the integers, that is, a polynomial which we cannot write as the product of two smaller integral polynomials. The objects corresponding to such prime ideals should be thought of as ‘horizontal’ curves in the plane.

The doodles depicted correspond to the prime ideal $~(x) $, containing all polynomials divisible by $x $ so when we divide it out we get, as expected, a domain $\mathbb{Z}[x]/(x) \simeq \mathbb{Z} $, and the one corresponding to the ideal $~(x^2+1) $, containing all polynomials divisible by $x^2+1 $, which can be proved to be a prime ideals of $\mathbb{Z}[x] $ by observing that after factoring out we get $\mathbb{Z}[x]/(x^2+1) \simeq \mathbb{Z}[i] $, the domain of all Gaussian integers $\mathbb{Z}[i] $. The corresponding doodles (the ‘generic points’ of the curvy-objects) have a predominant horizontal component as they have the express the fact that they depict horizontal curves in the plane. It is no coincidence that the doodle of $~(x^2+1) $ is somewhat bulkier than the one of $~(x) $ as the later one must only depict the fact that all points lying on the straight line to its left belong to it, whereas the former one must claim inclusion of all points lying on the ‘quadric’ it determines.

Apart from these ‘horizontal’ curves, there are also ‘vertical’ lines corresponding to the principal prime ideals $~(p) $, containing the polynomials, all of which coefficients are divisible by the prime number $p $. These are indeed prime ideals of $\mathbb{Z}[x] $, because their quotients are
$\mathbb{Z}[x]/(p) \simeq (\mathbb{Z}/p\mathbb{Z})[x] $ are domains, being the ring of polynomials over the finite field $\mathbb{Z}/p\mathbb{Z} = \mathbb{F}_p $. The doodles corresponding to these prime ideals have a predominant vertical component (depicting the ‘vertical’ lines) and have a uniform thickness for all prime numbers $p $ as each of them only has to claim ownership of the points lying on the vertical line under them.

Right! So far we managed to depict the zero prime ideal (the whole plane) and the principal prime ideals of $\mathbb{Z}[x] $ (the horizontal curves and the vertical lines). Remains to depict the maximal ideals. These are all known to be of the form
$\mathfrak{m} = (p,f(x)) $
where $p $ is a prime number and $f(x) $ is an irreducible integral polynomial, which remains irreducible when reduced modulo $p $ (that is, if we reduce all coefficients of the integral polynomial $f(x) $ modulo $p $ we obtain an irreducible polynomial in $~\mathbb{F}_p[x] $). By the algebra/geometry mantra mentioned before, the geometric object corresponding to such a maximal ideal can be seen as the ‘intersection’ of an horizontal curve (the object corresponding to the principal prime ideal $~(f(x)) $) and a vertical line (corresponding to the prime ideal $~(p) $). Because maximal ideals do not contain any other prime ideals, there is no reason to have a doodle associated to $\mathfrak{m} $ and we can just depict it by a “point” in the plane, more precisely the intersection-point of the horizontal curve with the vertical line determined by $\mathfrak{m}=(p,f(x)) $. Still, Mumford’s treasure map doesn’t treat all “points” equally. For example, the point corresponding to the maximal ideal $\mathfrak{m}_1 = (3,x+2) $ is depicted by a solid dot $\mathbf{.} $, whereas the point corresponding to the maximal ideal $\mathfrak{m}_2 = (3,x^2+1) $ is represented by a fatter point $\circ $. The distinction between the two ‘points’ becomes evident when we look at the corresponding quotients (which we know have to be fields). We have

$\mathbb{Z}[x]/\mathfrak{m}_1 = \mathbb{Z}[x]/(3,x+2)=(\mathbb{Z}/3\mathbb{Z})[x]/(x+2) = \mathbb{Z}/3\mathbb{Z} = \mathbb{F}_3 $ whereas $\mathbb{Z}[x]/\mathfrak{m}_2 = \mathbb{Z}[x]/(3,x^2+1) = \mathbb{Z}/3\mathbb{Z}[x]/(x^2+1) = \mathbb{F}_3[x]/(x^2+1) = \mathbb{F}_{3^2} $

because the polynomial $x^2+1 $ remains irreducible over $\mathbb{F}_3 $, the quotient $\mathbb{F}_3[x]/(x^2+1) $ is no longer the prime-field $\mathbb{F}_3 $ but a quadratic field extension of it, that is, the finite field consisting of 9 elements $\mathbb{F}_{3^2} $. That is, we represent the ‘points’ lying on the vertical line corresponding to the principal prime ideal $~(p) $ by a solid dot . when their quotient (aka residue field is the prime field $~\mathbb{F}_p $, by a bigger point $\circ $ when its residue field is the finite field $~\mathbb{F}_{p^2} $, by an even fatter point $\bigcirc $ when its residue field is $~\mathbb{F}_{p^3} $ and so on, and on. The larger the residue field, the ‘fatter’ the corresponding point.

In fact, the ‘fat-point’ signs in Mumford’s treasure map are an attempt to depict the fact that an affine scheme contains a lot more information than just the set of all prime ideals. In fact, an affine scheme determines (and is determined by) a “functor of points”. That is, to every field (or even every commutative ring) the affine scheme assigns the set of its ‘points’ defined over that field (or ring). For example, the $~\mathbb{F}_p $-points of $\mathbf{spec}(\mathbb{Z}[x]) $ are the solid . points on the vertical line $~(p) $, the $~\mathbb{F}_{p^2} $-points of $\mathbf{spec}(\mathbb{Z}[x]) $ are the solid . points and the slightly bigger $\circ $ points on that vertical line, and so on.

This concludes our first attempt to decypher Mumford’s drawing, but if we delve a bit deeper, we are bound to find even more treasures… (to be continued).

My blogging is (at least for the moment) transfered to the F_un Mathematics blog which some prefer to call the “ceci n’est pas un corps”-blog, which is very fine with me.

Javier gave a talk at MPI on Soule’s approach to algebraic geometry over the elusive field with one element $\mathbb{F}_1 $ and wrote two posts about it The skeleton of Soule’s F_un geometry and Gadgets a la Soule. The rough idea being that a variety over the field with one element only acquires flesh after a base extension to $\mathbb{Z} $ and to cyclotomic integers.

I did some posts on a related (but conceptually somewhat easier) approach due to Alain Connes and Katia Consani. I’ve tried to explain their construction at the level of (mature) undergraduate students. So far, there are three posts part1, part2 and part3. Probably there is one more session to come in which I will explain why they need functors to graded sets.

In the weeks to come we plan to post about applications of this F_un-geometry to noncommutative geometry (the Bost-Connes system) and Grothendieck’s anabelian geometry (the theory of dessins d’enfant). I’ll try to leave a short account of the main posts here, but clearly you are invited to feed your feedreader this.

Perhaps I’ll return here for a week mid november to do some old-fashioned vacation blogging. I have to admit I did underestimate Numeo.fr. Rumours have it that our place is connected wirelessly to the web…

I’ll try to convert chunks of related posts in one go, so that I can at least give them correct self-references. Today’s work consisted in rewriting the posts of my virtual course, in march of this year, on dessins d’enfants and its connection to noncommutative geometry (a precursor of what Ive been blogging about recently). These posts were available through the PDF-archive but are from now on open to the internal search-function. Here are the internal links and a short description of their contents

• The best rejected proposal ever on Grothendieck’s programme and Belyi maps.
• Monsieur Mathieu on the dessins of $M_{12} $ and $M_{24} $.
• The cartographers’ groups 1 on the free product structure of $PSL_2(\mathbb{Z}) $ and related groups.
• The cartographers’ groups 2 continuation and a simpler proof due to Alperin.
• Anabelian geometry on sources of representations of the modular group.
• The noncommutative manifold of a Riemann surface a mental picture to get all finite dimensional representations of the coordinate ring of a curve.
• Noncommutative curves and their manifolds extension to twisted curves.
• Permutation representations of monodromy groups how dessins give such representations.

Besides, I’ve added a few scattered old posts, many more to follow…

$\mathbf{simp}_c~\hat{\Gamma} \subset \mathbf{simp}~\Gamma $

and based on the example $\mu_{\infty} = \mathbf{simp}_c~\hat{\mathbb{Z}} \subset \mathbf{simp}~\mathbb{Z} = \mathbb{C}^{\ast} $ we would like the above embedding to be dense in some kind of noncommutative analogon of the Zariski topology on $\mathbf{simp}~\Gamma $.

We use the Zariski topology on $\mathbf{simp}~\mathbb{C} \Gamma $ as in these two M-geometry posts (( already, I regret terminology, I should have just called it noncommutative geometry )). So, what’s this idea in this special case? Let $\mathfrak{g} $ be the vectorspace with basis the conjugacy classes of elements of $\Gamma $ (that is, the space of class functions). As explained here it is a consequence of the Artin-Procesi theorem that the linear functions $\mathfrak{g}^{\ast} $ separate finite dimensional (semi)simple representations of $\Gamma $. That is we have an embedding

$\mathbf{simp}~\Gamma \subset \mathfrak{g}^{\ast} $

and we can define closed subsets of $\mathbf{simp}~\Gamma $ as subsets of simple representations on which a set of class-functions vanish. With this definition of Zariski topology it is immediately clear that the image of $\mathbf{simp}_c~\hat{\Gamma} $ is dense. For, suppose it would be contained in a proper closed subset then there would be a class-function vanishing on all simples of $\hat{\Gamma} $ so, in particular, there should be a bound on the number of simples of finite quotients $\Gamma/N $ which clearly is not the case (just look at the quotients $PSL_2(\mathbb{F}_p) $).

But then, the same holds if we replace ‘simples of $\hat{\Gamma} $’ by ‘simple components of permutation representations of $\Gamma $’. This is the importance of Farey symbols to the representation problem of the modular group. They give us a manageable subset of simples which is nevertheless dense in the whole space. To utilize this a natural idea might be to ask what such a permutation representation can see of the modular group, or in geometric terms, what the tangent space is to $\mathbf{simp}~\Gamma $ in a permutation representation (( more precisely, in the ‘cluster’ of points making up the simple components of the representation representation )). We will call this the modular content of the permutation representation and to understand it we will have to compute the tangent quiver $\vec{t}~\mathbb{C} \Gamma $.

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(\overline{K}/K) $, where K is an extension of finite type of the prime field) on (profinite) geometric fundamental
groups of algebraic varieties (defined over K), and more particularly (breaking 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 profinite compactification of the modular group $SL_2(\mathbb{Z}) $, whose quotient by its centre
$\{ \pm 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).

The above text is taken from Alexander Grothendieck‘s visionary text Sketch of a Programme. He was interested in the permutation representations of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ as they correspond via Belyi-maps and his own notion of dessins d’enfants to smooth projective curves defined over $\overline{\mathbb{Q}} $. One can now study the action of the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ on these curves and their associated dessins. Because every permutation representation of $\Gamma $ factors over a finite quotient this gives an action of the absolute Galois group as automorphisms on the profinite compactification

$\hat{\Gamma} = \underset{\leftarrow}{lim}~\Gamma/N $

where the limit is taken over all finite index normal subgroups $N \triangleleft PSL_2(\mathbb{Z}) $. In this way one realizes the absolute Galois group as a subgroup of the outer automorphism group of the profinite group $\hat{\Gamma} $. As a profinite group is a compact topological group one should study its continuous finite dimensional representations which are precisely those factoring through a finite quotient. In the case of $\hat{\Gamma} $ the simple continuous representations $\mathbf{simp}_c~\hat{\Gamma} $ are precisely the components of the permutation representations of the modular group. So in a sense, anabelian geometry is the study of these continuous simples together wirth the action of the absolute Galois group on it.

In noncommutative geometry we are interested in a related representation theoretic problem. We would love to know the simple finite dimensional representations $\mathbf{simp}~\Gamma $ of the modular group as this would give us all simples of the three string braid group $B_3 $. So a natural question presents itself : how are these two ‘geometrical’ objects $\mathbf{simp}_c~\hat{\Gamma} $ (anabelian) and $\mathbf{simp}~\Gamma $ (noncommutative) related and can we use one to get information about the other?

This is all rather vague so far, so let us work out a trivial case to get some intuition. Consider the profinite completion of the infinite Abelian group

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

As all simple representations of an Abelian group are one-dimensional and because all continuous ones factor through a finite quotient $\mathbb{Z}/n\mathbb{Z} $ we see that in this case

$\mathbf{simp}_c~\hat{\mathbb{Z}} = \mu_{\infty} $

is the set of all roots of unity. On the other hand, the simple representations of $\mathbb{Z} $ are also one-dimensional and are determined by the image of the generator so

$\mathbf{simp}~\mathbb{Z} = \mathbb{C} – { 0 } = \mathbb{C}^* $

Clearly we have an embedding $\mu_{\infty} \subset \mathbb{C}^* $ and the roots of unity are even dense in the Zariski topology. This might look a bit strange at first because clearly all roots of unity lie on the unit circle which ‘should be’ their closure in the complex plane, but that’s because we have a real-analytic intuition. Remember that the Zariski topology of $\mathbb{C}^*$ is just the cofinite topology, so any closed set containing the infinitely many roots of unity should be the whole space!

Let me give a pedantic alternative proof of this (but one which makes it almost trivial that a similar result should be true for most profinite completions…). If $c $ is the generator of $\mathbb{Z} $ then the different conjugacy classes are precisely the singletons $c^n $. Now suppose that there is a polynomial $a_0+a_1x+\ldots+a_mx^m $ vanishing on all the continuous simples of $\hat{\mathbb{Z}} $ then this means that the dimensions of the character-spaces of all finite quotients $\mathbb{Z}/n\mathbb{Z} $ should be bounded by $m $ (for consider $x $ as the character of $c $), which is clearly absurd.

Hence, whenever we have a finitely generated group $G $ for which there is no bound on the number of irreducibles for finite quotients, then morally the continuous simple space for the profinite completion

$\mathbf{simp}_c~\hat{G} \subset \mathbf{simp}~G $

should be dense in the Zariski topology on the noncommutative space of simple finite dimensional representations of $G $. In particular, this should be the case for the modular group $PSL_2(\mathbb{Z}) $.

There is just one tiny problem : unlike the case of $\mathbb{Z} $ for which this space is an ordinary (ie. commutative) affine variety $\mathbb{C}^* $, what do we mean by the “Zariski topology” on the noncommutative space $\mathbf{simp}~PSL_2(\mathbb{Z}) $ ? Next time we will clarify what this might be and show that indeed in this case the subset

$\mathbf{simp}_c~\hat{\Gamma} \subset \mathbf{simp}~\Gamma $

will be a Zariski closed subset!

The rationality problem

The Manin-Marcolli cave

The taxicab curve

Anabelian geometry

Noncommutative curves and their manifolds

The noncommutative manifold of a Riemann surface

The best rejected proposal ever

Noncommutative geometry – a medieval science?

Master class 2007

2006 paper nominees

Coalgebras and non-geometry 3

Coalgebras and non-geometry 2

Coalgebras and non-geometry 1

Krull and Paris

Noncommutative at Newton

Noncommutative Fourier transforms

Non-(commutative) geometry

Non-geometry

A good day at the arXiv

Noncommutative geometry master class

Noncommutative complete intersections

Master program 2006

Noncommutative topology 4

Noncommutative topology 3

Noncommutative topology 2

Noncommutative topology 1

Alain Connes on … everything

Jacobian conjecture remains open

The Klein stack

Sexing-up curves

Why nag? 3

Granada Notes

Why nag? 2

Why nag? 1

$\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…

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.

From what I deduce from reading the first posts, Arup Pal set up a new blog
called Noncommutative
Geometry and subsequently handed it over to Masoud Khalkhali who then
got Alain Connes to post on it who, in turn, is asking people to submit
posts, such as todays post by David Goss.

Somehow, most people refer to the Noncommutative Geometry blog as
“Alain Connes’ blog” (for example Doctor Gero, Not Even
Wrong, the n-category cafe and
possibly many others).

David Corfield over at the n-category cafe stops
short of suggesting to rename (by analogy) NeverEndingBooks into
Kontsevich’s blog

A new blog Noncommutative Geometry
has begun, which appears to be of the Connesian variety. (Connes himself
has already commented there.) We mentioned a couple of weeks ago that
there are different flavours of noncommutative geometry. The
Kontsevichian variety, nongeometry, finds its blog voice in Lieven Le
Bruyn’s NeverEndingBooks. It would be interesting to see some
interaction.

I don’t think I will opt for a dialectic
response to the Noncommutative Geometry Blog, although I realize this would result
in more enjoyable reading for some of you…

For starters,
I’ve signed up to another flagship of noncommutative
_differential_ geometry : noncommutativegeometry.net, though it is unclear to me what action (if any) is
going on over there.

Further, I plan to move my talks at the master class
noncommutative geometry to the virtual lecture room of this blog,
hoping to get the desired interaction. We’ll start later this week and
the pace will be pretty easy going. A tentative title might be
“Anabelian versus Noncommutative Geometry”.