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Tag: modular

anabelian geometry

Last time we saw
that a curve defined over Q gives rise
to a permutation representation of PSL2(Z) or one
of its subgroups Γ0(2) (of index 2) or
Γ(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 SL2(Z), which is the inverse limit
of finite
groups lim SL2(Z)/N
where N ranges over all normalsubgroups of finite index. These
profinte completions are horrible beasts even for easy groups such as
Z. Its profinite completion
is

lim Z/nZ=pZ^p

where the right hand side
product of p-adic integers ranges over all prime numbers! The
_absolute Galois group_
G=Gal(Q/Q) acts on all curves
defined over 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
Q of an algebraic curve defined over
the prime field  Q itself, it follows that the
Galois group G of Q over
 Q acts on the category of these maps in a
natural way.
For instance, the operation of an automorphism
 γG on a spherical map given by the rational
function above is obtained by applying  γ 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=Γ0(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
π^0,3 of
(U0,3)Q where
U0,3 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 π^0,3 , there is the profinite
compactification of the modular group Sl2(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 SL2(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
Sl2(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 Sl2(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(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(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 SL2(Z)^ and in the
knowledge of a certain subgroup G of its group of outer
automorphisms!

To study the absolute
Galois group Gal(Q/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 Sd where d is
    the number of edges in the dessin generated by the partitions τ0
    and τ1 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 M12 though they are
not conjugated as subgroups of S12.

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 Q there are
other sources of semi-simple SL2(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
Γ(2)S and hence all their representations
are also semi-simple Γ(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 Γ0(2)=C2C and that at least 12 of then are epimorphic images
of PSL2(Z)=C2C3.

Rational conformal field theories Another
source of SL2(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
SL2(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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the cartographers’ groups (2)

Fortunately,
there is a drastic shortcut to the general tree-argument of last time, due to
Roger Alperin. Recall that the Moebius
transformations corresponding to u resp. v send z resp. to

1z and 11z

whence the Moebius transformation
corresponding to v1 send z to 11z.

Consider
the set P of all positive irrational real numbers and the
set N of all negative irrational real numbers and observe
that

u(P)N and
v±(N)P

We have to show
that no alternating word w=(u)v±uv±uv±(u) in
u and v± can be the identity in PSL2(Z).

If the
length of w is odd then either w(P)N or w(N)P depending on whether w starts with a u or with
a v± term. Either way, this proves that no odd-length word can
be the identity element in PSL2(Z).

If the length of
the word w is even we can assume that w=v±uv±uv±u (if necessary, after conjugating with u we get to this form).

There are two subcases, either w=v1uv±uv±u in which case w(P)v1(N)
and this latter set is contained in the set of all positive irrational
real numbers which are strictly larger than one .

Or, w=vuv±uv±u in which case
w(P)v(N) and this set is contained in
the set of all positive irrational real numbers strictly smaller than
one
.

Either way, this shows that w cannot be the identity
morphism on P so cannot be the identity element in
PSL2(Z).
Hence we have proved that

PSL2(Z)=C2C3=u,v:u2=1=v3

A
description of SL2(Z) in terms of generators and relations
follows

SL2(Z)=U,V:U4=1=V6,U2=V3

It is not true that SL2(Z) is the free
product C4C6 as there is the extra relation U2=V3.

This relation says that the cyclic groups C4=U
and C6=V share a common subgroup C2=U2=V3 and this extra condition is expressed by saying that
SL2(Z) is the amalgamated free product of C4 with
C6, amalgamated over the common subgroup C2 and denoted
as

SL2(Z)=C4C2C6

More
generally, if G and H are finite groups, then the free product GH consists of all words of the form  (g1)h1g2h2g3gnhn(gn1) (so alternating between non-identity
elements of G and H) and the group-law is induced by concatenation
of words (and group-laws in either G or H when end terms are
elements in the same group).

For example, take the dihedral groups D4=U,R:U4=1=R2,(RU)2=1 and D6=V,S:V6=1=S2,(SV)2=1 then the free product can be expressed
as

D4D6=U,V,R,S:U4=1=V6=R2=S2=(RV)2=(RU)2

This almost fits in with
our obtained description of
GL2(Z)

GL2(Z)=U,V,R:U4=1=V6=R2=(RU)2=(RV)2,U2=V3

except for the
extra relations R=S and U2=V3 which express the fact that we
demand that D4 and D6 have the same subgroup

D2=U2=V3,S=R

So, again we can express these relations by
saying that GL2(Z) is the amalgamated free product of
the subgroups D4=U,R and D6=V,R, amalgamated over the common subgroup D2=C2×C2=U2=V3,R. We write

GL2(Z)=D4D2D6

Similarly (but a bit easier) for
PGL2(Z) we have

$PGL_2(\mathbb{Z}) = \langle u,v,R
u^2=v^3=1=R^2 = (Ru)^2 = (Rv)^2 \rangle $

which can be seen as
the amalgamated free product of D2=u,R with D3=v,R, amalgamated over the common subgroup C2=R and therefore

PGL2(Z)=D2C2D3

Now let us turn to congruence subgroups of
the modular group
.
With Γ(n) one denotes the kernel of the natural
surjection

PSL2(Z)PSL2(Z/nZ)

that is all elements represented by a matrix

[abcd]

such that a=d=1 (mod n) and b=c=0
(mod n). On the other hand Γ0(n) consists of elements
represented by matrices such that only c=0 (mod n). Both are finite
index subgroups of PSL2(Z).

As we have seen that
PSL2(Z)=C2C3 it follows from general facts
on free products that any finite index subgroup is of the
form

C2C2C2C3C3C3CCC

that is the
free product of k copies of C2, l copies of C3 and m copies
of C where it should be noted that k,l and m are allowed
to be zero. There is an elegant way to calculate explicit generators of
these factors for congruence subgroups, due to Ravi S. Kulkarni (An
Arithmetic-Geometric Method in the Study of the Subgroups of the Modular
Group , American Journal of Mathematics, Vol. 113, No. 6. (Dec.,
1991), pp. 1053-1133) which deserves another (non-course) post.

Using this method one finds that Γ0(2) is generated by
the Moebius transformations corresponding to the
matrices

X=[1101] and
Y=[1121]

and that
generators for Γ(2) are given by the
matrices

A=[1021]
and B=[1223]

Next,
one has to write these generators in terms of the generating matrices
u and v of PSL2(Z) and as we know all relations between
u and v the relations of these congruence subgroups will follow.

We
will give the details for Γ0(2) and leave you to figure out
that Γ(2)=CC (that is that
there are no relations between the matrices A and
B).

Calculate that X=v2u and that Y=vuv2. Because the
only relations between u and v are v3=1=u2 we see that Y is an
element of order two as Y2=vuv3uv2=v3=1 and that no power of
X can be the identity transformation.

But then also none of the
elements  (Y)Xi1YXi2YYXin(Y) can be the identity
(write it out as a word in u and v) whence,
indeed

Γ0(2)=CC2

In fact,
the group Γ0(2) is staring you in the face whenever you come to
this site. I fear I’ve never added proper acknowledgements for the
beautiful header-picture

so I’d better do it now. The picture is due to Helena
Verrill
and she has a
page with
more pictures. The header-picture depicts a way to get a fundamental
domain for the action of Γ0(2) on the upper half plane. Such a
fundamental domain consists of any choice of 6 tiles with different
colours (note that there are two shades of blue and green). Helena also
has a
Java-applet
to draw fundamental domains of more congruence subgroups.

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The cartographers’ groups

Just as cartographers like
Mercator drew maps of
the then known world, we draw dessins
d ‘enfants
to depict the
associated algebraic curve defined over
Q.

In order to see that such a dessin
d’enfant determines a permutation representation of one of
Grothendieck’s cartographic groups, SL2(Z),Γ0(2) or Γ(2) we need to have realizations of these
groups (as well as their close relatives
PSL2(Z),GL2(Z) and PGL2(Z)) in
terms of generators and relations.

As this lesson will be rather
technical I’d better first explain what we will prove (so that you can
skip it if you feel comfortable with the statements) and why we want to
prove it. What we will prove in detail below is that these groups
can be written as free (or amalgamated) group products. We will explain
what this means and will establish that

PSL2(Z)=C2C3,Γ0(2)=C2C,Γ(2)=CC

SL2(Z)=C4C2C6,GL2(Z)=D4D2D6,PGL2(Z)=D2C2D3

where Cn resp.
Dn are the cyclic (resp. dihedral) groups. The importance of these
facts it that they will allow us to view the set of (isomorphism classes
of) finite dimensional representations of these groups as
noncommutative manifolds . Looking at the statements above we
see that these arithmetical groups can be build up from the first
examples in any course on finite groups : cyclic and dihedral
groups.

Recall that the cyclic group of order n, Cn is the group of
rotations of a regular n-gon (so is generated by a rotation r with
angle 2πn and has defining relation rn=1, where 1
is the identity). However, regular n-gons have more symmetries :
flipping over one of its n lines of symmetry

The dihedral group Dn is the group generated by the n
rotations and by these n flips. If, as before r is a generating
rotation and d is one of the flips, then it is easy to see that the
dihedral group is generated by r and d and satisfied the defining
relations

rn=1 and d2=1=(rd)2

Flipping twice
does nothing and to see the relation  (rd)2=1 check that doing twice a
rotation followed by a flip brings all vertices back to their original
location. The dihedral group Dn has 2n elements, the n-rotations
ri and the n flips dri.

In fact, to get at the cartographic
groups we will only need the groups D4,D6 and their
subgroups. Let us start by finding generators of the largest
group GL2(Z) which is the group of all invertible 2×2 matrices with integer coefficients.

Consider the
elements

U=[0110],V=[0111]/tex]andR =
[0110] $

and form the
matrices

X=UV=[1101],Y=VU=[1011]

By induction we prove the following relations in
GL2(Z)

Xn[abcd]=[ancbndcd]
and [abcd]Xn=[abnacdnc]

Yn[abcd]=[abc+nad+nb] and
[abcd]Yn=[a+nbbc+ndd]

The determinant ad-bc of
a matrix in GL2(Z) must be ±1 whence all rows and
columns of

[abcd]GL2(Z)

consist of coprime numbers and hence a and
c can be reduced modulo each other by left multiplication by a power
of X or Y until one of them is zero and the other is ±1. We
may even assume that a=±1 (if not, left multiply with U).

So,
by left multiplication by powers of X and Y and U we can bring any
element of GL2(Z) into the form

[±1β0±1]

and again by left
multiplication by a power of X we can bring it in one of the four
forms

[±100±1]=1,UR,RU,U2

This proves that GL2(Z) is
generated by the elements U,V and R.

Similarly, the group
SL2(Z) of all 2×2 integer matrices with
determinant 1 is generated by the elements U and V as using the
above method and the restriction on the determinant we will end up with
one of the two matrices

[1001],[1001]=1,U2

so we never need the matrix R. As for
relations, there are some obvious relations among the matrices U,V and
R, namely

U2=V3 and 1=U4=R2=(RU)2=(RV)2 $

The
real problem is to prove that all remaining relations are consequences
of these basic ones. As R clearly has order two and its commutation
relations with U and V are just RU=U1R and RV=V1R we can
pull R in any relation to the far right and (possibly after
multiplying on the right with R) are left to prove that the only
relations among U and V are consequences of U2=V3 and
U4=1=V6.

Because U2=V3 this element is central in the
group generated by U and V (which we have seen to be
SL2(Z)) and if we quotient it out we get the modular
group

Γ=PSL2(Z)

Hence in order to prove our claim
it suffices that

PSL2(Z)=U,V:U2=V3=1

Phrased differently, we have to show that
PSL2(Z) is the free group product of the cyclic groups of
order two and three (those generated by u=U and
v=V) C2C3

Any element of this free group
product is of the form  (u)va1uva2uuvak(u) where beginning and trailing u are optional and
all ai are either 1 or 2.

So we have to show that in
PSL2(Z) no such word can give the identity
element. Today, we will first sketch the classical argument based
on the theory of groups acting on trees due to Jean-Pierre
Serre
and Hyman Bass. Tomorrow, we will give a short elegant proof due to
Roger Alperin and draw
consequences to the description of the carthographic groups as
amalgamated free products of cyclic and dihedral groups.

Recall
that GL2(Z) acts via Moebius
transformations
on
the complex plane C=R2 (actually it is an
action on the Riemann sphere PC1) given by the
maps

[abcd].z=az+bcz+d

Note that the action of the
center of GL2(Z) (that is of ±[1001]) acts trivially, so it is really an action of
PGL2(Z).

As R interchanges the upper and lower half-plane
we might as well restrict to the action of SL2(Z) on the
upper-halfplane H. It is quite easy to see that a
fundamental domain
for this action is given by the greyed-out area

To see that any zH can be taken into this
region by an element of PSL2(Z) note the following two
Moebius transformations

[1101].z=z+1 and [0110].z=1z

The first
operation takes any z into a strip of length one, for example that
with Re(z) between 12 and 12 and the second
interchanges points within and outside the unit-circle, so combining the
two we get any z into the greyed-out region. Actually, we could have
taken any of the regions in the above tiling as our fundamental domain
as they are all translates of the greyed-out region by an element of
PSL2(Z).

Of course, points on the boundary of the
greyed-out fundamental region need to be identified (in order to get the
identification of H/PSL2(Z) with the
Riemann sphere S2=PC1). For example, the two
halves of the boundary by the unit circle are interchanged by the action
of the map z1z and if we take the translates under
PSL2(Z) of the indicated circle-part

we get a connected tree with fundamental domain the circle
part bounded by i and ρ=12+32i.
Calculating the stabilizer subgroup of i (that is, the subgroup of
elements fixing i) we get that this subgroup
is u=C2 whereas the stabilizer subgroup of
ρ is v=C3.

Using this facts and the general
results of Jean-Pierre Serres book Trees
one deduces that PSL2(Z)=C2C3
and hence that the obvious relations among U,V and R given above do
indeed generate all relations.

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