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Tag: modular

anabelian geometry

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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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

$-\frac{1}{z} $ and $\frac{1}{1-z} $

whence the Moebius transformation
corresponding to $v^{-1} $ send z to $1-\frac{1}{z} $.

Consider
the set $\mathcal{P} $ of all positive irrational real numbers and the
set $\mathcal{N} $ of all negative irrational real numbers and observe
that

$u(\mathcal{P}) \subset \mathcal{N} $ and
$v^{\pm}(\mathcal{N}) \subset \mathcal{P} $

We have to show
that no alternating word $w=(u)v^{\pm}uv^{\pm}u \dots v^{\pm}(u) $ in
u and $v^{\pm} $ can be the identity in $PSL_2(\mathbb{Z}) $.

If the
length of w is odd then either $w(\mathcal{P}) \subset
\mathcal{N} $ or $w(\mathcal{N}) \subset
\mathcal{P} $ depending on whether w starts with a u or with
a $v^{\pm} $ term. Either way, this proves that no odd-length word can
be the identity element in $PSL_2(\mathbb{Z}) $.

If the length of
the word w is even we can assume that $w = v^{\pm}uv^{\pm}u \dots
v^{\pm}u $ (if necessary, after conjugating with u we get to this form).

There are two subcases, either $w=v^{-1}uv^{\pm}u \dots
v^{\pm}u $ in which case $w(\mathcal{P}) \subset v^{-1}(\mathcal{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^{\pm}u \dots v^{\pm}u $ in which case
$w(\mathcal{P}) \subset v(\mathcal{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 $\mathcal{P} $ so cannot be the identity element in
$PSL_2(\mathbb{Z}) $.
Hence we have proved that

$PSL_2(\mathbb{Z}) = C_2 \ast C_3 = \langle u,v : u^2=1=v^3 \rangle $

A
description of $SL_2(\mathbb{Z}) $ in terms of generators and relations
follows

$SL_2(\mathbb{Z}) = \langle U,V : U^4=1=V^6, U^2=V^3 \rangle $

It is not true that $SL_2(\mathbb{Z}) $ is the free
product $C_4 \ast C_6 $ as there is the extra relation $U^2=V^3 $.

This relation says that the cyclic groups $C_4 = \langle U \rangle $
and $C_6 = \langle V \rangle $ share a common subgroup $C_2 = \langle
U^2=V^3 \rangle $ and this extra condition is expressed by saying that
$SL_2(\mathbb{Z}) $ is the amalgamated free product of $C_4 $ with
$C_6 $, amalgamated over the common subgroup $C_2 $ and denoted
as

$SL_2(\mathbb{Z}) = C_4 \ast_{C_2} C_6 $

More
generally, if G and H are finite groups, then the free product $G
\ast H $ consists of all words of the form $~(g_1)h_1g_2h_2g_3
\dots g_nh_n(g_{n-1}) $ (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 $D_4
= \langle U,R : U^4=1=R^2,(RU)^2=1 \rangle $ and $D_6 = \langle V,S :
V^6=1=S^2,(SV)^2=1 \rangle $ then the free product can be expressed
as

$D_4 \ast D_6 = \langle U,V,R,S :
U^4=1=V^6=R^2=S^2=(RV)^2=(RU)^2 \rangle $

This almost fits in with
our obtained description of
$GL_2(\mathbb{Z}) $

$GL_2(\mathbb{Z}) = \langle U,V,R :
U^4=1=V^6=R^2=(RU)^2=(RV)^2, U^2=V^3 \rangle $

except for the
extra relations $R=S $ and $U^2=V^3 $ which express the fact that we
demand that $D_4 $ and $D_6 $ have the same subgroup

$D_2 = \langle U^2=V^3,S=R \rangle $

So, again we can express these relations by
saying that $GL_2(\mathbb{Z}) $ is the amalgamated free product of
the subgroups $D_4 = \langle U,R \rangle $ and $D_6 = \langle V,R
\rangle $, amalgamated over the common subgroup $D_2 = C_2 \times C_2 =
\langle U^2=V^3,R \rangle $. We write

$GL_2(\mathbb{Z}) = D_4
\ast_{D_2} D_6 $

Similarly (but a bit easier) for
$PGL_2(\mathbb{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 $D_2 = \langle u,R \rangle $ with $D_3
= \langle v,R \rangle $, amalgamated over the common subgroup $C_2 =
\langle R \rangle $ and therefore

$PGL_2(\mathbb{Z}) = D_2
\ast_{C_2} D_3 $

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

$PSL_2(\mathbb{Z}) \rightarrow
PSL_2(\mathbb{Z}/n\mathbb{Z}) $

that is all elements represented by a matrix

$\begin{bmatrix} a
& b \\ c & d \end{bmatrix} $

such that a=d=1 (mod n) and b=c=0
(mod n). On the other hand $\Gamma_0(n) $ consists of elements
represented by matrices such that only c=0 (mod n). Both are finite
index subgroups of $PSL_2(\mathbb{Z}) $.

As we have seen that
$PSL_2(\mathbb{Z}) = C_2 \ast C_3 $ it follows from general facts
on free products that any finite index subgroup is of the
form

$C_2 \ast C_2 \ast \dots \ast C_2 \ast
C_3 \ast C_3 \ast \dots \ast C_3 \ast C_{\infty}
\ast C_{\infty} \dots \ast C_{\infty} $

that is the
free product of k copies of $C_2 $, l copies of $C_3 $ and m copies
of $C_{\infty} $ 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 $\Gamma_0(2) $ is generated by
the Moebius transformations corresponding to the
matrices

$X=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $ and
$Y=\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} $

and that
generators for $\Gamma(2) $ are given by the
matrices

$A=\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} $
and $B=\begin{bmatrix} 1 & -2 \\ 2 & -3 \end{bmatrix} $

Next,
one has to write these generators in terms of the generating matrices
u and v of $PSL_2(\mathbb{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 $\Gamma_0(2) $ and leave you to figure out
that $\Gamma(2) = C_{\infty} \ast C_{\infty} $ (that is that
there are no relations between the matrices A and
B).

Calculate that $X=v^2u $ and that $Y=vuv^2 $. Because the
only relations between u and v are $v^3=1=u^2 $ we see that Y is an
element of order two as $Y^2 = vuv^3uv^2= v^3 = 1 $ and that no power of
X can be the identity transformation.

But then also none of the
elements $~(Y)X^{i_1}YX^{i_2}Y \dots YX^{i_n}(Y) $ can be the identity
(write it out as a word in u and v) whence,
indeed

$\Gamma_0(2) = C_{\infty} \ast C_2 $

In fact,
the group $\Gamma_0(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 $\Gamma_0(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
$\overline{\mathbb{Q}} $.

In order to see that such a dessin
d’enfant determines a permutation representation of one of
Grothendieck’s cartographic groups, $SL_2(\mathbb{Z}),
\Gamma_0(2) $ or $\Gamma(2) $ we need to have realizations of these
groups (as well as their close relatives
$PSL_2(\mathbb{Z}),GL_2(\mathbb{Z}) $ and $PGL_2(\mathbb{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

$PSL_2(\mathbb{Z}) = C_2
\ast C_3, \Gamma_0(2) = C_2 \ast C_{\infty}, \Gamma(2)
= C_{\infty} \ast C_{\infty} $

$SL_2(\mathbb{Z}) =
C_4 \ast_{C_2} C_6, GL_2(\mathbb{Z}) = D_4 \ast_{D_2} D_6,
PGL_2(\mathbb{Z}) = D_2 \ast_{C_2} D_3 $

where $C_n $ resp.
$D_n $ 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, $C_n $ is the group of
rotations of a regular n-gon (so is generated by a rotation r with
angle $\frac{2 \pi}{n} $ and has defining relation $r^n = 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 $D_n $ 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

$r^n=1 $ and $d^2 = 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 $D_n $ has 2n elements, the n-rotations
$r^i $ and the n flips $dr^i $.

In fact, to get at the cartographic
groups we will only need the groups $D_4, D_6 $ and their
subgroups. Let us start by finding generators of the largest
group $GL_2(\mathbb{Z}) $ which is the group of all invertible $2
\times 2 $ matrices with integer coefficients.

Consider the
elements

$U = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},
V = \begin{bmatrix} 0 & 1 \\ -1 & 1 \end{bmatrix}/tex] and $R =
\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $

and form the
matrices

$X = UV = \begin{bmatrix} 1 & -1 \\ 0 & 1
\end{bmatrix}, Y = VU = \begin{bmatrix} 1 & 0 \\ 1 & 1
\end{bmatrix} $

By induction we prove the following relations in
$GL_2(\mathbb{Z}) $

$X^n \begin{bmatrix} a & b \\ c & d
\end{bmatrix} = \begin{bmatrix} a-nc & b-nd \\ c & d \end{bmatrix} $
and $\begin{bmatrix} a & b \\ c& d \end{bmatrix} X^n =
\begin{bmatrix} a & b-na \\ c & d-nc \end{bmatrix} $

$Y^n \begin{bmatrix} a & b \\ c & d \end{bmatrix} =
\begin{bmatrix} a & b \\ c+na & d+nb \end{bmatrix} $ and
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} Y^n = \begin{bmatrix}
a+nb & b \\ c+nd & d \end{bmatrix} $

The determinant ad-bc of
a matrix in $GL_2(\mathbb{Z}) $ must be $\pm 1 $ whence all rows and
columns of

$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \in
GL_2(\mathbb{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 $\pm 1 $. We
may even assume that $a = \pm 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 $GL_2(\mathbb{Z}) $ into the form

$\begin{bmatrix}
\pm 1 & \beta \\ 0 & \pm 1 \end{bmatrix} $

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

$\begin{bmatrix} \pm 1 & 0 \\ 0 & \pm 1 \end{bmatrix}
= { 1,UR,RU,U^2 } $

This proves that $GL_2(\mathbb{Z}) $ is
generated by the elements U,V and R.

Similarly, the group
$SL_2(\mathbb{Z}) $ of all $2 \times 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

${ \begin{bmatrix} 1 & 0 \\ 0 & 1
\end{bmatrix},\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} } =
{ 1,U^2 } $

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

$U^2=V^3 $ and $1=U^4=R^2=(RU)^2=(RV)^2$ $

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=U^{-1}R $ and $RV=V^{-1}R $ 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 $U^2=V^3 $ and
$U^4=1=V^6 $.

Because $U^2=V^3 $ this element is central in the
group generated by U and V (which we have seen to be
$SL_2(\mathbb{Z}) $) and if we quotient it out we get the modular
group

$\Gamma = PSL_2(\mathbb{Z}) $

Hence in order to prove our claim
it suffices that

$PSL_2(\mathbb{Z}) = \langle
\overline{U},\overline{V} : \overline{U}^2=\overline{V}^3=1
\rangle $

Phrased differently, we have to show that
$PSL_2(\mathbb{Z}) $ is the free group product of the cyclic groups of
order two and three (those generated by $u = \overline{U} $ and
$v=\overline{V} $) $C_2 \ast C_3 $

Any element of this free group
product is of the form $~(u)v^{a_1}uv^{a_2}u \ldots
uv^{a_k}(u) $ where beginning and trailing u are optional and
all $a_i $ are either 1 or 2.

So we have to show that in
$PSL_2(\mathbb{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 $GL_2(\mathbb{Z}) $ acts via Moebius
transformations
on
the complex plane $\mathbb{C} = \mathbb{R}^2 $ (actually it is an
action on the Riemann sphere $\mathbb{P}^1_{\mathbb{C}} $) given by the
maps

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}.z =
\frac{az+b}{cz+d} $

Note that the action of the
center of $GL_2(\mathbb{Z}) $ (that is of $\pm \begin{bmatrix} 1 & 0
\\ 0 & 1 \end{bmatrix} $) acts trivially, so it is really an action of
$PGL_2(\mathbb{Z}) $.

As R interchanges the upper and lower half-plane
we might as well restrict to the action of $SL_2(\mathbb{Z}) $ on the
upper-halfplane $\mathcal{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 $z \in \mathcal{H} $ can be taken into this
region by an element of $PSL_2(\mathbb{Z}) $ note the following two
Moebius transformations

$\begin{bmatrix} 1 & 1 \\ 0 & 1
\end{bmatrix}.z = z+1 $ and $\begin{bmatrix} 0 & 1 \\ -1
& 0 \end{bmatrix}.z = -\frac{1}{z} $

The first
operation takes any z into a strip of length one, for example that
with Re(z) between $-\frac{1}{2} $ and $\frac{1}{2} $ 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
$PSL_2(\mathbb{Z}) $.

Of course, points on the boundary of the
greyed-out fundamental region need to be identified (in order to get the
identification of $\overline{\mathcal{H}/PSL_2(\mathbb{Z})} $ with the
Riemann sphere $S^2=\mathbb{P}^1_{\mathbb{C}} $). For example, the two
halves of the boundary by the unit circle are interchanged by the action
of the map $z \rightarrow -\frac{1}{z} $ and if we take the translates under
$PSL_2(\mathbb{Z}) $ of the indicated circle-part

we get a connected tree with fundamental domain the circle
part bounded by i and $\rho = \frac{1}{2}+\frac{\sqrt{3}}{2} i $.
Calculating the stabilizer subgroup of i (that is, the subgroup of
elements fixing i) we get that this subgroup
is $\langle u \rangle = C_2 $ whereas the stabilizer subgroup of
$\rho $ is $\langle v \rangle = C_3 $.

Using this facts and the general
results of Jean-Pierre Serres book Trees
one deduces that $PSL_2(\mathbb{Z}) = C_2 \ast C_3 $
and hence that the obvious relations among U,V and R given above do
indeed generate all relations.

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