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 $w=(u)v^{\pm }u{v}^{\pm }u\dots v^{\pm }(u)$ in
u and $v^{\pm }$ can be the identity in $PS{L}_2(\mathbb{Z})$.

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 $PS{L}_2(\mathbb{Z})$.

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

There are two subcases, either $w=v^{-1}u{v}^{\pm }u\dots v^{\pm }u$ 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, $w=vu{v}^{\pm }u\dots v^{\pm }u$ 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
$PS{L}_2(\mathbb{Z})$.
Hence we have proved that

$PS{L}_2(\mathbb{Z})=C_2\ast C_3=⟨u,v:u^2=1=v^3⟩$

A
description of $S{L}_2(\mathbb{Z})$ in terms of generators and relations
follows

$S{L}_2(\mathbb{Z})=⟨U,V:U^4=1=V^6,U^2=V^3⟩$

It is not true that $S{L}_2(\mathbb{Z})$ 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 $C_4=⟨U⟩$
and $C_6=⟨V⟩$ share a common subgroup $C_2=⟨U^2=V^3⟩$ and this extra condition is expressed by saying that
$S{L}_2(\mathbb{Z})$ is the amalgamated free product of $C_4$ with
$C_6$, amalgamated over the common subgroup $C_2$ and denoted
as

$S{L}_2(\mathbb{Z})=C_4{\ast }_{C_2}{C}_6$

More
generally, if G and H are finite groups, then the free product $G\ast H$ consists of all words of the form $(g_1)h_1{g}_2{h}_2{g}_3\dots g_n{h}_n(g_{n-1})$ (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 $D_4=⟨U,R:U^4=1=R^2,(RU{)}^2=1⟩$ and $D_6=⟨V,S:V^6=1=S^2,(SV{)}^2=1⟩$ then the free product can be expressed
as

$D_4\ast D_6=⟨U,V,R,S:U^4=1=V^6=R^2=S^2=(RV{)}^2=(RU{)}^2⟩$

This almost fits in with
our obtained description of
$G{L}_2(\mathbb{Z})$

$G{L}_2(\mathbb{Z})=⟨U,V,R:U^4=1=V^6=R^2=(RU{)}^2=(RV{)}^2,U^2=V^3⟩$

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

$D_2=⟨U^2=V^3,S=R⟩$

So, again we can express these relations by
saying that $G{L}_2(\mathbb{Z})$ is the amalgamated free product of
the subgroups $D_4=⟨U,R⟩$ and $D_6=⟨V,R⟩$, amalgamated over the common subgroup $D_2=C_2\times C_2=⟨U^2=V^3,R⟩$. We write

$G{L}_2(\mathbb{Z})=D_4{\ast }_{D_2}{D}_6$

Similarly (but a bit easier) for
$PG{L}_2(\mathbb{Z})$ 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 $D_2=⟨u,R⟩$ with $D_3=⟨v,R⟩$, amalgamated over the common subgroup $C_2=⟨R⟩$ and therefore

$PG{L}_2(\mathbb{Z})=D_2{\ast }_{C_2}{D}_3$

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

$PS{L}_2(\mathbb{Z})\to PS{L}_2(\mathbb{Z}/n\mathbb{Z})$

that is all elements represented by a matrix

$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

such that a=d=1 (mod n) and b=c=0
(mod n). On the other hand ${\mathrm{\Gamma }}_0(n)$ consists of elements
represented by matrices such that only c=0 (mod n). Both are finite
index subgroups of $PS{L}_2(\mathbb{Z})$.

As we have seen that
$PS{L}_2(\mathbb{Z})=C_2\ast C_3$ it follows from general facts
on free products that any finite index subgroup is of the
form

$C_2\ast C_2\ast \cdots \ast C_2\ast C_3\ast C_3\ast \cdots \ast C_3\ast C_{\mathrm{\infty }}\ast C_{\mathrm{\infty }}\cdots \ast C_{\mathrm{\infty }}$

that is the
free product of k copies of $C_2$, l copies of $C_3$ and m copies
of $C_{\mathrm{\infty }}$ 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 ${\mathrm{\Gamma }}_0(2)$ is generated by
the Moebius transformations corresponding to the
matrices

$X=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ and
$Y=\left[\begin{array}{cc}1& -1\\ 2& -1\end{array}\right]$

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

$A=\left[\begin{array}{cc}1& 0\\ -2& 1\end{array}\right]$
and $B=\left[\begin{array}{cc}1& -2\\ 2& -3\end{array}\right]$

Next,
one has to write these generators in terms of the generating matrices
u and v of $PS{L}_2(\mathbb{Z})$ and as we know all relations between
u and v the relations of these congruence subgroups will follow.

We
will give the details for ${\mathrm{\Gamma }}_0(2)$ and leave you to figure out
that $\mathrm{\Gamma }(2)=C_{\mathrm{\infty }}\ast C_{\mathrm{\infty }}$ (that is that
there are no relations between the matrices A and
B).

Calculate that $X=v^{2}u$ and that $Y=vu{v}^2$. 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 $Y^2=vu{v}^{3}u{v}^2=v^3=1$ and that no power of
X can be the identity transformation.

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

${\mathrm{\Gamma }}_0(2)=C_{\mathrm{\infty }}\ast C_2$

In fact,
the group ${\mathrm{\Gamma }}_0(2)$ 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 ${\mathrm{\Gamma }}_0(2)$ 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.