the cartographers’ groups (2)

By lieven

Dessins d'enfants

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

$PSL_2(\mathbb{Z}) = C2 \ast C_3 = \langle u,v : u^2=1=v^3 \rangle$

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 $C4 \ast C_6$ as there is the extra relation U^2=V^3 .

This relation says that the cyclic groups $C_4 = \langle U \rangle$ and $C6 = \langle V \rangle$ share a common subgroup $C_2 = \langle U^2=V^3 \rangle$ and this extra condition is expressed by saying that $SL2(\mathbb{Z})$ is the amalgamated free product of C_4 with C6 , amalgamated over the common subgroup C_2 and denoted as

$SL_2(\mathbb{Z}) = C4 \ast_{C2} C_6$

More generally, if G and H are finite groups, then the free product $G \ast H$ consists of all words of the form $~(g_1)h1g_2h2g_3 \dots gnh_n(g{n-1})$ (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 $D6 = \langle V,S : V^6=1=S^2,(SV)^2=1 \rangle$ then the free product can be expressed as

$D_4 \ast D6 = \langle U,V,R,S : U^4=1=V^6=R^2=S^2=(RV)^2=(RU)^2 \rangle$

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 D6 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 $D4 = \langle U,R \rangle$ and $D_6 = \langle V,R \rangle$ , amalgamated over the common subgroup $D2 = C_2 \times C2 = \langle U^2=V^3,R \rangle$ . We write

$GL_2(\mathbb{Z}) = D4 \ast_{D2} D_6$

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 $D3 = \langle v,R \rangle$ , amalgamated over the common subgroup $C_2 = \langle R \rangle$ and therefore

$PGL_2(\mathbb{Z}) = D2 \ast_{C2} D_3$

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

$PSL_2(\mathbb{Z}) \rightarrow PSL2(\mathbb{Z}/n\mathbb{Z})$

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 $PSL2(\mathbb{Z})$ .

As we have seen that $PSL_2(\mathbb{Z}) = C2 \ast C_3$ it follows from general facts on free products that any finite index subgroup is of the form

$C_2 \ast C2 \ast \dots \ast C_2 \ast C3 \ast C_3 \ast \dots \ast C3 \ast C_{\infty} \ast C{\infty} \dots \ast C_{\infty}$

that is the free product of k copies of C_2 , l copies of C3 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 $~(Y)X^{i_1}YX^{i2}Y \dots YX^{i_n}(Y)$ 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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2 comments
Posted in groups, modular
Written on Thu, 15 March 2007 at 12:31 pm
Tags: apple, groups, modular
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2 Responses to “the cartographers’ groups (2)”

John Baez Says:
April 8th, 2008 at 8:31 am
Great stuff!

The sentence beginning “Using this method one finds that…” start off in teletype font for some reason, and below that, “Gamma(2)” is written out rather than appearing as a Greek letter.

I’m too sleepy to say anything more interesting now, but this is a very good account of some very beautiful mathematics.
lieven Says:
April 8th, 2008 at 8:40 am
thanks John! Ive corrected it now.