The cartographers’ groups

By lieven

Dessins d'enfants

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

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

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 $~(u)v^{a_1}uv^{a_2}u \hdots uv^{a_k}(u)$ 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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Grothendieck, groups, modular, noncommutative, permutation representation, representations, Riemann, symmetry

3 comments
Posted in groups, modular
Written on Tue, 13 March 2007 at 10:44 pm
Tags: Grothendieck, groups, modular, noncommutative, permutation representation, representations, Riemann, symmetry
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3 Responses to “The cartographers’ groups”

Says:
December 26th, 2007 at 9:31 pm
[...] 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 [...]
permutation representations of monodromy groups at neverendingbooks Says:
December 27th, 2007 at 3:25 pm
[...] we will explain how curves defined over determine permutation representations of the carthographic groups. We have seen that any smooth projective curve (a Riemann surface) defined over the algebraic [...]
balm or poison | neverendingbooks Says:
January 8th, 2008 at 8:20 am
[...] don’t try to follow the suggestions of my Tracks post. It certainly is not enough to get tracks running under Tiger (and probably also not under Panther [...]