Monstrous dessins 2

Let’s try to identify the $\mathrm{\Psi }(n)=n\prod _{p|n}(1+\frac{1}{p})$ points of ${\mathbb{P}}^1(\mathbb{Z}/n\mathbb{Z})$ with the lattices $L_{M\frac{g}{h}}$ at hyperdistance $n$ from the standard lattice $L_1$ in Conway’s big picture.

Here are all $24=\mathrm{\Psi }(12)$ lattices at hyperdistance $12$ from $L_1$ (the boundary lattices):

You can also see the $4=\mathrm{\Psi }(3)$ lattices at hyperdistance $3$ (those connected to $1$ with a red arrow) as well as the intermediate $12=\mathrm{\Psi }(6)$ lattices at hyperdistance $6$.

The vertices of Conway’s Big Picture are the projective classes of integral sublattices of the standard lattice ${\mathbb{Z}}^2=\mathbb{Z}{e}_1\oplus \mathbb{Z}{e}_2$.

Let’s say our sublattice is generated by the integral vectors $v=(v_1,v_2)$ and $w=(w_1.w_2)$. How do we determine its class $L_{M,\frac{g}{h}}$ where $M\in {\mathbb{Q}}_{+}$ is a strictly positive rational number and $0\le \frac{g}{h}<1$?

Here’s an example: the sublattice (the thick dots) is spanned by the vectors $v=(2,1)$ and $w=(1,4)$

Well, we try to find a basechange matrix in $S{L}_2(\mathbb{Z})$ such that the new 2nd base vector is of the form $(0,z)$. To do this take coprime $(c,d)\in {\mathbb{Z}}^2$ such that $c{v}_1+d{w}_1=0$ and complete with $(a,b)$ satisfying $ad-bc=1$ via Bezout to a matrix in $S{L}_2(\mathbb{Z})$ such that
$$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}{v}_1& v_2\\ w_1& w_2\end{array}\right]=\left[\begin{array}{cc}x& y\\ 0& z\end{array}\right]$$
then the sublattice is of class $L_{\frac{x}{z},\frac{y}{z}mod1}$.

In the example, we have
$$\left[\begin{array}{cc}0& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}2& 1\\ 1& 4\end{array}\right]=\left[\begin{array}{cc}1& 4\\ 0& 7\end{array}\right]$$
so this sublattice is of class $L_{\frac{1}{7},\frac{4}{7}}$.

Starting from a class $L_{M,\frac{g}{h}}$ it is easy to work out its hyperdistance from $L_1$: let $d$ be the smallest natural number making the corresponding matrix integral
$$d.\left[\begin{array}{cc}M& \frac{g}{h}\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}u& v\\ 0& w\end{array}\right]\in M_2(\mathbb{Z})$$
then $L_{M,\frac{g}{h}}$ is at hyperdistance $u.w$ from $L_1$.

Now that we know how to find the lattice class of any sublattice of ${\mathbb{Z}}^2$, let us assign a class to any point $[c:d]$ of ${\mathbb{P}}^1(\mathbb{Z}/n\mathbb{Z})$.

As $gcd(c,d)=1$, by Bezout we can find a integral matrix with determinant $1$
$$S_{[c:d]}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$$
But then the matrix
$$\left[\begin{array}{cc}a.n& b.n\\ c& d\end{array}\right]$$
has determinant $n$.

Working backwards we see that the class $L_{[c:d]}$ of the sublattice of ${\mathbb{Z}}^2$ spanned by the vectors $(a.n,b.n)$ and $(c,d)$ is of hyperdistance $n$ from $L_1$.

This is how the correspondence between points of ${\mathbb{P}}^1(\mathbb{Z}/n\mathbb{Z})$ and classes in Conway’s big picture at hyperdistance $n$ from $L_1$ works.

Let’s do an example. Take the point $[7:3]\in {\mathbb{P}}^1(\mathbb{Z}/12\mathbb{Z})$ (see last time), then
$$\left[\begin{array}{cc}-2& -1\\ 7& 3\end{array}\right]\in S{L}_2(\mathbb{Z})$$
so we have to determine the class of the sublattice spanned by $(-24,-12)$ and $(7,3)$. As before we have to compute
$$\left[\begin{array}{cc}-2& -7\\ 7& 24\end{array}\right]\left[\begin{array}{cc}-24& -12\\ 7& 3\end{array}\right]=\left[\begin{array}{cc}-1& 3\\ 0& -12\end{array}\right]$$
giving us that the class $L_{[7:3]}=L_{\frac{1}{12}\frac{3}{4}}$ (remember that the second term must be taken $mod1$).

If you do this for all points in ${\mathbb{P}}^1(\mathbb{Z}/12\mathbb{Z})$ (and ${\mathbb{P}}^1(\mathbb{Z}/6\mathbb{Z})$ and ${\mathbb{P}}^1(\mathbb{Z}/3\mathbb{Z})$) you get this version of the picture we started with

You’ll spot that the preimages of a canonical coordinate of ${\mathbb{P}}^1(\mathbb{Z}/m\mathbb{Z})$ for $m|n$ are the very same coordinate together with ‘new’ canonical coordinates in ${\mathbb{P}}^1(\mathbb{Z}/n\mathbb{Z})$.

To see that this correspondence is one-to-one and that the index of the congruence subgroup
$${\mathrm{\Gamma }}_0(n)=\{\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]|n|r\text{and}ps-qr=1\}$$
in the full modular group $\mathrm{\Gamma }=PS{L}_2(\mathbb{Z})$ is equal to $\mathrm{\Psi }(n)$ it is useful to consider the action of $PG{L}_2(\mathbb{Q}{)}^{+}$ on the right on the classes of lattices.

The stabilizer of $L_1$ is the full modular group $\mathrm{\Gamma }$ and the stabilizer of any class is a suitable conjugate of $\mathrm{\Gamma }$. For example, for the class $L_n$ (that is, of the sublattice spanned by $(n,0)$ and $(0,1)$, which is of hyperdistance $n$ from $L_1$) this stabilizer is
$$Stab(L_n)=\{\left[\begin{array}{cc}a& \frac{b}{n}\\ c.n& d\end{array}\right]|ad-bc=1\}$$
and a very useful observation is that
$$Stab(L_1)\cap Stab(L_n)={\mathrm{\Gamma }}_0(n)$$
This is the way Conway likes us to think about the congruence subgroup ${\mathrm{\Gamma }}_0(n)$: it is the joint stabilizer of the classes $L_1$ and $L_n$ (as well as all classes in the ‘thread’ $L_m$ with $m|n$).

On the other hand, $\mathrm{\Gamma }$ acts by rotations on the big picture: it only fixes $L_1$ and maps a class to another one of the same hyperdistance from $L_1$.The index of ${\mathrm{\Gamma }}_0(n)$ in $\mathrm{\Gamma }$ is then the number of classes at hyperdistance $n$.

To see that this number is $\mathrm{\Psi }(n)$, first check that the classes at hyperdistance $p^k$ for $p$ a prime number and for all $k$ for the $p+1$ free valent tree with root $L_1$, so there are exactly $p^{k-1}(p+1)$ classes as hyperdistance $p^k$.

To get from this that the number of hyperdistance $n$ classes is indeed $\mathrm{\Psi }(n)=\prod _{p|n}{p}^{v_p(n)-1}(p+1)$ we have to use the prime- factorisation of the hyperdistance (see this post).

The fundamental domain for the action of ${\mathrm{\Gamma }}_0(12)$ by Moebius tranfos on the upper half plane must then consist of $48=2\mathrm{\Psi }(12)$ black or white hyperbolic triangles

Next time we’ll see how to deduce the ‘monstrous’ Grothendieck dessin d’enfant for ${\mathrm{\Gamma }}_0(12)$ from it