Roots of unity and the Big Picture

All lattices in the moonshine picture are number-like, that is of the form $M\frac{g}{h}$ with $M$ a positive integer and $0\le g<h$ with $(g,h)=1$. To understand the action of the Bost-Connes algebra on the Big Picture it is sometimes better to view the lattice $M\frac{g}{h}$ as a primitive $h$-th root of unity, centered at $hM$.

The distance from $M$ to any of the lattices $M\frac{g}{h}$ is equal to $2log(h)$, and the distances from $M$ and $M\frac{g}{h}$ to $hM$ are all equal to $log(h)$.

For a prime value $h$, these $h$ lattices are among the $h+1$ lattices branching off at $hM$ in the $h$-adic tree (the remaining one being $h^{2}M$).

For general $h$ the situation is more complex. Here’s the picture for $h=6$ with edges in the $2$-adic tree painted blue, those in the $3$-adic tree red.

To describe the moonshine group $(n|h)+e,f,\dots$ (an example was worked out in the tetrahedral snake post), we need to study the action of base-change with the matrix
$$x=\left[\begin{array}{cc}1& \frac{1}{h}\\ 0& 1\end{array}\right]$$
which sends a lattice of the form $M\frac{g}{h}$ with $0\le g<h$ to $M\frac{g+M}{h}$, so is a rotation over $\frac{2\pi M}{h}$ around $hM$. But, we also have to describe the base-change action with the matrix $$y=\left[\begin{array}{cc}1& 0\\ n& 1\end{array}\right]$$ and for this we better use the second description of the lattice as $M\frac{g}{h}=(\frac{g^{\prime }}{h},\frac{1}{h^{2}M})$ with $g^{\prime }$ the multiplicative inverse of $g$ modulo $h$. Under the action by $y$, the second factor $\frac{1}{h^{2}M}$ will be fixed, so this time we have to look at all lattices of the form $(\frac{g}{h},\frac{1}{h^{2}M})$ with $0\le g<h$, which again can be considered as another set of $h$-th roots of unity, centered at $hM$. Here's this second interpretation for $h=6$: Under $x$ the first set of $h$-th roots of unity centered at $hM$ is permuted, whereas $y$ permutes the second set of $h$-th roots of unity.
These interpretations can be used to spot errors in computing the finite groups ${\mathrm{\Gamma }}_0(n|h)/{\mathrm{\Gamma }}_0(n.h)$.

Here’s part of the calculation of the action of $y$ on the $(360|1)$-snake (which consists of $60$-lattices).

First I got a group of order roughly $600.000$. After correcting some erroneous cycles, the order went down to 6912.

Finally I spotted that I mis-numbered two lattices in the description of $x$ and $y$, and the order went down to $48$ as it should, because I knew it had to be equal to $C_2\times C_2\times A_4$.