A tetrahedral snake

A tetrahedral snake, sometimes called a Steinhaus snake, is a collection of tetrahedra, linked face to face.

Steinhaus showed in 1956 that the last tetrahedron in the snake can never be a translation of the first one. This is a consequence of the fact that the group generated by the four reflexions in the faces of a tetrahedron form the free product $C_2\ast C_2\ast C_2\ast C_2$.

For a proof of this, see Stan Wagon’s book The Banach-Tarski paradox, starting at page 68.

The tetrahedral snake we will look at here is a snake in the Big Picture which we need to determine the moonshine group $(3|3)$ corresponding to conjugacy class 3C of the Monster.

The thread $(3|3)$ is the spine of the $(9|1)$-snake which involves the following lattices

It is best to look at the four extremal lattices as the vertices of a tetrahedron with the lattice $3$ corresponding to its point of gravity.

The congruence subgroup ${\mathrm{\Gamma }}_0(9)$ fixes each of these lattices, and the arithmetic group ${\mathrm{\Gamma }}_0(3|3)$ is the conjugate of ${\mathrm{\Gamma }}_0(1)$
$${\mathrm{\Gamma }}_0(3|3)=\{\left[\begin{array}{cc}\frac{1}{3}& 0\\ 0& 1\end{array}\right].\left[\begin{array}{cc}a& b\\ c& d\end{array}\right].\left[\begin{array}{cc}3& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}a& \frac{b}{3}\\ 3c& 1\end{array}\right]|ad-bc=1\}$$
We know that ${\mathrm{\Gamma }}_0(3|3)$ normalizes the subgroup ${\mathrm{\Gamma }}_0(9)$ and we need to find the moonshine group $(3|3)$ which should have index $3$ in ${\mathrm{\Gamma }}_0(3|3)$ and contain ${\mathrm{\Gamma }}_0(9)$.

So, it is natural to consider the finite group $A={\mathrm{\Gamma }}_0(3|3)/{\mathrm{\Gamma }}_9(0)$ which is generated by the co-sets of
$$x=\left[\begin{array}{cc}1& \frac{1}{3}\\ 0& 1\end{array}\right]\text{and}y=\left[\begin{array}{cc}1& 0\\ 3& 0\end{array}\right]$$
To determine this group we look at the action of it on the lattices in the $(9|1)$-snake. It will fix the central lattice $3$ but will move the other lattices.

Recall that it is best to associate to the lattice $M.\frac{g}{h}$ the matrix
$${\alpha }_{M,\frac{g}{h}}=\left[\begin{array}{cc}M& \frac{g}{h}\\ 0& 1\end{array}\right]$$
and then the action is given by right-multiplication.

$$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right].x=\left[\begin{array}{cc}1& \frac{1}{3}\\ 0& 1\end{array}\right],\left[\begin{array}{cc}1& \frac{1}{3}\\ 0& 1\end{array}\right].x=\left[\begin{array}{cc}1& \frac{2}{3}\\ 0& 1\end{array}\right],\left[\begin{array}{cc}1& \frac{2}{3}\\ 0& 1\end{array}\right].x=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$
That is, $x$ corresponds to a $3$-cycle $1\to 1\frac{1}{3}\to 1\frac{2}{3}\to 1$ and fixes the lattice $9$ (so is rotation around the axis through the vertex $9$).

To compute the action of $y$ it is best to use an alternative description of the lattice, replacing the roles of the base-vectors ${\overrightarrow{e}}_1$ and ${\overrightarrow{e}}_2$. These latices are projectively equivalent
$$\mathbb{Z}(M{\overrightarrow{e}}_1+\frac{g}{h}{\overrightarrow{e}}_2)\oplus \mathbb{Z}{\overrightarrow{e}}_2\text{and}\mathbb{Z}{\overrightarrow{e}}_1\oplus \mathbb{Z}(\frac{g^{\prime }}{h}{\overrightarrow{e}}_1+\frac{1}{h^{2}M}{\overrightarrow{e}}_2)$$
where $g.g^{\prime }\equiv 1(modh)$. So, we have equivalent descriptions of the lattices
$$M,\frac{g}{h}=(\frac{g^{\prime }}{h},\frac{1}{h^{2}M})\text{and}M,0=(0,\frac{1}{M})$$
and we associate to the lattice in the second normal form the matrix
$${\beta }_{M,\frac{g}{h}}=\left[\begin{array}{cc}1& 0\\ \frac{g^{\prime }}{h}& \frac{1}{h^{2}M}\end{array}\right]$$
and then the action is again given by right-multiplication.

In the tetrahedral example we have
$$1=(0,\frac{1}{3}),1\frac{1}{3}=(\frac{1}{3},\frac{1}{9}),1\frac{2}{3}=(\frac{2}{3},\frac{1}{9}),9=(0,\frac{1}{9})$$
and
$$\left[\begin{array}{cc}1& 0\\ \frac{1}{3}& \frac{1}{9}\end{array}\right].y=\left[\begin{array}{cc}1& 0\\ \frac{2}{3}& \frac{1}{9}\end{array}\right],\left[\begin{array}{cc}1& 0\\ \frac{2}{3}& \frac{1}{9}\end{array}\right].y=\left[\begin{array}{cc}1& 0\\ 0& \frac{1}{9}\end{array}\right],\left[\begin{array}{cc}1& 0\\ 0& \frac{1}{9}\end{array}\right].y=\left[\begin{array}{cc}1& 0\\ \frac{1}{3}& \frac{1}{9}\end{array}\right]$$
That is, $y$ corresponds to the $3$-cycle $9\to 1\frac{1}{3}\to 1\frac{2}{3}\to 9$ and fixes the lattice $1$ so is a rotation around the axis through $1$.

Clearly, these two rotations generate the full rotation-symmetry group of the tetrahedron
$${\mathrm{\Gamma }}_0(3|3)/{\mathrm{\Gamma }}_0(9)\simeq A_4$$
which has a unique subgroup of index $3$ generated by the reflexions (rotations with angle ${180}^o$ around axis through midpoints of edges), generated by $x.y$ and $y.x$.

The moonshine group $(3|3)$ is therefore the subgroup generated by
$$(3|3)=⟨{\mathrm{\Gamma }}_0(9),\left[\begin{array}{cc}2& \frac{1}{3}\\ 3& 1\end{array}\right],\left[\begin{array}{cc}1& \frac{1}{3}\\ 3& 2\end{array}\right]⟩$$