Tag: Tits

Conway’s Big Picture consists of all pairs of rational numbers $M,\frac{g}{h}$ with $M>0$ and $0\le \frac{g}{h}<1$ with $(g,h)=1$. Recall from last time that $M,\frac{g}{h}$ stands for the lattice
$$\mathbb{Z}(M{\overrightarrow{e}}_1+\frac{g}{h}{\overrightarrow{e}}_2)\oplus \mathbb{Z}{\overrightarrow{e}}_2\subset {\mathbb{Q}}^2$$
and we associate to it the rational $2\times 2$ matrix
$${\alpha }_{M,\frac{g}{h}}=\left[\begin{array}{cc}M& \frac{g}{h}\\ 0& 1\end{array}\right]$$

If $M$ is a natural number we write $M\frac{g}{h}$ and call the corresponding lattice number-like, if $g=0$ we drop the zero and write $M$.

The Big Picture carries a wealth of structures. Today, we will see that it can be factored as the product of Bruhat-Tits buildings for $G{L}_2({\mathbb{Q}}_p)$, over all prime numbers $p$.

Here’s the factor-building for $p=2$, which is a $3$-valent tree:

To see this, define the distance between lattices to be
$$d(M,\frac{g}{h}|N,\frac{i}{j})=logDet(q({\alpha }_{M,\frac{g}{h}}.{\alpha }_{N,\frac{i}{j}}^{-1}))$$
where $q$ is the smallest strictly positive rational number such that $q({\alpha }_{M,\frac{g}{h}}.{\alpha }_{N,\frac{i}{j}}^{-1})\in G{L}_2(\mathbb{Z})$.

We turn the Big Picture into a (coloured) graph by drawing an edge (of colour $p$, for $p$ a prime number) between any two lattices distanced by $log(p)$.

The $p$-coloured subgraph is $p+1$-valent.

The $p$-neighbours of the lattice $1=\mathbb{Z}{\overrightarrow{e}}_1\oplus \mathbb{Z}{\overrightarrow{e}}_2$ are precisely these $p+1$ lattices:

$$p\text{and}\frac{1}{p},\frac{k}{p}\text{for}0\le k<p$$ And, multiplying the corresponding matrices with ${\alpha }_{M,\frac{g}{h}}$ tells us that the $p$-neighbours of $M,\frac{g}{h}$ are then these $p+1$ lattices: $$pM,\frac{pg}{h}mod1\text{and}\frac{M}{p},\frac{1}{p}(\frac{g}{h}+k)mod1\text{for}0\le k<p$$ Here's part of the $2$-coloured neighbourhood of $1$

To check that the $p$-coloured subgraph is indeed the Bruhat-Tits building of $G{L}_2({\mathbb{Q}}_p)$ it remains to see that it is a tree.

For this it is best to introduce $p+1$ operators on lattices

$$p\ast \text{and}\frac{k}{p}\ast \text{for}0\le k<p$$ defined by left-multiplying ${\alpha }_{M,\frac{g}{h}}$ by the matrices $$\left[\begin{array}{cc}p& 0\\ 0& 1\end{array}\right]\text{and}\left[\begin{array}{cc}\frac{1}{p}& \frac{k}{p}\\ 0& 1\end{array}\right]\text{for}0\le k<p$$ The lattice $p\ast M,\frac{g}{h}$ lies closer to $1$ than $M,\frac{g}{h}$ (unless $M,\frac{g}{h}=M$ is a number) whereas the lattices $\frac{k}{p}\ast M,\frac{g}{h}$ lie further, so it suffices to show that the $p$ operators $$\frac{0}{p}\ast ,\frac{1}{p}\ast ,\dots ,\frac{p-1}{p}\ast$$ form a free non-commutative monoid.
This follows from the fact that the operator
$$(\frac{k_n}{p}\ast )\circ \cdots \circ (\frac{k_2}{p}\ast )\circ (\frac{k_1}{p}\ast )$$
is given by left-multiplication with the matrix
$$\left[\begin{array}{cc}\frac{1}{p^n}& \frac{k_1}{p^n}+\frac{k_2}{p^{n-1}}+\cdots +\frac{k_n}{p}\\ 0& 1\end{array}\right]$$
which determines the order in which the $k_i$ occur.

A lattice at distance $nlog(p)$ from $1$ can be uniquely written as
$$(\frac{k_{n-l}}{p}\ast )\circ \cdots \circ (\frac{k_{l+1}}{p}\ast )\circ (p^l\ast )1$$
which gives us the unique path to it from $1$.

The Big Picture itself is then the product of these Bruhat-Tits trees over all prime numbers $p$. Decomposing the distance from $M,\frac{g}{h}$ to $1$ as
$$d(M,\frac{g}{h}|1)=n_{1}log(p_1)+\cdots +n_{k}log(p_k)$$
will then allow us to find minimal paths from $1$ to $M,\frac{g}{h}$.

But we should be careful in drawing $2$-dimensional cells (or higher dimensional ones) in this ‘product’ of trees as the operators
$$\frac{k}{p}\ast \text{and}\frac{l}{q}\ast$$
for different primes $p$ and $q$ do not commute, in general. The composition
$$(\frac{k}{p}\ast )\circ (\frac{l}{q}\ast )\text{with matrix}\left[\begin{array}{cc}\frac{1}{pq}& \frac{kq+l}{pq}\\ 0& 1\end{array}\right]$$
has as numerator in the upper-right corner $0\le kq+l<pq$ and this number can be uniquely(!) written as $$kq+l=up+v\text{with}0\le u<q,0\le v<p$$ That is, there are unique operators $\frac{u}{q}\ast$ and $\frac{v}{p}\ast$ such that $$(\frac{k}{p}\ast )\circ (\frac{l}{q}\ast )=(\frac{u}{q}\ast )\circ (\frac{v}{p}\ast )$$ which determine the $2$-cells These give us the commutation relations between the free monoids of operators corresponding to different primes.
For the primes $2$ and $3$, relevant in the description of the Moonshine Picture, the commutation relations are

$$(\frac{0}{2}\ast )\circ (\frac{0}{3}\ast )=(\frac{0}{3}\ast )\circ (\frac{0}{2}\ast ),(\frac{0}{2}\ast )\circ (\frac{1}{3}\ast )=(\frac{0}{3}\ast )\circ (\frac{1}{2}\ast ),(\frac{0}{2}\ast )\circ (\frac{2}{3}\ast )=(\frac{1}{3}\ast )\circ (\frac{0}{2}\ast )$$

$$(\frac{1}{2}\ast )\circ (\frac{0}{3}\ast )=(\frac{1}{3}\ast )\circ (\frac{1}{2}\ast ),(\frac{1}{2}\ast )\circ (\frac{1}{3}\ast )=(\frac{2}{3}\ast )\circ (\frac{0}{2}\ast ),(\frac{1}{2}\ast )\circ (\frac{2}{3}\ast )=(\frac{2}{3}\ast )\circ (\frac{1}{2}\ast )$$