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Category: groups

Moonshine’s green anaconda

Published January 18, 2018 by lievenlb

The largest snake in the moonshine picture determines the moonshine group $(24|12)$ and is associated to conjugacy class $24J$ of the monster.

It contains $70$ lattices, about one third of the total number of lattices in the moonshine picture.

The anaconda’s backbone is the $(288|1)$ thread below (edges in the $2$-tree are black, those in the $3$-tree red and coloured numbers are symmetric with respect to the $(24|12)$-spine and have the same local snake-structure.

\[
\xymatrix{9 \ar@{-}[r] \ar@[red]@{-}[d] & \color{green}{18} \ar@{-}[r] \ar@[red]@{-}[d] & \color{yellow}{36} \ar@{-}[r] \ar@[red]@{-}[d] & \color{yellow}{72} \ar@{-}[r] \ar@[red]@{-}[d] & \color{green}{144} \ar@{-}[r] \ar@[red]@{-}[d] & 288 \ar@[red]@{-}[d] \\
3 \ar@{-}[r] \ar@[red]@{-}[d] & \color{blue}{6} \ar@{-}[r] \ar@[red]@{-}[d] & \color{red}{12} \ar@{-}[r] \ar@[red]@{-}[d] & \color{red}{24} \ar@{-}[r] \ar@[red]@{-}[d] & \color{blue}{48} \ar@{-}[r] \ar@[red]@{-}[d] & 96 \ar@[red]@{-}[d] \\
1 \ar@{-}[r] & \color{green}{2} \ar@{-}[r] & \color{yellow}{4} \ar@{-}[r] & \color{yellow}{8} \ar@{-}[r] & \color{green}{16} \ar@{-}[r] & 32 } \]

These are the only number-lattices in the anaconda. The remaining lattices are number-like, that is of the form $M \frac{g}{h}$ with $M$ an integer and $1 \leq g < h$ with $(g,h)=1$.
There are

– $12$ with $h=2$ and $M$ a divisor of $72$.

– $12$ with $h=3$ and $M$ a divisor of $32$.

– $12$ with $h=4$ and $M$ a divisor of $18$.

– $8$ with $h=6$ and $M$ a divisor of $8$.

– $8$ with $h=12$ and $M=1,2$.

The non-number lattices in the snake are locally in the coloured numbers:

In $2,16,18,144=2M$

\[
\xymatrix{& \color{green}{2M} \ar@{-}[r] & M \frac{1}{2}} \]

In $4,8,36,72=4M$

\[
\xymatrix{M \frac{1}{2} \ar@{-}[d] & & M \frac{1}{4} \ar@{-}[d] \\
2M \ar@{-}[r] & \color{yellow}{4M} \ar@{-}[r] & 2M \frac{1}{2} \ar@{-}[d] \\
& & M \frac{3}{4}} \]

In $6,48=6M$

\[
\xymatrix{M \frac{1}{3} \ar@[red]@{-}[d] & 2M \frac{1}{3} & M \frac{1}{6} \ar@[red]@{-}[d] & \\
3M \ar@{-}[r] \ar@[red]@{-}[d] & \color{blue}{6M} \ar@{-}[r] \ar@[red]@{-}[u] \ar@[red]@{-}[d] & 3M \frac{1}{2} \ar@[red]@{-}[r] \ar@[red]@{-}[d] & M \frac{5}{6} \\
M \frac{2}{3} & 2M \frac{2}{3} & M \frac{1}{2} & } \]

In $12,24=12M$ the local structure looks like

Here, we used the commutation relations to reach all lattices at distance $log(6)$ and $log(12)$ by first walking the $2$-adic tree and postpone the last step for the $3$-tree.

Perhaps this is also a good strategy to get a grip on the full moonshine picture:

First determine subsets of the moonshine thread with the same local structure, and then determine for each class this local structure.

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Coxeter on Escher’s Circle Limits

Published January 15, 2018 by lievenlb

Conway’s orbifold notation gives a uniform notation for all discrete groups of isometries of the sphere, the Euclidian plane as well as the hyperbolic plane.

This includes the groups of symmetries of Escher’s Circle Limit drawings. Here’s Circle Limit III

And ‘Angels and Devils’ aka Circle Limit IV:

If one crawls along a mirror of this pattern until one hits another mirror and then turns right along this mirror and continues like this, you get a quadrilateral path with four corners $\frac{\pi}{3}$, whose center seems to be a $4$-fold gyration point. So, it appears to have symmetry $4 \ast 3$.


(image credit: MathCryst)

However, looking more closely, every fourth figure (either devil or angel) is facing away rather than towards us, so there’s no gyration point, and the group drops to $\ast 3333$.

Harold S. M. Coxeter met Escher in Amsterdam at the ICM 1954.

The interaction between the two led to Escher’s construction of the Circle Limits, see How did Escher do it?

Here’s an old lecture by Coxeter on the symmetry of the Circle Limits:



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Roots of unity and the Big Picture

Published January 14, 2018 by lievenlb

All lattices in the moonshine picture are number-like, that is of the form $M \frac{g}{h}$ with $M$ a positive integer and $0 \leq g < h$ 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 $2 log(h)$, 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 $h^2M$).

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.

\[
\xymatrix{& & M \frac{1}{2} \ar@[blue]@{-}[d] & \\
& M \ar@[blue]@{-}[r] \ar@[red]@{-}[d] & 2M \ar@[red]@{-}[d] & M \frac{1}{6} \ar@[red]@{-}[d] \\
M \frac{1}{3} \ar@[red]@{-}[r] & 3M \ar@[blue]@{-}[r] \ar@[red]@{-}[d] & \boxed{6 M} \ar@[blue]@{-}[r] & 3M \frac{1}{2} \ar@[red]@{-}[d] \\ & M \frac{2}{3} & & M \frac{5}{6}} \]

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
\[
x = \begin{bmatrix} 1 & \frac{1}{h} \\ 0 & 1 \end{bmatrix} \]
which sends a lattice of the form $M \frac{g}{h}$ with $0 \leq g < h$ to $M \frac{g+M}{h}$, so is a rotation over $\frac{2 \pi M}{h}$ around $h M$. But, we also have to describe the base-change action with the matrix \[ y = \begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix} \] and for this we better use the second description of the lattice as $M \frac{g}{h}=(\frac{g'}{h},\frac{1}{h^2M})$ with $g'$ the multiplicative inverse of $g$ modulo $h$. Under the action by $y$, the second factor $\frac{1}{h^2M}$ will be fixed, so this time we have to look at all lattices of the form $(\frac{g}{h},\frac{1}{h^2M})$ with $0 \leq g < h$, which again can be considered as another set of $h$-th roots of unity, centered at $hM$. Here's this second interpretation for $h=6$: \[ \xymatrix{M \frac{5}{6} \ar@[red]@{-}[d] & & 4M \frac{1}{3} \ar@[red]@{-}[d] & \\ 3M \frac{1}{2} \ar@[blue]@{-}[r] \ar@[red]@{-}[d] & \boxed{6M} \ar@[blue]@{-}[r] \ar@[red]@{-}[d] & 12 M \ar@[red]@{-}[r] \ar@[red]@{-}[d] & 4 M \frac{2}{3} \\ M \frac{1}{6} & 18 M \ar@[blue]@{-}[r] \ar@[blue]@{-}[d] & 36 M & \\ & 9M \frac{1}{2} & & } \] 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 $\Gamma_0(n|h)/\Gamma_0(n.h)$.

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 $C_2 \times C_2 \times A_4$.

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nc-geometry and moonshine?

Published January 12, 2018 by lievenlb

A well-known link between Conway’s Big Picture and non-commutative geometry is given by the Bost-Connes system.

This quantum statistical mechanical system encodes the arithmetic properties of cyclotomic extensions of $\mathbb{Q}$.

The corresponding Bost-Connes algebra encodes the action by the power-maps on the roots of unity.

It has generators $e_n$ and $e_n^*$ for every natural number $n$ and additional generators $e(\frac{g}{h})$ for every element in the additive group $\mathbb{Q}/\mathbb{Z}$ (which is of course isomorphic to the multiplicative group of roots of unity).

The defining equations are
\[
\begin{cases}
e_n.e(\frac{g}{h}).e_n^* = \rho_n(e(\frac{g}{h})) \\
e_n^*.e(\frac{g}{h}) = \Psi^n(e(\frac{g}{h}).e_n^* \\
e(\frac{g}{h}).e_n = e_n.\Psi^n(e(\frac{g}{h})) \\
e_n.e_m=e_{nm} \\
e_n^*.e_m^* = e_{nm}^* \\
e_n.e_m^* = e_m^*.e_n~\quad~\text{if $(m,n)=1$}
\end{cases}
\]

Here $\Psi^n$ are the power-maps, that is $\Psi^n(e(\frac{g}{h})) = e(\frac{ng}{h}~mod~1)$, and the maps $\rho_n$ are given by
\[
\rho_n(e(\frac{g}{h})) = \sum e(\frac{i}{j}) \]
where the sum is taken over all $\frac{i}{j} \in \mathbb{Q}/\mathbb{Z}$ such that $n.\frac{i}{j}=\frac{g}{h}$.

Conway’s Big Picture has as its vertices the (equivalence classes of) lattices $M,\frac{g}{h}$ with $M \in \mathbb{Q}_+$ and $\frac{g}{h} \in \mathbb{Q}/\mathbb{Z}$.

The Bost-Connes algebra acts on the vector-space with basis the vertices of the Big Picture. The action is given by:
\[
\begin{cases}
e_n \ast \frac{c}{d},\frac{g}{h} = \frac{nc}{d},\rho^m(\frac{g}{h})~\quad~\text{with $m=(n,d)$} \\
e_n^* \ast \frac{c}{d},\frac{g}{h} = (n,c) \times \frac{c}{nd},\Psi^{\frac{n}{m}}(\frac{g}{h})~\quad~\text{with $m=(n,c)$} \\
e(\frac{a}{b}) \ast \frac{c}{d},\frac{g}{h} = \frac{c}{d},\Psi^c(\frac{a}{b}) \frac{g}{h}
\end{cases}
\]

This connection makes one wonder whether non-commutative geometry can shed a new light on monstrous moonshine?

This question is taken up by Jorge Plazas in his paper Non-commutative geometry of groups like $\Gamma_0(N)$

Plazas shows that the bigger Connes-Marcolli $GL_2$-system also acts on the Big Picture. An intriguing quote:

“Our interest in the $GL_2$-system comes from the fact that its thermodynamic properties encode the arithmetic theory of modular functions to an extend which makes it possible for us to capture aspects of moonshine theory.”

Looks like the right kind of paper to take along when I disappear next week for some time in the French mountains…

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Everything’s wrappable to a sphere

Published January 11, 2018 by lievenlb

One of the better opening quotes of a paper:

“Even quite ungainly objects, like chairs and tables, will become almost spherical if you wrap them in enough newspaper.”

The paper in question is The orbifold notation for surface groups by John Conway.

Here’s Conway talking leisurely about Thurston’s idea to capture the acting group via the topology of the orbifold space and his own notation for such orbifolds.



Here’s another version of the paper, with illustrations: The orbifold notation for two-dimensional groups, by Conway and Daniel H. Huson.

A very accessible account are these lecture notes:

A field guide to the orbifolds, notes from class on “Geometry and the Imagination” in Minneapolis, with John Conway, Peter Doyle, Jane Gilman and Bill Thurston, on June 17–28, 1991.

And, here are notes by Thurston on The Geometry and Topology of Three-Manifolds, including stuff about orbifolds.

I came across these papers struggling my way through On the discrete groups of moonshine by Conway, McKay and Sebbar.

On the genus $0$ property of moonshine groups they have this to say:

“As for groups of the form $(n|h)+e,f,\dots$, the genus can be determined from the fundamental regions using the Riemann-Hurwitz formula. Since most of the groups are not subgroups of the modular group, the calculations of the genus, which cannot be produced here because of their length, are carried out by finding the elliptic fixed points and the cone points in the orbifolds attached to the fundamental regions. The Euler characteristic of the orbifold determines the genus of the group. See [paper] for more details on orbifold techniques.”

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A tetrahedral snake

Published January 7, 2018 by lievenlb

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
\[
\xymatrix{& & 1 \frac{1}{3} \ar@[red]@{-}[dd] & & \\
& & & & \\
1 \ar@[red]@{-}[rr] & & 3 \ar@[red]@{-}[rr] \ar@[red]@{-}[dd] & & 1 \frac{2}{3} \\
& & & & \\
& & 9 & &} \]
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 $\Gamma_0(9)$ fixes each of these lattices, and the arithmetic group $\Gamma_0(3|3)$ is the conjugate of $\Gamma_0(1)$
\[
\Gamma_0(3|3) = \{ \begin{bmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{bmatrix}.\begin{bmatrix} a & b \\ c & d \end{bmatrix}.\begin{bmatrix} 3 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & \frac{b}{3} \\ 3c & 1 \end{bmatrix}~|~ad-bc=1 \} \]
We know that $\Gamma_0(3|3)$ normalizes the subgroup $\Gamma_0(9)$ and we need to find the moonshine group $(3|3)$ which should have index $3$ in $\Gamma_0(3|3)$ and contain $\Gamma_0(9)$.

So, it is natural to consider the finite group $A=\Gamma_0(3|3)/\Gamma_9(0)$ which is generated by the co-sets of
\[
x = \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1 \end{bmatrix} \qquad \text{and} \qquad y = \begin{bmatrix} 1 & 0 \\ 3 & 0 \end{bmatrix} \]
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
\[
\alpha_{M,\frac{g}{h}} = \begin{bmatrix} M & \frac{g}{h} \\ 0 & 1 \end{bmatrix} \]
and then the action is given by right-multiplication.

\[
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.x = \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1 \end{bmatrix}.x = \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & 1 \end{bmatrix}.x=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
That is, $x$ corresponds to a $3$-cycle $1 \rightarrow 1 \frac{1}{3} \rightarrow 1 \frac{2}{3} \rightarrow 1$ 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 $\vec{e}_1$ and $\vec{e}_2$. These latices are projectively equivalent
\[
\mathbb{Z} (M \vec{e}_1 + \frac{g}{h} \vec{e}_2) \oplus \mathbb{Z} \vec{e}_2 \quad \text{and} \quad \mathbb{Z} \vec{e}_1 \oplus \mathbb{Z} (\frac{g’}{h} \vec{e}_1 + \frac{1}{h^2M} \vec{e}_2) \]
where $g.g’ \equiv~1~(mod~h)$. So, we have equivalent descriptions of the lattices
\[
M,\frac{g}{h} = (\frac{g’}{h},\frac{1}{h^2M}) \quad \text{and} \quad M,0 = (0,\frac{1}{M}) \]
and we associate to the lattice in the second normal form the matrix
\[
\beta_{M,\frac{g}{h}} = \begin{bmatrix} 1 & 0 \\ \frac{g’}{h} & \frac{1}{h^2M} \end{bmatrix} \]
and then the action is again given by right-multiplication.

In the tetrahedral example we have
\[
1 = (0,\frac{1}{3}), \quad 1\frac{1}{3}=(\frac{1}{3},\frac{1}{9}), \quad 1\frac {2}{3}=(\frac{2}{3},\frac{1}{9}), \quad 9 = (0,\frac{1}{9}) \]
and
\[
\begin{bmatrix} 1 & 0 \\ \frac{1}{3} & \frac{1}{9} \end{bmatrix}.y = \begin{bmatrix} 1 & 0 \\ \frac{2}{3} & \frac{1}{9} \end{bmatrix},\quad
\begin{bmatrix} 1 & 0 \\ \frac{2}{3} & \frac{1}{9} \end{bmatrix}. y = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{9} \end{bmatrix}, \quad
\begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{9} \end{bmatrix}. y = \begin{bmatrix} 1 & 0 \\ \frac{1}{3} & \frac{1}{9} \end{bmatrix} \]
That is, $y$ corresponds to the $3$-cycle $9 \rightarrow 1 \frac{1}{3} \rightarrow 1 \frac{2}{3} \rightarrow 9$ 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
\[
\Gamma_0(3|3)/\Gamma_0(9) \simeq A_4 \]
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
\[
(3|3) = \langle \Gamma_0(9),\begin{bmatrix} 2 & \frac{1}{3} \\ 3 & 1 \end{bmatrix},\begin{bmatrix} 1 & \frac{1}{3} \\ 3 & 2 \end{bmatrix} \rangle \]

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the 171 moonshine groups

Published January 6, 2018 by lievenlb

Monstrous moonshine associates to every element of order $n$ of the monster group $\mathbb{M}$ an arithmetic group of the form
\[
(n|h)+e,f,\dots \]
where $h$ is a divisor of $24$ and of $n$ and where $e,f,\dots$ are divisors of $\frac{n}{h}$ coprime with its quotient.

In snakes, spines, and all that we’ve constructed the arithmetic group
\[
\Gamma_0(n|h)+e,f,\dots \]
which normalizes $\Gamma_0(N)$ for $N=h.n$. If $h=1$ then this group is the moonshine group $(n|h)+e,f,\dots$, but for $h > 1$ the moonshine group is a specific subgroup of index $h$ in $\Gamma_0(n|h)+e,f,\dots$.

I’m sure one can describe this subgroup explicitly in each case by analysing the action of the finite group $(\Gamma_0(n|h)+e,f,\dots)/\Gamma_0(N)$ on the $(N|1)$-snake. Some examples were worked out by John Duncan in his paper Arithmetic groups and the affine E8 Dynkin diagram.

But at the moment I don’t understand the general construction given by Conway, McKay and Sebbar in On the discrete groups of moonshine. I’m stuck at the last sentence of (2) in section 3. Nothing a copy of Charles Ferenbaugh Ph. D. thesis cannot fix.

The correspondence between the conjugacy classes of the Monster and these arithmetic groups takes up 3 pages in Conway & Norton’s Monstrous Moonshine. Here’s the beginning of it.

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Snakes, spines, threads and all that

Published January 5, 2018 by lievenlb

Conway introduced his Big Picture to make it easier to understand and name the groups appearing in Monstrous Moonshine.

For $M \in \mathbb{Q}_+$ and $0 \leq \frac{g}{h} < 1$, $M,\frac{g}{h}$ denotes (the projective equivalence class of) the lattice \[ \mathbb{Z} (M \vec{e}_1 + \frac{g}{h} \vec{e}_2) \oplus \mathbb{Z} \vec{e}_2 \] which we also like to represent by the $2 \times 2$ matrix \[ \alpha_{M,\frac{g}{h}} = \begin{bmatrix} M & \frac{g}{h} \\ 0 & 1 \end{bmatrix} \] A subgroup $G$ of $GL_2(\mathbb{Q})$ is said to fix $M,\frac{g}{h}$ if
\[
\alpha_{M,\frac{g}{h}}.G.\alpha_{M,\frac{g}{h}}^{-1} \subset SL_2(\mathbb{Z}) \]
The full group of all elements fixing $M,\frac{g}{h}$ is the conjugate
\[
\alpha_{M,\frac{g}{h}}^{-1}.SL_2(\mathbb{Z}).\alpha_{M,\frac{g}{h}} \]
For a number lattice $N=N,0$ the elements of this group are all of the form
\[
\begin{bmatrix} a & \frac{b}{N} \\ cN & d \end{bmatrix} \qquad \text{with} \qquad \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in SL_2(\mathbb{Z}) \]
and the intersection with $SL_2(\mathbb{Z})$ (which is the group of all elements fixing the lattice $1=1,0$) is the congruence subgroup
\[
\Gamma_0(N) = \{ \begin{bmatrix} a & b \\ cN & d \end{bmatrix}~|~ad-Nbc = 1 \} \]
Conway argues that this is the real way to think of $\Gamma_0(N)$, as the joint stabilizer of the two lattices $N$ and $1$!

The defining definition of 24 tells us that $\Gamma_0(N)$ fixes more lattices. In fact, it fixes exactly the latices $M \frac{g}{h}$ such that
\[
1~|~M~|~\frac{N}{h^2} \quad \text{with} \quad h^2~|~N \quad \text{and} \quad h~|~24 \]
Conway calls the sub-graph of the Big Picture on these lattices the snake of $(N|1)$.

Here’s the $(60|1)$-snake (note that $60=2^2.3.5$ so $h=1$ or $h=2$ and edges corresponding to the prime $2$ are coloured red, those for $3$ green and for $5$ blue).

\[
\xymatrix{& & & 15 \frac{1}{2} \ar@[red]@{-}[dd] & & \\
& & 5 \frac{1}{2} \ar@[red]@{-}[dd] & & & \\
& 15 \ar@[red]@{-}[rr] \ar@[blue]@{-}[dd] & & 30 \ar@[red]@{-}[rr] \ar@[blue]@{-}[dd] & & 60 \ar@[blue]@{-}[dd] \\
5 \ar@[green]@{-}[ru] \ar@[blue]@{-}[dd] \ar@[red]@{-}[rr] & & 10 \ar@[green]@{-}[ru] \ar@[red]@{-}[rr] \ar@[blue]@{-}[dd] & & 20 \ar@[green]@{-}[ru] \ar@[blue]@{-}[dd] & \\
& 3 \ar@[red]@{-}[rr] & & 6 \ar@[red]@{-}[rr] \ar@[red]@{-}[dd] & & 12 \\
1 \ar@[green]@{-}[ru] \ar@[red]@{-}[rr] & & 2 \ar@[green]@{-}[ru] \ar@[red]@{-}[rr] \ar@[red]@{-}[dd] & & 4 \ar@[green]@{-}[ru] & \\
& & & 3\frac{1}{2} & & \\
& & 1 \frac{1}{2} & & &} \]

The sub-graph of lattices fixed by $\Gamma_0(N)$ for $h=1$, that is all number-lattices $M=M,0$ for $M$ a divisor of $N$ is called the thread of $(N|1)$. Here’s the $(60|1)$-thread

\[
\xymatrix{
& 15 \ar@[red]@{-}[rr] \ar@[blue]@{-}[dd] & & 30 \ar@[red]@{-}[rr] \ar@[blue]@{-}[dd] & & 60 \ar@[blue]@{-}[dd] \\
5 \ar@[green]@{-}[ru] \ar@[blue]@{-}[dd] \ar@[red]@{-}[rr] & & 10 \ar@[green]@{-}[ru] \ar@[red]@{-}[rr] \ar@[blue]@{-}[dd] & & 20 \ar@[green]@{-}[ru] \ar@[blue]@{-}[dd] & \\
& 3 \ar@[red]@{-}[rr] & & 6 \ar@[red]@{-}[rr] & & 12 \\
1 \ar@[green]@{-}[ru] \ar@[red]@{-}[rr] & & 2 \ar@[green]@{-}[ru] \ar@[red]@{-}[rr] & & 4 \ar@[green]@{-}[ru] &
} \]

If $N$ factors as $N = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}$ then the $(N|1)$-thread is the product of the $(p_i^{e_i}|1)$-threads and has a symmetry group of order $2^k$.

It is generated by $k$ involutions, each one the reflexion in one $(p_i^{e_i}|1)$-thread and the identity on the other $(p_j^{e_j}|1)$-threads.
In the $(60|1)$-thread these are the reflexions in the three mirrors of the figure.

So, there is one involution for every divisor $e$ of $N$ such that $(e,\frac{N}{e})=1$. For such an $e$ there are matrices, with $a,b,c,d \in \mathbb{Z}$, of the form
\[
W_e = \begin{bmatrix} ae & b \\ cN & de \end{bmatrix} \quad \text{with} \quad ade^2-bcN=e \]
Think of Bezout and use that $(e,\frac{N}{e})=1$.

Such $W_e$ normalizes $\Gamma_0(N)$, that is, for any $A \in \Gamma_0(N)$ we have that $W_e.A.W_e^{-1} \in \Gamma_0(N)$. Also, the determinant of $W_e^e$ is equal to $e^2$ so we can write $W_e^2 = e A$ for some $A \in \Gamma_0(N)$.

That is, the transformation $W_e$ (left-multiplication) sends any lattice in the thread or snake of $(N|1)$ to another such lattice (up to projective equivalence) and if we apply $W_e^2$ if fixes each such lattice (again, up to projective equivalence), so it is the desired reflexion corresponding with $e$.

Consider the subgroup of $GL_2(\mathbb{Q})$ generated by $\Gamma_0(N)$ and some of these matrices $W_e,W_f,\dots$ and denote by $\Gamma_0(N)+e,f,\dots$ the quotient modulo positive scalar matrices, then
\[
\Gamma_0(N) \qquad \text{is a normal subgroup of} \qquad \Gamma_0(N)+e,f,\dots \]
with quotient isomorphic to some $(\mathbb{Z}/2\mathbb{Z})^l$ isomorphic to the subgroup generated by the involutions corresponding to $e,f,\dots$.

More generally, consider the $(n|h)$-thread for number lattices $n=n,0$ and $h=h,0$ such that $h | n$ as the sub-graph on all number lattices $l=l,0$ such that $h | l | n$. If we denote with $\Gamma_0(n|h)$ the point-wise stabilizer of $n$ and $h$, then we have that
\[
\Gamma(n|h) = \begin{bmatrix} h & 0 \\ 0 & 1 \end{bmatrix}^{-1}.\Gamma_0(\frac{n}{h}).\begin{bmatrix} h & 0 \\ 0 & 1 \end{bmatrix} \]
and we can then denote with
\[
\Gamma_0(n|h)+e,f,\dots \]
the conjugate of the corresponding group $\Gamma_0(\frac{n}{h})+e,f,\dots$.

If $h$ is the largest divisor of $24$ such that $h^2$ divides $N$, then Conway calls the spine of the $(N|1)$-snake the subgraph on all lattices of the snake whose distance from its periphery is exactly $log(h)$.

For $N=60$, $h=2$ and so the spine of the $(60|1)$-snake is the central piece connected with double black edges

\[
\xymatrix{& & & 15 \frac{1}{2} \ar@[red]@{-}[dd] & & \\
& & 5 \frac{1}{2} \ar@[red]@{-}[dd] & & & \\
& 15 \ar@[red]@{-}[rr] \ar@[blue]@{-}[dd] & & 30 \ar@[red]@{-}[rr] \ar@[black]@{=}[dd] & & 60 \ar@[blue]@{-}[dd] \\
5 \ar@[green]@{-}[ru] \ar@[blue]@{-}[dd] \ar@[red]@{-}[rr] & & 10 \ar@[black]@{=}[ru] \ar@[red]@{-}[rr] \ar@[black]@{=}[dd] & & 20 \ar@[green]@{-}[ru] \ar@[blue]@{-}[dd] & \\
& 3 \ar@[red]@{-}[rr] & & 6 \ar@[red]@{-}[rr] \ar@[red]@{-}[dd] & & 12 \\
1 \ar@[green]@{-}[ru] \ar@[red]@{-}[rr] & & 2 \ar@[black]@{=}[ru] \ar@[red]@{-}[rr] \ar@[red]@{-}[dd] & & 4 \ar@[green]@{-}[ru] & \\
& & & 3\frac{1}{2} & & \\
& & 1 \frac{1}{2} & & &} \]

which is the $(30|2)$-thread.

The upshot of all this is to have a visual proof of the Atkin-Lehner theorem which says that the full normalizer of $\Gamma_0(N)$ is the group $\Gamma_0(\frac{N}{h}|h)+$ (that is, adding all involutions) where $h$ is the largest divisor of $24$ for which $h^2|N$.

Any element of this normalizer must take every lattice in the $(N|1)$-snake fixed by $\Gamma_0(N)$ to another such lattice. Thus it follows that it must take the snake to itself.
Conversely, an element that takes the snake to itself must conjugate into itself the group of all matrices that fix every point of the snake, that is to say, must normalize $\Gamma_0(N)$.

But the elements that take the snake to itself are precisely those that take the spine to itself, and since this spine is just the $(\frac{N}{h}|h)$-thread, this group is just $\Gamma_0(\frac{N}{h}|h)+$.

Reference: J.H. Conway, “Understanding groups like $\Gamma_0(N)$”, in “Groups, Difference Sets, and the Monster”, Walter de Gruyter-Berlin-New York, 1996

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The Big Picture is non-commutative

Published January 3, 2018 by lievenlb

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

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 $GL_2(\mathbb{Q}_p)$, 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
\[
d(M,\frac{g}{h}~|~N,\frac{i}{j}) = log~Det(q(\alpha_{M,\frac{g}{h}}.\alpha_{N,\frac{i}{j}}^{-1})) \]
where $q$ is the smallest strictly positive rational number such that $q(\alpha_{M,\frac{g}{h}}.\alpha_{N,\frac{i}{j}}^{-1}) \in GL_2(\mathbb{Z})$.

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 $log(p)$.

\[
\xymatrix{M,\frac{g}{h} \ar@[red]@{-}[rr]|p & & N,\frac{i}{j}} \qquad~\text{iff}~\qquad d(M,\frac{g}{h}~|~N,\frac{i}{j})=log(p) \]

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

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

\[
p \qquad \text{and} \qquad \frac{1}{p},\frac{k}{p} \qquad \text{for} \qquad 0 \leq k < p \] 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: \[ pM,\frac{pg}{h}~mod~1 \qquad \text{and} \qquad \frac{M}{p},\frac{1}{p}(\frac{g}{h}+k)~mod~1 \qquad \text{for} \qquad 0 \leq k < p \] Here's part of the $2$-coloured neighbourhood of $1$

To check that the $p$-coloured subgraph is indeed the Bruhat-Tits building of $GL_2(\mathbb{Q}_p)$ it remains to see that it is a tree.

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

\[
p \ast \qquad \text{and} \qquad \frac{k}{p} \ast \qquad \text{for} \qquad 0 \leq k < p \] defined by left-multiplying $\alpha_{M,\frac{g}{h}}$ by the matrices \[ \begin{bmatrix} p & 0 \\ 0 & 1 \end{bmatrix} \qquad \text{and} \qquad \begin{bmatrix} \frac{1}{p} & \frac{k}{p} \\ 0 & 1 \end{bmatrix} \qquad \text{for} \qquad 0 \leq k < p \] 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
\[
(\frac{k_n}{p} \ast) \circ \dots \circ (\frac{k_2}{p} \ast) \circ (\frac{k_1}{p} \ast) \]
is given by left-multiplication with the matrix
\[
\begin{bmatrix} \frac{1}{p^n} & \frac{k_1}{p^n}+\frac{k_2}{p^{n-1}}+\dots+\frac{k_n}{p} \\ 0 & 1 \end{bmatrix} \]
which determines the order in which the $k_i$ occur.

A lattice at distance $n log(p)$ from $1$ can be uniquely written as
\[
(\frac{k_{n-l}}{p} \ast) \circ \dots \circ (\frac{k_{l+1}}{p} \ast) \circ (p^l \ast) 1 \]
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
\[
d(M,\frac{g}{h}~|~1) = n_1 log(p_1) + \dots + n_k log(p_k) \]
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
\[
\frac{k}{p} \ast \qquad \text{and} \qquad \frac{l}{q} \ast \]
for different primes $p$ and $q$ do not commute, in general. The composition
\[
(\frac{k}{p} \ast) \circ (\frac{l}{q} \ast) \qquad \text{with matrix} \qquad \begin{bmatrix} \frac{1}{pq} & \frac{kq+l}{pq} \\ 0 & 1 \end{bmatrix} \]
has as numerator in the upper-right corner $0 \leq kq + l < pq$ and this number can be uniquely(!) written as \[ kq+l = up+v \qquad \text{with} \qquad 0 \leq u < q,~0 \leq v < p \] 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 \[ \xymatrix{ \bullet \ar@[blue]@{-}[rr]^{\frac{u}{q} \ast} \ar@[red]@{-}[dd]_{\frac{v}{p} \ast} & & \bullet \ar@[red]@{-}[dd]^{\frac{k}{p} \ast} \\ & & \\ \bullet \ar@[blue]@{-}[rr]_{\frac{l}{q} \ast} & & \bullet} \] 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

\[
(\frac{0}{2} \ast) \circ (\frac{0}{3} \ast) = (\frac{0}{3} \ast) \circ (\frac{0}{2} \ast), \quad
(\frac{0}{2} \ast) \circ (\frac{1}{3} \ast) = (\frac{0}{3} \ast) \circ (\frac{1}{2} \ast),
\quad
(\frac{0}{2} \ast) \circ (\frac{2}{3} \ast) = (\frac{1}{3} \ast) \circ (\frac{0}{2} \ast) \]

\[
(\frac{1}{2} \ast) \circ (\frac{0}{3} \ast) = (\frac{1}{3} \ast) \circ (\frac{1}{2} \ast), \quad
(\frac{1}{2} \ast) \circ (\frac{1}{3} \ast) = (\frac{2}{3} \ast) \circ (\frac{0}{2} \ast),
\quad
(\frac{1}{2} \ast) \circ (\frac{2}{3} \ast) = (\frac{2}{3} \ast) \circ (\frac{1}{2} \ast) \]

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The defining property of 24

Published January 2, 2018 by lievenlb

From Wikipedia on 24:

“$24$ is the only number whose divisors, namely $1, 2, 3, 4, 6, 8, 12, 24$, are exactly those numbers $n$ for which every invertible element of the commutative ring $\mathbb{Z}/n\mathbb{Z}$ is a square root of $1$. It follows that the multiplicative group $(\mathbb{Z}/24\mathbb{Z})^* = \{ \pm 1, \pm 5, \pm 7, \pm 11 \}$ is isomorphic to the additive group $(\mathbb{Z}/2\mathbb{Z})^3$. This fact plays a role in monstrous moonshine.”

Where did that come from?

In the original “Monstrous Moonshine” paper by John Conway and Simon Norton, section 3 starts with:

“It is a curious fact that the divisors $h$ of $24$ are precisely those numbers $h$ for which $x.y \equiv 1~(mod~h)$ implies $x \equiv y~(mod~h)$.”

and a bit further they even call this fact:

“our ‘defining property of $24$'”.

The proof is pretty straightforward.

We want all $h$ such that every unit in $\mathbb{Z}/h \mathbb{Z}$ has order two.

By the Chinese remainder theorem we only have to check this for prime powers dividing $h$.

$5$ is a unit of order $4$ in $\mathbb{Z}/16 \mathbb{Z}$.

$2$ is a unit of order $6$ in $\mathbb{Z}/ 9 \mathbb{Z}$.

A generator of the cyclic group $(\mathbb{Z}/p\mathbb{Z})^*$ is a unit of order $p-1 > 2$ in $\mathbb{Z}/p \mathbb{Z}$, for any prime number $p \geq 5$.

This only leaves those $h$ dividing $2^3.3=24$.

But, what does it have to do with monstrous moonshine?

Moonshine assigns to elements of the Monster group $\mathbb{M}$ a specific subgroup of $SL_2(\mathbb{Q})$ containing a cofinite congruence subgroup

\[
\Gamma_0(N) = \{ \begin{bmatrix} a & b \\ cN & d \end{bmatrix}~|~a,b,c,d \in \mathbb{Z}, ad-Nbc = 1 \} \]

for some natural number $N = h.n$ where $n$ is the order of the monster-element, $h^2$ divides $N$ and … $h$ is a divisor of $24$.

To begin to understand how the defining property of $24$ is relevant in this, take any strictly positive rational number $M$ and any pair of coprime natural numbers $g < h$ and associate to $M \frac{g}{h}$ the matrix \[ \alpha_{M\frac{g}{h}} = \begin{bmatrix} M & \frac{g}{h} \\ 0 & 1 \end{bmatrix} \] We say that $\Gamma_0(N)$ fixes $M \frac{g}{h}$ if we have that
\[
\alpha_{M\frac{g}{h}} \Gamma_0(N) \alpha_{M\frac{g}{h}}^{-1} \subset SL_2(\mathbb{Z}) \]

For those in the know, $M \frac{g}{h}$ stands for the $2$-dimensional integral lattice
\[
\mathbb{Z} (M \vec{e}_1 + \frac{g}{h} \vec{e}_2) \oplus \mathbb{Z} \vec{e}_2 \]
and the condition tells that $\Gamma_0(N)$ preserves this lattice under base-change (right-multiplication).

In “Understanding groups like $\Gamma_0(N)$” Conway describes the groups appearing in monstrous moonshine as preserving specific finite sets of these lattices.

For this, it is crucial to determine all $M\frac{g}{h}$ fixed by $\Gamma_0(N)$.

\[
\alpha_{M\frac{g}{h}}.\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}.\alpha_{M\frac{g}{h}}^{-1} = \begin{bmatrix} 1 & M \\ 0 & 1 \end{bmatrix} \]

so we must have that $M$ is a natural number, or that $M\frac{g}{h}$ is a number-like lattice, in Conway-speak.

\[
\alpha_{M\frac{g}{h}}.\begin{bmatrix} 1 & 0 \\ N & 1 \end{bmatrix}.\alpha_{M\frac{g}{h}}^{-1} = \begin{bmatrix} 1 + \frac{Ng}{Mh} & – \frac{Ng^2}{Mh^2} \\ \frac{N}{M} & 1 – \frac{Ng}{Mh} \end{bmatrix} \]

so $M$ divides $N$, $Mh$ divides $Ng$ and $Mh^2$ divides $Ng^2$. As $g$ and $h$ are coprime it follows that $Mh^2$ must divide $N$.

Now, for an arbitrary element of $\Gamma_0(N)$ we have

\[
\alpha_{M\frac{g}{h}}.\begin{bmatrix} a & b \\ cN & d \end{bmatrix}.\alpha_{M\frac{g}{h}}^{-1} = \begin{bmatrix} a + c \frac{Ng}{Mh} & Mb – c \frac{Ng^2}{Mh^2} – (a-d) \frac{g}{h} \\ c \frac{N}{M} & d – c \frac{Ng}{Mh} \end{bmatrix} \]
and using our divisibility requirements it follows that this matrix belongs to $SL_2(\mathbb{Z})$ if $a-d$ is divisible by $h$, that is if $a \equiv d~(mod~h)$.

We know that $ad-Nbc=1$ and that $h$ divides $N$, so $a.d \equiv 1~(mod~h)$, which implies $a \equiv d~(mod~h)$ if $h$ satisfies the defining property of $24$, that is, if $h$ divides $24$.

Concluding, $\Gamma_0(N)$ preserves exactly those lattices $M\frac{g}{h}$ for which
\[
1~|~M~|~\frac{N}{h^2}~\quad~\text{and}~\quad~h~|~24 \]

A first step towards figuring out the Moonshine Picture.

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