Tag: Conway

the monstrous moonshine picture – 2

Published January 19, 2018 by lievenlb

Time to wrap up my calculations on the moonshine picture, which is the subgraph of Conway’s Big Picture needed to describe all 171 moonshine groups.

No doubt I’ve made mistakes. All corrections are welcome. The starting point is the list of 171 moonshine groups which are in the original Monstrous Moonshine paper.

The backbone is given by the $97$ number lattices, which are closed under taking divisors and were found by looking at all divisors of the numbers $N=n \times h$ for the 171 moonshine groups of the form $N+e,f,\dots$ or $(n|h)+e,f,\dots$.

The Hasse-diagram of this poset (under division) is here (click on the image to get a larger version)

There are seven types of coloured numbers, each corresponding to number-lattices which have the same local structure in the moonshine picture, as in the previous post.

The white numbered lattices have no further edges in the picture.

The yellow number lattices (2,10,14,18,22,26,32,34,40,68,80,88,90,112,126,144,180,208 = 2M) have local structure

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

The green number lattices (3,15,21,39,57,93,96,120 = 3M) have local structure

\[
\xymatrix{M \frac{1}{3} \ar@[red]@{-}[r] & \color{green}{3M} \ar@[red]@{-}[r] & M \frac{2}{3}} \]

The blue number lattices (4,16,20,28,36,44,52,56,72,104 = 4M) have as local structure

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

where the leftmost part is redundant as they are already included in the yellow-bit.

The purple number lattices (6,30,42,48,60 = 6M) have local structure

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

where again the lefmost part is redundant, and I forgot to add the central part in the previous post… (updated now).

The unique brown number lattice 8 has local structure

\[
\xymatrix{& & 1 \frac{1}{4} \ar@{-}[d] & & 1 \frac{1}{8} \ar@{-}[d] & \\
& 1 \frac{1}{2} \ar@{-}[d] & 2 \frac{1}{2} \ar@{-}[r] \ar@{-}[d] & 1 \frac{3}{4} & 2 \frac{1}{4} \ar@{-}[r] & 1 \frac{5}{8} \\
1 \ar@{-}[r] & 2 \ar@{-}[r] & 4 \ar@{-}[r] & \color{brown}{8} \ar@{-}[r] & 4 \frac{1}{2} \ar@{-}[d] \ar@{-}[u] & \\
& & & 1 \frac{7}{8} \ar@{-}[r] & 2 \frac{3}{4} \ar@{-}[r] & 1 \frac{3}{8}} \]

The local structure in the two central red number lattices (not surprisingly 12 and 24) looks like the image in the previous post, but I have to add some ‘forgotten’ lattices.

That’ll have to wait…

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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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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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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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Aaron Siegel on transfinite number hacking

Published May 16, 2012 by lievenlb

One of the coolest (pure math) facts in Conway’s book ONAG is the explicit construction of the algebraic closure $\overline{\mathbb{F}_2}$ of the field with two elements as the set of all ordinal numbers smaller than $(\omega^{\omega})^{\omega}$ equipped with nimber addition and multiplication.

Some time ago we did run a couple of posts on this. In transfinite number hacking we recalled Cantor’s ordinal arithmetic and in Conway’s nim arithmetics we showed that Conway’s simplicity rules for addition and multiplication turns the set of all ordinal numbers into a field of characteristic zero : $\mathbb{On}_2$ (pronounced ‘Onto’).

In the post extending Lenstra’s list we gave Hendrik Lenstra’s effective construction of the mystery elements $\alpha_p$ (for prime numbers $p$) needed to do actual calculations in $\mathbb{On}_2$. We used SAGE to check the values for $p \leq 41$ and solved the conjecture left in Lenstra’s paper Nim multiplication that $(\omega^{\omega^{13}})^{43} = \omega^{\omega^7} + 1$ and determined $\alpha_p$ for $p \leq 67$.

Aaron Siegel has now dramatically extended this and calculated the $\alpha_p$ for all primes $p \leq 181$. He mails :

“thinking about the problem I figured it shouldn’t be too hard to write a dedicated program for it. So I threw together some Java code and… pushed the table up to p = 181! You can see the results below. Q(f(p)), excess, and alpha_p are all as defined by Lenstra. The “t(sec)” column is the number of seconds the calculation took, on my 3.4GHz iMac. The most difficult case, by far, was p = 167, which took about five days.

I’m including results for all p < 300, except for p = 191, 229, 263, and 283. p = 263 and 283 are omitted because they involve computations in truly enormous finite fields (exponent 102180 for p = 263, and 237820 for p = 283). I'm confident that if I let my computer grind away at them for long enough, we'd get an answer... but it would take several months of CPU time at least. p = 191 and 229 are more troubling cases. Consider p = 191: it's the first prime p such that p-1 has a factor with excess > 1. (190 = 2 x 5 x 19, and alpha_19 has excess 4.) This seems to have a significant effect on the excess of alpha_191. I’ve tried it for every excess up to m = 274, and for all powers of 2 up to m = 2^32. No luck.”

Aaron is writing a book on combinatorial game theory (to be published in the AMS GSM series, hopefully later this year) and will include details of these computations. For the impatient, here’s his list

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n-dimensional and transfinite Nimbers

Published January 8, 2011 by lievenlb

Today, we will expand the game of Nimbers to higher dimensions and do some transfinite Nimber hacking.

In our identification between $\mathbb{F}_{16}^* $ and 15-th roots of unity, the number 8 corresponds to $\mu^6 $, whence $\sqrt{8}=\mu^3=14 $. So, if we add a stone at the diagonal position (14,14) to the Nimbers-position of last time

we get a position of Nim-value 0, that is, winnable for the second player. In fact, this is a universal Nimbers-truth :

Either the 2nd player wins a Nimbers-position, or one can add one stone to the diagonal such that it becomes a 2nd player win.

The proof is elementary : choose a Fermat 2-power such that all stones have coordinates smaller than $2^{2^n} $. If the Nim-value of the position isn’t zero, it corresponds to a unit $\alpha \in \mathbb{F}_{2^{2^n}}^* $. Now,the Frobenius map $x \mapsto x^2 $ is an automorphism of any finite field of characteristic two, so the square root $\sqrt{\alpha} $ also belongs to $\mathbb{F}_{2^{2^n}} $, done!

3-dimensional Nimbers is played in the first octant of the integral lattice $\mathbb{Z}^3 $ by placing a finite number of balls at places $~(a,b,c) \in \mathbb{N}^3 $ with $abc \geq 1 $.

Moves are defined by replacing the rectangular-rule of the two-dimensional version by the cuboid-rule : take a cuboid with faces parallel to the coordinate planes whose corner of maximal distance from the origin is one of the balls in the position. Remove that ball and add new balls to the unoccupied corners and remove balls at occupied corners.

Here, we allow the corner-points to have zero as some of its coordinates, but these balls are considered dead in the game. As in the two-dimensional game, this cuboid-rule encompasses several legal moves depending on the number of corners in the cuboid having zero-coordinates.

Again, it follows by induction that the Nim-value of a ball placed at position $~(a,b,c) $ is equal to the Nim-multiplication $a \otimes b \otimes c $ and we can calculate the Nim-value of a 3-dimensional Nimbers-position by computing in a suitable field $\mathbb{F}_{2^{2^n}} $. (The extension to higher dimensions is now obvious)

Does 3-dimensional Nimbers satisfy the ‘universal truth’, that is, can one make any position a 2nd player win by adding at most one stone to the body-diagonal?

The previous argument fails. As $\mathbb{F}_4^* $ is the cyclic group of order three, the 3rd roots of unity in $\overline{\mathbb{F}_2} $ correspond to the numbers 1,2 and 3, so the map $x \mapsto x^3 $ cannot be a bijection on any of the finite fields $\mathbb{F}_{2^{2^n}} $.

But then, perhaps, a third root is added by going to a larger such field $\mathbb{F}_{2^{2^N}} $? Well, not quite. Take for example 2, then $\sqrt[3]{2} \notin \mathbb{F}_{2^{2^N}} $. (2 has order 3 in $\mathbb{F}_4 $ and so its 3rd root must have order 9, but 9 does not divide any number of the form $2^{2^N}-1 $ as the Fermat-powers mod 9 can only be 4 or 7).

In fact one can show that this also holds for any number not in the image of the cubing-map in some $\mathbb{F}_{2^{2^n}} $ as

$\mathbb{N}={ 0,1,2,\ldots } = \cup_N \mathbb{F}_{2^{2^N}} $

with Nim-addition and multiplication is the quadratic closure of $\mathbb{F}_2 $ (see for example ONAG).

The situation changes if we allow ourself to play transfinite Nimbers, with the same rules as before but now we allow the stone, balls etc. to be placed at points of which the coordinates are not restricted to $\mathbb{N}_+ = \{ 1,2,3,\ldots \} $ but may vary over $[\beta]_+ $ for some ordinal $\beta $ where $[\beta]_+=\{ 1,2,\ldots,\omega,\omega+1,\ldots \} $ is the set of all ordinals smaller than $\beta $.

In transfinite 3-dimensional Nimbers the ‘universal truth’ still holds, provided we play it on a cube of sizes $[\omega^{\omega}]_+ $. In particular we have that $\sqrt[3]{2}=\omega $ by the simplicity rule (see ONAG or the Conway’s nim-arithmetic post)

In general, n-dimensional transfinite Nimbers played on an n-gid of sizes $[\omega^{\omega^{\omega}}]_+ $ satisfies the universal truth : either a position is a 2nd player win or it becomes one by adding one n-ball to a diagonal position! (this follows immediately because $[\omega^{\omega^{\omega}}] $ with Nim-addition and multiplication is isomorphic to the algebraic closure of $\mathbb{F}_2 $).

2-dimensional transfinite Nimbers is still pretty playable. Below a position on a $\omega.2 $-board with stones as positions $~(2,2),(4,\omega),(\omega+2,\omega+3) $ and $~(\omega+4,\omega+1) $

Give a winning move for the first player!

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