# Tag: Norton

SNORT, invented by Simon NORTon is a map-coloring game, similar to COL. Only, this time, neighbours may not be coloured differently.

SNORTgo, similar to COLgo, is SNORT played with go-stones on a go-board. That is, adjacent stones must have the same colour.

SNORT is a ‘hot’ game, meaning that each player is eager to move as most moves will improve your position. In COL players are reluctant to move, because a move limits your next moves.

For this reason, SNORT positions are much harder to evaluate, and one needs the full force of Conways’s ONAG.

Here’s a SNORTgo endgame. Who has a winning strategy?, and what is the first move in that strategy?

The method to approach such an endgame is similar to that in COLgo. First we remove all dead spots from the board.

What remains, are a 4 spots available only to Right (white) and 5 spots available only to Left (bLack). Further, there a 3 ‘live’ regions: the upper righthand corner and the two lower corners.

The value of these corners must be computed inductively.

For example, Right’s best option in the left-most game (corresponding to the upper righthand corner of the endgame) is to put his stone on N12, resulting in a game in which neither player can move (the zero game).

On the other hand, Left can put a stone at either N11, N12 or N13 leaving a game in which she has two more moves, whereas Right han none (the $2$ game).

The other positions are computed similarly.

To get the value of the endgame we have to sum up all these values.

This can either be done using the addition rule given in ONAG, or by using programs in combinatorial game theory.

There’s Combinatorial Game Suite, developed by Aaron Siegel. But, for some reason I can no longer use it on macOS High Sierra.

Fortunately, the older program David Wolfe’s toolkit is still available, and runs on my MacBook.

The sum game evaluates to $\{ \{3|2 \}|-1 \}$, which is a ‘fuzzy’ game, that is, its value is confused with $0$.

This means that the first player to move has a winning strategy in the endgame.

Can you spot the (unique) winning move for Right (white) and one (of two) winning move for Left (bLack)?

At a seminar at the College de France in 1975, Tits wrote down the order of the monster group

$\# \mathbb{M} = 2^{46}.3^{20}.5^9.7^6.11^2.13^3.17·19·23·29·31·41·47·59·71$

Andrew Ogg, who attended the talk, noticed that the prime divisors are precisely the primes $p$ for which the characteristic $p$ super-singular $j$-invariants are all defined over $\mathbb{F}_p$.

Here’s Ogg’s paper on this: Automorphismes de courbes modulaires, Séminaire Delange-Pisot-Poitou. Théorie des nombres, tome 16, no 1 (1974-1975).

Ogg offered a bottle of Jack Daniels for an explanation of this coincidence.

Even Richard Borcherds didn’t claim the bottle of Jack Daniels, though his proof of the monstrous moonshine conjecture is believed to be the best explanation, at present.

A few years ago, John Duncan and Ken Ono posted a paper “The Jack Daniels Problem”, in which they prove that monstrous moonshine implies that if $p$ is not one of Ogg’s primes it cannot be a divisor of $\# \mathbb{M}$. However, the other implication remains mysterious.

Duncan and Ono say:

“This discussion does not prove that every $p ∈ \text{Ogg}$ divides $\# \mathbb{M}$. It merely explains how the first principles of moonshine suggest this implication. Monstrous moonshine is the proof. Does this then provide a completely satisfactory solution to Ogg’s problem? Maybe or maybe not. Perhaps someone will one day furnish a map from the characteristic $p$ supersingular $j$-invariants to elements of order $p$ where the group structure of $\mathbb{M}$ is apparent.”

I don’t know whether they claimed the bottle, anyway.

But then, what is the non-commutative Jack Daniels Problem?

A footnote on the first page of Conway and Norton’s ‘Monstrous Moonshine’ paper says:

“Very recently, A. Pizer has shown these primes are the only ones that satisfy a certain conjecture of Hecke from 1936 relating modular forms of weight $2$ to quaternion algebra theta-series.”

Pizer’s paper is “A note on a conjecture of Hecke”.

Maybe there’s a connection between monstrous moonshine and the arithmetic of integral quaternion algebras. Some hints:

The commutation relations in the Big Picture are reminiscent of the meta-commutation relations for Hurwitz quaternions, originally due to Conway in his booklet on Quaternions and Octonions.

The fact that the $p$-tree in the Big Picture has valency $p+1$ comes from the fact that the Brauer-Severi of $M_2(\mathbb{F}_p)$ is $\mathbb{P}^1_{\mathbb{F}_p}$. In fact, the Big Picture should be related to the Brauer-Severi scheme of $M_2(\mathbb{Z})$.

Then, there’s Jorge Plazas claiming that Connes-Marcolli’s $GL_2$-system might be related to moonshine.

One of the first things I’ll do when I return is to run to the library and get our copy of Shimura’s ‘Introduction to the arithmetic theory of automorphic functions’.

Btw. the bottle in the title image is not a Jack Daniels but the remains of a bottle of Ricard, because I’m still in the French mountains.

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.

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.