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Tag: Conway

The defining property of 24

Published January 2, 2018 by lieven

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 lieven

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 lieven

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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