# Tag: Conway

Dedekind’s Psi-function $\Psi(n)= n \prod_{p |n}(1 + \frac{1}{p})$ pops up in a number of topics:

• $\Psi(n)$ is the index of the congruence subgroup $\Gamma_0(n)$ in the modular group $\Gamma=PSL_2(\mathbb{Z})$,
• $\Psi(n)$ is the number of points in the projective line $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$,
• $\Psi(n)$ is the number of classes of $2$-dimensional lattices $L_{M \frac{g}{h}}$ at hyperdistance $n$ in Conway’s big picture from the standard lattice $L_1$,
• $\Psi(n)$ is the number of admissible maximal commuting sets of operators in the Pauli group of a single qudit.

The first and third interpretation have obvious connections with Monstrous Moonshine.

Conway’s big picture originated from the desire to better understand the Moonshine groups, and Ogg’s Jack Daniels problem
asks for a conceptual interpretation of the fact that the prime numbers such that $\Gamma_0(p)^+$ is a genus zero group are exactly the prime divisors of the order of the Monster simple group.

Here’s a nice talk by Ken Ono : Can’t you just feel the Moonshine?

For this reason it might be worthwhile to make the connection between these two concepts and the number of points of $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$ as explicit as possible.

Surely all of this is classical, but it is nicely summarised in the paper by Tatitscheff, He and McKay “Cusps, congruence groups and monstrous dessins”.

The ‘monstrous dessins’ from their title refers to the fact that the lattices $L_{M \frac{g}{h}}$ at hyperdistance $n$ from $L_1$ are permuted by the action of the modular groups and so determine a Grothendieck’s dessin d’enfant. In this paper they describe the dessins corresponding to the $15$ genus zero congruence subgroups $\Gamma_0(n)$, that is when $n=1,2,3,4,5,6,7,8,9,10,12,13,16,18$ or $25$.

Here’s the ‘monstrous dessin’ for $\Gamma_0(6)$

But, one can compute these dessins for arbitrary $n$, describing the ripples in Conway’s big picture, and try to figure out whether they are consistent with the Riemann hypothesis.

We will get there eventually, but let’s start at an easy pace and try to describe the points of the projective line $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$.

Over a field $k$ the points of $\mathbb{P}^1(k)$ correspond to the lines through the origin in the affine plane $\mathbb{A}^2(k)$ and they can represented by projective coordinates $[a:b]$ which are equivalence classes of couples $(a,b) \in k^2- \{ (0,0) \}$ under scalar multiplication with non-zero elements in $k$, so with points $[a:1]$ for all $a \in k$ together with the point at infinity $[1:0]$. When $n=p$ is a prime number we have $\# \mathbb{P}^1(\mathbb{Z}/p\mathbb{Z}) = p+1$. Here are the $8$ lines through the origin in $\mathbb{A}^2(\mathbb{Z}/7\mathbb{Z})$

Over an arbitrary (commutative) ring $R$ the points of $\mathbb{P}^1(R)$ again represent equivalence classes, this time of pairs
$(a,b) \in R^2~:~aR+bR=R$
with respect to scalar multiplication by units in $R$, that is
$(a,b) \sim (c,d)~\quad~\text{iff}~\qquad \exists \lambda \in R^*~:~a=\lambda c, b = \lambda d$
For $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$ we have to find all pairs of integers $(a,b) \in \mathbb{Z}^2$ with $0 \leq a,b < n$ with $gcd(a,b)=1$ and use Cremona’s trick to test for equivalence:
$(a,b) = (c,d) \in \mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})~\quad \text{iff}~\quad ad-bc \equiv 0~mod~n$
The problem is to find a canonical representative in each class in an efficient way because this is used a huge number of times in working with modular symbols.

Perhaps the best algorithm, for large $n$, is sketched in pages 145-146 of Bill Stein’s Modular forms: a computational approach.

For small $n$ the algorithm in $\S 1.3$ in the Tatitscheff, He and McKay paper suffices:

• Consider the action of $(\mathbb{Z}/n\mathbb{Z})^*$ on $\{ 0,1,…,n-1 \}=\mathbb{Z}/n\mathbb{Z}$ and let $D$ be the set of the smallest elements in each orbit,
• For each $d \in D$ compute the stabilizer subgroup $G_d$ for this action and let $C_d$ be the set of smallest elements in each $G_d$-orbit on the set of all elements in $\mathbb{Z}/n \mathbb{Z}$ coprime with $d$,
• Then $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})= \{ [c:d]~|~d \in D, c \in C_d \}$.

Let’s work this out for $n=12$ which will be our running example (the smallest non-squarefree non-primepower):

• $(\mathbb{Z}/12\mathbb{Z})^* = \{ 1,5,7,11 \} \simeq C_2 \times C_2$,
• The orbits on $\{ 0,1,…,11 \}$ are
$\{ 0 \}, \{ 1,5,7,11 \}, \{ 2,10 \}, \{ 3,9 \}, \{ 4,8 \}, \{ 6 \}$
and $D=\{ 0,1,2,3,4,6 \}$,
• $G_0 = C_2 \times C_2$, $G_1 = \{ 1 \}$, $G_2 = \{ 1,7 \}$, $G_3 = \{ 1,5 \}$, $G_4=\{ 1,7 \}$ and $G_6=C_2 \times C_2$,
• $1$ is the only number coprime with $0$, giving us $[1:0]$,
• $\{ 0,1,…,11 \}$ are all coprime with $1$, and we have trivial stabilizer, giving us the points $[0:1],[1:1],…,[11:1]$,
• $\{ 1,3,5,7,9,11 \}$ are coprime with $2$ and under the action of $\{ 1,7 \}$ they split into the orbits
$\{ 1,7 \},~\{ 3,9 \},~\{ 5,11 \}$
giving us the points $[1:2],[3:2]$ and $[5:2]$,
• $\{ 1,2,4,5,7,8,10,11 \}$ are coprime with $3$, the action of $\{ 1,5 \}$ gives us the orbits
$\{ 1,5 \},~\{ 2,10 \},~\{ 4,8 \},~\{ 7,11 \}$
and additional points $[1:3],[2:3],[4:3]$ and $[7:3]$,
• $\{ 1,3,5,7,9,11 \}$ are coprime with $4$ and under the action of $\{ 1,7 \}$ we get orbits
$\{ 1,7 \},~\{ 3,9 \},~\{ 5,11 \}$
and points $[1:4],[3:4]$ and $[5,4]$,
• Finally, $\{ 1,5,7,11 \}$ are the only coprimes with $6$ and they form a single orbit under $C_2 \times C_2$ giving us just one additional point $[1:6]$.

This gives us all $24= \Psi(12)$ points of $\mathbb{P}^1(\mathbb{Z}/12 \mathbb{Z})$ (strangely, op page 43 of the T-H-M paper they use different representants).

One way to see that $\# \mathbb{P}^1(\mathbb{Z}/n \mathbb{Z}) = \Psi(n)$ comes from a consequence of the Chinese Remainder Theorem that for the prime factorization $n = p_1^{e_1} … p_k^{e_k}$ we have
$\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z}) = \mathbb{P}^1(\mathbb{Z}/p_1^{e_1} \mathbb{Z}) \times … \times \mathbb{P}^1(\mathbb{Z}/p_k^{e_k} \mathbb{Z})$
and for a prime power $p^k$ we have canonical representants for $\mathbb{P}^1(\mathbb{Z}/p^k \mathbb{Z})$
$[a:1]~\text{for}~a=0,1,…,p^k-1~\quad \text{and} \quad [1:b]~\text{for}~b=0,p,2p,3p,…,p^k-p$
which shows that $\# \mathbb{P}^1(\mathbb{Z}/p^k \mathbb{Z}) = (p+1)p^{k-1}= \Psi(p^k)$.

Next time, we’ll connect $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$ to Conway’s big picture and the congruence subgroup $\Gamma_0(n)$.

David Singmaster‘s “Notes on Rubik’s magic cube” are a collectors item, but it is still possible to buy a copy. I own a fifth edition (august 1980).

These notes capture the Rubik craze of those years really well.

Here’s a Conway story, from Siobhan Roberts’ excellent biography Genius at Play.

The ICM in Helsinki in 1978 was Conway’s last shot to get the Fields medal, but this was the last thing on his mind. He just wanted a Rubik cube (then, iron-curtain times, only sold in Hungary), so he kept chasing Hungarians at the meeting, hoping to obtain one. Siobhan writes (p. 239):

“The Fields Medals went to Pierre Deligne, Charles Fefferman, Grigory Margulis, and Daniel Quillen. The Rubik’s cube went to Conway.”

After his Notes, David Singmaster produced a follow-up newsletter “The Cubic Circular”. Only 5 magazines were published, of which 3 were double issues, between the Autumn of 1981 and the summer of 1985.

The one thing harder than to start blogging after a long period of silence is to stop when you think you’re still in the flow.

(image credit Putnam Consulting)

The Januari 1st post a math(arty) 2018 was an accident. I only wanted to share this picture, of a garage-door with an uncommon definition of prime numbers, i saw the night before.

I had been working on a better understanding of Conway’s Big Picture so I had material for a few follow-up posts.

It was never my intention to start blogging on a daily basis.

I had other writing plans for 2018.

For years I’m trying to write a math-book for a larger audience, or at least to give it an honest try.

My pet peeve with such books is that most of them are either devoid of proper mathematical content, or focus too much on the personal lives of the mathematicians involved.

An inspiring counter-example is ‘Closing the gap’ by Vicky Neal.

From the excellent review by Colin Beveridge on the Aperiodical Blog:

“Here’s a clever way to structure a maths book (I have taken copious notes): follow the development of a difficult idea or discovery chronologically, but intersperse the action with background that puts the discovery in context. That’s not a new structure – but it’s tricky to pull off: you have to keep the difficult idea from getting too difficult, and keep the background at a level where an interested reader can follow along and either say “yes, that’s plausible” or better “wait, let me get a pen!”. This is where Closing The Gap excels.”

So it is possible to publish a math-book worth writing. Or at least, some people can pull it off.

Problem was I needed to kick myself into writing mode. Feeling forced to post something daily wouldn’t hurt.

Anyway, I was sure this would have to stop soon. I had plans to disappear for 10 days into the French mountains. Our place there suffers from frequent power- and cellphone-cuts, which can last for days.

Thank you Orange.fr for upgrading your network to the remotest of places. At times, it felt like I was working from home.

I kept on blogging.

Even now, there’s material lying around.

I’d love to understand the claim that non-commutative geometry may offer some help in explaining moonshine. There was an interesting question on an older post on nimber-arithmetic I feel I should be following up. I’ve given a couple of talks recently on $\mathbb{F}_1$-material, parts of which may be postable. And so on.

Problem is, I would stick to the same (rather dense) writing style.

Perhaps it would make more sense to aim for a weekly (or even monthly) post over at Medium.

Medium offers no MathJax support forcing me to write differently about maths, and for a broader potential audience.

I may continue to blog here (or not), stick to the current style (or try something differently). I have not the foggiest idea right now.

The monstrous moonshine picture is a sub-graph of Conway’s Big Picture on 218 vertices. These vertices are the classes of lattices needed in the construction of the 171 moonshine groups. That is, moonshine gives us the shape of the picture.

(image credit Friendly Monsters)

But we can ask to reverse this process. Is the shape of the picture dictated by group-theoretic properties of the monster?

That is, can we reconstruct the 218 lattices and their edges starting from say the conjugacy classes of the monster and some simple rules?

Look at the the power maps for the monster. That is, the operation on conjugacy classes sending the class of $g$ to that of $g^k$ for all divisors $k$ of the order of $g$. Or, if you prefer, the $\lambda$-ring structure on the representation ring.

Rejoice die-hard believers in $\mathbb{F}_1$-theory, rejoice!

Here’s the game to play.

Let $g$ be a monster element of order $n$ and take $d=gcd(n,24)$.

(1) : If $d=8$ and a power map of $g$ gives class $8C$ add $(n|4)$ to your list.

(2) : Otherwise, look at the smallest power of $g$ such that the class is one of $12J,8F,6F,4D, 3C,2B$ or $1A$ and add $(n|e)$ where $e$ is the order of that class, or, if $n > 24$ and $e$ is even add $(n | \frac{e}{2})$.

A few examples:

For class 20E, $d=4$ and the power maps give classes 4D and 2B, so we add $(20|2)$.

For class 32B, $d=8$ but the power map gives 8E so we resort to rule (2). Here the power maps give 8E, 4C and 2B. So, the best class is 4C but as $32 > 24$ we add $(32|2)$.

For class 93A, $d=3$ and the power map gives 3C and even though $93 > 24$ we add $(93|3)$.

This gives us a list of instances $(n|e)$ with $n$ the order of a monster element. For $N=n \times e$ look at all divisors $h$ of $24$ such that $h^2$ divides $N$ and add to your list of lattices those of the form $M \frac{g}{h}$ with $g$ strictly smaller than $h$ and $(g,h)=1$ and $M$ a divisor of $\frac{N}{h^2}$.

This gives us a list of lattices $M \frac{g}{h}$, which is an $h$-th root of unity centered as $L=M \times h$ (see this post). If we do this for all lattices in the list we can partition the $L$’s in families according to which roots of unity are centered at $L$.

This gives us the moonshine picture. (modulo mistakes I made)

The operations we have to do after we have our list of instances $(n|e)$ is pretty straightforward from the rules we used to determine the lattices needed to describe a moonshine group.

Perhaps the oddest part in the construction are the rules (1) and (2) and the prescribed conjugacy classes used in them.

One way to look at this is that the classes $8C$ and $12J$ (or $24J$) are special. The other classes are just the power-maps of $12J$.

Another ‘rationale’ behind these classes may come from the notion of harmonics (see the original Monstrous moonshine paper page 312) of the identity element and the two classes of involutions, 2A (the Fischer involutions) and 2B (the Conway involutions).

For 1A these are : 1A,3C

For 2A these are : 2A,4B,8C

For 2B these are : 2B,4D,6F,8F,12J,24J

These are exactly the classes that we used in (1) and (2), if we add the power-classes of 8C.

Perhaps I should take some time to write all this down more formally.

The monstrous moonshine picture is the finite piece of Conway’s Big Picture needed to understand the 171 moonshine groups associated to conjugacy classes of the monster.

Last time I claimed that there were exactly 7 types of local behaviour, but I missed one. The forgotten type is centered at the number lattice $84$.

Locally around it the moonshine picture looks like this
$\xymatrix{42 \ar@{-}[dr] & 28 \frac{1}{3} \ar@[red]@{-}[d] & 41 \frac{1}{2} \ar@{-}[ld] \\ 28 \ar@[red]@{-}[r] & \color{grey}{84} \ar@[red]@{-}[r] \ar@[red]@{-}[d] \ar@{-}[rd] & 28 \frac{2}{3} \\ & 252 & 168}$

and it involves all square roots of unity ($42$, $42 \frac{1}{2}$ and $168$) and $3$-rd roots of unity ($28$, $28 \frac{1}{3}$, $28 \frac{2}{3}$ and $252$) centered at $84$.

No, I’m not hallucinating, there are indeed $3$ square roots of unity and $4$ third roots of unity as they come in two families, depending on which of the two canonical forms to express a lattice is chosen.

In the ‘normal’ expression $M \frac{g}{h}$ the two square roots are $42$ and $42 \frac{1}{2}$ and the three third roots are $28, 28 \frac{1}{3}$ and $28 \frac{2}{3}$. But in the ‘other’ expression
$M \frac{g}{h} = (\frac{g’}{h},\frac{1}{h^2M})$
(with $g.g’ \equiv 1~mod~h$) the families of $2$-nd and $3$-rd roots of unity are
$\{ 42 \frac{1}{2} = (\frac{1}{2},\frac{1}{168}), 168 = (0,\frac{1}{168}) \}$
and
$\{ 28 \frac{1}{3} = (\frac{1}{3},\frac{1}{252}), 28 \frac{2}{3} = (\frac{2}{3},\frac{1}{252}), 252 = (0 , \frac{1}{252}) \}$
As in the tetrahedral snake post, it is best to view the four $3$-rd roots of unity centered at $84$ as the vertices of a tetrahedron with center of gravity at $84$. Power maps in the first family correspond to rotations along the axis through $252$ and power maps in the second family are rotations along the axis through $28$.

In the ‘normal’ expression of lattices there’s then a total of 8 different local types, but two of them consist of just one number lattice: in $8$ the local picture contains all square, $4$-th and $8$-th roots of unity centered at $8$, and in $84$ the square and $3$-rd roots.

Perhaps surprisingly, if we redo everything in the ‘other’ expression (and use the other families of roots of unity), then the moonshine picture has only 7 types of local behaviour. The forgotten type $84$ appears to split into two occurrences of other types (one with only square roots of unity, and one with only $3$-rd roots).

I wonder what all this has to do with the action of the Bost-Connes algebra on the big picture or with Plazas’ approach to moonshine via non-commutative geometry.