Snakes, spines, threads and all that

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

Brancusi’s advice : avoid vampires

My one and only resolution for 2018: ban vampires from my life!

Here’s the story.

In the 1920’s, Montparnasse was at the heart of the intellectual and artistic life in Paris because studios and cafés were inexpensive.

Artists including Picasso, Matisse, Zadkine, Modigliani, Dali, Chagall, Miro, and the Romanian sculptor Constantin Brancusi all lived there.

You’ll find many photographs of Picasso in the company of others (here center, with Modigliani and Salmon), but … not with Brancusi.

From A Life of Picasso: The Triumphant Years, 1917-1932 (Vol 3) by John Richardson:

“Brancusi disapproved of one of of Picasso’s fundamental characteristics—one that was all too familiar to the latter’s fellow artists and friends—his habit of making off not so much with their ideas as with their energy. “Picasso is a cannibal,” Brancusi said. He had a point. After a pleasurable day in Picasso’s company, those present were apt to end up suffering from collective nervous exhaustion. Picasso had made off with their energy and would go off to his studio and spend all night living off it. Brancusi hailed from vampire country and knew about such things, and he was not going to have his energy or the fruits of his energy appropriated by Picasso.”

I learned this story via Austin Kleon who made this video about it:


Show Your Work! Episode 1: Vampires from Austin Kleon on Vimeo.

The Big Picture is non-commutative

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

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

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

The defining property of 24

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.

A Math(Art)y 2018

Last night, on our way to the fireworks in Antwerp, we walked by this definition of prime numbers:

“The numbers, only divisible by $1$ and itself are: $2,3$ and every number before or after a multiple of $6$, without their squares or products.” (Peter Wynen)

True enough.

And a lot more user-friendly than: the generators of the multiplicative monoid of all natural numbers which are $\pm 1$ modulo $6$ are the prime numbers, except for $2$ and $3$.

I wish you a 2018 full of math (and artistic) pleasures.

Arithmetic topology in Quanta

Consider subscribing to the feed of the mathematics section of Quantamagazine.

The articles there are invariably of high quality and quite informative.

Their latest is Secret Link Uncovered Between Pure Math and Physics by Kevin Hartnett.

It features the work by number-theorist Minhyong Kim of Oxford University.



In it, Minhyong Kim comes out of the closet, revealing that many of his results on rational points of algebraic curves were inspired by analogies he sees between number theory and physics.

So far he refrained from mentioning this inspiration in papers because “Number theorists are a pretty tough-minded group of people,” he said.

Yesterday, Peter Woit had a post on this on his blog ‘Not Even Wrong’, stuffed with interesting links to recent talks and papers by Minhyong Kim.

Minhyong Kim’s ideas grew out the topic of arithmetic topology, that is, the analogy between number rings and $3$-dimensional compact manifolds and between their prime ideals and embedded knots.

In this analogy, which is based on the similarity between finite connected covers of manifolds on the one hand and connected etale extensions of rings on the other, the prime spectrum of $\mathbb{Z}$ should correspond (due to Minkowski’s result on discriminants and Perelman’s proof of the Poincare-conjecture) to the $3$-sphere $S^3$.

I’ve written more about this analogy here:

Mazur’s knotty dictionary.

What is the knot associated to a prime?

Who dreamed up the knots=primes analogy?

The birthday of the primes=knots analogy.

And probably I’ll mention it later this month when I give a couple of talks at the $\mathbb{F}_1$-seminar in Ghent.