From the Da Vinci code to Habiro

The Fibonacci sequence reappears a bit later in Dan Brown’s book ‘The Da Vinci Code’ where it is used to login to the bank account of Jacques Sauniere at the fictitious Parisian branch of the Depository Bank of Zurich.

Last time we saw that the Hankel matrix of the Fibonacci series $F=(1,1,2,3,5,\dots)$ is invertible over $\mathbb{Z}$
H(F) = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} \in SL_2(\mathbb{Z}) \]
and we can use the rule for the co-multiplication $\Delta$ on $\Re(\mathbb{Q})$, the algebra of rational linear recursive sequences, to determine $\Delta(F)$.

For a general integral linear recursive sequence the corresponding Hankel matrix is invertible over $\mathbb{Q}$, but rarely over $\mathbb{Z}$. So we need another approach to compute the co-multiplication on $\Re(\mathbb{Z})$.

Any integral sequence $a = (a_0,a_1,a_2,\dots)$ can be seen as defining a $\mathbb{Z}$-linear map $\lambda_a$ from the integral polynomial ring $\mathbb{Z}[x]$ to $\mathbb{Z}$ itself via the rule $\lambda_a(x^n) = a_n$.

If $a \in \Re(\mathbb{Z})$, then there is a monic polynomial with integral coefficients of a certain degree $n$

f(x) = x^n + b_1 x^{n-1} + b_2 x^{n-2} + \dots + b_{n-1} x + b_n \]

such that for every integer $m$ we have that

a_{m+n} + b_1 a_{m+n-1} + b_2 a_{m+n-2} + \dots + b_{n-1} a_{m+1} + a_m = 0 \]

Alternatively, we can look at $a$ as defining a $\mathbb{Z}$-linear map $\lambda_a$ from the quotient ring $\mathbb{Z}[x]/(f(x))$ to $\mathbb{Z}$.

The multiplicative structure on $\mathbb{Z}[x]/(f(x))$ dualizes to a co-multiplication $\Delta_f$ on the set of all such linear maps $(\mathbb{Z}[x]/(f(x)))^{\ast}$ and we can compute $\Delta_f(a)$.

We see that the set of all integral linear recursive sequences can be identified with the direct limit
\Re(\mathbb{Z}) = \underset{\underset{f|g}{\rightarrow}}{lim}~(\frac{\mathbb{Z}[x]}{(f(x))})^{\ast} \]
(where the directed system is ordered via division of monic integral polynomials) and so is equipped with a co-multiplication $\Delta = \underset{\rightarrow}{lim}~\Delta_f$.

Btw. the ring structure on $\Re(\mathbb{Z}) \subset (\mathbb{Z}[x])^{\ast}$ comes from restricting to $\Re(\mathbb{Z})$ the dual structures of the co-ring structure on $\mathbb{Z}[x]$ given by
\Delta(x) = x \otimes x \quad \text{and} \quad \epsilon(x) = 1 \]

From this description it is clear that you need to know a hell of a lot number theory to describe this co-multiplication explicitly.

As most of us prefer to work with rings rather than co-rings it is a good idea to begin to study this co-multiplication $\Delta$ by looking at the dual ring structure of
\Re(\mathbb{Z})^{\ast} = \underset{\underset{ f | g}{\leftarrow}}{lim}~\frac{\mathbb{Z}[x]}{(f(x))} \]
This is the completion of $\mathbb{Z}[x]$ at the multiplicative set of all monic integral polynomials.

This is a horrible ring and very little is known about it. Some general remarks were proved by Kazuo Habiro in his paper Cyclotomic completions of polynomial rings.

In fact, Habiro got interested is a certain subring of $\Re(\mathbb{Z})^{\ast}$ which we now know as the Habiro ring and which seems to be a red herring is all stuff about the field with one element, $\mathbb{F}_1$ (more on this another time). Habiro’s ring is

\widehat{\mathbb{Z}[q]} = \underset{\underset{n|m}{\leftarrow}}{lim}~\frac{\mathbb{Z}[q]}{(q^n-1)} \]

and its elements are all formal power series of the form
a_0 + a_1 (q-1) + a_2 (q^2-1)(q-1) + \dots + a_n (q^n-1)(q^{n-1}-1) \dots (q-1) + \dots \]
with all coefficients $a_n \in \mathbb{Z}$.

Here’s a funny property of such series. If you evaluate them at $q \in \mathbb{C}$ these series are likely to diverge almost everywhere, but they do converge in all roots of unity!

Some people say that these functions are ‘leaking out of the roots of unity’.

If the ring $\Re(\mathbb{Z})^{\ast}$ is controlled by the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, then Habiro’s ring is controlled by the abelianzation $Gal(\overline{\mathbb{Q}}/\mathbb{Q})^{ab} \simeq \hat{\mathbb{Z}}^{\ast}$.

The Langlands program and non-commutative geometry

The Bulletin of the AMS just made this paper by Julia Mueller available online: “On the genesis of Robert P. Langlands’ conjectures and his letter to Andre Weil” (hat tip +ChandanDalawat and +DavidRoberts on Google+).

It recounts the story of the early years of Langlands and the first years of his mathematical career (1960-1966)leading up to his letter to Andre Weil in which he outlines his conjectures, which would become known as the Langlands program.

Langlands letter to Weil is available from the IAS.

The Langlands program is a vast net of conjectures. For example, it conjectures that there is a correspondence between

– $n$-dimensional representations of the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, and

– specific data coming from an adelic quotient-space $GL_n(\mathbb{A}_{\mathbb{Q}})/GL_n(\mathbb{Q})$.

For $n=1$ this is essentially class field theory with the correspondence given by Artin’s reciprocity law.

Here we have on the one hand the characters of the abelianised absolute Galois group

Gal(\overline{\mathbb{Q}}/\mathbb{Q})^{ab} \simeq Gal(\mathbb{Q}(\pmb{\mu}_{\infty})/\mathbb{Q}) \simeq \widehat{\mathbb{Z}}^{\ast} \]

and on the other hand the connected components of the idele class space

GL_1(\mathbb{A}_{\mathbb{Q}})/GL_1(\mathbb{Q}) = \mathbb{A}_{\mathbb{Q}}^{\ast} / \mathbb{Q}^{\ast} = \mathbb{R}_+^{\ast} \times \widehat{\mathbb{Z}}^{\ast} \]

For $n=2$ it involves the study of Galois representations coming from elliptic curves. A gentle introduction to the general case is Mark Kisin’s paper What is … a Galois representation?.

One way to look at some of the quantum statistical systems studied via non-commutative geometry is that they try to understand the “bad” boundary of the Langlands space $GL_n(\mathbb{A}_{\mathbb{Q}})/GL_n(\mathbb{Q})$.

Here, the Bost-Connes system corresponds to the $n=1$ case, the Connes-Marcolli system to the $n=2$ case.

If $\mathbb{A}’_{\mathbb{Q}}$ is the subset of all adeles having almost all of its terms in $\widehat{\mathbb{Z}}_p^{\ast}$, then there is a well-defined map

\pi~:~\mathbb{A}’_{\mathbb{Q}}/\mathbb{Q}^{\ast} \rightarrow \mathbb{R}_+ \qquad (x_{\infty},x_2,x_2,\dots) \mapsto | x_{\infty} | \prod_p | x_p |_p \]

The inverse image of $\pi$ over $\mathbb{R}_+^{\ast}$ are exactly the idele classes $\mathbb{A}_{\mathbb{Q}}^{\ast}/\mathbb{Q}^{\ast}$, so we can view them as the nice locus of the horrible complicated quotient of adele-classes $\mathbb{A}_{\mathbb{Q}}/\mathbb{Q}^*$. And we can view the adele-classes as a ‘closure’ of the idele classes.

But, the fiber $\pi^{-1}(0)$ has horrible topological properties because $\mathbb{Q}^*$ acts ergodically on it due to the fact that $log(p)/log(q)$ is irrational for distinct primes $p$ and $q$.

This is why it is better to view the adele-classes not as an ordinary space (one with bad topological properties), but rather as a ‘non-commutative’ space because it is controlled by a non-commutative algebra, the Bost-Connes algebra.

For $n=2$ there’s a similar story with a ‘bad’ quotient $M_2(\mathbb{A}_{\mathbb{Q}})/GL_2(\mathbb{Q})$, being the closure of an ‘open’ nice piece which is the Langlands quotient space $GL_2(\mathbb{A}_{\mathbb{Q}})/GL_2(\mathbb{Q})$.

From the Da Vinci code to Galois

In The Da Vinci Code, Dan Brown feels he need to bring in a French cryptologist, Sophie Neveu, to explain the mystery behind this series of numbers:

13 – 3 – 2 – 21 – 1 – 1 – 8 – 5

The Fibonacci sequence, 1-1-2-3-5-8-13-21-34-55-89-144-… is such that any number in it is the sum of the two previous numbers.

It is the most famous of all integral linear recursive sequences, that is, a sequence of integers

a = (a_0,a_1,a_2,a_3,\dots) \]

such that there is a monic polynomial with integral coefficients of a certain degree $n$

f(x) = x^n + b_1 x^{n-1} + b_2 x^{n-2} + \dots + b_{n-1} x + b_n \]

such that for every integer $m$ we have that

a_{m+n} + b_1 a_{m+n-1} + b_2 a_{m+n-2} + \dots + b_{n-1} a_{m+1} + a_m = 0 \]

For the Fibonacci series $F=(F_0,F_1,F_2,\dots)$, this polynomial can be taken to be $x^2-x-1$ because
F_{m+2} = F_{m+1}+F_m \]

The set of all integral linear recursive sequences, let’s call it $\Re(\mathbb{Z})$, is a beautiful object of great complexity.

For starters, it is a ring. That is, we can add and multiply such sequences. If

a=(a_0,a_1,a_2,\dots),~\quad \text{and}~\quad a’=(a’_0,a’_1,a’_2,\dots)~\quad \in \Re(\mathbb{Z}) \]

then the sequences

a+a’ = (a_0+a’_0,a_1+a’_1,a_2+a’_2,\dots) \quad \text{and} \quad a \times a’ = (a_0.a’_0,a_1.a’_1,a_2.a’_2,\dots) \]

are again linear recursive. The zero and unit in this ring are the constant sequences $0=(0,0,\dots)$ and $1=(1,1,\dots)$.

So far, nothing terribly difficult or exciting.

It follows that $\Re(\mathbb{Z})$ has a co-unit, that is, a ring morphism

\epsilon~:~\Re(\mathbb{Z}) \rightarrow \mathbb{Z} \]

sending a sequence $a = (a_0,a_1,\dots)$ to its first entry $a_0$.

It’s a bit more difficult to see that $\Re(\mathbb{Z})$ also has a co-multiplication

\Delta~:~\Re(\mathbb{Z}) \rightarrow \Re(\mathbb{Z}) \otimes_{\mathbb{Z}} \Re(\mathbb{Z}) \]
with properties dual to those of usual multiplication.

To describe this co-multiplication in general will have to await another post. For now, we will describe it on the easier ring $\Re(\mathbb{Q})$ of all rational linear recursive sequences.

For such a sequence $q = (q_0,q_1,q_2,\dots) \in \Re(\mathbb{Q})$ we consider its Hankel matrix. From the sequence $q$ we can form symmetric $k \times k$ matrices such that the opposite $i+1$-th diagonal consists of entries all equal to $q_i$
H_k(q) = \begin{bmatrix} q_0 & q_1 & q_2 & \dots & q_{k-1} \\
q_1 & q_2 & & & q_k \\
q_2 & & & & q_{k+1} \\
\vdots & & & & \vdots \\
q_{k-1} & q_k & q_{k+1} & \dots & q_{2k-2} \end{bmatrix} \]
The Hankel matrix of $q$, $H(q)$ is $H_k(q)$ where $k$ is maximal such that $det~H_k(q) \not= 0$, that is, $H_k(q) \in GL_k(\mathbb{Q})$.

Let $S(q)=(s_{ij})$ be the inverse of $H(q)$, then the co-multiplication map
\Delta~:~\Re(\mathbb{Q}) \rightarrow \Re(\mathbb{Q}) \otimes \Re(\mathbb{Q}) \]
sends the sequence $q = (q_0,q_1,\dots)$ to
\Delta(q) = \sum_{i,j=0}^{k-1} s_{ij} (D^i q) \otimes (D^j q) \]
where $D$ is the shift operator on sequence
D(a_0,a_1,a_2,\dots) = (a_1,a_2,\dots) \]

If $a \in \Re(\mathbb{Z})$ is such that $H(a) \in GL_k(\mathbb{Z})$ then the same formula gives $\Delta(a)$ in $\Re(\mathbb{Z})$.

For the Fibonacci sequences $F$ the Hankel matrix is
H(F) = \begin{bmatrix} 1 & 1 \\ 1& 2 \end{bmatrix} \in GL_2(\mathbb{Z}) \quad \text{with inverse} \quad S(F) = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} \]
and therefore
\Delta(F) = 2 F \otimes ~F – DF \otimes F – F \otimes DF + DF \otimes DF \]
There’s a lot of number theoretic and Galois-information encoded into the co-multiplication on $\Re(\mathbb{Q})$.

To see this we will describe the co-multiplication on $\Re(\overline{\mathbb{Q}})$ where $\overline{\mathbb{Q}}$ is the field of all algebraic numbers. One can show that

\Re(\overline{\mathbb{Q}}) \simeq (\overline{\mathbb{Q}}[ \overline{\mathbb{Q}}_{\times}^{\ast}] \otimes \overline{\mathbb{Q}}[d]) \oplus \sum_{i=0}^{\infty} \overline{\mathbb{Q}} S_i \]

Here, $\overline{\mathbb{Q}}[ \overline{\mathbb{Q}}_{\times}^{\ast}]$ is the group-algebra of the multiplicative group of non-zero elements $x \in \overline{\mathbb{Q}}^{\ast}_{\times}$ and each $x$, which corresponds to the geometric sequence $x=(1,x,x^2,x^3,\dots)$, is a group-like element
\Delta(x) = x \otimes x \quad \text{and} \quad \epsilon(x) = 1 \]

$\overline{\mathbb{Q}}[d]$ is the universal Lie algebra of the $1$-dimensional Lie algebra on the primitive element $d = (0,1,2,3,\dots)$, that is
\Delta(d) = d \otimes 1 + 1 \otimes d \quad \text{and} \quad \epsilon(d) = 0 \]

Finally, the co-algebra maps on the elements $S_i$ are given by
\Delta(S_i) = \sum_{j=0}^i S_j \otimes S_{i-j} \quad \text{and} \quad \epsilon(S_i) = \delta_{0i} \]

That is, the co-multiplication on $\Re(\overline{\mathbb{Q}})$ is completely known. To deduce from it the co-multiplication on $\Re(\mathbb{Q})$ we have to consider the invariants under the action of the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ as
\Re(\overline{\mathbb{Q}})^{Gal(\overline{\mathbb{Q}}/\mathbb{Q})} \simeq \Re(\mathbb{Q}) \]

Unlike the Fibonacci sequence, not every integral linear recursive sequence has an Hankel matrix with determinant $\pm 1$, so to determine the co-multiplication on $\Re(\mathbb{Z})$ is even a lot harder, as we will see another time.

Reference: Richard G. Larson, Earl J. Taft, ‘The algebraic structure of linearly recursive sequences under Hadamard product’

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

& 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

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.