Skip to content →

Category: geometry

Monstrous dessins 2

Let’s try to identify the $\Psi(n) = n \prod_{p|n}(1+\frac{1}{p})$ points of $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$ with the lattices $L_{M \frac{g}{h}}$ at hyperdistance $n$ from the standard lattice $L_1$ in Conway’s big picture.

Here are all $24=\Psi(12)$ lattices at hyperdistance $12$ from $L_1$ (the boundary lattices):

You can also see the $4 = \Psi(3)$ lattices at hyperdistance $3$ (those connected to $1$ with a red arrow) as well as the intermediate $12 = \Psi(6)$ lattices at hyperdistance $6$.

The vertices of Conway’s Big Picture are the projective classes of integral sublattices of the standard lattice $\mathbb{Z}^2=\mathbb{Z} e_1 \oplus \mathbb{Z} e_2$.

Let’s say our sublattice is generated by the integral vectors $v=(v_1,v_2)$ and $w=(w_1.w_2)$. How do we determine its class $L_{M,\frac{g}{h}}$ where $M \in \mathbb{Q}_+$ is a strictly positive rational number and $0 \leq \frac{g}{h} < 1$?

Here’s an example: the sublattice (the thick dots) is spanned by the vectors $v=(2,1)$ and $w=(1,4)$



Well, we try to find a basechange matrix in $SL_2(\mathbb{Z})$ such that the new 2nd base vector is of the form $(0,z)$. To do this take coprime $(c,d) \in \mathbb{Z}^2$ such that $cv_1+dw_1=0$ and complete with $(a,b)$ satisfying $ad-bc=1$ via Bezout to a matrix in $SL_2(\mathbb{Z})$ such that
\[
\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} v_1 & v_2 \\ w_1 & w_2 \end{bmatrix} = \begin{bmatrix} x & y \\ 0 & z \end{bmatrix} \]
then the sublattice is of class $L_{\frac{x}{z},\frac{y}{z}~mod~1}$.

In the example, we have
\[
\begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 0 & 7 \end{bmatrix} \]
so this sublattice is of class $L_{\frac{1}{7},\frac{4}{7}}$.

Starting from a class $L_{M,\frac{g}{h}}$ it is easy to work out its hyperdistance from $L_1$: let $d$ be the smallest natural number making the corresponding matrix integral
\[
d. \begin{bmatrix} M & \frac{g}{h} \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} u & v \\ 0 & w \end{bmatrix} \in M_2(\mathbb{Z}) \]
then $L_{M,\frac{g}{h}}$ is at hyperdistance $u . w$ from $L_1$.

Now that we know how to find the lattice class of any sublattice of $\mathbb{Z}^2$, let us assign a class to any point $[c:d]$ of $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$.

As $gcd(c,d)=1$, by Bezout we can find a integral matrix with determinant $1$
\[
S_{[c:d]} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \]
But then the matrix
\[
\begin{bmatrix} a.n & b.n \\ c & d \end{bmatrix} \]
has determinant $n$.

Working backwards we see that the class $L_{[c:d]}$ of the sublattice of $\mathbb{Z}^2$ spanned by the vectors $(a.n,b.n)$ and $(c,d)$ is of hyperdistance $n$ from $L_1$.

This is how the correspondence between points of $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$ and classes in Conway’s big picture at hyperdistance $n$ from $L_1$ works.

Let’s do an example. Take the point $[7:3] \in \mathbb{P}^1(\mathbb{Z}/12\mathbb{Z})$ (see last time), then
\[
\begin{bmatrix} -2 & -1 \\ 7 & 3 \end{bmatrix} \in SL_2(\mathbb{Z}) \]
so we have to determine the class of the sublattice spanned by $(-24,-12)$ and $(7,3)$. As before we have to compute
\[
\begin{bmatrix} -2 & -7 \\ 7 & 24 \end{bmatrix} \begin{bmatrix} -24 & -12 \\ 7 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 3 \\ 0 & -12 \end{bmatrix} \]
giving us that the class $L_{[7:3]} = L_{\frac{1}{12}\frac{3}{4}}$ (remember that the second term must be taken $mod~1$).

If you do this for all points in $\mathbb{P}^1(\mathbb{Z}/12\mathbb{Z})$ (and $\mathbb{P}^1(\mathbb{Z}/6\mathbb{Z})$ and $\mathbb{P}^1(\mathbb{Z}/3 \mathbb{Z})$) you get this version of the picture we started with



You’ll spot that the preimages of a canonical coordinate of $\mathbb{P}^1(\mathbb{Z}/m\mathbb{Z})$ for $m | n$ are the very same coordinate together with ‘new’ canonical coordinates in $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$.

To see that this correspondence is one-to-one and that the index of the congruence subgroup
\[
\Gamma_0(n) = \{ \begin{bmatrix} p & q \\ r & s \end{bmatrix}~|~n|r~\text{and}~ps-qr=1 \} \]
in the full modular group $\Gamma = PSL_2(\mathbb{Z})$ is equal to $\Psi(n)$ it is useful to consider the action of $PGL_2(\mathbb{Q})^+$ on the right on the classes of lattices.

The stabilizer of $L_1$ is the full modular group $\Gamma$ and the stabilizer of any class is a suitable conjugate of $\Gamma$. For example, for the class $L_n$ (that is, of the sublattice spanned by $(n,0)$ and $(0,1)$, which is of hyperdistance $n$ from $L_1$) this stabilizer is
\[
Stab(L_n) = \{ \begin{bmatrix} a & \frac{b}{n} \\ c.n & d \end{bmatrix}~|~ad-bc = 1 \} \]
and a very useful observation is that
\[
Stab(L_1) \cap Stab(L_n) = \Gamma_0(n) \]
This is the way Conway likes us to think about the congruence subgroup $\Gamma_0(n)$: it is the joint stabilizer of the classes $L_1$ and $L_n$ (as well as all classes in the ‘thread’ $L_m$ with $m | n$).

On the other hand, $\Gamma$ acts by rotations on the big picture: it only fixes $L_1$ and maps a class to another one of the same hyperdistance from $L_1$.The index of $\Gamma_0(n)$ in $\Gamma$ is then the number of classes at hyperdistance $n$.

To see that this number is $\Psi(n)$, first check that the classes at hyperdistance $p^k$ for $p$ a prime number and for all $k$ for the $p+1$ free valent tree with root $L_1$, so there are exactly $p^{k-1}(p+1)$ classes as hyperdistance $p^k$.

To get from this that the number of hyperdistance $n$ classes is indeed $\Psi(n) = \prod_{p|n}p^{v_p(n)-1}(p+1)$ we have to use the prime- factorisation of the hyperdistance (see this post).

The fundamental domain for the action of $\Gamma_0(12)$ by Moebius tranfos on the upper half plane must then consist of $48=2 \Psi(12)$ black or white hyperbolic triangles



Next time we’ll see how to deduce the ‘monstrous’ Grothendieck dessin d’enfant for $\Gamma_0(12)$ from it



Leave a Comment

Monstrous dessins 1

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 $\Gamma_0(n)$.

Leave a Comment

Grothendieck’s gribouillis (4)

Fortunately, there are a few certainties left in life:

In spring, you might expect the next instalment of Connes’ and Consani’s quest for Gabriel’s topos. Here’s the latest: $\overline{\mathbf{Spec}(\mathbb{Z})}$ and the Gromov norm.

Every half year or so, Mochizuki’s circle-of-friends tries to create some buzz announcing the next IUTeich-workshop. I’ll spare you the link, if you are still interested, follow math_jin or IUTT_bot_math_jin on Twitter.

And then, there’s the never-ending story of Grothendieck’s griboullis, kept alive by the French journalist and author Philippe Douroux.

Here are some recent links:

Alexandre Grothendieck : une mathématique en cathédrale gothique, an article (in French) by Philippe Douroux in Le Monde, May 6th (behind paywall).

L’histoire étonnante des archives du mathématicien Alexandre Grothendieck, an article (in French) on France Inter by Mathieu Vidar, based on info from Philippe Douroux.

Les archives mystérieuses de Alexandre Grothendieck, a podcast of a broadcast on France Inter on June 10th. Interesting interview (in French) with Philippe Douroux and the French mathematician Etienne Ghys (with a guest appearance by Luc Illusie).

El enigmático legado de un genio de las matemáticas, an article (in Spanish) in El Pais, May 13th, with 8 photos of some of the Gribouillis. The two pictures in this post are taken from this article.

So, what’s the latest on the 70.000+ pages left by Grothendieck?

As far as i know, the Mormoiron part of the gribouillis is still at the University of Montpellier, and has been made available online at the Grothendieck archives.

The Lasserre part of the gribouillis is still in a cellar in Paris’ Saint-Germain-des-Prés, belonging to Jean-Bernard Gillot. The French national library cannot take possession of the notes before a financial agreement is reached with Grothendieck’s children (French law does not allow children to be disinherited).

And there’s a dispute about the price to be paid. The notes were estimated at 45.000 Euros, but some prefer to believe that they may be worth several millions of dollars.

It all depends on their mathematical content.

Unfortunately, pictures claimed to be of the Lasserre notes (such as the one above) are in fact from the Mormoiron/Montpellier notes, which do indeed contain interesting mathematics.

But, it is very unlikely that the Lasserre notes contain (math) surprises. Probably, most of them look like this one

endless lists of people deported by the Nazis to extermination camps in WW2.

Or, as Philippe Douroux is quoted in the El Pais piece: “I think it’s a treasure, maybe not a mathematical one, but a human one. It’s a descent into the hell of one the best organised brains in the world.”



The film made by Catherine Aira and Yves Le Pestipon “Alexandre Grothendieck: On the Paths of a Genius” (on the quest for G’s last hideout in the French Pyrenees) can now be watched on YouTube (with English subtitles)

Leave a Comment

RH and the Ishango bone

“She simply walked into the pond in Kensington Gardens Sunday morning and drowned herself in three feet of water.”

This is the opening sentence of The Ishango Bone, a novel by Paul Hastings Wilson. It (re)tells the story of a young mathematician at Cambridge, Amiele, who (dis)proves the Riemann Hypothesis at the age of 26, is denied the Fields medal, and commits suicide.

In his review of the novel on MathFiction, Alex Kasman casts he story in the 1970ties, based on the admission of the first female students to Trinity.

More likely, the correct time frame is in the first decade of this century. On page 121 Amiele meets Alain Connes, said to be a “past winner of the Crafoord Prize”, which Alain obtained in 2001. In fact, noncommutative geometry and its interaction with quantum physics plays a crucial role in her ‘proof’.



The Ishango artefact only appears in the Coda to the book. There are a number of theories on the nature and grouping of the scorings on the bone. In one column some people recognise the numbers 11, 13, 17 and 19 (the primes between 10 and 20).

In the book, Amiele remarks that the total number of lines scored on the bone (168) “happened to be the exact total of all the primes between 1 and 1000” and “if she multiplied 60, the total number of lines in one side column, by 168, the grand total of lines, she’d get 10080,…,not such a far guess from 9592, the actual total of primes between 1 and 100000.” (page 139-140)

The bone is believed to be more than 20000 years old, prime numbers were probably not understood until about 500 BC…



More interesting than these speculations on the nature of the Ishango bone is the description of the tools Amiele thinks to need to tackle the Riemann Hypothesis:

“These included algebraic geometry (which combines commutative algebra with the language and problems of geometry); noncommutative geometry (concerned with the geometric approach to associative algebras, in which multiplication is not commutative, that is, for which $x$ times $y$ does not always equal $y$ times $x$); quantum field theory on noncommutative spacetime, and mathematical aspects of quantum models of consciousness, to name a few.” (page 115)

The breakthrough came two years later when Amiele was giving a lecture on Grothendieck’s dessins d’enfant.

“Dessin d’enfant, or ‘child’s drawing’, which Amiele had discovered in Grothendieck’s work, is a type of graph drawing that seemed technically simple, but had a very strong impression on her, partly due to the familiar nature of the objects considered. (…) Amiele found subtle arithmetic invariants associated with these dessins, which were completely transformed, again, as soon as another stroke was added.” (page 116)

Amiele’s ‘disproof’ of RH is outlined on pages 122-124 of “The Ishango Bone” and is a mixture of recognisable concepts and ill-defined terms.

“Her final result proved that Riemann’s Hypothesis was false, a zero must fall to the east of Riemann’s critical line whenever the zeta function of point $q$ with momentum $p$ approached the aelotropic state-vector (this is a simplification, of course).” (page 123)

More details are given in a footnote:

“(…) a zero must fall to the east of Riemann’s critical line whenever:

\[
\zeta(q_p) = \frac{( | \uparrow \rangle + \Psi) + \frac{1}{2}(1+cos(\Theta))\frac{\hbar}{\pi}}{\int(\Delta_p)} \]

(…) The intrepid are invited to try the equation for themselves.” (page 124)

Wilson’s “The Ishango Bone” was published in 2012. A fair number of topics covered (the Ishango bone, dessin d’enfant, Riemann hypothesis, quantum theory) also play a prominent role in the 2015 paper/story by Michel Planat “A moonshine dialogue in mathematical physics”, but this time with additional story-line: monstrous moonshine

Such a paper surely deserves a separate post.



Leave a Comment

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 $GL_2(\mathbb{A}_{\mathbb{Q}})/GL_2(\mathbb{Q})$.

Leave a Comment