‘Gabriel’s topos’ (see here) is the conjectural, but still elusive topos from which the validity of the Riemann hypothesis would follow.

It is the latest attempt in Alain Connes’ 20 year long quest to tackle the RH (before, he tried the tools of noncommutative geometry and later those offered by the field with one element).

If you look at the points of these toposes you get horribly complicated ‘non-commutative’ spaces, such as the finite adele classes $\mathbb{Q}^*_+ \backslash \mathbb{A}^f_{\mathbb{Q}} / \widehat{\mathbb{Z}}^{\ast}$ (in case of the arithmetic site) and the full adele classes $\mathbb{Q}^*_+ \backslash \mathbb{A}_{\mathbb{Q}} / \widehat{\mathbb{Z}}^{\ast}$ (for the scaling site).

In Vienna, Connes gave a nice introduction to the arithmetic site in two lectures. The first part of the talk below also gives an historic overview of his work on the RH

However, not everyone is as optimistic about the topos-approach as he seems to be. Here’s an insightful answer on MathOverflow by Will Sawin to the question “What is precisely still missing in Connes’ approach to RH?”.

About a month ago, from May 10th till 14th Alain Connes gave a series of lectures at Ohio State University with title “The Riemann-Roch strategy, quantizing the Scaling Site”.

Especially interesting is section 2 “The geometry behind the zeros of $\zeta$” in which they explain how looking at the zeros locus inevitably leads to the space of adele classes and why one has to study this space with the tools from noncommutative geometry.

Perhaps further developments will be disclosed in a few weeks time when Connes is one of the speakers at Toposes in Como.

No kidding, this is the final sentence of Le spectre d’Atacama, the second novel by Alain Connes (written with Danye Chéreau (IRL Mrs. AC) and his former Ph.D. advisor Jacques Dixmier).

The book has a promising start. Armand Lafforet (IRL AC) is summoned by his friend Rodrigo to the Chilean observatory Alma in the Altacama desert. They have observed a mysterious spectrum, and need his advice.

Armand drops everything and on the flight he lectures the lady sitting next to him on proofs by induction (breaking up chocolate bars), and recalls a recent stay at the La Trappe Abbey, where he had an encounter with (the ghost of) Alexander Grothendieck, who urged him to ‘Follow the motif!’.

“Comment était-il arrivé là? Il possédait surement quelques clés. Pourquoi pas celles des songes?” (How did he get
there? Surely he owned some keys, why not those of our dreams?)

A few pages further there’s this on the notion of topos (my attempt to translate):

“The notion of space plays a central role in mathematics. Traditionally we represent it as a set of points, together with a notion of neighborhood that we call a ‘topology’. The universe of these new spaces, ‘toposes’, unveiled by Grothendieck, is marvellous, not only for the infinite wealth of examples (it contains, apart from the ordinary topological spaces, also numerous instances of a more combinatorial nature) but because of the totally original way to perceive space: instead of appearing on the main stage from the start, it hides backstage and manifests itself as a ‘deus ex machina’, introducing a variability in the theory of sets.”

So far, so good.

We have a mystery, tidbits of mathematics, and allusions left there to put a smile on any Grothendieck-aficionado’s face.

But then, upon arrival, the story drops dead.

Rodrigo has been taken to hospital, and will remain incommunicado until well in the final quarter of the book.

As the remaining astronomers show little interest in Alain’s (sorry, Armand’s) first lecture, he decides to skip the second, and departs on a hike to the ocean. There, he takes a genuine sailing ship in true Jules Verne style to the lighthouse at he end of the world.

All this drags on for at least half a year in time, and two thirds of the book’s length. We are left in complete suspense when it comes to the mysterious Atacama spectrum.

Perhaps the three authors deliberately want to break with existing conventions of story telling?

I had a similar feeling when reading their first novel Le Theatre Quantique. Here they spend some effort to flesh out their heroine, Charlotte, in the first part of the book. But then, all of a sudden, their main character is replaced by a detective, and next by a computer.

Anyway, when Armand finally reappears at the IHES the story picks up pace.

The trio (Armand, his would-be-lover Charlotte, and Ali Ravi, Cern’s computer guru) convince CERN to sell its main computer to an American billionaire with the (fake) promise of developing a quantum computer. Incidentally, they somehow manage to do this using Charlotte’s history with that computer (for this, you have to read ‘Le Theatre Quantique’).

By their quantum-computing power (Shor and quantum-encryption pass the revue) they are able to decipher the Atacame spectrum (something to do with primes and zeroes of the zeta function), send coded messages using quantum entanglement, end up in the Oval Office and convince the president to send a message to the ‘Riemann sphere’ (another fun pun), and so on, and on.

The book ends with a twist of the classic tale of the mathematician willing to sell his soul to the devil for a (dis)proof of the Riemann hypothesis:

After spending some time in purgatory, the mathematician gets a meeting with God and asks her the question “Is the Riemann hypothesis true?”.

“Of course”, God says.

“But how can you know that all non-trivial zeroes of the zeta function have real part 1/2?”, Armand asks.

And God replies:

“Simple enough, I can see them all at once. But then, don’t forget I’m God. I can see the disappointment in your face, yes I can read in your heart that you are frustrated, that you desire an explanation…

Well, we’re going to fix this. I will call archangel Gabriel, the angel of geometry, he will make you a topos!”

If you feel like running to the nearest Kindle store to buy “Le spectre d’Atacama”, make sure to opt for a package deal. It is impossible to make heads or tails of the story without reading “Le theatre quantique” first.

But then, there are worse ways to spend an idle week than by binge reading Connes…

Edit (February 28th). A short video of Alain Connes explaining ‘Le spectre d’Atacama’ (in French)

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 $n=2$ 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

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

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.

All lattices in the moonshine picture are number-like, that is of the form $M \frac{g}{h}$ with $M$ a positive integer and $0 \leq g < h$ with $(g,h)=1$.
To understand the action of the Bost-Connes algebra on the Big Picture it is sometimes better to view the lattice $M \frac{g}{h}$ as a primitive $h$-th root of unity, centered at $hM$.

The distance from $M$ to any of the lattices $M \frac{g}{h}$ is equal to $2 log(h)$, and the distances from $M$ and $M \frac{g}{h}$ to $hM$ are all equal to $log(h)$.

For a prime value $h$, these $h$ lattices are among the $h+1$ lattices branching off at $hM$ in the $h$-adic tree (the remaining one being $h^2M$).

For general $h$ the situation is more complex. Here’s the picture for $h=6$ with edges in the $2$-adic tree painted blue, those in the $3$-adic tree red.

\[
\xymatrix{& & M \frac{1}{2} \ar@[blue]@{-}[d] & \\
& M \ar@[blue]@{-}[r] \ar@[red]@{-}[d] & 2M \ar@[red]@{-}[d] & M \frac{1}{6} \ar@[red]@{-}[d] \\
M \frac{1}{3} \ar@[red]@{-}[r] & 3M \ar@[blue]@{-}[r] \ar@[red]@{-}[d] & \boxed{6 M} \ar@[blue]@{-}[r] & 3M \frac{1}{2} \ar@[red]@{-}[d] \\ & M \frac{2}{3} & & M \frac{5}{6}} \]

To describe the moonshine group $(n|h)+e,f,\dots$ (an example was worked out in the tetrahedral snake post), we need to study the action of base-change with the matrix
\[
x = \begin{bmatrix} 1 & \frac{1}{h} \\ 0 & 1 \end{bmatrix} \]
which sends a lattice of the form $M \frac{g}{h}$ with $0 \leq g < h$ to $M \frac{g+M}{h}$, so is a rotation over $\frac{2 \pi M}{h}$ around $h M$.
But, we also have to describe the base-change action with the matrix
\[
y = \begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix} \]
and for this we better use the second description of the lattice as $M \frac{g}{h}=(\frac{g'}{h},\frac{1}{h^2M})$ with $g'$ the multiplicative inverse of $g$ modulo $h$.
Under the action by $y$, the second factor $\frac{1}{h^2M}$ will be fixed, so this time we have to look at all lattices of the form $(\frac{g}{h},\frac{1}{h^2M})$ with $0 \leq g < h$, which again can be considered as another set of $h$-th roots of unity, centered at $hM$.
Here's this second interpretation for $h=6$:
\[
\xymatrix{M \frac{5}{6} \ar@[red]@{-}[d] & & 4M \frac{1}{3} \ar@[red]@{-}[d] & \\
3M \frac{1}{2} \ar@[blue]@{-}[r] \ar@[red]@{-}[d] & \boxed{6M} \ar@[blue]@{-}[r] \ar@[red]@{-}[d] & 12 M \ar@[red]@{-}[r] \ar@[red]@{-}[d] & 4 M \frac{2}{3} \\
M \frac{1}{6} & 18 M \ar@[blue]@{-}[r] \ar@[blue]@{-}[d] & 36 M & \\
& 9M \frac{1}{2} & & }
\]
Under $x$ the first set of $h$-th roots of unity centered at $hM$ is permuted, whereas $y$ permutes the second set of $h$-th roots of unity.
These interpretations can be used to spot errors in computing the finite groups $\Gamma_0(n|h)/\Gamma_0(n.h)$.

Here’s part of the calculation of the action of $y$ on the $(360|1)$-snake (which consists of $60$-lattices).

First I got a group of order roughly $600.000$. After correcting some erroneous cycles, the order went down to 6912.

Finally I spotted that I mis-numbered two lattices in the description of $x$ and $y$, and the order went down to $48$ as it should, because I knew it had to be equal to $C_2 \times C_2 \times A_4$.