# Archangel Gabriel will make you a topos

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)

# Rotations of Klein’s quartic

The usual argument to show that the group of all orientation-preserving symmetries of the Klein quartic is the simple group $L_2(7)$ of order $168$ goes like this:

There are two families of $7$ truncated cubes on the Klein quartic. The triangles of one of the seven truncated cubes in the first family have as center the dots, all having the same colour. The triangles of one of the truncated cubes in the second family correspond to the squares all having the same colour.

If you compare the two colour schemes, you’ll see that every truncated cube in the first family is disjoint from precisely $3$ truncated cubes in the second family.

That is, we can identify the truncated cubes of the first family with the points in the Fano plane $\mathbb{P}^2(\mathbb{F}_2)$, and those of the second family with the lines in that plane.

The Klein quartic consists of $24$ regular heptagons, so its rotation symmetry group consists of $24 \times 7 = 168$ rotations,each preserving the two families of truncated cubes. This is exactly the same number as there are isomorphisms of the Fano plane, $PGL_3(\mathbb{F}_2) = L_2(7)$. Done!

For more details, check John Baez’ excellent page on the Klein quartic, or the Buckyball curve post.

Here’s another ‘look-and-see’ proof, starting from Klein’s own description of his quartic.

Look at the rotation $g$, counter-clockwise with angle $2\pi / 7$ fixing the center of the central blue heptagon, and a similar rotation $h$ fixing the center of one of the neighbouring red heptagons.

The two vertices of the edge shared by the blue and red heptagon are fixed by $g.h$ and $h.g$, respectively, so these rotations must have order three (there are $3$ heptagons meeting in the vertex).

That is, the rotation symmetry group $G$ of the Klein quartic has order $168$, and contains two elements $g$ and $h$ of order $7$, such that the subgroup generated by them contains elements of order $3$.

This is enough to prove that the $G$ must be simple and therefore isomorphic to $L_2(7)$!

The following elegant proof is often attributed to Igor Dolgachev.

If $G$ isn’t simple there is a maximal normal subgroup $N$ with $G/N$ simple .

The only non-cyclic simple group having less elements that $168$ is $A_5$ but this cannot be $G/N$ as $60$ does not divide $168$.

So, $G/N$ must be cyclic of order $2,3$ or $7$ (the only prime divisors of $168=2^3.3.7$).

Order $2$ is not possible as any group $N$ of order $84=2^2.3.7$ can just have one Sylow $7$-subgroup. Remember that the number of $7$-Sylows of $N$ must divide $2^2.3=12$ and must be equal to $1$ modulo $7$. And $G$ (and therefore $N$) has at least two different cyclic subgroups of order $7$.

Order $3$ is impossible as this would imply that the normal subgroup $N$ of order $2^3.7=56$ must contain all $7$-Sylows of $G$, and thus also an element of order $3$. But, $3$ does not divide $56$.

Order $7$ is a bit more difficult to exclude. This would mean that there is a normal subgroup $N$ of order $2^3.3=24$.

$N$ (being normal) must contain all Sylow $2$-subgroups of $G$ of which there are either $1$ or $3$ (the order of $N$ is $2^3.3=24$).

If there is just one $S$ it should be a normal subgroup with $G/S$ (of order $21$) having a (normal) unique Sylow $7$-subgroup, but then $G$ would have a normal subgroup of index $3$, which we have excluded.

The three $2$-Sylows are conjugated and so the conjugation morphism
$G \rightarrow S_3$
is non-trivial and must have image strictly larger than $C_3$ (otherwise, $G$ would have a normal subgroup of index $3$), so must be surjective.

But, $S_3$ has a normal subgroup of index $2$ and pulling this back, $G$ must also have a normal subgroup of index two, which we have excluded. Done!

# Rarer books: Singmaster’s notes

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.

These magazines were reproduced on this page.

# Grothendieck, at the theatre

A few days ago, the theatre production “Rêves et Motifs” (Dreams and Motives) was put on stage in Argenteuil by la Compagnie Les Rémouleurs.

The stage director Anne Bitran only discovered Grothendieck’s life by reading the front pages of French newspapers, the day after Grothendieck passed away, in November 2014.

« Rêves et Motifs » is a piece inspired by Récoltes et Semailles.

Anne Bitran: ” In Récoltes et semailles we meet a scientist who has his feet on the ground and shares our curiosity about the world around us, with a strong political engagement. This is what I wanted to share with this piece.”

Some of Grothendieck’s dessins d’enfant make their appearance. Is that one Monsieur Mathieu in the center? And part of the Hexenkuche top left? (no, see Vimeo below)

And, does this looks like the sculpture ‘Grothendieck as Shepherd’ by Nina Douglas?

More information about the production can be found at the Les Remouleurs website (in French).

In case you are interested, make sure to be in Lunéville, November 29th or 30th.

# Heyting Smullyanesque problems

Raymond Smullyan brought Knights and Knaves puzzles to a high art in his books. Here’s the setting:

On Smullyan’s islands there are Knights, who always tell true statements, Knaves, who always lie, and sometimes also Normals, who sometimes tell the truth and sometimes lie.

(image credit MikeKlein)

Problems of this sort can be solved by classical propositional logic, replacing every sentence $P$ told by inhabitant $A$ by the formula $k_A \leftrightarrow P$ where $k_A$ is the sentence “A is a Knight”.

Some time ago I asked for Smullyanesque problems in a non-classical logic, where we replace the usual truth-values $T$ (true) and $F$ (false) of propositional logic by elements of an Heyting algebra.

Jason Rosenhouse wrote a paper Knights, Knaves, Normals, and Neutrals for The College Mathematics Journal, containing more information than in his blog post, as well as some nice challenging puzzles. Here’s Rosenhouse’s setting:

Apart from Knights, Knaves and Normals there’s also the tribe of Neutrals (which he describes as a sort of trans-Knights or trans-Knaves) who can only tell sentences with truth value $N$ in the Heyting algebra $\{ T,N,F \}$ with logical connectives

A typical sentence of truth value $N$ is “A is a Knight” or “A is a Knave” whereas, in reality, $A$ is Neutral. Here are the truth values of sentences about a person (the column headings) with respect to the actual situation (the row headings)

A good way into such problems is to focus on sentence like “A is a Neutral” as they have only classical truth values $T$ or $F$ and to remember that Neutrals can never say a sentence with classical truth value. Here’s an example (only Knights,Knaves or Neutrals present):

Problem 1: Suppose you meet three people, named Dave, Evan and Ford. They make the following statements:

Dave: Evan is a knight.
Evan: Ford is a knave.
Ford: Dave is a neutral.

Can you determine the types of all three people?

Solution: Ford’s sentence has value $T$ or $F$, so Ford cannot be Neutral so must be a Knight or Knave. But then Evan’s sentence also has classical truth value, so he can’t be Neutral either, and by the same reasoning also Dave cannot be neutral. So, we know that all three people must be either Knights or Knaves and then it follows at once that Ford is a Knave (because he lied) and that Evan and Dave must be knights.

These puzzles become more interesting once we use logical connectives.

Problem 2: What can you conclude from this dialog?

Mimi: Olaf is a Knight and Olaf is not a Knight.
Nate: Olaf is a Knave or Olaf is not a Knave.
Olaf: Mimi is a Knight or Nate is a Knave or I am Neutral.

Solution: Mimi’s sentence can only have truth value $F$ or $N$ so she can’t be a Knight. Nate’s sentence has value $T$ or $N$ so he cannot be a Knave.
The two first parts of Olaf’s line cannot be $T$ so must be either $N$ or $F$. So, Olav’s line has total value $T$ only if the last part ‘I am Neutral’ is $T$. So, Olaf can neither be a Knight (because then the last part is $F$) nor Neutral (because then the line would have value $T$ and Neutrals can only speak $N$-sentences). So, Olaf must be a knave. Nate must then be a Knight and Mimi a Knave (because the first part of her line is $F$ and so must be the whole sentence).

The situation becomes even more complicated when we allow Normals (who sometimes lie and sometimes tell the truth but never say sentences with value $N$) to be present. Here’s Rosenhouse’s ‘Grand Finale’:

Problem 3: One day a visitor encountered eight people, among them exactly two Knights, two Knaves, two Normals and two Neutrals. What can you conclude from this dialog:

Sara: Walt is a Neutral or Vera is a Neutral.
Todd: Xara is a Knave and Yoav is a Knave.
Ursa: If Sara is a Knight, then Todd and Yoav are Normals.
Vera: I am not a Neutral.
Walt: If Vera is not a Neutral, then neither is Todd.
Xara: Todd is a Knight if and only if Zack is a Neutral.
Zack: Xara is a Neutral if and only if Walt is not a Normal.

You will solve this problem much quicker than to read through the long explanation in Rosenhouse’s paper.

These puzzles beg to be generalised to more complicated Heyting algebras.

What about a book on “Knights, Knaves and Knols”? A Knol (or $X$-Knol) has knowledge limited to sentences with truth value $X$, an element in the Heyting algebra.