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Closure

Exactly 20 years ago I wrote my first blogpost, ‘a blogging 2004’. I wasn’t using WordPress yet (but something called pMachine), and this blog was not called ‘neverendingbooks’, but ‘matrix.ua.ac.be’ (the URL of the mac still running this blog).

At the time I wanted to find out whether blogging was something for me. “I’m just starting out. Give me a couple of weeks/months to develop my own style and topics and I’ll change the layout accordingly.”

Well, after 20 years I know what I can, and more important, what I cannot do within this framework. Time to move on.

There are other reasons why this might be the right time to pull the plug.

– I’m on retirement since October 1st and soon I’ll have to vacate my office, containing the webserver on which NeB runs.

– My days are filled with more activities now, and I don’t think you want to read here for example about my struggles with chestnut-farming.

– I like to explore other channels to talk about mathematics. This may happen on Mathstodon, MathOverflow or YouTube. Or it might be through teaching or writing a book, perhaps even a children’s book.

NeB will remain reachable until mid 2024. I’ll check out options to preserve its content after that (suggestions are welcome).

I wish you a better 2024.

WBM

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Grothendieck’s gribouillis (6)

After the death of Grothendieck in November 2014, about 30.000 pages of his writings were found in Lasserre.



Since then I’ve been trying to follow what happened to them:

So, what’s new?

Well, finally we have closure!

Last Friday, Grothendieck’s children donated the 30.000 Laserre pages to the Bibliotheque Nationale de France.

Via Des manuscrits inédits du génie des maths Grothendieck entrent à la BnF (and Google-translate):

“The singularity of these manuscripts is that they “cover many areas at the same time” to form “a whole, a + cathedral work +, with undeniable literary qualities”, analyzes Jocelyn Monchamp, curator in the manuscripts department of the BnF.

More than in “Récoltes et semailles”, very autobiographical, the author is “in a metaphysical retreat”, explains the curator, who has been going through the texts with passion for a month. A long-term task as the writing, in fountain pen, is dense and difficult to decipher. “I got used to it… And the advantage for us was that the author had methodically paginated and dated the texts.” One of the parts, entitled “Structures of the psyche”, a book of enigmatic diagrams translating psychology into algebraic language. In another, “The Problem of Evil”, he unfolds over 15,000 pages metaphysical meditations and thoughts on Satan. We sense a man “caught up by the ghosts of his past”, with an adolescence marked by the Shoah, underlines Johanna Grothendieck whose grandfather, a Russian Jew who fled Germany during the war, died at Auschwitz.

The deciphering work will take a long time to understand everything this genius wanted to say.

On Friday, the collection joined the manuscripts department of the Richelieu site, the historic cradle of the BnF, alongside the writings of Pierre and Marie Curie and Louis Pasteur. It will only be viewable by researchers.“This is a unique testimony in the history of science in the 20th century, of major importance for research,” believes Jocelyn Monchamp.

During the ceremony, one of the volumes was placed in a glass case next to a manuscript by the ancient Greek mathematician Euclid.”

Probably, the recent publication of Récoltes et Semailles clinched the deal.

Also, it is unclear at this moment whether the Istituto Grothendieck, which harbours The centre for Grothendieck studies coordinated by Mateo Carmona (see this post) played a role in the decision making, nor what role the Centre will play in the further studies of Grothendieck’s gribouillis.

For other coverage on this, see Hermit ‘scribblings’ of eccentric French math genius unveiled.

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A question of loyalty

On the island of two truths, statements are either false (truth-value $0$), Q-true (value $Q$) or K-true (value $K$).

The King and Queen of the island have an opinion on all statements which may differ from their actual truth-value. We say that the Queen believes a statement $p$ is she assigns value $Q$ to it, and that she knows $p$ is she believes $p$ and the actual truth-value of $p$ is indeed $Q$. Similarly for the King, replacing $Q$’s by $K$’s.

All other inhabitants of the island are loyal to the Queen, or to the King, or to both. This means that they agree with the Queen (or King, or both) on all statements they have an opinion on. Two inhabitants are said to be loyal to each other if they agree on all statements they both have an opinion of.

Last time we saw that Queen and King agree on all statements one of them believes to be false, as well as the negation of such statements. This raised the question:

Are the King and Queen loyal to each other? That is, do Queen and King agree on all statements?

We cannot resolve this issue without the information Oscar was able to extract from Pointex in Karin Cvetko-Vah‘s post Pointex:

“Oscar was determined to get some more information. “Could you at least tell me whether the queen and the king know that they’re loyal to themselves?” he asked.
“Well, of course they know that!” replied Pointex.
“You said that a proposition can be Q-TRUE, K-TRUE or FALSE,” Oscar said.
“Yes, of course. What else!” replied Pointex as he threw the cap high up.
“Well, you also said that each native was loyal either to the queen or to the king. I was just wondering … Assume that A is loyal to the queen. Then what is the truth value of the statement: A is loyal to the queen?”
“Q, of course,” answered Pointex as he threw the cap up again.
“And what if A is not loyal to the queen? What is then the truth value of the statement: A is loyal to the queen?”
He barely finished his question as something fell over his face and covered his eyes. It was the funny cap.
“Thanx,” said Pointex as Oscar handed him the cap. “The value is 0, of course.”
“Can the truth value of the statement: ‘A is loyal to the queen’ be K in any case?”
“Well, what do you think? Of course not! What a ridiculous thing to ask!” replied Pointex.”

Puzzle : Show that Queen and King are not loyal to each other, that is, there are statements on which they do not agree.



Solution : ‘The King is loyal to the Queen’ must have actual truth-value $0$ or $Q$, and the sentence ‘The Queen is loyal to the King’ must have actual truth-value $0$ or $K$. But both these sentences are the same as the sentence ‘The Queen and King are loyal to each other’ and as this sentence can have only one truth-value, it must have value $0$ so the statement is false.

Note that we didn’t produce a specific statement on which the Queen and King disagree. Can you find one?

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the strange island of two truths

Last time we had a brief encounter with the island of two truths, invented by Karin Cvetko-Vah. See her posts:

On this island, false statements have truth-value $0$ (as usual), but non-false statements are not necessarily true, but can be given either truth-value $Q$ (statements which the Queen on the island prefers) or $K$ (preferred by the King).

Think of the island as Trump’s paradise where nobody is ever able to say: “Look, alternative truths are not truths. They’re falsehoods.”



Even the presence of just one ‘alternative truth’ has dramatic consequences on the rationality of your reasoning. If we know the truth-values of specific sentences, we can determine the truth-value of more complex sentences in which we use logical connectives such as $\vee$ (or), $\wedge$ (and), $\neg$ (not), and $\implies$ (then) via these truth tables:

\[
\begin{array}{c|ccc}
\downarrow~\bf{\wedge}~\rightarrow & 0 & Q & K \\
\hline
0 & 0 & 0 & 0 \\
Q & 0 & Q & Q \\
K & 0 & K & K
\end{array} \quad
\begin{array}{c|ccc}
\downarrow~\vee~\rightarrow & 0 & Q & K \\
\hline
0 & 0 & Q & K \\
Q & Q & Q & K \\
K & K & Q & K
\end{array} \]
\[
\begin{array}{c|ccc}
\downarrow~\implies~\rightarrow & 0 & Q & K \\
\hline
0 & Q & Q & K \\
Q & 0 & Q & K \\
K & 0 & Q & K
\end{array} \quad
\begin{array}{c|c}
\downarrow & \neg~\downarrow \\
\hline
0 & Q \\
Q & 0 \\
K & 0
\end{array}
\]

Note that the truth-values $Q$ and $K$ are not completely on equal footing as we have to make a choice which one of them will stand for $\neg 0$.

Common tautologies are no longer valid on this island. The best we can have are $Q$-tautologies (giving value $Q$ whatever the values of the components) or $K$-tautologies.

Here’s one $Q$-tautology (check!) : $(\neg p) \vee (\neg \neg p)$. Verify that $p \vee (\neg p)$ is neither a $Q$- nor a $K$-tautology.

Can you find any $K$-tautology at all?

Already this makes it incredibly difficult to adapt Smullyan-like Knights and Knaves puzzles to this skewed island. Last time I gave one easy example.



Puzzle : On an island of two truths all inhabitants are either Knaves (saying only false statements), Q-Knights (saying only $Q$-valued statements) or K-Knights (who only say $K$-valued statements).

The King came across three inhabitants, whom we will call $A$, $B$ and $C$. He asked $A$: “Are you one of my Knights?” $A$ answered, but so indistinctly that the King could not understand what he said.

He then asked $B$: “What did he say?” $B$ replies: “He said that he is a Knave.” At this point, $C$ piped up and said: “That’s not true!”

Was $C$ a Knave, a Q-Knight or a K-Knight?

Solution : Q- and K-Knights can never claim to be a Knave. Neither can Knaves because they can only say false statements. So, no inhabitant on the island can ever claim to be a Knave. So, $B$ lies and is a Knave, so his stament has truth-value $0$. $C$ claims the negation of what $B$ says so the truth-value of his statement is $\neg 0 = Q$. $C$ must be a Q-Knight.

As if this were not difficult enough, Karin likes to complicate things by letting the Queen and King assign their own truth-values to all sentences, which may coincide with their actual truth-value or not.

Clearly, these two truth-assignments follow the logic of the island of two truths for composed sentences, and we impose one additional rule: if the Queen assigns value $0$ to a statement, then so does the King, and vice versa.

I guess she wanted to set the stage for variations to the island of two truths of epistemic modal logical puzzles as in Smullyan’s book Forever Undecided (for a quick summary, have a look at Smullyan’s paper Logicians who reason about themselves).

A possible interpretation of the Queen’s truth-assignment is that she assigns value $Q$ to all statements she believes to be true, value $0$ to all statements she believes to be false, and value $K$ to all statements she has no fixed opinion on (she neither believes them to be true nor false). The King assigns value $K$ to all statements he believes to be true, $0$ to those he believes to be false, and $Q$ to those he has no fixed opinion on.

For example, if the Queen has no fixed opinion on $p$ (so she assigns value $K$ to it), then the King can either believe $p$ (if he also assigns value $K$ to it) or can have no fixed opinion on $p$ (if he assigns value $Q$ to it), but he can never believe $p$ to be false.



Puzzle : We say that Queen and King ‘agree’ on a statement $p$ if they both assign the same value to it. So, they agree on all statements one of them (and hence both) believe to be false. But there’s more:

  • Show that Queen and King agree on the negation of all statements one of them believes to be false.
  • Show that the King never believes the negation of whatever statement.
  • Show that the Queen believes all negations of statements the King believes to be false.

Solution : If one of them believes $p$ to be false (s)he will assign value $0$ to $p$ (and so does the other), but then they both have to assign value $Q$ to $\neg p$, so they agree on this.

The value of $\neg p$ can never be $K$, so the King does not believe $\neg p$.

If the King believes $p$ to be false he assigns value $0$ to it, and so does the Queen, but then the value of $\neg p$ is $Q$ and so the Queen believes $\neg p$.

We see that the Queen and King agree on a lot of statements, they agree on all statements one of them believes to be false, and they agree on the negation of such statements!

Can you find any statement at all on which they do not agree?

Well, that may be a little bit premature. We didn’t say which sentences about the island are allowed, and what the connection (if any) is between the Queen and King’s value-assignments and the actual truth values.

For example, the Queen and King may agree on a classical ($0$ or $1$) truth-assignments to the atomic sentences for the island, and replace all $1$’s with $Q$. This will give a consistent assignment of truth-values, compatible with the island’s strange logic. (We cannot do the same trick replacing $1$’s by $K$ because $\neg 0 = Q$).

Clearly, such a system may have no relation at all with the intended meaning of these sentences on the island (the actual truth-values).

That’s why Karin Cvetko-Vah introduced the notions of ‘loyalty’ and ‘sanity’ for inhabitants of the island. That’s for next time, and perhaps then you’ll be able to answer the question whether Queen and King agree on all statements.

(all images in this post are from Smullyan’s book Alice in Puzzle-Land)

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some skew Smullyan stumpers

Raymond Smullyan‘s logic puzzles are legendary. Among his best known are his Knights (who always tell the truth) and Knaves (who always lie) puzzles. Here’s a classic example.

“On the day of his arrival, the anthropologist Edgar Abercrombie came across three inhabitants, whom we will call $A$, $B$ and $C$. He asked $A$: “Are you a Knight or a Knave?” $A$ answered, but so indistinctly that Abercrombie could not understand what he said.

He then asked $B$: “What did he say?” $B$ replies: “He said that he is a knave.” At this point, $C$ piped up and said: “Don’t believe that; it’s a lie!”

Was $C$ a Knight or a Knave?”

If you are stumped by this, try to figure out what kind of inhabitant can say “I am a Knave”.

Some years ago, my friend and co-author Karin Cvetko-Vah wrote about a much stranger island, the island of two truths.

“The island was ruled by a queen and a king. It is important to stress that the queen was neither inferior nor superior to the king. Rather than as a married couple one should think of the queen and the king as two parallel powers, somewhat like the Queen of the Night and the King Sarastro in Mozart’s famous opera The Magic Flute. The queen and the king had their own castle each, each of them had their own court, their own advisers and servants, and most importantly each of them even had their own truth value.

On the island, a proposition p is either FALSE, Q-TRUE or K-TRUE; in each of the cases we say that p has value 0, Q or K, respectively. The queen finds the truth value Q to be superior, while the king values the most the value K. The queen and the king have their opinions on all issues, while other residents typically have their opinions on some issues but not all.”

The logic of the island of two truths is the easiest example of what Karin and I called a non-commutative frame or skew Heyting algebra (see here), a notion we then used, jointly with Jens Hemelaer, to define the notion of a non-commutative topos.

If you take our general definitions, and take Q as the distinguished top-element, then the truth tables for the island of two truths are these ones (value of first term on the left, that of the second on top):

\[
\begin{array}{c|ccc}
\wedge & 0 & Q & K \\
\hline
0 & 0 & 0 & 0 \\
Q & 0 & Q & Q \\
K & 0 & K & K
\end{array} \quad
\begin{array}{c|ccc}
\vee & 0 & Q & K \\
\hline
0 & 0 & Q & K \\
Q & Q & Q & K \\
K & K & Q & K
\end{array} \quad
\begin{array}{c|ccc}
\rightarrow & 0 & Q & K \\
\hline
0 & Q & Q & K \\
Q & 0 & Q & K \\
K & 0 & Q & K
\end{array} \quad
\begin{array}{c|c}
& \neg \\
\hline
0 & Q \\
Q & 0 \\
K & 0
\end{array}
\]

Note that on this island the order of statements is important! That is, the truth value of $p \wedge q$ may differ from that of $q \wedge p$ (and similarly for $\vee$).

Let’s reconsider Smullyan’s puzzle at the beginning of this post, but now on an island of two truths, where every inhabitant is either of Knave, or a Q-Knight (uttering only Q-valued statements), or a K-Knight (saying only K-valued statements).

Again, can you determine what type $C$ is?

Well, if you forget about the distinction between Q- and K-valued sentences, then we’re back to classical logic (or more generally, if you divide out Green’s equivalence relation from any skew Heyting algebra you obtain an ordinary Heyting algebra), and we have seen that then $B$ must be a Knave and $C$ a Knight, so in our new setting we know that $C$ is either a Q-Knight or a K-Knight, but which of the two?

Now, $C$ claims the negation of what $B$ said, so the truth value is $\neg 0 = Q$, and therefore $C$ must be a Q-Knight.

Recall that in Karin Cvetko-Vah‘s island of two truths all sentences have a unique value which can be either $0$ (false) or one of the non-false values Q or K, and the value of combined statements is given by the truth tables above. The Queen and King both have an opinion on all statements, which may or may not coincide with the actual value of that statement. However, if the Queen assigns value $0$ to a statement, then so does the King, and conversely.

Other inhabitants of the island have only their opinion about a subset of all statements (which may be empty). Two inhabitants agree on a statement if they both have an opinion on it and assign the same value to it.

Now, each inhabitant is either loyal to the Queen or to the King (or both), meaning that they agree with the Queen (resp. King) on all statements they have an opinion of. An inhabitant loyal to the Queen is said to believe a sentence when she assigns value $Q$ to it (and symmetric for those loyal to the King), and knows the statement if she believes it and that value coincides with the actual value of that statement.

Further, if A is loyal to the Queen, then the value of the statement ‘A is loyal to the Queen’ is Q, and if A is not loyal to the Queen, then the value of the sentence ‘A is loyal to the Queen’ is $0$ (and similarly for statements about loyalty to the King).

These notions are enough for the first batch of ten puzzles in Karin’s posts

Just one example:

Show that if anybody on the island knows that A is not loyal to the Queen, then everybody that has an opinion about the sentence ‘A is loyal to the Queen’ knows that.

After these two posts, Karin decided that it was more fun to blog about the use of non-commutative frames in data analysis.

But, she once gave me a text containing many more puzzles (as well as all the answers), so perhaps I’ll share these in a follow-up post.

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A projective plain (plane) of order ten

A projective plane of order $n$ is a collection of $n^2+n+1$ lines and $n^2+n+1$ points satisfying:

  • every line contains exactly $n+1$ points
  • every point lies on exactly $n+1$ lines
  • any two distinct lines meet at exactly one point
  • any two distinct points lie on exactly one line

Clearly, if $q=p^k$ is a pure prime power, then the projective plane over $\mathbb{F}_q$, $\mathbb{P}^2(\mathbb{F}_q)$ (that is, all nonzero triples of elements from the finite field $\mathbb{F}_q$ up to simultaneous multiplication with a non-zero element from $\mathbb{F}_q$) is a projective plane of order $q$.

The easiest example being $\mathbb{P}^2(\mathbb{F}_2)$ consisting of seven points and lines

But, there are others. A triangle is a projective plane of order $1$, which is not of the above form, unless you believe in the field with one element $\mathbb{F}_1$…

And, apart from $\mathbb{P}^2(\mathbb{F}_{3^2})$, there are three other, non-isomorphic, projective planes of order $9$.

It is clear then that for all $n < 10$, except perhaps $n=6$, a projective plane of order $n$ exists.

In 1938, Raj Chandra Bose showed that there is no plane of order $6$ as there cannot be $5$ mutually orthogonal Latin squares of order $6$, when even the problem of two orthogonal squares of order $6$ (see Euler’s problem of the $36$ officers) is impossible.

Yeah yeah Bob, I know it has a quantum solution.

Anyway by May 1977, when Lenstra’s Festschrift ‘Een pak met een korte broek’ (a suit with shorts) was published, the existence of a projective plane of order $10$ was still wide open.

That’s when Andrew Odlyzko (probably known best for his numerical work on the Riemann zeta function) and Neil Sloane (probably best known as the creator of the On-Line Encyclopedia of Integer Sequences) joined forces to publish in Lenstra’s festschrift a note claiming (jokingly) the existence of a projective plane of order ten, as they were able to find a finite field of ten elements.



Here’s a transcript:

A PROJECTIVE PLAIN OF ORDER TEN

A. M. Odlyzko and N.J.A. Sloane

This note settles in the affirmative the notorious question of the existence of a projective plain of order ten.

It is well-known that if a finite field $F$ is given containing $n$ elements, then the projective plain of order $n$ can be immediately constructed (see M. Hall Jr., Combinatorial Theory, Blaisdell, Waltham, Mass. 1967 and D.R. Hughes and F.C. Piper, Projective Planes, Springer-Verlag, N.Y., 1970).

For example, the points of the plane are represented by the nonzero triples $(\alpha,\beta,\gamma)$ of elements of $F$, with the convention that $(\alpha,\beta,\gamma)$ and $(r\alpha, r\beta, r\gamma)$ represent the same point, for all nonzero $r \in F$.

Furthermore this plain even has the desirable property that Desargues’ theorem holds there.

What makes this note possible is our recent discovery of a field containing exactly ten elements: we call it the digital field.

We first show that this field exists, and then give a childishly simple construction which the reader can easily verify.

The Existence Proof

Since every real number can be written in the decimal system we conclude that

\[
\mathbb{R} = GF(10^{\omega}) \]

Now $\omega = 1.\omega$, so $1$ divides $\omega$. Therefore by a standard theorem from field theory (e.g. B. L. van der Waerden, Modern Algebra, Ungar, N.Y., 1953, 2nd edition, Volume 1, p. 117) $\mathbb{R}$ contains a subfield $GF(10)$. This completes the proof.

The Construction

The elements of this digital field are shown in Fig. 1.

They are labelled $Left_1, Left_2, \dots, Left_5, Right_1, \dots, Right_5$ in the natural ordering (reading from left to right).



Addition is performed by counting, again in the natural way. An example is shown in Fig. 2, and for further details the reader can consult any kindergarten student.

In all digital systems the rules for multiplication can be written down immediately once addition has been defined; for example $2 \times n = n+n$. The reader will easily verify the rest of the details.

Since this field plainly contains ten elements (see Fig. 1) we conclude that there is a projective plain of order ten.

So far, the transcript.

More seriously now, the non-existence of a projective plane of order ten was only established in 1988, heavily depending on computer-calculations. A nice account is given in

C. M. H. Lam, “The Search for a Finite Projective Plane of Order 10”.

Now that recent iPhones nearly have the computing powers of former Cray’s, one might hope for easier proofs.

Fortunately, such a proof now exists, see A SAT-based Resolution of Lam’s Problem by Curtis Bright, Kevin K. H. Cheung, Brett Stevens, Ilias Kotsireas, Vijay Ganesh

David Roberts, aka the HigherGeometer, did a nice post on this
No order-10 projective planes via SAT
.

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A suit with shorts

I’m retiring in two weeks so I’m cleaning out my office.

So far, I got rid of almost all paper-work and have split my book-collection in two: the books I want to take with me, and those anyone can grab away.

Here’s the second batch (math/computer books in the middle, popular science to the right, thrillers to the left).



If you’re interested in some of these books (click for a larger image, if you want to zoom in) and are willing to pay the postage, leave a comment and I’ll try to send them if they survive the current ‘take-away’ phase.

Here are two books I definitely want to keep. On the left, an original mimeographed version of Mumford’s ‘Red Book’.

On the right, ‘Een pak met een korte broek’ (‘A suit with shorts’), a collection of papers by family and friends, presented to Hendrik Lenstra on the occasion of the defence of his Ph.D. thesis on Euclidean number-fields, May 18th 1977.

If the title intrigues you, a photo of young Hendrik in suit and shorts is included.

This collection includes hilarious ‘papers’ by famous people including

  • ‘A headache-causing problem’ by Conway (J.H.), Paterson (M.S.), and Moscow (U.S.S.R.)
  • ‘A projective plain of order ten’ by A.M. Odlyzko and N.J.A. Sloane
  • ‘La chasse aux anneaux principaux non-Euclidiens dans l’enseignement’ by Pierre Samuel
  • ‘On time-like theorems’ by Michiel Hazewinkel
  • ‘She loves me, she loves me not’ by Richard K. Guy
  • ‘Theta invariants for affine root systems’ by E.J.N. Looijenga
  • ‘The prime of primes’ by F. Lenstra and A.J. Oort
  • (and many more, most of them in Dutch)

Perhaps I can do a couple of posts on some of these papers. It might break this clean-up routine.

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the L-game

In 1982, the BBC ran a series of 10 weekly programmes entitled de Bono’s Thinking Course. In the book accompanying the series Edward de Bono recalls the origin of his ‘L-Game’:



Many years ago I was sitting next to the famous mathematician, Professor Littlewood, at dinner in Trinity College. We were talking about getting computers to play chess. We agreed that chess was difficult because of the large number of pieces and different moves. It seemed an interesting challenge to design a game that was as simple as possible and yet could be played with a degree of skill.

As a result of that challenge I designed the ‘L-Game’, in which each player has only one piece (the L-shape piece). In turn he moves this to any new vacant position (lifting up, turning over, moving across the board to a vacant position, etc.). After moving his L-piece he can – if he wishes – move either one of the small neutral pieces to any new position. The object of the game is to block your opponent’s L-shape so that no move is open to it.

It is a pleasant exercise in symmetry to calculate the number of possible L-game positions.

The $4 \times 4$ grid has $8$ symmetries, making up the dihedral group $D_8$: $4$ rotations and $4$ reflections.

An L-piece breaks all these symmetries, that is, it changes in form under each of these eight operations. That is, using the symmetries of the $4 \times 4$-grid we can put one of the L-pieces (say the Red one) on the grid as a genuine L, and there are exactly 6 possibilities to do so.

For each of these six positions one can then determine the number of possible placings of the Blue L-piece. This is best done separately for each of the 8 different shapes of that L-piece.

Here are the numbers when the red L is placed in the left bottom corner:



In total there are thus 24 possibilities to place the Blue L-piece in that case. We can repeat the same procedure for the remaining Red L-positions. Here are the number of possibilities for Blue in each case:



That is, there are 82 possibilities to place the two L-pieces if the Red one stands as a genuine L on the board.

But then, the L-game has exactly $18368 = 8 \times 82 \times 28$ different positions, where the factor

  • $8$ gives the number of symmetries of the square $4 \times 4$ grid.
  • Using these symmetries we can put the Red L-piece on the grid as a genuine $L$ and we just saw that this leaves $82$ possibilities for the Blue L-piece.
  • This leaves $8$ empty squares and so $28 = \binom{8}{2}$ different choices to place the remaining two neutral pieces.

The $2296 = 82 \times 28$ positions in which the red L-piece is placed as a genuine L can then be analysed by computer and the outcome is summarised in Winning Ways 2 pages 384-386 (with extras on pages 408-409).

Of the $2296$ positions only $29$ are $\mathcal{P}$-positions, meaning that the next player (Red) will loose. Here are these winning positions for Blue




Here, neutral piece(s) should be put on the yellow square(s). A (potential) remaining neutral piece should be placed on one of the coloured squares. The different colours indicate the remoteness of the $\mathcal{P}$-position:

  • Pink means remoteness $0$, that is, Red has no move whatsoever, so mate in $0$.
  • Orange means remoteness $2$: Red still has a move, but will be mated after Blue’s next move.
  • Purple stands for remoteness $4$, that is, Blue mates Red in $4$ moves, Red starting.
  • Violet means remoteness $6$, so Blue has a mate in $6$ with Red starting
  • Olive stands for remoteness $8$: Blue mates within eight moves.

Memorising these gives you a method to spot winning opportunities. After Red’s move image a board symmetry such that Red’s piece is a genuine L, check whether you can place your Blue piece and one of the yellow pieces to obtain one of the 29 $\mathcal{P}$-positions, and apply the reverse symmetry to place your piece.

If you don’t know this, you can run into trouble very quickly. From the starting position, Red has five options to place his L-piece before moving one of the two yellow counters.



All possible positions of the first option loose immediately.



For example in positions $a,b,c,d,f$ and $l$, Blue wins by playing



Here’s my first attempt at an opening repertoire for the L-game. Question mark means immediate loss, question mark with a number means mate after that number of moves, x means your opponent plays a sensible strategy.









Surely I missed cases, and made errors in others. Please leave corrections in the comments and I’ll try to update the positions.

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9 Bourbaki founding members, really?

The Clique (Twenty Øne Piløts fanatic fanbase) is convinced that the nine Bishops of Dema were modelled after the Bourbaki-group.

It is therefore of no surprise to see a Photoshopped version circulating of this classic picture of some youthful Bourbaki-members (note Jean-Pierre Serre poster-boying for Elon Musk’s site),

replacing some of them with much older photos of other members. Crucial seems to be that there are just nine of them.

I don’t know whether the Clique hijacked Bourbaki’s Wikipedia page, or whether they were inspired by its content to select those people, but if you look at that Wikipedia page you’ll see in the right hand column:

Founders

  • Henri Cartan
  • Claude Chevalley
  • Jean Coulomb
  • Jean Delsarte
  • Jean Dieudonné
  • Charles Ehresmann
  • René de Possel
  • André Weil

Really? Come on.

We know for a fact that Charles Ehresmann was brought in to replace Jean Leray, and Jean Coulomb to replace Paul Dubreil. Surely, replacements can’t be founders, can they?

Well, unfortunately it is not quite that simple. There’s this silly semantic discussion: from what moment on can you call someone a Bourbaki-member…

The collective name ‘Nicolas Bourbaki’ was adopted only at the Bourbaki-congress in Besse in July 1935 (see also this post).

But, before the Besse-meeting there were ten ‘proto-Bourbaki’ meetings, the first one on December 10th, 1934 in Cafe Capoulade. These meetings have been described masterly by Liliane Beaulieu in A Parisian Cafe and Ten Proto-Bourbaki Meetings (1934-35) (btw. if you know a direct link to the pdf, please drop it in the comments).

During these early meetings, the group called itself ‘The Committee for the Treatise on Analysis’, and not yet Bourbaki, whence the confusion.

Do we take the Capoulade-1934 meeting as the origin of the Bourbaki group (in which case the founding-members would be Cartan, Chevalley, De Possel, Delsarte, Dieudonne, and Weil), or was the Bourbaki-group founded at the Besse-congress in 1935 (when Cartan, Chevalley, Coulomb, De Possel, Dieudonne, Mandelbrojt, and Weil were present)?

Here’s a summary of which people were present at all meetings from December 1934 until the second Chancay-congress in September 1939, taken from Gatien Ricotier ‘Projets collectifs et personnels autour de Bourbaki dans les années 1930 à 1950′:

07-1935 is the Besse-congress, 09-1936 is the ‘Escorial’-congress (or Chancay 1) and 09-1937 is the second Chancay-congress. The ten dates prior to July 1935 are the proto-Bourbaki meetings.

Even though Delsarte was not present at the Besse-1935 congress, and De Possel moved to Algiers and left Bourbaki in 1941, I assume most people would agree that the six people present at the first Capoulade-meeting (Cartan, Chevalley, De Possel, Delsarte, Dieudonne, and Weil) should certainly be counted among the Bourbaki founding members.

What about the others?

We can safely eliminate Dubreil: he was present at just one proto-Bourbaki meeting and left the group in April 1935.

Also Leray’s case is straightforward: he was even excluded from the Besse-meeting as he didn’t contribute much to the group, and later he vehemently opposed Bourbaki, as we’ve seen.

Coulomb’s role seems to restrict to securing a venue for the Besse-meeting as he was ‘physicien-adjoint’ at the ‘Observatoire Physique du Globe du Puy-de-Dome’.



Because of this he could rarely attend the Julia-seminar or Bourbaki-meetings, and his interest in mathematical physics was a bit far from the themes pursued in the seminar or by Bourbaki. It seems he only contributed one small text, in the form of a letter. Due to his limited attendance, even after officially been asked to replace Dubreil, he can hardly be counted as a founding member.

This leaves Szolem Mandelbrojt and Charles Ehresmann.

We’ve already described Mandelbrojt as the odd-man-out among the early Bourbakis. According to the Bourbaki archive he only contributed one text. On the other hand, he also played a role in organising the Besse-meeting and in providing financial support for Bourbaki. Because he was present already early on (from the second proto-Bourbaki meeting) until the Chancay-1937 meeting, some people will count him among the founding members.

Personally I wouldn’t call Charles Ehresmann a Bourbaki founding member because he joined too late in the process (March 1936). Still, purists (those who argue that Bourbaki was founded at Besse) will say that at that meeting he was put forward to replace Jean Leray, and later contributed actively to Bourbaki’s meetings and work, and for that reason should be included among the founding members.

What do you think?

How many Bourbaki founding members are there? Six (the Capoulade-gang), seven (+Mandelbrojt), eight (+Mandelbrojt and Ehresmann), or do you still think there were nine of them?

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TØP PhotoShop mysteries

Suppose you’re writing a book, and for the duration of that project you keep a certain photo as your desktop-background. I guess we might assume that picture to be inspirational for your writing process.

If you PhotoShopped it to add specific elements, might we assume these extra bits to play a crucial role in your story?

Now, let’s turn to Twenty One Pilots and the creation process of their album Trench, released on October 5, 2018

We know from this tweet (from August 19th, 2018) that Tyler Joseph’s desktop-background picture was a photoshopped version of the classic Bourbaki-1938 photo on the left below, given it Trench-yellow, and added a bearded man in the doorway (photo on the right)




And we know from this interview (from September 5th, 2018) that, apart from the bearded man, he also replaced in the lower left corner the empty chair by a sitting person (lower photo).

The original photo features on the Wikipedia page on Nicolas Bourbaki, and as Tyler Joseph has revealed that Blurryface‘s real name is Nicolas Bourbaki (for whatever reason), and that he appears in the lyrics of Morph on Trench, this may make some sense.

But, of the seven people in the picture only three were founding members of Bourbaki: Weil, Diedonne and Delsarte. Ehresmann entered later, replacing Jean Leray, and Pison and Chabauty were only guinea pigs at that moment (they later entered Bourbaki, Chabauty briefly and Pison until 1950), and finally, Simonne Weil never was a member.

There’s another strange thing about the original picture. All of them, but Andre and Simone Weil, look straight into the camera, the Weil’s seem to be more focussed on something happening to the right.

Now, TØP has something with the number 9. There are nine circles on the cover of Blurryface (each representing one of a person’s insecurities, it seems), there are nine towers in the City of Dema, nine Bishops, etc.



So, from their perspective it makes sense to Photoshop two extra people in, and looking at the original there are two obvious places to place them: in the empty doorway, and on the empty chair.

But, who are they, and what is their significance?

1. The bearded man in the doorway

As far as I know, nobody knows who he is. From a Bourbaki point of view it can only be one person: Elie Cartan.

We know he was present at the 1938 Bourbaki Dieulefit/Beauvallon meeting, and that he was kind of a father figure to Bourbaki. Among older French mathematicians he was one of few (perhaps the only one) respected by all of Bourbaki.

But, bearded man is definitely not Elie Cartan…

If bearded man exists and has a Wikipedia page, the photo should be on that page. So, if you find him, please leave a comment.

Previous in this series I made a conjecture about him, but I’m not at all sure.

2. Why, of all people, Szolem Mandelbrojt?

We know from this Twentyonepilots subReddit post that the man sitting on the previously empty chair in none other than Bourbaki founding member Szolem Mandelbrojt, shopped in from this other iconic early Bourbaki-photo from the 1937 Chancay-meeting.

Let me tell you why this surprises me.

Szolem Mandelbrojt was atypical among the first Bourbaki-gang in many ways: he was the only one who didn’t graduate from the ENS, he was a bit older than the rest, he was the only one who was a full Professor (at Clermont-Ferrand) whereas the others were ‘maitre de conference’, he was the only one who didn’t contribute actively in the Julia seminar (the proto-Bourbaki seminar) nor much to the Bourbaki-congresses either for that matter, etc. etc.

Most of all, I don’t think he would feel particularly welcome at the 1938 congress. Here’s why.



(Jacques Hadamard (left), and Henri Lebesgue (right))

From Andre Weil’s autobiography (page 120):

Hadamard’s retirement left his position open. I thought myself not unworthy of succeeding him; my friends, especially Cartan and Delsarte, encouraged me to a candidate. It seemed to me that Lebesgue, who was the only mathematician left at the College de France, did not find my candidacy out of place. He even let me know that it was time to begin my ‘campaign visits’.

But the Bourbaki-campaign against a hierarchy of scientific prizes instituted by Jean Perrin (the so called ‘war of the medals’) interfered with his personal campaign. (Perhaps more important was that Mandelbrojt did his Ph.D. under Hadamard…)

Again from Weil’s autobiography (page 121):

Finally Lebesque put an end to my visits by telling me that he had decided on Mandelbrojt. It seemed to me that my friends were more disappointed than I at this outcome.

In the spring of 1938, Mandelbrojt succeeded Hadamard at the College de France.

There’s photographic evidence that Mandelbrojt was present at the 1935 Besse-congress and clearly at the 1937-Chancay meeting, but I don’t know that he was even present at Chancay-1936.

The only picture I know of that meeting is the one below. Standing on bench: Chevalley’s nephews, seated Andre Weil and Chevalley’s mother; standing, left to right: Ninette Ehresmann, Rene de Possel, Claude Chavalley, Jacqueline Chavalley, Mirles, Jean Delsarte and Charles Ehresmann.

Of all possible people, Szolem Mandelbrojt would be the miscast at the 1938-meeting. So, why did they shop him in?

– convenience: they had an empty chair in the original picture, another Bourbaki-photo with a guy sitting on such a chair, so why not shop him in?

– mistaken identity: in the subReddit post the sitting guy was mistakenly identified as Claude Chevalley. Now, there is a lot to say about wishing to add Chevalley to the original. He is by far the most likeable of all Bourbakis, so if these nine were ever supposed to be the nine Bishops of Dema, he most certainly would be Keons. But, Chevalley was already in the US at that time, and was advised by the French consul to remain there in view of the situation in Europe. As a result, Chevalley could not obtain a French professorship before the early 50ties.

– a deep hidden clue: remember all that nonsense about Josh Dun’s ‘alma mater’ being that Ukrainian building where Nico and the niners was shot? Well, Szolem Mandelbrojt’s alma mater was the University of Kharkiv in Ukraine. See this post for more details.

3. Is it all about Simone Weil?

If you super-impose the two photographs, pinning Mandelbrojt in both, the left border of the original 1938-picture is an almost perfect mirror for both appearances of Simone Weil. Can she be more important in all of this than we think?

Previous in the Bourbaki&TØP series:

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