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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.


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:

\downarrow~\bf{\wedge}~\rightarrow & 0 & Q & K \\
0 & 0 & 0 & 0 \\
Q & 0 & Q & Q \\
K & 0 & K & K
\end{array} \quad
\downarrow~\vee~\rightarrow & 0 & Q & K \\
0 & 0 & Q & K \\
Q & Q & Q & K \\
K & K & Q & K
\end{array} \]
\downarrow~\implies~\rightarrow & 0 & Q & K \\
0 & Q & Q & K \\
Q & 0 & Q & K \\
K & 0 & Q & K
\end{array} \quad
\downarrow & \neg~\downarrow \\
0 & Q \\
Q & 0 \\
K & 0

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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