# Tag: books

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?

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)

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.

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.

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.

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?

In December last year, there was the official opening of the Istituto Grothendieck in the little town of Mondovi in Northern Italy.The videos of the talks given at that meeting are now online.

The Institute houses two centres, the Centre for topos theory and its applications with mission statement:

The Centre for Topoi Theory and its Applications carries out highly innovative research in the field of Grothendieck’s topos theory, oriented towards the development of the unifying role of the concept of topos across different areas of mathematics.

Particularly relevant to these aims is the theory of topos-theoretic ‘bridges’ of Olivia Caramello, coordinator of the Centre and principal investigator of the multi-year project “Topos theory and its applications”.

and the Centre for Grothendiecian studies with mission:

The Centre for Grothendiecian Studies is dedicated to honoring the memory of Alexander Grothendieck through extensive work to valorize his work and disseminate his ideas to the general public.

In particular, the Centre aims to carry out historical/philosophical and editorial work to promote the publication of the unpublished works of A. Grothendieck, as well as to promote the production of translations of already published works in various languages.

No comment on the first. You can look up the Institute’s Governance page, contemplate recent IHES-events, and conjure up your own story.

More interesting is the Centre of Grothendiec(k)ian studies. Here’s the YouTube-clip of the statement made by Johanna Grothendieck (daughter of) at the opening.

She hopes for two things: to find money and interested persons to decrypt and digitalise Grothendieck’s Lasserre gribouillis, and to initiate the re-edition of the complete mathematical works of Grothendieck.

So far, Grothendieck’s family was withholding access to the Lasserre writings. Now they seem to grant access to the Istituto Grothendieck and authorise it to digitalise the 30.000 pages.

Further good news is that a few weeks ago Mateo Carmona was appointed as coordinator of the Centre of grothendieckian studies.

You may know Mateo from his Grothendieck Github Archive. A warning note on that page states: “This site no longer updates (since Feb. 2023) and has been archived. Please visit [Instituto Grothendieck] or write to Mateo Carmona at mateo.carmona@csg.igrothendieck.org”. So probably the site will be transferred to the Istituto.

Mateo Carmona says:

As Coordinator of the CSG, I will work tirelessly to ensure that the Centre provides comprehensive resources for scholars, students, and enthusiasts interested in Grothendieck’s original works and modern scholarship. I look forward to using my expertise to coordinate and supervise the work of the international group of researchers and volunteers who will promote Grothendieck’s scientific and cultural heritage through the CSG.

It looks as if Grothendieck’s gribouillis are in good hands, at last.

The French anthropologist and ethnologist Claude Levi-Strauss once observed

“In Paris, intellectuals need a new toy every 15 years.”

Some pointers to applications of their toy of choice for the past ten years:

How do Parisian mathematicians with a lifelong interest in topos theory react to this hype?

With humour!

Here’s an ‘exposé parodique’ (parodical lecture) by Stéphane Dugowson on “Contre les topos” (against toposes).

A month ago, Stephen Wolfram put out a little booklet (140 pages) What Is ChatGPT Doing … and Why Does It Work?.

It gives a gentle introduction to large language models and the architecture and training of neural networks.

The entire book is freely available:

The advantage of these online texts is that you can click on any of the images, copy their content into a Mathematica notebook, and play with the code.

This really gives a good idea of how an extremely simplified version of ChatGPT (based on GPT-2) works.

Downloading the model (within Mathematica) uses about 500Mb, but afterwards you can complete any prompt quickly, and see how the results change if you turn up the ‘temperature’.

You should’t expect too much from this model. Here’s what it came up with from the prompt “The major results obtained by non-commutative geometry include …” after 20 steps, at temperature 0.8:

 NestList[StringJoin[#, model[#, {"RandomSample", "Temperature" -> 0.8}]] &, "The major results obtained by non-commutative geometry include ", 20]

 

The major results obtained by non-commutative geometry include vernacular accuracy of math and arithmetic, a stable balance between simplicity and complexity and a relatively low level of violence. 

Lol.

In the more philosophical sections of the book, Wolfram speculates about the secret rules of language that ChatGPT must have found if we want to explain its apparent succes. One of these rules, he argues, must be the ‘logic’ of languages:

But is there a general way to tell if a sentence is meaningful? There’s no traditional overall theory for that. But it’s something that one can think of ChatGPT as having implicitly “developed a theory for” after being trained with billions of (presumably meaningful) sentences from the web, etc.

What might this theory be like? Well, there’s one tiny corner that’s basically been known for two millennia, and that’s logic. And certainly in the syllogistic form in which Aristotle discovered it, logic is basically a way of saying that sentences that follow certain patterns are reasonable, while others are not.

Something else ChatGPT may have discovered are language’s ‘semantic laws of motion’, being able to complete sentences by following ‘geodesics’:

And, yes, this seems like a mess—and doesn’t do anything to particularly encourage the idea that one can expect to identify “mathematical-physics-like” “semantic laws of motion” by empirically studying “what ChatGPT is doing inside”. But perhaps we’re just looking at the “wrong variables” (or wrong coordinate system) and if only we looked at the right one, we’d immediately see that ChatGPT is doing something “mathematical-physics-simple” like following geodesics. But as of now, we’re not ready to “empirically decode” from its “internal behavior” what ChatGPT has “discovered” about how human language is “put together”.

So, the ‘hidden secret’ of successful large language models may very well be a combination of logic and geometry. Does this sound familiar?

If you prefer watching YouTube over reading a book, or if you want to see the examples in action, here’s a video by Stephen Wolfram. The stream starts about 10 minutes into the clip, and the whole lecture is pretty long, well over 3 hours (about as long as it takes to read What Is ChatGPT Doing … and Why Does It Work?).

This week, I was hit hard by synchronicity.

Lately, I’ve been reading up a bit on psycho-analysis, tried to get through Grothendieck’s La clef des songes (the key to dreams) and I’m in the process of writing a series of blogposts on how to construct a topos of the unconscious.

And then I read Cormac McCarthy‘s novels The passenger and Stella Maris, and got hit.

Stella Maris is set in 1972, when the math-prodigy Alicia Western, suffering from hallucinations, admits herself to a psychiatric hospital, carrying a plastic bag containing forty thousand dollars. The book consists entirely of dialogues, the transcripts of seven sessions with her psychiatrist Dr. Cohen (nomen est omen).

Alicia is a doctoral candidate at the University Of Chicago who got a scholarship to visit the IHES to work with Grothendieck on toposes.

During the psychiatric sessions, they talk on a wide variety of topics, including the nature of mathematics, quantum mechanics, music theory, dreams, and the unconscious (and its role in doing mathematics).

The core question is not how you do math but how does the unconscious do it. How it is that it’s demonstrably better at it than you are? You work on a problem and then you put it away for a while. But it doesnt go away. It reappears at lunch. Or while you’re taking a shower. It says: Take a look at this. What do you think? Then you wonder why the shower is cold. Or the soup. Is this doing math? I’m afraid it is. How is it doing it? We dont know. How does the unconscious do math? (page 99)

Before going to the IHES she had to send Grothendieck a paper (‘It was an explication of topos theory that I thought he probably hadn’t considered.’ page 136, and ‘while it proved three problems in topos theory it then set about dismantling the mechanism of the proofs.’ page 151). At the IHES ‘I met three men that I could talk to: Grothendieck, Deligne, and Oscar Zariski.’ (page 136).

I don’t know whether Zariski visited the IHES in the early 70ties, and while most historical allusions (to Grothendieck’s life, his role in Bourbaki etc.) are correct, Alicia mentions the ‘Langlands project’ (page 66) which may very well have been the talk of town at the IHES in 1972, but the mention of Witten ‘Grothendieck writes everything down. Witten nothing.’ (page 100) raised an eyebrow.

The book also contains these two nice attempts to capture some of the essence of topos theory:

When you get to topos theory you are at the edge of another universe.
You have found a place to stand where you can look back at the world from nowhere. It’s not just some gestalt. It’s fundamental. (page 13)

You asked me about Grothendieck. The topos theory he came up with is a witches’ brew of topology and algebra and mathematical logic.
It doesnt even have a clear identity. The power of the theory is still speculative. But it’s there.
You have a sense that it is waiting quietly with answers to questions that nobody has asked yet. (page 68)

I did read ‘The passenger’ first, which is probably better as then you’d know already some of the ghosts haunting Alicia, but it’s not a must if you are only interested in their discussions about the nature of mathematics. Be warned that it is a pretty dark book, better not read when you’re already feeling low, and it should come with a link to a suicide prevention line.

Here’s a more considered take on Stella Maris:

Before ChatGPT, the hype among productivity boosters was a PKMs or Personal knowledge management system.

It gained popularity through Tiago Forte’s book ‘Building a second brain’, and (for academics perhaps a more useful read) ‘How to take smart notes’ by Sönke Ahrens.

These books promote new techniques for note-taking (and for storing these notes) such as the PARA-method, the CODE-system, and Zettelkasten.

Unmistakable Creative has some posts on the principles behing the ‘second brain’ approach.

Your brain isn’t like a hard drive or a dropbox, where information is stored in folders and subfolders. None of our thoughts or ideas exist in isolation. Information is organized in a series of non-linear associative networks in the brain.

Networked thinking is not just a more efficient way to organize information. It frees your brain to do what it does best: Imagine, invent, innovate, and create. The less you have to remember where information is, the more you can use it to summarize that information and turn knowledge into action.

and

A network has no “correct” orientation and thus no bottom and no top. Each individual, or “node,” in a network functions autonomously, negotiating its own relationships and coalescing into groups. Examples of networks include a flock of birds, the World Wide Web, and the social ties in a neighborhood. Networks are inherently “bottom-up” in that the structure emerges organically from small interactions without direction from a central authority.

-Tiago Forte, Tagging for Personal Knowledge Management

There are several apps you can use to start building your second brain, the more popular seem to be Roam Research, LogSeq, and Obsidian.

These systems allow you to store, link and manipulate a large collection of notes, query them as a database, modify them in various ways via plugins or scripts, and navigate the network created via graph-views.

Exactly the kind of things we need to modify the simple system from the shape of languages-post into a proper topos of the unconscious.

I’ve been playing around with Obsidian which I like because it has good LaTeX plugins, powerful database tools via the Dataview plugin, and one can execute codeblocks within notes in almost any programming language (python, haskell, lean, Mathematica, ruby, javascript, R, …).

Most of all it has a vibrant community of users, an excellent forum, and a well-documented Obsidian hub.

There’s just one problem, I’m a terrible note-taker, so how can I begin to load my ‘second brain’?

Obsidian has several plugins to import data, such as your Kindle highlights, your Twitter feed, your Readwise-data, and many others, but having been too lazy in the past, I cannot use any of them.

In fact, the only useful collection of notes I have are my blog-posts. So, I’ve uploaded NeverEndingBooks into Obsidian, one note per post (admittedly, not very Zettelkasten-like), half a million words in total.

Fortunately, I did tag most of these posts at the time. Together with other meta-data this results in the Graph view below (under ‘Files’ toggled tags, under ‘Groups’ three tag-colours, and under ‘Display’ toggled arrows). One can add colour-groups based on tags or other information (here, red dots are posts tagged ‘Grothendieck’, the blue ones are tagged ‘Conway’, the purple ones tagged ‘Connes’, just for the sake of illustration). In Obsidian you can zoom into this graph, place a pointer on a node to highlight the connecting dots, and much more.

Because I tend to forget such things, and as it may be useful to other people running a WordPress-blog making heavy use of MathJax, here’s the procedure I followed:

1. Follow the instructions from Convert wordpress articles to markdown.

In the wizard I’ve opted to go only for yearly folders, to prefix posts with the date, and to save all images.

2. This gives you a directory with one folder per year containing markdown versions of your posts, and in each year-folder a subfolder ‘img’ containing all images.

Turn this directory into an Obsidian-vault by opening Obsidian, click on the ‘open another vault’ icon (third from bottom-left), select ‘Open folder as vault’ and navigate to your directory.

3. You will notice that most of your LaTeX cannot be parsed because during the markdown-process backslashes are treaded as special character, resulting in two backslashes for every LaTeX-command…

A remark before trying to solve this: another option might be to use the wordpress-to-hugo-exporter, resulting in clean LaTeX, but lacking the possibility to opt for yearly-folders (it dumps all posts into one folder), and it makes a mess of the image-files.

4. So, we will need to do a lot of search&replaces in all files, and need a convenient tool for this.

First option was the Sublime Text app, which is free and does the search&replaces quickly. The problem is that you have to save each of the files, one at a time! This may take hours.

I’ve done it using the Search and Replace app (3\$) which allows you to make several searches/replaces at the same time (I messed up LaTeX code in previous exports, so needed to do many more changes). It warns you that it is dangerous to replace strings in all files (which is the reason why Sublime Text makes it difficult), you can ignore it, but only after you put the ‘img’ folders away in a safe place. Otherwise it will also try to make the changes to these files, recognise that they are not text-files, and drop them altogether…

That’s it.

I now have a backup network-version of this blog.

As we mentioned in the previous post a first attempt to construct the ‘topos of the unconscious’ might be to start with a collection of notes (the ‘conscious’) and work on the semantics of text-snippets to unravel (a part of) the unconscious underpinning of these notes. We also mentioned that the poset-structure in that post should be replaced by a more involved network structure.

What interests me most is whether such an approach might be doable ‘in practice’, and Obsidian looks like the perfect tool to try this out.

What we need is a sufficiently large set of notes, of independent interest, to inject into Obsidian. The more meta it is, the better…

(tbc)

Previously in this series:

Next:

The enriched vault