On categories, go and the book $\in$

A nice interview with Jacques Roubaud (the guy responsible for Bourbaki’s death announcement) in the courtyard of the ENS. He talks about go, categories, the composition of his book $\in$ and, of course, Grothendieck and Bourbaki.

Clearly there are pop-math books like dedicated to $\pi$ or $e$, but I don’t know just one novel having as its title a single mathematical symbol : $\in$ by Jacques Roubaud, which appeared in 1967.

The book consists of 361 small texts, 180 for the white stones and 181 for the black stones in a game of go, between Masami Shinohara (8th dan) and Mitsuo Takei (2nd Kyu). Here’s the game:

In the interview, Roubaud tells that go became quite popular in the mid sixties among French mathematicians, or at least those in the circle of Chevalley, who discovered the game in Japan and became a go-envangelist on his return to Paris.

In the preface to $\in$, the reader is invited to read it in a variety of possible ways. Either by paying attention to certain groupings of stones on the board, the corresponding texts sharing a common theme. Or, by reading them in order of how the go-game evolved (the numbering of white and black stones is not the same as the texts appearing in the book, fortunately there’s a conversion table on pages 153-155).

Or you can read them by paragraph, and each paragraph has as its title a mathematical symbol. We have $\in$, $\supset$, $\Box$, Hilbert’s $\tau$ and an imagined symbol ‘Symbole de la réflexion’, which are two mirrored and overlapping $\in$’s. For more information, thereader should consult the “Dictionnaire de la langue mathématique” by Lachatre and … Grothendieck.

According to the ‘bibliographie’ below it is number 17 in the ‘Publications of the L.I.T’.

Other ‘odd’ books in the list are: Bourbaki’s book on set theory, the thesis of Jean Benabou (who is responsible for Roubaud’s conversion from solving the exercises in Bourbaki to doing work in category theory. Roubaud also claims in the interview that category theory inspired him in the composition of the book $\in$) and there’s also Guillaume d’Ockham’s ‘Summa logicae’…

Scottish solids, final(?) comments

In the spring of 2009 I did spend a fortnight dog-sitting in a huge house in the countryside, belonging to my parents-in-law, who both passed away the year before.

That particular day it was raining and thundering heavily. To distract myself from the sombre and spooky atmosphere in the house I began to surf the web looking for material for a new series of blogposts (yes, in those days I was still thinking in ‘series’ of posts…).

Bookmarks for that day tell me that the first picture grasping my attention was Salvador Dali’s Sacrament of the last supper, in particular the depicted partial dodecahedron

I did compare it with Leonardo’s last supper and in the process stumbled upon Leonardo’s drawings of polyhedra, among which these two dodecahedra


From there it went on and on : the Mystery of the 2nd and 3rd Century Roman Dodecahedron and its posible use ‘casting dodecahedra’ in Tarot Divination Without Tarot Cards or as an astronomical instrument, a text on polyhedra and plagiarism in the renaissance, the history of the truncated icosahedron, a Bosnian pyramid and its stone balls, the sacred geometry of the dodecahedron, mathematics in the Vatican library, and on and on and on…

By noon, I felt I had enough material to post for a couple of weeks on “platonic solids through the ages”.

In between two rain showers, I walked the dog, had a quick lunch, and started writing.

I wanted to approach the topic in chronological order, and as I had done already a quicky on Scottish solids, the first post of the series would have to extend on this picture of five stone balls from the Ashmolean museum (or so it was claimed).

So, I hunted for extra pictures of these stone balls from the Ashmolean, and when comparing the two, clearly something had to be wrong…

It took me a couple of hours to catch up with the scientific literature on these Scottish balls, their cataloguing system and the museums of Scotland and England that house them.

Around 4pm I had compiled a list of all potential dodecahedra and icosahedra Scottish balls: ‘there are only 8 possible candidates for a Scottish dodecahedron (below their catalogue numbers, indicating to the knowledgeable which museum owns them and where they were found)

NMA AS 103 : Aberdeenshire
AS 109 : Aberdeenshire
AS 116 : Aberdeenshire (prob)
AUM 159/9 : Lambhill Farm, Fyvie, Aberdeenshire
Dundee : Dyce, Aberdeenshire
GAGM 55.96 : Aberdeenshire
Montrose = Cast NMA AS 26 : Freelands, Glasterlaw, Angus
Peterhead : Aberdeenshire

The case for a Scottish icosahedron looks even worse. Only two balls have exactly 20 knobs

NMA AS 110 : Aberdeenshire
GAGM 92 106.1. : Countesswells, Aberdeenshire’

About an hour later I’d written the post, clicked the ‘Publish’ button and The Scottish solids hoax, began to live a life of its own!

From the numerous reactions let me single out 3 follow-ups which I believe to be most important.

John McKay and Tom Leinster did some legwork, tracking down resp. photographer and one of the 20 knobs balls.

John Baez gave a talk at an AMS meeting dedicated to the history of mathematics on Who discovered the icosahedron? mentioning my post and extending it by:

“And here is where I did a little research of my own. The library at UC Riverside has a copy of Keith Critchlow’s 1979 book Time Stands Still. In this book, we see the same photo of stones with ribbons that appears in Lawlor’s book – the photo that Atiyah and Suttcliffe use. In Critchlow’s book, these stones are called “a full set of Neolithic ‘Platonic solids'”. He says they were photographed by one Graham Challifour – but he gives no information as to where they came from!

And Critchlow explicitly denies that the Ashmolean has an icosahedral stone! He writes:

… the author has, during the day, handled five of these remarkable objects in the Ashmolean museum…. I was rapt in admiration as I turned over these remarkable stone objects when another was handed to me which I took to be an icosahedron…. On careful scrutiny, after establishing apparent fivefold symmetry on a number of the axes, a count-up of the projections revealed 14! So it was not an icosahedron.”

And now there is even a published paper out!

Bob Lloyd wrote How old are the Platonic solids?, published in BSHM Bulletin: Journal of the British Society for the History of Mathematics. The full article is behind a paywall but Bob graciously send me a copy.

Bob believes the balls in the picture to belong to the Scottish ‘National Museum of Antiquities’ (NMA in the Marshall list), now the National Museum of Scotland (NMS) in Edinburgh.

He believes the third and fourth ball to be two pictures of the same object “recorded as having been discovered in Aberdeenshire” so it should be NMA AS 103 : Aberdeenshire in the above list. (Or, the other one may be NMA AS 26?).

He also attempts to identify the other 3 balls with objects in the NMS-collection. In short, he gives compelling evidence that the picture must have been taking in Edinburg and exists of genuine artifacts.

Perhaps even more important is that he finally puts the case of a Scottish icosahedron to rest. As mentioned above, there are just two candidates NMA AS 110 (Edinburg) and GAGM 92 106.1 (Glasgow). He writes:

“According to the Marshall list, there are only two balls known which have 20K; one of these is at the NMS. Alan Saville, Senior Curator for Earliest Prehistory at this Museum, has provided a photograph which shows that this object is complex, and certainly not a dodecahedron. It could be considered as a modified octahedron, with five large knobs in the usual positions, but with the sixth octahedral position occupied by twelve small knobs, and in addition there are also three small triangles carved at some of the interstices, the three-fold positions of the ‘octahedron’. These make up a total of twenty ‘protrusions’, though the word ‘knobs’ is hard to justify.

The other 20K object is at the Kelvingrove museum in Glasgow. Photographs taken by Tracey Hawkins, assistant curator, show that this also is very far from being a dodecahedron, though this time there are twenty clearly defined knobs of roughly the same size. The shape is somewhat irregular, but two six-sided pyramids can be picked out, and much of the structure, though not all, is deltahedral in form, with sets of three balls at the corners of equilateral triangles.”

So, sadly for John McKay, there is no Scottish icosahedron out there!

One final comment. Both John Baez (in a comment) and Bob Lloyd (in a comment and in his paper) argue that I shouldn’t have used the term “hoax” for something that is merely a ‘matter of sloppy scholarship’.

My apologies.

Given Bob’s evidence that the balls in the picture are genuine artifacts, I have deleted the ‘fabrication or falsification’-phrase in the original post.

Summarizing : the Challifour photograph is not taken at the Ashmolean museum, but at the National Museum of Scotland in Edinburgh and consists of 5 of their artifacts (or 4 if ball 3 and 4 are identical) vaguely resembling cube, tetrahedron, dodecahedron (twice) and octahedron. The fifth Platonic solid, the icosahedron, remains elusive.

What is the knot associated to a prime?

Sometimes a MathOverflow question gets deleted before I can post a reply…

Yesterday (New-Year) PD1&2 were visiting, so I merely bookmarked the What is the knot associated to a prime?-topic, promising myself to reply to it this morning, only to find out that the page no longer exists.

From what I recall, the OP interpreted one of my slides of the April 1st-Alumni talk

as indicating that there might be a procedure to assign to a prime number a specific knot. Here’s the little I know about this :

Artin-Verdier duality in etale cohomology suggests that $Spec(\mathbb{Z}) $ is a 3-dimensional manifold, as Barry Mazur pointed out in this paper

The theory of discriminants shows that there are no non-trivial global etale extensions of $Spec(\mathbb{Z}) $, whence its (algebraic) fundamental group should be trivial. By Poincare-Perelman this then implies that one should view $Spec(\mathbb{Z}) $ as the three-sphere $S^3 $. Note that there is no ambiguity in this direction. However, as there are other rings of integers in number fields having trivial fundamental group, the correspondence is not perfect.

Okay, but then primes should correspond to certain submanifolds of $S^3 $ and as the algebraic fundamental group of $Spec(\mathbb{F}_p) $ is the profinite completion of $\mathbb{Z} $, the first option that comes to mind are circles

Hence, primes might be viewed as circles embedded in $S^3 $, that is, as knots! But which knots? Well, as far as I know, nobody has a procedure to assign a knot to a prime number, let alone one having p crossings. What is known, however, is that different primes must correspond to different knots

because the algebraic fundamental groups of $Spec(\mathbb{Z})- { p } $ differ for distinct primes. This was the statement I wanted to illustrate in the first slide.

But, the story goes a lot further. Knots may be linked and one can detect this by calculating the link-number, which is symmetric in the two knots. In number theory, the Legendre symbol, plays a similar role thanks to quadratic reciprocity

and hence we can view the Legendre symbol as indicating whether the knots corresponding to different primes are linked or not. Whereas it is natural in knot theory to investigate whether collections of 3, 4 or 27 knots are intricately linked (or not), few people would consider the problem whether one collection of 27 primes differs from another set of 27 primes worthy of investigation.

There’s one noteworthy exception, the Redei symbol which we can now view as giving information about the link-behavior of the knots associated to three different primes. For example, one can hunt for prime-triples whose knots link as the Borromean rings

(note that the knots corresponding to the three primes are not the unknot but more complicated). Here’s where the story gets interesting : in number-theory one would like to discover ‘higher reciprocity laws’ (for collections of n prime numbers) by imitating higher-link invariants in knot-theory. This should be done by trying to correspond filtrations on the fundamental group of the knot-complement to that of the algebraic fundamental group of $Spec(\mathbb{Z})-{ p } $ This project is called arithmetic topology

Perhaps I should make a pod- or vod-cast of that 20 minute talk, one day…

The odd knights of the round table

Here’s a tiny problem illustrating our limited knowledge of finite fields : “Imagine an infinite queue of Knights ${ K_1,K_2,K_3,\ldots } $, waiting to be seated at the unit-circular table. The master of ceremony (that is, you) must give Knights $K_a $ and $K_b $ a place at an odd root of unity, say $\omega_a $ and $\omega_b $, such that the seat at the odd root of unity $\omega_a \times \omega_b $ must be given to the Knight $K_{a \otimes b} $, where $a \otimes b $ is the Nim-multiplication of $a $ and $b $. Which place would you offer to Knight $K_{16} $, or Knight $K_n $, or, if you’re into ordinals, Knight $K_{\omega} $?”

What does this have to do with finite fields? Well, consider the simplest of all finite field $\mathbb{F}_2 = { 0,1 } $ and consider its algebraic closure $\overline{\mathbb{F}_2} $. Last year, we’ve run a series starting here, identifying the field $\overline{\mathbb{F}_2} $, following John H. Conway in ONAG, with the set of all ordinals smaller than $\omega^{\omega^{\omega}} $, given the Nim addition and multiplication. I know that ordinal numbers may be intimidating at first, so let’s just restrict to ordinary natural numbers for now. The Nim-addition of two numbers $n \oplus m $ can be calculated by writing the numbers n and m in binary form and add them without carrying. For example, $9 \oplus 1 = 1001+1 = 1000 = 8 $. Nim-multiplication is slightly more complicated and is best expressed using the so-called Fermat-powers $F_n = 2^{2^n} $. We then demand that $F_n \otimes m = F_n \times m $ whenever $m < F_n $ and $F_n \otimes F_n = \frac{3}{2}F_n $. Distributivity wrt. $\oplus $ can then be used to calculate arbitrary Nim-products. For example, $8 \otimes 3 = (4 \otimes 2) \otimes (2 \oplus 1) = (4 \otimes 3) \oplus (4 \otimes 2) = 12 \oplus 8 = 4 $. Conway’s remarkable result asserts that the ordinal numbers, equipped with Nim addition and multiplication, form an algebraically closed field of characteristic two. The closure $\overline{\mathbb{F}_2} $ is identified with the subfield of all ordinals smaller than $\omega^{\omega^{\omega}} $. For those of you who don’t feel like going transfinite, the subfield $~(\mathbb{N},\oplus,\otimes) $ is identified with the quadratic closure of $\mathbb{F}_2 $.

The connection between $\overline{\mathbb{F}_2} $ and the odd roots of unity has been advocated by Alain Connes in his talk before a general public at the IHES : “L’ange de la géométrie, le diable de l’algèbre et le corps à un élément” (the angel of geometry, the devil of algebra and the field with one element). He describes its content briefly in this YouTube-video

At first it was unclear to me which ‘coupling-problem’ Alain meant, but this has been clarified in his paper together with Caterina Consani Characteristic one, entropy and the absolute point. The non-zero elements of $\overline{\mathbb{F}_2} $ can be identified with the set of all odd roots of unity. For, if x is such a unit, it belongs to a finite subfield of the form $\mathbb{F}_{2^n} $ for some n, and, as the group of units of any finite field is cyclic, x is an element of order $2^n-1 $. Hence, $\mathbb{F}_{2^n}- { 0 } $ can be identified with the set of $2^n-1 $-roots of unity, with $e^{2 \pi i/n} $ corresponding to a generator of the unit-group. So, all elements of $\overline{\mathbb{F}_2} $ correspond to an odd root of unity. The observation that we get indeed all odd roots of unity may take you a couple of seconds (( If m is odd, then (2,m)=1 and so 2 is a unit in the finite cyclic group $~(\mathbb{Z}/m\mathbb{Z})^* $ whence $2^n = 1 (mod~m) $, so the m-roots of unity lie within those of order $2^n-1 $ )).

Assuming we succeed in fixing a one-to-one correspondence between the non-zero elements of $\overline{\mathbb{F}_2} $ and the odd roots of unity $\mu_{odd} $ respecting multiplication, how can we recover the addition on $\overline{\mathbb{F}_2} $? Well, here’s Alain’s coupling function, he ties up an element x of the algebraic closure to the element s(x)=x+1 (and as we are in characteristic two, this is an involution, so also the element tied up to x+1 is s(x+1)=(x+1)+1=x. The clue being that multiplication together with the coupling map s allows us to compute any sum of two elements as $x+y=x \times s(\frac{y}{x}) = x \times (\frac{y}{x}+1) $.
For example, all information about the finite field $\mathbb{F}_{2^4} $ is encoded in this identification with the 15-th roots of unity, together with the pairing s depicted as

Okay, we now have two identifications of the algebraic closure $\overline{\mathbb{F}_2} $ : the smaller ordinals equipped with Nim addition and Nim multiplication and the odd roots of unity with complex-multiplication and the Connes-coupling s. The question we started from asks for a general recipe to identify these two approaches.

To those of you who are convinced that finite fields (LOL, even characteristic two!) are objects far too trivial to bother thinking about : as far as I know, NOBODY knows how to do this explicitly, even restricting the ordinals to merely the natural numbers!

Please feel challenged! To get you started, I’ll show you how to place the first 15 Knights and give you a procedure (though far from explicit) to continue. Here’s the Nim-picture compatible with that above

To verify this, and to illustrate the general strategy, I’d better hand you the Nim-tables of the first 16 numbers. Here they are

It is known that the finite subfields of $~(\mathbb{N},\oplus,\otimes) $ are precisely the sets of numbers smaller than the Fermat-powers $F_n $. So, the first one is all numbers smaller than $F_1=4 $ (check!). The smallest generator of the multiplicative group (of order 3) is 2, so we take this to correspond to the unit-root $e^{2 \pi i/3} $. The next subfield are all numbers smaller than $F_2 = 16 $ and its multiplicative group has order 15. Now, choose the smallest integer k which generates this group, compatible with the condition that $k^{\otimes 5}=2 $. Verify that this number is 4 and that this forces the identification and coupling given above.

The next finite subfield would consist of all natural numbers smaller than $F_3=256 $. Hence, in this field we are looking for the smallest number k generating the multiplicative group of order 255 satisfying the extra condition that $k^{\otimes 17}=4 $ which would fix an identification at that level. Then, the next level would be all numbers smaller than $F_4=65536 $ and again we would like to find the smallest number generating the multiplicative group and such that the appropriate power is equal to the aforementioned k, etc. etc.

Can you give explicit (even inductive) formulae to achieve this? I guess even the problem of placing Knight 16 will give you a couple of hours to think about… (to be continued).